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https://www.cgtk.co.uk/metalwork/reference/screwmeasurement | [
"# Three-Wire Screw Measurement\n\nThis tool can be used to calculate the correct measurement for a given thread using the three-wire method. It is based in concept on Screwmez by Pete Worden (described in Model Engineer's Workshop issue 246), but has been written from scratch to be a web-based application usable on any operating system. I don't have a Windows PC upon which to run Screwmez, so if you notice any discrepancies, please use the contact page to let me know!\n\nAll the fields that expect a number (diameter, pitch etc), will accept fractions (e.g. 1 1/4) and numbers followed by an explicit unit (\" or mm), so if you chose to do so you could specify an imperial thread with a diameter of 25 1/2 mm and a pitch of 0.1\" (rather than specifying a TPI). More usefully, if you have measuring wires that are specified in imperial units, you can select a metric thread but still enter (e.g.) 0.040\" as the actual wire size.\n\nNote that the minimum and maximum wire size fields may be wrong as I haven't yet managed to find consistent equations for all thread angles (in particular 47.5°). If anyone can help with this, then please do get in touch.\n\n Thread Class: Metric Coarse Metric Fine Custom Metric BA UNC UNF BSW BSF Custom Imperial Selected Thread: Thread Nominal Diameter (mm): mm Thread Pitch (mm): mm Thread Angle (degrees) 47.5° 55° 60° Thread Pitch Diameter (mm): mm Depth of Thread (mm): mm Helix Angle (degrees) Preferred Wire Size (mm): mm ??? mm Maximum Wire Size (mm): mm ??? mm Minimum Wire Size (mm): mm Actual Wire Size (mm): mm Dimension Over Wires (mm): mm\n\n## Background\n\nThis calculator uses the following equations - let me know if you think they're wrong! Regardless of thread type, all entered data is converted into millimetres prior to calculating the results; if the thread type is an imperial one, the result is then converted back into inches for display (although both inches and millimetres are displayed in the right-hand column regardless of the thread unit).\n\n### Key\n\n$$\\qquad D$$ is the Thread Diameter (e.g. 6 mm for M6).\n\n$$\\qquad C_p$$ is the Pitch Circumference (defined below).\n\n$$\\qquad D_p$$ is the Pitch Diameter (defined below).\n\n$$\\qquad P$$ is the Thread Pitch (for imperial threads, $$P = 1/TPI$$).\n\n$$\\qquad \\theta$$ is the Thread Angle.\n\n$$\\qquad \\alpha$$ is the Helix Angle.\n\n$$\\qquad \\delta$$ is the depth of thread.\n\n$$\\qquad M$$ is the dimension over the wires.\n\n$$\\qquad W_p$$ is the preferred wire size.\n\n$$\\qquad W_a$$ is the actual wire size.\n\n$$\\qquad W_{max}$$ is the maximum wire size.\n\n$$\\qquad W_{min}$$ is the minimum wire size.\n\n$$\\qquad k_1$$, $$k_2$$ and $$k_3$$ are constants, defined below.\n\n### Thread Depth\n\n$$\\delta = k_1 \\cdot P$$\n\nwhere $$k_1$$ depends on thread type as follows:\n\nMetric:\n\n$$k_1 = \\frac{5}{8} \\cdot \\cos(30)$$\n\nUNC:\n\n$$k_1 = 0.625 \\cdot \\cos(30)$$\n\nWhitworth/BSF:\n\n$$k_1 = 0.640327$$\n\nBA:\n\n$$k_1 = 0.6$$\n\n### Helix Angle\n\n$$\\alpha = \\frac{\\arctan(P)}{C_p}$$\n\nwhere,\n\n$$C_p = D_p \\cdot \\pi$$ $$D_p = D - \\left( 0.75 \\cdot P \\cdot \\cos \\left( \\frac{\\theta}{2} \\right) \\right)$$\n\n### Preferred Wire Size\n\n$$W_p = \\frac{0.5 \\cdot P}{\\cos\\left(\\frac{\\theta}{2}\\right)}$$\n\n### Minimum / Maximum Wire Size\n\n$$W_{max} = \\frac{0.873 \\cdot P}{\\cos\\left(\\frac{\\theta}{2}\\right)}$$ $$W_{min} = \\frac{0.485 \\cdot P}{\\cos\\left(\\frac{\\theta}{2}\\right)}$$\n\n### Dimension Over Wires\n\n$$M = D_p + \\left( k_2 \\cdot W_a \\right) - \\left( k_3 \\cdot P \\right)$$\n\nwhere $$k_2$$ and $$k_3$$ depend on the thread angle ($$\\theta$$):\n\n60° thread angles:\n\n$$k_2 = 3.0, k_3 = \\cos(30)$$\n\n55° thread angles:\n\n$$k_2 = 3.1657, k_3 = 0.9605$$\n\n47.5° thread angles:\n\n$$k_2 = 3.4829, k_3 = 1.1363$$"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.76904655,"math_prob":0.99969804,"size":3548,"snap":"2019-35-2019-39","text_gpt3_token_len":1098,"char_repetition_ratio":0.14221218,"word_repetition_ratio":0.024590164,"special_character_ratio":0.34470123,"punctuation_ratio":0.120815136,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999664,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-22T21:55:25Z\",\"WARC-Record-ID\":\"<urn:uuid:dc7d4a61-676e-46bc-b166-86518680b608>\",\"Content-Length\":\"20615\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e2e0bd5-6215-4fe1-8ef5-ce52b484e1be>\",\"WARC-Concurrent-To\":\"<urn:uuid:1a3b64f8-fefe-428b-8f98-f432970202dd>\",\"WARC-IP-Address\":\"172.217.164.179\",\"WARC-Target-URI\":\"https://www.cgtk.co.uk/metalwork/reference/screwmeasurement\",\"WARC-Payload-Digest\":\"sha1:KEOBF6HWNQUZL32MIILWECRAU6B6KLVU\",\"WARC-Block-Digest\":\"sha1:PUUC5P44LTEIUL6N3XYPH72JVUS63R2C\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027317516.88_warc_CC-MAIN-20190822215308-20190823001308-00358.warc.gz\"}"} |
http://www.binaryoptionsexplained.com/fibonacci-time-zones-strategy-binary-options | [
"# Fibonacci Time Zones Strategy for Binary Options\n\nThe Fibonacci Time Zone indicator is another tool out of the Fibonacci stable which is based on the Fibo ratios and works just like the Fibonacci retracement and expansion tools. However, the Fibonacci Time Zone indicator plots the areas where reversal of price action is to be expected as vertical lines as opposed to the horizontal lines of the retracement and expansion tool.\n\nThe starting point of the Fibonacci time zone strategy is to be able to draw it correctly on the charts, understand what the vertical lines signify and then use the information derived therein to create a strategy for trading the binary options market. Since this is a reversal strategy, it is best suited to trade the Call/Put strategy.\n\n## Drawing the Fibonacci Time Zones on the Chart\n\nThe Fibonacci Time Zones tool is based on a sequence of numbers known as the Fibonacci ratios. These number sequences are derived by adding the previous two numbers to get a third number. The sequence is thus listed as follows: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, etc). You can see that 1+2 = 3, 3 + 5 = 8, and so on. This will be reflected in the appearance and spacing of the time zone lines as they are plotted on the chart.\n\nThe Fibonacci time zone tool is drawn either from a swing high to low, or a swing low to high depending on what the trend of the asset is. The indicator will then draw the vertical lines that represent each time zone automatically. The lines tend to cluster at first, reflecting the closeness of the numbers that constitute the number sequence, and from there the lines tend to become less clustered. These lines are potential reversal points. After the lines are plotted, they will look something like what is shown below:",
null,
"The chart shows the Fibonacci Time Zone ratios drawn on the daily chart of the EURUSD. We will see here that the tool was drawn from a swing high to swing low, and we can see possible reversal points at time zone ratios 1 and 8. Many authorities will advise ignoring the first 5 time zone ratios, but then if you use a long term chart such as the daily chart as shown above, you can still find opportunities between the clustered areas.\n\nUsually, the Fibonacci Time Zone tool is not super-accurate when it comes to picking out these reversal points. The tool must be combined with other indicators of technical analysis, as well as candlestick patterns to pick out these areas of reversal as accurately as possible.\n\n## How to Trade Binary Options with Fibonacci Time Zones\n\nThe trades to take on the binary options platforms using the Fibonacci Time Zone tool are the Call option and Put option contracts. The expiry times of the contract is taken to be the time distance between time zone lines to be used for the analysis.\n\nWhen you trace the Fibo Time Zone tool from a swing low to high and vice versa, note that you are doing this in real time and so the ratios that will follow will appear on the chart in areas where the price action is yet to form, i.e. the parallel vertical ratio lines will form in the future. These will therefore form areas of projections of what the price action will do at those areas. In addition, it is also pertinent to state that the initial trace must be done between a high and low that is close together, unlike in the Fibonacci retracements/expansions where the low and high used for the trace are a bit apart from each other.",
null,
"The trace for the trade can be made either from swing high to low, or from swing low to high. This is because the Fibonacci time zone ratios are not very reliable indicators for predicting the direction of price movement. Hence we will combine this trace described with our colour-coded MACD histogram indicator.\n\nAfter the trace, note the colour of the MACD indicator where the time ratio line marked “2” is located. Then draw a horizontal line from the low of the candle on line 2 to the time ratio line marked “3”.\n\nIf the MACD histogram is blue at line 2, then the bias for the trade is upwards. Wait for the price action to bounce on the horizontal line between time zone line 2 and 3. If this bounce occurs on the line and at the same time the MACD histogram is blue, place a CALL option on this bounce with an 8-hour expiry time.",
null,
"Trace the time zone tool either from a swing low to swing high, or vice versa. Note the colour of the MACD histogram at the number 2 time zone line. Then trace a horizontal line from the high of the candle on line 2 across to line 3."
]
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"http://www.binaryoptionsexplained.com/wp-content/uploads/2014/03/fibonacci.jpg",
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"http://www.binaryoptionsexplained.com/wp-content/uploads/2014/03/fibonacci-call-trade.jpg",
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"http://www.binaryoptionsexplained.com/wp-content/uploads/2014/03/fibonacci-put-option.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.93026453,"math_prob":0.94434935,"size":5081,"snap":"2021-21-2021-25","text_gpt3_token_len":1088,"char_repetition_ratio":0.14417963,"word_repetition_ratio":0.030075189,"special_character_ratio":0.21078528,"punctuation_ratio":0.0801564,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98705536,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,6,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-12T01:43:57Z\",\"WARC-Record-ID\":\"<urn:uuid:763faf37-c6e9-485a-a641-83896599bf0d>\",\"Content-Length\":\"32768\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3e20a8f2-04e8-4182-8a09-19aab8a97785>\",\"WARC-Concurrent-To\":\"<urn:uuid:5b39bd77-74e9-4076-a400-7dd84c96ce96>\",\"WARC-IP-Address\":\"129.121.5.186\",\"WARC-Target-URI\":\"http://www.binaryoptionsexplained.com/fibonacci-time-zones-strategy-binary-options\",\"WARC-Payload-Digest\":\"sha1:KA7WSSWFQI6T6A5WT7AVW6VGM54BINBQ\",\"WARC-Block-Digest\":\"sha1:DWDKPHWH6HQU65BYRLOHU25LFJPCVLOJ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991693.14_warc_CC-MAIN-20210512004850-20210512034850-00296.warc.gz\"}"} |
https://tamref.com/48 | [
"# 레프네 약방\n\nKeith Nicholson - Introduction to Abstract Algebra, Good Exercises\n\n2018. 1. 11. 22:12수학 이론/추상대수학\n\nKeith의 책으로 현대대수를 공부하고 있는데, 연습문제 중에 괜찮은 개념을 담고 있거나 난도가 있는 문제들을 따로 정리해 두기로 한다.\n\nCH 2.4. Cyclic Groups and Order of an element\n\n24.\n\n(a) $$h$$ is the only element of $$\\text{order 2}$$ in a group $$G$$. show that $$h \\in Z(G)$$, where $$Z(G)$$ is the center of group $$G$$.\n\n(b) $$k$$ is the only element of $$\\text{order 3}$$ in a group $$G$$. What can you say about $$k$$?\n\n35.\n\n(a) Let $$a,b$$ are elements of a group $$G$$, and let $$m,n$$ be $$\\text{ord}(a)$$ and $$\\text{ord}(b)$$, respectively.\n\nIf $$ab = ba$$, show that $$G$$ has an element $$c$$, such that $$\\text{ord}(c) = \\text{lcm}(m,n)$$.\n\n(b) Let $$G$$ be an abelian. And assume that $$G$$ has an element of maximal order $$n$$.\n\nShow that $$\\forall g \\in G, \\ g^n = 1$$.\n\n37.\n\nFaro shuffle of a deck which contains $$2n$$ cards can be represented by the permutation $$\\phi$$ :\n\n$$\\phi = \\begin{pmatrix} 1 & 2 & 3 & 4 & \\cdots & 2n \\\\ 1 & n+1 & 2 & n+2 & \\cdots & 2n \\end{pmatrix}$$\n\nLet $$m \\ge 1$$ be a minimum integer such that $$\\iota = \\phi^{m}$$, where $$\\iota$$ is the identity permutation.\n\nExpress $$m$$ by terms about $$n$$.\n\n#### '수학 이론 > 추상대수학' 카테고리의 다른 글\n\n Normal Subgroup 이야기 (4) - Butterfly Lemma (Zassenhaus Lemma) (0) 2019.01.08 2019.01.08 2019.01.05 2019.01.04 2019.01.03 2018.01.11"
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https://groupprops.subwiki.org/wiki/Odd-order_implies_solvable | [
"# Odd-order implies solvable\n\nThis article gives the statement and possibly, proof, of an implication relation between two group properties. That is, it states that every group satisfying the first group property (i.e., odd-order group) must also satisfy the second group property (i.e., solvable group)\nView all group property implications | View all group property non-implications\nThis article states and (possibly) proves a fact that is true for certain kinds of odd-order groups. The analogous statement is not in general true for groups of even order.\nView other such facts for finite groups\n\nThis fact is useful in work leading up to the Classification of finite simple groups\n\n## History\n\n### Original proof\n\nThis result was proved by Feit and Thompson, and is called the Feit-Thompson Theorem or the Odd order theorem.\n\n### Computer verification of proof\n\nIn September 2012, it was announced that the Feit-Thompson theorem proof had been completely verified using the proof assistant Coq.\n\n## Statement\n\nThere are two versions of the statement:\n\n1. There is no finite simple non-abelian group that has odd order.\n2. Every odd-order group is a solvable group, i.e., all its composition factors are abelian groups (and hence, cyclic of prime order).\n\n## Applications\n\nThere are several applications of the result that given two groups of coprime order, one of them is solvable. In fact, many theorems in group theory can be proved modulo the assumption that among two given groups of coprime order, one is solvable. For a list of such facts, refer:\n\n## Proof\n\n### Proof that (1) implies (2)\n\nSuppose (1) holds, i.e., there is no finite simple non-abelian group of odd order. Then, we want to show that (2) holds. Consider a minimal counterexample to (2), i.e., an odd-order group",
null,
"$G$ that is not solvable and has the minimum possible order among such groups.\n\nStep no. Assertion/construction Facts used Given data/assumptions used Previous steps used Explanation\n1",
null,
"$G$ is nontrivial. the trivial group is solvable\n2",
null,
"$G$ is not simple non-abelian. (1) holds, i.e., there is no simple non-abelian group of odd order direct\n3",
null,
"$G$ must has a proper nontrivial normal subgroup",
null,
"$N$ Steps (1), (2) direct\n4 Both",
null,
"$N$ and",
null,
"$G/N$ have odd order. Fact (1) Step (3) Step-fact direct\n5 Both",
null,
"$N$ and",
null,
"$G/N$ have order strictly smaller than",
null,
"$G$. Step (3)",
null,
"$N$ is proper and nontrivial in",
null,
"$G$.\n6 Both",
null,
"$N$ and",
null,
"$G/N$ are solvable.",
null,
"$G$ is a counterexample of minimal order Steps (4), (5) direct\n7",
null,
"$G$ is solvable, leading to a contradiction to our assumption that it is a counterexample. Fact (2) Step (6) Step-fact direct.\n\n### Proof of (1)\n\nThe complete proof (which forms the subject of a 255-page paper) is beyond the scope of this page. However, we will attempt to describe the key idea.\n\nThe idea is to attempt to construct a \"minimal counterexample\" to (1), i.e., a simple non-abelian group of odd order that has the smallest possible order among such groups. Any such minimal counterexample must in particular be a minimal simple group: every proper subgroup is solvable (note therefore that the classification of finite minimal simple groups would settle this question; however, such a classification is itself conditional to first proving this fact, hence it does not help). In particular, it is a N-group: every local subgroup (the normalizer of a nontrivial solvable subgroup) is solvable. Or equivalently, for every prime",
null,
"$p$, every p-local subgroup is solvable (this follows from the fact that local subgroup of finite group is contained in p-local subgroup for some prime p).\n\n### The CN-group case\n\nAn example that is relatively easy to follow is the proof that odd-order and CN implies solvable.\n\nA CN-group is a group where the centralizer of every non-identity element is nilpotent. The structure of CN-groups allows us to define an equivalence relation on the prime divisors of the group order based on commuting (see commuting of non-identity elements defines an equivalence relation between prime divisors of the order of a finite CN-group). For each equivalence class",
null,
"$\\omega$ under the equivalence relation on the prime divisors of the order of a finite CN-group",
null,
"$G$,",
null,
"$G$ has a nilpotent",
null,
"$\\omega$-Hall subgroup.\n\nWhat we have said so far applies to all finite CN-groups. Restricting to the \"minimal odd-order counterexample\" that we want to show does not exist, we can obtain more structural restrictions on the nature of the Hall subgroups. The structural restrictions obtained from the proof mimic those used in the proof that finite non-abelian and every proper subgroup is abelian implies not simple. However, they are not quite as strong, and ultimately, we need to use ideas from character theory to complete the proof."
]
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https://de.maplesoft.com/support/help/maple/view.aspx?path=define_external/CustomWrapper | [
"",
null,
"define_external(MAPLE) - Maple Programming Help\n\nHome : Support : Online Help : Programming : Calling External Routines : define_external/CustomWrapper\n\ndefine_external(MAPLE)\n\nconnect to a custom external wrapper library\n\n Calling Sequence define_external( functionName, MAPLE, LIB=libName)\n\nParameters\n\n functionName - name of externally defined function MAPLE - keyword indicating the wrapper library is custom built libName - name of external library containing the function\n\nDescription\n\n • For complete control over data conversions, Maple allows modification of existing wrappers and creation of custom wrappers. There is an extensive API of functions available to external programs called from Maple using define_external.\n • If instead of the arg parameters, the single word MAPLE is specified, the external function is assumed to accept the raw Maple data structures passed without conversion. This assumes that the custom wrapper has been manually generated and compiled into a DLL.\n • When declaring a wrapper external function in Maple via define_external, there is no need to specify the parameter types. The wrapper always gets the same two arguments -- a kernel handle and an expression sequence (dag) of arguments.\n • C wrappers need to include the file maplec.h. This file is located in the extern/include subdirectory of your Maple installation.\n • The main entry point in your custom wrapper should accept two parameters, the MKernelVector handle and something of type ALGEB. The entry point prototype will look like the following.\n\n ALGEB MWRAP_mult( MKernelVector kv, ALGEB fn_args );\n\n • Note that the generated wrapper uses a slightly different argument list than the custom wrapper. The custom wrapper does not get a pointer to the main external function. The generated wrapper main entry point for a function that takes two int arguments and returns an int has the following prototype.\n\n ALGEB MWRAP_mult( MKernelVector kv, INTEGER32 (*fn) ( INTEGER32 a1, INTEGER32 a2 ), ALGEB fn_args );\n\n • When Maple calls this main entry point, it passes the kernel handle and an EXPSEQ dag. Index directly into the EXPSEQ dag to get at the raw Maple arguments. Use the NumArgs function available in maplec.dll to determine the number of arguments.\n • For more information on the external API, refer to the Maple Programming Guide section External Function Interface in the Advanced Connectivity chapter. Alternately, refer directly to the C header file, extern/include/maplec.h\n\nExamples\n\nConsider the following C wrapper function saved in mylib.dll. It takes three arguments, args is a Matrix, args is an integer, and args is a string. The function finds all the values in args that satisfy the relation \"<\", \">\", \"=\", or \"!=\" args. The relation is given in args. The number of entries found is displayed, and a vector result is returned.\n\n ;include \"maplec.h\" ALGEB ExtractElems( MKernelVector kv, ALGEB args ) { char relation; int n, elems, result_elems, count, i; ALGEB Matrix, result; RTableSettings settings; M_INT bounds; /* basic argument checking */ if( MapleNumArgs(kv,args) != 3 || !IsMapleRTable(kv,(ALGEB)args) || !IsMapleInteger32(kv,(ALGEB)args) || !IsMapleString(kv,(ALGEB)args) ) { MapleRaiseError(kv,\"usage: ExtractElems(Matrix,integer,string)\"); } /* find more about the Matrix */ Matrix = (ALGEB)args; RTableGetSettings(kv,&settings,Matrix); if( settings.data_type != RTABLE_INTEGER32 || settings.storage != RTABLE_RECT || !IsMapleNULL(kv,settings.index_functions) ) { /* not the format we wanted -- make a copy */ settings.data_type = RTABLE_INTEGER32; settings.storage = RTABLE_RECT; settings.index_functions = ToMapleNULL(kv); settings.foreign = FALSE; Matrix = RTableCopy(kv,&settings,Matrix); } /* get a pointer to the actual Matrix data */ elems = (int)RTableDataBlock(kv,Matrix); /* translate the other arguments */ n = MapleToInteger32(kv,(ALGEB)args); relation = MapleToString(kv,(ALGEB)args); /* scan over the elements linearly -- we don't need to look at RTableNumDimensions, RTableLowerBound, and RTableUpperBound because we don't care where the elements are, just what they are. */ /* first count the elements that satisfy the relation */ count = 0; switch( relation ) { case '>': for( i=0; i n ) count++; break; case '<': for( i=0; i': for( i=0; i n ) result_elems[count++] = elems[i]; break; case '<': for( i=0; i\n\nAfter compiling this function into a .dll, it can be used directly in Maple as follows.\n\n > $\\mathrm{extract}≔\\mathrm{define_external}\\left('\\mathrm{ExtractElems}','\\mathrm{MAPLE}',\\mathrm{LIB}=\"mylib.dll\"\\right):$\n > $M≔\\mathrm{Matrix}\\left(4,4,\\left(i,j\\right)↦4\\cdot i+j-12\\right):$\n > $\\mathrm{extract}\\left(M,0,\"<\"\\right)$\n $\\left[{-7}{,}{-3}{,}{-6}{,}{-2}{,}{-5}{,}{-1}{,}{-4}\\right]$ (1)\n\n7 elements satisfied the relation ${\\mathrm{elem}}_{i}<0$.\n\n > $\\mathrm{extract}\\left(M,0,\">\"\\right)$\n $\\left[{1}{,}{5}{,}{2}{,}{6}{,}{3}{,}{7}{,}{4}{,}{8}\\right]$ (2)\n\n8 elements satisfied the relation $0<{\\mathrm{elem}}_{i}$.\n\n > $\\mathrm{extract}\\left(M,0,\"=\"\\right)$\n $\\left[{0}\\right]$ (3)\n\n1 element satisfied the relation ${\\mathrm{elem}}_{i}=0$.\n\n > $\\mathrm{extract}\\left(M,0,\"!=\"\\right)$\n $\\left[{-7}{,}{-3}{,}{1}{,}{5}{,}{-6}{,}{-2}{,}{2}{,}{6}{,}{-5}{,}{-1}{,}{3}{,}{7}{,}{-4}{,}{4}{,}{8}\\right]$ (4)\n\n15 elements satisfied the relation ${\\mathrm{elem}}_{i}\\ne 0$."
]
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"https://bat.bing.com/action/0",
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http://www.powerfromthesun.net/calculators/LocalToSolarTime.html | [
"# Local to Solar Time Converter\n\nPROGRAM LocalToSolarTime - This program uses the equations developed in this book to calculate the solar time (Eqn. 3.5) at a specified location, if you know the local (clock) time. In order to do this, you need to know whether or not it is daylight savings time, the local longitude, the day number (1 to 366) and your time zone. Time zones are specified as the number of hours east (+) or west (-) of Greenwich Mean Time zone. Examples are: Western Europe = +1 hours, Greenwich Mean Time = 0 hours, Eastern Standard Time = -5 hours, Central Standard Time = -6 hours, Mountain Standard Time = -7 hours and Pacific Standard Time is -8 hours. In this program, the simplified Equation of Time (Eqn. 3.2) is used.\n\n Longitude (degrees): (negative west, and positive east of Prime Meridian) Day Number: Time Zone (hours from GMT): (negative west of GMT and positive east of GMT) Daylight Savings Time? : (0 = no, 1 = yes) Local (clock) Time (hr): (24-hour decimal format) or Local (clock) Time (hh:mm:ss) : :"
]
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http://www.numbersaplenty.com/420562573 | [
"Search a number\nBaseRepresentation\nbin11001000100010…\n…100011010001101\n31002022100210101121\n4121010110122031\n51330131000243\n6105422040541\n713264502644\noct3104243215\n91068323347\n10420562573\n111a643a982\n12b8a19151\n1369190ac2\n143dbd835b\n1526dc61ed\nhex1911468d\n\n420562573 has 4 divisors (see below), whose sum is σ = 420611488. Its totient is φ = 420513660.\n\nThe previous prime is 420562559. The next prime is 420562603. The reversal of 420562573 is 375265024.\n\nIt is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also a brilliant number, because the two primes have the same length.\n\nIt is a cyclic number.\n\nIt is a de Polignac number, because none of the positive numbers 2k-420562573 is a prime.\n\nIt is a Duffinian number.\n\nIt is a congruent number.\n\nIt is not an unprimeable number, because it can be changed into a prime (420562073) by changing a digit.\n\nIt is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 7761 + ... + 30022.\n\nIt is an arithmetic number, because the mean of its divisors is an integer number (105152872).\n\nIt is a 8598-hyperperfect number.\n\nAlmost surely, 2420562573 is an apocalyptic number.\n\nIt is an amenable number.\n\n420562573 is a deficient number, since it is larger than the sum of its proper divisors (48915).\n\n420562573 is a wasteful number, since it uses less digits than its factorization.\n\n420562573 is an evil number, because the sum of its binary digits is even.\n\nThe sum of its prime factors is 48914.\n\nThe product of its (nonzero) digits is 50400, while the sum is 34.\n\nThe square root of 420562573 is about 20507.6223146419. The cubic root of 420562573 is about 749.2214575901.\n\nThe spelling of 420562573 in words is \"four hundred twenty million, five hundred sixty-two thousand, five hundred seventy-three\".\n\nDivisors: 1 11131 37783 420562573"
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https://www.jpost.com/international/royal-iran-is-not-to-be-trusted | [
"(function (a, d, o, r, i, c, u, p, w, m) { m = d.getElementsByTagName(o), a[c] = a[c] || {}, a[c].trigger = a[c].trigger || function () { (a[c].trigger.arg = a[c].trigger.arg || []).push(arguments)}, a[c].on = a[c].on || function () {(a[c].on.arg = a[c].on.arg || []).push(arguments)}, a[c].off = a[c].off || function () {(a[c].off.arg = a[c].off.arg || []).push(arguments) }, w = d.createElement(o), w.id = i, w.src = r, w.async = 1, w.setAttribute(p, u), m.parentNode.insertBefore(w, m), w = null} )(window, document, \"script\", \"https://95662602.adoric-om.com/adoric.js\", \"Adoric_Script\", \"adoric\",\"9cc40a7455aa779b8031bd738f77ccf1\", \"data-key\");\nvar domain=window.location.hostname; var params_totm = \"\"; (new URLSearchParams(window.location.search)).forEach(function(value, key) {if (key.startsWith('totm')) { params_totm = params_totm +\"&\"+key.replace('totm','')+\"=\"+value}}); var rand=Math.floor(10*Math.random()); var script=document.createElement(\"script\"); script.src=`https://stag-core.tfla.xyz/pre_onetag?pub_id=34&domain=\\${domain}&rand=\\${rand}&min_ugl=0\\${params_totm}`; document.head.append(script);"
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https://investinganswers.com/dictionary/d/dividend-discount-model-ddm | [
"## What is the DDM?\n\nThe discount model (DDM) is a method for assessing the present value of a stock based on the growth rate of dividends.\n\n## How the DDM Works\n\nThe dividend discount model (DDM) seeks to estimate the current value of a given stock on the basis of the spread between projected dividend growth and the associated discount rate. The DDM calculates this present value in the following manner:\n\nPresent Stock Value = DividendShare / (RDiscount – RDividend Growth)\n\nIn the DDM, a present stock value that is higher than a stock's market value indicates that the stock is undervalued and that it is a good time to purchase shares.\n\nTo illustrate, suppose stock XYZ declares a dividend of two dollars per share and is currently valued at \\$125 in the market. Based on the stock's dividend history, a broker determines a dividend growth rate for the stock of five percent per year and a discount rate of seven percent. The present stock value is calculated as follows:\n\nPresent Stock Value = \\$2.00 per share / (0.07 discount – 0.05 dividend growth)\n= \\$2.00 / 0.02\n= \\$100\n\nWith a calculated present value of \\$100 against a market value of \\$125, stock XYZ is overvalued in this instance and represents an opportunity to sell.\n\n## Why the DDM Matters\n\nThe DDM is a tool used by many investors and analysts as a simple method to choosing stocks. The greatest disadvantage of the DDM is that it is inapplicable to companies which do not pay dividends."
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https://www.nigerialocalnews.com/converter/kelvin.html | [
"Kelvin to celsius Calculator - Use this Kelvin to celsius calculator to determine your k to C based on the formula of k to C conversion, which is 1 Kelvin is equivalent to -272.15 celsius.\n\nTo calculate just use formula -272.15 celsius * Kelvin. The calculator below uses that principle to derive its answer. Use it to do your calculations faster.\n\nKELVIN TO CELSIUS CONVERSION\n\nKelvin in celsius\n\nConversion chart for Kelvin to celsius",
null,
"Examples for Kelvin to celsius\n\n100*-272.15= -27215\n\n200*-272.15= -54430\n\n300*-272.15= -81645\n\n400*-272.15= -108860\n\n500*-272.15= -136075\n\n600*-272.15= -163290\n\n700*-272.15= -190505\n\n800*-272.15= -217720\n\n900*-272.15= -244935\n\n1000*-272.15= -272150\n\n1100*-272.15= -299365\n\n1200*-272.15= -326580\n\n1300*-272.15= -353795\n\n1400*-272.15= -381010\n\n1500*-272.15= -408225\n\n1600*-272.15= -435440\n\n1700*-272.15= -462655\n\n1800*-272.15= -489870\n\n1900*-272.15= -517085\n\n2000*-272.15= -544300\n\nThe example above illustrate simple calculations with different numbers. We provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, currency, etc.\n\nHow many Kelvin are there in x celsius ?\n\nHow do i convert k to C?\n\nHow many Kelvin make 1 celsius ?\n\nHow can I convert k to C?\n\nTo find out the answer to any of these questions, simply select the appropriate unit from each 'select' box above, enter your figure (x) into the 'value to convert' box and click the 'Convert!' button.\n\nWhilst every effort has been made in building this tool, I am not to be held liable for any special, incidental,indirect or consequential damages or monetary losses of any kind arising out of or in connection with the use of the converter tools and information derived from the web site. This is here purely as a service to you, please use it at your own risk. Do not use calculations for anything where loss of life, money, property, etc or anything that could result to damage from inaccurate conversions.\n\nIf you spot an error on this site, we would be grateful if you could report it to us by using the contact email provided at the bottom of this page and we will endeavour to correct it as soon as possible. send email to contact at nigerialocalnews.com\n\nThis site provides an online conversion calculator for all types of measurement units. You can find metric conversion tables for SI units, as well as English units, currency, and other data. Type in unit symbols, abbreviations, or full names for units of length, area, mass, pressure, and other types.\n\n• Base64encode\n• Bbcodetohtml\n• Bbcodetotext\n• Bcryptcheck\n• Bcryptencrypt\n• Binarytoip\n• Changedelimiter\n• Cssminify\n• Cssprettify\n• Csvappendcolumn\n• Csvchangedelimiter\n• Csvcolumnstorows\n• Csvdeletecolumn\n• Csvextractcolumn\n• Csvinsertcolumn\n• Csvprependcolumn\n• Csvreplacecolumn\n• Csvrowstocolumns\n• Csvswapcolumns\n• Csvtojson\n• Csvtotext\n• Csvtotsv\n• Csvtoxml\n• Csvtoyaml\n• Csvtranspose\n• Decimaltoip\n• Deletecolumn\n• Extractcolumn\n• Hextoip\n• Htmldecode\n• Htmlencode\n• Htmlminify\n• Htmlprettify\n• Htmlstrip\n• Htmltomarkdown\n• Htmltotext\n• Index\n• Ipsort\n• Iptobinary\n• Iptodecimal\n• Iptohex\n• Iptooctal\n• Jsminify\n• Jsprettify\n• Jsvalidate\n• Jsonescape\n• Jsonminify\n• Jsonprettify\n• Jsontocsv\n• Jsontotext\n• Jsontotsv\n• Jsontoxml\n• Jsontoyaml\n• Jsonunescape\n• Jsonvalidate\n• Markdowntohtml\n• Metataggenerator\n• Octaltoip\n• Randomlines\n• Replacecolumn\n• Swapcolumns\n• Textcolumnstorows\n• Textrowstocolumns\n• Texttocsv\n• Texttohtmlentities\n• Texttotsv\n• Texttranspose\n• Tsvappendcolumn\n• Tsvchangedelimiter\n• Tsvcolumnstorows\n• Tsvdeletecolumn\n• Tsvextractcolumn\n• Tsvinsertcolumn\n• Tsvprependcolumn\n• Tsvreplacecolumn\n• Tsvrowstocolumns\n• Tsvswapcolumns\n• Tsvtocsv\n• Tsvtojson\n• Tsvtotext\n• Tsvtoxml\n• Tsvtoyaml\n• Tsvtranspose\n• Unixtoutc\n• Urldecode\n• Urlencodedecode\n• Urlencode\n• Urlparse\n• Utctounix\n• Xmlminify\n• Xmlprettify\n• Xmltocsv\n• Xmltojson\n• Xmltotext\n• Xmltotsv\n• Xmltoyaml\n• Yamltocsv\n• Yamltojson\n• Yamltotsv"
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https://math.stackexchange.com/questions/133185/explaining-horizontal-shifting-and-scaling/133348 | [
"# Explaining Horizontal Shifting and Scaling\n\nI always find myself wanting for a clear explanation (to a college algebra student) for the fact that horizontal transformations of graphs work in the opposite way that one might expect.\n\nFor example, $f(x+1)$ is a horizontal shift to the left (a shift toward the negative side of the $x$-axis), whereas a cursory glance would cause one to suspect that adding a positive amount should shift in the positive direction. Similarly, $f(2x)$ causes the graph to shrink horizontally, not expand.\n\nI generally explain this by saying $x$ is getting a \"head start\". For example, suppose $f(x)$ has a root at $x = 5$. The graph of $f(x+1)$ is getting a unit for free, and so we only need $x = 4$ to get the same output before as before (i.e. a root). Thus, the root that used to be at $x=5$ is now at $x=4$, which is a shift to the left.\n\nMy explanation seems to help some students and mystify others. I was hoping someone else in the community had an enlightening way to explain these phenomena. Again, I emphasize that the purpose is to strengthen the student's intuition; a rigorous algebraic approach is not what I'm looking for.\n\n• Some time ago I took an example of $\\alpha\\cdot x+\\beta = c$, $\\alpha > 0$, and explained that $x$ gets smaller as $\\alpha$ or $\\beta$ gets bigger, all to keep the balance enforced by equality sign \"$=$\". I have no idea if this is intuitive, but it helped the people I was talking to. – dtldarek Apr 17 '12 at 23:02\n• Perhaps this should be community wiki? – Jim Belk Apr 17 '12 at 23:12\n• I too lack a great explanation, but I find that the example of $f(x)=x$ aids in remembering the behavior. There you can show that $f(x+1)=f(x)+1$ is both a positive vertical shift and a negative horizontal shift, and $f(2x)=2f(x)$ is both a vertical expansion and a horizontal shrinking. This helps cement the idea that the transformations work in opposite directions (but may cause other misconceptions if you don't follow up quickly with other examples). – AMPerrine Apr 17 '12 at 23:17\n• This is a very difficult question to answer. I'm trying to compose an answer from my perspective, as I too dealt with this oddity (as a student). – 000 Apr 17 '12 at 23:20\n• Perhaps the question should be from who's perspective does the graph shift? The graph of $f$ shifted to the right $1$ 'uses' $x-1$ to get back to the original function? – john w. Apr 18 '12 at 1:40\n\nHere is a setting where the students will have experience, having nothing to do with skills in algebra: daylight savings time (since you're in the USA). When we advance clocks ahead by an hour we lose an hour of sleep. Thus replacing $t$ with $t+1$ shifts your schedule for the day back by one hour.\n\nHere, $t$ is the old \"function\" and $t + 1$ is the new \"function\". $t$ is one hour behind $t + 1$, but after this transformation, $t + 1$ shortens the day by one hour. So in the end, we lose an hour. Point out to the students that it also works the other way when we turn the clocks back ($t$ is replaced with $t-1$) and then gain an hour in our schedule.\n\n• That's very clever! – Austin Mohr Apr 17 '12 at 23:37\n• I think looking at shifts of $y=x^2$ is easier to use than shifts of the line $y=x$, because one's attention focuses on where the parabola touches the $x$-axis instead of the crossing. – KCd Apr 19 '12 at 2:53\n• I love this, especially because there is an intuitive way for people with weak math backgrounds to understand why this happens (without a need for actually writing down a specific wake-up time to see what happens, analogous to setting x=5 in the original post): if you add an hour to the clocks, you will have to skip an hour (i.e., cut one out of the day) at some time. Then, of course, if everyone skips that hour, people have less time for sleep, work, etc. – Barry Smith Apr 20 '12 at 18:20\n• There is a way to make this into an intuitive explanation for students (although probably too complicated to be useful) --- imagine traversing the graph of $f(x)$ from left to right, and then at some specific value $x=c$, begin traversing $f(x+1)$ instead. You will have \"skipped\" the part of the graph of $f$ with $x$ in the interval $(c,c+1)$, but you didn't skip any part of the $x$-axis. So the graph you create would be the graph of $f$ with the part between $(c,c+1)$ snipped out and the part of the graph to the right of $x=c$ ''pulled to the left''. – Barry Smith Apr 20 '12 at 18:28\n• I shall also mention daylight savings in explaining re-indexing series. – Barry Smith Apr 20 '12 at 18:30\n\nEnter, the Function Monkey ...\n\nAs described in the link, to plot a graph, simply imagine the $x$-axis covered in coconuts, one for every $x$ value, like this:\n\n$$\\cdots \\quad (-3\\;) \\quad \\color{#9509A5}{(-2\\;)} \\quad \\color{green}{(\\;-1\\;)} \\quad \\color{red}{[\\;0\\;]} \\quad \\color{ #2B87CD}{(\\;1\\;)} \\color{#D86907}{\\quad (\\;2\\;)} \\quad (\\;3\\;) \\quad \\cdots$$\n\nwith \"$\\color{red}{[\\;\\cdot\\;]}$\" indicating the coconut at the origin. The Function Monkey strolls along the axis, picks up each $x$ coconut, evaluates the corresponding $y$ value (as is his wont), and throws the coconut to the appropriate height (or depth). Graph plotted!\n\nBut, wait! No one ever said that the values of the coconuts were required to match the values of the $x$ coordinates which have been coloured for convenience. Hmmmm ...\n\nIf the coconut at (the coloured) location $x$ has value $x+1$, then the row of coconuts looks like this: $$\\cdots \\quad (-2\\;) \\quad \\color{#9509A5}{(\\;-1\\;)} \\quad \\color{green}{(\\;0\\;)} \\quad \\color{red}{[\\;1\\;]} \\quad \\color{ #2B87CD}{(\\;2\\;)} \\color{#D86907}{\\quad (\\;3\\;)} \\quad (\\;4\\;) \\quad \\cdots$$\n\nImportantly, the coconut locations in colour have not changed. The coconuts still sit on the axis at locations $x = \\cdots, -3, \\color{#9509A5}{-2},\\color{green}{-1}, \\color{red}{0},\\color{#2B87CD}{1}, \\color{#D86907}{2}, 3, \\cdots$ still indicating the coconut at the origin.\n\nSo far as the Function Monkey is concerned, adding $1$ to each coconut value has effectively shifted the row of coconuts to the left. (Likewise, adding $-1$ to each coconut value effectively shifts the row of coconuts to the right.) Since the Function Monkey throws coconuts vertically, the plotted graph shifts the same way.\n\nIn a similar manner, multiplying each (original) coconut value by $2$ yields this row of coconuts:\n\n$$\\cdots \\quad (-6\\;) \\quad \\color{#9509A5}{(-4\\;)} \\quad \\color{green}{(-2\\;)} \\quad \\color{red}{[\\;0\\;]} \\quad \\color{ #2B87CD}{(\\;2\\;)} \\color{#D86907}{\\quad (\\;4\\;)} \\quad (\\;6\\;) \\quad \\cdots$$\n\nAgain, the locations of the coconuts haven't changed, but we see that the span of coconut values from $-6$ to $6$ has been compressed into the space between locations $x=-3$ and $x=3$. The graph will exhibit the same, horizontal, compression.\n\nA plot-less way of thinking about this, which is close to your \"head start\" interpretation, is that \"$f(x+1)$\" fools Function Monkey $f$ into thinking that each input value is one unit bigger than it really is. Similarly \"$f(2x)$\" fools him into thinking that each input is twice its actual size. The effective changes in output values correspond to the Function Monkey's altered perceptions.\n\n• The Function Monkey: the metaphor that keeps on giving... – J. M. is a poor mathematician Apr 18 '12 at 11:48\n• @J.M.: \"To a man with a hammer ...\" or, as the case may be, a hominid ... – Blue Apr 18 '12 at 15:34\n\nYou should change variables when changing the coordinate system! x + 1 is confusing because you're re-using the variable. x is already used for representing our \"home base\" frame of reference.\n\nFor instance, pick the symbols s and t, and change to the coordinate system:\n\nlet s = x + 1\nlet t = y + 3\n\n\nNow this does not mean we are moving up and to the right!\n\nThink: where is the origin of the <s, t> coordinate system in the <x, y> coordinate system? It is at <-1, -3>, because we have to solve by setting <s, t> to <0, 0>; i.e. solve x + 1 = 0 and y + 3 = 0.\n\nThe <x, y> system must remain our frame of reference if we are to visualize the motion, and so what we do is rework the equations to isolate x and y:\n\nx = s - 1\ny = t - 3\n\n\nNow you can see that it's a move left and down and this is consistent with the minus signs.\n\nThe \"backwards\" issue arises when you put yourself into the frame of reference of the transformation, but continue using <x, y> to think about that frame. You have to think about the backwards mapping: how do you recover <x, y> from that other frame of reference?\n\nWhen you see x + 1, you must not imagine that x is moving. (This may be \"brain damage\" from working in imperative computer programming languages, especially ones like C and BASIC where the = sign is used for assignment: x = x + 1 moves coordinate x to the right.) Rather, x is used as the input to a calculation that produces some other value. And so x is that other value, minus 1.\n\nIf I say my money = your money + 1 are you richer than me? No, you're poorer by a dollar! That's easy to see because when the variables have these names, of course you instantly and intuitively put yourself into the your money frame of reference, and easily see that although 1 is being added to your money, the result of that formula expresses my money and not yours! Haha.\n\nWhen I shift the graph of $f$ one unit to the right I create a new function $g$. To find out the value $g(x)$ I have to look what the value of $f$ was one unit to the left of $x$. therefore $g(x):=f(x-1)$.\n\nIntuitively, the graph of $f(x+1)$ is horizontally shifted to the left because the coordinate system itself is shifted to the right. Likewise, the graph of $f(2x)$ is shrunk horizontally because the coordinate system is stretched by a factor of $2$.\n\n• I've heard this explanation offered before, but, in my experience, it only serves to make vertical transformations confusing. \"If $f(x+1)$ shifts the axis to the right, why doesn't $f(x) + 1$ shift the axis upward?\" – Austin Mohr Apr 17 '12 at 23:08\n• @AustinMohr: Roughly, a transformation on the argument of $f(x)$ is a transformation on the coordinate system. A transformation on $f(x)$ is a transformation on the graph. A function and its argument play very different roles! – user26872 Apr 17 '12 at 23:14\n• Also, $y=f(x)+1$ is equivalent to $y-1=f(x)$. So, the effect on $y$ is exactly analogous to the effect on $x$: subtracting $1$ from the coordinate variable corresponds to a \"positive\" translation along the coordinate axis, adding $1$ corresponds to a \"negative\" translation. Likewise, with $y=2f(x)$ the equivalent of $y/2=f(x)$. – Blue Apr 18 '12 at 12:18\n• @DayLateDon: Well put! – user26872 Apr 18 '12 at 14:52\n• @Blue What does 'coordinate variable' mean please? – Greek - Area 51 Proposal Nov 22 '18 at 18:06\n\nThe map $$f$$ already assigns $$\\color{Purple}{\\sigma x\\mapsto f(\\sigma x)}$$. In order to construct the assignment $$\\color{DarkBlue}x\\mapsto \\color{DarkOrange}{f(\\sigma x)}$$, we must apply the inverse $$\\sigma^{-1}$$ in order to put $$f(\\sigma x)$$ (originally located at $$\\sigma x$$) \"back\" to $$x$$, that is\n\n$$(\\sigma^{-1},\\,\\mathrm{Id}):\\big(\\sigma x,f(\\sigma x)\\big)\\mapsto \\big(x,f(\\sigma x)\\big).$$\n\nThe above is, of course, too algebraic. So here's a colorful depiction of what's going on:\n\n$$\\hskip 1in$$",
null,
"As we can see, \"putting it back over the $$x$$-value\" does the inverse of the transform to the actual graph (squiggly line) of $$f$$.\n\n• No offense intended, but I don't think this explanation will do well with a college algebra student. (This reminds me of Category Theory and frankly terrifies me.) – 000 Apr 17 '12 at 23:34\n• @Limitless: What if I substituted $\\sigma, f$ and $\\sigma^{-1}$ respectively with the words \"transform,\" \"apply function,\" and \"put it back over $x$-value\"? – anon Apr 17 '12 at 23:37\n• Oh, wow.... Bravo, Sir. It makes perfect sense now. Thank you. I think I was more terrified by the sense abstraction and symbols than its actual degree of difficulty. I'll have to keep this abstract principle of ignoring the intimidating aspects of mathematics in mind. ;) – 000 Apr 17 '12 at 23:50\n\nIt may help to try to comprehend the abridged version of Kaz's answer: https://math.stackexchange.com/a/133348/53259.\n\nBecause we want to work with $(x,y)$, the key is to remain focused on the $xy$-axis in black. It truly helps to use new variables for the new coordinate axes. So define $\\color{green}{\\left\\{ \\begin{array}{rcl} s = x + 1 \\\\ t = y + 3 \\end{array}\\right.} \\iff \\left\\{ \\begin{array}{rcl} x = s - 1 \\\\ y = t - 3 \\end{array}\\right.$. Thus, this shows the source of the bewilderment. We should remain thinking of the black and not the green.",
null,
"What does the graph of $g(x) = f(x+1)$ look like? Well, $g(0)$ is $f(1)$, $g(1)$ is $f(2)$, and so on. Put another way, the point $(1,f(1))$ on the graph of $f(x)$ has become the point $(0,g(0))$ on the graph of $g(x)$, and so on. At this point, drawing an actual graph and showing how the points on the graph of $f(x)$ move one unit to the left to become points on the graph of $g(x)$ helps the student understand the concept. Whether the student absorbs the concept well enough to utilize it correctly later is quite another matter.\n\nWell we replaced everywhere $x$ by $x'=x+1$, $x'=2x$ or whatever so that the 'old' variable $x$ was shifted by $-1$ ($x=x'-1$), divided by $2$ or whatever. Of course you'll have to convince yourself first! :-)\n\nTo expand a bit (and considering Oenamen's comment) I'll add that you may use the physical concept of Active and passive transformation :\n\n• $f(x)\\to f(x)+1$ is an active transformation : in a fixed frame the curve became higher of a unit\n\n• $f(x)\\to f(x+1)$ is a passive transformation : the frame of reference changed (that's the case of interest for you : the old frame moved one unit left)\n\nAfter whacking my head for a rather irksome period, I've concluded this is an error in our perception. It's deceptively natural to see the $+$ and immediately think that the graph tends toward the positive, rather than the negative. I think this is one of those cases of mathematical notation gone wrong. Rather than encouraging correct patterns, it is encouraging incorrect patterns.\n\nIt's just like how, without knowing any better, one may presume that: $\\lim_{x \\to 0^{+}}f(x)$ is intuitively the value that $f(x)$ approaches as one approaches $0$ from the right to the left (i.e. it gets big as it comes toward $0$, or more positive) rather than the fact that it actually comes from the left to the right (i.e. it gets smaller as it comes toward $0$ or more negative*). Pardon me if that choice of words is rather confusing; it's more emphatic of how confusing our notation is with regard to 'left' and 'right'.\n\nIn short, I don't think there is an adequate intuitive justification for this deceptive notation.\n\n(*=Don't take the phrases \"more postive\" and \"more negative\" too seriously in this context. They respectively indicate, \"closer to positive numbers\" and \"closer to negative numbers\".)\n\nNot sure if this works in every situation but...\n\nSay you have function f(x) = X^2 with domain [-1,1]\n\nand function g(x) = f(x + 1)...\n\nThe correct domain for g(x) should be [-2,0]\n\nOne can arrive at this conclusion using the domain of f(x)\n\nx + 1 = -1 => x = -1 - 1 => x = -2.\n\nand similarly...\n\nx + 1 = 1 => x = 1 - 1 => x = 0\n\nGiving you the domain for g(x) = [-2, 0]\n\nI try to explain this concept several different ways, using some of the more mathematical approaches above. Sometimes I run into students who still just don't get it, so I offer them this analogy:\n\nPretend your car represents a mathematical function. The inputs are the gas you put in the tank, and the output is how far you can drive. Let's say that you fill up your gas tank, which means you can then drive a set distance. Further suppose that one of your jerk friends comes by in the night and siphons off 5 gallons of gas. If you want to drive the same distance (i.e. achieve the same outputs), you then need to add back that 5 gallons of gas. So if we have to add back those inputs, it's going to shift the graph to the right. Conversely, if you had a nice friend who gave you 5 gallons of gas, you could take out those 5 gallons from your tank and still drive the same distance. The analogy also works for horizontal stretches/shrinks.\n\nIt's not a perfect analogy, but for those kids who are really confused and just can't seem to wrap their minds around the mathematics, it seems to help.\n\n$$\\hspace{1cm}$$",
null,
"When the $$x$$-scale is increased by $$R$$, the function's argument must be decreased by $$R$$ to compensate.\n\nWhen the $$x$$-scale is decreased by $$L$$, the function's argument must be increased by $$L$$ to compensate.\n\nIn a basic way, I'm not sure that there's much that you can do here beyond just plotting points. For example, you could draw a graph of $y=\\sqrt{x}$ by plotting a few points, and then draw a graph of $y=\\sqrt{x-3}$ by plotting the corresponding points. After you draw the graphs, you can observe that one is obtained by translating the other.\n\nUltimately, I don't think this has a simple conceptual explanation -- it's just a phenomenon that you observe when you start drawing graphs. It's almost closer in character to a scientific fact than a mathematical fact.\n\nThat doesn't mean that there's nothing to teach here. The important thing is that students should understand how the result was obtained, and should be able to re-derive it for themselves if they need it. If I were teaching precalculus, I would give the students challenges like drawing graphs of $f(x^2)$ and $f(x)^2$ from the graph of $f(x)$. The reasoning is essentially the same as for shifting and scaling, and it will help them to improve their overall understanding of graphs.\n\nTo answer simply: when assessing the vertical shift, one isolates the Y variable. For example: y= x^2 + 3. This results in a vertical shift up. However When determining a horizontal shift in standard form, such as y=(x-1)^2 - 3 It appears to be opposite because again the Y variable is isolated. This shift would be to the right by one unit. If we isolated the X variable, the equation would appear as +/- root 3 + 1 = x\n\nHere you see that with x isolated, the horizontal shift is to the right by 1 unit.\n\nCheers :)\n\n• Welcome to the site! Check out the math notation guide. You may want to improve the appearance of your answer by editing it. – user147263 Jul 20 '14 at 0:45\n• This is a bit confusingly worded, you may want to explain what you mean by your last sentence (for example) – Adam Hughes Jul 20 '14 at 1:07\n\nI'm not sure how effective this will be for a kid these days, but this was how I managed to grasp it when I was once one.\n\nI find that scaling and shifting is a lot more natural when one considers parametric equations. Given some curve\n\n\\begin{align*}x&=f(t)\\\\y&=g(t)\\end{align*}\n\nwe find adding a nonnegative quantity to either component does the expected shift behavior:\n\n\\begin{align*}x&=h+f(t)\\\\y&=k+g(t)\\end{align*}\n\nwith $h,k \\geq 0$ shifts $h$ units to the right and $k$ units upwards (negative $h,k$ of course does the opposite direction); and\n\n\\begin{align*}x&=c\\,f(t)\\\\y&=c\\,g(t)\\end{align*}\n\nwith $c > 0$ scales the curve as expected.\n\n(Note that the matrix-vector treatment also works here; the first bit is adding a translation vector, and the second bit is multiplication of a vector by a diagonal scaling matrix.)\n\nNow, letting $f(t)=t$ (thus yielding a trivial set of parametric equations for $y=g(x)$), and eliminating variables, we then have this strange artifact of this natural transformation:\n\n\\begin{align*}x&=h+t\\\\y&=k+g(t)\\end{align*}\n\n\\begin{align*}x-h&=t\\\\y&=k+g(t)\\end{align*}\n\nand eliminating the parameter gives\n\n$$y=k+g(x-h)$$\n\nand that's where the minus sign in shifting an explicit form horizontally comes from.\n\nFor scaling, we have\n\n\\begin{align*}x&=ct\\\\y&=c\\,g(t)\\end{align*}\n\n\\begin{align*}\\frac{x}{c}&=t\\\\y&=c\\,g(t)\\end{align*}\n\nfrom which we can obtain\n\n$$y=c\\,g\\left(\\frac{x}{c}\\right)$$\n\nand that's why we have to divide instead of multiplying when scaling horizontally in explicit form.\n\n• Exactly, go based on what the transformation is to the relevant variable. So y = f(x) + 1 could be rewritten as y - 1 = f(x) and it all makes sense. – Christian Mann Apr 18 '12 at 4:00\n\nJust as an extension to the KCd's answer, we can think of this not only as daylight saving problem, but also as a timezone problem. For example, if you leave New York for Singapore you have to shift time on your watch $+12$ hours in order to keep up with local Singapore time. But if you want to express Singapore time in terms of New York time (due to the jet lag, perhaps) you could say \"it's $x - 12$ hours in Singapore now in terms of New York time\" or \"it's $x - 12$ hours in New York now\".\n\n• I feel like this explanation is more likely to confuse than to help, In your analogy, it is not obvious what function is being shifted (since the choice coordinates---either New York is at zero, or Singapore is at zero---is unclear), and people have enough trouble with communicating clock-time (for example, if I move a noon meeting \"up two hours\", does that mean that we are now meeting at 10:00 am, or at 2:00 pm?). – Xander Henderson Jun 16 '18 at 17:50\n• Honestly, I don't grasp where's the confusion. For instance, if you leave New York, which is your starting point, I assume, that it's a zero, and Singapore, while a destination, must be +12 on the coordinate plane. But you're right, that time analogy is always confusing. I think at the end the best way to form an intuition for this shifting problem, is to remember, that you're evaluating the original and the shifted function. You can ask yourself \"what should I do to the shifted function in order it to be equal to the original function\". Although, this is still not crystal clear. – Tim Nikitin Jun 16 '18 at 18:08\n\n## protected by Community♦Sep 5 '18 at 1:20\n\nThank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count)."
]
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"https://i.stack.imgur.com/BO3Hr.png",
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"https://i.stack.imgur.com/FDUPU.png",
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"https://i.stack.imgur.com/er6fW.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.78383785,"math_prob":0.99435633,"size":2647,"snap":"2019-26-2019-30","text_gpt3_token_len":813,"char_repetition_ratio":0.1888006,"word_repetition_ratio":0.026666667,"special_character_ratio":0.33471856,"punctuation_ratio":0.17635658,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9982426,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-21T19:19:28Z\",\"WARC-Record-ID\":\"<urn:uuid:abd7dc1f-6d94-4a1e-b0d7-c0cabb12807c>\",\"Content-Length\":\"266628\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b55e490-4c9c-4313-bc5f-3764ac489355>\",\"WARC-Concurrent-To\":\"<urn:uuid:386d0340-71c6-43a5-88ee-14a8604a4c2f>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://math.stackexchange.com/questions/133185/explaining-horizontal-shifting-and-scaling/133348\",\"WARC-Payload-Digest\":\"sha1:SCJMXI5W75NBQU3OUCSG2JFQRDDVGJBN\",\"WARC-Block-Digest\":\"sha1:LOIGTVUQLSVEQTBFLOKABVNLGBSDENRE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195527196.68_warc_CC-MAIN-20190721185027-20190721211027-00145.warc.gz\"}"} |
https://www.studentsguide360.com/12th-physics-important-questions-2022/ | [
"",
null,
"# 12th Physics Important Questions 2022\n\n## 12th Physics Important Questions Reduced Syllabus 2022\n\n12th Physics Public Exam Important Questions Volume 1 and Volume 2 based on reduced syllabus 2022 Tamil Medium and English Medium Download PDF. 12th Physics Public Exam Model Questions. 12th Physics Free Online Test.\n\n## Volume 1\n\n### 12th Physics Important 2 Marks\n\n1. Define ‘Electric dipole and dipole moment?\n2. Define ‘electrostatic potential”.\n3. Define ‘electric flux’. Mention its unit.\n4. Define ‘capacitance’. Give its unit.\n5. What is corona discharge?\n6. State Coulomb’s law\n7. Distinguish between drift velocity and mobility.\n8. Define electrical resistivity. Mention its unit.\n9. Define temperature coefficient of resistance. Mention its unit.\n10. Distinguish electric power and electric energy?\n11. What is the Thomson effect?\n12. What is the Peltier effect?\n13. Define magnetic flux. Mention its unit.\n14. State Coulomb’s inverse law in magnetism.\n15. State Ampere’s circuital law.\n16. State Flemmings left-hand rule\n17. How is a galvanometer converted into i)ammeter ii) voltmeter\n18. State Fleming’s right-hand rule.\n19. Mention the ways of producing induced emf.\n20. Define electric resonance.\n21. How will you define Q-factor?\n22. Write down the integral form of modified Ampere’s circuital law.\n23. What is meant by Fraunhofer lines?\n24. Why are em waves non-mechanical?\n\n### 12th Physics Important 3 Marks\n\n1. Derive an expression for the torque experienced by a dipole due to a uniform electric field.\n2. Derive an expression for electrostatic potential due to a point charge.\n3. Obtain the expression for capacitance for a parallel plate capacitor.\n4. Obtain the expression for energy stored in the parallel plate capacitor.\n5. Obtain the macroscopic form of Ohm’s law from its microscopic form and discuss its limitation.\n6. State and explain Kirchhoff’s rules.\n7. Explain the equivalent resistance of a series and parallel resistor network\n8. Explain the determination of the internal resistance of a cell using a voltmeter.\n9. Find the magnetic induction due to a long straight conductor using Ampere’s circuital law.\n10. Give an account of magnetic Lorentz force.\n11. Explain how a galvanometer is converted into an ammeter.\n12. Explain how a galvanometer is converted into a voltmeter.\n13. How will you induce an emf by changing the area enclosed by the coil?\n14. Mention the various energy losses in a transformer.\n15. Find out the phase relationship between voltage and current in a pure resistive circuit.\n16. Assuming that the length of the solenoid is large when compared to its diameter, find the equation for its inductance\n17. What are emission spectra? Give their types.\n18. What are absorption spectra? Give their types.\n19. Write down the properties of electromagnetic waves\n\n### 12th Physics Important 5 Marks\n\n1. Calculate the electric field due to a dipole on its axial line.\n2. Derive an expression for electrostatic potential due to an electric dipole.\n3. Obtain the expression for electric field due to an infinitely long charged wire.\n4. Derive the expression for resultant capacitance, when capacitors are connected in series and in parallel.\n5. Explain in detail the construction and working of a Van de Graaff generator.\n6. Describe the microscopic model of current and obtain the general form of Ohm’s law.\n7. Obtain the condition for bridge balance in Wheatstone’s bridge\n8. Deduce the relation for the magnetic induction at a point due to an infinitely long straight conductor carrying current.\n9. Obtain an expression for the force between two long parallel current-carrying conductors.\n10. Obtain an expression for the force on a current-carrying conductor placed in a magnetic field.\n11. Show mathematically that the rotation of a coil in a magnetic field over one rotation induces an alternating emf of one cycle.\n12. Explain the construction and working of the transformer.\n13. Find out the phase relationship between voltage and current in a pure inductive circuit.\n14. Derive an expression for the phase angle between the applied voltage and current in a series RLC circuit.\n\n## Volume 2\n\n### 12th Physics Important 2 Marks\n\n1. State the laws of refraction (OR) Snells law.\n2. What is critical angle and total internal reflection?\n3. What is Rayleigh’s scattering?\n4. Why does the sky appear blue?\n5. What is the reason for the reddish appearance of the sky during sunset and sunrise?\n6. Why do clouds appear white?\n7. What is Huygens’ principle?\n8. What is the bandwidth of the interference pattern?\n9. What is plane polarised, unpolarized and partially polarised light?\n10. State Brewster’s law\n11. Define the work function of a metal. Give its unit.\n12. What is the photoelectric effect?\n13. How will you define threshold frequency?\n14. Define stopping potential.\n15. What is meant by excitation energy.\n16. Define the ionization energy and ionization potential.\n17. Define impact parameter.\n18. What is mass defect?\n19. What is binding energy of a nucleus? Give its expression.\n20. Calculate the energy equivalent of 1 atomic mass unit.\n21. What is mean life of nucleus? Give the expression.\n22. What is meant by activity or decay rate? Give its unit.\n23. Define curie.\n24. What do you mean by doping?\n25. Distinguish between intrinsic and extrinsic semiconductors.\n26. What is rectification?\n27. What is an integrated circuit?\n28. What is modulation?\n29. Define center frequency or resting frequency in frequency modulation.\n\n### 12th Physics Important 3 Marks\n\n1. Derive the relation between f and R for a spherical mirror.\n2. What is an optical path? Obtain the equation for the optical path of a medium of thickness d and refractive index n.\n3. Obtain the equation for apparent depth.\n4. Obtain the equation for the critical angle\n5. Mention the differences between interference and diffraction\n6. State and obtain Malus’ law\n7. List the uses of polaroids.\n8. What is the angle of polarisation and obtain the equation for angle of polarisation?\n9. State and obtain Brewster’s law.\n10. List out the laws of the photoelectric effect.\n11. Give the construction and working of photoemissive cell.\n12. Derive an expression for de Broglie wavelength of electrons.\n13. Write the applications fo photocells.\n14. Write the properties of cathode rays.\n15. Write down the postulates of Bohr atom model.\n16. Derive the energy expression for hydrogen atoms using the Bohr atom model.\n17. What are the properties of nuclear force?\n18. Discuss the alpha decay process for example.\n19. Discuss the gamma decay process for an example.\n20. Discuss the properties of neutrino and its role in beta decay\n21. Transistor functions as a switch. Explain.\n22. State and prove De Morgan’s First and Second theorems.\n23. What are the advantages and limitations of FM?\n\n### 12th Physics Important 5 Marks\n\n1. Derive the mirror equation and the equation for lateral magnification.\n2. Describe Fizeau’s method to determine the speed of light.\n3. Obtain the lens maker’s formula and mention its significance.\n4. Derive the equation for the angle of deviation produced by a prism and thus obtain the equation for the refractive index of the material of the prism\n5. What is dispersion? Obtain the equation for the dispersive power of a medium.\n6. Obtain the equation for bandwidth in Young’s double-slit experiment.\n7. Discuss a simple microscope and obtain the equations for magnification for near point focusing and normal focusing.\n8. Obtain Einstein’s photoelectric equation with the necessary explanation.\n9. Describe briefly Davisson – the Germer experiment which demonstrated the wave nature of electrons\n10. Explain the J.J. Thomson experiment to determine the specific charge of electrons.\n11. Discuss the spectral series of the hydrogen atoms.\n12. Obtain the law of radioactivity.\n13. Obtain an expression for the radius and velocity of the electron in the nth orbit.\n14. Draw the circuit diagram of a half-wave rectifier and explain its working.\n15. Explain the construction and working of a full-wave rectifier."
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"https://i0.wp.com/www.studentsguide360.com/wp-content/uploads/2022/05/12th-Physics-Important-Questions-2022.png",
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https://www.easycalculation.com/engineering/civil/cantilever-beam.php | [
"# Cantilever Beam Calculator\n\nSimple online calculator to calculate the stiffness of the cantilever beam from the Young's Modulus, area moment of inertia and length.\n\n## Cantilever Beam Calculatoion\n\nNm-2\nm4\nm\nNm-1\n\nSimple online calculator to calculate the stiffness of the cantilever beam from the Young's Modulus, area moment of inertia and length.\n\nCode to add this calci to your website",
null,
"",
null,
"#### Formula Used:\n\nStiffness (k) = (3 × E × I ) / l3 Where, E - Young's Modulus I - Area Moment of Inertia l - Length\n\nCantilever beam stiffness calculation is made easier here."
]
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null,
"https://www.easycalculation.com/images/embed-plus.gif",
null,
"https://www.easycalculation.com/images/embed-minus.gif",
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https://pypi.org/project/orthorectification/ | [
"No project description provided\n\n# Orthorectification Using RPCs",
null,
"This repository houses some methods and utilities to help perform orthorectification on raw satellite imagery such as worldview2/3/4(RIP) panchromatic imagery.\n\nThese algorithms were presented as a deep dive in the Charlottesville Data Science Meetup, February 27th, 2020. The original presentation material is included under /presentation\n\nNote that all of this functionality already exists in libraries like GDAL and others. The goal of this codebase was to present and deep dive into these subroutines.\n\nNearly all of the code is Numba JIT-able for maximum performance. There is also a pure C++ implementation of the orthorectification routine that can be optionally imported using cppyy.\n\n# Functions\n\n### ortho_tools.py\n\ndef unpack_rpc_parameters(dataset: gdal.Dataset) -> RPCCoeffs:\n\nReturns RPC coefficients collection as a NamedTuple when provided with a GDAL dataset if that dataset contains RPCs\n\ndef retrieve_dem( min_lon: float, min_lat: float, degrees_lon: float, degrees_lat: float, sampling_rate: int = 1, output_path: str = \"/tmp/elevation.dem\", ) -> Tuple[np.ndarray, np.array]:\n\nLoads up a DEM tile given an upper-left world coordinate (min_lon, min_lat) and a width/height in degrees.\n\nThe return object is a tuple containing the DEM image and its GeoTransform respectively.\n\ndef lon_lat_to_pixel(lon: float, lat: float, geot: np.array) -> Tuple[float, float]:\n\nReprojects a world coordinate into pixel space using a GeoTransform, (Array of 6 floats) This is useful for querying DEM information, or finding an image pixel coordinate for an image that has already been orthorectified.\n\ndef linear_interp(x: float, y: float, source: np.ndarray, source_height: int) -> int: Given a pixel space coordinate x,y and a source image as a flattened array (with the stride!), returns a bilinearly interpolated measurement. Useful for DEM interpolation and also interpolation during the orthorectification process.\n\ndef lon_lat_alt_to_xy( lon: float, lat: float, alt: float, rpcs: RPCCoeffs, ) -> Tuple[float, float]:\n\nReturns an image pixel coordinate (x, y) corresponding to a provided world coordinate (lon, lat, alt) using provided RPC coefficients\n\ndef make_ortho( x1: float, x2: float, y1: float, y2: float, width: int, source: np.ndarray, rpcs: RPCCoeffs, dem: np.ndarray, dem_geot: np.array, ) -> Tuple[np.array, float, float, float]: Creates an ortho with using the provided bounding box (x1, x2, y1, y2) which is min_lon, max_lon, min_lat, max_lat, a desired width (number of pixels), a source image, that source image's RPCs, a DEM, and the DEM's affine GeoTransform.\n\n### scaling_tools.py\n\ndef fracture_polygon_north_up(poly: Polygon, factor_x: int, factor_y) -> Sequence[Polygon]: Breaks a Shapely polygon into small north up rectangles, attempts to fill as much area as possible with smaller squares. The factors control the number of rectangles per dimension (resolution)\n\ndef fracture_parallelogram(poly: Polygon, factor: int) -> Sequence[Polygon]:\n\nFractures a parallelogram into smaller parallelogram with a non-regular orientation. The orientation of each individual small parallelogram with respect to the original parallelogram is maintained.\n\ndef rescale_elevation_data(elevation_data: np.ndarray) -> np.ndarray: Maps elevation data into a 0-255 8-bit representation that is suitable for viewing\n\ndef reproject_with_affine( coords: Sequence[Tuple[float, float]], geo_transform: np.array, resize_factor: float = 1.0, ) -> Sequence[Tuple[float, float]]: Basically the same thing as ortho_tools.lon_lat_to_pixel, but works on a sequence of coordinates, which can be exactly what you get out of a shapely.Polygon.exterior.coords for example. You can optionally pass in a resize_factor that rescales your result as desired.\n\ndef overlay_polygon( img: np.ndarray, polygon: Polygon, color: Tuple[int, int, int, int] = (0, 255, 0, 0), opacity: float = 0.2 ) -> np.ndarray:\n\nBurns a polygon onto an image using OpenCV. Example:",
null,
"def generate_triangle_mesh(elevation_data: np.ndarray, reach: int = 10) -> Tuple[np.array, np.array, np.array]: Generates a triangle mesh from elevation data using a very simple and non-optimal but extremely fast procedure Produces a 3D vertex list as a Tuple of np.array. The resulting data can be visualized using OpenCV or something like that.",
null,
"### io_tools.py\n\ndef save_raster_as_geotiff(ortho: np.ndarray, ul_lon: float, ul_lat: float, gsd: float, filename: str) -> None: You provide an image, it's upper left world coordinate (ul_lon, ul_lat), and its GSD (ground sampling distance) in degrees, and it writes out a mapping compatible GeoTIFF file. This can be imported into QGIS/ArcGIS or whatever to visualize the ortho product on a map.\n\n### scaling_tools.py\n\ndef resize(img: np.array, factor: int) -> np.array: OpenCV resize\n\ndef convert_to_8bit(img: np.array) -> np.array: Jams higher bitness image pixels into 0-255 range... not elegant.\n\ndef gaussian_rescale(img: np.array, bitness=11, stdev_bound=3) -> np.array: Much more elegant way to rescale a >8 bitness image into 0-255 range. Bitness of the image must be provided. WV3 is usually 11 or 12 bit.\n\nSee a before and after example:",
null,
"### experimental.py\n\nOnly use this if you are brave. Requires cppyy which can be a pain to install, so it is not marked as a required package.\n\ndef ortho_cpp( x1: float, x2: float, y1: float, y2: float, width: int, rpcs: RPCCoeffs, source: np.ndarray, dem: np.ndarray, dem_geot: np.array, ):\n\nThis is the exact same method signature as ortho_tools.make_ortho, but it uses pure C++ code as an accelerated version. Generally about twice as fast as even the Numba JITted version of ortho_tools.make_ortho\n\nComing soon to experimental.py: Orthorectify your RPC images with CUDA!\n\n## Release history Release notifications | RSS feed\n\nThis version",
null,
"0.0.4",
null,
"0.0.3",
null,
"0.0.2",
null,
"0.0.1\n\nUploaded source"
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null,
"https://warehouse-camo.ingress.cmh1.psfhosted.org/744a416f701292f1083f9ede9e68ade3d1f8caca/2e2f696d616765732f70686f746f6772616d6d65747269635f76697375616c697a6174696f6e2e706e67",
null,
"https://warehouse-camo.ingress.cmh1.psfhosted.org/70d294673d0efd2d8fa79f0d1ce8fe198e2cd96b/2e2f696d616765732f706f6c79676f6e5f6275726e696e2e706e67",
null,
"https://warehouse-camo.ingress.cmh1.psfhosted.org/66ad902fe9be01ae98ed33b3f0f9d9605181b84c/2e2f696d616765732f64656d33642e706e67",
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http://math.au.dk/aktuelt/aktiviteter/event/item/4924/ | [
"# Institut for Matematik",
null,
"# Group von Neumann-algebras\n\nJohannes Christensen\nForedrag for studerende\nFredag, 14 februar, 2014, at 15:15-16:00, in Aud. D4 (1531-219)\nAbstrakt:\nGiven a group $G$, one can construct a von Neumann algebra by considering certain bounded operators on the Hilbert space $l^{2}(G)$. These von Neumann algebras are called group von Neumann algebras, and if one works with the restriction of only considering ICC groups, these von Neumann algebras all become factors of type $II_{1}$.\n\nIt has been known for a long time, that not all von Neumann algebras constructed using ICC groups are isomorphic, but to which extend the von Neumann algebra inherits the properties of the underlying group is still a rather open question. One example of this is the case where the group is the free group $F_{n}$ on $n>1$ generators. Mathematicians have tried to determine whether these are isomorphic or not for decades, and trying to answer this question, a new branch of mathematics has been constructed - free probability.\n\nIn my lecture I will very briefly define a von Neumann algebra and state its basic properties before I begin constructing the group von Neumann algebra, so it should be possible for an audience with no knowlegde of operator-algebras to attend the lecture.\nKontaktperson: Thomas Lundsgaard Schmidt"
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"https://www.aucdn.dk/2016/assets/img/au_segl-inv.svg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8846939,"math_prob":0.8911059,"size":1189,"snap":"2019-35-2019-39","text_gpt3_token_len":280,"char_repetition_ratio":0.16962026,"word_repetition_ratio":0.0,"special_character_ratio":0.22708158,"punctuation_ratio":0.0969163,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9905152,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-17T11:10:16Z\",\"WARC-Record-ID\":\"<urn:uuid:8135444e-c2d6-4496-a640-7348fdefc7ca>\",\"Content-Length\":\"22357\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b7053194-6ed0-4c79-827c-316194d8bc71>\",\"WARC-Concurrent-To\":\"<urn:uuid:618b5739-420d-4466-9b23-de43d7b92b1b>\",\"WARC-IP-Address\":\"185.45.20.48\",\"WARC-Target-URI\":\"http://math.au.dk/aktuelt/aktiviteter/event/item/4924/\",\"WARC-Payload-Digest\":\"sha1:XQTTVMORVRY5IG75E6SG5GXU3YWPXVKC\",\"WARC-Block-Digest\":\"sha1:YXVFZD2ONRDXDHSMV3WXBWU2FF5VMF4K\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573070.36_warc_CC-MAIN-20190917101137-20190917123137-00290.warc.gz\"}"} |
https://www.jiskha.com/questions/1000060/write-an-equation-in-slope-intercept-form-point-slope-or-standard-form-for-the-line-with | [
"# Math\n\nWrite an equation in slope-intercept form, point-slope, or standard form for the line with the given information. Explain why you chose the form you used. a. Passes through (-1, 4) and (-5, 2)\n\n1. 👍 0\n2. 👎 1\n3. 👁 536\n1. What is your question on this assignment?\n\nI would use point slope, but it could be any of the forms.\n\n1. 👍 0\n2. 👎 2\n2. (-1,4) (-5,2)\n\nSlope: m = y2 - y1/x2 - x1\n\n2 - 4 / -5 - (-1) = -2/-4\n\nm = 1/2\n\nEquation: y - y1 = m(x - x1)\n\ny - 4 = 1/2(x - (-1))\n\ny - 4 = 1/2(x + 1)\n\ny - 4 + 4 = 1/2x + 1/2 + 4\n\ny = 1/2x + 9/2\n\n1. 👍 0\n2. 👎 2\n\n## Similar Questions\n\n1. ### Algebra\n\n1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-2);y=-x-2 A.y=-2x*** B.y=2x C.y=1/2x D.y=-x 2.Write an equation in slope-\n\n2. ### Math(check)\n\n1)Find the range of the relation [(-1,4)},(2,5),(3,5)}.Then determine whether the relation is a funtion. 4,5,5 is the range 4,5 is a function 2)Find f(-1),if f(x)= x^2-6x/x+2 (-1)^2-6(-1)/-1+2 1+6/-1+2 7/1 =7 3)Find f(a),if f(t)=\n\n3. ### Mathematics\n\nCan someone check my answers? 1. For the data in the table, does y vary directly with X? If it does write an equation for the direct variation.(X,y) (8,11) (16,22) (24,33) Yes y=2.75x Yes y=0.6875x*** Yes y=1.375x No y does not\n\n4. ### math\n\nThis is for the pretest. What is the solution to the system of equations pictured below? A.(-3,5) B.(-3,-2) C(3,4) D.(1,2) E(6,-2) When writing in slope intercept form what does the 'm\" Represent? What is the 'b'? A. M is the x\n\n1. ### Math\n\n1. Use point-slope form to write the equation of a line that has a slope of 2/3 and passes through (-3, -1). Write your final equation in slope-intercept form. 2. Write the equation in standard form using integers (no fractions or\n\n2. ### algebra2\n\nwrite the slope-intercept form of the equation of the line through the given point with the given slope. through:(-1,-2), slope=6\n\nGiven A(-4,-2), B(44), and C(18,-8, answer the following questions Write the equations of the line containing the altitude the passes through B in standard form. Write the equation of the line containing the median that passes\n\n4. ### Algebra\n\nFind the slope and y- intercept of a line by writing the equation in slope- intercept form.Hint: Convert the given equation form into the slope-intercept form as deemed necessary: a. y=4-2/3 x b. 3x+7y-10=0 c. y-4=2x d. 2y-10=2x\n\n1. ### math\n\nwrite an equation for the line in point slope form...when it passes through (9,1) and the x intercept is 5 and then slope intercept form...im so confused\n\n2. ### Algebra\n\n1.Tell whether the sequence is arithmetic. If it is, what is the common difference? -19,-11,-3,5... A.yes;5 B.yes;6 C.yes;8 D.no 2.What is the slope of that passes through the pair of points? (-6,8),(2,3) A.-3/8 B.-1/8 C.-7/8\n\n3. ### algebra\n\nFind an equation for the line parallel to y=2x-7 with y-intercept(0,6). Write the answer in slope-intercept form. The equation of the line in slope-intercrpt form form is\n\n4. ### Math\n\nWrite an equation in the indicated form for each statement described below. (A graph shows two lines intersecting at x=-2 and y= 1 in a tilted X shape) A. Write an equation in slope-intercept form for the line parallel to the red"
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https://grams-to-kilograms.appspot.com/916-grams-to-kilograms.html | [
"Grams To Kilograms\n\n# 916 g to kg916 Grams to Kilograms\n\ng\n=\nkg\n\n## How to convert 916 grams to kilograms?\n\n 916 g * 0.001 kg = 0.916 kg 1 g\nA common question is How many gram in 916 kilogram? And the answer is 916000.0 g in 916 kg. Likewise the question how many kilogram in 916 gram has the answer of 0.916 kg in 916 g.\n\n## How much are 916 grams in kilograms?\n\n916 grams equal 0.916 kilograms (916g = 0.916kg). Converting 916 g to kg is easy. Simply use our calculator above, or apply the formula to change the length 916 g to kg.\n\n## Convert 916 g to common mass\n\nUnitMass\nMicrogram916000000.0 µg\nMilligram916000.0 mg\nGram916.0 g\nOunce32.3109491458 oz\nPound2.0194343216 lbs\nKilogram0.916 kg\nStone0.1442453087 st\nUS ton0.0010097172 ton\nTonne0.000916 t\nImperial ton0.0009015332 Long tons\n\n## What is 916 grams in kg?\n\nTo convert 916 g to kg multiply the mass in grams by 0.001. The 916 g in kg formula is [kg] = 916 * 0.001. Thus, for 916 grams in kilogram we get 0.916 kg.\n\n## 916 Gram Conversion Table",
null,
"## Alternative spelling\n\n916 Gram to Kilograms, 916 Gram in Kilograms, 916 Grams to Kilograms, 916 Grams in Kilograms, 916 g to Kilograms, 916 g in Kilograms, 916 Gram to Kilogram, 916 Gram in Kilogram, 916 Grams to Kilogram, 916 Grams in Kilogram, 916 g to Kilogram, 916 g in Kilogram, 916 Gram to kg, 916 Gram in kg"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.80591655,"math_prob":0.94947505,"size":776,"snap":"2023-40-2023-50","text_gpt3_token_len":241,"char_repetition_ratio":0.2746114,"word_repetition_ratio":0.0,"special_character_ratio":0.37371135,"punctuation_ratio":0.15384616,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9774712,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-03T02:21:08Z\",\"WARC-Record-ID\":\"<urn:uuid:c7eba333-1fbe-415b-9089-cb0ee7e2e8b4>\",\"Content-Length\":\"29093\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:19d7d3c3-0c3f-4285-86ed-4de666edc834>\",\"WARC-Concurrent-To\":\"<urn:uuid:5dbd0f53-c118-4ab7-b3bf-c1c48a6bd297>\",\"WARC-IP-Address\":\"172.253.122.153\",\"WARC-Target-URI\":\"https://grams-to-kilograms.appspot.com/916-grams-to-kilograms.html\",\"WARC-Payload-Digest\":\"sha1:A2RLWHEU27SSWFYACBHMOPFUFFDI2LCF\",\"WARC-Block-Digest\":\"sha1:QJ73MFR2PF5IJ234OM7O3IPMXYWYOAAN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100476.94_warc_CC-MAIN-20231202235258-20231203025258-00679.warc.gz\"}"} |
https://fr.slideserve.com/avery/alpha-decay | [
"1 / 26\n\n# Alpha Decay\n\nAlpha Decay. Readings Nuclear and Radiochemistry: Chapter 3 Modern Nuclear Chemistry : Chapter 7 Energetics of Alpha Decay Theory of Alpha Decay Hindrance Factors Heavy Particle Radioactivity Proton Radioactivity Identified at positively charged particle by Rutherford",
null,
"Télécharger la présentation",
null,
"## Alpha Decay\n\nE N D\n\n### Presentation Transcript\n\n1. Alpha Decay • Readings • Nuclear and Radiochemistry: Chapter 3 • Modern Nuclear Chemistry: Chapter 7 • Energetics of Alpha Decay • Theory of Alpha Decay • Hindrance Factors • Heavy Particle Radioactivity • Proton Radioactivity • Identified at positively charged particle by Rutherford • Helium nucleus (4He2+) based on observed emission bands • Energetics • Alpha decay energies 4-9 MeV • Originally thought to be monoenergetic, fine structure discovered • AZ(A-4)(Z-2) + 4He + Qa\n\n2. Fine Structure for 228Th decay • Different alpha decay energies for same isotope • Relative intensities vary • Coupled with gamma decay\n\n3. Energetics • Over 350 artificially produced alpha emitting nuclei • Alpha energy variations used to develop decay schemes • All nuclei with mass numbers greater than A of 150 are thermodynamically unstable against alpha emission (Qα is positive) • However alpha emission is dominant decay process only for heaviest nuclei, A≥210 • Energy ranges 1.8 MeV (144Nd) to 11.6 MeV (212mPo) • half-life of 144Nd is 5x1029 times longer then 212mPo Alpha decay observed for negative binding energies\n\n4. Energetics • Q values generally increase with A • variation due to shell effects can impact trend increase • Peaks at N=126 shell • For isotopes decay energy generally decreases with increasing mass • 82 neutron closed shell in the rare earth region • increase in Qα • α-decay for nuclei with N=84 as it decays to N=82 daughter • short-lived α-emitters near doubly magic 100Sn • 107Te, 108Te, 111Xe • alpha emitters have been identified by proton dripline above A=100\n\n5. Alpha Decay Energetics • Q value positive for alpha decay • Q value exceeds alpha decay energy • maTa = mdTd • md and Td represent daughter • From semiempirical mass equation • emission of an α-particle lowers Coulomb energy of nucleus • increases stability of heavy nuclei while not affecting the overall binding energy per nucleon • tightly bound α-particle has approximately same binding energy/nucleon as the original nucleus • Emitted particle must have reasonable energy/nucleon • Energetic reason for alpha rather than proton • Energies of alpha particles generally increase with atomic number of parent\n\n6. Energetics • Calculation of Q value from mass excess • 238U234Th + a + Q • Isotope Δ (MeV) 238U 47.3070 234Th 40.612 4He 2.4249 • Qa=47.3070 – (40.612 + 2.4249) = 4.270 MeV • Q energy divided between the α particle and the heavy recoiling daughter • kinetic energy of the alpha particle will be slightly less than Q value • Conservation of momentum in decay, daughter and alpha are equal rd=r • recoil momentum and the -particle momentum are equal in magnitude and opposite in direction • p2=2mT where m= mass and T=kinetic energy • 238Ualpha decay energy\n\n7. Energetics • Kinetic energy of the emitted particle is less than Coulomb barrier α-particle and daughter nucleus • Equation specific of alpha • Particles touching • For 238 U decay • Alpha decay energies are small compared to the required energy for the reverse reaction • Alpha particle carries as much energy as possible from Q value, • For even-even nuclei, alpha decay leads to the ground state of the daughter nucleus • as little angular momentum as possible • ground state spins of even-even parents, daughters and alpha particle are l=0\n\n8. Energetics • Some decays of odd-A heavy nuclei populate low-lying daughter excited states that match spin of parent • Leads to fine structure of alpha decay energy • Orbital angular momentum of α particle can be zero • 83% of alpha decay of 249Cf goes to 9th excited state of 245Cm • lowest lying state with same spin and parity as parent • Long range alpha decay • Decay from excited state of parent nucleus to ground state of daughter • 212mPo • 2.922 MeV above 212Po ground state • Decays to ground state of 208Pb with emission of 11.65 MeV alpha particle • Systematics result from • Coulomb potential • Higher mass accelerates products • larger mass • daughter and alpha particle start further apart • mass parabolas from semiempirical mass equation • cut through the nuclear mass surface at constant A • Explains beta decay in decay chain\n\n9. Mass parabolas: 235U decay series Beta Decay to Energy minimum, then Alpha decay to different A Branched decay observed (red circles)\n\n10. Alpha decay theory • Distance of closest approach for scattering of a 4.2 MeV alpha particle is ~62 fm • Distance at which alpha particle stops moving towards the daughter • Repulsion from Coulomb barrier • An alpha particle should not get near the nucleus • Alpha particle should be trapped behind a potential energy barrier Vc Alpha decay energy\n\n11. Alpha decay theory • Wave functions are only completely confined by potential energy barriers that are infinitely high • With finite size barrier wave function has different behavior • main component inside the barrier • finite piece outside barrier • Tunneling • classically trapped particle has component of wave function outside the potential barrier • Some probability to go through barrier • Closer the energy of the particle to the top of the barrier more likely the particle will penetrate barrier • Increase probability of barrier penetration • Higher alpha decay energy, higher probability to penetrate barrier • Shorter half life with higher alpha decay energy\n\n12. Alpha Decay Theory • Geiger Nuttall law of alpha decay • Log t1/2=A+B/(Qa)0.5 • constants A and B have a Z dependence. • simple relationship describes the data on α-decay • over 20 orders of magnitude in decay constant or half-life • 1 MeV change in -decay energy results in a change of 105 in the half-life\n\n13. Even-Even Nuclei (235U comparison)\n\n14. Expanded Alpha Half Life Calculation • More accurate determination of half life from Hatsukawa, Nakahara and Hoffman • Theoretical description of alpha emission based on calculating the rate in terms of two factors • rate at which an alpha particle appears at the inside wall of the nucleus • probability that the alpha particle tunnels through the barrier • a=P*f • f is frequency factor • P is transmission coefficient Outside of closed shells 78Z82; 100N126 82Z90; 100N126\n\n15. Alpha Decay Theory • Alternate expression includes an additional factor that describes probability of preformation of alpha particle inside parent nucleus • No clear way to calculate such a factor • empirical estimates have been made • theoretical estimates of the emission rates are higher than observed rates • preformation factor can be estimated for each measured case • uncertainties in the theoretical estimates that contribute to the differences • Frequency for an alpha particle to reach edge of a nucleus • estimated as velocity divided by the distance across the nucleus • twice the radius • lower limit for velocity could be obtained from the kinetic energy of emitted alpha particle • However particle is moving inside a potential energy well and its velocity should be larger and correspond to the well depth plus the external energy • On the order of 1021 s-1\n\n16. Alpha Decay Calculations • Alpha particle barrier penetration from Gamow • T=e-2G • Determination of decay constant from potential information • Using the square-well potential, integrating and substituting • Z daughter, z alpha\n\n17. Gamow calculations • From Gamow • Log t1/2=A+B/(Qa)0.5 • Calculated emission rate typically one order of magnitude larger than observed rate • observed half-lives are longer than predicted • Observation suggest probability to find a ‘preformed’ alpha particle on order of 10-1\n\n18. Alpha Decay Theory • Even-even nuclei undergoing l=0 decay • average preformation factor is ~ 10-2 • neglects effects of angular momentum • Assumes α-particle carries off no orbital angular momentum (ℓ = 0) • If α decay takes place to or from excited state some angular momentum may be carried off by the α-particle • Results in change in the decay constant when compared to calculated\n\n19. Hindered -Decay • Previous derivation only holds for even-even nuclei • odd-odd, even-odd, and odd-even nuclei have longer half-lives than predicted due to hindrance factors • Assumes the existence of pre-formed -particles • a ground-state transition from a nucleus containing an odd nucleon in highest filled state can take place only if that nucleon becomes part of the -particle and therefore if another nucleon pair is broken • less favorable situation than formation of an -particle from already existing pairs in an even-even nucleus and may give rise to the observed hindrance. • if -particle is assembled from existing pairs in such a nucleus, the product nucleus will be in an excited state, and this may explain the “favored” transitions to excited states • Hindrance from difference between calculation and measured half-life • Hindrance factors between 1 and 3E4 • Determine by ratio of measured alpha decay half life over calculated alpha decay half life\n\n20. Hindrance Factors • Transition of 241Am (5/2-) to 237Np • states of 237Np (5/2+) ground state and (7/2+) 1st excited state have hindrance factors of about 500 (red circle) • Main transition to 60 keV above ground state is 5/2-, almost unhindered\n\n21. Hindrance Factors • 5 classes of hindrance factors based on hindrance values • Between 1 and 4, the transition is called a “favored” • emitted alpha particle is assembled from two low lying pairs of nucleons in the parent nucleus, leaving the odd nucleon in its initial orbital • Hindrance factor of 4-10 indicates a mixing or favorable overlap between the initial and final nuclear states involved in the transition • Factors of 10-100 indicate that spin projections of the initial and final states are parallel, but the wave function overlap is not favorable • Factors of 100-1000 indicate transitions with a change in parity but with projections of initial and final states being parallel • Hindrance factors of >1000 indicate that the transition involves a parity change and a spin flip\n\n22. Heavy Particle Decay • Possible to calculate Q values for the emission of heavier nuclei • Is energetically possible for a large range of heavy nuclei to emit other light nuclei. • Q-values for carbon ion emission by a large range of nuclei • calculated with the smooth liquid drop mass equation without shell corrections • Decay to doubly magic 208Pb from 220Ra for 12C emission • Actually found 14C from 223Ra • large neutron excess favors the emission of neutron-rich light products • emission probability is much smaller than the alpha decay • simple barrier penetration estimate can be attributed to the very small probability to preform 14C residue inside the heavy nucleus\n\n23. Proton Decay • For proton-rich nuclei, the Q value for proton emission can be positive • Line where Qp is positive, proton drip line • Describes forces holding nuclei together • Similar theory to alpha decay • no preformation factor for the proton • proton energies, even for the heavier nuclei, are low (Ep~1 to 2 MeV) • barriers are large (80 fm) • Long half life\n\n24. Topic Review • Understand and utilize systematics and energetics involved in alpha decay • Calculate Q values for alpha decay • Relate to alpha energy and fine structure • Correlate Q value and half-life • Models for alpha decay constant • Tunneling and potentials • Hindered of alpha decay • Understand proton and other charged particle emission\n\n25. Homework Questions • Calculate the alpha decay Q value and Coulomb barrier potential for the following, compare the values • 212Bi, 210Po, 238Pu, 239Pu, 240Am, 241Am • What is the basis for daughter recoil during alpha decay? • What is the relationship between Qa and the alpha decay energy (Ta) • What are some general trends observed in alpha decay? • Compare the calculated and experimental alpha decay half life for the following isotopes • 238Pu, 239Pu, 241Pu, 245Pu • Determine the hindrance values for the odd A Pu isotopes above • What are the hindrance factor trends? • How would one predict the half-life of an alpha decay from experimental data?\n\n26. Pop Quiz • Calculate the alpha decay energy for 252Cf and 254Cf from the mass excess data below. • Which is expected to have the shorter alpha decay half-life and why? • Calculate the alpha decay half-life for 252Cf and 254Cf from the data below. (use % alpha decay)\n\nMore Related"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.83910024,"math_prob":0.8921455,"size":12451,"snap":"2023-40-2023-50","text_gpt3_token_len":2720,"char_repetition_ratio":0.1583514,"word_repetition_ratio":0.00723589,"special_character_ratio":0.22560437,"punctuation_ratio":0.033021193,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99207723,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,3,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T08:51:17Z\",\"WARC-Record-ID\":\"<urn:uuid:f62e9e11-aaf8-41ee-a422-4ce36950ca35>\",\"Content-Length\":\"96367\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c6bffa43-0bde-42e5-880e-796c6fd782f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:0595095e-52fd-4172-80b8-34bc9ab4e701>\",\"WARC-IP-Address\":\"52.24.166.104\",\"WARC-Target-URI\":\"https://fr.slideserve.com/avery/alpha-decay\",\"WARC-Payload-Digest\":\"sha1:JBVRBOOZX3GYKD6SEJOLOS6A6SXJC5KB\",\"WARC-Block-Digest\":\"sha1:MLCGHHHLSYP74ESIXMXMOF55HELQDRLO\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100172.28_warc_CC-MAIN-20231130062948-20231130092948-00767.warc.gz\"}"} |
https://www.impan.pl/en/publishing-house/journals-and-series/annales-polonici-mathematici/all/89/3/85030/the-bic-of-a-singular-foliation-defined-by-an-abelian-group-of-isometries | [
"A+ CATEGORY SCIENTIFIC UNIT\n\n# Publishing house / Journals and Serials / Annales Polonici Mathematici / All issues\n\n## The BIC of a singular foliation defined by an abelian group of isometries\n\n### Volume 89 / 2006\n\nAnnales Polonici Mathematici 89 (2006), 203-246 MSC: 53C12, 57R30, 55N33, 58A35, 22Fxx. DOI: 10.4064/ap89-3-1\n\n#### Abstract\n\n% We study the cohomology properties of the singular foliation $\\cal F$ determined by an action ${\\mit\\Phi} \\colon G \\times M\\to M$ where the abelian Lie group $G$ preserves a riemannian metric on the compact manifold $M$. More precisely, we prove that the basic intersection cohomology $\\mathbb H^{*}_{\\overline{p}}{(M/\\mathcal F)}$ is finite-dimensional and satisfies the Poincaré duality. This duality includes two well known situations:\n\n$\\bullet$ Poincaré duality for basic cohomology (the action ${\\mit\\Phi}$ is almost free).\n\n$\\bullet$ Poincaré duality for intersection cohomology (the group $G$ is compact and connected).\n\n#### Authors\n\n• Martintxo Saralegi-ArangurenLaboratoire de Mathématiques de Lens EA 2462\nFédération CNRS Nord-Pas-de-Calais FR 2956\nFaculté des Sciences Jean Perrin\nUniversité d'Artois\nRue Jean Souvraz S.P. 18\n62 307 Lens Cedex, France\ne-mail\n• Robert WolakInstitute of Mathematics\nJagiellonian University\nReymonta 4\n30-059 Kraków, Poland\ne-mail\n\n## Search for IMPAN publications\n\nQuery phrase too short. Type at least 4 characters.\n\n## Rewrite code from the image",
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"https://www.impan.pl/en/publishing-house/journals-and-series/annales-polonici-mathematici/all/89/3/85030/the-bic-of-a-singular-foliation-defined-by-an-abelian-group-of-isometries",
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https://mcqslearn.com/o-level/physics/quizzes/quiz-questions-and-answers.php?page=29 | [
"Cambridge O Level Courses Online\n\nO Level Physics Quizzes\n\nO Level Physics Quiz Answers - Complete\n\n# Introduction to Light Multiple Choice Questions PDF p. 29\n\nIntroduction to Light multiple choice questions and answers, introduction to light quiz answers PDF 29 to learn O Level Physics course for college certification. Learn Light in Physics MCQ trivia questions, introduction to light Multiple Choice Questions (MCQ) for online college degrees. Introduction to Light Interview Questions PDF: measurement of time, heat capacity: physics, work and energy, latent heat, introduction to light test prep for online bachelor degree programs.\n\n\"_______ is a form of energy.\" MCQ PDF with choices acceleration, force, light, and speed for ACT test. Solve light in physics questions and answers to improve problem solving skills for ACT subject tests.\n\n## Introduction to Light Questions and Answers MCQs\n\nMCQ: _______ is a form of energy.\n\nForce\nAcceleration\nLight\nSpeed\n\nMCQ: Specific latent heat of fusion is\n\nthe amount of heat required to raise the temperature of a 1 kg of a substance by 1 K\nthe amount of heat required to change the phase of a substance from solid to liquid without any change in temperature\nthe amount of heat required to change the phase of a 1 kg of substance from solid to liquid without any change in temperature\nthe amount of heat required to change the phase of a 1 kg of substance from liquid to gas without any change in temperature\n\nMCQ: An object is lifted 5 m above the levelled ground, the mass of the object is 20 kg and the gravitational pull is 10 N kg-2, the Ep of the object is\n\n40 J\n1000 J\n2.5 J\n0.4 J\n\nMCQ: A piece of carbon of mass 0.40 kg requires thermal energy of 20 J to get heated from a temperature of 5 °C to 9 °C, the specific heat capacity of the carbon would be\n\n1.25 J kg-1 °C-1\n12.5 J kg-1 °C-1\n125 J kg-1 °C-1\n1250 J kg-1 °C-1\n\nMCQ: Periodic time of a simple pendulum depends on\n\nThe mass of the pendulum bob\nThe collective mass of the pendulum bob and thread\nThe length of the pendulum\nNone of the above"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.81117874,"math_prob":0.83963805,"size":1230,"snap":"2022-40-2023-06","text_gpt3_token_len":275,"char_repetition_ratio":0.123980425,"word_repetition_ratio":0.0,"special_character_ratio":0.22276422,"punctuation_ratio":0.11297071,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9892021,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T12:43:19Z\",\"WARC-Record-ID\":\"<urn:uuid:10b4f776-c875-4d0b-9981-f233865073ff>\",\"Content-Length\":\"102623\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a4499c98-85b4-433b-a477-edeaa5d981a1>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e7e413e-295d-4d76-a5a3-c5ea58a55c83>\",\"WARC-IP-Address\":\"184.168.100.182\",\"WARC-Target-URI\":\"https://mcqslearn.com/o-level/physics/quizzes/quiz-questions-and-answers.php?page=29\",\"WARC-Payload-Digest\":\"sha1:4PTOJ6XIW24V3HHEYFDFY4KC5BQTCLP3\",\"WARC-Block-Digest\":\"sha1:T6M7MTIIYSAHODZAJ2GXULT2OE2YC63G\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335254.72_warc_CC-MAIN-20220928113848-20220928143848-00116.warc.gz\"}"} |
http://tigernt.com/math/ | [
"# Free Math Practice Pages\n\n Resources Math for Kindergarten and Grade 1, such as A + B, A - B Math for Grade 2, 3, 4, such as A * B - C, A + B * C, with option of decimal",
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"Math for Grade 2, 3, division Math for Grade 3, 4, such as (A + B) * C, A * ( B - C) Math for Grade 4, 5, such as (A + B) * (A - B) = A*A - B*B, A*C +B*C = (A + B ) *C, A*C - B*C = (A-B) * C Math for Grade 3,4,5, fraction addition/subtraction with option of decimal",
null,
"Math for Grade 4,5, decimal addition/subtraction, multiplication Math for Grade 4, 5, 1-variable Algebra, More 1-variable Algebra Math for Grade 5, 6, 2-variable Algebra Math for Grade 4,5, division with remaining Math for Grade 4,5, Calculate GCD & LCM with option of algebra variables",
null,
"Math for Grade 3,4,5, pattern, added a(2)=a(1)+a(0)",
null,
"Math for Grade 3,4,5, fraction sorting Math for Grade 5, 6, negative numbers, absolute values, exponents",
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"Math for Grade 5, 6, absolute value equation",
null,
"Math for Grade 5, 6, inequality algebra",
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"Math for Grade 5, 6, equation factoring",
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"Suggested books for Math Competition Suggested books for SAT Preparation More coming soon... How to use those pages One refresh will create new set of exercises. Every exercise is fresh and random and is created on the fly. The answer for each exercise is hidden until you click the 'Flip Answers' on the top.",
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"To set different difficulty of the exercises, you can set your favorable mininum and maxinum values. Some answers have \"Graph\" links",
null,
"which point to the powerful WolframeAlpha site. The equation of the exercise will be passed in as the parameter in the link. User are able to view the answer and graph of the equation. Motivation to create those pages Save me tons of time and energy to search or prepare math practice exercises/worksheets for my kids Answer is always provided and saves me time to solve and verify. I am adding more from time to time in order to keep pace with my kids' acedemic needs. Your feedback will be welcomed and helpful. Last modified: 12/11/2011 Any questions or comments please contact us"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.801675,"math_prob":0.9533124,"size":1112,"snap":"2021-31-2021-39","text_gpt3_token_len":378,"char_repetition_ratio":0.25631768,"word_repetition_ratio":0.17937219,"special_character_ratio":0.3561151,"punctuation_ratio":0.19615385,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99567467,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T21:56:44Z\",\"WARC-Record-ID\":\"<urn:uuid:8a5a4910-74a4-48ae-bd03-1810a4d1b198>\",\"Content-Length\":\"7503\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b87d4a71-5bce-430f-93eb-bdc94346f4f5>\",\"WARC-Concurrent-To\":\"<urn:uuid:204fcfc2-b9dc-47d0-ba51-1bc9fa4d2e51>\",\"WARC-IP-Address\":\"74.208.236.33\",\"WARC-Target-URI\":\"http://tigernt.com/math/\",\"WARC-Payload-Digest\":\"sha1:6PS6Z2KFWLIMYTUAWUXKCLVDRX6QNZCE\",\"WARC-Block-Digest\":\"sha1:YIXWYJODWRLQ3NU7P5OLNGKXML7ZSNT2\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057580.39_warc_CC-MAIN-20210924201616-20210924231616-00351.warc.gz\"}"} |
https://www.ijert.org/modelling-uncertain-spatial-data-sets-using-uncertain-partitioning-clustering | [
"# Modelling Uncertain Spatial Data Sets using Uncertain Partitioning Clustering\n\nText Only Version\n\n#### Modelling Uncertain Spatial Data Sets using Uncertain Partitioning Clustering\n\nAssistant Professor, Department of Computer Applications Shri Vishnu Engineering College for Women, Bhimavaram [email protected]\n\n2Aruna Kumari M Department of Computer Applications\n\nShri Vishnu Engineering College for Women, Bhimavaram [email protected]\n\nAbstract- Uncertainty in spatial data set is due to various causes like imprecision, inconsistency and inaccuracy in the information acquired through data collection and amalgamation from the instruments and infrastructures on Geo- Informatics. Such uncertain data is usually represented in terms of uncertain regions over which the Probability Density Function (PDF) is defined. In the present paper, the problem of clustering uncertain data has been addressed by proposing UK-Medoids algorithm. The proposed algorithm employs a similarity function DIE deviation to find the distance between uncertain objects. Experiments have shown that UK- Medoids outperforms conventional algorithm from an accuracy view point while achieving reasonably good and efficient results.\n\nKeywords: Information Entropy, UK-Medoids clustering, Uncertain Spatial Data.\n\n1. Introduction\n\nHandling uncertainly in data management is of paramount importance in a wide range of application contexts. Indeed, data uncertainty ,\n\n naturally arises from implicit randomness in a process of data generation/ acquisition, imprecision in physical measurements, and loss of data freshness.\n\nThe uncertainty-based spatial data mining is to extract knowledge from the vast repositories of practical spatial data under the umbrella of uncertainties with the given perspectives and parameters. With different granularities, scales, mining-angles, and uncertain parameters, it discovers the collective attribute distribution of spatial entities by perceiving various variations of spatial data and their combinations in the data space.\n\nVarious notations of uncertainty have been defined depending on the application domain. Attribute level uncertainty , has been considered in this paper, to model according to a probability model. The uncertain object is usually represented by means of probability density functions PDFs, which describe the likelihood that the object appears at each position in a multidimensional space rather than by a traditional vector form of deterministic values.\n\nClustering is useful in exploratory data analysis. Cluster analysis organizes data by grouping individuals into a population in order to discover structure or clusters. Many partitioned algorithms\n\n have been proposed, some based on k-centroid, some based on K-Medoid, some based on fuzzy analysis, etc. K-Medoid clustering is similar to k- Centroid clustering. Both of them attempt to partition the data by assigning each object to a representative and then optimizing a statistical homogeneity criterion – namely, the total expected squared dissimilarity. However, K-Medoid clustering only allows objects to be chosen as representatives. In comparison with K-Centroid, the use of Medoids for clustering has several advantages. Firstly, this method has been shown to be robust to the existence of noise or outliers and generally produces clusters of high quality. Secondly, this method can be used not only on points or vectors for which the mean is defined but also on any objects for which a similarity measure between two objects is given.\n\nThis paper aims to apply uncertain K-medoids partitioning clustering to model uncertain spatial data and evaluate the pattern after clustering is applied. The remained sections will be organized as follows. Section 3 presents the Literature survey. The accessible method is presented in Section 4, Section 5 gives the experimental results. Conclusion is drawn finally in Section 6.\n\n2. Literature Survey\n\nUncertain data is ubiquitous in real world applications due to various causes. In recent years, clustering uncertain data has been paid more attention by the research community and the classical clustering algorithm based on partition, density and hierarchy have been extended to handle the uncertain data.\n\nClustering is known as very useful tool in many fields for data mining. The structure of data sets is found through the clustering methods . Now, the more the ability of computers increase, the more works for uncertainties have been studied. In the past, data handled by the computers was approximately represented as one point or value because of poor ability of the computers. However, the ability is now enough to handle uncertain data.\n\nTherefore, a lot of researchers have tried to handle original data from the viewpoint that the data should been represented as not one point approximately but some range exactly in a data space. Uncertainty may have an influence on the confidential level, supportable level, and interesting level of spatial data mining .\n\nK-Medoids is a clustering algorithm that is very much like K-means. The main difference between the two algorithms is the cluster center being used. K-means uses the average of all instances in a cluster, while K-Medoids uses the instance that is the closest to the mean, i.e. the most 'central' point of the cluster. Using an actual point of the data set to cluster makes the K-Medoids algorithm more robust to outliers than the K-means algorithm.\n\n3. Proposed Work\n\n1. Uncertain Partitioning based Clustering Approaches: In UK-Medoids, as every object has the same cluster, the expected distance based approach in UK-Means , cannot distinguish the two sets of objects having different distributions. In UK-Mediods clustering of uncertain objects is made according to the similarity between their probability distribution. In information theory, the similarity between two distributions can be measured by the Information Entropy deviation (IE). This IE is used to measure the similarity between distributions, and demonstrate the effectiveness of similarity. The PDFs , is used over the entire data domain and the difference is captured using the IE deviation.\n\n2. Similarity using IE Deviation: In general, IE between the two probability density functions is defined as follows: In the discrete case, let A and B be two probability distribution functions in a discrete domain D with a finite number of values.\n\nThe IE deviation between A and B is\n\n. In the continuous case, let\n\nA and B be two probability density functions in a continuous domain D with a continuous range of values. The IE deviation between A and B is\n\nIn both\n\ndiscrete and continuous cases, IE deviation is defined only in the case where for any x in domain D if A(x) > 0 then B(x) > 0, by\n\nconvention, , for any p 0 and the\n\nbase of log is 2.\n\n3. Partitioning Clustering Methods on Uncertain data: A partitioning clustering method organizes a set of k uncertain objects O into n clusters\n\nsimilarity, a partitioning clustering method tries to partition objects into k clusters and chooses the best n representatives, one for each cluster. To minimize the total IE deviation the computation\n\nis ). For an object\n\nx in cluster , the IE deviation\n\nbetween p and the representative\n\nmeasures the extra information required to construct p given . Therefore, captures the total extra information required to construct the whole cluster using its\n\nrepresentative . Summing over all n clusters, the total IE deviation ths measures the quality of the partitioning clustering. The smaller the value of TIE, the better the clustering is performed.\n\nThe pseudo code of core uncertain K-Medoids is presented initially with IE deviations as a distance function to calculate the distances between any two uncertain objects. Further refinement is made by incorporate Min-Max bounding box pruning technique into UK-Mediods and final sophistication to the algorithm is made by integrating R*-Tree indexing in to UK-Mediods.\n\n4. Uncertain K-Medoids Method\n\nThe Uncertain K-Medoids method consists of two phases, the building phase and the swapping phase. In the building phase, the uncertain K-Medoids method obtains an initial clustering by selecting k representatives one after another. The first representatives C1 is the one which has the smallest sum of the IE deviation to all other objects in O.\n\nThat is, The rest of the n-1representatives are selected iteratively. In the ith (2 i n) iteration, the\n\nalgorithm selects the representative which decreases the total IE deviation as much as possible. For each object x which has not been selected, a test is made to select the current round. For any other non-selected object x, x will be assigned to the new representative x if the\n\ndivergence is smaller than the divergence between x and any previously selected representatives. Therefore, a calculation is made for the contribution of p to the decrease of the total IE deviation by selecting x as\n\n.\n\nThe total decrease of the IE deviation is calculated by selecting x as the sum over the contribution of the non-selected objects, denoted by DIE(x). Then, objects to be selected in the ith iteration is the one\n\nthat can incur the largest decrease that\n\n. In the swapping phase, the uncertain K-Medoids method\n\nUsing IE deviation as\n\niteratively improves the clustering by swapping a\n\nnon-representative object with the representative to which it is assigned. For a non-representative\n\nobject x, suppose it is assigned to cluster C whose representative is c. The effect of swapping x and c in two cases for all non-selected object x other than x is considered. If x currently belongs to c, when c is replaced by x, a reassignment of x to x or one of the other n 1 existing representatives, to which x is the most similar is considered. If x currently belongs to a representative c other than\n\nc, and , x is reassigned to x.\n\nWhen a reassignment happens, the decrease in the total KL deviation by swapping x and c is recorded. After all non-representative objects are examined; the object is selected by swapping x and c. Then,\n\nobject is selected by object which can make the largest decrease. That is,\n\nA check is\n\nmade on swapping to improve the clusters, i.e., . If so, the swapping is carried into execution. Otherwise, the method\n\nterminates and reports the final clustering. The\n\nfollowing algorithm presents the pseudo code of the uncertain K-Medoids method.\n\nAlgorithm 1: Uncertain K-Medoids Algorithm Input: a set of uncertain objects, the number of clusters n; Output: n clusters ; 1:\n\n2: i=2;\n\n3: while do\n\n4: for\n\n5:\n\n6:\n\n7:\n\n8:\n\n9: repeat\n\n10: for\n\n11: .\n\n12: ,\n\n13: for do\n\n14: assume\n\n15:\n\n22: replace the center to which is assigned in the last iteration by\n\n23: until\n\n24: return\n\n4. Experimental Results\n\nThe algorithm described in the previous section have been implemented on java using JDK 1.6.0 and a series of experiments were performed on a PC with Intel(R) corei3, 2.93 GHz and 4GB of main memory, running on windows 7 operating system. This section focuses on the algorithm runtime performance and cluster generation in generation of a pattern.\n\nThe spatial dataset Forest Cover Type is used for all the experiments to evaluate the performance of data uncertainty. Forest Cover Type is a benchmark quandary extracted from http://kdd.ics.uci.edu\n\n/databases/covertype/covertype.html UCI KDD Archive. This problem relates to the actual forest cover type for given observation that was determined from US Forest Service (USFS) Region to Resource Information System (RIS). Forest Cover Type includes 581,012 samples represented in point valued data with 7 cover type; each sample has 54 attributes including 10 remotely sensed data and 44 cartographic data. Data preprocessing is applied to this data set and is transformed into many uncertain data sets by replacing each data point with an MBR and also generates the PDF. The proposed algorithm in the previous section is experimented with only ten percent of the available data object due to main memory constriction.\n\nFig. 1 Execution Time Effectiveness\n\n1. Execution Time: One of an obligatory constraint for any algorithm was to assess its execution time recital when applied over its implanted datasets or databases. To evaluate the\n\n16: )\n\n17: for do\n\n18: if then\n\n19:\n\n20:\n\n21: if then\n\nalgorithms competence proposed in this paper, Fig.\n\n1 is symbolized with the number of uncertain spatial data objects on horizontal axis and execution time measured in milliseconds on vertical axis. It is evident from Fig. 1 that as the number of data objects are increased the execution time also increases. It is one of substantiation that UKMdd is praiseworthy to model the spatial data objects.\n\nFig. 2 Cluster Generation on Uncertain Spatial Data Objects\n\n2. Cluster Generation: It is observed from Fig. 2 that the cluster arrangement in UKMdd seen to be consistently growing with the whole range of total percentage of DIE deviations. This is because with a large number of clusters, cluster representatives are generally less spread out. Hence total DIE deviations will have to be computed to determine the cluster assignment. The total percentage of DIE deviations in the algorithms increases when there is more number of clusters.\n\nThis substantiation is evidently revealed in Fig. 2 with number of clusters on horizontal axis and percentage of total DIE deviations in vertical axis. It is also verified in the anticipated algorithms as the number of cluster patterns increases the total percentage of DIE deviations.\n\nJustification for Results: All the selected datasets originally contain deterministic values; hence uncertainty was synthetically generated for each object in the dataset. It is prominent that anticipated algorithms performed similarly for all the PDFs computations, since they employ similar clustering scheme. The new techniques bring down the execution of UKMdd and also reduce the computation time of PDF. A consistent demise pattern is observed in DIE deviations which substantiate that the integration being practical on uncertain spatial data and affirms the completeness in assortment of the anticipated algorithms.\n\nConclusion\n\nThis paper proposes a realistic way to model uncertainties in spatial data under the umbrella of uncertain spatial data mining with given perspective and parameter. The novel approach proposed in this paper accommodates tuples having numerical attributes with uncertainty described by arbitrary PDF. Performance is an issue to provoke privileged number of cluster progression even if more complicated computations of Information entropy deviations are involved.\n\nUncertain K-Medoids algorithm is primarily designed to handle tremendous uncertain spatial data. This algorithm is experientially verified to be highly effective and empirically good in lessening the execution time. This anticipated algorithm is pragmatic in exploiting spatial data uncertainties with remarkable higher accuracies. The execution times are of an order of magnitude comparable to classical algorithms.\n\nFuture work may address the issues involved in modeling uncertain data with soft computing ased partitioned clustering.\n\nReferences\n\n1. B. Kao, S.D. Lee, D.W. Cheung, W.-S. Ho, and K.F. Chan, \"Clustering Uncertain Data Using Voronoi Diagrams,\" Proc. IEEE Int'l Conf. Data Mining (ICDM), pp. 333-342, Dec. 2008.\n\n2. Beckmann, N., Kriegel, H.P., Schneider, R., Seeger, B.: The R*-tree: An efficient and robust access method for points and rectangles. In: SIGMOD. (1990) 322331.\n\n3. C. Bohm, P. Kunath, A. Pryakhin, and M. Schubert, Querying Objects Modeled by Arbitrary Probability Distributions, Proc. 10th Intl Symp. Spatial and Temporal Databases (SSTD), 2007.\n\n4. C.C. Aggarwal and P.S. Yu, A Framework for Clustering Uncertain Data Streams, Proc. 24th IEEE Intl Conf. Data Eng. (ICDE), 2008.\n\n5. Ester, M., Kriegel, H.P., Xu, X.: Knowledge discovery in large spatial databases: Focusing techniques for efficient class identification. In: SSD. (1995) 6782.\n\n6. Francesco Gullo, Giovanni Ponti, Andrea Tagarelli: Clustering Uncertain Data Via K-Medoids. SUM 2008: 229-242.\n\n7. G. Cormode and A. McGregor, Approximation algorithms for clustering uncertain data, in PODS, M. Lenzerini and D. Lembo, Eds. Vancouver, BC, Canada: ACM, 9tp1th Jun. 2008, pp. 191200.\n\n8. Jaiwei Han, Micheline Kamber, Book Title Data Mining Concept and Techniques, Morgan Kaufmann (An Imprint of ELSEVIER) Publication, ISBN: 1-55860-489-8, 2003.\n\n9. M. Chau, Reynold Cheng, B. Kao and J. Ng. Data with uncertainty Mining: An Example in Clustering Location Data. In the Methodologies for Knowledge Discovery and Data Mining, Pacific-Asia Conference (PAKDD 2006), Singapore, 2006.\n\n10. Ng R.T. and Han J. 1994. Efficient and Effective Clustering Methods for Spatial Data Mining, Proc. 20th Int. Conf. on Very Large Data Bases, 144-155. Santiago, Chile.\n\n11. Ng, R.T., Han, J.: Efficient and effective clustering methods for spatial data mining. In: VLDB. (1994) 144 155.\n\n12. Ramachandra Rao Kurada, Durga Sri B, A Novel Approach by Applying Partitioning Clustering over Uncertain Spatial Data Objects using Pruning Techniques and R*-tree Indexing in ICCCE 2012, 12 & 13 April 2012, Dr. MGR University, Chennai, Published by Coimbatore Institute of Information Technology, ISBN No: 978-1-4675-2248-9.\n\n13. Ramachandra Rao Kurada, Durga Sri B, Unsupervised Classification of Uncertain Data Objects in Spatial Databases Using Computational Geometry and Indexing Techniques, IJERA, Vol.2 Issue 2, March-April 2012, pp. 806-814.ISSN No.: 2248-9622."
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https://www.drdawnwright.com/tag/normal-approximation/ | [
"# Normal Approximation\n\n## Do I need to know the Normal Approximation?\n\nSection 5.5: Normal Approximation to the Binomial Distribution The premise in our Larson textbook for using the normal approximation is that finding the binomial probability of an “less than x” or “greater than x” problem would be onerous – having to manually calculate each of the discrete values and then sum them up. They give …\n\n## Normal Approximation & Continuity Correction Excel Worksheet\n\nFor those who prefer to work with Excel, here is a worksheet that may help. Enter your data into the blue cells. You may enter an “x” as well as the a and the b if you are asked for a between probability. The answers will be in the yellow cells. There is a link …",
null,
"## Continuity Correction – Filling the cracks in the Normal Approximation to the Binomial\n\nWhen we approximate a discrete distribution, such as the binomial, by a continuous distribution, such as the normal, we need to make adjustments so “things don’t fall in the cracks.” From the Central Limit Theorem, we know that a sample distribution from a population, even a non-normal one, becomes normal if the sample size is …"
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https://studyres.com/doc/899576/yess-2010-work-and-energy-ii--small-group-problems-problem-1 | [
"# Download YESS 2010 Work and Energy II: Small Group Problems Problem 1\n\nSurvey\n\n* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project\n\nDocument related concepts\nno text concepts found\nTranscript\n```YESS 2010\nWork and Energy II: Small Group Problems\nProblem 1: Escape Velocity\na) Calculate the speed required for a rocket to escape Earth’s gravitational field. Assume the rocket’s\ninitial position is on the surface of Earth. After the escape, assume that the rocket is infinitely far from\nEarth and that its final velocity is approximately zero. (We want to calculate the escape speed necessary\nto just barely escape Earth’s gravity.) Does escape speed depend on the mass of the rocket?\n√\nb) A rocket is accelerated to speed v = 2 gR near Earth’s surface and then it coasts upward. Show that\nthis rocket will escape from Earth. Determine the speed of the rocket very far from Earth.\nSOLUTION:\na) Set up an energy balance. Be sure to use the general form of gravitational potential energy.\n∆K + ∆U = 0\n1\n1\n−GM m −GM m\nmv22 − mv12 +\n−\n=0\n2\n2\nr2\nr1\nFrom the problem statement: v2 = 0, r1 = R, and r2 = ∞. So we have:\n−GM m\n1\n=0\n0 − mv12 + 0 −\n2\nR\nSolve for v1 :\nr\nv1 = vesc =\n2GM\n= 11.18 km/s\nR\nThe escape speed does not depend on the mass of the rocket.\np\nb) From part (a), we know that a rocket must be accelerated to vesc p\n= 2GM/R near Earth’s surface\n√\nin order to escape. Since g =p\nGM/R2 , we have that v2 = 2 gR = 2 GM/R. This is larger than the\nrequired escape speed, vesc = 2GM/R, and therefore, the rocket will escape Earth.\nSince the rocket’s initial velocity exceeds the escape velocity, it will have some residual velocity at infinity.\n∆K + ∆U = 0\n1\n−GM m −GM m\n1\n2\n2\nmv − mv +\n−\n=0\n2 2 2 1\nr2\nr1\np\nFrom the problem statement: v1 = 2 GM/R, r1 = R, and r2 = ∞. So we have:\n1\n2GM m\n−GM m\n2\nmv −\n+ 0−\n=0\n2 2\nR\nR\nSolve for v2 to obtain:\nr\nv2 =\n2GM\nR\n1\nProblem 2: Friction\nConsider a block on a plane with coefficient of friction, µ1 , inclined at an angle, θ. Determine its final\nspeed, v, when it reaches the bottom of the slope, a distance d from its starting point. Solve this problem\ntwo ways: 1) using energy principles, and 2) using force and motion. Compare your results.\nSOLUTION USING ENERGY:\n∆K + ∆U = WN C\n1\n1\n2\n2\nmv − mv + (mgh2 − mgh1 ) = WN C\n2 2 2 1\nLet state 1 correspond to the initial position of the mass at the top of the incline. Let state 2 correspond\nto the final position of the mass at the bottom of the incline.\nThe work due to friction is: Wf = F~f · d~ = −Ff d, where d is the length of the path that the block\ntraveled. Also recall that Ff = µFN .\nOur energy balance becomes:\n1\n2\nmv − 0 + (0 − mgh) = −µ1 FN d\n2\n1\nmv 2 − mgd sin θ + µ1 mgd cos θ = 0\n2\nSolve the above equation for v:\np\nv = 2gd(sin θ − µ1 cos θ)\nSOLUTION USING FORCE/MOTION:\nLet the positive x direction be along the incline. Set up a force balance:\nΣF~ = m~a\nConsider the x component of the force:\nFgx − Ff = max\nmg sin θ − µ1 FN = max\nmg sin θ − µ1 mg cos θ = max\nSolve for ax :\nax = g(sin θ − µ1 cos θ)\nTo solve for the final speed, use the following kinematic equation:\n2\nWe know that v1 = 0 so we have:\nSubstitute in for a and solve for v:\np\nv = 2gd(sin θ − µ1 cos θ)\nThe results are the same for both methods as expected.\nProblem 3: Air Resistance\nIn early test flights for the space shuttle using a “glider” (mass of 1000 kg including the pilot), it was\nnoted that after a horizontal launch at 500 km/h at a height of 3500 m, the glider eventually landed at a\nspeed of 200 km/h. a) What would the glider’s landing speed have been in the absence of air resistance?\nb) What was the average force of air resistance exerted on the glider if it came in at a constant glide angle\nof 10◦ to the Earth? (The angle is measured from the horizontal.) Use energy principles to calculate\nSOLUTION:\na) Set up an energy balance:\n∆K + ∆U = WN C\n1\n1\n−GM m −GM m\nmv22 − mv12 +\n−\n= WN C\n2\n2\nr2\nr1\nTake r = 0 to be at the center of the Earth. Using this convention, we have that r1 = R + h =\n(6378000 + 3500) m = 6381500 m, and r2 =6378000 m. Recall that M is the mass of the Earth.\nIf there is no air resistance, then WN C = 0. Substitute the values given in the problem. Be sure to\nconvert speeds from km/h to m/s:\n3500 km/h · 1000 m/km\n= 138.89 m/s\n3600 s/h\nv2 = 200 km/h = 55.56 m/s\n1 2 1\n−(6.67 × 10−11 )(5.9742 × 1024 ) −(6.67 × 10−11 )(5.9742 × 1024 )\n2\nv − (138.89 ) +\n−\n=0\n2 2 2\n6378000\n6381500\nv1 = 500 km/h =\nSolve for the final velocity, v2 :\nv2 = 296 m/s = 1066 km/h\nb) First determine the work done due to air resistance:\n1\n1\n−GM m −GM m\nmv22 − mv12 +\n−\n= WN C\n2\n2\nr2\nr1\n1\n1000\n1000\n(1000) 55.562 − 138.892 + (−6.67 × 10−11 )(5.9742 × 1024 )\n−\n= Wair\n2\n6378000 6381500\nWair = −4.24 × 107 Nm\nThe work due to air resistance is given by: Wair = −Fair d.\n3\nSo we have that:\nWair = −Fair\nh\nsin θ\nSolve for force:\nsin θ\nh\n(4.24 × 107 ) sin 10◦\n=\n3500\nFair = −Wair\nFair\nFair = 2100 N\n4\n```"
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https://trac.haskell.org/trac/ghc/ticket/15583 | [
"# Treating Constraint as Type when using (type C = Constraint)\n\nReported by: Owned by: Iceland_jack normal 8.6.1 Compiler 8.5 TypeInType Unknown/Multiple Unknown/Multiple None/Unknown #15412\n\nThis may be similar/same as #15412? I can't get to the latest GHC right now. Again the \"culprit\" is using `type C = Constraint`.\n\nThe code example should not compile, but the error it gives is unexpected. If we define the associated type family `Ob_ (cat :: Cat ob) :: ob -> C` using `C`\n\n```{-# Language KindSignatures, PolyKinds, DataKinds, TypeInType, TypeFamilies, FlexibleInstances #-}\n\nimport Data.Kind\n\ntype T = Type\ntype C = Constraint\n\ntype Cat ob = ob -> ob -> T\n\nclass Category_ (cat :: Cat ob) where\ntype Ob_ (cat :: Cat ob) :: ob -> C\n\nid_ :: Ob_ cat a => cat a a\n\nclass NoCls (a::k)\ninstance NoCls (a::k)\n\ninstance Category_ (->) where\ntype Ob_ (->) = NoCls\n\n-- id_ :: NoCls a => (a -> a) -XInstanceSigs\nid_ = ()\n```\n\nthe definition of `id_` fails (as we expect), but the expected type (`Ob_ (->) a -> a -> a`) is wrong:\n\n```\\$ ghci -ignore-dot-ghci 348.hs\nGHCi, version 8.5.20180128: http://www.haskell.org/ghc/ :? for help\n[1 of 1] Compiling Main ( 348.hs, interpreted )\n\n348.hs:22:9: error:\n• Couldn't match expected type ‘Ob_ (->) a -> a -> a’\nwith actual type ‘()’\n• In the expression: ()\nIn an equation for ‘id_’: id_ = ()\nIn the instance declaration for ‘Category_ (->)’\n• Relevant bindings include\nid_ :: Ob_ (->) a -> a -> a (bound at 348.hs:22:3)\n|\n22 | id_ = ()\n| ^^\nPrelude>\n```\n\nIf we simply replace `Ob_` with `type Ob_ (cat :: Cat ob) :: ob -> Constraint`, the expected type (`a -> a`) matches my intuition:\n\n```348.hs:22:9: error:\n• Couldn't match expected type ‘a -> a’ with actual type ‘()’\n• In the expression: ()\nIn an equation for ‘id_’: id_ = ()\nIn the instance declaration for ‘Category_ (->)’\n• Relevant bindings include id_ :: a -> a (bound at 348.hs:22:3)\n|\n22 | id_ = ()\n| ^^\n```\n\n### comment:1 Changed 15 months ago by Iceland_jack\n\nDescription: modified (diff)\n\n### comment:2 Changed 15 months ago by RyanGlScott\n\nResolution: → duplicate new → closed\n\nYep, this is a duplicate of #15412. Now that that's been fixed, you get the error message you'd expect on the original program:\n\n```\\$ /opt/ghc/8.6.1/bin/ghci Bug.hs\nGHCi, version 8.6.0.20180823: http://www.haskell.org/ghc/ :? for help"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.70959693,"math_prob":0.8226512,"size":2549,"snap":"2019-43-2019-47","text_gpt3_token_len":761,"char_repetition_ratio":0.105697446,"word_repetition_ratio":0.23648648,"special_character_ratio":0.34994116,"punctuation_ratio":0.20948617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9545004,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-22T12:02:43Z\",\"WARC-Record-ID\":\"<urn:uuid:e9d2a0f0-d330-4d15-95d1-0abf9db059dc>\",\"Content-Length\":\"28948\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:791ab920-b254-4070-8d11-92d840ed5f45>\",\"WARC-Concurrent-To\":\"<urn:uuid:bdeb0364-226e-4372-8068-4c8b96711ac8>\",\"WARC-IP-Address\":\"162.242.166.62\",\"WARC-Target-URI\":\"https://trac.haskell.org/trac/ghc/ticket/15583\",\"WARC-Payload-Digest\":\"sha1:52OSMFVWN56N4J3I7TSEMWFA3LNUTSDN\",\"WARC-Block-Digest\":\"sha1:R7ZRZBYDA2MDUDTFDLCDCU7J7GHVE3NK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496671260.30_warc_CC-MAIN-20191122115908-20191122143908-00066.warc.gz\"}"} |
https://www.mezzo-piano.nl/2021-01-31/755.html | [
"",
null,
"# ball mill performance calculation\n\n•",
null,
"### Design, Construction and Performance Analysis of a 5\n\nSince for the ball mill design we are using 80 passing, the required value of C2 for the ball mill will be equal to1.20. C3 is the correction factor for mill diameter and is given as 𝐶𝐶3 = 2.44 𝐷𝐷 0.2 (3) However, it is important to note that C3 =0.914 vessel used in producing the ball mill was got from a The ball mill is the most common ore grinding technology today, and probably more than 50 of the total world energy consumption for ore grinding is consumed in ball mills.\n\n•",
null,
"### (PDF) Effect of circulating load and classification\n\nThe ball mill is the most common ore grinding technology today, and probably more than 50 of the total world energy consumption for ore grinding is consumed in ball mills. Since for the ball mill design we are using 80 passing, the required value of C2 for the ball mill will be equal to1.20. C3 is the correction factor for mill diameter and is given as 𝐶𝐶3 = 2.44 𝐷𝐷 0.2 (3) However, it is important to note that C3 =0.914 vessel used in producing the ball mill was got from a\n\n•",
null,
"### Page 1 Ball Milling Theoryfreeshell\n\ninvolve grinding). With Lloyd s ball milling book having sold over 2000 copies, there are probably over 1000 home built ball mills operating in just America alone. This article borrows from Lloyd s research, which was obtained from the commercial ball milling industry, and explains some of the key design criteria for making your own ball mill. to calculate ball mill drive hp . ball mill performance calculation. to calculate ball mill drive hp gold ore crusher ball mill drive motor choices for presentation at the the time to accelerate the mill to the design speed is given by the following calcul. ball mill load calculations ball mill load .\n\n•",
null,
"### CALCULATION OF BALL MILL GRINDING EFFICIENCYPage 1\n\nMar 08, 2013 · calculation of ball mill grinding efficiency. dear experts . please tell me how to calculate the grinding efficiency of a closed ckt open ckt ball mill. in literatures it is written that the grinding efficiency of ball mill is very less less than 10 . please expalin in a n excel sheet to calcualte the same. thanks. sidhant. reply in a l6 in. diameter, 16 in. long Denver Ball Mill having a grate discharge. The ball mill, driven by a 1.5 hp motor, operated at a fixed speed of 53.8 rpm. This speed was 78.5 percent of the critical speed. The normal operating ball load was set at 230 pounds, and was comprised of 2.5, 2.0, 1.5, 1.0, and\n\n•",
null,
"### Ball Mill Finish CalculatorMartin Chick Associates\n\nThe Ball Mill Finish Calculator can be used when an end mill with a full radius (a ball mill) is used on a contoured surface. The tool radius on each side of the cut will leave stock referred to as a scallop. The finish of the part will be determined by the height of the scallop, amd the scallop will be determined by the stepover distance In this module, you will learn how to characterize the performance of ball mill circuits. Specifically, after completing this module, you will be able to • List and describe the four elements of the functional performance equation for ball mill circuits. • Define and calculate the classification system efficiency of a ball mill circuit.\n\n•",
null,
"### Comparing ball and vertical mills performance An\n\nthe ball mills. It was expected a loss on mills performance as predicted by grinding tests, showed on Figure 8. Figure 8Comparison to standard charge and balls scraps. Thankfully, the performance of the vertical mill was still better than the ball mills, which reinforces the use of this technology. Calculation Of Ball Mill Description. Optimization of mill performance by using. Optimization of mill performance by using online ball and pulp measurements by B. Clermont and B. de Haas Synopsis Ball mills are usually the largest consumers of energy within a mineral concentrator. Comminution is responsible for 50 of the total mineral\n\n•",
null,
"### MODULE #5 FUNCTIONAL PERFOMANCE OF BALL MILLING\n\nIn this module, you will learn how to characterize the performance of ball mill circuits. Specifically, after completing this module, you will be able to • List and describe the four elements of the functional performance equation for ball mill circuits. • Define and calculate the classification system efficiency of a ball mill circuit. in a l6 in. diameter, 16 in. long Denver Ball Mill having a grate discharge. The ball mill, driven by a 1.5 hp motor, operated at a fixed speed of 53.8 rpm. This speed was 78.5 percent of the critical speed. The normal operating ball load was set at 230 pounds, and was comprised of 2.5, 2.0, 1.5, 1.0, and\n\n•",
null,
"### Ball Mill Finish CalculatorMartin Chick Associates\n\nBall Mill Finish Calculator The Ball Mill Finish Calculator can be used when an end mill with a full radius (a ball mill) is used on a contoured surface. The tool radius on each side of the cut will leave stock referred to as a scallop. Ball Nose Effective Feed rate is the speed of the end mill s movement correspondent to the workpiece. The feed rate is measured in inches per minute (IPM) COMMON EQUATIONS FOR OPTIMAL PERFORMANCE Too high of a speed or too light of a feed leads to reduction in tool life.\n\n•",
null,
"Dec 12, 2016 · If P is less than 80 passing 70 microns, power consumption will be. Ball Mill Power Calculation Example. A wet grinding ball mill in closed circuit is Aug 30, 2019 · 1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and humidity), the degree of\n\n•",
null,
"### End Mill Speed and Feed CalculatorMartin Chick Associates\n\nSpeed And Feed Calculators Ball Mill Finish Calculator Part Spacing Calculator G And M Code Characters Standard End Mill Sizes Standard Drill Sizes Drill And Counterbore Sizes. Contact. End Mill Speed Feed Calculator. Tool Dia. In. Radial (Side) Depth of Cut. This will adjust the feedrate if less than the tool rad. In. Num of Flutes Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to . CALCULATION OF BALL MILL GRINDING EFFICIENCY IN LITERATURES IT . grinding 350 for dry grinding 355) S the grind material bulk mass, g/cc. How to Size a Ball Mill Design Calculator\n\n•",
null,
"### Estimating Ball ConsumptionMolycop\n\nSep 17, 2018 · The work performed included analysis of operating performance data from 49 mills and included extensive determinations of the Bond Abrasion Index for different ore types. For an improved understanding, the wear rate constant is directly related to the grinding media consumption rate expressed in grams of steel per kWh drawn by the mill, using observed that these parameters influence the performance of the ball mill significantly. The performance of a ball mill is measured with reference to the quantity of undersize or fines (amount of grounded material passing through120 mesh screen) per revolution of the mill, collected for variation of each of the individual parameters.\n\n•",
null,
"### Calculating Ball Mill Throughput\n\ncement ball mill throughput calculation. calculate ball mill throughput in closed circuit. Get more info of silica sand mining process equipment for Advanced process control for the cement industry. Aug 30, 2019 · 1 Calculation of ball mill capacity. The production capacity of the ball mill is determined by the amount of material required to be ground, and it must have a certain margin when designing and selecting. There are many factors affecting the production capacity of the ball mill, in addition to the nature of the material (grain size, hardness, density, temperature and humidity), the degree of\n\n•",
null,
"### Estimating Ball ConsumptionMolycop\n\nSep 17, 2018 · The work performed included analysis of operating performance data from 49 mills and included extensive determinations of the Bond Abrasion Index for different ore types. For an improved understanding, the wear rate constant is directly related to the grinding media consumption rate expressed in grams of steel per kWh drawn by the mill, using A numerical dynamic-mechanical model of a planetary ball-mill is developed to study the dependence of process efficiency on milling parameters like ball size and number, jar geometry and velocity\n\n•",
null,
"Dec 12, 2016 · If P is less than 80 passing 70 microns, power consumption will be. Ball Mill Power Calculation Example. A wet grinding ball mill in closed circuit is observed that these parameters influence the performance of the ball mill significantly. The performance of a ball mill is measured with reference to the quantity of undersize or fines (amount of grounded material passing through120 mesh screen) per revolution of the mill, collected for variation of each of the individual parameters.\n\n•",
null,
"### Ball Mill Design/Power Calculation\n\nThe basic parameters used in ball mill design (power calculations), rod mill or any tumbling mill sizing are material to be ground, characteristics, Bond Work Index, bulk density, specific density, desired mill tonnage capacity DTPH, operating solids or pulp density, feed size as F80 and maximum chunk size , product size as P80 and maximum and finally the type of circuit open/closed Ball Mill Power Calculation Example A wet grinding ball mill in closed circuit is to . CALCULATION OF BALL MILL GRINDING EFFICIENCY IN LITERATURES IT . grinding 350 for dry grinding 355) S the grind material bulk mass, g/cc. How to Size a Ball Mill Design Calculator\n\n•",
null,
"### Feed/Speed Calculator Nachi America\n\nCalculator Adjusted Feedrate Chip Thinning Calculator Launch Tool Ball End Mill Effective Diameter Launch Tool Aqua V Mill Speed/Feed Calculator L9702-L9704, L9706-L9710 Launch Tool Aqua Hard Mill Speed/Feed Calculator L9711-L9714 Launch Tool Aqua Hard Ball Mill Speed/Feed Calculator L9615, L9616 Launch Tool ALH Mill Speed/Feed Calculating Ball Mill Throughput. The reactivity of the coal, Throughput, tons/hr. Grindability Several performance parameters are calculated for the pulverizer train.\n\n•",
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"•",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.88306123,"math_prob":0.938378,"size":11367,"snap":"2021-21-2021-25","text_gpt3_token_len":2456,"char_repetition_ratio":0.17125759,"word_repetition_ratio":0.69214547,"special_character_ratio":0.21201724,"punctuation_ratio":0.09721567,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9500186,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38],"im_url_duplicate_count":[null,null,null,2,null,3,null,1,null,1,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null,1,null,1,null,1,null,2,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-06T05:56:24Z\",\"WARC-Record-ID\":\"<urn:uuid:20bc510a-0e8d-4695-b0ef-c8ee10754fde>\",\"Content-Length\":\"31002\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:fb91c7b0-e7f3-444c-8312-b992d55831f1>\",\"WARC-Concurrent-To\":\"<urn:uuid:01553922-4e7f-49c2-a461-d3a0c0e8d3e5>\",\"WARC-IP-Address\":\"172.67.194.24\",\"WARC-Target-URI\":\"https://www.mezzo-piano.nl/2021-01-31/755.html\",\"WARC-Payload-Digest\":\"sha1:LFO5FSTUXN446UZBQWPJTVFGM7X2IVC7\",\"WARC-Block-Digest\":\"sha1:2SPHJUVS6AJT6MYQRKOMHCDU34Y6XLKP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243988741.20_warc_CC-MAIN-20210506053729-20210506083729-00296.warc.gz\"}"} |
https://www.science20.com/alpha_meme/empirical_probability_versus_classical_fair_metarandomness-81195 | [
"Lets gamble; participating costs you only 50 cents per game. The odds are in your favor! Two out of three times, you win and get a dollar. So we start playing, and it seems as if we walk along time, every game we get to a point in the road where it splits into three paths, two of them are winning one-dollar branches, one is the zero-dollar branch, there we toss a three sided fair die, then we ‘find ourselves’ in one of the three branches. However, I cheated.\n\nI used a quantum die. Although the classical probabilities are V0 = 1/3 and V1 = 2/3, the zero-dollar branch itself branches into three more. If you believe in classical physics, then such branching does not make any difference. You go with probability 2/3 into one of the winning branches. The zero dollar branch may as well split into a million more; since we only get there with probability 1/3 in the first place, it does not matter.",
null,
"You happen to lose maybe, but then you want to play again of course – after all, the game seems in your favor. After two runs, you have a good chance of having won two dollars already and the ticket fee is only one dollar for two trials. The classical probability of getting two dollars is V11 = 4/9, and the other classical probabilities are V10 = 2/9, V01 = 2/9, and V00 = 1/9. Thus, you expect to break even or win in 8 out of 9 cases. Except for one little problem: the die is a quantum die, and the zero-dollar branch always splits into three more.\n\nAfter a while, you will maybe believe in having bad luck (or preferably accept quantum mechanics). What happened? Well, we do not walk along time and then come to junctions in the road and then have some meta-randomness with meta-fairness that lets us go this or that path.\n\nSpace as such does not reside inside some meta-space. If time is put down as a t-axis, it should not be discussed as if there is a ‘now-moment’ creeping along the axis, as if there is a meta-time that allowed such movement. One rejects such meta-levels (this is discussed generally in detail here), because describing them would require another ‘meta-meta’ level, leading to infinite regress or at least regress without definite termination.\n\nWith the zero-dollar path being there three times, there are after the first run already five outcomes, and three of them got zero dollars, only two branches won. After two trials, the outcome numbers are N11 = 4, N10 = 6, N01 = 6, and N00 = 9 (total 25, as you can easily check in the picture). In twelve of these parallel worlds, you just break even, but in nine you paid the ticket fee and I gave you nothing in return.\n\nIt does almost not matter what the classical probabilities V are. We find probabilities by repeating an experiment many times and then judging from the past record. In the overwhelming number of world branches in which you remember to have played the above game many times, you will also find that the probabilities are the quantum probabilities P0 = 3/5 and P1 = 2/5, not the classical V.\n\nConsider the branching tree of the potential outcomes of coin tosses. There is no meta level on which we throw a fair meta-coin whenever we reach a branching point. Classical “meta-probability” can be represented as a (phase space) volume V. A random vector Real may then select the actual outcome without bias for any points in that space: The more volume V a branch has, the more likely it will be selected. Statistical mechanics similarly assumes fair meta-probability via the ergodic hypothesis. Instead of the whole being already fully described by the tree alone, the meta-probability V makes Real behave properly. Such meta-coin tosses are unnecessary and lead to difficulties especially in cosmology, where also space-in-space and time-of-time are most problematic.\n\nIn a true many-worlds model, all outcomes are actualized relative to their own branch. You do not advance into the ‘heads’ instead of the ‘tails’ branch with meta-probability Vheads = 50%; both futures exist equally. Most outcome branches of several tosses will observe close to 50% heads. The probability P of an outcome is proportional to the number N of branches with that outcome, which can only in a classical many worlds model be replaced by V.\n\nNothing selects any branches or needs to count the parallel branches in order to establish P. The branches remember their past, that is enough.\n\nTo understand the above is the next step on the path of making the Many Worlds Wiener Sausage resolve the Einstein-Podolsky-Rosen paradox. Remember, it can produce the correct quantum factors, but we also saw that although it is a many worlds model with parallel universes and all that, it is not a quantum world! Simply introducing the “Real” direction turned it completely classical. Now we see how to fix what is wrong: The volume V of the sausage is the classical meta-probability and Real does the meta-fair meta-coin toss, i.e. it lands somewhere inside V without bias. If we want to make it a mature quantum model, we necessarily need to grow new worlds, not just cut the one that is there into small pieces.",
null,
"Adding a single little arrow “Real” (short \"DR\") clarifies that the sausage model is fundamentally a local realism and thus cannot violate Bell’s inequality. No such model can possibly describe quantum physics.\n\nS. Vongehr: \"Many Worlds Model resolving the Einstein Podolsky Rosen paradox via a Direct Realism to Modal Realism Transition that preserves Einstein Locality\" arXiv:1108.1674v1 [quant-ph] (2011) UPDATE: This reference is the first paper on the possibility of such models, but the models have now actually been constructed and are much better explained in S. Vongehr: “Against Absolute Actualization: Three \"Non-Localities\" and Failure of Model-External Randomness made easy with Many-Worlds Models including Stronger Bell-Violation and Correct QM Probability” http://arxiv.org/abs/1311.5419 (2013)\n\n--------------------------------------\n\nSascha Vongehr's Articles Topic by Topic"
]
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"https://www.science20.com/files/images/VPgame_0.JPG",
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"https://www.science20.com/files/images/SausageDR.JPG",
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https://mimoformatlab.com/docs/channel/ | [
"The channel object represents a generic MIMO channel. It does not hold any value in representing a particular channel model but rather acts as the foundation for all other channel models within MFM.\n\nMFM currently only supports frequency-flat channels, meaning the channel can be largely characterized by a single matrix $\\mathbf{H}$. Support for frequency-selectivity may exist in future releases of MFM.\n\n### Creating a Generic Channel\n\nWhile the channel object on its own isn’t all that useful, it can be created via\n\nc = channel.create()\n\n\nAs mentioned, more useful channel models are subclasses of the channel object, meaning they will inherit the properties and methods of the channel object. Moreover, the channel object includes some of the key setup functions that most—if not all—other channel objects will use.\n\n### Key Properties\n\nThe channel object (and its subclasses) contain the following important properties.\n\nEach channel requires knowledge of the transmit array at the channel input, which is captured in\n\nc.array_transmit\n\n\nSimilarly, the receive array at the channel output is captured in\n\nc.array_receive\n\n\nThe number of antennas at the transmit array and receive array, respectively, are captured by the properties\n\nc.num_antennas_transmit\n\n\nThe carrier frequency (in Hertz), carrier wavelength (in wavelengths), and propagation velocity (in meters per second) of the channel—more accurately, of the signals we are interested in—are captured by\n\nc.carrier_frequency\nc.carrier_wavelength\nc.propagation_velocity\n\n\nWhile not all channel models require knowledge of these three properties (e.g., a Rayleigh-faded channel), several channels do and thus, for simplicity, MFM includes them in the generic channel object.\n\nFinally, the channel matrix $\\mathbf{H}$ is stored in\n\nc.channel_matrix\n\n\n### Setting the Transmit and Receive Arrays\n\nTo set the transmit and receive arrays of a channel object c, simply use\n\nc.set_arrays(atx,arx)\n\n\nwhere atx and arx are array objects corresponding to the antenna arrays at the channel input and output, respectively.\n\nTo set the transmit and receive arrays individually, one can use\n\nc.set_array_transmit(atx)\n\n\nUpon setting the transmit array and receive array, the number of transmit antennas (c.num_antennas_transmit) and the number of receive antennas (c.num_antennas_receive) are automatically updated based on the number of elements in the array objects.\n\nWhile the channel is aware of the transmit and receive arrays, it is important to note that the channel represents solely the over-the-air propagation and does not account for any phase profiles inherent to the arrays themselves. At the very least, the channel object will merely take note of the number of antennas at the transmit array and receive array. Some channel models (e.g., the ray/cluster channel model) use the transmit and receive array objects to obtain array response vectors.\n\n### Setting the Carrier Frequency\n\nTo set the carrier frequency of the channel, use\n\nc.set_carrier_frequency(fc)\n\n\nwhere fc is the carrier frequency (in Hz) of the signals into the channel. This will automatically update the carrier wavelength based on the current propagation velocity.\n\n### Setting the Propagation Velocity\n\nTo set the propagation velocity of the channel, use\n\nc.set_propagation_velocity(vel)\n\n\nwhere vel is the propagation velocity (in m/s) of the signals into the channel.\n\nIt is common to use vel = 3e8 for electromagnetic signals, for example.\n\nSince this parameter can be set by the user, MFM may support other communication systems and/or other propagation media, such as underwater acoustics where the propagation velocity is commonly approximated to vel = 1.5e3 for oceanic environments.\n\n### Example Setup\n\nA typical channel object setup looks something similar to\n\nc = channel.create()\nc.set_carrier_frequency(fc)\nc.set_propagation_velocity(vel)\nc.set_arrays(atx,arx)\n\n\n### Forcing Channel Energy Normalization\n\nSometimes, channel matrices are normalized to a specific energy (Frobenius norm squared). To force MFM to normalize channel matrices to a fixed value, use\n\nc.set_force_channel_energy_normalization(true)\n\n\nand use\n\nc.set_normalized_channel_energy(val)\n\n\nwhere val is the desired Frobenius norm squared the channel matrices. If not set, the normalized channel energy is equal to the number of elements in the channel matrix (i.e., the product of the number of transmit antennas and the number of receive antennas).\n\nWith this, all realized channel matrices $\\mathbf{H}$ will satisfy $\\lVert\\mathbf{H}\\rVert_{\\mathrm{F}}^2 =$ val.\n\nBy default channels do not normalize the channel matrices.\n\n### Shorthand Methods\n\nTo provide convenient ways to retrieve common MIMO-related quantities from a channel object c, there exist the following so-called shorthand methods.\n\n• c.H — Returns the channel matrix.\n• c.Nt — Returns the number of transmit antennas at the channel input.\n• c.Nr — Returns the number of receive antennas at the channel output.\n\n### List of Properties\n\nThe channel object contains the following properties:\n\n• channel.name\n• channel.channel_matrix\n• channel.num_antennas_transmit\n• channel.num_antennas_receive\n• channel.array_transmit\n• channel.array_receive\n• channel.carrier_frequency\n• channel.carrier_wavelength\n• channel.propagation_velocity\n• channel.normalized_channel_energy\n• channel.force_channel_energy_normalization\n\n### List of Methods\n\nThe channel object contains the following methods:\n\n### Methods Documentation\n\n#### H()\n\nReturns the channel matrix.\n\nUsage:\nval = H()\nReturn Values:\nval — the channel matrix\n\nBack to methods\n\n#### Nr()\n\nReturns the number of receive antennas out of the channel.\n\nUsage:\nval = Nr()\nReturn Values:\nval — the number of receive antennas out of the channel\n\nBack to methods\n\n#### Nt()\n\nReturns the number of transmit antennas into the channel.\n\nUsage:\nval = Nt()\nReturn Values:\nval — the number of transmit antennas into the channel\n\nBack to methods\n\n#### channel(name)\n\nCreates a MIMO channel object.\n\nUsage:\nobj = channel()\nobj = channel(name)\nInput Arguments:\nname — an optional name for the object\nReturn Values:\nobj — an object representing a MIMO channel\n\nBack to methods\n\n#### create(type)\n\nCreates a channel object of a specific type.\n\nUsage:\nc = channel.create()\nc = channel.create(type)\nInput Arguments:\ntype — (optional) a string specifying what type of channel to create\nReturn Values:\nc — a channel object of the type specified\n\nBack to methods\n\n#### enforce_channel_energy_normalization(H)\n\nNormalizes the channel matrix so that its total energy (squared Frobenius norm) is equal the current normalized channel energy property. The default normalized channel energy is the product of the number of transmit antenans and the number of receive antennas.\n\nUsage:\nG = enforce_channel_energy_normalization()\nG = enforce_channel_energy_normalization(H)\nInput Arguments:\nH — (optional) a channel matrix; if not passed, the current channel matrix will be used and overwritten with the normalized version; if passed, the channel matrix property will not be set\nReturn Values:\nG — the normalized channel matrix\n\nBack to methods\n\n#### get_channel_matrix()\n\nReturns the channel matrix.\n\nUsage:\nH = get_channel_matrix()\nReturn Values:\nH — channel matrix\n\nBack to methods\n\n#### initialize()\n\nInitializes a channel.\n\nUsage:\ninitialize()\n\nBack to methods\n\n#### set_array_receive(array)\n\nSets the receive array object. Also sets the number of receive antennas accordingly.\n\nUsage:\nset_array_receive(array)\nInput Arguments:\narray — an array object\n\nBack to methods\n\n#### set_array_transmit(array)\n\nSets the transmit array object. Also sets the number of transmit antennas accordingly.\n\nUsage:\nset_array_transmit(array)\nInput Arguments:\narray — an array object\n\nBack to methods\n\n#### set_arrays(array_transmit,array_receive)\n\nSets the transmit and receive arrays at the input and output of the channel.\n\nUsage:\nset_arrays(array_transmit,array_receive)\nInput Arguments:\narray_transmit — an array object at the channel input\narray_receive — an array object at the channel output\n\nBack to methods\n\n#### set_carrier_frequency(fc)\n\nSets the carrier frequency of the channel.\n\nUsage:\nset_carrier_frequency(fc)\nInput Arguments:\nfc — carrier frequency (Hz)\nNotes:\n\nBack to methods\n\n#### set_channel_matrix(H)\n\nSets the channel matrix.\n\nUsage:\nset_channel_matrix(H)\nInput Arguments:\nH — channel matrix\n\nBack to methods\n\n#### set_force_channel_energy_normalization(force)\n\nSets the enforcement of channel energy normalization. If true, the channel matrix will always be normalized such that its energy is of the desired value.\n\nUsage:\nset_force_channel_energy_normalization(force)\nInput Arguments:\nforce — a boolean indicating if the channel matrix should be normalized or not\n\nBack to methods\n\n#### set_name(name)\n\nSets the name of the channel.\n\nUsage:\nset_name()\nset_name(name)\nInput Arguments:\nname — (optional) a string; if not passed, ‘channel’ is the default name used\n\nBack to methods\n\n#### set_normalized_channel_energy(E)\n\nSets the normalized energy of the channel.\n\nUsage:\nset_normalized_channel_energy()\nset_normalized_channel_energy(E)\nInput Arguments:\nE — (optional) the desired normalized channel energy; if not passed, the product of the number of transmit antennas and the number of receive antennas will be used\n\nBack to methods\n\n#### set_propagation_velocity(val)\n\nSets the propagation velocity of the channel.\n\nUsage:\nset_propagation_velocity(val)\nInput Arguments:\nval — propagation velocity (meters/sec)\n\nBack to methods\n\n#### set_receive_array(array)\n\nSets the receive array object (LEGACY).\n\nUsage:\nset_receive_array(array)\nInput Arguments:\narray — an array object\n\nBack to methods\n\n#### set_transmit_array(array)\n\nSets the transmit array object (LEGACY).\n\nUsage:\nset_transmit_array(array)\nInput Arguments:\narray — an array object\n\nBack to methods"
]
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https://askfilo.com/user-question-answers-physics/due-to-variation-of-direction-scalar-quantity-does-no-change-33393437373830 | [
"",
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"World's only instant tutoring platform",
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"Question",
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"# Due to variation of direction, scalar quantity does nò change, while vector quantity changes. ii. Temperature, distance, mass, speed, density, energy, power, pressure, electric charge and time are scalar quantities, iii. Scalar quantity obeys the law of ordinary algebra. iv. Scalar quantity may obey the law of vector algebra. v. Displacement, velocity, acceleration, force, linear momentum, torque, angular momentum and impulse are vector quantities. vi. Vector does not obey the law of ordinary algebra. vii. A vector may be equal to a scalar. viii. A scalar can never be added to or subtracted from a vector. 4. Find displacement of a person when he moving on regular hexagon of side from to . (via 5. A vector may change if :\n\n1. frame of reference is translated\n2. frame of reference is rotated\n3. vector is translated parallel to itself\n4. vector is rotated",
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"## Filo tutor solutions (1)\n\nLearn from their 1-to-1 discussion with Filo tutors.\n\n3 mins",
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"Connect instantly with this tutor\n\nConnect now\n\nTaught by",
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"Trusted by 1 crore+ students\n Question Text Due to variation of direction, scalar quantity does nò change, while vector quantity changes. ii. Temperature, distance, mass, speed, density, energy, power, pressure, electric charge and time are scalar quantities, iii. Scalar quantity obeys the law of ordinary algebra. iv. Scalar quantity may obey the law of vector algebra. v. Displacement, velocity, acceleration, force, linear momentum, torque, angular momentum and impulse are vector quantities. vi. Vector does not obey the law of ordinary algebra. vii. A vector may be equal to a scalar. viii. A scalar can never be added to or subtracted from a vector. 4. Find displacement of a person when he moving on regular hexagon of side from to . (via 5. A vector may change if : Updated On Jan 24, 2023 Topic Electrostats Subject Physics Class Class 12 Answer Type Video solution: 1 Upvotes 93 Avg. Video Duration 3 min"
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https://www.jiskha.com/archives/2012/07/27 | [
"# Questions Asked onJuly 27, 2012\n\n1. ## math\n\nWhat is the probability that a randomly selected three-digit number has the property that one digit is equal to the product of the other two? Express your answer as a common fraction.\n\n2. ## math\n\nThe diagonal of a rectangular room is 52 ft long. one wall measures 28 ft longer than the adjacent wall. Find the dimensions of the room.\n\n3. ## algebra\n\na function is defined by the equation y=8x-3. If the domain is 2≤x≤4, find the minimum value in the range of the function.\n\n4. ## Physics\n\nWhen uranium 92U235 decays, it emits a ray. If the emitted ray has a wavelength of 1.30 x 10-11 m, determine the energy (in MeV) of the ray\n\n5. ## geometry\n\nIn a triangle ABC,right angled at B, BD is drawn perpendicular to AC. prove that: (i) angle ABD = angle c (ii) angle CBD = angle A\n\n6. ## Math- STATS 218\n\nWhen the following hypotheses are being tested at a level of significance of aH0: m 100 Ha: m < 100 the null hypothesis will be rejected if the test statistic Z is a. >Z b. >Z c.\n\n7. ## math\n\nIn a triangle ABC,right angled at B, BD is drawn perpendicular to AC. prove that: (i) angle ABD = angle c (ii) angle CBD = angle A\n\n8. ## math\n\nRichard is standing between two buildings in a town house development.The building on the left is 9 m away and the angle of elevation to its security spotlight A is 68 degrees. The building on the right is 6m away and the angle of elevation to its security\n\n9. ## Math\n\nThe length of a rectangle is 2 more than twice its width. If the area of the rectangle is 24 sq. m., find its dimensions.\n\n10. ## Math\n\nA square painting is surrounded by a 3-cm-wide frame. If the total area of the painting plus the frame is 81 cm2, find the dimensions of the painting.\n\n11. ## math\n\nIn a triangle ABC , the internal bisectors of angles B and C meet at P and the external bisectors of the angles B and C meet at Q. prove that: angle BPC + angle BQC = 2 rt. angles.\n\n12. ## to ms.sue- writing project\n\nwhat i wrote: every monday, my aunt would come and visit me.everytime my aunt visits me, she would always have something special for me. This monday was different.when my aunt came to visit me today, she was holding a huge box. I heard strange noises\n\n13. ## accounting\n\nAssume that a company purchases land for \\$100,000, paying \\$20,000 cash and borrowing the remainder with a long-term note payable. How should this transaction be reported on a statement of cash flows\n\n14. ## algebra\n\nFind the product: (8x-3y)^2\n\n15. ## Chemistry\n\nCalculate the solubility of ZnCO3 in .050 M Zn(NO3)2. Ksp= 3.0*10^-8 solubility= 1.7*10^-4\n\n16. ## statistics\n\nA rental agent claims that the mean monthly rent, u , for apartments on the east side of town is less than \\$675. A random sample of 21 monthly rents for apartments on the east side has a mean of \\$674, with a standard deviation of \\$18. If we assume that the\n\n17. ## algebra\n\nHannah invests \\$3850 dollars at an annual rate of 6% compounded continuously, according to the formula A=Pe^rt, where A is the amount, P is the principal, e=2.718, r is the rate of interest, and t is the time, in years. (a) Determine, to the nearest dollar\n\n18. ## Finance\n\nDuring a job interview, Pam Thomson is offered a salary of \\$28,000. The Company gives annual raises of 6 percent. What would be Pam's salary during her fifth year on the job?\n\n19. ## Mathsw\n\nA man 6 feet tall is standing 9 feet from a 18 feet tall light pole. His shadow cast 15 degree. What is the angle formed from this equation?\n\n20. ## AP Chemistry\n\nIf you were to draw diagrams (such as that shown on the right) representing aqueous solutions of each of the following ionic compounds, how many anions would you show if the diagram contained six cations? (a) NiSO4, (b) Ca(NO3)2, (c) Na3PO4, (d) Al2(SO4)3\n\n21. ## math\n\nRound all answers to the nearest hundredth square root of b^2-4ac;b=3,a=-1,c=-2?\n\n22. ## Math\n\nI don't understand how u can get 3454 from mutiplying 1,100(3.14) When i did it i got 1414 The question: is 2(3.14r2 + 2(3.14)rh 2(3.14)(10)2 + 2(3.14)(10)(45) and when i got done with the promblem the answer in book said approximately 3454in2 when i got\n\n23. ## Math\n\nThere are 6 pairs of black socks and 6 pairs of white socksWhat is the probability to pick a pair of black or white socks when 2 socks are selected randomly in darkness?M\n\n24. ## College Algebra\n\nGraph the following function using transformations. Be sure to graph all of the stages on one graph. State the domain and range. y = -SQRT x - 6. My instructor is saying that I have to do more than one graph. Can anybody help me I am lost!\n\n25. ## design\n\n1. A person with a square face shape and a rectangle body shape has what bodyline? A. Straight B. Sharp-straight.(yes) C. Soft-straight D. Soft-straight II 2. A suit with rounded lapels and a shawl collar would work best for a woman with a _______\n\n26. ## Physics\n\nDiffraction grating #1 has half the lines/cm that diffraction grating #2 does. When used with a certain wavelength of light, both gratings give first order maxima. Which of the following statements is true for the same wavelength of light.\n\n27. ## math\n\nA fair, six-sided die is rolled eight times, to form an eight-digit number. What is the probability that the resulting number is a multiple of 8? Express your answer as a common fraction.\n\n28. ## algebra\n\nFind the product (x+5)(5x+25)\n\n29. ## algbra\n\nFind the product: (8x-3y)^2\n\n30. ## Algebra\n\nFind the product: (5^2-3y)(5x^2+3y)\n\n31. ## Physics\n\nA)What is the equation for equivalent resistance for resistors wired in series? What about series wiring makes resistors in series add the way they do? B)What is the equation for capacitors wired in parallel? What about parallel wiring makes capacitors add\n\n32. ## Algebra\n\nFind product: (5x^2-3y)(5x^23y)\n\n33. ## Algebra\n\nEssay: Show all work. A gardner wants to create wants to create a triangle garden in the shape of a right triangle with skortesr length x-5yft. and the middle length side x+6yft. what is an algebraic expression for the area of the garden? Be sure to\n\n34. ## algebra\n\nEssay: Show all work. Find the following product. (x^2+7)(x^3-x7)\n\n35. ## math\n\n250 is number of sales, which is down 12% of the previous year. What is last year's total?\n\n36. ## Algebra\n\nEssay: Show all work. A school designer wants to create a white board with the ptimal dimensions to enhance if one side of the white board x^-5xy+8y inches. Write an algebra expression for the area of the area of the area of such a white board, simplify\n\n37. ## geometry\n\nIn a triangle ABC , the internal bisectors of angles B and C meet at P and the external bisectors of the angles B and C meet at Q. prove that: angle BPC + angle BQC = 2 rt. angles.\n\n38. ## math\n\nthe mean of a list of seven numbers is 45. if i add two numbers to this list the mean of he list becomes 37 what is the mean of the two numbers that i added to the list\n\n39. ## algebra\n\na function is defined by the equation y=8x-3. If the domain is 2≤x≤4, find the minimum value in the range of the function.\n\n40. ## Algebra\n\nThe kinetic energy(K) of an object is dependent on its mass(m) and velocity(v) and is given by the equation K=.5mv^2. If an object has a velocity of 5 and a kinetic energy of 75, what is its mass?\n\n41. ## algebra\n\n1.if f(x)=8^x, find the value of f(2/3) 2. If 8^2x=2^(x+5), what is the value of x?\n\n42. ## Microeconomics\n\nAssume that the marginal utilities for the first three units of a good consumed are 200, 150, and 125 respectively. What is the total utility for the first unit?\n\n43. ## philosophy\n\ni have to write an essay comparing Continental, Pragmatic, and Analytic philosophies, focusing on one particular concept that is treated differently by all three schools of thought. what three schools are they talk in about?\n\n44. ## relative frequency\n\nhelp with relative frequency\n\n45. ## physics\n\nif a brick of mass 7 kg is thrown from a height of 24 ft at an angle of 45 degrees, what impact force will it land with on the ground? what will be the impact of the force on the brick if it lands on. a) concrete b) soft surface\n\n46. ## Math\n\nStalls in a particular market occupy a 3 m by 2m spaces. The stall owners requested that the area of their stalls be doubled. The market owner granted this request by increasing the length and the width equally. What are the new dimensions of the stalls?\n\nthe facility\n\n48. ## algebra\n\nPlease Answer the following: What do you call the Situation when the Rubber man Runs into the Bearded lady? What did Mergatroid get for losing 20 pounds?\n\n49. ## biology\n\nwhat is anterior tibial translation?\n\n50. ## physics\n\nA 1.50 kg mass attached to a spring oscillates with a period of 0.365 s and an amplitude of 15.5 cm. (a) Find the total mechanical energy of the system\n\n51. ## English 002\n\nTell us why it is important to use good grammar, proper punctuation and clear senteces when you need to le someone know are displeased.\n\n52. ## Math\n\nConsider the weighted voting system [75: 31, 29, 23, 16, 8, 7]. Find each: a. The total number of players b. The total number of votes c. The weight of P3 d. The minimum percentage of the votes needed to pass a motion 9rounded to the next whole percent)\n\n53. ## science\n\nHow many outlets can you place on one curcuit in your house\n\n54. ## physics\n\ntwo equal masses of 5kg are placed on two vertex of an an equilateral triangle and 20 kg mass on the third.each side is 20 cm long. find the force acting on 1 kg mass kept at the centroid of the triangle.\n\n55. ## math\n\nfind the quantities specified ð ~3.14 find the area of a triangle with a base of 12 meters and an altitude of 14 meters\n\n56. ## math\n\nRound all answers to the nearest hundredth -5a^2;a=-2?\n\n57. ## Trigonometry\n\nwhat is the frequency of y=tan(Pi/3 x)?\n\n58. ## Trigonometry\n\nHow many solutions does the system {y=x, y=tanx have?\n\n59. ## Trigonometry\n\nwhat is the amplitude of y=1/5sin(sqr root5x + 1/sqrroot5)?\n\n60. ## physics\n\nIn the circuit below, at time t = 0, the switch is closed, causing the capacitors to charge. The voltage across the battery is 12 V The circuit: flickr. com/photos/ 68751220@N03 /6848519773/ A)Calculate the time constant for the RC circuit B)Calculate the\n\n61. ## to ms.sue-writing project\n\nplease check: Narrative Prompts (choose only one) • Your aunt arrives with a huge box. You hear strange noises coming from the box. Write a story about what’s inside. what i wrote: every monday, my aunt would come and visit me.everytime my aunt visits\n\n62. ## english\n\nhe the attention of a mathematician {frame a wh-type question to get the underlined part as answer}\n\n63. ## law/421\n\nWhat are some examples of arbitration that can occur in your professional and personal life?\n\n64. ## math\n\n2sec(11)sec(19)-2cot(71)\n\n65. ## English\n\ncan you please identify the nouns and verbs in this sentence. Don't be impressed with your own wisdom. Instead, fear the Loard and turn byour back on evil.\n\n66. ## Chem 2\n\nUsing the reaction HB H+ + B- and the Ka for that expression, calculate the Ka given the initial concentration of weak acid to be 0.150 M and the pH of the solution to be 2.79.\n\n67. ## chem\n\nIf I am placing intermolecular Forces in order by boiling point. weakest to strongest C3H8 CH4 LiF HBr I'm not sure about the last two, is this in right order?\n\n68. ## chem\n\nusing gas laws using P1,V1,T1,P2,V2,T2 soling for P2 P1,V1/T1 =P2,V2/T2 Divide by V2 =T2/V2 (P1)(V1)(T2)/(T1)(V2)=P2 If I'm solving for P2, is this the right set up? Thanks\n\n69. ## chem\n\nI am not sure how to even set this question up. I need to find out how many moles of C2 gas is contained in a 2.0L container that has a pressure of 125 atmoshpers and a temp of 10 C. I have P1-125 Atom V1-2.0L T1-10 C + 273 = 283K N1 ? I'm not sure where\n\n70. ## chem\n\nlabeling form highest to lowest IMF and why HBr, LiF,CH4 HBr- Dipole, covalent LiF-Dipoole, Ionic CH4, Hydrogen bonding, -4 Is this right?\n\n71. ## algebra\n\n\\$400 invested at 3% compounded daily after a period of 3 years=\\$?(round to nearest cent)\n\n72. ## math\n\nif my cross is 3 3/8 inches tall and 2 1/2 inches wide how wide should it be if its 7 1/2 inches tall\n\n73. ## Math\n\nI don't understand how u can get 3454 from mutiplying 1,100(3.14)\n\n74. ## chemistry\n\nDetermine the activity of 1.11 g of 88Ra226 (a) in disintegrations per second (Bq) (b) and in Curies (Ci). The half-life is 1600 yr (1 yr = 3.156 x 10^7 s). 1 Ci is equal to 3.7 x 10^10 Bq.\n\n75. ## Physics\n\nHow many half-lives are required for the number of radioactive nuclei to decrease to 1/10^4 of the initial number?\n\n76. ## Chemistry\n\nWhat volume (in liters) is occupied by 3.85×10^22 nitrogen molecules at 105 C and 255 mmHg?\n\n77. ## Physics\n\nWhen any radioactive dating method is used, experimental error in the measurement of the sample's activity leads to error in the estimated age. In an application of the radiocarbon dating technique to certain fossils, an activity of 0.15 Bq per gram of\n\n78. ## SPANISH\n\nChange the following constructions to an adjetivo posesivo and noun according to the example: Example: la mesa de ella su mesa 1. el gato de Juan y María 2. las sillas de ti 3. el hermano de mí 4. la computadora de tu y yo 5. la mesa de usted 6. la mesa\n\n79. ## math\n\ndoes anyone know the formula for finding the base of a square based pyramid when u have the volume and height ?\n\n80. ## Algebra\n\nFind the product -7y^2(y^4-5y^3+8y^2-6y+5)\n\n81. ## Chem\n\nWhat volume (L) of a 3M KOH solution can be prepared by diluting 0.5 L of a 5M KOH solution?\n\n82. ## algebra\n\nFind the product (x+5)(x^2-5x+25)"
]
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https://answers.everydaycalculation.com/multiply-fractions/56-4-times-5-27 | [
"# Answers\n\nSolutions by everydaycalculation.com\n\n## Multiply 56/4 with 5/27\n\n1st number: 14 0/4, 2nd number: 5/27\n\nThis multiplication involving fractions can also be rephrased as \"What is 56/4 of 5/27?\"\n\n56/4 × 5/27 is 70/27.\n\n#### Steps for multiplying fractions\n\n1. Simply multiply the numerators and denominators separately:\n2. 56/4 × 5/27 = 56 × 5/4 × 27 = 280/108\n3. After reducing the fraction, the answer is 70/27\n4. In mixed form: 216/27\n\nMathStep (Works offline)",
null,
"Download our mobile app and learn to work with fractions in your own time:\nAndroid and iPhone/ iPad\n\n#### Multiply Fractions Calculator\n\n×\n\n© everydaycalculation.com"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8105395,"math_prob":0.97004133,"size":448,"snap":"2021-21-2021-25","text_gpt3_token_len":178,"char_repetition_ratio":0.1599099,"word_repetition_ratio":0.0,"special_character_ratio":0.45535713,"punctuation_ratio":0.08490566,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98051184,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-06-21T00:20:37Z\",\"WARC-Record-ID\":\"<urn:uuid:60afc613-48e3-4110-8fc4-88dbb6543625>\",\"Content-Length\":\"7480\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9f1a9d79-2ca4-4cdd-9b63-df8d64756f67>\",\"WARC-Concurrent-To\":\"<urn:uuid:5e73ef17-62b9-4acc-b806-cfabb722fe81>\",\"WARC-IP-Address\":\"96.126.107.130\",\"WARC-Target-URI\":\"https://answers.everydaycalculation.com/multiply-fractions/56-4-times-5-27\",\"WARC-Payload-Digest\":\"sha1:ALAK3MJ2XOUYVFWITDESW3NRJDHKBMSK\",\"WARC-Block-Digest\":\"sha1:HU4XVAKCYVKKC4IKVGIL4CRL4RHJQFGG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-25/CC-MAIN-2021-25_segments_1623488259200.84_warc_CC-MAIN-20210620235118-20210621025118-00127.warc.gz\"}"} |
https://www.medentry.edu.au/ucat/entry/umat-test-tactics-and-preparation-part-3 | [
"### MedEntry\n\nTrusted UCAT prep.\n\nBy accepting you will be accessing a service provided by a third-party external to https://www.medentry.edu.au/\n\nFont size: +\n\n# UMAT test tactics and preparation part 3: UMAT questions involving percentages\n\n## What you need to know about percentage % questions before starting your UMAT preparation:\n\nProblems involving percent are common on the UMAT. The word percent means “divided by one hundred.”\n\nWhen you see the word “percent,” or the symbol %, remember it means 1/100. For example,\n\n 25 Percent 25 x 1/100 = 1/4\n\nTo convert a decimal into a percent, move the decimal point two places to the right. For example,\n\n0.25 = 25%\n\n0.023 = 2.3%\n\n1.3 = 130%\n\nConversely, to convert a percent into a decimal, move the decimal point two places to the left. For example,\n\n47% = .47\n\n3.4% = .034\n\n175% = 1.75\n\nTo convert a fraction into a percent, first change it into a decimal (by dividing the denominator [bottom] into the numerator [top]) and then move the decimal point two places to the right. For example,\n\n7/8 = 0.875 = 87.5%\n\nConversely, to convert a percent into a fraction, first change it into a decimal and then change the decimal into a fraction. For example,\n\n80%=.80 = 80/100 = 4/5\n\nFollowing are the most common fractional equivalents of percents:\n\n 33.33% = 1/3 20% = 1/5 66.67% = 2/3 40% = 2/5 25% = 1/4 60% = 3/5 50% = 1/2 80% = 4/5\n\n# Sample UMAT percentage % problems with worked answers.\n\nPercent problems often require you to translate a sentence into a mathematical equation.\n\nExample 1: What percent of 25 is 5?\n\n(A) 10% (B) 20% (C) 30% (D) 35%\n\nTranslate the sentence into a mathematical equation as follows:\n\n What percent of 25 is 5 x 1/100 * 25 = 5\n\n(25/100)x = 5\n\n(1/4)x = 5\n\nx=20\n\nExample 2: 2 is 10% of what number?\n\n(A) 10 (B) 12 (C) 20 (D) 24\n\nTranslate the sentence into a mathematical equation as follows:\n\n 2 is 10 % of what number 2 = 10 1/100 * x\n\n2 = (10/100)x\n\n2 = (1/10)x\n\n20 = x\n\nExample 3: What percent of is 3a ?\n\n(A) 100% (B) 150% (C) 200% (D) 300%\n\nTranslate the sentence into a mathematical equation as follows:\n\n What percent of a is 3a x 1/100 * a - 3a\n\n(x/100)a = 3a\n\nx/100 = 3\n\nx= 300\n\nExample 4: If there are 15 boys and 25 girls in a class, what percent of the class is boys?\n\n(A) 15%\n\n(B) 18%\n\n(C) 25%\n\n(D) 37.5%\n\nThe total number of students in the class is 15 + 25 = 40. Now, translate the main part of the sentence into a mathematical equation:\n\n What percent of the class is boys x 1/100 * 40 = 15\n\n(40/100)x = 15\n\n(2/5)x = 15\n\n2x = 75\n\nx = 37.5%\n\nOften you will need to find the percent of increase (or decrease). To find it, calculate the increase (or decrease) and divide it by the original amount:\n\nPercent of change: (Amount of change/Original amount) xl00%\n\nExample 5: The population of a town was 12,000 in 1980 and 16,000 in 1990. What was the percent increase in the population of the town during this period?\n\n(A) 33 1/3%\n\n(B) 50%\n\n(C) 75%\n\n(D) 120%\n\nThe population increased from 12,000 to 16,000. Hence, the change in population was 4,000. Now, translate the main part of the sentence into a mathematical equation:\n\nPercent of change: (Amount of change/Original amount) * l00% =\n\n(4000/12000) * l00% =\n\n1/3 * 100% = (by canceling 4000)\n\n= 33+1/3%"
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https://www.assignmentexpert.com/homework-answers/mathematics/other/question-91613 | [
"# Answer to Question #91613 in Other Math for Nasreen akhtar\n\nQuestion #91613\nFind ratio of temperature of second and third week of june 2019\n1\n2019-07-15T09:48:50-0400\n\nIn my region the temperatures in second and third week of June measured in Celsius scale were\n\n24, 25, 28, 28, 24, 25, 26, 25, 24, 19, 12, 11, 21, 18.\n\nThe average of the second week is\n\n(24 + 25 + 28 + 28 + 24 + 25 + 26)/7 = 25.71.\n\nThe average of the third week is\n\n(25 + 24 + 19 + 12 + 11 + 21 + 18)/7 = 18.57.\n\nThe ratio is 18.57/25.71 = 0.72.\n\nNeed a fast expert's response?\n\nSubmit order\n\nand get a quick answer at the best price\n\nfor any assignment or question with DETAILED EXPLANATIONS!"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9220292,"math_prob":0.96359324,"size":366,"snap":"2020-34-2020-40","text_gpt3_token_len":153,"char_repetition_ratio":0.12154696,"word_repetition_ratio":0.0,"special_character_ratio":0.5601093,"punctuation_ratio":0.22,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97673917,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-10T02:38:34Z\",\"WARC-Record-ID\":\"<urn:uuid:f588ca2e-b814-4efa-a015-1ceca16e57c0>\",\"Content-Length\":\"1049100\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5e9bf7e1-3846-4656-a039-b12913354ac5>\",\"WARC-Concurrent-To\":\"<urn:uuid:28a0df6e-3913-4afe-a5ba-7cde702e454a>\",\"WARC-IP-Address\":\"52.24.16.199\",\"WARC-Target-URI\":\"https://www.assignmentexpert.com/homework-answers/mathematics/other/question-91613\",\"WARC-Payload-Digest\":\"sha1:PGG6AQI2PBZEQAMEWNYSY7UX66MDFB5L\",\"WARC-Block-Digest\":\"sha1:UJFSCOLORRHMT7XOYB7ECGUPCGL5N4XR\",\"WARC-Truncated\":\"length\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738603.37_warc_CC-MAIN-20200810012015-20200810042015-00426.warc.gz\"}"} |
http://www.thecoalminetour.com/ChronometerWatch/digital-clock-wikipedia | [
"",
null,
"# Digital clock Wikipedia\n\nAs of 2008, the most common binary clocks sold are designed by Anelace Inc., and use six columns of LEDs to represent zeros and ones. Each column represents a single decimal digit, a format known as binary-coded decimal (BCD). The bottom row in each column represents 1 (or 20), with each row above representing higher powers of two, up to 23 (or 8).\n\nTo read each individual digit in the time, the user adds the values that each illuminated LED represents, then reads these from left to right. The first two columns represent the hour, the next two represent the minute and the last two represent the second. Since zero digits are not illuminated, the positions of each digit must be memorized if the clock is to be usable in the dark.\n\n### Binary-coded sexagesimal clocks\n\nTime Technology's Samui Moon binary-coded sexagesimal wristwatch. This clock reads 3:25.\n\nBinary clocks that display time in binary-coded sexagesimal also exist. Instead of representing each digit of traditional sexagesimal time with one binary number, each component of traditional sexagesimal time is represented with one binary number, that is, using up to 6 bits instead of only 4.\n\nFor 24-hour binary-coded sexagesimal clocks, there are 11 or 17 LED lights to show us the time. There are 5 LEDs to show the hours, there are 6 LEDs to show the minutes, and there are 6 LEDs to show the seconds. 6 LEDs to show the seconds are not needed in 24-hour binary-coded sexagesimal clocks with 11 LED lights.\n\n1 1 0 0 1 1 0 0 0 1 0 0 1 1\nHours Minutes Seconds\n32\n16\n10 37 49\nA binary clock circuit displaying 13:49:37\n\nThe above display uses three binary number columns, one for each of the units (hours, minutes and seconds) of the conventional time system.\n\nSource: en.wikipedia.org\n\n### Related posts\n\n#### Mechanical digital clock\n\nJANUARY 21, 2020\n\nSchematic of a split-flap display in a digital clock display A Flip Clock is an electromechanical, digital time keeping device…"
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"http://www.thecoalminetour.com//img/digital_alarm_3d_drawing_ideas.jpg",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8627858,"math_prob":0.8800137,"size":2051,"snap":"2019-51-2020-05","text_gpt3_token_len":474,"char_repetition_ratio":0.14802149,"word_repetition_ratio":0.020114943,"special_character_ratio":0.22866894,"punctuation_ratio":0.092269324,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9578938,"pos_list":[0,1,2],"im_url_duplicate_count":[null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-21T08:08:22Z\",\"WARC-Record-ID\":\"<urn:uuid:ecad0614-a5fd-4dfe-a462-11cf0b91886a>\",\"Content-Length\":\"30585\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5993e22e-e18b-47a3-9f56-a4f37d1e7c9b>\",\"WARC-Concurrent-To\":\"<urn:uuid:1338ab21-69dc-4e30-ae1c-ea832c77ae89>\",\"WARC-IP-Address\":\"174.127.97.47\",\"WARC-Target-URI\":\"http://www.thecoalminetour.com/ChronometerWatch/digital-clock-wikipedia\",\"WARC-Payload-Digest\":\"sha1:CT5TU65HZM7LPQWRZLCGPIER7DG7QTQO\",\"WARC-Block-Digest\":\"sha1:I4ZRYG22U66LOM6ZE27MBWQM6SBI33CP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250601628.36_warc_CC-MAIN-20200121074002-20200121103002-00286.warc.gz\"}"} |
http://vaguery.com/words/finishing-the-interval-in-klapaucius | [
"Draft\n\n# Finishing the Interval Push type in Klapaucius\n\nDraft of 2016.07.09\n\nMay include: GPPushlearning in public&c.\n\nI’ll get straight to work this morning. I’m thinking of using this Wikipedia page on Interval Arithmetic as a simple guide, and creating pairs of instructions in which either two :interval items or an :interval and a :scalar are involved, like :interval-add and :interval-offset for addition, and so on. Although there really isn’t a good reason to have a separate function adding a :scalar to an :interval, and another subtracting a :scalar from an :interval….\n\nI’ll get to that. I have a suspicion, going in, that division (as ever) will be tricky and not especially fun, what with its contingencies and the description on the Wikipedia page of how Infinity is used explicitly in the calculations of the most common algorithms. But I think I can manage.\n\nInterval addition is relatively simple. The sum is a new :interval with :min equal to the sum of the :min of each argument, and the result’s :min-open? will be true if either argument’s was true. And, of course, the same for :max throughout.\n\nAs a side note, I realize this only works because I changed the unsorted :start and :end of Span to a sorted :min and :max in Interval. Otherwise I’d need to to all kinds of checks (as I will with :interval-multiply, shortly).\n\n(def interval-add\n(build-instruction\n\":interval-add pops the top two :interval items and pushes a new :interval which is the sum of the two. If either :min (or :max) is open, the result :min (or :max) is also open.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i2)\n(consume-top-of :interval :as :i1)\n(calculate [:i1 :i2]\n#(interval/make-interval\n(+' (:min %1) (:min %2))\n(+' (:max %1) (:max %2))\n:min-open? (or (:min-open? %1) (:min-open? %2))\n:max-open? (or (:max-open? %1) (:max-open? %2))) :as :result)\n(push-onto :interval :result)))\n\n\nI write some tests, and realize an interesting thing about interval arithmetic: it’s not immediately obvious or intuitive what Interval I should add to $$[2,3]$$ to “zero it out”. The “negative” interval, which I innocently wrote as $$[-2,-3]$$ in my test, is actually $$[-3,-2]$$ (because the constructor’s arguments are sorted in creating the interval proper). I don’t think there’s any single interval I can add to $$[2,3]$$ to “zero it out”. I can subtract an identical interval to get $$[0,0]$$, but apparently I can’t add one. Which strikes me as freaky somehow, and an interesting peek into some Group Theory stuff I didn’t know was lurking here.\n\nHere are my tests for :interval-add\n\n(tabular\n(fact \":interval-add returns the sum of two intervals\"\n(register-type-and-check-instruction\n?set-stack ?items interval-type ?instruction ?get-stack) => ?expected)\n\n?set-stack ?items ?instruction ?get-stack ?expected\n\n:interval (list (s/make-interval 2 3)\n(s/make-interval 2 3))\n:interval (list\n(s/make-interval 4 6))\n;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\n:interval (list (s/make-interval 2 3)\n(s/make-interval -3 -1))\n:interval (list\n(s/make-interval -1 2))\n;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\n:interval (list (s/make-open-interval 2 3)\n(s/make-interval 2 3))\n:interval (list\n(s/make-open-interval 4 6))\n;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\n:interval (list (s/make-open-interval 2 3)\n(s/make-interval -3 -2))\n:interval (list\n(s/make-open-interval -1 1))\n;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\n:interval (list (s/make-interval 3 3 :min-open? true)\n(s/make-interval 2 2 :max-open? true))\n:interval (list\n(s/make-open-interval 5 5))\n;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\n)\n\n\n## Subtraction\n\nI decide to forge ahead and build subtraction, since (it seems at first glance that) it’s essentially identical to addition, only “backwards”. As soon as I start checking, it turns out that it isn’t though. Instead of addition, in which\n\n$[a,b]+[c,d]=[a+c,b+d]$\n\nsubtraction is\n\n$[a,b]-[c,d]=[a-d,b-c].$\n\nWhaaaaaat?\n\nOK, I can go with that. I’d feel a bit better if, for example, $$A-B=C$$ implied that $$C+B=A$$, but that is apparently not a thing I get to assume. So it’s not a group or a field or one of those math things I vaguely remember from our over-ambitious 11th-grade pre-calculus exposure to Group Theory at all in the sense I think of them. Ah well. I trust Wikipedia enough to copy from them. I suppose this has something to do with… well, something I can’t quite visualize.\n\nAnyway, once I discover my confusion, I quickly work this out, which is quite similar to :interval-addition:\n\n(def interval-subtract\n(build-instruction\ninterval-subtract\n\":interval-subtract pops the top two :interval items and pushes a new :interval which is the difference of the two. If either :min (or :max) is open, the result :min (or :max) is also open.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i2)\n(consume-top-of :interval :as :i1)\n(calculate [:i1 :i2]\n#(interval/make-interval\n(-' (:min %1) (:max %2))\n(-' (:max %1) (:min %2))\n:min-open? (or (:min-open? %1) (:max-open? %2))\n:max-open? (or (:max-open? %1) (:min-open? %2))) :as :result)\n(push-onto :interval :result)))\n\n\nAs for the openness of bounds, I assume it works more or less like addition, and try to keep things together. I’ve found a few IEEE standards and academic papers on how bounds are handled (and division, which is starting to worry me to be frank), but I literally can’t read them when they use symbols like upside-down-triangle-surrounding-plus and so forth. So I’m punting here.\n\nAt least now I have glanced at multiplication enough to understand what to expect.\n\n## Multiplication\n\nMultiplication is relatively simple, at least for determining the values of the bounds. The lower and upper bounds are the min and max, respectively, of {$$ac$$, $$ad$$, $$bc$$, $$bd$$}, for two intervals $$[a,b]$$ and $$[c,d]$$.\n\nThe first part, the bounds calculation itself, is relatively straightforward:\n\n(def interval-multiply\n(build-instruction\ninterval-multiply\n\":interval-multiply pops the top two :interval items and pushes a new :interval which is the product of the two. If either :min (or :max) is open, the result :min (or :max) is also open.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i2)\n(consume-top-of :interval :as :i1)\n(calculate [:i1 :i2]\n#(let [a (:min %1)\nb (:max %1)\nc (:min %2)\nd (:max %2)]\ninterval/make-interval\n(min (*' a c) (*' a d) (*' b c) (*' b d))\n(max (*' a c) (*' a d) (*' b c) (*' b d))\n;; :min-open? ???\n;; :max-open? ???\n) :as :result)\n(push-onto :interval :result)))\n\n\nbut as you can see, I really don’t know what to set as :min-open? and :max-open? in the result.\n\nThis starts to feel like yet another place where the association between a numerical bounds value and whether it’s open or not is important. Basically I’d like the result’s :min-open? to be true if either of the bounds used to calculate it were open. Which means I need to know which of the four options was chosen….\n\nI decide to associate the value and its openness up front, and see how messy the calculations get. Clojure seems to like me to fiddle complex structures, so let’s see what Clojure offers me as a mechanism for this task.\n\nLater…\n\nAll I can say at this point is that I seem to have it working, based on my tests. But sheesh.\n\n(defn product-of-bounds-from-pair\n\"helper function for interval-multiply, which takes two [number, open?] tuples and returns the product of the number values\"\n[t1 t2]\n(*' (first t1) (first t2)))\n\n(defn disjunction-of-bounds-from-pair\n\"helper function for interval-multiply, which takes two [number, open?] tuples and returns the OR of the openness values\"\n[t1 t2]\n(or (second t1) (second t2)))\n\n(def interval-multiply\n(build-instruction\ninterval-multiply\n\":interval-multiply pops the top two :interval items and pushes a new :interval which is the product of the two. If either :min (or :max) is open, the result :min (or :max) is also open.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i2)\n(consume-top-of :interval :as :i1)\n(calculate [:i1 :i2]\n#(let [a [(:min %1) (:min-open? %1)]\nb [(:max %1) (:max-open? %1)]\nc [(:min %2) (:min-open? %2)]\nd [(:max %2) (:max-open? %2)]\npairs [[a c] [a d] [b c] [b d]]\nmin-choice (apply min-key\n(fn [x] (apply product-of-bounds-from-pair x))\npairs)\nmax-choice (apply max-key\n(fn [x] (apply product-of-bounds-from-pair x))\npairs)]\n(interval/make-interval\n(apply product-of-bounds-from-pair min-choice)\n(apply product-of-bounds-from-pair max-choice)\n:min-open? (apply disjunction-of-bounds-from-pair min-choice)\n:max-open? (apply disjunction-of-bounds-from-pair max-choice))\n) :as :result)\n(push-onto :interval :result)\n))\n\n\nSo what I’m doing here is once again building a temporary association tuple with the value of a bound (:min or :max) and its openness (:min-open? or :max-open?). Then I create an explicit collection holding the four pairs of comparisons between those tuples; I could have possibly used some combinatoric library stuff to get the powerset here, but I already know this will never be larger than four elements, so easier and tidier to make it explicit.\n\nThen I invoke an extracted “helper function” I’ve called product-of-bounds-from-pair, which reaches back into those two tuples being compared and calculates the product of their numerical values. The new min-choice is thus the pair of tuples of the lowest-valued product, and the new max-choice is the pair of tuples with the highest-valued product.\n\nFinally, in constructing the resulting Interval, I use that same product-of-bounds-from-pair function to set :min and :max, and then invoke another “helper” disjunction-of-bounds-from-pair to or the openness values.\n\nI hates it. But it works.\n\nI spend some time trying to refactor it. But as before, I can’t really see a place to put a tool (as it were): no clean seams along which I could break off and extract a function, nothing repeated very often. If nothing else, there may be a problem with the verbosity of the PushDSL I’m responding too. But the function itself is essentially a relatively long calculation of the base numbers, and a lot of that is creating and juggling those [number, open?] tuples.\n\nI’d like to get done. Maybe division, with its particular oddities, will spark an insight that leads to a general simplification? We’ll see shortly.\n\n## Reciprocals and Interval Division\n\nAt least based on the Wikipedia entry, the mathematical definition is broken into two steps. For $$[a,b]÷[c,d]$$, when $$0 \\notin [c,d]$$ the result is $$[a,b] \\times [c,d]^{-1}$$, and $$[c,d]^{-1} = [d^{-1},c^{-1}]$$. For $$[a,b]÷[c,d]$$ when $$0 \\in [c,d]$$, we have a problem of diviing by zero, so we make a couple of adjustments. We define division by 0 as resulting in positive or negative infinity (depending on the sign of the dividend), and we split the interval $$[c,d]$$ into two intervals, $$[c,0)$$ and $$[0,d]$$.\n\nAn example might be good around now.\n\n$[2,3] \\div [-2,3] \\\\ [2,3] \\div [-2,0) \\cup [2,3] \\div [0,3] \\\\ [2,3] \\times [-2,0)^{-1} \\cup [2,3] \\times [0,3]^{-1} \\\\ [2,3] \\times (-\\infty, -1/2] \\cup [2,3] \\times [1/3,\\infty] \\\\ \\ldots$\n\nAnd so on.\n\nSo I see a few ways to approach this.\n\nFirst I realize it wouldn’t be a bad thing to have an :interval-reciprocal function on hand. Second—and more conveniently for me—I am reminded that in “idiomatic Push” it might be better to let the intermediate product of the dividend and the reciprocal be calculated by the actual application of :interval-multiply. In other words, I am thinking that :interval-divide will produce a continuation form that invokes :interval-multiply and :interval-reciprocal, and let those instructions do the subsequent work.\n\nI won’t bore you with the wrestling and mistakes (and there were a few), but here’s the function I finally wrote for interval-reciprocal in push.type.definitions.interval:\n\n(defn interval-reciprocal\n\"Takes one Interval record, and returns its reciprocal. If the argument contains 0, then two intervals are returned in a list\"\n[i]\n(let [s (:min i)\ne (:max i)\nso (:min-open? i)\neo (:max-open? i)\ninfty Double/POSITIVE_INFINITY\nninfty Double/NEGATIVE_INFINITY]\n(cond\n(and (zero? s) (zero? e))\n(make-interval ninfty infty)\n(zero? (:min i))\n(make-interval (/ 1 (:max i)) infty\n:min-open? (:max-open? i))\n(zero? (:max i))\n(make-interval ninfty (/ 1 (:min i))\n:max-open? (:min-open? i))\n(interval-include? i 0)\n(list\n(interval-reciprocal (make-interval s 0 :min-open? so))\n(interval-reciprocal (make-interval 0 e :max-open? eo)))\n:else\n(make-interval (/ 1 e) (/ 1 s)\n:min-open? eo\n:max-open? so)\n)))\n\n\nI realized there is a “hidden” behavior tucked into my glib reading of the Wikipedia example, which is what happens when an interval ends in 0. Technically, “ending in zero” is “containing zero”, but really I am enforcing a more stringent criterion: that when the interval includes 0 but not as an end, then I should split it. Testing revealed another “edge case”, which is the interval $$[0,0]$$, and also made me wonder whether I should bother preserving the “openness” of an infinite bound. As a result, all bounds that are infinite are closed.\n\nThe nice thing, to me, was the realization that when splitting an interval I could re-process the components recursively. I always get a little thrill when I find a recursive solution like that. Your mileage may vary.\n\nOK, so having some confidence (since I tested extensively in the process of writing it) of my interval-reciprocal function, I will now implement the Push instruction, and then I think I have all the pieces for the Push :interval-divide instruction as well.\n\nHere’s the Push instruction I design:\n\n(def interval-reciprocal\n(build-instruction\ninterval-reciprocal\n\":interval-reciprocal pops the top :scalar item and pushes its reciprocal to :exec. If the span strictly covers zero, then a code block containing _two_ spans is pushed.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(calculate [:i] #(interval/interval-reciprocal %1) :as :results)\n(push-onto :exec :results)\n))\n\n\nI’ve shunted most of the work to the definition of interval-reciprocal, and it’s tempting to do the same with the :interval-multiply instruction that bugged me earlier. I may still do that; don’t know.\n\nOne design decision I’ve made here is a common Push idiom that’s developed in the course of the project: When there’s any ambiguity in the return values of an instruction—for instance when the number of items or the type of the result might be contingent on the arguments—I prefer to return a code block containing all the arguments to the :exec stack and let the interpreter’s router push those results to their correct homes. I have a sense (because of another feature coming soon) that this may become more of a rule than an occasional quirk, but for now I’m happy enough to notice it.\n\nI think I have everything in place now for a version of the :interval-divide instruction. Let’s see….\n\nHere’s what the instruction definition ends up looking like:\n\n(def interval-divide\n(build-instruction\ninterval-divide\n\":interval-divide pops the top two :interval items (call them B and A, respectively) and pushes a continuation that will calculate their quotient(s) A÷B onto :exec. If B strictly covers zero, then two continuations are pushed: one for the positive and one for the negative regions.\"\n:tags #{:interval}\n(consume-top-of :interval :as :divisor)\n(consume-top-of :interval :as :dividend)\n(calculate [:divisor] #(interval/interval-reciprocal %1) :as :inverses)\n(calculate [:dividend :inverses]\n#(if (seq? %2)\n(list %1 (first %2) :interval-multiply\n%1 (second %2) :interval-multiply)\n(list %1 %2 :interval-multiply)) :as :results)\n(push-onto :exec :results)\n))\n\n\nThat is, if the result of interval-reciprocal are a collection, produce a long continuation that includes both partial calculations in turn; otherwise just one.\n\nThere’s a bit of a mathematical glitch I sense here, which is that the reciprocal of a zero-spanning Interval will always produce two Interval items. But I don’t really want to go through the trouble for now of creating some kind of multi-interval arithmetic, which retains the result as the union of two continuous regions. Besides, we’re dividing by zero—I should be happy there is any result at all, and that it makes some sense.\n\nThat said, there’s another thing that catches my attention, and that may be more salient when I start stress-testing these new instructions: the appearance of $$\\infty$$ values in the numeric values of Push items. If $$\\infty$$ is to be a valid :scalar value, there will almost certainly be consequences somewhere down the line in genetic programming land….\n\nBut consequences are something evolutionary search is intended to work around. So let’s just remember this and move along.\n\nHere’s the test that really exercises this “continuation form” approach:\n\n(tabular\n(fact \":interval-divide works for zero-spanning intervals\"\n(register-type-and-check-instruction\n?set-stack ?items interval-type ?instruction ?get-stack) => ?expected)\n\n?set-stack ?items ?instruction ?get-stack ?expected\n\n:interval (list (s/make-interval -2 3)\n(s/make-interval 2 3))\n:interval-divide\n:exec (list\n(list\n(s/make-interval 2 3)\n(s/make-interval Double/NEGATIVE_INFINITY -1/2)\n:interval-multiply\n(s/make-interval 2 3)\n(s/make-interval 1/3 Double/POSITIVE_INFINITY)\n:interval-multiply))\n;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;\n)\n\n\nAnd that passes. It manages to break the zero-spanning interval across the origin, and results in a sufficient continuation that after a few more steps will produce two Interval items that represent the two parts of the correct answer.\n\nI think I’m done with arithmetic. I am still tempted to move some of the logic written explicitly into the :interval-multiply instruction up to an interval-multiply function in push.type.definitions.interval, but I think I can save that for another day.\n\nLooking over my list from several days ago, the remaining work is making sure the Push :interval type is connected sufficiently, through conversion and manipulation instructions, to the basic types that comprise it. I can see a few of those already, and they should be (compared to the core arithmetic) relatively simple. Here’s what I think I can and should finish this morning:\n\n• :interval-recenter move the center of the interval to zero\n• :interval-min push the :min to :scalar\n• :interval-max push the :max to :scalar\n• :interval-scale multiply :min and :max by a :scalar\n• :interval-shift add a :scalar to :min and :max\n• :interval-negate multiply :min and :max by -1\n\nThere are a few other pieces of functionality that I think should wait a few days. For instance, one of the reasons for having these numerical interval objects at all is to emulate Clojure’s sensible idiom of using a set as a filter for a collection. Eventually I can see using something like $$[2,3]$$ as an argument in a suite of functional collection-munging instructions, something like :scalars-retain, which would keep only the items in a :scalars item that fall within a given :interval; or :tagspace-remove, which removes all entries from a given :tagspace with keys falling within a given :interval, and so forth.\n\nBut those are not for today.\n\n## Wiring it up\n\nBefore I write those “simple” instructions I’ve just listed, I realize I should register the :interval type I’ve been working on, and check that a Push interpreter can integrate these new instructions into the ecosystem of existing types and instructions.\n\nThat process is not yet very polished, and as I move towards a first “finished” release of the Klapaucius interpreter as semantic version 1.0.0 I will want to improve that functionality. At the moment it boils down to three steps:\n\n• add the new type to push.interpreter.templates.one-with-everything\n• fix a few badly-written tests that will fail because they check (explicitly) for the list of registered types, which I know is a bad thing but which didn’t I just say I was hoping to improve? Like literally in the preceding paragraph?\n• run a few thousand stress tests\n\nAs with the embarrassing tests I still need to clean up (which explicitly ask what the known types for a new Interpreter instance are, and which of course fail as soon as you add a new type), the stress test I’ve built is rather primitive as of this writing. Think of it as an acceptance test that basically reads something like “There should be no exceptions at all when I construct 10000 random Push programs, with 20 random inputs each, and run them all 3000 steps each.”\n\nThe business of concocting “random programs” is necessarily slapdash in this library, since that responsibility really should reside in the search system one develops: there’s no reason for an interpreter made to run programs to know anything about writing programs. So in this case what I’ve done is simply bashed together the simplest possible “random program” maker. For most (maybe all?) of the defined types, it has a little function that can make a “random one”. A “random :scalar” is some number in a given range, a “random :boolean” is true or false with equal chance, a “random :instruction” is any of the registered instruction names for a new Interpreter, chosen uniformly, and so on.\n\nBecause I want to make relatively sure there are Interval items being made and exercised, and especially because those Infinity values floating around are starting to worry me, I add a little “random :interval” function to my stress-testing harness before I trust the results:\n\n(defn random-interval\n[]\n(interval/make-interval\n(random-float)\n(random-rational)\n:min-open? (random-boolean)\n:max-open? (random-boolean)))\n\n\nI’m not showing the context because I hate it and I want to clean it all up before sharing more of it, but I think you can infer what’s happening here. I’m just slapping together some numbers—random-float returns a float from 0 to 100000, and random-rational is the ratio of two integers somewhere in [1,200], and setting the bounds open with uniform coin flips of random-boolean. Literally no idea what the results will be, except they will be valid Interval items because that’s how I defined them!\n\nWhen I run this, my computer gets Very Hot (I am partly embarrassed by the code because I never even bothered to make this a parallel algorithm), but all of the 10000 random programs—which include both Interval literals and all my new :interval instructions—run to completion. Which is almost disturbing, so I do some poking around to check, but yes it seems things are working well. I think the recent change in the way numbers are handled (leading to the :scalar unified type in Push, which contains all of Clojure’s integer, double, float, long, rational, biginteger and bigdecimal in one huge pile) shook out a lot of the arithmetic errors I was expecting. I hope so fervently, at least.\n\nSo that actually makes :interval an official Push type. Before I merge this branch with master I’d like to add those “few extra” instructions. Then I believe we’re good!\n\n## Leftovers and connections\n\nI feel I’m on a roll, so I’ll just report what happens for each instruction on my list in turn\n\n### :interval-recenter\n\nThe tricky thing here is the possibility that an Interval has one or both bounds infinite. I decide that if either bound is infinite, “centering” an infinite span should result in the span $$[-\\infty, \\infty]$$. Makes sense to me, at least.\n\nIn working through this, I realize I want a new predicate function, and it seems useful and general enough to be placed in push.util.numerics for use elsewhere:\n\n(defn infinite?\n[number]\n(or (= number Double/NEGATIVE_INFINITY)\n(= number Double/POSITIVE_INFINITY)))\n\n\nThen the instruction itself (after this utility file has been added to the namespace) becomes\n\n(def interval-recenter\n(build-instruction\ninterval-recenter\n\":interval-recenter pops the top :interval item and pushes a new :interval with center at 0. If either bound of the argument is infinite, the result will be infinite in both directions.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(calculate [:i]\n#(if (or (infinite? (:min %1)) (infinite? (:max %1)))\n(interval/make-interval Double/NEGATIVE_INFINITY Double/POSITIVE_INFINITY)\n(let [c (/ (-' (:max %1) (:min %1)) 2)]\n(interval/make-interval\n(- c)\nc\n:min-open? (:min-open? %1)\n:max-open? (:max-open? %1)))) :as :result)\n(push-onto :interval :result)))\n\n\n### :interval-min\n\nThese two are simple enough getters (if we were in Object Oriented territory). Here they are, with no fuss:\n\n(def interval-min\n(build-instruction\ninterval-min\n\":interval-min pops the top :interval item and pushes its :min value to :scalar.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(calculate [:i] #(:min %1) :as :result)\n(push-onto :scalar :result)))\n\n(def interval-max\n(build-instruction\ninterval-max\n\":interval-max pops the top :interval item and pushes its :max value to :scalar.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(calculate [:i] #(:max %1) :as :result)\n(push-onto :scalar :result)))\n\n\n### :interval-scale\n\n(def interval-scale\n(build-instruction\ninterval-scale\n\":interval-scale pops the top :interval item and top :scalar item, and pushes a new :interval with the original :max and :min multiplied by the :scalar.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(consume-top-of :scalar :as :factor)\n(calculate [:i :factor] #(\ninterval/make-interval\n(*' %2 (:min %1))\n(*' %2 (:max %1))\n:min-open? (:min-open? %1)\n:max-open? (:max-open? %1)) :as :result)\n(push-onto :interval :result)\n))\n\n\n### :interval-shift\n\nAnd:\n\n(def interval-shift\n(build-instruction\ninterval-shift\n\":interval-shift pops the top :interval item and top :scalar item, and pushes a new :interval with the original :max and :min added to the :scalar.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(consume-top-of :scalar :as :factor)\n(calculate [:i :factor] #(\ninterval/make-interval\n(+' %2 (:min %1))\n(+' %2 (:max %1))\n:min-open? (:min-open? %1)\n:max-open? (:max-open? %1)) :as :result)\n(push-onto :interval :result)\n))\n\n\n### :interval-negate\n\nI decide to call this one :interval-reflect, since what I’d like it to do is literally flip the :interval across zero. Thus, if the :min of the argument is open, the :max of the result will be, and so forth.\n\n(def interval-reflect\n(build-instruction\ninterval-reflect\n\":interval-reflect pops the top :interval item and pushes a new one with the signs of the :min and :max reversed, and also the boundedness of the ends.\"\n:tags #{:interval}\n(consume-top-of :interval :as :i)\n(calculate [:i] #(\ninterval/make-interval\n(- (:max %1))\n(- (:min %1))\n:min-open? (:max-open? %1)\n:max-open? (:min-open? %1)) :as :result)\n(push-onto :interval :result)\n))\n\n\n## Cleanup\n\nThat seems to be it. I’ve constructed (finally) an Interval record, added a bunch of functions to that, implemented it as a Push type, and added the requisite instructions to the Push interpreter definition. That’s registered (in the branch I am working in) with the “one-with-everything” interpreter template, and it seems to be passing the “stress” acceptance test.\n\nI go over it all one last time, and notice some things to clean up. The fact that I dropped interval-reciprocal into the “helper” functions, but not interval-multiply bothers me, so I want to shift things over and make the Push instructions simply call the library that handles the type. That seems safer and more elegant, to be honest. Something about separation of responsibility and all, too.\n\nThat works simply, and I don’t even bother adding additional tests for the “new” integer-multiply and its helpers. I’m comfortable enough seeing that the instruction, which is the only thing exercising interval-multiply, produces correct results.\n\nI say we ship it. Push now has an :interval type.\n\n## Reflections\n\nI realize I should clean up and formalize the process at “both ends”. It should be easier to construct a new “empty” Push type (possibly with a generator function of some kind?), and it should definitely be simpler and easier to add all the types (and modules) to the universal one-with-everything interpreter when I’m done.\n\nWhen I (the best proxy so far for the proverbial “end user”) want to add a new type to the already-crowded Push ecosystem embodied in the Klapaucius interpreter, there’s a necessary but difficult bit of research that needs to be done. That is, one needs to be able to “look” in some sense at the interconnectedness of the old types and the new ones, as the instruction set transforms them into one another. A good deal of this might be solved with a dynamic visualization of the graph of transformations of arguments to results by all the instructions acting on all the types. But at least a little bit is a reminder that getters, setters, and the things that take structured types apart are important.\n\nFor some time I’ve felt that the types in Push should be defined structurally rather than in this sort of ad hoc way. For example, I think :interval should be automatically recognized and detected—in some way I can’t quite put together yet—as a record containing two :scalar and two :boolean values, with appropriate labels. The labels are right there inside the Clojure record, and it seems reasonable and appropriate to let Push also see the keys and do some introspection. But the next step, it seems to me, is to define something like an Interval with this (pseudocode macro warning)\n\n(defpushstructure :interval\n:min :scalar\n:max :scalar\n:min-open? :boolean\n:max-open? :boolean)\n\n\nand automatically get instructions like\n\n• :interval-min\n• :interval-max\n• :interval-min-open?\n• :interval-max-open?\n• :interval-setmin\n• :interval-setmax\n• :interval-setmin-open?\n• :interval-setmax-open?\n• :interval-new\n• :interval-parts\n• … and so on\n\nI don’t think I should expect everybody two write those on their own, especially for more complex Clojure record items I have planned, like :graph or :time-series.\n\nBut that is definitely for another day.\n\nFinally, there are a few philosophical concerns I should check, involving the notion of Infinity and numbers and how this plays out consistently in the Push language. At the moment, division by zero produces an :error result for a :scalar but a complex assortment of non-:error results for an :interval. But the results from dividing by an :interval that spans zero will include bounds at Infinity.\n\nSpecifically I’ve been relying on Clojure’s (and in turn, Java’s) Double/POSITIVE_INFINITY and Double/NEGATIVE_INFINITY constants. These are consistently handled by most of the Clojure functions mathematical functions I’ve checked so far; (Math/sin ∞) is NaN (reasonably, because what’s the phase at ∞?), (/ 1 ∞) produces the java.lang/Double value 0.0, and (-' 7812318273M -∞) produces ∞… also reasonable.\n\nWhat isn’t so clear is whether this will hold up over time. There will shortly be higher-order functions floating around, and instead of simply worrying about :scalar and :scalars, :complex and :complexes, :interval and :intervals, there will be a sort of combinatorial… well, not “explosion”, but “rapid low-temperature expansion” when those arise. Lots of stuff will be able to interact with lots of other stuff.\n\nThis makes me wonder whether I should make my own Infinity, and what that might be, and how it would propagate. I think at the very least I should let Push pay attention to the Java/Clojure infinities when they crop up. The deeper question, though, is whether the behavior of :scalar-divide and :integer-divide are mathematically consistent now. I don’t know.\n\nI commit and merge the changes, and I open some issues in the project repository on GitHub. I think we’re good to go."
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https://basisschooldeark.com/education/basic-math-skills-to-review-for-the-gmat.html | [
"# Basic Math Skills to Review for the GMAT\n\nThe four sections of the GMAT namely: Analytical Writing Assessment (AWA), Integrated Reasoning (IR), Quantitative Aptitude (QA) and Verbal Reasoning (VR); and among all, the quantitative is considered to be highest scoring section. On an average, if a candidate is able to score in between 50-51 in it leads to hit the 700+ GMAT quant score and covers up the gap of a less score secured in VR.\n\nCertainly, there are vast number of GMAT test-takers for whom quant is not less than a nightmare. The long word problems demand extensive thought process for its solving within stipulated time may break down in cold sweat.To excel in quant you need to lay a strong foundation of basics. It enables to grasp the problem quickly and requires less time to solve. GMAT quant revolves around the three topics:",
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"• Arithmetic\n• Algebra\n• Geometry\n\nLet’s discuss what the constituents of these topics are:\n\n• Arithmetic\n\nIt includes the basic mathematical topics like Integers, Prime Number, Units Digit, Tens Digit, Hundreds Digit, Ratio, Percentage, Mean, Median, And Standard Deviation. To solve these kind of questions you should be impeccable in quick addition, subtraction, division and multiplications. Brush up these; it will pay you back in GMAT test. We have solved one arithmetic question for your easy understanding.\n\nQuestion: H, I, J, K and L are playing a game of cards. So, H says I that if you give me three of your cards then you will have exact numbers of cards that L has, but if I give you three of my cards then your number of cards will be equal to that of K.The combined number of cards of H and I is 10 more than that of the total cards K and L have. If it is given that I have 2 more cards compared to J and the total number of cards they are playing with is 137, then how many cards H and J have?\n\nSolution: According to question,\n\n• I – 3 = L …………. (1)\n• And I + 3 = K ………….. (2)\n• H + I = K + L + 10 ……… (3)\n• I = J + 2 ……………………. (4)\n• H+ I + J + K + L = 137 …….. (5)\n\nFrom equation (1) and (2);\n\n• 2 I = L + K ………….. (6)\n\nFrom equations (3) and (4),\n\n• H = I + 10 ………….. (7)\n\nUsing the values of equation (4), (6), (7) in equation (5) yields,\n\nI + 10 + I + I + 2 + 2 I = 137\n\n5 I + 12 = 137\n\n5 I = 125\n\nI = 25\n\nSo, number of cards H has = I + 10 = 35\n\nNumbers of cards in J’s possession = I – 2 = 25 – 2 = 23\n\n• Algebra\n\nAlgebraic equations are fundamentals of the GMAT. Questions often integrate one or more variables.\n\nQuestion: There are two examinations rooms P and Q. Suppose, if 10 students are sent from P to Q, then the number of students in both the room will be equal. Though, if 20 candidates are sent from Q to P, then the number of students in P becomes double of the number of students in Q. So what is the number of students in P and Q?\n\n1. 100, 160\n2. 100, 80\n3. 80, 100\n4. 160, 100\n\nSolution: Let the number of students in room P and Q is p and q respectively.\n\nThen according to question, p – 10 = q + 10\n\n• p – q = 20 …………………….. (1)\n• And p + 20 = 2 (q – 20)\n• p – 2q = – 60 ………………….. (2)\n\nSubtracting (1) and (2),\n\n• q = 80\n• Putting value of q in equation (1)\n• p = 20 + q\n• p = 20 +80\n• p = 100\n\nHence the right option is B.\n\n(iii) Geometry\n\nDon’t fall in the trap of highly complex geometrical figures. Start your GMAT geometry prep with lines, angles, polygons and coordinate planes on a regular basis. Slowly and gradually increase the questions difficulty level.",
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"Question: If each side of a rectangle is increased by 30%, then what will be the percentage increase in the area of that rectangle?\n\nSolution: Let the original length of the rectangle is a meters and breadth is b meters\n\nSo, original area = (ab) meters2.\n\nNew Length = () meters = ()\n\nNew Breadth = () meters = ()\n\nNew area = ()() = () meters2\n\nDifference between new and original area = () –\n\n• Difference = meters2\n• So, percentage increase =\n• Increase = 69 %\n\nGet the complete GMAT syllabus here. You can ask for assistance related to GMAT and MBA from us by just giving a missed call at 08033172797, or you can drop an SMS. You can write to us at [email protected]."
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http://www.ques10.com/p/652/engineering-maths-2-question-paper-jan-2014-firs-1/ | [
"Question Paper: Engineering Maths 2 : Question Paper Jan 2014 - First Year Engineering (C Cycle) (Semester 2) | Visveswaraya Technological University (VTU)\n0\n\n## Engineering Maths 2 - Jan 2014\n\n### First Year Engineering (C Cycle) (Semester 2)\n\nTOTAL MARKS: 100\nTOTAL TIME: 3 HOURS\n(1) Question 1 is compulsory.\n(2) Attempt any four from the remaining questions.\n(3) Assume data wherever required.\n(4) Figures to the right indicate full marks.\n1 (a) Choose the correct answer for the following:\n\n( i) Suppose the equation to be solved is of the form, y=f(x, φ) then differentiating x we get equation of the form,\n$$(a) \\ \\phi \\left (x,p, \\dfrac{dp}{dy} \\right )= 0 \\$$b) \\ \\phi \\left ( y, p , \\dfrac {dp}{dx} \\right )= 0 \\$$c) \\ \\phi (x,yp)=0 \\$$d)\\ \\phi (x,y,0)= 0 $$(ii) The general solution of the equation p2-3p+2=0 is, (a) (y+x-c)y+2x-c) (b) (y-x-c)(y-2x-c)=0 (c) (-y-x-c)(y-2x-c)=0 (y-x-c)(y+x-c)=0 (iii) Clairaut's equation is of the form, (a) x=py+f(p) (b) y=p2+f(p) (c) y=px+f(p) (d) None of these (iv) Singular solution of y=px+2p2 is, (a) y2+8y=0 (b) x2-8y=0 (c) x2+8y-c=0 (d) x2+8y=0 (4 marks) 1 (b) Solve p2+2p cosh x+1=0.(4 marks) 1 (c) Find singular solution of p=sin(y-xp).(6 marks) 1 (d) Solve the equation y2(y-xp)=x4p2 using substitution$$ X=\\dfrac {1}{x} and Y=\\dfrac {1}{y} $$(6 marks) 2 (a) Choose the correct answer for the following: (i) A second order linear differential equation has, (a) two arbitary solution (b) One arbitary solution (c) no arbitary solution (d) None of these (ii) If 2, 4i and -4i are the roots of A.E of a homogeneous linear differential equation then its solution is,$$ (a) \\ e^x+ e^x (\\cos 4x+\\sin 4x) \\\\ (b) \\ C_1 e^{2x}+ C_2 \\cos 4x + C_3 \\sin 4x \\$$c) \\ C_1 e^{2x} + C_2 e^x \\cos 4x+C_3 e^x \\sin 4x \\$$d) \\ C_1 e^{2x}\\cos 4x+ C_2e^{2x}\\sin 4x $$(iii) P.I. of (D+1)2 y=e-x+3$$ (a)\\ \\dfrac {x^2}{2} \\\\ (b) \\ x^3 e^x \\\\ (c) \\ \\dfrac {x^3}{3} e^{-x=3} \\\\ (d)\\ \\dfrac {x^2}{2}e^{-x+3}$$(iv) Particular integral of f(D)y=eax V(x) is,$$ (a)\\ \\dfrac {e^{ax}V(x)}{f(D)} \\\\ (b) \\ e^{ax}= \\dfrac {1}{f(D)}[V(x)] \\\\ (c) \\ e^{ax} \\dfrac {1}{f(D+a)}[V(x)] \\\\ (d) \\ \\dfrac {1}{f(D+a)} [e^{ax}V(x)] $$(4 marks) 2 (b)$$ Solve \\ \\dfrac {d^3y}{dx^3}- 3 \\dfrac {d^2y}{dx^2}+ 3 \\dfrac {dy}{dx}- y = 0 $$(4 marks) 2 (c) Solve y\"-3y'+2y=2 sin x cos x(6 marks) 2 (d) Solve the system of equation,$$ \\dfrac {dx}{dt}- 2y = \\cos 2t, \\ \\dfrac{dy}{dt} + 2x =\\sin 2t $$(6 marks) 3 (a) Choose the correct answer for the following: (i) In x2y\"+ xy'-y=0 if et=x then we get x2y\" as, (a) (D-1)y (b) (D+1)y (c) D(D+1)y (d) None of these (ii) In second order homogeneous differential equation P0(x)y\"+P1(x)y'+P2(x)y=0 x=a is a singular point if, (a) P0(a)>0 (b) P0(a)?0 (c) P0(a)=0 (d) P0(a)<0 (iii) The general solution of$$ x^2 \\dfrac {d^2 y}{dx^2}+ x\\dfrac{dy}{dx}-y = 0 \\ is, \\\\ (a) \\ y=C_1x-C_2 \\dfrac {1}{x} \\\\ (b) \\ C_1x + C_2 \\dfrac {1}{x} \\\\ (c) \\ C_1x+C_2 x \\\\ (d) \\ C_1 x- C_2 x $$(iv) Frobenius series solution of second order linear differential equation is of the form,$$ (a) \\ x^{m} \\sum^{\\infty}_{r=0}a_rx^r \\\\ (b) \\ \\sum^{\\infty}_{r=0}a_rx^r \\\\ (c) \\ \\sum^{\\infty}_{r=a}a_rx^{m-r} \\\\ None \\ of \\ these $$(4 marks) 3 (b) Solve y\"+a2y=sec ax by the method of variation of parameters.(4 marks) 3 (c)$$ Solve \\ x^2 \\dfrac {d^2 y}{dx^2}+ 4x \\dfrac {dy}{dx}+ 2 y = e^x $$(6 marks) 3 (d) Obtain the series solution of$$ \\dfrac {dy}{dx}- 2xy=0 $$(6 marks) 4 (a) Choose the correct answer for the following: (i) PDE of az+b=a2x+y is,$$ (a) \\ \\dfrac {\\partial z}{\\partial x} \\cdot \\dfrac{\\partial z}{\\partial y}= 1 \\\\ (b) \\dfrac {\\partial z} {\\partial x} \\cdot \\dfrac {\\partial z}{\\partial y} = 0 \\\\ (c) \\ \\dfrac {\\partial z}{\\partial x} + \\dfrac {\\partial z}{\\partial y} = 0 \\\\ (d)\\ \\dfrac{\\partial z}{\\partial x}+ \\dfrac {\\partial z}{\\partial y}=1 $$(ii) The solution of PDE Zxx=2 y2 is, (a) z=x2+xf(y)+ g(y) (b) z=x2y2+xf(y)+g(y) (c) z=x2y2+f(x)+g(x) (d) z=y2+xf(y)+g(y) iii) The subsidiary equations of (y2+z2)p+x(yq-z)=0 are,$$ (a)\\ \\dfrac {dx}{p}= \\dfrac {dy}{q} = \\dfrac {dz}{R} \\\\ (b) \\ \\dfrac {dx}{y^2 + z^2} = \\dfrac {dy}{x} = \\dfrac {dz}{xz} \\\\ (c) \\ \\dfrac {dx}{y^2 + z^2} = \\dfrac {dy}{xy} = \\dfrac {dz}{xz} \\\\ (d) \\ None \\ of \\ these $$(iv) In the method of seperation of variable to solve xzn+zt=0 the assumed solution is of the form, (a) X(x)Y(x) (b) X(y)Y(y) (c) X(t)Y(t) (d) X(x)T(t) (4 marks) 4 (b)$$ Solve \\ \\dfrac {\\partial ^3 z}{\\partial x^2 \\partial y}= cos (2x+3y)$$(4 marks) 4 (c) Solve xp-yq=y2-x2(6 marks) 4 (d) Solve 3ux+2uy=0 by the seperation of variable method given that u=4e-x when y=0(6 marks) 5 (a) Choose the correct answer for the following:$$ \\int^{1}_0 \\int^{x^2}_0 e^{y/x}dy dx = \\_\\_\\_\\_\\_\\_\\_ \\\\ (a) \\ 1 \\ \\ (b) \\ -1/2 \\ \\ (c) \\ 1/2 \\ \\ (d) \\ None\\ of \\ these $$(ii) The integral$$ \\iint_R f(x,y) dxdy $$by changing to polar form becomes,$$ (a) \\ \\iint_R \\phi (r, \\theta) drd\\theta \\\\ (b) \\ \\iint_R f(r, \\theta)drd\\theta \\\\ (c) \\ \\iint_R f(r,\\theta)rdrd\\theta \\\\ (d)\\ \\iint_R \\phi (r, \\theta)rdrd \\theta $$(iii) For a real positive number n, the Gamma function ?(n)= _________$$ (a) \\ \\int^{\\infty}_0 x^{n-1}e^{-x}dx \\\\ (b) \\ \\int^1_0 x^{n-1}e^{-x}dx \\\\ (c) \\ \\int^{x}_0 x^ne^{-x}dx \\\\ (d) \\ \\int^1 _0 x^n e^{-x}dx $$(iv) The Beta and Gamma functions relation for B(,n)= _______$$ (a) \\ \\dfrac {\\Gamma (m )\\Gamma (n)} {\\Gamma (m-n)} \\\\ (b) \\ \\dfrac {\\Gamma (m)\\Gamma (n)}{\\Gamma (m+n)} \\\\ (c) \\ \\Gamma (m)\\Gamma(n) \\\\ (d) \\ \\Gamma(mn) $$(4 marks) 5 (b) By changing the order of integration evaluate,$$ \\int^a_0 \\int^{\\sqrt{x/a}}_{x/a}(x^2 + y^2)dydx, \\ a>0 $$(4 marks) 5 (c)$$ \\displaystyle Evaluate \\ \\int^a_0 \\int^{x}_0 \\int^{x-y}_0 e^{x+y+z}dzdydx $$(6 marks) 5 (d) Express the integral$$ \\int^1_0 \\dfrac{dx}{\\sqrt{1-x^n}}$$in terms of the Gamma function, Hence evaluate$$ \\int^1_0 \\dfrac {dx}{\\sqrt{1-x^{2/3}}} $$(6 marks) 6 (a) Choose the correct answer for the following: (i) The scalar surface integral of$$ \\overrightarrow{f} $$over s, where s is a surface in a three-dimensional region R is given by,$$ \\int \\overrightarrow{f}.nds= \\_\\_\\_\\_\\_\\_\\_ $$by using Gauss divergence theorem$$ (a) \\ \\iiint_v \\nabla\\cdot \\overrightarrow{f}dV \\\\ (b)\\ \\iint_s \\nabla\\cdot \\overrightarrow{t}dx dy \\\\ (c) \\ \\iiint_v \\nabla \\cdot \\overrightarrow{F}dV \\\\ (d) \\ None \\ of \\ these $$(ii) If all the surface are closed in a region containing volume V then the following theorem is applicable. (a) Stroke's theorem (b) Green's theorem (c) Gauss divergence theorem (d) None of these (iii) The value of$$ \\int \\left \\{ (2xy-x^2)dx + (x^2 + y^2)dx \\right \\} $$by using Green's theorem is, (a) Zeron (b) One (c) Two (d) Three (iv)$$ \\iint_s f.nds = \\_\\_\\_\\_\\_\\_\\_ $$where f=xi+yj+2k and S is the surface of the sphere x2y2+z2=a2 (a) 4πa (b) 4πa2 (c) 4πa3 (d) 4π (4 marks) 6 (b) Find the work done by a force f=(2y-x2)i+ 6yzj-8xz2k from the point (0, 0, 0) to the point (1, 1, 1) along the straight-line joining these points.(4 marks) 6 (c) If C is a simple closed curve in the xy-plane, prove by using Green's theorem that the integral$$ \\int_C \\dfrac {1}{2} (xdy-ydx) $$represent the area A enclosed by . Hence evaluate$$ \\dfrac {x^2}{a^2} + \\dfrac {y^2}{b^2} = 1 $$(6 marks) 6 (d) Verify Stoke's theorem for$$ \\overrightarrow{f} = (2x-y)i - yz^2 j- y^2 zk $$for the upper half of the sphere x2+y2+z2=1(6 marks) 7 (a) Choose the correct answer for the following: (i) L[tn]= ________$$ (a) \\ \\dfrac {n}{s^{n+1}} \\\\ (b) \\ \\dfrac {n}{s^{n-1}} \\\\ (c) \\ \\dfrac {n!}{s^{n-1}} \\\\ (d) \\ \\dfrac {n!}{s^{n+1}} $$(ii) L[e-3t]= _______$$ (a) \\ \\dfrac {3}{s-3} \\\\ (b) \\ \\dfrac {3}{s+3} \\\\ (c) \\ \\dfrac {1}{s+3} \\\\ (d) \\ \\dfrac {1}{s-3} $$iii) L{f(t-a)H(t-a)} is equal to,$$ (a) \\ \\dfrac{3!}{(s+2)^4} \\\\ (b) \\ \\dfrac{3!}{(s-2)^4} \\\\ (c) \\ \\dfrac{3}{(s-2)^4} \\\\ (d) \\ \\dfrac{3}{(s-2)} $$(iv) L{δ(t-1)}= _______ (a) e-s (b) e5 (c) eaS (d) e-aS (4 marks) 7 (b) Evaluate L{sin3 2t}(6 marks) 7 (c) Find L{f(t)} given that$$f(t)= \\begin{cases}2 &3>t>0 \\\\t &t>3 \\end{cases}$$(6 marks) 7 (d) Express$$f(t) = \\begin{cases}t^2 &2>t>0 \\\\4t &4\\ge t>2 \\\\8 &t>4 \\end{cases}$$in terms of unit step function and hence find their Laplace transform.(4 marks) 8 (a) Choose the correct answer for the following: (i) L-1 {cos at}= _______$$ (a)\\ \\dfrac {s}{s^2 + a^2} \\\\ (b) \\ \\dfrac {s}{s^2 - a^2} \\\\ (c) \\ \\dfrac {1}{s^2 + a^2} \\\\ (d) \\ \\dfrac {1}{s^2 - a^2} $$(ii) L-1 {F (s-a)}= ________ (a) etf(t) (b) eatf(t) (c) e-atf(t) (d) None of these$$ L^{-1} \\left \\{ \\cot^{-1} \\left ( \\dfrac {2}{s^2} \\right ) \\right \\} = \\_\\_\\_\\_\\_\\_ \\\\ (a) \\ \\dfrac {\\sin t}{t} \\\\ (b) \\ \\dfrac {\\sinh at}{t} \\\\ (c) \\ \\dfrac{\\sin at }{t} \\\\ (d) \\ \\dfrac {\\sinh t}{t} $$(iv) For the function f(t)=1, convolution theorem condition, (a) Not satisfied (b) Satisfied with some condition (c) Satisfied (d) None of these (4 marks) 8 (b) Find the inverse Laplace transform of$$ \\dfrac {2s^2 - 6s + 5}{(s-1)(s-2)(s-3)} $$(4 marks) 8 (c) Find$$ L^{-1} \\left(\\dfrac{s}{(s-1)(s^2 + 4)}\\right) $$using convolution theorem(6 marks) 8 (d) Solve differential equation y\"(t) + y = F(t) where$$F(t)= \\begin{cases} 0 & 1>t>0 \\\\2 &t>1 \\end{cases} Given that y(0)=0=y'(0)(6 marks)\n\n written 3.0 years ago by",
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https://www.colorhexa.com/06126f | [
"# #06126f Color Information\n\nIn a RGB color space, hex #06126f is composed of 2.4% red, 7.1% green and 43.5% blue. Whereas in a CMYK color space, it is composed of 94.6% cyan, 83.8% magenta, 0% yellow and 56.5% black. It has a hue angle of 233.1 degrees, a saturation of 89.7% and a lightness of 22.9%. #06126f color hex could be obtained by blending #0c24de with #000000. Closest websafe color is: #000066.\n\n• R 2\n• G 7\n• B 44\nRGB color chart\n• C 95\n• M 84\n• Y 0\n• K 56\nCMYK color chart\n\n#06126f color description : Very dark blue.\n\n# #06126f Color Conversion\n\nThe hexadecimal color #06126f has RGB values of R:6, G:18, B:111 and CMYK values of C:0.95, M:0.84, Y:0, K:0.56. Its decimal value is 397935.\n\nHex triplet RGB Decimal 06126f `#06126f` 6, 18, 111 `rgb(6,18,111)` 2.4, 7.1, 43.5 `rgb(2.4%,7.1%,43.5%)` 95, 84, 0, 56 233.1°, 89.7, 22.9 `hsl(233.1,89.7%,22.9%)` 233.1°, 94.6, 43.5 000066 `#000066`\nCIE-LAB 13.344, 34.293, -53.121 3.16, 1.619, 15.184 0.158, 0.081, 1.619 13.344, 63.229, 302.845 13.344, -4.28, -46.62 12.723, 22.069, -61.851 00000110, 00010010, 01101111\n\n# Color Schemes with #06126f\n\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #6f6306\n``#6f6306` `rgb(111,99,6)``\nComplementary Color\n• #06466f\n``#06466f` `rgb(6,70,111)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #2f066f\n``#2f066f` `rgb(47,6,111)``\nAnalogous Color\n• #466f06\n``#466f06` `rgb(70,111,6)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #6f2f06\n``#6f2f06` `rgb(111,47,6)``\nSplit Complementary Color\n• #126f06\n``#126f06` `rgb(18,111,6)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #6f0612\n``#6f0612` `rgb(111,6,18)``\nTriadic Color\n• #066f63\n``#066f63` `rgb(6,111,99)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #6f0612\n``#6f0612` `rgb(111,6,18)``\n• #6f6306\n``#6f6306` `rgb(111,99,6)``\nTetradic Color\n• #020626\n``#020626` `rgb(2,6,38)``\n• #030a3f\n``#030a3f` `rgb(3,10,63)``\n• #050e57\n``#050e57` `rgb(5,14,87)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #071687\n``#071687` `rgb(7,22,135)``\n• #091a9f\n``#091a9f` `rgb(9,26,159)``\n• #0a1eb8\n``#0a1eb8` `rgb(10,30,184)``\nMonochromatic Color\n\n# Alternatives to #06126f\n\nBelow, you can see some colors close to #06126f. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #062c6f\n``#062c6f` `rgb(6,44,111)``\n• #06246f\n``#06246f` `rgb(6,36,111)``\n• #061b6f\n``#061b6f` `rgb(6,27,111)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #06096f\n``#06096f` `rgb(6,9,111)``\n• #0c066f\n``#0c066f` `rgb(12,6,111)``\n• #14066f\n``#14066f` `rgb(20,6,111)``\nSimilar Colors\n\n# #06126f Preview\n\nText with hexadecimal color #06126f\n\nThis text has a font color of #06126f.\n\n``<span style=\"color:#06126f;\">Text here</span>``\n#06126f background color\n\nThis paragraph has a background color of #06126f.\n\n``<p style=\"background-color:#06126f;\">Content here</p>``\n#06126f border color\n\nThis element has a border color of #06126f.\n\n``<div style=\"border:1px solid #06126f;\">Content here</div>``\nCSS codes\n``.text {color:#06126f;}``\n``.background {background-color:#06126f;}``\n``.border {border:1px solid #06126f;}``\n\n# Shades and Tints of #06126f\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010312 is the darkest color, while #fefeff is the lightest one.\n\n• #010312\n``#010312` `rgb(1,3,18)``\n• #020625\n``#020625` `rgb(2,6,37)``\n• #030937\n``#030937` `rgb(3,9,55)``\n• #040c4a\n``#040c4a` `rgb(4,12,74)``\n• #050f5c\n``#050f5c` `rgb(5,15,92)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #071582\n``#071582` `rgb(7,21,130)``\n• #081894\n``#081894` `rgb(8,24,148)``\n• #091ba7\n``#091ba7` `rgb(9,27,167)``\n• #0a1eb9\n``#0a1eb9` `rgb(10,30,185)``\n• #0b21cc\n``#0b21cc` `rgb(11,33,204)``\n• #0c24df\n``#0c24df` `rgb(12,36,223)``\n• #0d27f1\n``#0d27f1` `rgb(13,39,241)``\nShade Color Variation\n• #1f37f3\n``#1f37f3` `rgb(31,55,243)``\n• #3248f4\n``#3248f4` `rgb(50,72,244)``\n• #4458f5\n``#4458f5` `rgb(68,88,245)``\n• #5769f6\n``#5769f6` `rgb(87,105,246)``\n• #697af7\n``#697af7` `rgb(105,122,247)``\n• #7c8af8\n``#7c8af8` `rgb(124,138,248)``\n• #8f9bf9\n``#8f9bf9` `rgb(143,155,249)``\n• #a1abfa\n``#a1abfa` `rgb(161,171,250)``\n• #b4bcfb\n``#b4bcfb` `rgb(180,188,251)``\n• #c7cdfc\n``#c7cdfc` `rgb(199,205,252)``\n• #d9ddfd\n``#d9ddfd` `rgb(217,221,253)``\n• #eceefe\n``#eceefe` `rgb(236,238,254)``\n• #fefeff\n``#fefeff` `rgb(254,254,255)``\nTint Color Variation\n\n# Tones of #06126f\n\nA tone is produced by adding gray to any pure hue. In this case, #38383e is the less saturated color, while #010f74 is the most saturated one.\n\n• #38383e\n``#38383e` `rgb(56,56,62)``\n• #333542\n``#333542` `rgb(51,53,66)``\n• #2f3147\n``#2f3147` `rgb(47,49,71)``\n• #2a2e4b\n``#2a2e4b` `rgb(42,46,75)``\n• #262a50\n``#262a50` `rgb(38,42,80)``\n• #212754\n``#212754` `rgb(33,39,84)``\n• #1d2359\n``#1d2359` `rgb(29,35,89)``\n• #18205d\n``#18205d` `rgb(24,32,93)``\n• #141c62\n``#141c62` `rgb(20,28,98)``\n• #0f1966\n``#0f1966` `rgb(15,25,102)``\n• #0b156b\n``#0b156b` `rgb(11,21,107)``\n• #06126f\n``#06126f` `rgb(6,18,111)``\n• #010f74\n``#010f74` `rgb(1,15,116)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #06126f is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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https://schoollearningcommons.info/question/the-product-of-two-rational-numbers-is-9-if-one-of-the-numbers-is-12-find-the-others-add-file-24698018-56/ | [
"## The product of two rational numbers is -9. If one of the numbers is -12. Find the others? Add file \n\nQuestion\n\nThe product of two rational numbers is -9. If one of the numbers is -12. Find the others?\n\nin progress 0\n4 weeks 2021-11-03T21:10:44+00:00 2 Answers 0 views 0\n\n1. sum of 2 rational numbers= -9\n\nOne number= -12\n\nlet other number = x\n\nA.T.Q\n\nx+ (-12) = -9\n\n=> x-12 = -9\n\n=>x= -9 – 12\n\n=>x= -21",
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"https://schoollearningcommons.info/wp-content/ql-cache/quicklatex.com-c6a74d2c3b55ad1ad8e17119e729805d_l3.png",
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http://www.edparrish.net/cs11/19f/lesson02.php | [
"# 2: Basic Coding Skills\n\n## Continuations\n\n#### Announcements\n\n1. If not enrolled, make sure you add right away: Registration Help\n2. SI is worth extra credit for assignment 1\n3. What to do when Canvas has problems\n4. Reminder: your last submittal to Canvas must include all your assignment files\n5. Ignore the extra `-1` (or any other number) added to the end of files in Canvas\n6. Room Policies\n7. Student Behavior Policies\n8. Connecting to Hawknet Wireless Internet\n\n#### The Assignment Cycle\n\nMost assignments (after the first) have three parts:\n\n1. Preparation:\n• Helps prepare for the problem-solving portion of the assignment\n• First make sure to complete the exercises from the prior lesson\n• Then complete several short review exercises, usually in CodeLab\n• CodeLab will tell you if you answered correctly and give hints if you make a mistake\n• If you make a mistake, try again\n• If you get stuck, get help\n2. Programming Projects:\n• Lets your explore the concepts in a more challenging way than the lesson exercises\n• We are given one or more problems for which we need to develop a solution\n• Acceptable solutions must meet a list of specifications\n\nspecification: A condition or restriction that is insisted upon; precise requirement\n\n3. Tutorial:\n• Introduces the new material\n• First read the assigned reading in the textbook to understand how to solve the problems\n• Then complete the tutorial exercises as specified\n• Also, refer to the online lecture notes as they become available\n• If stuck on CodeLab, look at CodeLab answers in the Solutions tab\n\n#### Homework Questions?\n\n• Quiz 1 (Canvas) (9/3/19)\nA1-Getting Started (9/5/19)\n• Who has installed g++? How did it go?\n• Any problems compiling the `hello.cpp` program?\n• Any problems with CodeLab?\n\n#### Questions from last class or the Reading?\n\n1. Computers: CPU, Memory, etc.\n2. Binary Numbers\n3. Developing a C++ Program \"Hello World\"\n4. Compiling and Programs\n\n## 2.1: Elements of a C++ Program\n\n### Learner Outcomes\n\nAt the end of the lesson the student will be able to:\n\n• Include libraries and use namespaces in programs\n• Identify statements in programs\n• Code `main()` functions\n\n### 2.1.1: Programming Languages and C++\n\nRecall how a computer is organized:",
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"",
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"Processor (CPU)\n\n• The heart of a computer is the Central Processing Unit (CPU or processor)\n• A processor executes machine instructions, which are extremely primitive operations like:\n1. Move memory location 40000 into register `eax`\n2. Subtract the value 100\n3. If the result is positive, start processing at location 11280\n• These instructions are encoded as numbers such as:\n`161 40000 45 100 127 11280`\n• While the computer only understands binary, we use decimal numbers for convenience\n\n#### Assemblers\n\n• Each processor has its own set of machine instructions\n• Looking up numeric codes for instructions like these is tedious and error prone\n• To make programming easier, computer scientists first developed a program called an assembler\n• An assembler allows a programmer to assign short names for the machine codes like:\n```movw r30, r24 (copy register word - word is 4 bytes)\nsubi r30, 0x12 (subtract immediate)\nbreq .+8 (branch if equal)\n```\n• The assembler translates the names into the correct machine codes\n• While easier for humans to use, assemblers still have two problems:\n1. We still have to enter a great many primitive instructions to create a program\n2. The instructions change from one processor to another\n\n#### Higher-level Languages\n\n• To make programming easier, computer scientists developed higher-level languages\n• Higher-level languages like C++ and Java let us write more readable instructions like:\n```if (interestRate > 100) {\ncout << \"Interest rate error\";\n}\n```\n• Programmers call these instructions source code\n• To translate these high-level instructions to machine instructions we use a compiler\n\ncompiler: a program that translates source code into machine instructions\n\n• Another advantage of higher-level languages is we can move programs to another processor more easily\n• Moving programs to another processor requires someone to write a compiler\n• After the new compiler is available, we can recompile our program with (hopefully) no changes\n• There are thousands of higher-level languages\n• The one we use in this course is C++\n\n#### Check Yourself\n\n1. True or false: computers can directly carry out C++ source code instructions.\n2. True or false: an assembler translates high-level language programs to a set of computer instructions.\n3. To translate source code to machine code we use a ________.\n4. For a high-level language like C++, which of the following statements are true?\n1. easier to program than machine code\n2. easier to program than assembler code\n3. easier for a human to read\n4. easier to move from one type of processor to another\n5. True or false: the ANSI standard for C++ was released after the ANSI standard for C.\n\n### 2.1.2: Example Program\n\n• Here is an example program like we looked at before ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 ``` ```/** CS-11 Asn 0, helloworld.cpp Purpose: Prints a message to the screen. @author Ed Parrish @version 1.0 8/30/05 */ #include using namespace std; int main() { cout << \"Hello, World!\\n\"; return 0; } // end of main function ```\n\n#### Brief Explanation by Line Number\n\n• Lines 1-7: comments -- notes to programmers\n• Line 8: adds a library (pre-written code) to our program\n• Line 9: all the standard libraries use the `std` namespace\n• Line 10: a blank line that we can use anywhere in our programs\n• Line 11: the `main()` function where all C++ programs start\n• Line 12-13: programming statements that give instructions to the computer\n• Line 14: the end of the `main()` function followed by another comment\n\n• Use comments to document blocks of code and to describe unusual code\n• One form of comments starts with `//` and lasts to end of the line\n`// this is a comment`\n• Another form affects a section of code: `/*` ... `*/`\n• This form can span multiple lines:\n```/* This is a multi-line comment\nwhich can be split\nover many lines or a portion of one line. */\n```\nor just a portion of one line\n```x = /* 41 + */ 1 ;\n```\n\n• We use block comments like the following at the start of a program file:\n```/**\nCS-11 Asn 0, helloworld.cpp\nPurpose: Prints a message to the screen.\n\n@author Emma Programmer\n@version 1.0 8/20/15\n*/\n```\n• The grading of most programming projects includes checking the comments of our code\n• Put a comment like this at the top of the source code file for all programming projects assigned\n• On the other hand, there is usually no need to put comments in lesson exercises\n\n#### Check Yourself\n\n1. True or false: comments are notes to the compiler.\n2. Of the following, the symbols that indicate a single line of commentary will follow is ________.\n1. /*\n2. /**\n3. //\n4. */\n3. To make C++ code more explanatory for others ________.\n1. use more English statements in a code.\n3. avoid usage of complex calculations in code\n4. always enclose code statements in curly braces\n4. Of the following, the type of comments to use at the top of each programming project file is ________.\n1. /*...*/\n2. /**...*/\n3. //...\n4. #...\n5. True or false: comments are generally required in lesson exercises.\n\n### 2.1.4: Statements and Whitespace\n\n• Statements are commands we give a computer in a programming language\n• A statement is similar to a sentence in English\n• Most statements end in a semicolon (;)\n\n#### Example Statement: `cout`\n\n• One of the many statements of C++ is: `cout << something`\n• This statement is one way to send output to a console\n• For example, the statement:\n`cout << \"Hello, World!\";`\n• Sends \"Hello, World!\" to the console like this:\n```Hello, World!\n```\n\n#### Preprocessor Directives\n\n• Some commands we place in our source code start with a `#` sign\n```#include <iostream>\n```\n• Technically, this command is known as a preprocessor directive\n• A preprocessor directive is a command to the compiler rather than a program command\n• Find a statement and preprocessor directive in the code below\n```#include <iostream>\nusing namespace std;\n\nint main() {\ncout << \"Hello, World!\\n\";\nreturn 0;\n}\n```\n\n#### Whitespace\n\n• Whitespace: blank lines, spaces, and tabs\n• The compiler ignores extra white space\n• This lets us add extra whitespace to our code, making it easier to read\n\n#### Programming Style: Line Length\n\n• Do not make your lines longer than 80 characters so code is easier to read\n• Longer lines can cause problems in many terminal windows, text editors and other programming tools\n\n#### Check Yourself\n\n1. Statements usually end in a ________.\n2. To make the code easier to read, always limit line lengths to ________ characters.\n3. True or false: the compiler ignores extra whitespace.\n\n### 2.1.5: The `main()` Function and Blocks\n\n• C++ programs are structured into subprograms called functions\n\nFunction - a named block of code that executes a series of statements to perform a task\n\n• Every C++ program has at least one function, `main`, defined like this:\n```int main() {\n// program statements go here\n}\n```\n• Programs begin executing at the first line of the `main()` function\n• The `int` means `main` returns a value of type `int` (more on this later)\n• For now, mimic the first line of main\n\n#### Sequence\n\n• Within a function, code runs from top to bottom in a continuous sequence\n• Example code sequence:\n```cout << \"Hello World!\\n\";\ncout << \"Goodbye World!\\n\";\nreturn 0;\n```\n• For a program to run correctly, we must place statements in the correct order\n\n#### Blocks\n\n• C++ is known as a block-structured language\n• A block in C++ is a section of code grouped together with `{ }`\n• This means that most source code is grouped within pairs of matching `{ }`\n• Left brace `{` begins every block\n• Right brace `}` ends every block\n• All functions have associated blocks\n• However, as we will see later in the course, we group statements into blocks in other places as well\n\n#### Programming Style: Indentation Inside Braces\n\n• Always indent statements inside braces to make code easier to read\n• After an opening curly brace \"`{`\", start indenting\n• After a closing brace \"`}`\", remove the level of indentation as shown below\n```#include <iostream>\nusing namespace std;\n\nint main() { // start indenting after opening {\ncout << \"Hello, World!\\n\";\nreturn 0;\n} // stop indenting after closing }\n```\n\n#### Check Yourself\n\n1. True or false: every C++ program has at least one function.\n2. A C++ function is ________\n1. A mapping of a domain value to a codomain value\n2. A named block of code that executes a series of statements\n3. A named block of code that returns a value for a given argument\n4. Any procedure that returns a value\n3. True or false: every C++ program has a `main()` function.\n4. Every C++ program starts executing ________.\n1. at the top of the file\n2. at the start of the main function\n3. immediately after the program compiles\n4. when the program is finished\n5. True or false: the default behavior for program statements within a function is to execute sequentially.\n6. A block is a section of ________ grouped together with curly braces.\n7. True or false: always indent within curly braces.\n\n### 2.1.6: Using Libraries and Namespaces\n\n• In programming terms, a library is a collection of prewritten code we can use in our programs\n• This saves us the effort of writing our own code for commonly-used functions\n• C++ has a number of standard libraries\n• These libraries place their code in what is called the `std` namespace\n\n#### Libraries and `include` Directives\n\n• To use a library, we add the `include` preprocessor directive to our programs\n`#include <libraryName>`\n• Most of our programs begin with an include statement like:\n`#include <iostream>`\n• The `iostream` library is for console input and output (I/O)\n• This library allows us to use the word `cout` for sending data to a terminal screen\n• Other libraries exist for math, strings and more (cover later)\n• Some compilers are picky about spaces in include directives\n• Do not put spaces before or after the `#` sign\n• Do not put spaces inside the angle brackets\n\n#### Namespaces\n\n• Namespace: a set of name definitions where all the names are unique\n• Names in a namespace cannot have more than one meaning\n• As an example, when we use the word `cout` we expect the command to send data to a terminal screen\n• Most of the prewritten commands we use are part of the standard (`std`) namespace\n• Thus, most of our programs will begin with two statements:\n```#include <iostream>\nusing namespace std;\n```\n• For now, define the namespace as above\n• Later in the course we will look at cases that use namespaces differently\n\n#### Check Yourself\n\n1. A library is a collection of ________ code.\n2. We use libraries in our programs to ________.\n3. We tell the compiler to use the library named iostream by writing ________.\n4. To tell the compiler to use the standard namespace by default write ________.\n\n### Exercise 2.1: Elementary C++\n\nIn this exercise we examine the basic elements of a C++ program.\n\n#### Specifications Part A: Experimentation (5m)\n\n1. Start your text editor and enter the following code:\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 ``` ```/** CS-11 Asn 0, helloworld.cpp Purpose: Prints a message to the screen. @author Ed Parrish @version 1.0 8/30/05 */ #include using namespace std; int main() { cout << \"Hello, World!\\n\"; return 0; } // end of main function ```\n2. Save the file as \"`hellome.cpp`\".\n3. Compile the code using:\n```g++ -Wall -Wextra -Wpedantic -o hellome hellome.cpp\n```\n\nIf you have problems, ask a classmate or the instructor for help as needed.\n\n4. Run the code and verify you get the message, \"Hello, World!\"\n```\\$ ./hellome\nHello, World!\n```\n5. Try changing the message to personally greet you with your own name, like the following:\n```Hello, Ed Parrish!\n```\n6. The code that starts with `/**` and ends with `*/` is known as a block comment. Comments are parts of code that are ignored by the compiler. Delete the entire block comment and then recompile and rerun your code.\n\nYou should see no change in how your code compiles or runs. If you see a difference, ask a classmate or the instructor for help as needed.\n\n7. Look at the following line of the code:\n`} // end of main function`\nThe last part of the line is another type of comment that starts with `//` and lasts until the end of the line. Delete the comment and then recompile and rerun your code.\n\nYou should see no change in how your code compiles or runs. If you see a difference, ask a classmate or the instructor for help as needed.\n\n8. Remove the directive `using namespace std;` and then try to recompile the code.\n\nYour program should not compile and you should get an error message. If you have a different experience, ask a classmate or the instructor for help as needed.\n\n9. Restore the directive `using namespace std;` back into your program and verify that it compiles.\n\nIf you have problems, ask a classmate or the instructor for help as needed.\n\nWhen finished, move on to Part B.\n\n#### Specifications Part B: Reflection (3m)\n\n1. Create a second text file named \"syntax.txt\" and record your answers to the following questions:\n1. Does a comment change the way that a program compiles or runs?\n2. What error message does the compiler report when you leave out the following command?\n`using namespace std;`\n2. Submit both your `hellome.cpp` and `syntax.txt` files to Canvas as part of assignment 2.\n\n### 2.1.7: Summary\n\n• C++ has two styles of comments:\n• `//`\n• `/* ... */`\n• Comments help document what a program does:\n• Use block comments at beginning of a file and before functions\n• Otherwise, use comments them sparingly\n• Statements are the commands we give a computer in a program\n• C++ has standard libraries of prewritten code\n• Definitions for these libraries are collected in a namespace called: `std`\n• Thus, most of our programs will begin with two statements:\n```#include <iostream>\nusing namespace std;\n```\n• Every C++ application starts with a function named `main()`\n\n#### Self Reflection\n\nAnswer these questions to check your understanding. If you are not sure, then follow the links to the section and review the material.\n\n1. What is the purpose of a comment? (2.1.3)\n2. What styles of comments are allowed in C++? (2.1.3)\n3. What is a statement? (2.1.4)\n4. How can you tell which lines of a program are statements? (2.1.4)\n5. What statement do you use to print a message to the console window? (2.1.4)\n6. What is meant by the term \"whitespace\"? (2.1.4)\n7. Where does every C++ program start? (2.1.5)\n8. What code do you write for the `main()` function? (2.1.5)\n9. What are the rules for indenting code? (2.1.5)\n10. How do you include libraries in your C++ programs? (2.1.6)\n\n## 2.2: Memory Concepts\n\n### Learner Outcomes\n\nAt the end of the lesson the student will be able to:\n\n• Describe why data is important in a computer program\n• Write code for declaring variables and assigning them values\n• Create variable names according to the rules of C++\n• Identify the literal values of basic data types\n• Write code to get input from a user and display the value of variables\n\n### 2.2.1: Importance of Memory\n\n• Let us pretend that we have a friend, named Grace, and we want to remember her phone number: 555-2368\n• We can store our friend's phone number in our memory\n• We even label our friend's phone number, like \"Grace's phone number\"\n• We do not really know where in our brain we store Grace's phone number\n• However, whenever we need her phone number, we say to our self, \"What is Grace's phone number\" and out pops 555-2368\n• Just like we store our friend's number in our memory, we can store it in a computer's memory\n• We store data in a computer program using a variable\n\nvariable: the name of a place to store data in a computer's memory\n\n• Just like we do not know where in our brain we store a phone number, we do not know where in a computer's memory we store data\n• We simply give it a name and let the compiler decide where to store the data\n\n#### Why Data Matters\n\n• Why should we care about variables or storing data?\n• Variables are the most important part of any computer program\n• Just like in real life, it is hard to do anything without memory\n• Consider a simple algorithm like adding two numbers:\n1. Get the first number\n2. Get the second number\n3. Add the first and second number and assign it to sum\n4. Display that `sum` is the result\n• How many variables did we need for this algorithm?\n• To find out, let us do some role playing\n• Imagine a conversation between Hal and Grace:\nHal: Hey Grace, I just learned to add two numbers together.\nGrace: w00t!\nHal: Give me the first number.\nGrace: 2\nHal: OK, give me the second number.\nGrace: 3\nHal: OK, the answer for 2 + 3 is 5\n• After Grace says, \"2\", Hal has to store the number in his memory\n• The same things happens with the number, \"3\"\n• Even if the numbers were given in the same sentence, Hal would have to store the numbers somewhere in his memory\n• After adding the two numbers together, Hal has to store the result of the addition, at least temporarily, so he can state the answer\n• If we were to write a program to add two numbers together, the computer would have to use memory just like Hal\n\n#### Check Yourself\n\n1. The name of a location to store data in a computer's memory is known as a(n) ________.\n2. True or false: remembering data is rarely important when processing information.\n3. To add two numbers, we need to store at least ________ pieces of information.\n1. 0\n2. 1\n3. 2\n4. 3",
null,
"### 2.2.2: Introduction to Variables\n\n• Recall how a computer is organized\n• Main memory is organized as a long list of memory locations\n• Each location stores one byte and is identified by an address number\n• 1 gigabyte (GB) is about 1 billion bytes\nIn RAM memory terms, 1 GB is 230, 10243 or 1,073,741,824 bytes\n\n#### Storing Data\n\n• A key part of a computer is memory\n• To store and retrieve data in a program we use a variable\n• A variable is a name for a location in a computer's memory",
null,
"#### Variable Definition\n\n• Here is an example C++ variable definition:\n```int num1;\n```\n• When we declare a variable we tell the computer to set aside space to store data in its memory\n• Notice that a variable definition has two parts:\n• int: the type of data the variable will store, integer values in this case\n• num1: the name of the variable, which we make up while coding\n• Variable names are a sequence of letters, numbers and the underscore ( `_` )\n• However, variable names cannot start with a number and cannot contain spaces\n\n#### Data Types\n\n• Like human memory, a computer must encode information before storage\n• As programmers we tell the computer how to encode information using data types\n• A commonly used numerical data type is `int`, which is shorthand for integer\n```int num1;\n```\n• An `int` specifies a whole number with no fractions or decimal points\n• If we want to store numbers with a decimal point, we use a floating point type like `double`\n```double pi;\n```\n\n#### Some Commonly Used Numerical Data Types\n\nType Bytes Use\n`int` 4 integers (whole numbers) like `-1`, `0` and `123`\n`double` 8 double-precision floating-point numbers like `-1.23` and `3.14159`\n\n#### Assignment\n\n• After we declare a variable, we may store a value in the variable\n• One way to store values is with an assignment operator, which is the \"equals sign\" (`=`)\n• We assign a value to the variable with an assignment statement in this form:\n`variable = expression;`\n• Where:\n• `variable`: the name of the variables\n• `expression`: the data to store in the variable\n• An assignment statement stores the value of expression (right side) in the variable (left side)\n• Switching sides is a common error\n• An expression is a group of operators, numbers and variables that compute a value\n• An assignment statement is like a simple equation where an expression is assigned to a variable\n• Examples:\n```int num1; // variable definition\nnum1 = 45; // assignment\nint num2 = 12; // definition + assignment\nint total = num1 + num2; // more complex expression\n```\n• Notice that we can combine variable definition with assignment\n• A common error is to forget to assign a value to a variable!!\n• Avoid this error by assigning a value whenever declaring a variable\n\n#### Variables by Analogy",
null,
"• Variables are like boxes made of computer memory (images source: David Goodger)\n• A computer memory location can hold one value at a time, thus our box can hold only one value\n• We put names on boxes to tell one box from another, like \"`a`\":\n• This is like declaring the variable \"`a`\"\n```int a;\n```\n• When we first declare a variable the box is empty but has a name\n• Storing values in a variable is like putting values in a box",
null,
"• For example, we assign the variable \"a\" the value `1`\n```a = 1;\n```\n• We put names on a box\n• We put values inside of boxes\n• When we want to see the value inside a box we look at it with code like:\n```cout << a << \"\\n\";\n```\n• If we change the value of a variable, we are putting a new value in the same box, like:",
null,
"```a = 2;\n```\n• Making a new variable and assigning one variable to another makes a copy of the value and puts it into the new box:\n```int b = a;\n```",
null,
"",
null,
"• We now have two different boxes that have independent values\n\n#### Activity: Code Variables (3m)\n\n1. Start your text editor and open the hellome.cpp from the last exercise.\n2. Inside the curly braces of `main()`, declare an integer variable named \"`favNum`\" and assign it the value of your favorite number, like:\n```int favNum = 42;\n```\n3. After displaying your name, print the value of your favorite number like:\n```cout << \"Your favorite number is \" << favNum << \".\\n\";\n```\n4. Compile and run the code and verify your program shows output like the following:\n```Hello, Ed Parrish!\n```\n\nWhen finished, please help those around you. Help is available during break or after class if you cannot finish in the allotted time.\n\n#### Check Yourself\n\n1. The name of a location to store data in a computer's memory is known as a(n) ________.\n2. To specify the type of data stored in a variable, variable definitions include a(n) ________ ________.\n3. True or false: the \"equals sign\" (=) is the assignment operator in C++.\n4. The following code prints the value ________.\n```int x = 42;\ncout << x << \"\\n\";\n```\n5. After executing the following statement, the value of `number` is ________.\n`int number;`",
null,
"### 2.2.3: Input and Output\n\n• Recall the main parts of a computer\n• Input sends information to the computer\n• Output is the computer sending information\n• We now look at how to code input and output\n\n#### Output\n\n• We have been using `cout` to display information on our video monitors\n• The command `cout` sends data to standard output, usually the terminal (console) window\n• The \"`<<`\" takes data from the right-hand side and sends it to the console\n```cout << \"Hello, World!\" << endl;\n```\n• The `endl` prints a newline so the next item displayed goes on a new line.\n• Notice the last letter of `endl` is an \"el\"--not a one!\n• Most basic data can be output to the console including\n• Variables (like `num1`)\n• Literals (like `12.34`)\n• Expressions (like `num1` + `12.34`)\n• We can display multiple values in one `cout`\n• However, each data item must be separated with a `<<` operator\n• For example:\n```int numberOfDragons = 3;\ncout << numberOfDragons << \" dragons.\\n\";\n```\n• Three values are output in order from left to right:\n1. The value of the variable `numberOfDragons`\n2. A literal string \"` dragons.`\"\n3. An end-of-line (newline) character\n• Prints out: `3 dragons.`\n• Notice that \"`\\n`\" may be used instead of `endl` to print a newline\n\n#### User Input\n\n• So far we have used an assignment operator \"=\" to assign a value to a variable\n```int numberOfDragons = 3;\n```\n• Another way to assign a value to a variable is to read it from the console\n• The keyboard input console is called `cin` (console input)\n• We use the `>>` operator with `cin` to send data to a variable\n• For example:\n`cin >> numberOfDragons;`\n• In this example, whatever valid integer number the user types is stored in the variable `numberOfDragons`\n\n#### Prompting Users\n\n• Good programming practice is to always \"prompt\" users for input like:\n```cout << \"Enter number of dragons: \";\nint numberOfDragons;\ncin >> numberOfDragons;\ncout << \"You entered \" << numberOfDragons\n<< \" dragons\\n\";\n```\n• Note that we do not put a newline after the prompt\n• The prompt waits on the same line for keyboard input like:\n```Enter number of dragons: _\n```\n• Where the underbar above denotes where keyboard entry is made\n• Every `cin` should have a `cout` prompt before it so users know what to enter\n\n#### Check Yourself\n\n1. The following code outputs ________.\n```int numberOfGames = 12;\ncout << numberOfGames << \" games played.\\n\";\n```\n2. Before getting input from a user, common practice is to display a(n) ________.\n3. Of the following statements, the one that places input into the variable \"value\" is ________.\n1. `value >> cin;`\n2. `cin >> value;`\n3. `value << cin;`\n4. `cin << value;`\n4. True or false: the << and >> always point in the direction the data is flowing.\n\n### Exercise 2.2: Adding Two Numbers (8m)\n\nIn this exercise we will write a program to add two numbers together. When it runs, the program acts like this:\n\n```Enter the first number: 2\nEnter the second number: 3\nThe sum of 2 and 3 is 5.\n```\n\nMake sure to compile after each step so you know where an error is located if you make a mistake.\n\n#### Specifications\n\n1. Copy the following program into a text editor, save it as `variables.cpp`, and then compile and run the starter program to make sure you copied it correctly.\n```#include <iostream>\nusing namespace std;\n\nint main() {\n// Input first variable\n\n// Input second variable\n\n// Calculate sum\n\nreturn 0;\n}\n```\n2. In `main()`, declare an `int` variable named `num1` and assign it a value of `0`:\n`int num1 = 0;`\n\nFor more information on declaring variables, see section: 2.2.2: Introduction to Variables\n\n3. Add code to display a prompt to the screen:\n`cout << \"Enter the first number: \";`\n\nFor more information on prompting users, see section: 2.2.5: Input and Output\n\n4. Add a statement to input a new value for the variable and store it in memory:\n`cin >> num1;`\n\n5. Declare a second variable of type `double` named `num2` and assign it a value of 0.0:\n`double num2 = 0.0;`\n\nFor more information on data types, including the `double` data types, see section: 2.2.4: More About Data Types\n\n6. Add code to display a prompt to the screen:\n`cout << \"Enter the second number: \";`\n7. Add a statement to input a new value for the variable and store it in memory:\n`cin >> num2;`\n8. Declare a third variable of type double named `total` and assign it the result of adding the two variable together:\n`double total = num1 + num2;`\n9. Write another statement that displays the result of adding the two numbers together:\n```cout << \"The sum of \" << num1 << \" and \"\n<< num2 << \" is \" << total << \".\\n\";\n```\n10. Compile and run your program to make sure it works correctly.\n11. Submit your program source code to Canvas as part of assignment 2.\n\n#### Completed Program\n\nWhen finished, check your source code by clicking here.",
null,
"Reviewing another solution after you have completed your own is often helpful in learning how to improve your programming skills.\n\n### 2.2.4: Summary\n\n• Variables are how we can store data in our programs\n• Variables must be declared before use like:\n```int x;\n```\n• Notice that the variable definition has two parts:\n• int: the type of data the variable will store, integer values in this case\n• x: the name of the variable, which we make up while coding\n• Once declared, we assign a value to the variable like:\n```x = 42;\n```\n• Simple assignment statements have a variable, equals sign and an expression:\n`variable = expression;`\n\n#### User Input\n\n• Another way to store data in a variable is to read it from the console\n• `cin` is an input stream bringing data from the keyboard\n• The \"`>>`\" point towards where the data goes\n• When designing I/O:\n• Prompt the user for input\n• Echo the input by displaying what was input\n• `cout` is an output stream sending data to the monitor\n• The insertion operator \"`<<`\" inserts data into `cout`\n• There are two ways to output a newline character:\n```cout << \"Hello World!\\n\";\ncout << \"Hello World!\" << endl;\n```\n\n#### Self Reflection\n\nAnswer these questions to check your understanding. If you are not sure, then follow the links to the section and review the material.\n\n1. Why do you need to store data when creating a program? (2.2.1)\n2. How do you store information in a computer's main memory? (2.2.2)\n3. What code would you write to declare an `int` variable named `foo` and assign it a value of `10`? (2.2.2)\n4. What code would you write to display a variable named `foo` to standard output? (2.2.3)\n5. What code would you write to input data from the keyboard and store it in a variable named `foo`? (2.2.3)\n6. Why is it a good practice to prompt users before input? (2.2.3)\n7. What are two ways to assign a value to a variable? (2.2.3)\n\n## 2.3: Numbers and Arithmetic\n\n### Learner Outcomes\n\nAt the end of the lesson the student will be able to:\n\n• Distinguish between an integer and a floating-point number\n• Write C++ code for arithmetic expressions\n• Infer the type returned from a mixed-mode arithmetic expression\n• Construct expressions that use mathematical functions\n\n### 2.3.1: Computations Using Numbers and Arithmetic\n\n• Many problems can be solved using mathematical formulas and numbers\n• Consider the simple problem of adding up all the coins in a pocket to get the total value in dollars\n• If we know the number of each coin type what would be the formula to calculate the dollar value?\n```int pennies = 8;\nint nickels = 5;\nint dimes = 4;\nint quarters = 3;\n```\n• To implement this formula, we will need to use numbers and arithmetic\n• Remember that C++ has two general types of numbers: integers and floating-point\n\n#### Integers\n\n• An integer number is zero or any positive or negative number without a decimal point\n• Examples of integers include:\n```0 1 -1 +5 -27 1000 -128\n```\n• We call plain numbers like these literal integers because they stand for what they look like\n• By comparison, a variable may contain an integer value but is not a literal integer\n```int num1 = 3;\n```\n• Literal numbers are constant and do not change while a program executes\n• The compiler usually stores integers in four (4) bytes of computer memory\n\n#### Floating-Point Value\n\n• A floating-point number is any signed or unsigned number with a decimal point\n• A floating point number approximates a real number and has a trade-off between range and precision\n• For example:\n`0.0 1.0 -1.1 +5. -6.3 3234.56 0.33`\n• Note that `0.0`, `1.0` and `+5.` are floating-point numbers, but could be rewritten as integers\n• In C++, both integers and floating-point numbers cannot have any commas or special symbols\n• The compiler usually stores floating-point numbers in eight (8) bytes of computer memory\n\n#### Program coins.cpp to Sum Coins\n\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ``` ```#include using namespace std; int main() { int pennies = 8; int nickels = 5; int dimes = 4; int quarters = 3; double total = pennies * 0.01 + nickels * .05 + dimes * 0.10 + quarters * 0.25; cout << \"Total value = \" << total << \"\\n\"; return 0; } ```\n\n#### Check Yourself\n\n1. The two types of numbers in C++ are ________ and ________.\n2. Of the following literal numbers, the single integer is ________.\n1. `1`\n2. `1.2`\n3. `1.23`\n4. `1.234`\n3. For the following code, the literal number is ________.\n```int pennies = 8;\n```\n\n### 2.3.2: Arithmetic\n\n• C++ uses the following operators for arithmetic:\n• - for subtraction\n• * for multiplication\n• / for division\n• % for modulus (remainder after integer division)\n• Modulus is a type of division operation for integer numbers\n• The first four operators should be familiar and we will discuss modulus (`%`) today\n\n#### Precedence Rules\n\n• Precedence: what gets done first\n• Arithmetic operators are processed in algebraic order:\n1. Parenthesis: `( )`\n2. Unary operators: `+, -`\n3. Multiplication, division, modulus: `*, /, %`\n4. Addition, subtraction: `+, -`\n• Binary operators of same precedence are evaluated from left to right\n• As in algebra, multiplication and division (including modulus) are performed before addition and subtraction\n• To change the order of operation, we use parentheses\n• For example:",
null,
"is written as: `a + b / 2`",
null,
"is written as: `(a + b) / 2`\n\n• Notice that we cannot use parenthesis to indicate multiplication, but must explicitly use the `'*'` operator in C++\n\n#### Examples of Expressions\n\nAlgebra Expression C++ Expression\n2(10 + 5) 2 * (10 + 5)\n1\n\n12.2 + 3 · 7.3\n1 / (12.2 + 3 * 7.3)\n10 - 7\n\n3.2 + 9 · 1.6\n(10 - 7) / (3.3 + 9 * 1.6)\n2 · 42 2 * 4 * 4\n\n#### Programming Style\n\n• Programming style: add spaces around binary operators\n• `2 + 3`, not `2+3`\n• Programming style: no spacing after opening or before closing parenthesis\n• `(2 / 3)`, not `( 2/3 )`\n\n#### Check Yourself\n\n1. The five arithmetic operators in C++ are ________.\n1. `+, -, /, *, %`\n2. `+, -, \\, *, %`\n3. `+, -, /, *, ^`\n4. `+, -, \\, *, ^`\n2. The first operation performed in the following arithmetic expression is ________.\n```1 + 2 * 3 / 4 % 5\n```\n3. If we wanted a different ordering of operations in the above example, we add ________ to the expression.\n\n### 2.3.3: Mixed-Mode Expressions\n\n• Recall that different data types are stored in different forms\n• An integer (`int`) is usually stored in 4 bytes\n• A floating-point number (`double`) is usually stored in 8 bytes\n• The format of the types is different as well\n• The computer needs both operands in the same form before it can perform an operation\n• If one operand is different than the other, the compiler converts it to the wider of the two types\n• For example:\n`2 + 2.3`\n• First number (`2`) is an `int`\n• Second number (`2.3`) is a `double`\n• C++ will automatically convert an `int` to a `double`\n• Then the arithmetic operation can take place to produce a result of `4.3`\n• Remember that the result of arithmetic with an `int` and a `double` is a `double`\n\n#### Check Yourself\n\n1. The result of adding an integer with a double in the following expression is ________.\n`1 + 2.4`\n2. In the above expression, C++ converts the integer `1` to type ________.\n3. The data type of the number returned by the following expression is ________.\n`3 + 4.5`\n\n### Exercise 2.3a: Calculator Basic (7m)\n\nThrough the miracles of computer science, we will now convert your \\$500 computer into a \\$5 calculator! Along the way, we learn how to work with arithmetic using C++.\n\n#### Specifications\n\n1. Type the following program into a text editor, save it as arithmetic.cpp, and then compile and run the starter program to make sure you typed it correctly.",
null,
"2. Within the curly braces of the `main()` function, declare two `double` variables named `a` and `b`, and assign them a value of `5` and `2` respectively. For instance:\n```double a = 5, b = 2;\n```\n3. Add a line of code to display the arithmetic expression `(a + b)` and then recompile and run the program.\n```cout << \"a + b = \" << a + b << endl;\n```\n\nNotice that the last letter on `endl` is a lower-case \"L\", NOT a one. The output when you run the program should look like this:\n\n```a + b = 7\n```\n\nIf you do not see this output, please ask a classmate or the instructor for help.\n\n4. Add three more lines of code like the previous one that computes the expressions: `a - b`, `a * b` and `a / b`. Compile and run your program again and make sure your program now displays the following output:\n```a + b = 7\na - b = 3\na * b = 10\na / b = 2.5\n```\n5. The order of operations matters in C++ just like it does in algebra. Multiplication and division are performed before addition and subtraction. Add the following two statements to your program:\n```cout << \"a + b / 2 = \" << a + b / 2 << endl;\ncout << \"(a + b) / 2 = \" << (a + b) / 2 << endl;\n```\n6. Compile and run your program again and compare the output. Your program should now display the following output:\n```a + b = 7\na - b = 3\na * b = 10\na / b = 2.5\na + b / 2 = 6\n(a + b) / 2 = 3.5\n```\n\nNote how the output of the two statements is different. You can change the order of operation using parenthesis, just like in algebra. For more information on the order of operations see section: 2.3.2: Arithmetic.\n\nAs you can see, arithmetic in C++ works much like you would expect. However, there are some mysteries when working with integer variables which we will explore in the next section:\n\n• Truncation in integer division\n• Modulus (%) operator\n7. Save your `arithmetic.cpp` file as we will add to it in the following sections.\n\n#### Discussion Questions\n\n1. What is wrong with the following code?\n```cout << \"a + b = a + b\" << endl;\n```\n2. How should the above code be written?\n\n### 2.3.4: Integer Division and Modulus\n\n• Dividing two integers can produce unexpected results for the unwary\n• In division, if at least one of the numbers is a floating-point number, the result is a floating point number:\n```7.0 / 2.0 // 3.5\n7 / 2.0 // 3.5\n7.0 / 2 // 3.5\n```\n• However, if both numbers are integers, then the result is an integer:\n```7 / 2 // 3\n```\n• The decimal remainder is truncated (cut short, discarded, thrown away)\n• To get the integer remainder of division between two integers, we use the modulus operator: %\n```7 % 2 // 1 (remainder)\n```\n• `7 % 2` returns `1` because `1` is the remainder when `7` is divided by `2`:\n``` 3 r 1\n2 ) 7\n-6\n1 remainder\n```\n• For a refresher on remainders see: Long Division with Remainders\n• The modulus operator (%) can only be used with integer type operands and has an integer type result\n• C++ allows a negative modulus result from a modulo operation\n\n#### Uses of Modulus and Division\n\n• We can use modular arithmetic for operations that roll over like clock arithmetic\n• For example, if the time is 9:00 now, in 4 hours it is 1:00\n```int hours = 9;\nhours = (9 + 4) % 12;\ncout << hours << endl;\n```",
null,
"• We can extend the same idea to work with weeks, minutes and seconds\n• Modulus is also useful when working with different units like yards, feet and inches\n• For example, 49 inches is 1 yard, 1 foot and 1 inch which is calculated like:\n```int lengthInches = 49;\nint yards = lengthInches / 36;\nint remainder = lengthInches % 36;\nint feet = remainder / 12;\nint inches = lengthInches % 12;\ncout << yards << \" yards, \"\n<< feet << \" feet and \"\n<< inches << \" inches\\n\";\n```\n• Another use of the modulus operator is to split the last digit off of an `int`\n```int num = 123;\nint lastDigit = num % 10;\ncout lastDigit << endl;\n```\n• We can get the first digit using integer division\n```int firstDigit = num / 100;\n```\n• The middle digit(s) need a combination of integer division and modulus\n```int secondDigit = num / 10 % 10;\n```\n\n#### Program using Division and Modulus to \"Split\" Digits\n\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ``` ```#include using namespace std; int main( ) { int num = 1234; cout << num << \": \"; int digit1 = num % 10; // rightmost digit int digit2 = num / 10 % 10; int digit3 = num / 100 % 10; int digit4 = num / 1000 % 10; cout << digit1 << \" \" << digit2 << \" \" << digit3 << \" \" << digit4 << endl; return 0; } ```\n• We can use modulus and integer division to translate integers to different bases as well\n• For example, to convert from decimal to binary we divide by 2 and take the remainder of 2 instead of 10\n\n#### Program Changing Decimal to Binary\n\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 ``` ```#include using namespace std; int main() { int decimalNum = 6, newNum = 0; //what is 6 in binary? int firstBit = 0, secondBit = 0, thirdBit = 0; firstBit = decimalNum % 2; //rightmost bit newNum = decimalNum / 2; secondBit = newNum % 2; newNum = newNum / 2; thirdBit = newNum % 2; cout << decimalNum << \" in binary is \" << thirdBit << secondBit << firstBit << endl; return 0; } ```\n\n#### Check Yourself\n\n1. In division between two integer numbers, the remainder is ________.\n1. rounded up\n2. rounded off\n3. averaged\n4. truncated\n2. Of the following, the expression that returns an integer value is ________.\n1. `22.0 / 7`\n2. `22 / 7.0`\n3. `22.0 / 7.0`\n4. `22 / 7`\n3. To compute the integer remainder we use the operator ________ .\n4. What is the result of the following arithmetic operations?\n\n#### Discussion Questions\n\n1. What happens to a remainder in integer division?\n2. What is one use for the modulus operator?\n3. The last three operations produced the binary number `101`. What is the decimal equivalent?\n\n### 2.3.5: Mathematical Functions\n\n• Operators provide only the simplest mathematical operations\n• For more complex operations, we use mathematical functions\n• A C++ function is like a mathematical function that takes an argument (\"input\") and returns or produces a value (\"output\")\n• To make use of mathematical functions we need to store or display the returned value\n• For example:\n`cout << sqrt(9.0) << endl;`\n• In the above example, the input is `9.0` and the `sqrt()` function returns the square root of the argument\n• C++ has a standard library named cmath that contains many such functions\n```#include <cmath>\n```\n• Some of the functions are listed below\n• Note that the `pow()` function needs a floating-point type like `double` as the first argument\n\n#### Some Commonly Used Math Functions\n\nName Description Example Result\nabs absolute value `abs(-3.9)`\n`abs(3.9)`\n`3.9`\n`3.9`\nexp exponent of e (ex) `exp(1.0)` `2.71828`\npow powers (xy) `pow(2.0, 3)` `8`\nsqrt square root `sqrt(4.0)` `2`\nsin sine `sin(0.0)` `0`\ncos cosine `cos(0.0)` `1`\n\n#### Some Commonly Used Nearest Integer Functions\n\nName Description Example Result\nceil ceiling: round up `ceil(3.456)`\n`ceil(3.543)`\n`4`\n`4`\nfloor floor: round down `floor(3.456)`\n`floor(3.543)`\n`3`\n`3`\nround rounding: round off `round(3.456)`\n`round(3.543)`\n`3`\n`4`\n• Nearest Integer Functions return floating point numbers that are the closest to an integer value\n\n#### Using Mathematical Functions\n\n• How are mathematical functions evaluated?\n• Whatever is within the parenthesis of the function call is evaluated first\n• Thus, in the following example, we get the square root of `9.0`\n`cout << sqrt(3.0 * 3) << endl;`\n• If the function is used in an arithmetic expression, the expression uses the returned value\n• For example, in the following, the value `4.0` is stored in the double variable `num`:\n```double num = 1 + sqrt(3.0 * 3);\ncout << num << endl;\n```\n• Note that the function evaluates the `sqrt(3.0 * 3)` before adding it to `1.0`\n• Thus functions have a higher precedence than arithmetic operators\n\n#### Check Yourself\n\n1. C++ mathematical functions take an ________ and return a value.\n2. The name of the library that contains many common mathematical functions is ________\n\n4. To round up a number, call the mathematical function ________.\n5. Use a ________ to save the value returned from a mathematical function.\n\n### Exercise 2.3b: Calculator Deluxe (3m)\n\nIn this exercise we use integer division, modulus, and mathematical functions to create a deluxe calculator.\n\n#### Specifications\n\n1. Start with your `arithmetic.cpp` code from the last Exercise.\n2. Integer Division: Modify your `arithmetic.cpp` code from the last Exercise by changing the data type of the two variables from `double` to `int`, like this:\n```int a = 5, b = 2;\n```\n3. Compile and run your program again and compare the output. Note how the result of the division operation changed. What happened to the decimal part of the result?\n\nIn programming terms, we say that the decimal part is truncated (cut short). We have to watch out for this in C++ programming or we may get unexpected results in our calculations.\n\n4. Modulus (%) operator: Sometimes we want the integer remainder from an integer division. To see the integer remainder, we use the modulus (%) operator. Add the following statements to your program:\n```cout << \"a % b = \" << a % b << endl;\ncout << \"a / b % b = \" << a / b % b << endl;\ncout << \"a / (b * b) = \" << a / (b * b) << endl;\n```\n5. Compile and run your program again with this added statement. Your program should now display the following output:\n```a + b = 7\na - b = 3\na * b = 10\na / b = 2\na + b / 2 = 6\n(a + b) / 2 = 3\na % b = 1\na / b % b = 0\na / (b * b) = 1\n```\n6. Mathematical functions: More complex mathematical operations require the use of a function in C++. One such function is `sqrt(number)` which calculates the square root of the number inside the parenthesis.\n```cout << \"sqrt(a + b) = \" << sqrt(a + b) << endl;\n```\n8. You program will not compile with this new statement because you must include a library of the mathematical functions. Add the statement: `#include <cmath>` to the top of your program like this:\n```#include <iostream>\n#include <cmath> // math function library\nusing namespace std;\n```\n9. Compile and run your program again with this added statement. Your program should now compile and display the following output when run:\n```a + b = 7\na - b = 3\na * b = 10\na / b = 2\na + b / 2 = 6\n(a + b) / 2 = 3\na % b = 1\na / b % b = 0\na / (b * b) = 1\nsqrt(a + b) = 2.64575\n```\n10. Save your program source code that displays all ten (10) calculator operations so you can submit it to Canvas as part of assignment 2.\n\nWhen completed, please help those around you. Then compare your code to the example below. Your code need not be exactly the same but it is helpful to see other solutions after you have solved the problem yourself.\n\n#### Completed Program",
null,
"#### Discussion Questions\n\n1. How natural does arithmetic seem in C++ compared to what you normally use?\n2. A mathematical function has an input and an output. What is the input and output of the C++ `sqrt()` function?\n3. What is the value of 2.64575 squared?\n\n### 2.3.6: Summary\n\n• C++ uses the following operators for arithmetic:\n• `+` for addition\n• `-` for subtraction\n• `*` for multiplication\n• `/` for division\n• `%` for modulus (remainder)\n• The dash is also used for negation (minus sign)\n• We write arithmetic expressions using combinations of numbers, variables and operators, like:\n```double area = 0.0, radius = 4.3;\n```\n• As in algebra, multiplication and division are performed before addition and subtraction\n• To change the order of operation, we use parenthesis\n• Recall that the results of integer division are truncated\n`7 / 2 // 3`\n• We must use modulus operator (`%`) to get the remainder value\n`7 % 2 // 1 (remainder)`\n• For more complex operations we use mathematical functions from libraries such as cmath\n`cout << sqrt(3.0 * 3) << endl;`\n• C++ processes arithmetic expressions in the same order (precedence) as algebra:\n1. Parenthesis: `( )`\n2. Function calls\n3. Unary operators: `+, -`\n4. Multiplication, division, modulus: `*, /, %`\n5. Addition, subtraction: `+, -`\n• Since parenthesis are processed first, we use parenthesis to change the order of operations\n\n#### Self Reflection\n\nAnswer these questions to check your understanding. If you are not sure, then follow the links to the section and review the material.\n\n1. How can you tell the difference between an integer and a floating-point number? (2.3.1)\n2. What are the five operators C++ provides for arithmetic? (2.3.2)\n3. What is the first operation performed in the expression: `1 + 2 * 3 / 4 % 5`? (2.3.2)\n4. What is the data type returned by the following expression? (2.3.3)\n`3 + 4.5`\n5. What happens to the remainder during integer division? (2.3.4)\n6. What operator can you use to compute the integer remainder? (2.3.4)\n7. What is the result of the following arithmetic expressions? (2.3.4)\n```5 / 4\n5 % 4\n```\n8. What is a function? (2.3.5)\n9. What code do you write to calculate the square root of the number `49`? (2.3.5)\n\n## 2.4: Developing Computer Programs\n\n### Learner Outcomes\n\nAt the end of the lesson the student will be able to:\n\n• Describe various kinds of programming errors\n• Find the syntax errors in programs\n\n### 2.4.1: Solving Problems with Code\n\n• Creating programs is like solving puzzles: they take some work to figure out\n• In this lesson we discuss a general approach to solving problems with computer programs\n\n#### Parts of a Computer Program\n\n• Remember that all computers have four parts\n1. Input\n2. Processing\n3. Storage\n4. Output\n• Similarly, a computer program has at least three parts\n1. Input and store data\n2. Process and store data\n3. Output information\n• Each part may be repeated many times\n• Most programs we write have all these parts\n\n#### Four Steps to Solving a Problem\n\n• A computer program is an algorithm implemented in computer code\n\nAlgorithm: A sequence of precise instructions leading to a solution\n\n• To write a program we must develop a computer algorithm\n• One approach is based on Polya's Four Step Problem Solving Process and has four parts\n1. Understand the problem\n2. Create a plan and check it\n3. Translate to code and perfect the plan\n4. Test and debug the code\n• The end result is a computer program that solves a problem\n\n#### Check Yourself\n\n1. A sequence of precise instructions to accomplish a task is called a(n) ________.\n2. Most all computer programs include steps to ________.\n1. input data\n2. store data\n3. process data\n4. output information\n5. do all of these\n3. The first step to solve a problem with a computer program is to ________.\n1. create a plan and check it\n2. test and debug the code\n3. translate to code and perfect the plan\n4. understand the problem\n\n### 2.4.2: Understanding the Problem\n\n• The first step in solving a problem is to understand it\n• This section suggests ways to analyze and understand a problem\n\n#### Starting Out\n\n• When analyzing a problem, start by reading the directions\n• After you read the directions, ask yourself, \"what is the goal of the problem?\"\n• Also ask, \"what are the inputs and outputs?\"\n• Then try restating the problem in your own words\n1. What does it do?\n2. Why do you think it's there?\n\n• As we analyze a problem, we naturally ask questions like: What? When? Why? Where? How?\n• Here are some questions that are often useful to ask when faced with a programming problem:\n1. What are you asked to find or output?\n2. Can you restate the problem in your own words?\n3. Is this problem similar to something covered in the lesson?\n4. Can you think of a picture or a diagram that might help you understand the problem?\n5. What input is needed or provided?\n6. What calculation must be performed?\n7. What data is needed for the calculations?\n• Start by reviewing this list and answering the questions that seem pertinent to the problem\n• Ask more questions as the analysis continues until you have a solution\n\n#### Exercise: Understanding the Problem (2m)\n\n1. To help us understand the development process we have an example problem\n\nWrite a program that asks the user for a number. The program then displays twice that number.\n\n2. The input and output of the program looks like:\n```Enter a number: 10\nTwice that number is: 20\n```\n\nIn the above example run, the users entered the value shown in aqua italics (for emphasis) to produce the output. Your program does NOT print the characters in aqua italics, nor does the user input appear in aqua italics.\n\n3. As a first step, spend two minutes to open a text editor and rewrite the problem in your own words.\n4. Save the text of your example problem rewrite in a file named \"plan.txt\".\n5. Be prepared to share your problem rewrite with the class.\n\n#### Check Yourself\n\n1. The first step to understand a problem, as assigned in this course, is to ________.\n1. restate the problem in your own words\n2. identify the inputs and outputs\n3. determine if starter code was provided\n2. The first step in problem solving is ________.\n1. To write the expression that calculates the answer\n2. To understand the problem and its inputs and outputs\n3. To do examples by hand that confirm the solution will work\n4. To write C++ code that can be executed and tested\n3. Useful questions to ask while trying to understand a problem include ________.\n1. What are you asked to find or output?\n2. Is this problem similar to something covered in the lesson?\n3. What input is needed?\n4. all of these\n4. The reason to restate the problem in your own words is because ________.\n1. of grammar errors in the problem statement\n2. the original words may not be an optimal statement of the problem\n3. restating the problem in your own words guarantees that the code will be correct in most cases\n4. if you cannot restate a problem, it is unlikely you will be able to write code to solve it\n\n### 2.4.3: Creating a Plan\n\n• Our goal for the plan is to design an algorithm for a computer\n• An example algorithm:\n```Computer algorithm to square any number:\n1. output \"Enter a number\"\n2. input number\n3. x = number * number\n4. output \"The square of the number is \"\n5. output x\n```\n• Notice that the steps are combinations of input, processing (calculation), storage, and output\n• There are many reasonable ways to solve problems and generate algorithms\n• The following is a partial list of strategies\n\n#### Problem Solving Strategies\n\n• Make an orderly list of steps\n• Draw a diagram or picture\n• Look for a pattern\n• Use a formula\n• Compare to a previously solved problem\n• Eliminate possibilities\n• Work backwards (output, processing and storage, input)\n• Break into smaller pieces and solve one small part at a time\n• Solve an equivalent problem and translate the solution\n• Guess and check\n\n#### Writing and Checking the Plan\n\n• Once we have a plan we write down the steps\n• Write each step on its own line like the example algorithm above\n• Make sure the plan has steps for:\n1. input\n2. storage\n3. processing\n4. output\n• Check the plan by working out an example\n```Computer algorithm to square any number:\n1. output \"Enter a number\"\n2. input number (example: 10)\n3. x = number * number (10 x 10 is 100)\n4. output \"The square of the number is \"\n5. output x (x is 100)\n```\n• If you do not get the correct result then revise the algorithm\n\n#### Exercise: Make a Plan (5m)\n\n1. In the same file as the previous activity, prepare a plan and list the steps needed to solve the problem.\n2. To test your plan, work out an example.\n3. Be prepared to share your plan.\n4. Save your `plan.txt` file to turn in with the other lesson exercises for the week.\n\n#### Check Yourself\n\n1. Of the following, a computer cannot directly ________.\n1. input data\n2. store data\n3. output information\n4. devise a plan\n2. The goal of the planning phase is a(n) ________.\n3. True or false: a computer can perform any algorithm step we can write down.\n4. The reason to check your algorithm \"by hand\" is because ________.\n1. C++ code is not able to capture the subtleties of complex problems\n2. it is faster to do computations by hand than to do them by computer\n3. checking a problem \"by hand\" guarantees that programs will be correct\n4. if you cannot compute a solution by hand, it is unlikely you will be able to write a program that can do it\n\n### 2.4.4: Translating to Code\n\n• In this section we look at translating to code and perfecting the plan\n• When writing software, always start with something that compiles and runs\n• As an example, start with the following code that compiles and runs\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 ``` ```/** CS-11 Asn 0, program.cpp Purpose: Your problem restatement here. @author Your name @version 1.0 Today's date */ #include using namespace std; int main() { cout << \"Hello, World!\\n\"; return 0; } ```\n\n#### Exercise: Translate to C++ (5m)\n\n1. Start a new file, copy the above starter code into the file, and save the file as `twice.cpp`.\n2. Compile to make sure you are starting out right.\n3. Fill in the following information in the comment block\n1. program name\n4. today's date\n4. Now copy your algorithm steps from your plan (from the last activity) into `main()` between the curly braces `{ }`.\n5. Before each step of the algorithm, add two slashes (`//`) to make each line a comment.\n\nFor example if one line of your algorithm was:\n\n```output \"Enter a number\"\n```\n```// output \"Enter a number\"\n```\n6. After copying and commenting all the algorithm steps, compile to make sure the code has no setup errors.\n\nAsk for help, if needed, to find setup errors.\n\n7. After each commented algorithm step, add the C++ statements needed to implement the algorithm step.\n```// output \"Hello out there.\"\ncout << \"Hello out there.\" << endl;\n```\n8. Compile after adding each statement (or so) of code.\n\nIf the code does not compile then you know where the problem exists.\n\n9. After translating all the steps, run your code and verify it produces the right output.\n```Enter a number: 10\nTwice that number is: 20\n```\n10. Save `twice.cpp` to turn in with the other lesson exercises for the week.\n\n• Verify that your program does what you expected so far\n• If not, revise your algorithm to create the desired result\n• If you start to get frustrated, take a deep breath, or leave your screen for a minute\n• When you come back, you may see what was causing the trouble!\n• If you are still stuck, ask questions and try another approach\n• If you are still stuck after three attempts, consider getting help\n• Maybe one of your friends, a tutor or the instructor can point out where your plan goes awry\n• Be prepared to discuss your plan and show the steps you have already taken\n\n#### Reflection\n\n• After solving the problem it is time to reflect and look back at what you have done\n• Look at what worked well and what did not work\n• Doing this will help you to predict what strategy to use to solve similar problems in the future\n• Does your program solve the problem?\n• Did you reach all of the goals of this problem?\n• Now that you have one way to solve the program, is there an easier way to do it?\n• If you change this solution a little, will it work for any other programs?\n• Could you explain your solution to another person?\n\n#### Check Yourself\n\n1. True or false: you should compile your code after adding every line or so to check for errors.\n2. True or false: after finishing your program, run the program to test it.\n3. True or false: the purpose of reflection after developing a program is to improve your problem solving ability.\n\n### 2.4.5: Debugging Code\n\n• As we develop programs, we make errors\n• These errors are known euphemistically as \"bugs\" (See the first computer bug)\n• Grace Hopper, an early computer pioneer, popularized the term debugging\n• Often the error is because we left out a character or misspelled a word\n• In programming, one small mistake means the program does not work\n• The following are some types of errors we may encounter and what to do about them\n\n#### Syntax Errors\n\n• One type of error is the syntax error or compile-time error\n• This means our source code violates the language rules\n• Some examples:\n```cout << \"Hello, World!\\n\" // missing semicolon (;)\ncot << \"Hello, World!\\n\"; // mispelled letters in cout command\ncout << \"Hello, World!\\n; // missing closing \"\n```\n• The good news: our compiler finds the error and reports an error message like:\n```fiddle.cpp: In function 'int main()':\nfiddle.cpp:13:5: error: expected ';' before 'cot'\ncot << \"Hello, World!\\n\";\n^~~\n```\n• The message can tell us several import things but the most immediately useful is the line number\n```fiddle.cpp:13:5: error: expected ';' before 'cot'\n```\n• This tells us the approximate location of the error in the source code\n• Another useful part of the message is the error message itself\n```fiddle.cpp:6: error: expected ';' before 'cot'\n```\n• The message is trying to explain the problem but is often highly technical\n• You will get a sense of their meaning over time\n• In general, we should look at the first error or warning and ignore the rest\n• Often the first error or warning causes all the rest of the errors and warnings\n• For step-by-step instructions on debugging syntax errors, see: How To Debug Syntax Errors\n\n#### Logic Errors\n\n• Another type of error is the logic error:\n• Logic errors are mistakes in the program's algorithm\n• Our program does not do what it is supposed to do\n• For example:\n`cout << \"Hell, World\\n\";`\n• The compiler does not find these types of errors for us\n• Instead, we must run our program and look for errors in how it operates\n• Running a program and looking for errors is known as testing\n\n#### Run-time Errors\n\n• A third type of error that we may encounter is the run-time error\n```int x = 0;\ncout << 1 / x;\n```\n• Run-time errors cause our program to exit with an abnormal error (\"crash\")\n• To find a run-time error we look in the first line of the `stackdump` file\n```Exception: STATUS_INTEGER_DIVIDE_BY_ZERO at eip=00401228\n```\n\n#### Errors Versus Warnings\n\n• If we violate a syntax rule, the compiler gives us an error message\n• However, sometimes the compiler will give us a warning message instead\n• For example:\n`erroneous.cpp:10:1: warning: \"/*\" within comment`\n• This indicates that our code is technically correct\n• However, the code is unusual enough that it may be a mistake or is otherwise undesirable\n• Thus, you lose points for warning messages in your programming assignments\n\n#### Preventing Bugs\n\n• The best way to debug is not to make bugs in the first place!\n• Think about what you're writing before you write it--design first, code later\n• Compile frequently!\n• Execute (run the program) frequently!\n• Review before testing: Walk through your code mentally and try different values\n• Write readable code following the professional coding guidelines\n\n#### Check Yourself\n\n1. True or false: one of the goals of testing is to find and correct errors.\n2. The program that tests if your program has a syntax error is known as a(n) ________.\n3. True or false: the hardest error to find is a syntax error.\n4. To test for logic or runtime errors, you must ________ your program.\n5. True or false: warnings are technically not errors so it is safe to ignore them.\n6. True or false: the best way to debug is not to write bugs in the first place.\n7. We should always compile after writing every line or two of code because ________.\n1. the compiler will tell you about syntax errors\n2. we know the location of any errors reported\n3. it is easier to prevent problems than correct them\n4. of all of these\n8. True or false: following a professional programming style makes it easier to see and prevent bugs.\n\n### Exercise 2.4: Debugging a Program\n\nIn this exercise we practice finding and fixing errors with a partner. Make sure you finished the previous activities as well so you can turn in:\n\n#### Specifications\n\nThe following program was written by a person in a hurry. During the writing process a large number of errors were made.\n\n#### Erroneous Program\n\n ```1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 ``` ```/** A very erroneous program. @author B. A. Ware @version 1.0 1/25/05 #include using namespace standard; /** * The main functoin for the program. */ int main() { cout <<< \"Hell out there.\\\"; cout << \"Enter a number as a percent: \" cin << ware; Cout << \"As a double the number is: \"; cout << ware / 00 << end1; return; }} / / end of main ```\n##### Part A: Peer Review (3m)\n1. Find a partner for this exercise and create a text file named `errors.txt`\n2. Add a comment at the top of the file that contains the name of the person(s) with whom you reviewed the code, like:\n```// Reviewed with Emma Programmer\n```\n3. With your partner, review the program, finding and listing in your `errors.txt` file the line number and description of as many errors as you can within 3 minutes. For example:\n```// Reviewed with Emma Programmer\nLine 6: missing closing comment symbols\nLine 7: space in iostream\n(more errors listed)\n```\n4. Submit your list of errors to Canvas as part of next weeks assignment.\n##### Part B: Syntax Errors (5m)\n\nFind and correct the errors reported by the compiler by following the process described in How To Debug Syntax Errors.\n\n1. Copy and save the program code with the name: `erroneous.cpp`\n2. Find and correct the syntax errors or warnings reported by the compiler one at a time in order they appear.\n3. Submit your corrected program source code to Canvas as part of next weeks assignment.\n\nBe prepared to answer the Discussion Questions when called upon.\n\n#### Discussion Questions\n\n1. Did having someone review your code help find errors?\n2. What was the cause of the message, \"warning: \"/*\" within comment\"?\n3. Could the message, \"warning: \"/*\" within comment\", safely be ignored since it was just a warning?\n4. Did the list of errors reported by the compiler change dramatically as you corrected each one?\n5. Were the errors always on the line reported by the compiler?\n6. Did the compiler always find errors in the order in which the program code was listed?\n\n### 2.4.6: Summary\n\n• Before starting to code you need a plan\n• The result of the plan is an algorithm that you translate to code\n• Even with a good plan, you will make errors as you develop code\n• One type of error is the syntax error or compile-time error where code violates the rules of the language\n• Another type of error is the logic error, where a program does not do what it is supposed to do\n• Yet another type of error is the run-time error\n• Run-time errors cause programs to exit with an abnormal error (\"crash\")\n• Sometimes the compiler will issue a warning if your code is unusual but still correct\n• You should treat this warning as an error because it is probably a logic error\n• Syntax errors are the easiest errors to find and correct\n• For instructions on solving syntax errors see How To Debug `g++` Compile-time Errors\n\n#### Review Questions\n\nAnswer these questions to check your understanding. If you are not sure, then follow the links to the section and review the material.\n\n1. What are the four parts of a computer? (2.4.1)\n2. What are the four parts of a computer program? (2.4.1)\n3. A sequence of precise instructions which leads to a solution is called an ________________. (2.4.1)\n4. What is the first thing we should do when developing a computer program? (2.4.2)\n1. Start typing C++ code into a text editor.\n2. Check our compiler to verify it is fully functional and up-to-date..\n3. Search the Internet to see if someone else has already solved the problem.\n4. Investigate and understand the problem we are trying to solve.\n5. What are two techniques we can use to help us understand a problem? (2.4.2)\n6. What are three good questions we might ask to help us understand a problem? (2.4.2)\n7. What is the result of developing a plan? (2.4.3)\n8. What four types of statements belong in most all programs? (2.4.3)\n9. What are three ways to think of how to solve a problem? (2.4.3)\n10. What is the relationship between an algorithm and a computer program? (2.4.4)\n11. What is the purpose of testing code that compiles without compile-time errors? (2.4.4)\n12. What are two kinds of programming errors? (2.4.5)\n13. What is a warning? (2.4.5)\n14. True or false: the best way to remove errors is not to write them in the first place (2.4.5)\n15. What is the importance of following professional programming style conventions? (2.4.5)\n16. In the following error message, what is the most immediately useful information? (2.4.5)\n```fiddle.cpp: In function `int main()':\nfiddle.cpp:6: error: parse error before `return'\n```\n17. Why should you ignore errors after the first one? (2.4.5)"
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https://review.gerrithub.io/plugins/gitiles/cue-lang/cue/+/0a0a3ac9cde699a0ab045298969dfb1035de0a8b%5E%21/ | [
"```doc/ref: introduce regular expression syntax\n\nThis is one sorely missing feature of a data constraint\nlanguage. Now we have unary constraints, the\nsyntax and semantics for it naturally arises.\n\nAlso fixed a comment about the role of null.\n\nFixes cuelang/cue#12.\n\n```\n```diff --git a/doc/ref/spec.md b/doc/ref/spec.md\nindex 0b56d40..dd1df4d 100644\n--- a/doc/ref/spec.md\n+++ b/doc/ref/spec.md\n```\n```@@ -269,14 +269,15 @@\nThe following character sequences represent operators and punctuation:\n\n```\n-+ div && == != ( )\n-- mod || < <= [ ]\n-* quo ! > >= { }\n-/ rem & : <- ; ,\n-% _|_ | = ... .\n++ div && == < . ( )\n+- mod || != > : { }\n+* quo & =~ <= = [ ]\n+/ rem | !~ >= <- ... ,\n+% _|_ ! ;\n```\n<!-- :: for \"is-a\" definitions -->\n\n+\n### Integer literals\n\nAn integer literal is a sequence of digits representing an integer value.\n@@ -1586,12 +1587,18 @@\n<= less or equal\n> greater\n>= greater or equal\n+=~ matches regular expression\n+!~ does not match regular expression\n```\n+<!-- regular expression operator inspired by Bash, Perl, and Ruby. -->\n\n-In any comparison, the types of the two operands must unify.\n+In any comparison, the types of the two operands must unify or one of the\n+operands must be null.\n\nThe equality operators `==` and `!=` apply to operands that are comparable.\nThe ordering operators `<`, `<=`, `>`, and `>=` apply to operands that are ordered.\n+The matching operators `=~` and `!~` apply to a string and regular\n+expression operand.\nThese terms and the result of the comparisons are defined as follows:\n\n- Null is comparable with itself and any other type.\n@@ -1605,11 +1612,25 @@\nnormalization to Unicode normal form NFC.\n- Struct are not comparable.\n- Lists are not comparable.\n+- The regular expression syntax is the one accepted by RE2,\n+ except for `\\C`.\n+- `s =~ r` is true if `s` matches the regular expression `r`.\n+- `s !~ r` is true if `s` does not match regular expression `r`.\n+<!-- TODO: Implementations should adopt an algorithm that runs in linear time? -->\n+<!-- Consider implementing Level 2 of Unicode regular expresssion. -->\n+\n```\n-a: 3 < 4 // true\n-b: null == 2 // false\n-c: null != {} // true\n-d: {} == {} // _|_: structs are not comparable against structs\n+3 < 4 // true\n+null == 2 // false\n+null != {} // true\n+{} == {} // _|_: structs are not comparable against structs\n+\n+\"Wild cats\" =~ \"cat\" // true\n+\"Wild cats\" !~ \"dog\" // true\n+\n+\"foo\" =~ \"^[a-z]{3}\\$\" // true\n+\"foo\" =~ \"^[a-z]{4}\\$\" // false\n```\n\n<!-- jba\n```"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.6943746,"math_prob":0.9797251,"size":2540,"snap":"2022-40-2023-06","text_gpt3_token_len":716,"char_repetition_ratio":0.12263407,"word_repetition_ratio":0.061269145,"special_character_ratio":0.38897637,"punctuation_ratio":0.19555555,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9703875,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-03T07:15:55Z\",\"WARC-Record-ID\":\"<urn:uuid:6212479c-597a-47b3-a2d2-67103e05ebcb>\",\"Content-Length\":\"9228\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:f5fdca13-0385-4ca1-b570-a9514f06146d>\",\"WARC-Concurrent-To\":\"<urn:uuid:57c519a4-cbc9-48e1-afb4-9f4575450999>\",\"WARC-IP-Address\":\"144.217.182.195\",\"WARC-Target-URI\":\"https://review.gerrithub.io/plugins/gitiles/cue-lang/cue/+/0a0a3ac9cde699a0ab045298969dfb1035de0a8b%5E%21/\",\"WARC-Payload-Digest\":\"sha1:G5P3WYBQ3JV3PTFVUJWS7DT2YOTN6CGJ\",\"WARC-Block-Digest\":\"sha1:7QQPWPX4GUAWPHLRDUVPQWVREGLL226D\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764500044.16_warc_CC-MAIN-20230203055519-20230203085519-00060.warc.gz\"}"} |
https://www.tutorialspoint.com/increasing-alternate-element-pattern-in-list-in-python | [
"# Increasing alternate element pattern in list in Python\n\nIn this tutorial, we are going to write a program that includes increasing order of an element to the given list after each element. Let's see an example to understand it clearly.\n\n## Input\n\nalphabets = ['a', 'b', 'c']\n\n## Output\n\n['a', '#', 'b', '##', 'c', '###']\n\nFollow the below steps to solve the problem.\n\n• Initialize a list.\n• 3Create an empty list.\n• Iterate over the initial list.\n• Add the current element and the respective number of hashes to the empty list.\n• Print the resultant list.\n\n## Example\n\nLive Demo\n\n# initializing the list\nalphabets = ['a', 'b', 'c']\n# empty list\nresult = []\n# iterating over the alphabets\nfor i in range(len(alphabets)):\n# appending the current element\nresult.append(alphabets[i])\n# appending the (i + 1) number of hashes\nresult.append((i + 1) * '#')\n# printing the result\nprint(result)\n\n## Output\n\nIf you run the above code, then you will get the following result.\n\n['a', '#', 'b', '##', 'c', '###']\n\n## Conclusion\n\nIf you have doubts regarding the tutorial, mention them in the comment section."
]
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https://docs.nosleepcreative.com/after-effects/expressions/utilities | [
"# Utilities\n\n## Accessing sub-objects\n\n### Get layer's pixel size\n\nfunction getSize(layer){\nvar src = layer.sourceRectAtTime();\nvar s = layer.transform.scale\nvar size = [src.width*s,src.height*s]/100;\nreturn size;\n}\n\n### Get layer's top left coordinates\n\nfunction getSrcRectTopLeft(layer){\ns = layer.sourceRectAtTime();\nlocalPos = [s.left,s.top];\nworldPos = layer.toWorld(localPos);\nreturn worldPos;\n}\n\n### Get effect index number\n\n// effect index\nthisProperty.propertyGroup(1).propertyIndex;\n\n### Get number of effects applied to a layer\n\n// referencing locally\nthisLayer(\"Effects\").numProperties\n// referencing other layer\nthisComp.layer(\"layer\")(\"Effects\").numProperties\n// other method\nthisProperty.propertyGroup(2).numProperties\n\n## Working with Array\n\n• https://dmitripavlutin.com/operations-on-arrays-javascript/\n\n### Shuffling elements\n\nfunction shuffle(a) {\nvar j, x, i;\nfor (i = a.length - 1; i > 0; i--) {\nj = Math.floor(Math.random() * (i + 1));\nx = a[i];\na[i] = a[j];\na[j] = x;\n}\nreturn a;\n}\n\n### Check how many array is empty\n\nconst arr = [[], [1,4,5], [], [4,6,7];\n// method 1: conditional -> sum\nnumbers.map(x=> x.length>0: 1:0).reduce((a, b) => a + b, 0)\n// method 2\narr.filter(x => x.length>0).length // get array of true results -> length\n\n### Finding minimum or maximum element\n\nIn ES6, you can use the `...` operator to spread an array and take the minimum or maximum element.\nvar myArray = [1, 2, 3, 4, 99, 20];\nvar maxValue = Math.max(...myArray); // 99\nvar minValue = Math.min(...myArray); // 1\n\n### Get length of each element\n\n// get length of each array element\nelemLength = str.map(s=>s.length)\n\n### Sum of previous elements \n\n// method 1: using mapget character index\nlet array = [1,5,6,8,10],sum;\narray = array.map(elem => sum = (sum || 0) + elem);\n// method 2: using reduce and map\nlet array = [280,430,408,430,408];\narray = array.map((elem, index) => array.slice(0,index + 1).reduce((a, b) => a + b));\n// best method: using double arrow functions\nvar array =[280, 430, 408, 430, 408]\nresult = array.map((s => a => s += a)(0));\nresult;\n\n### Comparing a value to elements, filtering conditions\n\nlet array = [1, 3, 5, 7, 10];\nv = 6;\n// method 1: conditional -> sum of array\nrow = (array.map(c => v >= c ? 1 : 0)).reduce((a, b) => a + b, 0)\n// method 2: get index of true results, then get max value\nrow = Math.max(...array.map(c => v>=c? array.indexOf(c):0))\n// best method: filter\nrow = array.filter(c => v >= c).length\n\n## Increment by Index\n\n• Usage: tiling, valueAtTime offset\nfunction indexInc(mainlayer, offset, randRange) {\nvar startIndex = mainlayer.layer(index - 1).index\nvar myIndex = index - startIndex;\ntry {\nvar inc = offset * random(randRange, randRange)* myIndex;\n} catch (e) {\ninc = offset * myIndex;\n}\nreturn inc;\n}\n// usage for tile y\nmainLayer.transform.position + [0+indexInc(mainLayer,50]\n\n## Loops\n\n### Through every frame\n\nfor(t = 0; t < thisComp.duration; t = t + thisComp.frameDuration){\n// statements to execute\n}\n\n## Unsorted\n\n### Convert cartesian coordinates to polar\n\n//If you want it relative to the comp’s [0,0] (upper left corner)\n\n#### Explanation\n\nUsing simple trigonometry:\nImagine you have the point [960, 540];\nThose 2 values, 960 and 540, represent two adjacent sides of a triangle.\nNow, if you recall back to 9th Grade, you may remember that there is a simple equation to get the third side of that triangle. A² + B² = C². So, we can find the third side (the magnitude) by using an equation like:\nvar mag = Math.sqrt(Math.exp(position, 2) + Math.exp(position, 2)); Essentially saying take the square root of X² + Y².\nFortunately, AE gives us the handy little ‘length’ operator, which simply does all of these computations for us.\nAnother similar equation can be used to find the angle of a triangle when you know two other sides, you may remember SOH CAH TOA.\nImagine the angle at the top left of your comp, now notice that you have the opposite side (the Y coord) and the adjacent side (the X coord). Since TOA uses both opposite and adjacent sides, we can get that angle using an equation like this:\ntan(theta) = 540/960\ntheta = tan−1(5625)\n//or, in code\nvar angle = Math.atan(position / position);\n\n## Null / Camera Rigs\n\n### Add sliders with specific affix names to property\n\nexpression\nfunction\nvar n = thisLayer(\"Effects\").numProperties\nfor(i=1;i<=n;i++){\nif(effect(i).name.includes(str)){\nvalue+= effect(i)(1);\n}\n}\nn = thisLayer(\"Effects\").numProperties;\nfor(i=1;i<=n;i++){\nif(effect(i).name.indexOf(fxName)==0){\nvalue+= effect(i)(1);\n}\n}\nreturn value;\n};\n\n## Autonomous agents\n\n### Auto-capture ball by Aaron Cobb\n\nvar ball = thisLayer;\nvar cup = thisComp.layer(\"Cup\");\nvar captureDuration = .25; //time for ball to reach center of cup once capture beings\nvar captureRadius = 100; //radius around anchor point of cup at which ball will be captured.\nvar captureTime = thisComp.duration; //time at which capture begins, default to end of comp\nvar currentDistance;\nfor(t = 0; t < captureTime; t = t + thisComp.frameDuration){ //loop through frames\ncurrentDistance = length(ball.toComp(ball.anchorPoint.valueAtTime(t), t), cup.toComp(cup.anchorPoint.valueAtTime(t), t));\nif(currentDistance < captureRadius) captureTime = t; //if inside capture radius exit the loop\n}\n// execute\nease(time, captureTime, captureTime + captureDuration, value, cup.toComp(cup.anchorPoint.value));\n\n### Compounding ease() interpolation (source)\n\nYou can't control the influence directly, but you can compound the ease such as below.\nif (numKeys > 1){\nt1 = key(1).time;\nt2 = key(2).time;\nv1 = [0,540];\nv2 = [1080,540];\nt = easeOut(time,t1,t2,0,1); // using a normalized value to drive the 2nd ease\neaseOut(t,0,1,v1,v2);\n}else\nvalue"
]
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http://sites.science.oregonstate.edu/physics/portfolioswiki/whitepapers:representations:qm:start | [
"## Representations for a Spins-First Approach to Quantum Mechanics\n\nCorinne Manogue, Oregon State University Elizabeth Gire, University of Memphis David McIntyre, Oregon State University Janet Tate, Oregon State University\n\n### Submitted Abstract to PERC 2011\n\nIn the Paradigms in Physics Curriculum at Oregon State University, we take a spins-first approach to quantum mechanics using a java simulation of successive Stern-Gerlach experiments to explore the postulates. The experimental schematic is a diagrammatic representation that we use throughout our discussion of quantum measurements. With a spins-first approach, it is natural to start with Dirac bra-ket language for states, observables, and projection operators. We also use explicit matrix representations of operators and ask students to translate between the Dirac and matrix languages. The projection of the state onto a basis is represented with a histogram. When we subsequently introduce wave functions, the wave function attains a natural interpretation as the continuous limit of these discrete histograms or of a projection of a Dirac ket onto position or momentum eigenstates. We are able to test the students’ facility with moving between these representations in later modules.\n\n• Spins First: we use the experimental schematic as a diagrammatic representation.\n• Dirac Bra-ket language\n• students use spontaneously for wave functions\n• allows you to bypass explicit spatial integrals\n• Ethan's example on last Tuesday of 2011 periodic potentials: ask student to write Hamiltonian in matrix form. Students write matrix with entries $\\langle 1|H|1\\rangle$, etc. Then give alpha, beta and zero names to entries of approximately the same size. Allows you to see symmetries in the Hamiltonian explicitly. Do we have video of this?\n• Introduce the idea that $\\langle n|\\Psi\\rangle$ is a coefficient of a vector, eigenfunction expansions: make this a central EARLY idea in quantum. Bra-ket notation makes this representation easy and natural. What evidence do we have that students get this?\n• Janet's representation of these coefficients as a histogram. Really emphasize this/who else does it?\n• Different kinds of operators: rotations (classical), observalbles, projections,\n• bra-ket, abstract symbols, and explicit matrix representations\n• Activities where we explicitly ask students to move back and forth between representations, paired activities where they do the same calculations in different representations.\n• explicit matrix multiplication (matrices with just underscores) vs. Henri and David Roundy's representations with $A_{ij}$.\n• Wave functions are then an example of coefficients of vectors\n• $\\psi(x)=\\langle x\\vert \\psi\\rangle$\n• Talk explicitly about the transition from discrete to continuous representations.\n• histograms become wave function graphs.\n\nResearch questions:\n\n• Are students doing sense-making with these representations?\n• Look for video from 1-2 waves talking about wave functions.\n• What representations do students use to make sense out of quantum systems?\n• Look for changes in representation compared to the problem statement.\n• What kinds of discussions do students have when we are asking them to translate between representations?\n\nPossible approaches:\n\n• Pick one activity: look for evidence of student understanding\n• Grounded approach: let the research question arise from the data set.\n• Ampere's Law approach:\n• here are some of the issues\n• here are some things that you might do\n• here are affordances of the activities tha tmight help students come to a new understanding.\n• Formative assessment: assess student understanding as they are doing the activities.\n• Use “standard” quantum assessments\n• Chandralekha Singh\n• Sam McKagan\n• We should arrange a session where participants bring in quantum questions and we try to devise new parts to an assessment. People to invite: Chandralekha Singh, Sam McKagan, Mark Haugan, Liz Gire, Stamatis Vokos, David McIntyre, Corinne Manogue, (others??)\n\n##### Views\n\nNew Users\n\nCurriculum\n\nPedagogy\n\nInstitutional Change\n\nPublications\n\n##### Toolbox",
null,
""
]
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null,
"http://sites.science.oregonstate.edu/portfolioswiki/lib/exe/indexer.php",
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https://courses.lumenlearning.com/suny-osalgebratrig/chapter/non-right-triangles-law-of-cosines/ | [
"## Non-right Triangles: Law of Cosines\n\n### Learning Objectives\n\nIn this section, you will:\n\n• Use the Law of Cosines to solve oblique triangles.\n• Solve applied problems using the Law of Cosines.\n• Use Heron’s formula to find the area of a triangle.\n\nSuppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in (Figure). How far from port is the boat?",
null,
"Figure 1.\n\nUnfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases.\n\n### Using the Law of Cosines to Solve Oblique Triangles\n\nThe tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level.\n\nUnderstanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem, which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle$\\,ABC\\,$is placed in the coordinate plane with vertex$\\,A\\,$at the origin, side$\\,c\\,$drawn along the x-axis, and vertex$\\,C\\,$located at some point$\\,\\left(x,y\\right)\\,$in the plane, as illustrated in (Figure). Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted.",
null,
"Figure 2.\n\nWe can drop a perpendicular from$\\,C\\,$to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that\n\n$\\mathrm{cos}\\,\\theta =\\frac{x\\text{(adjacent)}}{b\\text{(hypotenuse)}}\\text{ and }\\mathrm{sin}\\,\\theta =\\frac{y\\text{(opposite)}}{b\\text{(hypotenuse)}}$\n\nIn terms of$\\,\\theta ,\\text{ }x=b\\mathrm{cos}\\,\\theta \\,$and$y=b\\mathrm{sin}\\,\\theta .\\text{ }$The$\\,\\left(x,y\\right)\\,$point located at$\\,C\\,$has coordinates$\\,\\left(b\\mathrm{cos}\\,\\theta ,\\,\\,b\\mathrm{sin}\\,\\theta \\right).\\,$Using the side$\\,\\left(x-c\\right)\\,$as one leg of a right triangle and$\\,y\\,$as the second leg, we can find the length of hypotenuse$\\,a\\,$using the Pythagorean Theorem. Thus,\n\n$\\begin{array}{llllll} {a}^{2}={\\left(x-c\\right)}^{2}+{y}^{2}\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\text{ }={\\left(b\\mathrm{cos}\\,\\theta -c\\right)}^{2}+{\\left(b\\mathrm{sin}\\,\\theta \\right)}^{2}\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Substitute }\\left(b\\mathrm{cos}\\,\\theta \\right)\\text{ for}\\,x\\,\\,\\text{and }\\left(b\\mathrm{sin}\\,\\theta \\right)\\,\\text{for }y.\\hfill \\\\ \\text{ }=\\left({b}^{2}{\\mathrm{cos}}^{2}\\theta -2bc\\mathrm{cos}\\,\\theta +{c}^{2}\\right)+{b}^{2}{\\mathrm{sin}}^{2}\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Expand the perfect square}.\\hfill \\\\ \\text{ }={b}^{2}{\\mathrm{cos}}^{2}\\theta +{b}^{2}{\\mathrm{sin}}^{2}\\theta +{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Group terms noting that }{\\mathrm{cos}}^{2}\\theta +{\\mathrm{sin}}^{2}\\theta =1.\\hfill \\\\ \\text{ }={b}^{2}\\left({\\mathrm{cos}}^{2}\\theta +{\\mathrm{sin}}^{2}\\theta \\right)+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Factor out }{b}^{2}.\\hfill \\\\ {a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\theta \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\end{array}$\n\nThe formula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion.\n\nKeep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve.\n\n### Law of Cosines\n\nThe Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in (Figure), with angles$\\,\\alpha ,\\beta ,$ and$\\,\\gamma ,$ and opposite corresponding sides$\\,a,b,$ and$\\,c,\\,$respectively, the Law of Cosines is given as three equations.\n\n$\\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\\,\\,\\mathrm{cos}\\,\\alpha \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\,\\,\\mathrm{cos}\\,\\beta \\\\ {c}^{2}={a}^{2}+{b}^{2}-2ab\\,\\,\\mathrm{cos}\\,\\gamma \\end{array}$",
null,
"Figure 3.\n\nTo solve for a missing side measurement, the corresponding opposite angle measure is needed.\n\nWhen solving for an angle, the corresponding opposite side measure is needed. We can use another version of the Law of Cosines to solve for an angle.\n\n$\\begin{array}{l}\\hfill \\\\ \\begin{array}{l}\\begin{array}{l}\\hfill \\\\ \\mathrm{cos}\\text{ }\\alpha =\\frac{{b}^{2}+{c}^{2}-{a}^{2}}{2bc}\\hfill \\end{array}\\hfill \\\\ \\mathrm{cos}\\text{ }\\beta =\\frac{{a}^{2}+{c}^{2}-{b}^{2}}{2ac}\\hfill \\\\ \\mathrm{cos}\\text{ }\\gamma =\\frac{{a}^{2}+{b}^{2}-{c}^{2}}{2ab}\\hfill \\end{array}\\hfill \\end{array}$\n\nGiven two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle.\n\n1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles.\n2. Apply the Law of Cosines to find the length of the unknown side or angle.\n3. Apply the Law of Sines or Cosines to find the measure of a second angle.\n4. Compute the measure of the remaining angle.\n\n### Finding the Unknown Side and Angles of a SAS Triangle\n\nFind the unknown side and angles of the triangle in (Figure).",
null,
"Figure 4.\n\n### Try It\n\nFind the missing side and angles of the given triangle:$\\,\\alpha =30°,\\,\\,b=12,\\,\\,c=24.$\n\n### Solving for an Angle of a SSS Triangle\n\nFind the angle$\\,\\alpha \\,$for the given triangle if side$\\,a=20,\\,$side$\\,b=25,\\,$and side$\\,c=18.$\n\n#### Analysis\n\nBecause the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method.\n\n### Try It\n\nGiven$\\,a=5,b=7,\\,$and$\\,c=10,\\,$find the missing angles.\n\n### Solving Applied Problems Using the Law of Cosines\n\nJust as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few.\n\n### Using the Law of Cosines to Solve a Communication Problem\n\nOn many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5050 feet from the first tower and 2420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway.\n\n### Calculating Distance Traveled Using a SAS Triangle\n\nReturning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in (Figure).",
null,
"Figure 8.\n\n### Using Heron’s Formula to Find the Area of a Triangle\n\nWe already learned how to find the area of an oblique triangle when we know two sides and an angle. We also know the formula to find the area of a triangle using the base and the height. When we know the three sides, however, we can use Heron’s formula instead of finding the height. Heron of Alexandria was a geometer who lived during the first century A.D. He discovered a formula for finding the area of oblique triangles when three sides are known.\n\n### Heron’s Formula\n\nHeron’s formula finds the area of oblique triangles in which sides$\\,a,b\\text{,}$and$\\,c\\,$are known.\n\n$\\text{Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}$\n\nwhere$\\,s=\\frac{\\left(a+b+c\\right)}{2}\\,$ is one half of the perimeter of the triangle, sometimes called the semi-perimeter.\n\n### Using Heron’s Formula to Find the Area of a Given Triangle\n\nFind the area of the triangle in (Figure) using Heron’s formula.",
null,
"Figure 9.\n\n### Try It\n\nUse Heron’s formula to find the area of a triangle with sides of lengths$\\,a=29.7\\,\\text{ft},b=42.3\\,\\text{ft},\\,$and$\\,c=38.4\\,\\text{ft}.$\n\n### Applying Heron’s Formula to a Real-World Problem\n\nA Chicago city developer wants to construct a building consisting of artist’s lofts on a triangular lot bordered by Rush Street, Wabash Avenue, and Pearson Street. The frontage along Rush Street is approximately 62.4 meters, along Wabash Avenue it is approximately 43.5 meters, and along Pearson Street it is approximately 34.1 meters. How many square meters are available to the developer? See (Figure) for a view of the city property.",
null,
"Figure 10.\n\n### Try It\n\nFind the area of a triangle given$\\,a=4.38\\,\\text{ft}\\,,b=3.79\\,\\text{ft,}\\,$and$\\,c=5.22\\,\\text{ft}\\text{.}$\n\nAccess these online resources for additional instruction and practice with the Law of Cosines.\n\n### Key Equations\n\n Law of Cosines $\\begin{array}{l}{a}^{2}={b}^{2}+{c}^{2}-2bc\\mathrm{cos}\\,\\alpha \\hfill \\\\ {b}^{2}={a}^{2}+{c}^{2}-2ac\\mathrm{cos}\\,\\beta \\hfill \\\\ {c}^{2}={a}^{2}+{b}^{2}-2abcos\\,\\gamma \\hfill \\end{array}$ Heron’s formula $\\begin{array}{l}\\text{ Area}=\\sqrt{s\\left(s-a\\right)\\left(s-b\\right)\\left(s-c\\right)}\\hfill \\\\ \\text{where }s=\\frac{\\left(a+b+c\\right)}{2}\\hfill \\end{array}$\n\n### Key Concepts\n\n• The Law of Cosines defines the relationship among angle measurements and lengths of sides in oblique triangles.\n• The Generalized Pythagorean Theorem is the Law of Cosines for two cases of oblique triangles: SAS and SSS. Dropping an imaginary perpendicular splits the oblique triangle into two right triangles or forms one right triangle, which allows sides to be related and measurements to be calculated. See (Figure) and (Figure).\n• The Law of Cosines is useful for many types of applied problems. The first step in solving such problems is generally to draw a sketch of the problem presented. If the information given fits one of the three models (the three equations), then apply the Law of Cosines to find a solution. See (Figure) and (Figure).\n• Heron’s formula allows the calculation of area in oblique triangles. All three sides must be known to apply Heron’s formula. See (Figure) and See (Figure).\n\n### Section Exercises\n\n#### Verbal\n\nIf you are looking for a missing side of a triangle, what do you need to know when using the Law of Cosines?\n\nIf you are looking for a missing angle of a triangle, what do you need to know when using the Law of Cosines?\n\nExplain what$\\,s\\,$represents in Heron’s formula.\n\nExplain the relationship between the Pythagorean Theorem and the Law of Cosines.\n\nWhen must you use the Law of Cosines instead of the Pythagorean Theorem?\n\n#### Algebraic\n\nFor the following exercises, assume$\\,\\alpha \\,$is opposite side$\\,a,\\beta \\,$ is opposite side$\\,b,\\,$and$\\,\\gamma \\,$ is opposite side$\\,c.\\,$If possible, solve each triangle for the unknown side. Round to the nearest tenth.\n\n$\\gamma =41.2°,a=2.49,b=3.13$\n\n$\\alpha =120°,b=6,c=7$\n\n$\\beta =58.7°,a=10.6,c=15.7$\n\n$\\gamma =115°,a=18,b=23$\n\n$\\alpha =119°,a=26,b=14$\n\n$\\gamma =113°,b=10,c=32$\n\n$\\beta =67°,a=49,b=38$\n\n$\\alpha =43.1°,a=184.2,b=242.8$\n\n$\\alpha =36.6°,a=186.2,b=242.2$\n\n$\\beta =50°,a=105,b=45{}_{}{}^{}$\n\nFor the following exercises, use the Law of Cosines to solve for the missing angle of the oblique triangle. Round to the nearest tenth.\n\n$\\,a=42,b=19,c=30;\\,$find angle$\\,A.$\n\n$\\,a=14,\\text{ }b=13,\\text{ }c=20;\\,$find angle$\\,C.$\n\n$\\,a=16,b=31,c=20;\\,$find angle$\\,B.$\n\n$\\,a=13,\\,b=22,\\,c=28;\\,$find angle$\\,A.$\n\n$a=108,\\,b=132,\\,c=160;\\,$find angle$\\,C.\\,$\n\nFor the following exercises, solve the triangle. Round to the nearest tenth.\n\n$A=35°,b=8,c=11$\n\n$B=88°,a=4.4,c=5.2$\n\n$C=121°,a=21,b=37$\n\n$a=13,b=11,c=15$\n\n$a=3.1,b=3.5,c=5$\n\n$a=51,b=25,c=29$\n\nFor the following exercises, use Heron’s formula to find the area of the triangle. Round to the nearest hundredth.\n\nFind the area of a triangle with sides of length 18 in, 21 in, and 32 in. Round to the nearest tenth.\n\nFind the area of a triangle with sides of length 20 cm, 26 cm, and 37 cm. Round to the nearest tenth.\n\n$a=\\frac{1}{2}\\,\\text{m},b=\\frac{1}{3}\\,\\text{m},c=\\frac{1}{4}\\,\\text{m}$\n\n$a=12.4\\text{ ft},\\text{ }b=13.7\\text{ ft},\\text{ }c=20.2\\text{ ft}$\n\n$a=1.6\\text{ yd},\\text{ }b=2.6\\text{ yd},\\text{ }c=4.1\\text{ yd}$\n\n#### Graphical\n\nFor the following exercises, find the length of side $x.$ Round to the nearest tenth.",
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"For the following exercises, find the measurement of angle$\\,A.$",
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"Find the measure of each angle in the triangle shown in (Figure). Round to the nearest tenth.",
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"Figure 11.\n\nFor the following exercises, solve for the unknown side. Round to the nearest tenth.",
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"For the following exercises, find the area of the triangle. Round to the nearest hundredth.",
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"#### Extensions\n\nA parallelogram has sides of length 16 units and 10 units. The shorter diagonal is 12 units. Find the measure of the longer diagonal.\n\nThe sides of a parallelogram are 11 feet and 17 feet. The longer diagonal is 22 feet. Find the length of the shorter diagonal.\n\nThe sides of a parallelogram are 28 centimeters and 40 centimeters. The measure of the larger angle is 100°. Find the length of the shorter diagonal.\n\nA regular octagon is inscribed in a circle with a radius of 8 inches. (See (Figure).) Find the perimeter of the octagon.",
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"Figure 12.\n\nA regular pentagon is inscribed in a circle of radius 12 cm. (See (Figure).) Find the perimeter of the pentagon. Round to the nearest tenth of a centimeter.",
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"Figure 13.\n\nFor the following exercises, suppose that$\\,{x}^{2}=25+36-60\\mathrm{cos}\\left(52\\right)\\,$represents the relationship of three sides of a triangle and the cosine of an angle.\n\nDraw the triangle.\n\nFind the length of the third side.\n\nFor the following exercises, find the area of the triangle.",
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"#### Real-World Applications\n\nA surveyor has taken the measurements shown in (Figure). Find the distance across the lake. Round answers to the nearest tenth.",
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"Figure 14.\n\nA satellite calculates the distances and angle shown in (Figure) (not to scale). Find the distance between the two cities. Round answers to the nearest tenth.",
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"Figure 15.\n\nAn airplane flies 220 miles with a heading of 40°, and then flies 180 miles with a heading of 170°. How far is the plane from its starting point, and at what heading? Round answers to the nearest tenth.\n\nA 113-foot tower is located on a hill that is inclined 34° to the horizontal, as shown in (Figure). A guy-wire is to be attached to the top of the tower and anchored at a point 98 feet uphill from the base of the tower. Find the length of wire needed.",
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"Figure 16.\n\nTwo ships left a port at the same time. One ship traveled at a speed of 18 miles per hour at a heading of 320°. The other ship traveled at a speed of 22 miles per hour at a heading of 194°. Find the distance between the two ships after 10 hours of travel.\n\nThe graph in (Figure) represents two boats departing at the same time from the same dock. The first boat is traveling at 18 miles per hour at a heading of 327° and the second boat is traveling at 4 miles per hour at a heading of 60°. Find the distance between the two boats after 2 hours.",
null,
"Figure 17.\n\nA triangular swimming pool measures 40 feet on one side and 65 feet on another side. These sides form an angle that measures 50°. How long is the third side (to the nearest tenth)?\n\nA pilot flies in a straight path for 1 hour 30 min. She then makes a course correction, heading 10° to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 680 miles per hour, how far is she from her starting position?\n\nLos Angeles is 1,744 miles from Chicago, Chicago is 714 miles from New York, and New York is 2,451 miles from Los Angeles. Draw a triangle connecting these three cities, and find the angles in the triangle.\n\nPhiladelphia is 140 miles from Washington, D.C., Washington, D.C. is 442 miles from Boston, and Boston is 315 miles from Philadelphia. Draw a triangle connecting these three cities and find the angles in the triangle.\n\nTwo planes leave the same airport at the same time. One flies at 20° east of north at 500 miles per hour. The second flies at 30° east of south at 600 miles per hour. How far apart are the planes after 2 hours?\n\nTwo airplanes take off in different directions. One travels 300 mph due west and the other travels 25° north of west at 420 mph. After 90 minutes, how far apart are they, assuming they are flying at the same altitude?\n\nA parallelogram has sides of length 15.4 units and 9.8 units. Its area is 72.9 square units. Find the measure of the longer diagonal.\n\nThe four sequential sides of a quadrilateral have lengths 4.5 cm, 7.9 cm, 9.4 cm, and 12.9 cm. The angle between the two smallest sides is 117°. What is the area of this quadrilateral?\n\nThe four sequential sides of a quadrilateral have lengths 5.7 cm, 7.2 cm, 9.4 cm, and 12.8 cm. The angle between the two smallest sides is 106°. What is the area of this quadrilateral?\n\nFind the area of a triangular piece of land that measures 30 feet on one side and 42 feet on another; the included angle measures 132°. Round to the nearest whole square foot.\n\nFind the area of a triangular piece of land that measures 110 feet on one side and 250 feet on another; the included angle measures 85°. Round to the nearest whole square foot.\n\n### Glossary\n\nLaw of Cosines\nstates that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle\nGeneralized Pythagorean Theorem\nan extension of the Law of Cosines; relates the sides of an oblique triangle and is used for SAS and SSS triangles"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8429394,"math_prob":0.9973957,"size":17122,"snap":"2021-43-2021-49","text_gpt3_token_len":4571,"char_repetition_ratio":0.18670405,"word_repetition_ratio":0.15331544,"special_character_ratio":0.26480553,"punctuation_ratio":0.14089912,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9998448,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72],"im_url_duplicate_count":[null,3,null,3,null,3,null,3,null,3,null,3,null,3,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,3,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,6,null,3,null,3,null,6,null,6,null,6,null,3,null,3,null,3,null,3,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-20T23:41:27Z\",\"WARC-Record-ID\":\"<urn:uuid:c1dbd246-61e9-4440-a85d-fd7892357ae5>\",\"Content-Length\":\"92776\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddf14389-7763-40f7-b7d0-2c5a35e7cbc5>\",\"WARC-Concurrent-To\":\"<urn:uuid:1b2c50c4-6b78-458e-8f4d-7b29555dcb2f>\",\"WARC-IP-Address\":\"23.185.0.1\",\"WARC-Target-URI\":\"https://courses.lumenlearning.com/suny-osalgebratrig/chapter/non-right-triangles-law-of-cosines/\",\"WARC-Payload-Digest\":\"sha1:VHKB355B5SNPZRQIP4PLKO5LSUYMSF5N\",\"WARC-Block-Digest\":\"sha1:44KJL5CIXRC4SIIKNOALBM6KJBSSMKZ5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585353.52_warc_CC-MAIN-20211020214358-20211021004358-00451.warc.gz\"}"} |
http://bba.pz10.com/2013/07/static-dynamic-multiplier.html | [
"## Static Dynamic Multiplier\n\nDepending on the purpose of analysis sometimes a distinction is made between the static multiplier and the dynamic multiplier. The static multiplier is also called comparative static multiple simultaneous multiplier, logical multiplier, timeless multiplier, legless multiplier and instant multiplier.\n\nStatic Multiplier\nThe concept of static multiplier implies that changes in investment causes change in income instantaneously. It means that there is no time lag between the change in investment and the change in income. It implies that the moment a rupee is spent on investment project, society’s income increases by a multiple. Let us explain the concept of the dynamic multiplier also known as period and sequence multiplier.\nDynamic Multiplier\nThe concept of dynamic multiplier recognizes the fact that the overall change in income as a result of the change in investment is not instantaneous. There is a gradual process by which income change as a result of change in investment or other determinants of income. The process of change in income involves a time lag. The multiplier process works through the process of income generation and consumption expenditure. The dynamic multiplier takes into account the dynamic process of the change in income and the change in consumption at different stages due to change in investment. The dynamic multiplier is essentially a stage-by stage computation of the change in income resulting from the change in investment till the full effect of the multiplier is realize."
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https://solvedlib.com/n/three-systems-of-linear-equations-are-shown-for-each-system,14226005 | [
"# Three systems of linear equations are shown For each system choose the best description of its solution_ If the system\n\n###### Question:\n\nThree systems of linear equations are shown For each system choose the best description of its solution_ If the system has exactly one solution, give its solution: System A System B System € Line 1:y= X+3 Line 1: V=- X+ 2 Line 1:y= X-4 Line 2: y= x-3 Line 2: y=-x+3 Line 2: -x+4y=-16 The system has exactly one solution. The system has exactly one solution_ The system has exactly one solution_ Solution: Solution: Solution: ( D) The system has infinitely many solutions_ The system has no solution. The system has infinitely many solutions_ The system has no solution. The system has infinitely many solutions_ The system has no solution.",
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"",
null,
"#### Similar Solved Questions\n\n##### QUESTION 34 Jim saw a decrease in the quantity demanded for his firm's product from 8000 to 6000 units a week when...\nQUESTION 34 Jim saw a decrease in the quantity demanded for his firm's product from 8000 to 6000 units a week when he raised the price of the product from $200 to$250 Based on this information, the price elasticity of demand for Jim's product i. b. less than 1 than...\n##### What is the correct IUPAC name of the compound shown? NHCH; 1-methyl aniline 1-methyl-2-aniline 2-methyl aniline N-methyl aniline 1-methyl-1-aniline\nWhat is the correct IUPAC name of the compound shown? NH CH; 1-methyl aniline 1-methyl-2-aniline 2-methyl aniline N-methyl aniline 1-methyl-1-aniline...\n##### ι υ.abe Χ2 x' a-e.-. El芴: AaBb( :i': d. , 4. You are the health education...\nι υ.abe Χ2 x' a-e.-. El芴: AaBb( :i': d. , 4. You are the health education specialist for a large city hospital. Your supervisor has asked you to develop a program on safer sex practices for people who identify as gay and lesbian. The program is to be made available to...\n##### Chow Company has gathered the following information on an investment project: ANNUAL CUSI LIILLOWS . ....\nChow Company has gathered the following information on an investment project: ANNUAL CUSI LIILLOWS . . . . . . . . . . . . . . . . . Initial investment .................... Annual cash inflows. Cost of capital ......... Life of project ............. ........ $169,809$ 30,000 20% 12 years Calculate ...\n##### In each scenario below, find the critical value that would be used in the confidence interval. Round each answer to EXACTLY THREE decimal places_A95% Confidence Interval for a population mean, using a sample of size 43 with a sample mean of 74.3 and a sample standard deviation of 7.2_A 90% Confidence Interval for a population mean_ using sample of size 68 with a sample mean of 62.7 and a population standard deviation of 3.9.fora population proportion using sample with of size Al9594 Confidence I\nIn each scenario below, find the critical value that would be used in the confidence interval. Round each answer to EXACTLY THREE decimal places_ A95% Confidence Interval for a population mean, using a sample of size 43 with a sample mean of 74.3 and a sample standard deviation of 7.2_ A 90% Confide...\n##### = H radicals derivallve L ol Iho funcllon ((xY) xy + 2y at Ihe polnt P( = 1 creclon Lolthe 1 '941\n= H radicals derivallve L ol Iho funcllon ((xY) xy + 2y at Ihe polnt P( = 1 creclon Lolthe 1 '941...\n##### Apply Concepts Consider a point on the edge of a rotating wheel.a. Under what conditions can the centripetal acceleration be zero?b. Under what conditions can the tangential (linear) acceleration be zero?c. Can the tangential acceleration be nonzero while the centripetal acceleration is zero? Explain.d. Can the centripetal acceleration be nonzero while the tangential acceleration is zero? Explain.\nApply Concepts Consider a point on the edge of a rotating wheel. a. Under what conditions can the centripetal acceleration be zero? b. Under what conditions can the tangential (linear) acceleration be zero? c. Can the tangential acceleration be nonzero while the centripetal acceleration is zero? Exp...\n##### Point) Write limits of integration for the integral Jw 9( €,y, 2) dV , where W is the quarter cylinder shown; if the length of the cylinde and its radius is 2,Jw 9(z,y,2) dV = J4 Jd f' g(z,y,2) d x where a b =d =~sqrt(4-Y^2) Lnand f = sqrt(4-y^2) Wa Ctn\npoint) Write limits of integration for the integral Jw 9( €,y, 2) dV , where W is the quarter cylinder shown; if the length of the cylinde and its radius is 2, Jw 9(z,y,2) dV = J4 Jd f' g(z,y,2) d x where a b = d = ~sqrt(4-Y^2) Ln and f = sqrt(4-y^2) Wa Ctn...\n##### QUESTION 3 Clear Sound Ltd. produces stereo speakers. The company is considering a modification to one...\nQUESTION 3 Clear Sound Ltd. produces stereo speakers. The company is considering a modification to one of its models - CSIIX, that would make it more attractive to customers. One unit of CSIIX currently costs RM250 and sells for RM300. Unit cost RM150 90 Variable manufacturing cost Fixed manufacturi..."
]
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https://www.matchfishtank.org/curriculum/math/8th-grade/solving-one-variable-equations/ | [
"# Solving One-Variable Equations\n\nStudents hone their skills of solving multi-step equations and inequalities, redefining their definition of \"solution\" to include results such as 4 = 5, and interpreting solutions in context.\n\n## Unit Summary\n\nIn Unit 2, eighth-grade students hone their skills of solving equations and inequalities. They encounter complex-looking, multi-step equations, and they discover that by using properties of operations and combining like terms, these equations boil down to simple one- and two-step equations. Students also discover that there are many different ways to approach solving a multi-step equation, and they spend time closely looking at their own work and the work of their peers. When solving an equation with variables on both sides of the equal sign, students are challenged with results such as 4=5, and they refine their definition of “solution” to include such examples. Throughout this unit, students use equations as models to capture real-world applications. They reason abstractly and quantitatively as they de-contextualize situations to represent them with symbols and then re-contextualize the numbers to make sense in context (MP.2).\n\nIn sixth grade, students developed the conceptual understanding of how the components of expressions and equations work. They learned how the distributive property can create equivalent forms of an expression and how combining like terms can turn an expression with three terms into an expression with one term. By the end of seventh grade, students fluently solved one- and two-step equations with rational numbers and used equations and inequalities to represent and solve word problems.\n\nStudents’ ability to manipulate and transform equations will be required again in Unit 5: Linear Relationships and Unit 6: Systems of Linear Equations. Furthermore, these skills will be needed throughout high school as students are introduced to new types of equations involving radicals, exponents, multiple variables, and more.\n\nPacing: 16 instructional days (12 lessons, 3 flex days, 1 assessment day)\n\nFor guidance on adjusting the pacing for the 2020-2021 school year due to school closures, see our 8th Grade Scope and Sequence Recommended Adjustments.",
null,
"• Expanded Assessment Package\n• Problem Sets for Each Lesson\n• Student Handout Editor\n• Vocabulary Package\n\n• Inequalities\n\n• Equations\n\n## Assessment\n\nThis assessment accompanies Unit 2 and should be given on the suggested assessment day or after completing the unit.\n\n## Unit Prep\n\n### Intellectual Prep\n\n?\n\n#### Internalization of Standards via the Unit Assessment\n\n• Take unit assessment. Annotate for:\n• Standards that each question aligns to\n• Strategies and representations used in daily lessons\n• Relationship to Essential Understandings of unit\n• Lesson(s) that assessment points to\n\n#### Internalization of Trajectory of Unit\n\n• Read and annotate “Unit Summary.”\n• Notice the progression of concepts through the unit using “Unit at a Glance”\n• Essential understandings\n• Connection to assessment questions\n• Identify key opportunities to engage students in academic discourse. Read through our Guide to Academic Discourse and refer back to it throughout the unit.\n\n### Essential Understandings\n\n?\n\n• Solving an equation or inequality involves changing the equation into equivalent, simpler forms in order to reveal possible values for the variable.\n• Solving an equation or inequality involves using properties of operations, combining like terms, and using inverse operations.\n• An equation in one-variable may have a unique solution for the variable, no possible solutions for the variable, or all possible values may be solutions for the variable.\n\n?\n\n• calculators\n\n### Vocabulary\n\n?\n\ninfinite solutions\n\ncombine like terms\n\nterm\n\ndistribute a negative\n\nproperties of operations\n\ninequality\n\nexpression\n\nsolution\n\nunique solution\n\nequation\n\ndistributive property\n\nno solution\n\nbreak even\n\n## Common Core Standards\n\nKey: Major Cluster Supporting Cluster Additional Cluster\n\n### Core Standards\n\n?\n\n##### Expressions and Equations\n• 8.EE.C.7 — Solve linear equations in one variable.\n\n• 8.EE.C.7.A — Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).\n\n• 8.EE.C.7.B — Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.\n\n##### Reasoning with Equations and Inequalities\n• A.REI.B.3 — Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters.\n\n?\n\n• 6.EE.A.1\n\n• 6.EE.A.2\n\n• 6.EE.A.3\n\n• 6.EE.B.5\n\n• 6.EE.B.7\n\n• 7.EE.A.1\n\n• 7.EE.B.4\n\n• 7.EE.B.4.A\n\n• 7.EE.B.4.B\n\n?\n\n• A.CED.A.1\n\n• A.CED.A.3\n\n• A.CED.A.4\n\n• 8.EE.C.8\n\n• A.REI.A.1\n\n• A.REI.B.3\n\n### Standards for Mathematical Practice\n\n• CCSS.MATH.PRACTICE.MP1 — Make sense of problems and persevere in solving them.\n\n• CCSS.MATH.PRACTICE.MP2 — Reason abstractly and quantitatively.\n\n• CCSS.MATH.PRACTICE.MP3 — Construct viable arguments and critique the reasoning of others.\n\n• CCSS.MATH.PRACTICE.MP4 — Model with mathematics.\n\n• CCSS.MATH.PRACTICE.MP5 — Use appropriate tools strategically.\n\n• CCSS.MATH.PRACTICE.MP6 — Attend to precision.\n\n• CCSS.MATH.PRACTICE.MP7 — Look for and make use of structure.\n\n• CCSS.MATH.PRACTICE.MP8 — Look for and express regularity in repeated reasoning."
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null
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https://github.global.ssl.fastly.net/software/square-roots-print-activity-answer-key.html | [
"# Square roots print activity answer key. Free Math Worksheets (pdfs) with answer keys on Algebra I, Geometry, Trigonometry, Algebra II, and Calculus\n\nSquare roots print activity answer key Rating: 8,1/10 1421 reviews\n\n## Estimating square roots worksheet, 7th grade math",
null,
"Square Root - Fractions Perfect Squares Each question contains perfect square numbers in both numerator and denominator. Square Root Charts Perfect Square Root Chart: Download this printable chart and keep it for your reference. Non Perfect Squares Simplify the numerator and denominator and then find the square root. Click on the different category headings to find out more. You can also change some of your preferences. Find printable worksheets, math activities, coloring pages for kids! This option is useful for algebra 1 and 2 courses. Remember to refer to the game below for more practice: This skill is an algebra 2 skill which comes handy for 7 th graders who will later have to solve much challenging problems.\n\nNext\n\n## Solving Square Root Answer Key Worksheets",
null,
"Perfect squares are squares of whole numbers i. Students begin their study of algebra in Books 1-4 using only integers. These don't differ much from radical equations. When we square a value, we are multiplying that value by the value itself. It is sometimes fun breaking apart numbers. Each one has model problems worked out step by step, practice problems, as well as challenge questions at the sheets end. It makes you into a mathematical hulk, of sorts! Please be aware that this might heavily reduce the functionality and appearance of our site.\n\nNext\n\n## Solving Square Root Answer Key Worksheets",
null,
"Next\n\n## Finding Squares and Square Roots Worksheets",
null,
"If you choose to allow non-perfect squares, the answer is typically an unending decimal that is rounded to a certain number of digits. Lowest common multiple worksheet for 6th grade children. Plus each one comes with an answer key. You can also make worksheets that include one or two other operations, besides taking a square root. Here are several worksheets covering various topics about square roots. Some of the worksheets displayed are Square root equations, Solving quadratic roots, Solving quadratic equations square root law, Square roots work, Math 154b name the square root property work the, Quadratic and square root functions, Math 154b name completing the square work, Square root property.\n\nNext\n\n## Square and Cube Roots Worksheets",
null,
"Learn how to find the value of a square root, evaluate perfect squares, and deal with squares of imaginary numbers. The answer key is automatically generated and is placed on the second page of the file. Math is one of the initial subjects that kids are taught and there is a particular reason for that. To download a worksheet image — just click on the worksheet, right click and choose the save image option. Solving Square Root Answer Key Answer Key Showing top 8 worksheets in the category - Solving Square Root Answer Key Answer Key.\n\nNext\n\n## Estimating Square Roots Worksheet",
null,
"All of your worksheets are now here on mathwarehouse. We can estimate the square roots of numbers between the square numbers. Parents: Work with your child to give them extra practice, to help them learn a math skill or to keep their skills fresh over school breaks. Also, like any other skills, when you practice math, you are sure to become perfect in it. This math worksheet will also serve as a math test or printable math quiz for teachers and parents of 7 th graders. Remember that square roots refer to the inverse operation of squaring a number. The kids will apply the basic operation skills and find the square root of each digit.\n\nNext\n\n## Free square root worksheets (PDF and html)",
null,
"In this worksheet, we will practise estimating square roots. There are square roots tables which students can use to review and also develop mental abilities for some commonly used numbers. You should know the squares from 1 to 15. Feel free to download and enjoy these free worksheets on functions and relations. It is an easy and interesting worksheet for children. Books 8-10 extend coverage to the real number system.\n\nNext\n\n## Square roots math worksheet for 6th grade children",
null,
"This is a square roots worksheet for 7 th grade. New concepts are explained in simple language, and examples are easy to follow. Answers in Hundredth Place Though the answers are hundredth places, the questions are easy to solve. You can also find the list of worksheets by grade levels on our website. For example, 5 squared equals 25, however the inverse of this i.\n\nNext\n\n## Square and Cube Roots Worksheets",
null,
"Method We know the square numbers from 1 2 to 12 2 i. At our website, you can create printable worksheets for math for a different of topics, including all the fundamental operations, money, clock, measuring, money, decimals, fractions, proportions, percent, factoring, ratios, expressions, equations, square roots, and geometry. Calculator Worksheets Use calculator or long division method to find the value of the square root rounding to two decimal places. Then ask the students to measure the , sides etc. Remember the second page contains an answer key to all the problems; hence referencing is easy. The kids will learn to calculate the root of each given number in this worksheet.\n\nNext\n\n## Estimating Square Roots Worksheet",
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"https://i.pinimg.com/originals/d4/2b/c9/d42bc9a04319df39fe750a25f931f6c4.png",
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https://text.123doc.net/document/4955607-mathmatical-method-and-algorithms-for-signal-processing.htm | [
"# Mathmatical method and algorithms for signal processing\n\nMathematical Methods and Algorithms\nfor\nSignal Processing\nTodd K. Moon\nUtah State University\n\nWynn С Stirling\nBrigham Young University\n\nPRENTICE HALL\n\nThis previously included a CD. The\nCD contents can now be accessed\nat www.prenhall.com/moon. Thank You.\n\nContents\n1\n1\n\nII\n2\n\nIntroduction and Foundations\n\n1\n\nIntroduction and Foundations\n1.1\nWhat is signal processing?\n1.2\nMathematical topics embraced by signal processing\n1.3\nMathematical models\n1.4\nModels for linear systems and signals\n1.4.1 Linear discrete-time models\n1.4.2 Stochastic MA and AR models\n1.4.3 Continuous-time notation\n1.4.4 Issues and applications\n1.4.5 Identification of the modes\n1.4.6 Control of the modes\n1.5\nfiltering\n1.5.1 System identification\n1.5.2 Inverse system identification\n1.5.4 Interference cancellation\n1.6\nGaussian random variables and random processes\n1.6.1 Conditional Gaussian densities\n1.7\nMarkov and Hidden Markov Models\n1.7.1 Markov models\n1.7.2 Hidden Markov models\n1.8\nSome aspects of proofs\n1.8.1 Proof by computation: direct proof\n\n1.8.3 Proof by induction\n1.9\nAn application: LFSRs and Massey's algorithm\n1.9.1 Issues and applications of LFSRs\n1.9.2 Massey's algorithm\n1.9.3 Characterization of LFSR length in Massey's algorithm\n1.10 Exercises\n1.11 References\n\n3\n3\n5\n6\n7\n7\n12\n20\n21\n26\n28\n28\n29\n29\n29\n30\n31\n36\n37\n37\n39\n41\n43\n45\n46\n48\n50\n52\n53\n58\n67\n\nVector Spaces and Linear Algebra\n\n69\n\nSignal Spaces\n2.1\nMetric spaces\n2.1.1 Some topological terms\n2.1.2 Sequences, Cauchy sequences, and completeness\n\n71\n72\n76\n78\n\nContents\n\n2.2\n\n2.3\n2.4\n2.5\n2.6\n2.7\n2.8\n2.9\n2.10\n2.11\n2.12\n2.13\n2.14\n2.15\n2.16\n2.17\n2.18\n\n2.1.3 Technicalities associated with the Lp and L^ spaces\nVector spaces\n2.2.1\nLinear combinations of vectors\n2.2.2\nLinear independence\n2.2.3\nBasis and dimension\n2.2.4\nFinite-dimensional vector spaces and matrix notation\nNorms and normed vector spaces\n2.3.1\nFinite-dimensional normed linear spaces\nInner products and inner-product spaces\n2.4.1\nWeak convergence\nInduced norms\nThe Cauchy-Schwarz inequality\nDirection of vectors: Orthogonality\nWeighted inner products\n2.8.1\nExpectation as an inner product\nHilbert and Banach spaces\nOrthogonal subspaces\nLinear transformations: Range and nullspace\nInner-sum and direct-sum spaces\nProjections and orthogonal projections\n2.13.1 Projection matrices\nThe projection theorem\nOrthogonalization of vectors\nSome final technicalities for infinite dimensional spaces\nExercises\nReferences\n\nRepresentation and Approximation in Vector Spaces\n3.1\nThe approximation problem in Hilbert space\n3.1.1\nThe Grammian matrix\n3.2\nThe orthogonality principle\n3.2.1\nRepresentations in infinite-dimensional space\n3.3\n3.4\nMatrix representations of least-squares problems\n3.4.1\nWeighted least-squares\n3.4.2\nStatistical properties of the least-squares estimate\n3.5\nMinimum error in Hilbert-space approximations\n\n82\n84\n87\n88\n90\n93\n93\n97\n97\n99\n99\n100\n101\n103\n105\n106\n107\n108\n110\n113\n115\n116\n118\n121\n121\n129\n130\n130\n133\n135\n136\n137\n138\n140\n140\n141\n\nApplications of the orthogonality theorem\n3.6\n3.7\n3.8\n3.9\n\n3.10\n3.11\n3.12\n3.13\n\nApproximation by continuous polynomials\nApproximation by discrete polynomials\nLinear regression\nLeast-squares\nfiltering\n3.9.1\nLeast-squares prediction and AR spectrum\nestimation\nMinimum mean-square estimation\nMinimum mean-squared error (MMSE)\nfiltering\nComparison of least squares and minimum mean squares\nFrequency-domain optimal\nfiltering\n3.13.1 Brief review of stochastic processes and\nLaplace transforms\n\n143\n145\n147\n149\n154\n156\n157\n161\n162\n162\n\nContents\n\n3.13.2\n\n3.14\n3.15\n3.16\n3.17\n3.18\n\n3.19\n\n3.20\n3.21\n4\n\nTwo-sided Laplace transforms and their\ndecompositions\n3.13.3 The Wiener-Hopf equation\n3.13.4 Solution to the Wiener-Hopf equation\n3.13.5 Examples of Wiener\nfiltering\n3.13.6 Mean-square error\n3.13.7 Discrete-time Wiener filters\nA dual approximation problem\nMinimum-norm solution of underdetermined equations\nIterative Reweighted LS (IRLS) for Lp optimization\nSignal transformation and generalized Fourier series\nSets of complete orthogonal functions\n3.18.1 Trigonometric functions\n3.18.2 Orthogonal polynomials\n3.18.3 Sine functions\n3.18.4 Orthogonal wavelets\nSignals as points: Digital communications\n3.19.1 The detection problem\n3.19.2 Examples of basis functions used in digital\ncommunications\n3.19.3 Detection in nonwhite noise\nExercises\nReferences\n\nLinear Operators and Matrix Inverses\n4.1\nLinear operators\n4.1.1\nLinear functionals\n4.2\nOperator norms\n4.2.1\nBounded operators\n4.2.2\nThe Neumann expansion\n4.2.3\nMatrix norms\n4.3\n4.3.1\nA dual optimization problem\n4.4\nGeometry of linear equations\n4.5\nFour fundamental subspaces of a linear operator\n4.5.1\nThe four fundamental subspaces with\nnon-closed range\n4.6\nSome properties of matrix inverses\n4.6.1\nTests for invertibility of matrices\n4.7\nSome results on matrix rank\n4.7.1\nNumeric rank\n4.8\nAnother look at least squares\n4.9\nPseudoinverses\n4.10 Matrix condition number\n4.11 Inverse of a small-rank adjustment\n4.11.1 An application: the RLS\n4.11.2 Two RLS applications\n4.12 Inverse of a block (partitioned) matrix\n4.12.1 Application: Linear models\n4.13 Exercises\n4.14 References\n\n165\n169\n171\n174\n176\n176\n179\n182\n183\n186\n190\n190\n190\n193\n194\n208\n210\n212\n213\n215\n228\n229\n230\n231\n232\n233\n235\n235\n237\n239\n239\n242\n\nfilter\n\n246\n247\n248\n249\n250\n251\n251\n253\n258\n259\n261\n264\n267\n268\n274\n\nContents\n\nviii\n\n5\n\nSome Important Matrix Factorizations\n5.1\nThe LU factorization\n5.1.1\nComputing the determinant using the LU factorization\n5.1.2\nComputing the LU factorization\n5.2\nThe Cholesky factorization\n5.2.1\nAlgorithms for computing the Cholesky factorization\n5.3\nUnitary matrices and the QR factorization\n5.3.1\nUnitary matrices\n5.3.2\nThe QR factorization\n5.3.3\nQR factorization and least-squares\nfilters\n5.3.4\nComputing the QR factorization\n5.3.5\nHouseholder transformations\n5.3.6\nAlgorithms for Householder transformations\n5.3.7\nQR factorization using Givens rotations\n5.3.8\nAlgorithms for QR factorization using Givens rotations\n5.3.9\nSolving least-squares problems using Givens rotations\n5.3.10 Givens rotations via CORDIC rotations\n5.3.11 Recursive updates to the QR factorization\n5.4\nExercises\n5.5\nReferences\n\n275\n275\n277\n278\n283\n284\n285\n285\n286\n286\n287\n287\n291\n293\n295\n296\n297\n299\n300\n304\n\n6\n\nEigenvalues and Eigenvectors\n6.1\nEigenvalues and linear systems\n6.2\nLinear dependence of eigenvectors\n6.3\nDiagonalization of a matrix\n6.3.1\nThe Jordan form\n6.3.2\n6.4\nGeometry of invariant subspaces\n6.5\nGeometry of quadratic forms and the minimax principle\n6.6\nExtremal quadratic forms subject to linear constraints\n6.7\nThe Gershgorin circle theorem\n\n305\n305\n308\n309\n311\n312\n316\n318\n324\n324\n\nApplication of Eigendecomposition methods\n6.8\n6.9\n\n6.10\n\n6.11\n6.12\n\n6.13\n6.14\n\nKarhunen-Loeve low-rank approximations and principal methods —\n6.8.1\nPrincipal component methods\nEigenfilters\n6.9.1\nEigenfilters for random signals\n6.9.2\nEigenfilter for designed spectral response\n6.9.3\nConstrained eigenfilters\nSignal subspace techniques\n6.10.1 The signal model\n6.10.2 The noise model\n6.10.3 Pisarenko harmonic decomposition\n6.10.4 MUSIC\nGeneralized eigenvalues\n6.11.1 An application: ESPRIT\nCharacteristic and minimal polynomials\n6.12.1 Matrix polynomials\n6.12.2 Minimal polynomials\nMoving the eigenvalues around: Introduction to linear control\nNoiseless constrained channel capacity\n\n327\n329\n330\n330\n332\n334\n336\n336\n337\n338\n339\n340\n341\n342\n342\n344\n344\n347\n\nix\n\n6.15\n\n6.16\n6.17\n\nComputation of eigenvalues and eigenvectors\n6.15.1 Computing the largest and smallest eigenvalues\n6.15.2 Computing the eigenvalues of a symmetric matrix\n6.15.3 The QR iteration\nExercises\nReferences\n\nThe Singular Value Decomposition\n7.1\nTheory of the SVD\n7.2\nMatrix structure from the SVD\n7.3\nPseudoinverses and the SVD\n7.4\nNumerically sensitive problems\n7.5\nRank-reducing approximations: Effective rank\nApplications of the SVD\n7.6\nSystem identification using the SVD\n7.7\nTotal least-squares problems\n7.7.1 Geometric interpretation of the TLS solution\n7.8\nPartial total least squares\n7.9\nRotation of subspaces\n7.10 Computation of the SVD\n7.11 Exercises\n7.12 References\n\n350\n350\n351\n352\n355\n368\n369\n369\n372\n373\n375\n377\n378\n381\n385\n386\n389\n390\n392\n395\n\nSome Special Matrices and Their Applications\n8.1\nModal matrices and parameter estimation\n8.2\nPermutation matrices\n8.3\nToeplitz matrices and some applications\n8.3.1\nDurbin's algorithm\n8.3.2\nPredictors and lattice filters\n8.3.3\nOptimal predictors and Toeplitz inverses\n8.3.4\nToeplitz equations with a general right-hand side\n8.4\nVandermonde matrices\n8.5\nCirculant matrices\n8.5.1\nRelations among Vandermonde, circulant, and\ncompanion matrices\n8.5.2\nAsymptotic equivalence of the eigenvalues of Toeplitz and\ncirculant matrices\n8.6\nTriangular matrices\n8.7\nProperties preserved in matrix products\n8.8\nExercises\n8.9\nReferences\n\n396\n396\n399\n400\n402\n403\n407\n408\n409\n410\n\n413\n416\n417\n418\n421\n\nKronecker Products and the Vec Operator\n9.1\nThe Kronecker product and Kronecker sum\n9.2\nSome applications of Kronecker products\n9.2.1\n9.2.2\nDFT computation using Kronecker products\n9.3 The vec operator\n9.4 Exercises\n9.5 References\n\n422\n422\n425\n425\n426\n428\n431\n433\n\n412\n\nX\n\nIII\n\nDetection, Estimation, and Optimal Filtering\n\n435\n\n10\n\nIntroduction to Detection and Estimation, and Mathematical Notation\n10.1 Detection and estimation theory\n10.1.1 Game theory and decision theory\n10.1.2 Randomization\n10.1.3 Special cases\n10.2 Some notational conventions\n10.2.1 Populations and statistics\n10.3 Conditional expectation\n10.4 Transformations of random variables\n10.5 Sufficient statistics\n10.5.1 Examples of sufficient statistics\n10.5.2 Complete sufficient statistics\n10.6 Exponential families\n10.7 Exercises\n10.8 References\n\n437\n437\n438\n440\n441\n442\n443\n444\n445\n446\n450\n451\n453\n456\n459\n\n11\n\nDetection Theory\n11.1 Introduction to hypothesis testing\n11.2 Neyman-Pearson theory\n11.2.1 Simple binary hypothesis testing\n11.2.2 The Neyman-Pearson lemma\n11.2.3 Application of the Neyman-Pearson lemma\n11.2.4 The likelihood ratio and the receiver operating\ncharacteristic (ROC)\n11.2.5 A Poisson example\n11.2.6 Some Gaussian examples\n11.2.7 Properties of the ROC\n11.3 Neyman-Pearson testing with composite binary hypotheses\n11.4 Bayes decision theory\n11.4.1 The Bayes principle\n11.4.2 The risk function\n11.4.3 Bayes\nrisk\n11.4.4 Bayes tests of simple binary hypotheses\n11.4.5 Posterior distributions\n11.4.6 Detection and sufficiency\n11.4.7 Summary of binary decision problems\n11.5 Some M-ary problems\n11.6 Maximum-likelihood detection\n11.7 Approximations to detection performance: The union bound\n11.8 Invariant Tests\n11.8.1 Detection with random (nuisance) parameters\n11.9\nDetection in continuous time\n11.9.1\nSome extensions and precautions\n11.10 Minimax Bayes decisions\n11.10.1 Bayes envelope function\n11.10.2 Minimax rules\n11.10.3 Minimax Bayes in multiple-decision problems\n\n460\n460\n462\n462\n463\n466\n467\n468\n469\n480\n483\n485\n486\n487\n489\n490\n494\n498\n498\n499\n503\n503\n504\n507\n512\n516\n520\n520\n523\n524\n\nxi\n\n11.11\n11.12\n\n11.10.4 Determining the least favorable prior\n11.10.5 A minimax example and the minimax theorem\nExercises\nReferences\n\n528\n529\n532\n541\n\nEstimation Theory\n12.1\nThe maximum-likelihood principle\n12.2\nML estimates and sufficiency\n12.3\nEstimation quality\n12.3.1\nThe score function\n12.3.2\nThe Cramer-Rao lower bound\n12.3.3\nEfficiency\n12.3.4\nAsymptotic properties of maximum-likelihood\nestimators\n12.3.5\nThe multivariate normal case\n12.3.6\nMinimum-variance unbiased estimators\n12.3.7\nThe linear statistical model\n12.4\nApplications of ML estimation\n12.4.1\nARMA parameter estimation\n12.4.2\nSignal subspace identification\n12.4.3\nPhase estimation\n12.5\nBayes estimation theory\n12.6\nBayes risk\n12.6.1\nMAP estimates\np 12.6.2\nSummary\n12.6.3\nConjugate prior distributions\n12.6.4\nConnections with minimum mean-squared\nestimation\n12.6.5\nBayes estimation with the Gaussian distribution\n12.7\nRecursive estimation\n12.7.1\nAn example of non-Gaussian recursive Bayes\n12.8\nExercises\n12.9\nReferences\n\n542\n542\n547\n548\n548\n550\n552\n\n577\n578\n580\n582\n584\n590\n\nThe Kaiman Filter\n13.1\nThe state-space signal model\n13.2\nKaiman filter I: The Bayes approach\n13.3\nKaiman filter II: The innovations approach\n13.3.1\nInnovations for processes with linear observation models.\n13.3.2\nEstimation using the innovations process\n,\n13.3.3\nInnovations for processes with state-space models\n13.3.4\nA recursion for P„ r _|\n13.3.5\nThe discrete-time Kaiman\nfilter\n13.3.6\nPerspective\n13.3.7\nComparison with the RLS adaptive filter algorithm\n13.4 Numerical considerations: Square-root\nfilters\n13.5 Application in continuous-time systems\n13.5.1 Conversion from continuous time to discrete time\n13.5.2 A simple kinematic example\n13.6 Extensions of Kaiman filtering to nonlinear systems\n\n591\n591\n592\n595\n596\n597\n598\n599\n601\n602\n603\n604\n606\n606\n606\n607\n\n553\n556\n559\n561\n561\n561\n565\n566\n568\n569\n573\n574\n574\n\nxii\n\nContents\n\nSmoothing\n13.7.1 The Rauch-Tung-Streibel fixed-interval smoother\n13.8 Another approach: Я«, smoothing\n13.9 Exercises\n13.10 References\n\nIV\n14\n\n15\n\n13.7\n\n613\n613\n616\n617\n620\n\nIterative and Recursive Methods in Signal Processing\n\n621\n\nBasic Concepts and Methods of Iterative Algorithms\n14.1 Definitions and qualitative properties of iterated\nfunctions\n14.1.1 Basic theorems of iterated functions\n14.1.2 Illustration of the basic theorems\n14.2 Contraction mappings\n14.3 Rates of convergence for iterative algorithms\n14.4 Newton's method\n14.5 Steepest descent\n14.5.1 Comparison and discussion: Other techniques\nSome Applications of Basic Iterative Methods\n14.6.1 An example LMS application\n14.6.2 Convergence of the LMS algorithm\n14.7 Neural networks\n14.7.1 The backpropagation training algorithm\n14.7.2 The nonlinearity function\n14.7.3 The forward-backward training algorithm\n14.7.5 Neural network code\n14.7.6 How many neurons?\n14.7.7 Pattern recognition: ML or NN?\n14.8 Blind source separation\n14.8.1 A bit of information theory\n14.8.2 Applications to source separation\n14.8.3 Implementation aspects\n14.9 Exercises\n14.10 References\nIteration by Composition of Mappings\n15.1 Introduction\n15.2 Alternating projections\n15.2.1 An applications: bandlimited reconstruction\n15.3 Composite mappings\n15.4 Closed mappings and the global convergence theorem\n15.5 The composite mapping algorithm\n15.5.1 Bandlimited reconstruction, revisited\n15.5.2 An example: Positive sequence determination\n15.5.3 Matrix property mappings\n15.6 Projection on convex sets\n15.7 Exercises\n15.8 References\n\n623\n624\n626\n627\n629\n631\n632\n637\n642\n643\n645\n646\n648\n650\n653\n654\n654\n655\n658\n659\n660\n660\n662\n664\n665\n668\n670\n670\n671\n675\n676\n677\n680\n681\n681\n683\n689\n693\n694\n\nContents\n\nxiii\n\n16\n\nOther Iterative Algorithms\n16.1 Clustering\n16.1.1 An example application: Vector quantization\n16.1.2 An example application: Pattern recognition\n16.1.3 к -means Clustering\n16.1.4 Clustering using fuzzy к -means\n16.2 Iterative methods for computing inverses of matrices\n16.2.1 The Jacobi method\n16.2.2 Gauss-Seidel iteration\n16.2.3 Successive over-relaxation (SOR)\n16.3 Algebraic reconstruction techniques (ART)\n16.4 Conjugate-direction methods\n16.7 Exercises\n16.8 References\n\n695\n695\n695\n697\n698\n700\n701\n702\n703\n705\n706\n708\n710\n713\n713\n715\n\n17\n\nThe EM Algorithm in Signal Processing\n17.1 An introductory example\n17.2 General statement of the EM algorithm\n17.3 Convergence of the EM algorithm\n17.3.1 Convergence rate: Some generalizations\nExample applications of the EM algorithm\n17.4 Introductory example, revisited\n17.5 Emission computed tomography (ЕСТ) image reconstruction\n17.6 Active noise cancellation (ANC)\n17.7 Hidden Markov models\n17.7.1 The E-and M-steps\nr,\nr\n17.7.2 The forward and backward probabilities\n17.7.3 Discrete output densities\n17.7.4 Gaussian output densities\n17.7.5 Normalization\n17.7.6 Algorithms for HMMs\n17.9 Summary\n17.10 Exercises\n17.11 References\n\n717\n718\n721\n723\n724\n725\n725\n729\n732\n734\n735\n736\n736\n737\n738\n740\n743\n744\n747\n\nV\n\nMethods of Optimization\n\n749\n\n18\n\nTheory of Constrained Optimization\n18.1 Basic definitions\n18.2 Generalization of the chain rule to composite functions\n18.3 Definitions for constrained optimization\n18.4 Equality constraints: Lagrange multipliers\n18.4.1 Examples of equality-constrained optimization\n18.5 Second-order conditions\n18.6 Interpretation of the Lagrange multipliers\n18.7 Complex constraints\n......\n18.8 Duality in optimization\n\n751\n751\n755\n757\n758\n764\n767\n770\n773\n773\n\nContents\n\nxiv\n\n19\n\n18.9\n\nInequality constraints: Kuhn-Tucker conditions\n18.9.1 Second-order conditions for inequality constraints\n18.9.2 An extension: Fritz John conditions\n18.10 Exercises\n18.11 References\n\n777\n783\n783\n784\n786\n\nShortest-Path Algorithms and Dynamic Programming\n19.1 Definitions for graphs\n19.2 Dynamic programming\n19.3 The Viterbi algorithm\n19.4 Code for the Viterbi algorithm\n19.4.1 Related algorithms: Dijkstra's and Warshall's\n19.4.2 Complexity comparisons of Viterbi and Dijkstra\n\n787\n787\n789\n791\n795\n798\n799\n\nApplications of path search algorithms\n19.5\n\nMaximum-likelihood sequence estimation\n19.5.1 The intersymbol interference (ISI) channel\n19.5.2 Code-division multiple access\n19.5.3 Convolutional decoding\nHMM likelihood analysis and HMM training\n19.6.1 Dynamic warping\nAlternatives to shortest-path algorithms\nExercises\nReferences\n\n800\n800\n804\n806\n808\n811\n813\n815\n817\n\nLinear Programming\n20.1 Introduction to linear programming\n20.2 Putting a problem into standard form\n20.2.1 Inequality constraints and slack variables\n20.2.2 Free variables\n20.2.3 Variable-bound constraints\n20.2.4 Absolute value in the objective\n20.3 Simple examples of linear programming\n20.4 Computation of the linear programming solution\n20.4.1 Basic variables\n20.4.2 Pivoting\n20.4.3 Selecting variables on which to pivot\n20.4.4 The effect of pivoting on the value of the problem\n20.4.5 Summary of the simplex algorithm\n20.4.6 Finding the initial basic feasible solution\n20.4.7 MATLAB® code for linear programming\n20.4.8 Matrix notation for the simplex algorithm\n20.5 Dual problems\n20.6 Karmarker's algorithm for LP\n20.6.1 Conversion to Karmarker standard form\n20.6.2 Convergence of the algorithm\n20.6.3 Summary and extensions\n\n818\n818\n819\n819\n820\n822\n823\n823\n824\n824\n826\n828\n829\n830\n831\n834\n835\n836\n838\n842\n844\n846\n\n19.6\n19.7\n19.8\n19.9\n20\n\nExamples and applications of linear programming\n20.7\n20.8\n\nLinear-phase FIR filter design\n20.7.1 Least-absolute-error approximation\nLinear optimal control\n\n846\n847\n849\n\nContents\n\nxv\n\n20.9 Exercises\n20.10 References\n\n850\n853\n\nA\n\nBasic Concepts and Definitions\nA.l\nSet theory and notation\nA.2\nMappings and functions\nA.3\nConvex functions\nA.4\nО and о Notation\nA.5\nContinuity\nA.6\nDifferentiation\nA.6.1 Differentiation with a single real variable\nA.6.2 Partial derivatives and gradients on W\"\nA.6.3 Linear approximation using the gradient\nA.6.4 Taylor series\nA.7\nBasic constrained optimization\nA.8\nThe Holder and Minkowski inequalities\nA.9\nExercises\nA. 10 References\n\n855\n855\n859\n860\n861\n862\n864\n864\n865\n867\n868\n869\n870\n871\n876\n\nВ\n\nCompleting the Square\nB. 1\nThe scalar case\nB.2\nThe matrix case\nB.3\nExercises\n\n877\n877\n879\n879\n\nС\n\nBasic Matrix Concepts\nC.l\nNotational conventions\nC.2\nMatrix Identity and Inverse\nC.3\nTranspose and trace\nC.4\nBlock (partitioned) matrices\nC.5\nDeterminants\nC.5.1 Basic properties of determinants\nC.5.2 Formulas for the determinant\nC.5.3 Determinants and matrix inverses\nC.6\nExercises\nC.7\nReferences\n\n880\n880\n882\n883\n885\n885\n885\n887\n889\n889\n890\n\nD\n\nRandom Processes\nD.l\nDefinitions of means and correlations\nг D.2\nStationarity\nD.3\nPower spectral-density functions\nD.4\nLinear systems with stochastic inputs\nD.4.1 Continuous-time signals and systems\nD.4.2 Discrete-time signals and systems\nD.5\nReferences\n\n891\n891\n892\n893\n894\n894\n895\n895\n\nE\n\nE. 1 Derivatives of vectors and scalars with respect to a real vector\nE.2 Derivatives of real-valued functions of real matrices\nE.3 Derivatives of matrices with respect to scalars, and vice versa\nE.4 The transformation principle\nE.5 Derivatives of products of matrices\n\n896\n896\n897\n899\n901\n903\n903\n\nxvi\n\nContents\n\nE.6\nE.7\nE.8\nE.9\nE.10\nF\n\nDerivatives of powers of a matrix\nDerivatives involving the trace\nModifications for derivatives of complex vectors and matrices\nExercises\nReferences\n\n904\n906\n908\n910\n912\n\nConditional Expectations of Multinomial and Poisson r.v.s\nF. 1 Multinomial distributions\nF.2 Poisson random variables\nF.3 Exercises\n\n913\n913\n914\n914\n\nBibliography\n\n915\n\nIndex\n\n929\n\n&",
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"### Tài liệu bạn tìm kiếm đã sẵn sàng tải về\n\nTải bản đầy đủ ngay\n\n×"
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https://edurev.in/course/quiz/attempt/-1_Sample-BITSAT-Logical-Reasoning-Test/df5a914b-537c-4d64-9abd-210fd004260b | [
"Courses\n\n# Sample BITSAT Logical Reasoning Test\n\n## 10 Questions MCQ Test BITSAT Subject Wise & Full Length Mock Tests | Sample BITSAT Logical Reasoning Test\n\nDescription\nThis mock test of Sample BITSAT Logical Reasoning Test for JEE helps you for every JEE entrance exam. This contains 10 Multiple Choice Questions for JEE Sample BITSAT Logical Reasoning Test (mcq) to study with solutions a complete question bank. The solved questions answers in this Sample BITSAT Logical Reasoning Test quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Sample BITSAT Logical Reasoning Test exercise for a better result in the exam. You can find other Sample BITSAT Logical Reasoning Test extra questions, long questions & short questions for JEE on EduRev as well by searching above.\nQUESTION: 1\n\n### Choose the word which is least like the other words in the group.\n\nSolution: Because mustard is not famous in all worlds it is not will grow.\nQUESTION: 2\n\n### Find the missing character.",
null,
"Solution:\n\nThe sequence in first column is x 5. Thus, 1 x 5 = 5, 5 x 5 = 25, 25 x 5 = 125.\nThe sequence in third column is x 2. Thusx 7 x 2 = 14, 14 x 2 = 28, 28 x 2 = 56.\nThe sequence in second column is x 4.\n∴ Missing number = 12 x 4 = 48.\n\nQUESTION: 3\n\n### In following number series, either one term is missing or is incorrect. 1,5,11,19,29,55\n\nSolution:\n\nThe correct sequence is + 4, + 6, + 8, + 10, ....\nSo, next term after 29 = 29 + 12 = 41.\nThe term after 41 will then be (41 + 14) i.e. 55.\n.'. 41 is missing.\n\nQUESTION: 4\n\nGroup the following figures into three classes on the basis of identical properties.",
null,
"Solution:\n\n1,5,6 have two similar elements, one inside the other.\n2,3,4 contain straight lines each dividing the figure into two equal parts.\n7,8,9 have one element placed inside a different element.\n\nQUESTION: 5\n\nComplete the analogous pair.\nVoter : 18 years :: Lok Sabha Member : ?\n\nSolution:\n\nThe minimum age required to vote is 18 similarly for Lok Sabha member it is 25 years.\n\nQUESTION: 6\n\nIn this letter series, some of the letters are missing. Choose the correct letter given below -\nabca _ bcaab _ ca _ bbc _ a\n\nSolution: The series is abc / aabc / aabbc / aabbcc / a.\nQUESTION: 7\n\nSelect the correct mirror-image of the Figure (X) from amongst the given alternatives.",
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"Solution:\nQUESTION: 8\n\nSelect the correct mirror-image of the Figure (X) from amongst the given alternatives.",
null,
"Solution: Because in mirror image right become left and left become right.\nQUESTION: 9\n\nEach of the following questions consists of five Problem figures marked 1,2,3,4 and 5 followed by five Answer figures marked A,B,C,D and E . Select a figure from the Answer figures which will continue the same series as given in the Problem figures.",
null,
"Solution:\nThe symbol gets vertically inverted and laterally inverted alternately. It also moves in ACW direction through distances equal to two half-sides (of square boundary) and three half-sides alternately.\nQUESTION: 10\n\nEach of the following questions consists of five Problem figures marked 1,2,3,4 and 5 followed by five Answer figures marked A,B,C,D and E. Select a figure from the Answer figures which will continue the same series as given in the Problem figures.",
null,
"Solution:"
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"https://cdn3.edurev.in/ApplicationImages/Temp/98544822-10f5-482a-8404-e9a3decb5f85_lg.jpg",
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http://basilisk.fr/src/test/adaptbug.c | [
"# Check that user flags are properly reset when adapting\n\n#include \"fractions.h\"\n#include \"utils.h\"\n\n#define LEVEL 12\n\nstatic bool check_flags()\n{\nforeach_cell()\nforeach_neighbor()\nif (allocated(0))\nfor (int i = user; i <= user + 7; i++)\nassert (!(cell.flags & (1 << i)));\nreturn true;\n}\n\nint main(int argc, char const *argv[])\n{\nsize (10);\norigin (-L0/2., 0.);\ninit_grid (256);\n\nInitial refinement.\n\n scalar f[];\nint iteration = 0;\ndo {\nfraction (f, sq(x) + sq(y) - sq(1.));\n}\nwhile (check_flags() &&\nadapt_wavelet ({f}, (double[]){1e-3}, LEVEL).nf != 0 &&\niteration++ <= 10);\n\nfor (int i = 0; i <= 5; i++) {\nscalar s[];\nforeach()\ns[] = noise();\nboundary ({s});\n\nWe adapt noise with zero tolerance i.e. the mesh should be refined everywhere down to LEVEl. The bug was triggered when minlevel is set.\n\n check_flags();\nadapt_wavelet ({s}, (double []){0}, maxlevel = LEVEL, minlevel = 5);\n\nThis should eventually give a uniform refinement.",
null,
"Refinement levels\n\n scalar l[];\nforeach()\nl[] = level;\noutput_ppm (l, min = 5, max = LEVEL, file = \"grid.gif\");\n}\n}"
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"http://basilisk.fr/src/test/adaptbug/grid.gif",
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https://www.educationquizzes.com/11-plus/maths/multiplication-and-division-2-easy/ | [
"",
null,
"14 x 12 = 3, whether it's oranges, numbers, or anything else.\n\n# Multiplication and Division 2 (Easy)\n\nIn the previous quiz we learned how to perform multiplication and division with fractions – I hope you didn’t find it too difficult! In this, the second of our easy level quizzes on the topic, we look at multiplying and dividing decimal numbers, as well as fractions and whole numbers.\n\nDecimal numbers don’t pose too much of a problem – just treat them the same as whole numbers and make sure you correctly place the decimal point. Making an estimate of the answer before you try any calculations should help with this. As far as fractions go, well, practice makes perfect! Keep playing these quizzes until you feel confident enough to move on to the next level.\n\nAs ever, be sure to read each question thoroughly – and make sure you understand it – before you choose your answer. And don’t forget the helpful comments after each question. You’ll find many useful tips hidden away in them.\n\n1.\nWhat is the correct answer to the given calculation?\n97 ÷ 9\n10.9\n1034\n10.7\n1079\n97 ÷ 9 = 10 remainder 7. As we are dealing with ninths, the remainder becomes 79\n2.\nWhat is the correct answer to the given calculation?\n18 × 2.4\n48.6\n36.4\n40.5\n43.2\nTreat decimals just like whole numbers, but remember to correctly place the decimal point. 18 x 2 = 36 and 18 x 3 = 54, so your answer must fall between these two values (i.e. 43.2 rather than 432)\n3.\nWhat is the correct answer to the given calculation?\n35 ÷ 212\n28\n87.5\n14\n56.7\nTo make this easier, first convert 212 into halves:\n35 ÷ 212 = 35 ÷ 52\nNext, invert (turn upside down) the fraction and multiply:\n35 ÷ 52 = 35 × 25 = (35 × 2) ÷ 5 = 70 ÷ 5 = 14\n4.\nWhat is the correct answer to the given calculation?\n24 ÷ 68\n16\n18\n32\n36\nTo divide by a fraction, invert (turn upside down) the fraction and multiply. LEARN this technique. So, 24 ÷ 68 = 24 × 86 = (24 × 8) ÷ 6 = 192 ÷ 6 = 32\n5.\nWhat is the correct answer to the given calculation?\n1.7 × 110\n170\n17\n1.7\n0.17\nRemember that multiplying by 1 over a number is the same as dividing that number: in this case, 10:\n1.7 ÷ 10 = 0.17\n6.\nWhat is the correct answer to the given calculation?\n19 × 13\n58\n613\n57\n623\nIn general, multiplying by 1 over a number is the same as dividing by that number: in this case, 3:\n19 ÷ 3 = 6.333 or 613\n7.\nWhat is the correct answer to the given calculation?\n10 ÷ 13\n30\n3.3\n313\n33\nRemember that dividing by 1 over a number is the same as multiplying by that number: in this case, 3:\n10 x 3 = 30\n8.\nWhat is the correct answer to the given calculation?\n142 x 13\n1,846\n1,704\n1,988\n1,820\nLooking at the last digit in a number can provide a clue:\n2 x 3 = 6 so the answer to 142 x 13 must end with a 6\n9.\nWhat is the correct answer to the given calculation?\n32 ÷ 34\n24\n4223\n48\n4213\nTo divide by a fraction, invert (turn upside down) the fraction and multiply. LEARN this technique. So, 32 ÷ 34 = 32 × 43 = (32 × 4) ÷ 3 = 128 ÷ 3 = 42.666 or 4223\n10.\nWhat is the correct answer to the given calculation?\n16 x 120\n60\n20\n30\n40\nAs you know, multiplying by 1 over a number is the same as dividing by that number: in this case, 6:\n120 ÷ 6 = 20\nAuthor: Frank Evans"
]
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https://www.colorhexa.com/0c9dca | [
"# #0c9dca Color Information\n\nIn a RGB color space, hex #0c9dca is composed of 4.7% red, 61.6% green and 79.2% blue. Whereas in a CMYK color space, it is composed of 94.1% cyan, 22.3% magenta, 0% yellow and 20.8% black. It has a hue angle of 194.2 degrees, a saturation of 88.8% and a lightness of 42%. #0c9dca color hex could be obtained by blending #18ffff with #003b95. Closest websafe color is: #0099cc.\n\n• R 5\n• G 62\n• B 79\nRGB color chart\n• C 94\n• M 22\n• Y 0\n• K 21\nCMYK color chart\n\n#0c9dca color description : Strong cyan.\n\n# #0c9dca Color Conversion\n\nThe hexadecimal color #0c9dca has RGB values of R:12, G:157, B:202 and CMYK values of C:0.94, M:0.22, Y:0, K:0.21. Its decimal value is 826826.\n\nHex triplet RGB Decimal 0c9dca `#0c9dca` 12, 157, 202 `rgb(12,157,202)` 4.7, 61.6, 79.2 `rgb(4.7%,61.6%,79.2%)` 94, 22, 0, 21 194.2°, 88.8, 42 `hsl(194.2,88.8%,42%)` 194.2°, 94.1, 79.2 0099cc `#0099cc`\nCIE-LAB 60.297, -17.892, -32.569 22.866, 28.454, 60.161 0.205, 0.255, 28.454 60.297, 37.16, 241.217 60.297, -41.304, -48.562 53.342, -16.831, -29.529 00001100, 10011101, 11001010\n\n# Color Schemes with #0c9dca\n\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #ca390c\n``#ca390c` `rgb(202,57,12)``\nComplementary Color\n• #0cca98\n``#0cca98` `rgb(12,202,152)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #0c3eca\n``#0c3eca` `rgb(12,62,202)``\nAnalogous Color\n• #ca980c\n``#ca980c` `rgb(202,152,12)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #ca0c3e\n``#ca0c3e` `rgb(202,12,62)``\nSplit Complementary Color\n• #9dca0c\n``#9dca0c` `rgb(157,202,12)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #ca0c9d\n``#ca0c9d` `rgb(202,12,157)``\nTriadic Color\n• #0cca39\n``#0cca39` `rgb(12,202,57)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #ca0c9d\n``#ca0c9d` `rgb(202,12,157)``\n• #ca390c\n``#ca390c` `rgb(202,57,12)``\nTetradic Color\n• #086582\n``#086582` `rgb(8,101,130)``\n• #09789a\n``#09789a` `rgb(9,120,154)``\n• #0b8ab2\n``#0b8ab2` `rgb(11,138,178)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #0db0e2\n``#0db0e2` `rgb(13,176,226)``\n• #18bef1\n``#18bef1` `rgb(24,190,241)``\n• #30c5f3\n``#30c5f3` `rgb(48,197,243)``\nMonochromatic Color\n\n# Alternatives to #0c9dca\n\nBelow, you can see some colors close to #0c9dca. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #0ccac8\n``#0ccac8` `rgb(12,202,200)``\n• #0cbdca\n``#0cbdca` `rgb(12,189,202)``\n• #0cadca\n``#0cadca` `rgb(12,173,202)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #0c8dca\n``#0c8dca` `rgb(12,141,202)``\n• #0c7dca\n``#0c7dca` `rgb(12,125,202)``\n• #0c6eca\n``#0c6eca` `rgb(12,110,202)``\nSimilar Colors\n\n# #0c9dca Preview\n\nText with hexadecimal color #0c9dca\n\nThis text has a font color of #0c9dca.\n\n``<span style=\"color:#0c9dca;\">Text here</span>``\n#0c9dca background color\n\nThis paragraph has a background color of #0c9dca.\n\n``<p style=\"background-color:#0c9dca;\">Content here</p>``\n#0c9dca border color\n\nThis element has a border color of #0c9dca.\n\n``<div style=\"border:1px solid #0c9dca;\">Content here</div>``\nCSS codes\n``.text {color:#0c9dca;}``\n``.background {background-color:#0c9dca;}``\n``.border {border:1px solid #0c9dca;}``\n\n# Shades and Tints of #0c9dca\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #010d11 is the darkest color, while #fdffff is the lightest one.\n\n• #010d11\n``#010d11` `rgb(1,13,17)``\n• #021b23\n``#021b23` `rgb(2,27,35)``\n• #032a36\n``#032a36` `rgb(3,42,54)``\n• #043848\n``#043848` `rgb(4,56,72)``\n• #05475b\n``#05475b` `rgb(5,71,91)``\n• #07556d\n``#07556d` `rgb(7,85,109)``\n• #086380\n``#086380` `rgb(8,99,128)``\n• #097292\n``#097292` `rgb(9,114,146)``\n• #0a80a5\n``#0a80a5` `rgb(10,128,165)``\n• #0b8fb7\n``#0b8fb7` `rgb(11,143,183)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #0dabdd\n``#0dabdd` `rgb(13,171,221)``\n• #0ebaef\n``#0ebaef` `rgb(14,186,239)``\nShade Color Variation\n• #1fc0f2\n``#1fc0f2` `rgb(31,192,242)``\n• #32c5f3\n``#32c5f3` `rgb(50,197,243)``\n• #44caf4\n``#44caf4` `rgb(68,202,244)``\n• #57d0f5\n``#57d0f5` `rgb(87,208,245)``\n• #69d5f6\n``#69d5f6` `rgb(105,213,246)``\n• #7cdaf7\n``#7cdaf7` `rgb(124,218,247)``\n• #8edff8\n``#8edff8` `rgb(142,223,248)``\n• #a1e4f9\n``#a1e4f9` `rgb(161,228,249)``\n• #b3eafb\n``#b3eafb` `rgb(179,234,251)``\n• #c6effc\n``#c6effc` `rgb(198,239,252)``\n• #d8f4fd\n``#d8f4fd` `rgb(216,244,253)``\n• #ebf9fe\n``#ebf9fe` `rgb(235,249,254)``\n• #fdffff\n``#fdffff` `rgb(253,255,255)``\nTint Color Variation\n\n# Tones of #0c9dca\n\nA tone is produced by adding gray to any pure hue. In this case, #676d6f is the less saturated color, while #04a1d2 is the most saturated one.\n\n• #676d6f\n``#676d6f` `rgb(103,109,111)``\n• #5e7278\n``#5e7278` `rgb(94,114,120)``\n• #567680\n``#567680` `rgb(86,118,128)``\n• #4e7a88\n``#4e7a88` `rgb(78,122,136)``\n• #467f90\n``#467f90` `rgb(70,127,144)``\n• #3d8399\n``#3d8399` `rgb(61,131,153)``\n• #3587a1\n``#3587a1` `rgb(53,135,161)``\n• #2d8ca9\n``#2d8ca9` `rgb(45,140,169)``\n• #2590b1\n``#2590b1` `rgb(37,144,177)``\n• #1c94ba\n``#1c94ba` `rgb(28,148,186)``\n• #1499c2\n``#1499c2` `rgb(20,153,194)``\n• #0c9dca\n``#0c9dca` `rgb(12,157,202)``\n• #04a1d2\n``#04a1d2` `rgb(4,161,210)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #0c9dca is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5651301,"math_prob":0.57137203,"size":3703,"snap":"2019-13-2019-22","text_gpt3_token_len":1704,"char_repetition_ratio":0.1216545,"word_repetition_ratio":0.011111111,"special_character_ratio":0.53929245,"punctuation_ratio":0.23634337,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9765484,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-26T21:52:43Z\",\"WARC-Record-ID\":\"<urn:uuid:196ffd36-7fc2-4fc1-8b4e-f264de6c2e42>\",\"Content-Length\":\"36436\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9981f2f9-241d-484d-9590-f555bbc80650>\",\"WARC-Concurrent-To\":\"<urn:uuid:349bdd86-33d9-4e98-95ce-d3c202421519>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/0c9dca\",\"WARC-Payload-Digest\":\"sha1:GR6A6JUUD6FTYJHIAQ5X3BNJIWOECACG\",\"WARC-Block-Digest\":\"sha1:OZ6D5JXTHQU7RDDR6VNIRHQT5DGONLHE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232259757.86_warc_CC-MAIN-20190526205447-20190526231447-00054.warc.gz\"}"} |
https://echomasaz.pl/7993-calcium_metal_molecular_mass__ca_in_ghana.html | [
"# calcium metal molecular mass ca in ghana\n\n#### Current and Future Theranostic Appliions of the Lipid …\n\n8. Li J, Yang Y, Huang L. Calcium phosphate nanoparticles with an asymmetric lipid bilayer coating for siRNA delivery to the tumor. J Control Release. 2012;158(1):108-14 9. Giger EV, Puigmarti-Luis J, Schlatter R, Castagner B, Dittrich PS, Leroux JC. .\n\n#### How to Find the Molecular Mass of a Compound\n\n2019/11/6· Here''s a more complied example: Find the molecular mass (molecular weight) of Ca 3 (PO 4) 2. From the periodic table, the atomic masses of each element are: Ca = 40…\n\n#### calcium polystyrene sulfonate | Drug Information, Uses, …\n\nTechnical details about calcium polystyrene sulfonate, learn more about the structure, uses, toxicity, action, side effects and more\n\n#### GCSE SCIENCE CHEMISTRY HIGH SCHOOL - Moles - …\n\nSo, one mole of carbon dioxide has a mass of 44 g. The Relative Formula Mass of carbon dioxide is 44. This may also be called the Relative Molecular Mass (), since carbon dioxide is a molecule. The mass of one mole of calcium carbonate (CaCO 3) is (1 x RAM 1\n\n#### The Impact of Thyme and Rosemary on Prevention of …\n\nAs presented in Table 3, supplementation with thyme and rosemary significantly increased plasma calcium levels in both T and R groups (8.92 ± 0.13 and 8.54 ± 0.11) compared to the positive control group (7.57 ± 0.12); on the other hand, magnesium and phosphorus levels had no significant changes among all groups.. Concomitantly, the inflammatory biomarkers TNF-α and CRP were also estimated\n\n#### EDTA Titrations 2: Analysis of Calcium in a Supplement Tablet; …\n\nmg of Calcium Mass of Calcium (mg) = 0.01365 L × 0.01000 mol / L × 1 mol Calcium 1 mo lEDTA × × 40.08 g Calcium Ca ciu 1000 mg 1 g × 100 mL Total Volume 2 mL Sampled Volume Mass of Calcium = 274 mg This Section Continues on the Next Page à\n\n#### Calcium and phosphate compatibility: • The influence of other …\n\nCOMMENTARy Calcium and phosphates Am J Health-Syst Pharm—Vol 65 Jan 1, 2008 73 COMMENTARy Calcium and phosphate compatibility: Revisited again Dav i D W. Ne W t o N a ND Dav i D F. Driscoll Am J Health-Syst Pharm. 2008; 65:73-80Dav i D W. N e W to N, B.S.Pharm., Ph.D.,\n\n#### Commack Schools\n\n(gram-formula mass = 58.3 grams/mole) in an 8.40-gram sample. 8. Base your answer to the following question on Given the compound C4H1008, a Calculate the molar mass compound. + b Calculate the nuer of moles in 17.7 grams of the compound. c What\n\n#### Licl Molecular Or Ionic\n\nWrite the molecular equation and the net ionic equation for each of the following aqueous reactions. asked by Yaekob on March 27, 2016; Science urgent. Molecular Physics 2014 , 112 (12) , 1710-1723. Or does hydrogen have similar properties to a non-metal\n\n#### Stoichiometry: Calculations with Chemical Formulas and Equations …\n\nMolar Mass •By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol). –The molar mass of an element is the mass nuer for the element that we find on the periodic table. –The formula weight (in amu’s) will be the same nuer as theg/mol).\n\n#### WebElements Periodic Table » Calcium » isotope data\n\nMass / Da Half-life Mode of decay Nuclear spin Nuclear magnetic moment 41 Ca 40.9622783 102000 y EC to 41 K 7 / 2-1.595 45 Ca 44.956186 162.7 d β-to 45 Sc 7 / 2-1.327 47 Ca 46.954546 4.536 d β-to 47 Sc 7 / 2-1.38 49 Ca 48.955673 8.72 m β-to 49 Sc 3 /\n\n#### Molar Mass / Molecular Weight of Ca(OH)2 : Calcium …\n\n2014/4/18· For molar mass for Ca(OH)2 (molar mass and molecular weight are the same thing) just add the atomic weights. Note that for Ca(OH)2 you need to …\n\n#### Help with Empirical Formulas? | Yahoo Answers\n\n2011/12/30· 1) In an experiment, a 2.514-g sample of calcium was heated in a stream of pure oxygen, and was found to increase in mass by 1.004 g. Calculate the empirical formula of calcium oxide. 2) A compound has the following percentages by mass: barium, 58.84%; sulfur, 13.74%; oxygen, 27.43% 3) If a 1.271-g sample of aluminum metal is heated on a chlorine gas atmosphere, the mass of aluminum …\n\n#### Calculating Calcium Hardness as CaCO3 - Water Chemistry\n\nFrank R. Spellman and it states that the answer is 124.8 mg/L Ca as CaCO3. Seems to follow the calculation I always knew of 2.5(Ca++) thus 2.5*51=127.5 mg/L Molar mass (CO3(2-)) = M(C) + 3 x M(O) = 12+ 3 x 16 = 60 g/mole. Molar mass (CaCO3) = M(Ca\n\n#### The Mole and Molar Masses - GitHub Pages\n\nSimilarly, the formula mass of calcium phosphate [Ca 3 (PO 4) 2] is 310.177 amu, so its molar mass is 310.177 g/mol. This is the mass of calcium phosphate that contains 6.022 × 10 23 formula units. Figure 3.1 \"Samples of 1 Mol of Some Common\n\n#### Calcium Cyclamate | Drug Information, Uses, Side Effects, …\n\nMolecular Weight 396.53676 g/mol Molecular Formula C 12 H 24 CaN 2 O 6 S 2 Hydrogen Bond Donor Count 2 Hydrogen Bond Acceptor Count 8 Rotatable Bond Count 2 Exact Mass 396.070169 g/mol Monoisotopic Mass 396.070169 g/mol Topological Polar\n\n#### Potassium, Magnesium, and Calcium: Their Role in Both …\n\nHypertension remains the leading cause of cardiovascular disease (CVD), affecting approximately 1 billion individuals worldwide. 1 Although more than 72 million Americans, or nearly 1 in 3 adults, are estimated to have hypertension, blood pressure (BP) control is achieved in only 35% 2; recent surveys estimate that a higher percentage of patients have BP levels reduced to <140/90 mm Hg.\n\n#### Chemistry Flashcards | Quizlet\n\n%Ca = 40.1gCa / 1moleCa - 111.1gCaCl2 / 1moleCaCl2 x 100% = 36.1%Ca A 9.56 g sample of CuS is found to contain 6.35 g of copper. What is the percent by mass of the copper and sulfur in this compound\n\n#### chem tutor: MOLE CONCEPT - 10 ICSE CHAPTER WISE …\n\n6. Calcium hydroxide reacts with ammonium chloride to give ammonia, according to the following equation: Ca(OH)2 + 2NH4Cl --> CaCl2 + 2NH3 + 2H2O . If 5.35g of ammonium chloride are used, calculate: a. The mass of Calcium chloride formed and b. The\n\n#### sodium sulfide | Sigma-Aldrich\n\nMolecular Weight: 40.00 CAS Nuer: 1310-73-2 415413 50% in H 2 O Sigma-Aldrich pricing SDS CH 3 NaO 3 S · xH 2 O (x ca. 2) CAS Nuer: 6035-47-8 8.17068 stabilized with sodium carbonate SAFC pricing 1 2 > >> Service & Support Web Help\n\n#### Chemical composition of seawater; Salinity and the major …\n\nMass concentration units 1. wt.% = “weight percent” (actually, mass percent) = g per 100 g •Used for solids 2. ‰ = parts per thousand = g/kg for liquids and solids = mL/L for gas mixtures 3. Per mil = parts per thousand •Term is analogous to \"per cent“ •Is used\n\n#### Chemistry Final All old tests, quizzes, practice tests …\n\nWhat mass of calcium metal contains the same nuer of atoms as 24.3 g of magnesium? 40.1 g The empirical formula of a compound is CH2O, and its mass is 90 amu/molecule.\n\n#### Molar Mass And Formula Mass Worksheet Answers\n\n14 · Moles & Stoichiometry Cheat Sheet Calculating Molar Mass 1. Find the formula mass of Ca(NO 3) 2 Ca: 1 x 40. Worksheet: More Molecular Formulas and Hydrates Name_____ CHEMISTRY: A Study of Matter the molar mass of the compound is 283. 5%O\n\n#### Relative Molecular Mass & Relative Formula Mass …\n\nChemistry, How to calculate relative molecular mass, relative formula mass, percentage mass, percent mass of an element in a compound, percent mass of water in a …\n\n#### 66905-23-5 - Calcium D-gluconate monohydrate, 98+% - …\n\n448.40 (430.37anhy) Melting point ca 195 dec. Storage & Sensitivity Aient temperatures. Solubility Soluble in water(30g/L). Appliions Calcium D-gluconate monohydrate is used as a soluble for developing the highly effective, durable, and cheap CaO\n\n#### What is the Relative Atomic Mass and Relative Molecular …\n\nTable: Relative molecular masses of some substances [Relative atomic mass: H, 1; C, 12; O, 16] The term ‘relative molecular mass’ can only be used for substances that are made up of molecules. For ionic compounds, the term ‘relative formula mass’ or F r is used instead.\n\n#### atoms and molecules - SlideShare\n\nAtomic masses of some elements :- Element Atomic mass (u) Element Atomic mass (u) Hydrogen 1 Magnesium 24 Carbon 12 Aluminium 27 Nitrogen 14 Sulphur 32 Oxygen 16 Chlorine 35.5 Sodium 23 Calcium 40 8. 6) Molecule :- A molecule is the smallest particle of an element or compound which exists independently and shows all the properties of that substance.\n\n#### ChemTeam: Empirical Formula\n\nNext we need to determine the molecular formula, knowing the empirical formula and the molecular weight. Here''s how: 1) Calculate the \"empirical formula weight.\" This is not a standard chemical term, but the ChemTeam believes it is understandable. C 4 H 6 O gives an \"EFW\" of 70.092."
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https://cs.stackexchange.com/questions/106732/djikstras-algorithm-to-compute-shortest-paths-using-at-least-k-edges | [
"# Djikstra's algorithm to compute shortest paths using at least k edges\n\nI have a graph G = (V, E) where each edge is bidirectional with positive weight. I want to find the shortest path from vertex s to vertex t using at least k edges but less than 20. No vertex may be repeated in a path.\n\nI am aware of this problem: Dijkstra's algorithm to compute shortest paths using k edges?\n\nThat problem wants to find the shortest paths using at most k edges.\n\nMy current idea is to create the product graph via V' = V x {0, 1, 2, ..., 20} and then ignore any shortest paths from (s, 0) to (t, n) where n < k. However, because the product construction essentially duplicates nodes, I am not aware of a way to ignore duplicate vertices in the path-finding algorithm.\n\nP.S. If k were very large, say, k = |V|, wouldn't this be the Hamiltonian path problem and therefore NP-complete?\n\n• For the PS: yes, it does include Hamiltonian Path Problem as a subproblem. – John Dvorak Apr 9 at 17:10\n• @John Dvorak the linked problem gave a worst case complexity of 𝑂(𝑘⋅(|𝑉|+|𝐸|)lg|𝐸) and it includes the Hamiltonian Path Problem if k = |V| and there are no paths for k less than this. Am I mistaken? – cs questions Apr 9 at 17:46\n• @csquestions, if you are not allowed cycles, then this is NP-complete. See this question, specifically the first and last bullet point. Also see last paragraph of this answer to that question. – ryan Apr 9 at 17:52\n• if you allow cycles, then you can either replicate the graph $k$ times, or apply some dynamic programming approach. Since you don't, it's NP-complete – lox Apr 9 at 17:57\n\nConsider we had a polynomial time solution for some $$k$$. For clarification; given an undirected graph $$G=(V,E)$$ and two vertices $$s,t$$, we can find a shortest path from $$s$$ to $$t$$ of length $$\\geq k$$ in polynomial time. Then, by extension we can solve in polynomial time for $$k=n-1$$ and thus find a path that visits eeach vertex in $$V$$ exactly once. In other words, we can \"solve\" Hamiltonian Path.\nIf you know in advance that $$k \\leq 20$$ then you can exhaustive search all paths of length at most 20, and therefore solve for small $$k$$.\nAn exhaustive search, at worst, is to search every vertex order starting from $$s$$ that ends in $$t$$ (corresponding to existing edges), so in total you will search: $$(n-1)(n-2)(n-3)...(n-20) = O(n^{20})$$"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.91197073,"math_prob":0.99095726,"size":797,"snap":"2019-35-2019-39","text_gpt3_token_len":204,"char_repetition_ratio":0.1185372,"word_repetition_ratio":0.0,"special_character_ratio":0.25846925,"punctuation_ratio":0.14772727,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99978775,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-09-18T16:27:48Z\",\"WARC-Record-ID\":\"<urn:uuid:a53dd584-e4ab-4630-be92-4f3eea2fcd75>\",\"Content-Length\":\"143149\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:3fb5052e-866e-4224-9a8a-1e4381471666>\",\"WARC-Concurrent-To\":\"<urn:uuid:8f921002-64f8-409e-8dbf-3a8ae9e1abe9>\",\"WARC-IP-Address\":\"151.101.129.69\",\"WARC-Target-URI\":\"https://cs.stackexchange.com/questions/106732/djikstras-algorithm-to-compute-shortest-paths-using-at-least-k-edges\",\"WARC-Payload-Digest\":\"sha1:CVM5G2522NES5XVEECOREVX6H7ZEFMCX\",\"WARC-Block-Digest\":\"sha1:KBJ7U3OQQHEMLSUT3ECKAEYOWWXRLSCI\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-39/CC-MAIN-2019-39_segments_1568514573309.22_warc_CC-MAIN-20190918151927-20190918173927-00344.warc.gz\"}"} |
https://www.reference.com/education/true-false-question-bebaa73d70b414d4 | [
"What Is a True or False Question?\n\nA true or false question makes a statement and asks the person taking the test if the statement is true or false. There are no other responses. Each answer has a 50 percent chance of being correct.\n\nMultiple choice questions follow a somewhat similar format, and do not require the person taking the test to create an answer of their own. Multiple choice questions ask a question, and the person taking the test must choose the correct answer from a list of four or five similar answers. There is a 20 to 25 percent chance of randomly choosing the correct answer.\n\nSimilar Articles"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.9174793,"math_prob":0.66591483,"size":562,"snap":"2019-43-2019-47","text_gpt3_token_len":112,"char_repetition_ratio":0.13799283,"word_repetition_ratio":0.030927835,"special_character_ratio":0.20284697,"punctuation_ratio":0.07339449,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95802474,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-10-19T05:34:19Z\",\"WARC-Record-ID\":\"<urn:uuid:eb66649d-7c47-4cf8-9081-465c3c22449b>\",\"Content-Length\":\"171412\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ddecef25-a391-48cf-b610-ab5e5eaf037f>\",\"WARC-Concurrent-To\":\"<urn:uuid:a22801f1-f0c6-475c-94f6-2bcf8dd674d1>\",\"WARC-IP-Address\":\"151.101.202.114\",\"WARC-Target-URI\":\"https://www.reference.com/education/true-false-question-bebaa73d70b414d4\",\"WARC-Payload-Digest\":\"sha1:2RZ7LZMBVEDAR4ZUSHLZWJMKC3XK36EZ\",\"WARC-Block-Digest\":\"sha1:6MJDASZFEQVC5SYWMFI6FPIAJZPTZPWV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-43/CC-MAIN-2019-43_segments_1570986688826.38_warc_CC-MAIN-20191019040458-20191019063958-00484.warc.gz\"}"} |
http://www.comtechpass.com/tag/homework/ | [
"# Tagged: Homework\n\n## Algorithms Homework Exercise 2.6\n\nExercise 2.6. Consider the following basic problem. You’re given an array A consisting of n integers A, A, …, A[n]. You’d like to output a two-dimensional n-by-n array B in which B[i, j] (for...\n\n## Algorithms Homework Exercise 2.4\n\nTake the following list of functions and arrange them in ascending order of growth rate. That is, if function g(n) immediately follows function f(n) in your list, then it should be the case that...\n\n## Algorithms Homework Exercise 2.2\n\nSuppose you have algorithms with the six running times listed below. (Assume these are the exact number of operations performed as a function of the input size n.) Suppose you have a computer that...\n\n## Why would a network manager benefit from having network management tools?\n\nBecause it is easy to control and find the problem; when you find the problem, with having network management tools, you can easily troubleshoot and verify the problem with replacing component with a new..."
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89512473,"math_prob":0.8509793,"size":1039,"snap":"2022-05-2022-21","text_gpt3_token_len":246,"char_repetition_ratio":0.109178744,"word_repetition_ratio":0.3529412,"special_character_ratio":0.24927816,"punctuation_ratio":0.17021276,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9864115,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-05-18T03:19:52Z\",\"WARC-Record-ID\":\"<urn:uuid:2af2eacc-bae7-473b-a49d-f7604b9c6c94>\",\"Content-Length\":\"80802\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5b1ec85f-3e4d-4d2e-965b-806c92c43344>\",\"WARC-Concurrent-To\":\"<urn:uuid:08fd24f4-ac7d-48ba-a672-2b27f6a33695>\",\"WARC-IP-Address\":\"107.180.47.9\",\"WARC-Target-URI\":\"http://www.comtechpass.com/tag/homework/\",\"WARC-Payload-Digest\":\"sha1:G5TA27KMGYSAJQX5Q3TK6E5QVGSMLDND\",\"WARC-Block-Digest\":\"sha1:RKQ4NIMSCZ5TQ5DIK72VRDEO2O3AL3PD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-21/CC-MAIN-2022-21_segments_1652662521041.0_warc_CC-MAIN-20220518021247-20220518051247-00174.warc.gz\"}"} |
https://sources.debian.org/src/evolver/2.70+ds-4/fe/mylarcube.fe/ | [
"## File: mylarcube.fe\n\npackage info (click to toggle)\nevolver 2.70+ds-4\n `123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177` ``````// mylarcube.fe // By Kenneth Brakke, Feb. 17, 2002 // Evolver data for mylar cube. Idea is to deform cube // without stretching to maximize volume. Implemented with // relaxed_elastic_A method, which permits a little stretching, but // shrinks freely. Can crank up the elastic modulus to get // effective no stretching. The relaxed_elastic_A method calculates // the Cauchy-Green strain tensor of a facet, and then uses // Poisson's ratio to get the stress tensor. Then it calculates // the stress eigenvalues, and assigns energy proportional to the // square of positive eigenvalues, but zero energy for negative // eigenvalues. So it permits effective wrinkling. The original // unstretched shape of a facet is remembered in an extra facet // attribute named form_factors, which has the components of the // Gram matrix of the unstretched facet. This datafile assumes the // initial surface is unstretched, and sets the form_factors in // the \"read\" section below. To recompute form factors after refining, // there is also a vertex attribute that holds the unstretched // vertex coordinates. The \"r\" command has been redefined in the // \"read\" section to properly adjust the form factors. DO NOT refine // or modify the triangulation in any other way than using the \"r\" // command, unless you want to take responsibility for fixing up // the form factors. // The Poisson ratio in this datafile is set to be 0.80 in the \"read\" // section. This was the Poisson ratio used in the project that // relaxed_elastic_A was originally developed for: modeling high-altitude // balloons. A couple of test runs with lower Poisson ratio led to // indefinite hessians at the end, so I am leaving the ratio as 0.80. // Evolving hints: See the sample evolution \"gogo\" below. The idea is // to ramp up the elastic modulus as you refine and approach the final // shape. For final convergence, use \"hessian_seek\" instead of \"hessian\", // since \"hessian\" has a tendency to blow up the surface here, and // \"hessian_seek\" prevents that by doing a minimum-energy search along // the direction to the putative hessian minimum. Note that the energy // contribution of the elastic is inversely proportional to the elastic // modulus. I'm not sure how high you can get the modulus effectively; // at least 1000000. Extrapolate to infinite modulus for unstretchable // mylar. // If you want confirmation at the end that there is little stretching // going on, the command \"echeck\" below will print out a histogram of // edge stretch factors. // Do not try to do quadratic or lagrange models; relaxed_elastic_A is // defined only for the linear model. /***************************************************************************/ // Set volume to be maximized. Do this by setting energy to negative // volume, since Evolver minimizes energy. quantity cube_volume energy modulus -1 method facet_volume global // Elastic energy. // Stuff to do relaxed_elastic_A method define facet attribute poisson_ratio real // For defining unstretched facet shape define facet attribute form_factors real // For resetting form_factors in refining define vertex attribute ref_coord real define vertex attribute old_vid integer define edge attribute old_eid integer // This is the key energy to permit wrinkling but discourage stretching. quantity stretch energy modulus 10 method relaxed_elastic_A global // For components of stress quantity stretch1 info_only method relaxed_elastic1_A global quantity stretch2 info_only method relaxed_elastic2_A global vertices 1 0.0 0.0 0.0 2 1.0 0.0 0.0 3 1.0 1.0 0.0 4 0.0 1.0 0.0 5 0.0 0.0 1.0 6 1.0 0.0 1.0 7 1.0 1.0 1.0 8 0.0 1.0 1.0 edges /* given by endpoints and attribute */ 1 1 2 2 2 3 3 3 4 4 4 1 5 5 6 6 6 7 7 7 8 8 8 5 9 1 5 10 2 6 11 3 7 12 4 8 faces /* given by oriented edge loop */ 1 1 10 -5 -9 tension 0 2 2 11 -6 -10 tension 0 3 3 12 -7 -11 tension 0 4 4 9 -8 -12 tension 0 5 5 6 7 8 tension 0 6 -4 -3 -2 -1 tension 0 bodies /* one body, defined by its oriented faces */ 1 1 2 3 4 5 6 read // refine original edges so better suited to detect creases refine edge where id >= 1 and id <=12 set facet poisson_ratio 0.80 set vertex ref_coord x; set vertex ref_coord y; set vertex ref_coord z; // Set form factors so initial cube is unstretched. set_form_factors := { foreach facet ff do { set ff.form_factors (ff.vertex.ref_coord - ff.vertex.ref_coord)^2 + (ff.vertex.ref_coord - ff.vertex.ref_coord)^2 + (ff.vertex.ref_coord - ff.vertex.ref_coord)^2; set ff.form_factors (ff.vertex.ref_coord - ff.vertex.ref_coord) *(ff.vertex.ref_coord - ff.vertex.ref_coord) + (ff.vertex.ref_coord - ff.vertex.ref_coord) *(ff.vertex.ref_coord - ff.vertex.ref_coord) + (ff.vertex.ref_coord - ff.vertex.ref_coord) *(ff.vertex.ref_coord - ff.vertex.ref_coord); set ff.form_factors (ff.vertex.ref_coord - ff.vertex.ref_coord)^2 + (ff.vertex.ref_coord - ff.vertex.ref_coord)^2 + (ff.vertex.ref_coord - ff.vertex.ref_coord)^2; } } set_form_factors; // redefine the \"r\" command to adjust form factors. r :::= { set vertex old_vid id; set edge old_eid id; 'r'; foreach vertex vv where old_vid == 0 do { vv.ref_coord := avg(vv.edge ee where old_eid != 0,sum(ee.vertex where old_vid != 0, ref_coord)); vv.ref_coord := avg(vv.edge ee where old_eid != 0,sum(ee.vertex where old_vid != 0, ref_coord)); vv.ref_coord := avg(vv.edge ee where old_eid != 0,sum(ee.vertex where old_vid != 0, ref_coord)); }; set_form_factors; recalc; } // check of how edge lengths stretched echeck := { histogram(edge ee, ee.length/sqrt( (ee.vertex.ref_coord - ee.vertex.ref_coord)^2 + (ee.vertex.ref_coord - ee.vertex.ref_coord)^2 + (ee.vertex.ref_coord - ee.vertex.ref_coord)^2)); } // Typical evolution gogo := { g 10; r; g 10; stretch.modulus := 100; g 10; r; g 10; conj_grad; g 20; r; g 50; stretch.modulus := 1000; g 50; stretch.modulus := 10000; g 150; g 200; hessian_seek; hessian_seek; hessian_seek; hessian_seek; } ``````"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5115799,"math_prob":0.9009474,"size":6713,"snap":"2019-51-2020-05","text_gpt3_token_len":2199,"char_repetition_ratio":0.17349829,"word_repetition_ratio":0.024215247,"special_character_ratio":0.39639506,"punctuation_ratio":0.15352112,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9927823,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-01-23T13:32:49Z\",\"WARC-Record-ID\":\"<urn:uuid:09299de9-a3e1-4b3e-aabb-634b66ca9ad1>\",\"Content-Length\":\"29469\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:69776831-b06b-44d2-a3c9-8f6124029561>\",\"WARC-Concurrent-To\":\"<urn:uuid:fa171c7d-8e32-4412-86be-19b765e6cdc0>\",\"WARC-IP-Address\":\"5.153.231.38\",\"WARC-Target-URI\":\"https://sources.debian.org/src/evolver/2.70+ds-4/fe/mylarcube.fe/\",\"WARC-Payload-Digest\":\"sha1:CLSUMRJ5L2SV7ALYEAB4HTZ5L72D4UK4\",\"WARC-Block-Digest\":\"sha1:7FAA7V5VWSC2WW6UOUIHJ7E6ZUBZS574\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-05/CC-MAIN-2020-05_segments_1579250610919.33_warc_CC-MAIN-20200123131001-20200123160001-00254.warc.gz\"}"} |
https://codereview.stackexchange.com/questions/152143/doctrine-symfony-where-clause-for-reports | [
"# Doctrine/Symfony WHERE clause for reports\n\nI'm trying to generate a report from two doctrine entities.\n\nThe fields to filter the report are optional. This is the code I've got now and it looks really bad to me. How can I improve this?\n\npublic function filter($accountNumber,$companyName, $branchName,$branchManager)\n{\n$em =$this->managerRegistry->getManager();\n$whereClause = 'WHERE '; if (!empty($accountNumber)) {\n$whereClause .= ' u.accountNumber = :acc_number'; } if (!empty($companyName)) {\n$whereClause .= ' AND um.companyName = :company_name'; } if (!empty($branchName)) {\n$whereClause .= ' AND um.branchName = :branch_name'; } if (!empty($branchManager)) {\n$whereClause .= ' AND um.branchAccountManager = :branch_manager'; }$query = $em->createQuery( 'SELECT u FROM CarbonUserBundle:User u JOIN CarbonUserBundle:UserMeta um WITh u.meta=um.id ' .$whereClause\n);\n\nif (!empty($accountNumber)) {$query->setParameter('acc_number', $accountNumber); } if (!empty($companyName)) {\n$query->setParameter('company_name',$companyName);\n}\nif (!empty($branchName)) {$query->setParameter('branch_name', $branchName); } if (!empty($branchManager)) {\n$query->setParameter('branch_manager',$branchManager);\n}\n\nreturn $query->getResult(); } ## 1 Answer I would suggest using the QueryBuilder $em = $this->managerRegistry->getManager();$qb = $em->createQueryBuilder() ->select('u') ->from(User::class, 'u') ->join('u.meta', 'um'); // doctrine generates the join condition automatically if (!empty($accountNumber)) {\n$qb ->andWhere('u.accountNumber = :acc_number') ->setParameter('acc_number',$accountNumber)\n}\nif (!empty($companyName)) {$qb\n->andWhere('u.companyName = :company_name')\n->setParameter('company_name', $companyName) } // and so on... return$qb->getQuery()->getResult();"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.51328665,"math_prob":0.79672104,"size":1191,"snap":"2023-40-2023-50","text_gpt3_token_len":298,"char_repetition_ratio":0.16849199,"word_repetition_ratio":0.02238806,"special_character_ratio":0.2854744,"punctuation_ratio":0.25945947,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9698514,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-12-01T15:39:18Z\",\"WARC-Record-ID\":\"<urn:uuid:a8500f8e-c3de-4211-a505-809e8cbed276>\",\"Content-Length\":\"160405\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7720fd7d-3b34-4402-8148-c8832c79ac72>\",\"WARC-Concurrent-To\":\"<urn:uuid:deda3bed-b348-4461-9df0-84aa58a5988a>\",\"WARC-IP-Address\":\"104.18.43.226\",\"WARC-Target-URI\":\"https://codereview.stackexchange.com/questions/152143/doctrine-symfony-where-clause-for-reports\",\"WARC-Payload-Digest\":\"sha1:FZKUZSD2QJ7TIUT4BRKWBUUQANDY4WPN\",\"WARC-Block-Digest\":\"sha1:TTHLZMS4IBU37J25ALYZY4QJCLD5ZYTE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100290.24_warc_CC-MAIN-20231201151933-20231201181933-00063.warc.gz\"}"} |
http://digitalaudiorock.com/cgi-bin/qrd.cgi | [
"QRD Diffuser Well Depth Calculator\n\nWelcome. This script will calculate the well depths of a QRD (Quadratic Residue Diffuser) given a specified well width and a maximum well depth. The width and max depth should be specified in inches, with or without decimal fractions (ie 9 or 8.5 for example). This can be used as a guide for building a diffuser. The sequence of well depths shown are simply repeated over and over.\n\nAt some point I may add a detailed description of the math used. Briefly, the residue is the modulo of the well sequence number squared and the number of wells per period. For example, with 7 wells per period, well number 5 will have a residue of 4 (the mod of 25 and 7). Note that the number of wells per period is always a prime number. The well depths are proportioned based on the residues.\n\nThe results show the well sequence number, the quadratic residue used to calculate the depth, and the well depth for each well in one period. It also shows the resulting frequency range that will be diffused. The minimum frequency is a function of the maximum well depth (the deeper the well, the lower the minimum frequency), while the maximum frequency is a function of the well width (the narrower the well, the higher the maximum frequency). Note that the practical frequency range will generally be wider, about a half octave below the minimum and a half octave above the maximum shown.\n\nIf anyone finds a problem with my math, please let me know. The calculations are based on equations I found on the net, which were from various books by F. Alton Everest. I'm not 100% sure on the minimum frequency calculation.\n\nNOTE: You may want to check out the QRDude Quadratic Residue Diffuser calculator. Among other things, it's been brought to my attention that my calculations don't factor in the width of fins at all. The QRdude calculator does this.\n\n Well Width (inches) Max Well Depth (inches) Wells per Period 5 7 11 13 17 19 23 29 31 37 41 43 47"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.89649343,"math_prob":0.9767089,"size":1828,"snap":"2022-05-2022-21","text_gpt3_token_len":413,"char_repetition_ratio":0.14638157,"word_repetition_ratio":0.018691588,"special_character_ratio":0.21115974,"punctuation_ratio":0.09944751,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9682495,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-25T01:22:38Z\",\"WARC-Record-ID\":\"<urn:uuid:c256e459-f6e5-41af-abde-a8340f084490>\",\"Content-Length\":\"3170\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9a7e5f32-15e5-45bb-be07-b7caa4f7b607>\",\"WARC-Concurrent-To\":\"<urn:uuid:a9a33242-3d63-4c10-ba76-79f357f39dd4>\",\"WARC-IP-Address\":\"67.227.130.24\",\"WARC-Target-URI\":\"http://digitalaudiorock.com/cgi-bin/qrd.cgi\",\"WARC-Payload-Digest\":\"sha1:ETKZWOEPWDBLWFD77BGIVYUQWFJF7NOR\",\"WARC-Block-Digest\":\"sha1:YEJSZDAQ6I5H7GQVF4O2CH7UDCJQ2KM5\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304749.63_warc_CC-MAIN-20220125005757-20220125035757-00018.warc.gz\"}"} |
https://brainmass.com/engineering/power-engineering/pg3 | [
"Explore BrainMass\n\n# Power Engineering\n\n## BrainMass Solutions Available for Instant Download\n\n### Electrical networks\n\nEngine consumes 5kw of power and a reactive power of 2kvar at current of (3+j2) Amps. Find the applied voltage in complex form V=(a+jb). Then use P=Re(VI*) and Q=IM(VI*) to form two simultaneous equations. Please show workings.\n\n### AM Wave and Power\n\nConsider the message signal: m(t) = 20 cos (2*pi*t) volts and the carrier wave: c(t) = 50 cost(100*pi*t) volts a. Sketch (to scale) the resulting AM wave for 75% of the modulation b. Find the power developed across a load of 100 ohms due to this AM wave.\n\n### Voltage in complex forms\n\nAn engine consumes 10kw of power and 4kvar reactive power at a current of (6+j4)Amps . Find the applied voltage in complex form using P=Re(VI) and Q=Im(VI) as simultaneous equations. Obviously V=(a+jb)\n\n### Solving for Power\n\nIf the voltage and current supplied to a circuit or load by a source are: Vs = 170<(-0.157) V Is = 13<0.28 A determine a) The power supplied by the source which is dissipated as heat or work in the circuit (load). b) The power stored in reactive components in the circuit (load). c) The power factor angle and power facto\n\n### Power Factor: Solving for Circuits\n\nDetermine C so that the plant power factor of Figure P7.25 (see attached file) is corrected to 1 (or the power factor angle to zero) so that Is is minimized and in phase with Vo. vs(t) = 450cos(wt) V w = 377 rad/s Z = 7 < 0.175 Ω\n\n### Maximum Power Theorem\n\nA) Find the Thevenin's circuit for the circuit shown in Fig. 5.1 (see attachment) similar to the circuit shown in Fig. 5.2 (see attachment). b) Calculate the output voltage for different load resistors and the corresponding power and enter these values in Table 1. c) Plot a graph of load resistance vs power using Matlab (\n\n### Complex power of electrical load\n\nAn electrical load operates at 240 Vrms. The load absorbs an average power of 8kW with 0.8 lagging power factor. How do I compute the complex power of the load and impedance?\n\n### Shunt resistance effect in a voltage divider\n\nWhat is the effect of the DMM input resistance on the measured results? Derive the voltage gain of the divider with a DMM resistance of Rm, which is also called shunt resistance (See Figure 2.3-attached file).\n\n### Current, Power and Total Power\n\nSee the attached file. For the circuit shown in Figure 2.24 (attached file), find: a) The currents i1 and i2. b) The power delivered by the 3-A current source and by the 12-V voltage source. c) The total power dissipated by the circuit. Let R1=25 ohms, R2= 10 ohms, R3= 5 ohms, R4= 7 ohms, and express i1 and i2 as function\n\n### Terminal Voltage & Power Supplied/Dissipated\n\nRefer to Figure 2.16 (see attached file): a) Find the total power supplied by the ideal source. b) Find the power dissipated and lost within the nonideal source. c) What is the power supplied by the source to the circuit as modeled by the load resistance? d) Plot the terminal voltage and power supplied to the circuit as a fu\n\n### Supplying and Dissipating Power\n\nDetermine which elements in the circuit of Figure 2.10 (attached file) are supplying power and which are dissipating power. Also, determine the amount of power dissipated and supplied.\n\n### An amplifier with a voltage gain of +30db, an input resistance of 8 k ohms, and an output resistance of 2 k ohms is used to drive\n\nAn amplifier with a voltage gain of +30db, an input resistance of 8 k ohms, and an output resistance of 2 k ohms is used to drive a 2 k ohm load. What is the value of Avo? a. 3.162 b. 31.62 c. 100 d. 1000\n\n### Star vs Delta for pf correction capacitors\n\nComment on the difference between pf correction capacitors connected in star and delta.\n\n### Capacitance\n\nPrior to t=0, a 100- uF capacitance is uncharged. Starting at t=0, the voltage across the capacitor is increased linearly with time to 100V in 2s. Then, the voltage remains constant at 100V. Sketch the voltage, current, power, and stored energy to scale versus time. Please show equations for the sketches.\n\n### Voltage through 3-uF Capacitance\n\nThe current through a 3-uF capacitance is shown in the attachment. At t=0, the voltage is v(0)=10v. Sketch the voltage, power, and stored energy to scale vs. time. In the sketch of each one, I need the equations worked out for each plot. (See attached file for full problem description).\n\n### Circuit Elements - Power Dissipated\n\nGiven the circuit shown in the attachment find: values and power dissipated. (Please see attachment for complete question and diagram. Thanks)\n\n### Circuit Variable Problem - Time and Power\n\nPlease see the attached. The voltage and current at the terminals of the circuit element in Fig. 1.5 are zero for... (continued in the attachment).\n\n### Circuit Analysis\n\nAn electric load operates at 240 volts. The load absorbs an average power of 8Kw at a lagging power factor of 0.8. a) Calculate the complex power of the load b) Calculate the impedance of the load\n\n### Rankine Cycle - Net Power, Thermal Efficiency and Heat Transfer\n\nWater is the working fluid in a vapor power cycle with reheat, superheat and reheat. Superheated steam enters the first turbine stage at 8 MPa, 480 C and expands to 0.7 MPa. It then is reheated to 480 C before entering the second turbine stage, where it expands to the condenser pressure of 8 KPa. The mass flow rate of steam e\n\n### Complex Power and Impedance\n\nQuestion: An electrical load operates at 240 volts rms. The load absorbs an average power of 8 kW at a lagging power factor of 0.8. a) Calculate the complex power of the load. b) Calculate the impedance of the load. Please view attachment for multiple choice options.\n\n### Dissipations of Power in Lightbulbs\n\nAn incandescent lightbulb rated at 100 W will dissipate 100 W as heat and light when connected across a 110-V ideal voltage source. If three of these bulbs are connected in series across the same source, determine the power each bulb will dissipate.\n\n### Reactive Power Absorbed by a Line\n\nTwo balanced, in parallel connected, three-phase loads are fed by a three-phase line with an impedance of (2+j4) Ohm per phase. The first load is delta-connected with an impedance of (60 ? j45) Ohm per phase, and the second load is Y-connected with an impedance of (30 + j40) Ohm per phase. The line is energized at the sending en\n\n### Power Plants - Burning Coal\n\nSee attached file.\n\n### Turbine Unit and Mechanical Output Generated\n\nAn area of an interconnected 60 hz power system has three turbine generator units rated 100MVA, 200MVA, and 600MVA respectively. Regulation constants are given for each unit, with the load suddenly decreasing by 100MW. With assumptions made and a frequency response of beta = 165 p.u., calculate the MW decrease in mechanical ou\n\n### A Three-Phase Overhead Transmission Line\n\nA three-phase overhead transmission line line has a per-phase resistance of 0.15 ohm/km and a per-phase inductance of 1.3263 mH/km. Assume that the shunt capacitance is negligible. The length f the line is 40 kilometers and it operates under 220 kV and 60-Hz. The line is supplying a three-phase load of 381 MVA at 0.8 power facto\n\n### Real Power Generated and Total Real Power Losses\n\nConsider the electric power system show {see attachment}. The power-flow solution of this system can be obtained without resorting to iterative techniques. The elements of the bus admittance matrix Ybus have been calculated as attached. If sigma_2 = 16.97 degrees and sigma_3 = 21 degrees, calculate the real power generated a\n\n### Power-Flow Solution (Calculate Phase Angle via Real and Reactive Power Equations)\n\nConsider the electric power system shown {see attachment}. The power-flow solution of this system can be obtained without resorting to iterative techniques. The elements of the bus admittance matrix Ybus have been calculated as: {see attachment}. Calculate the phase angle {see attachment} by using the real and reactive power\n\n### Electronics (about operational amplifier)\n\nShow how an ideal Operational Amplifier with an open loop gain = A can be used to provide: i) a non-inverting amplifier with gain= +6v/v. ii) an inverting amplifier with gain= -3v/v. iii) an integrator with gain= +3. Explain the purpose of every component in your circuits. Please see attached for full question.\n\n### Electrical Energy Conversion: Synchronous Generator\n\nSynchronous Generator A 250 kVA, 280 V, three-phase, four-pole, 60 Hz, synchronous generator with a synchronous reactance of 0.99 ohms per phase is operating at rated conditions and a power factor of 0.832 lagging. The magnetization curve for the generator is shown in Figure 1. a. Sketch a phasor diagram for the generator. b.\n\n### Probability of a False Alarm: Noise Power\n\nThe fixed threshold of a radar detector was set, assuming a known noise power, to yield P_FA = 10^-6. The actual noise power was 3 dB higher than expected. What will be the actual P_FA?"
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http://www.bzst.com/2006/02/data-partitioning.html | [
"## Tuesday, February 14, 2006\n\n### Data partitioning\n\nA central initial step in data mining is to partition the data into two or three partitions. The first partition is called the training set, the second is the validation set, and if there is a third, it is usually called the test set.\n\nThe purpose of data partitioning is to enable evaluating model predictive performance. In contrast to an explanatory goal, where we want to fit the data as closely as possible, good predictive models are those that have high predictive accuracy. Now, if we fit a model to data, then obviously the \"tighter\" the model, the better it will predict those data. But what about new data? How well will the model predict those?\n\nPredictive models are different from explanatory models in various aspects. But let's only focus on performance evaluation here. Indications of good model fit are usually high R-squared values, low standard-error-of-estimate, etc. These do not measure predictive accuracy.\n\nSo how does partioning help measure predictive performance? The training set is first used to fit a model (also called to \"train the model\".) The validation set is then used to evaluate model performance on new data that it did not \"see\". At this stage we compare the model predictions for the new validation data to the actual values and use different metrics to quantify predictive accuracy.\n\nSometimes, we actually use the validation set to tweak the original model. In other words, after seeing how the model performed on the validation data, we might go back and change the model. In that case we are \"using\" our validation data, and the model is no longer blind to them. This is when a third, test set, comes in handy. The final evaluation of predictive performance is then achieved by applying the model (which is based on the training data and tweaked using the validation data) to the test data that it never \"saw\"."
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https://www.winnerswire.com/how-are-horse-race-odds-payouts-calculated/ | [
"",
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"Home » How are Horse Race Odds Payouts Calculated?\n\n# How are Horse Race Odds Payouts Calculated?",
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"Horse racing is a thrilling and exciting sport that has the potential to bring in huge amounts of money if you make the right bets. However, the payouts can be confusing and intimidating, especially if you don’t know how they are calculated. In this article, we’ll discuss how horse race odds payouts are calculated and how understanding them can help you make better bets.\n\n## What are Horse Racing Odds?\n\nHorse racing odds are a representation of the probability of a horse winning a race. The higher the odds, the less likely it is for the horse to win. Odds can either be expressed in fractional or decimal form, with fractional odds being more popular in the UK and decimal odds being more popular in Europe and Australia.\n\nFractional odds are written as a fraction, with the numerator representing the profit you will make if you win the bet and the denominator representing the stake. For example, if you bet £5 at odds of 4/1, you would make a profit of £20 if you win (4 x £5 = £20).\n\nDecimal odds represent the total amount of money that you will receive if you win the bet, including your stake. For example, if you bet £5 at odds of 5.0, you will receive a total of £25 (5 x £5 = £25).\n\n## How are Horse Race Odds Payouts Calculated?\n\nHorse race odds payouts are calculated by taking the odds of the horse and subtracting 1 from them. For example, if a horse has odds of 7.0, the payout would be calculated as 7.0 – 1 = 6.0. This means that if you bet £5 at these odds, you would receive a total of £30 (6 x £5 = £30).\n\n## The Formula for Calculating Horse Race Odds Payouts\n\nThe formula for calculating horse race odds payouts is as follows:\n\nRelated content How to Always Win at Horse Racing\n\n## Odds Payout = (Odds – 1) x Stake\n\nWhere “Odds” is the odds of the horse and “Stake” is the amount of money you are betting.\n\nFor example, if you bet £5 at odds of 4.0, the payout would be calculated as (4.0 – 1) x £5 = £15.\n\n## Factors Affecting Horse Race Odds Payouts\n\nThere are a number of factors that can affect the odds payouts in horse racing, such as the number of horses in the race, the track conditions, the form of the horses, and the jockeys. All of these factors can influence the odds and the payouts, so it’s important to take them into consideration before making a bet.\n\n## The Type of Bet Placed\n\nThe type of bet placed can also affect the odds and the payouts. For example, a win bet pays out at the odds displayed, but a place bet pays out at a fraction of the odds and an each-way bet pays out at a fraction of the odds for both the win and the place.\n\n## The Bookmaker’s Margin\n\nThe bookmaker’s margin is also an important factor to consider when calculating horse race odds payouts. The bookmaker’s margin is the amount of money the bookmaker takes from each bet, which is usually around 5% to 10%. This means that the odds and payouts displayed by the bookmaker are not the actual odds and payouts, as they have been reduced to account for the bookmaker’s margin.\n\n## Conclusion\n\nHorse race odds payouts can be complicated and intimidating, but understanding how they are calculated can help you make better bets. The formula for calculating horse race odds payouts is (Odds – 1) x Stake, but it’s important to take into consideration other factors such as the type of bet placed, the number of horses in the race, the track conditions, the form of the horses, the jockeys, and the bookmaker’s margin. By taking these factors into consideration, you can increase your chances of winning and make more profitable bets."
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https://ems.press/books/elm/93/contents | [
"# Lectures on Duflo Isomorphisms in Lie Algebra and Complex Geometry\n\n• ### Damien Calaque\n\nETH Zurich, Switzerland\n• ### Carlo A. Rossi\n\nMax Planck Institute for Mathematics, Bonn, Germany",
null,
"A subscription is required to access this book.\n\n Frontmatterpp. i–iv Prefacepp. v–vi Contentspp. vii–viii 1 Lie algebra cohomology and the Duflo isomorphismpp. 1–9 2 Hochschild cohomology and spectral sequencespp. 11–18 3 Dolbeault cohomology and the Kontsevich isomorphismpp. 19–24 4 Superspaces and Hochschild cohomologypp. 25–31 5 The Duflo–Kontsevich isomorphism for $Q$-spacespp. 33–39 6 Configuration spaces and integral weightspp. 41–49 7 The map $\\mathcal{U}_\\mathcal{Q}$ and its propertiespp. 51–59 8 The map $\\mathcal{H}_\\mathcal{Q}$ and the homotopy argumentpp. 61–69 9 The explicit form of $\\mathcal{U}_\\mathcal{Q}$pp. 71–77 10 Fedosov resolutionspp. 79–87 Deformation-theoretical interpretation of the Hochschild cohomology of a complex manifoldpp. 89–99 Bibliographypp. 101–103 Indexpp. 105–106"
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http://tasks.illustrativemathematics.org/content-standards/HSS/ID/C/7/tasks/1554 | [
"# Olympic Men's 100-meter dash\n\nAlignments to Content Standards: S-ID.C.7 S-ID.B.6.a\n\nThe scatterplot below shows the finishing times for the Olympic gold medalist in the men's 100-meter dash for many previous Olympic games. The line of best fit is also shown. (Source: http://trackandfield.about.com/od/sprintsandrelays/qt/olym100medals.htm)",
null,
"1. Is a linear model a good fit for the data? Explain, commenting on the strength and direction of the association.\n2. The equation of the linear function that best fits the data (the line of best fit) is $$\\widehat{\\mbox{Finishing time}} = 10.878 - 0.0106 \\left( \\mbox{Year after 1900} \\right).$$ Given that the summer Olympic games only take place every four years, how should we expect the gold medalist's finishing time to change from one Olympic games to the next?\n3. What is the vertical intercept of the function's graph? What does it mean in context of the 100-meter dash?\n4. Note that the gold medalist finishing time for the 1940 Olympic games is not included in the scatterplot. Use the model to estimate the gold medalist's finishing time for that year.\n5. What is a realistic domain for the linear function? Comment on how your answer pertains to using this function to make predictions about future Olympic 100-m dash race times.\n\n## IM Commentary\n\nThe task asks students to identify when two quantitative variables show evidence of a linear association, and to describe the strength and direction of that association. Students then utilize a least-squares regression line to make predictions, and to make conjectures about the limitations of the model, which is a very important aspect of SMP4 - Model with Mathematics. They must apply their knowledge of slope and intercept of a linear function in context of the problem; i.e., understand that the slope of a regression line is the predicted change in the response variable per unit change of the explanatory variable, and that the vertical intercept corresponds to a value of zero in the explanatory variable.\n\nLinear models are a very nice connection between statistics and functions in high school mathematics. Coherence in high school mathematics means drawing connections between topics that use the same mathematical concept. In this case we use linear functions to model the relationship between two quantitative variables. We can use the context of investigating if there is an association between two variables to strengthen our understanding of slope and intercept of a linear function.\n\nThis task is probably most appropriate for use in instruction. Consider having students work together in pairs or small groups on parts a - d. Part e could then be the basis for a whole class discussion.\n\n## Solution\n\n1. The data in the scatterplot are from two quantitative variables (year and finishing time), and the overall pattern is linear. There are also no obvious outliers, so it is reasonable to fit a linear model to the data. The direction of the association is negative (finishing time decreases as the year after 1900 increases), and the association is strong because the points are tightly clustered in a linear form.\n2. The slope of the equation of the line of best fit is $-0.0106.$ This means for every 1 year that passes, we would predict that the finishing time for the 100-m dash decreases by $0.0106$ seconds. Since the Olympics take place every four years, we would expect the predicted gold medalist's finishing time to decrease by $4(0.0106) = 0.0424$ seconds from one Olympic games to the next.\n3. The vertical intercept of the line of best fit's equation is 10.878. In context, this would be the predicted finishing time (in seconds) for the 100-m dash gold medalist in the 1900 Olympic games.\n4. To predict the finishing time for the 1940 gold medalist, we would simply substitute $\\mbox{Years after 1900} = 40$ into the equation of the line of best fit to solve for $\\widehat{\\mbox{Finishing Time}}.$ This yields $$\\widehat{\\mbox{Finishing Time}} = 10.878 - 0.0106 (40) = 10.454.$$ The predicted finishing time for the 1940 gold medalist is 10.454 seconds.\n5. At the most basic level, we know that the model will fail to be realistic once we obtain predicted racing times of zero or less. Substituting $0$ into the equation for $\\widehat{\\mbox{Finishing Time}}$, we can solve for $\\mbox{Years after 1900} \\approx 1026.2.$ This equates to roughly the year 2926. If we take into account the current four-year rotation for the summer Olympic games, however, we see that the model will only provide a positive prediction up through the Olympic games in the year 2924. To be even more realistic, we should expect any 100-m dash to be completed in some positive amount of time; however, it may be difficult for students to put a specific value on a reasonable result. This discussion could also open up the topic of extrapolation versus interpolation when using linear models."
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"http://s3.amazonaws.com/illustrativemathematics/images/000/002/409/large/100m_Dash_Scatterplo_040e4049d135468a0a72fe9bafbd27b1.jpg",
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https://fr.mathworks.com/matlabcentral/cody/problems/109-check-if-sorted/solutions/1580622 | [
"Cody\n\n# Problem 109. Check if sorted\n\nSolution 1580622\n\nSubmitted on 11 Jul 2018 by madhan ravi\nThis solution is locked. To view this solution, you need to provide a solution of the same size or smaller.\n\n### Test Suite\n\nTest Status Code Input and Output\n1 Pass\nx = sort(rand(1,10^5)); y_correct = 1; assert(isequal(sortok(x),y_correct))\n\n2 Pass\nx = [1 5 4 3 8 7 3]; y_correct = 0; assert(isequal(sortok(x),y_correct))"
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.62757355,"math_prob":0.9687319,"size":426,"snap":"2020-34-2020-40","text_gpt3_token_len":133,"char_repetition_ratio":0.14691943,"word_repetition_ratio":0.0,"special_character_ratio":0.33333334,"punctuation_ratio":0.085365854,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97314966,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-05T02:57:59Z\",\"WARC-Record-ID\":\"<urn:uuid:2d2837da-46bd-43b7-a400-6ba72a33af7c>\",\"Content-Length\":\"72680\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e056a6af-c160-4c81-8384-a0aea597cec4>\",\"WARC-Concurrent-To\":\"<urn:uuid:8ea34536-d484-44f8-9d51-dbf6e6af6b42>\",\"WARC-IP-Address\":\"23.212.144.59\",\"WARC-Target-URI\":\"https://fr.mathworks.com/matlabcentral/cody/problems/109-check-if-sorted/solutions/1580622\",\"WARC-Payload-Digest\":\"sha1:ZKR3F4MMP73QUX3SSOICWTE5MELC75GV\",\"WARC-Block-Digest\":\"sha1:5PNYPESG4MJ26E5WUV7F3VA7TETFWNC6\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735906.77_warc_CC-MAIN-20200805010001-20200805040001-00043.warc.gz\"}"} |
https://ebooks.spiedigitallibrary.org/conference-proceedings-of-spie/9979/997907/Paraxial-polarized-waves-in-inhomogeneous-media/10.1117/12.2237332.short?SSO=1 | [
"Translator Disclaimer\n19 September 2016 Paraxial polarized waves in inhomogeneous media\nAbstract\nA paraxial equation for electromagnetic wave propagation in a random medium is extended to include the depolarization effects in the narrow-angle, forward-scattering setting. A system of two coupled parabolic equations describes propagation of the polarized wave through random medium. In the Cartesian coordinate formulation the coupling term is related to the second mixed derivative of the refractive index. Closed-form parabolic equation for propagation of the coherence tensor is derived under a Markov random process propagation model. The scattering term in this equation includes a rank-four tensor that contains derivatives of the correlation function of the refractive index up to the fourth order. This equation can be also formulated as vector equations for generalized Stokes or lexicographic vectors. In contrast to the scalar case, these equations do not have an analytical solution. For a general partially coherent and partially polarized beam wave, this equation can be reduced to a system of ordinary differential equations allowing a simple numeric solution. For a special case of statistically homogeneous waves an analytical solution exists. In the Stokes vector formulation this solution is described by a range-dependent Mueller matrix. For isotropic random medium this Mueller matrix is diagonal and describes a pure non-uniform depolarizer. Statistics of the random medium is wrapped in a single parameter – depolarization length which is proportional to the fourth derivative of the covariance function at zero. For propagation through atmospheric turbulence estimates based on the perturbation solution support the common knowledge that the depolarization at the optical frequencies is negligible.\nConference Presentation",
null,
"",
null,
""
]
| [
null,
"https://ebooks.spiedigitallibrary.org/conference-proceedings-of-spie/9979/997907/Paraxial-polarized-waves-in-inhomogeneous-media/10.1117/Content/themes/SPIEImages/DL_Icon_Video_Lock.png",
null,
"https://ebooks.spiedigitallibrary.org/Content/themes/SPIEImages/Share_white_icon.png",
null
]
| {"ft_lang_label":"__label__en","ft_lang_prob":0.88045394,"math_prob":0.97538996,"size":2790,"snap":"2021-04-2021-17","text_gpt3_token_len":515,"char_repetition_ratio":0.1328069,"word_repetition_ratio":0.0,"special_character_ratio":0.17670251,"punctuation_ratio":0.08388521,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99532235,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,1,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-01-22T12:42:53Z\",\"WARC-Record-ID\":\"<urn:uuid:8d462919-3aa3-421f-ad36-ae0d10283d7c>\",\"Content-Length\":\"174879\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:497dbe1a-b752-4927-b349-67e487c6fcc9>\",\"WARC-Concurrent-To\":\"<urn:uuid:a487617d-9ed3-46be-b67f-b2167705692e>\",\"WARC-IP-Address\":\"107.154.251.12\",\"WARC-Target-URI\":\"https://ebooks.spiedigitallibrary.org/conference-proceedings-of-spie/9979/997907/Paraxial-polarized-waves-in-inhomogeneous-media/10.1117/12.2237332.short?SSO=1\",\"WARC-Payload-Digest\":\"sha1:UKWRYITJMEAYHJVATZMFYLMEWOCM5N3T\",\"WARC-Block-Digest\":\"sha1:I7HOKE6ZBV7ANYIAOPRDSHNU2GP2EHTX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-04/CC-MAIN-2021-04_segments_1610703529331.99_warc_CC-MAIN-20210122113332-20210122143332-00276.warc.gz\"}"} |
https://www.kartzapper.com/collections/swing-windows | [
"## Swing Windows\n\nFilter\nView\nShowing 1 - 48 of 885 Items\nRs. 6,162.65\nRs. 3,813.40\nRs. 7,027.57\nRs. 7,130.53\nRs. 3,111.93\nRs. 8,050.21\nRs. 5,986.93\nRs. 6,708.68\nRs. 2,267.22\nRs. 4,352.50\nRs. 19,485.23\nRs. 6,181.85\nRs. 8,246.14\nRs. 1,171.61\nRs. 1,057.65\nRs. 9,995.02\nRs. 2,084.29\nRs. 5,676.03\nRs. 9,804.60\nRs. 5,365.14\nRs. 6,806.64\nRs. 7,016.57\nRs. 6,718.68\nRs. 7,947.95\nRs. 3,308.86\nRs. 13,186.42\nRs. 11,364.06\nRs. 6,138.37\nRs. 1,042.65\nRs. 3,746.60\nRs. 6,512.75\nRs. 1,475.80\nRs. 1,455.50\nRs. 6,213.85\nRs. 1,321.06\nRs. 6,693.68\nRs. 3,097.93\nRs. 3,938.64\nRs. 5,680.03\nRs. 3,116.93\nRs. 4,136.57\nRs. 18,545.55\nRs. 10,115.49\nRs. 51,516.08\nRs. 5,687.03\nRs. 5,159.62\nRs. 8,276.14\nRs. 7,739.32"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.5520348,"math_prob":0.99966824,"size":640,"snap":"2021-21-2021-25","text_gpt3_token_len":233,"char_repetition_ratio":0.21698113,"word_repetition_ratio":0.0,"special_character_ratio":0.35,"punctuation_ratio":0.21969697,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9589103,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-16T18:53:57Z\",\"WARC-Record-ID\":\"<urn:uuid:de55147f-4e71-44a5-b9b4-08694f6b4732>\",\"Content-Length\":\"453320\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d96dde83-c67e-411f-988d-bfe124097aac>\",\"WARC-Concurrent-To\":\"<urn:uuid:6395e359-9ce7-4d4a-b4eb-66ba4ff7e3a2>\",\"WARC-IP-Address\":\"23.227.38.74\",\"WARC-Target-URI\":\"https://www.kartzapper.com/collections/swing-windows\",\"WARC-Payload-Digest\":\"sha1:NOL2KPTH6T4QO447RA5PKDIJSBU6RV65\",\"WARC-Block-Digest\":\"sha1:27L2GWK2IFOZ52MADZGR5XSN3F4VT6LX\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243991178.59_warc_CC-MAIN-20210516171301-20210516201301-00369.warc.gz\"}"} |
https://toloka.ai/docs/crowd-kit/reference/crowdkit.aggregation.classification.gold_majority_vote.GoldMajorityVote.predict_proba/ | [
"# predict_proba\n\ncrowdkit.aggregation.classification.gold_majority_vote.GoldMajorityVote.predict_proba | Source code\n\npredict_proba(self, data: DataFrame)\n\nReturns probability distributions of labels for each task when the model is fitted.\n\n## Parameters description\n\nParametersTypeDescription\ndataDataFrame\n\nThe training dataset of workers' labeling results which is represented as the pandas.DataFrame data containing task, worker, and label columns.\n\n• Returns:\n\nProbability distributions of task labels. The pandas.DataFrame data is indexed by task so that result.loc[task, label] is the probability that the task true label is equal to label. Each probability is in he range from 0 to 1, all task probabilities must sum up to 1.\n\n• Return type:\n\nDataFrame\n\nCrowd-Kit\nOverview\nReference\nAggregation\nDatasets\nLearning\nMetrics\nPostprocessing"
]
| [
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7429477,"math_prob":0.7561867,"size":645,"snap":"2023-14-2023-23","text_gpt3_token_len":143,"char_repetition_ratio":0.14664586,"word_repetition_ratio":0.0,"special_character_ratio":0.18604651,"punctuation_ratio":0.15178572,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9839574,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-06-06T22:42:16Z\",\"WARC-Record-ID\":\"<urn:uuid:d858ed2a-a6e4-4da5-881f-6aa8eb772e99>\",\"Content-Length\":\"327673\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e2bae12-55b9-4db1-b9a2-dcc295ed45f4>\",\"WARC-Concurrent-To\":\"<urn:uuid:996cc71d-4071-4349-b4e7-65a4b55d3eb7>\",\"WARC-IP-Address\":\"20.229.88.228\",\"WARC-Target-URI\":\"https://toloka.ai/docs/crowd-kit/reference/crowdkit.aggregation.classification.gold_majority_vote.GoldMajorityVote.predict_proba/\",\"WARC-Payload-Digest\":\"sha1:DAREC4HGW2B25A7CA6GMPSF5WME73K24\",\"WARC-Block-Digest\":\"sha1:AVCOXUSOZKBOH6BYW4RGNVTUGQLHWTDM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-23/CC-MAIN-2023-23_segments_1685224653183.5_warc_CC-MAIN-20230606214755-20230607004755-00458.warc.gz\"}"} |
https://www.colorhexa.com/f78fa7 | [
"# #f78fa7 Color Information\n\nIn a RGB color space, hex #f78fa7 (also known as Pink Sherbet) is composed of 96.9% red, 56.1% green and 65.5% blue. Whereas in a CMYK color space, it is composed of 0% cyan, 42.1% magenta, 32.4% yellow and 3.1% black. It has a hue angle of 346.2 degrees, a saturation of 86.7% and a lightness of 76.5%. #f78fa7 color hex could be obtained by blending #ffffff with #ef1f4f. Closest websafe color is: #ff9999.\n\n• R 97\n• G 56\n• B 65\nRGB color chart\n• C 0\n• M 42\n• Y 32\n• K 3\nCMYK color chart\n\n#f78fa7 color description : Very soft red.\n\n# #f78fa7 Color Conversion\n\nThe hexadecimal color #f78fa7 has RGB values of R:247, G:143, B:167 and CMYK values of C:0, M:0.42, Y:0.32, K:0.03. Its decimal value is 16224167.\n\nHex triplet RGB Decimal f78fa7 `#f78fa7` 247, 143, 167 `rgb(247,143,167)` 96.9, 56.1, 65.5 `rgb(96.9%,56.1%,65.5%)` 0, 42, 32, 3 346.2°, 86.7, 76.5 `hsl(346.2,86.7%,76.5%)` 346.2°, 42.1, 96.9 ff9999 `#ff9999`\nCIE-LAB 71.017, 41.975, 4.673 55.156, 42.213, 41.8 0.396, 0.303, 42.213 71.017, 42.234, 6.352 71.017, 67.654, -1.353 64.971, 37.833, 7.335 11110111, 10001111, 10100111\n\n# Color Schemes with #f78fa7\n\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #8ff7df\n``#8ff7df` `rgb(143,247,223)``\nComplementary Color\n• #f78fdb\n``#f78fdb` `rgb(247,143,219)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #f7ab8f\n``#f7ab8f` `rgb(247,171,143)``\nAnalogous Color\n• #8fdbf7\n``#8fdbf7` `rgb(143,219,247)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #8ff7ab\n``#8ff7ab` `rgb(143,247,171)``\nSplit Complementary Color\n• #8fa7f7\n``#8fa7f7` `rgb(143,167,247)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #a7f78f\n``#a7f78f` `rgb(167,247,143)``\n• #df8ff7\n``#df8ff7` `rgb(223,143,247)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #a7f78f\n``#a7f78f` `rgb(167,247,143)``\n• #8ff7df\n``#8ff7df` `rgb(143,247,223)``\n• #f2486f\n``#f2486f` `rgb(242,72,111)``\n• #f45f82\n``#f45f82` `rgb(244,95,130)``\n• #f57794\n``#f57794` `rgb(245,119,148)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #f9a7ba\n``#f9a7ba` `rgb(249,167,186)``\n• #fabfcc\n``#fabfcc` `rgb(250,191,204)``\n• #fcd6df\n``#fcd6df` `rgb(252,214,223)``\nMonochromatic Color\n\n# Alternatives to #f78fa7\n\nBelow, you can see some colors close to #f78fa7. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #f78fc1\n``#f78fc1` `rgb(247,143,193)``\n• #f78fb8\n``#f78fb8` `rgb(247,143,184)``\n• #f78fb0\n``#f78fb0` `rgb(247,143,176)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #f78f9e\n``#f78f9e` `rgb(247,143,158)``\n• #f78f96\n``#f78f96` `rgb(247,143,150)``\n• #f7918f\n``#f7918f` `rgb(247,145,143)``\nSimilar Colors\n\n# #f78fa7 Preview\n\nThis text has a font color of #f78fa7.\n\n``<span style=\"color:#f78fa7;\">Text here</span>``\n#f78fa7 background color\n\nThis paragraph has a background color of #f78fa7.\n\n``<p style=\"background-color:#f78fa7;\">Content here</p>``\n#f78fa7 border color\n\nThis element has a border color of #f78fa7.\n\n``<div style=\"border:1px solid #f78fa7;\">Content here</div>``\nCSS codes\n``.text {color:#f78fa7;}``\n``.background {background-color:#f78fa7;}``\n``.border {border:1px solid #f78fa7;}``\n\n# Shades and Tints of #f78fa7\n\nA shade is achieved by adding black to any pure hue, while a tint is created by mixing white to any pure color. In this example, #100105 is the darkest color, while #fffdfd is the lightest one.\n\n• #100105\n``#100105` `rgb(16,1,5)``\n• #22020a\n``#22020a` `rgb(34,2,10)``\n• #35040f\n``#35040f` `rgb(53,4,15)``\n• #470514\n``#470514` `rgb(71,5,20)``\n• #59061a\n``#59061a` `rgb(89,6,26)``\n• #6c081f\n``#6c081f` `rgb(108,8,31)``\n• #7e0924\n``#7e0924` `rgb(126,9,36)``\n• #900a29\n``#900a29` `rgb(144,10,41)``\n• #a30c2e\n``#a30c2e` `rgb(163,12,46)``\n• #b50d34\n``#b50d34` `rgb(181,13,52)``\n• #c70e39\n``#c70e39` `rgb(199,14,57)``\n• #da103e\n``#da103e` `rgb(218,16,62)``\n• #ec1143\n``#ec1143` `rgb(236,17,67)``\n• #ef2151\n``#ef2151` `rgb(239,33,81)``\n• #f0335f\n``#f0335f` `rgb(240,51,95)``\n• #f2466d\n``#f2466d` `rgb(242,70,109)``\n• #f3587c\n``#f3587c` `rgb(243,88,124)``\n• #f46a8a\n``#f46a8a` `rgb(244,106,138)``\n• #f67d99\n``#f67d99` `rgb(246,125,153)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #f8a1b5\n``#f8a1b5` `rgb(248,161,181)``\n• #fab4c4\n``#fab4c4` `rgb(250,180,196)``\n• #fbc6d2\n``#fbc6d2` `rgb(251,198,210)``\n• #fcd8e1\n``#fcd8e1` `rgb(252,216,225)``\n• #feebef\n``#feebef` `rgb(254,235,239)``\n• #fffdfd\n``#fffdfd` `rgb(255,253,253)``\nTint Color Variation\n\n# Tones of #f78fa7\n\nA tone is produced by adding gray to any pure hue. In this case, #c4c2c2 is the less saturated color, while #fc8aa5 is the most saturated one.\n\n• #c4c2c2\n``#c4c2c2` `rgb(196,194,194)``\n• #c9bdc0\n``#c9bdc0` `rgb(201,189,192)``\n• #cdb9bd\n``#cdb9bd` `rgb(205,185,189)``\n• #d2b4bb\n``#d2b4bb` `rgb(210,180,187)``\n• #d7afb8\n``#d7afb8` `rgb(215,175,184)``\n• #dbabb6\n``#dbabb6` `rgb(219,171,182)``\n• #e0a6b3\n``#e0a6b3` `rgb(224,166,179)``\n• #e5a1b1\n``#e5a1b1` `rgb(229,161,177)``\n• #e99dae\n``#e99dae` `rgb(233,157,174)``\n• #ee98ac\n``#ee98ac` `rgb(238,152,172)``\n• #f294a9\n``#f294a9` `rgb(242,148,169)``\n• #f78fa7\n``#f78fa7` `rgb(247,143,167)``\n• #fc8aa5\n``#fc8aa5` `rgb(252,138,165)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #f78fa7 is perceived by people affected by a color vision deficiency. This can be useful if you need to ensure your color combinations are accessible to color-blind users.\n\nMonochromacy\n• Achromatopsia 0.005% of the population\n• Atypical Achromatopsia 0.001% of the population\nDichromacy\n• Protanopia 1% of men\n• Deuteranopia 1% of men\n• Tritanopia 0.001% of the population\nTrichromacy\n• Protanomaly 1% of men, 0.01% of women\n• Deuteranomaly 6% of men, 0.4% of women\n• Tritanomaly 0.01% of the population"
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.50431436,"math_prob":0.57621604,"size":3740,"snap":"2021-21-2021-25","text_gpt3_token_len":1693,"char_repetition_ratio":0.12821199,"word_repetition_ratio":0.010989011,"special_character_ratio":0.53368986,"punctuation_ratio":0.23652366,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9596011,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-13T12:23:13Z\",\"WARC-Record-ID\":\"<urn:uuid:07c8cf1f-a7c9-4564-8e4a-3696186f00ce>\",\"Content-Length\":\"36510\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d0375747-3017-4d83-afb5-5880355a5166>\",\"WARC-Concurrent-To\":\"<urn:uuid:5d2b5e91-971c-475a-a96f-44af538222df>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/f78fa7\",\"WARC-Payload-Digest\":\"sha1:CEGUOLZQ6XEVR2OX4GAA6H6YBXJJGKM5\",\"WARC-Block-Digest\":\"sha1:DIQZNPZ2BBTOYBYJWLC4EGXMMEY2LCIC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989916.34_warc_CC-MAIN-20210513111525-20210513141525-00591.warc.gz\"}"} |
https://mjrube94.savingadvice.com/2006/04/08/energy-savings_7093/ | [
"User Real IP - 3.218.67.1\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => 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Array\n(\n => 103.12.135.37\n)\n\n => Array\n(\n => 49.207.120.225\n)\n\n => Array\n(\n => 182.64.195.207\n)\n\n => Array\n(\n => 103.99.37.16\n)\n\n => Array\n(\n => 46.150.104.221\n)\n\n => Array\n(\n => 104.236.195.147\n)\n\n => Array\n(\n => 103.104.192.43\n)\n\n => Array\n(\n => 24.242.159.118\n)\n\n => Array\n(\n => 39.42.179.143\n)\n\n => Array\n(\n => 111.93.58.131\n)\n\n => Array\n(\n => 193.176.84.127\n)\n\n => Array\n(\n => 209.58.142.218\n)\n\n => Array\n(\n => 69.243.152.129\n)\n\n => Array\n(\n => 117.97.131.249\n)\n\n => Array\n(\n => 103.230.180.89\n)\n\n => Array\n(\n => 106.212.170.192\n)\n\n => Array\n(\n => 171.224.180.95\n)\n\n => Array\n(\n => 158.222.11.87\n)\n\n => Array\n(\n => 119.155.60.246\n)\n\n => Array\n(\n => 41.90.43.129\n)\n\n => Array\n(\n => 185.183.104.170\n)\n\n => Array\n(\n => 14.248.67.65\n)\n\n => Array\n(\n => 117.217.205.82\n)\n\n => Array\n(\n => 111.88.7.209\n)\n\n => Array\n(\n => 49.36.132.244\n)\n\n => Array\n(\n => 171.48.40.2\n)\n\n => Array\n(\n => 119.81.105.2\n)\n\n => Array\n(\n => 49.36.128.114\n)\n\n => Array\n(\n => 213.200.31.93\n)\n\n => Array\n(\n => 2.50.15.110\n)\n\n => Array\n(\n => 120.29.104.67\n)\n\n => Array\n(\n => 223.225.32.221\n)\n\n => Array\n(\n => 14.248.67.195\n)\n\n => Array\n(\n => 119.155.36.13\n)\n\n => Array\n(\n => 101.50.95.104\n)\n\n => Array\n(\n => 104.236.205.233\n)\n\n => Array\n(\n => 122.164.36.150\n)\n\n => Array\n(\n => 157.45.93.209\n)\n\n => Array\n(\n => 182.77.118.100\n)\n\n => Array\n(\n => 182.74.134.218\n)\n\n => Array\n(\n => 183.82.128.146\n)\n\n => Array\n(\n => 112.196.170.234\n)\n\n => Array\n(\n => 122.173.230.178\n)\n\n => Array\n(\n => 122.164.71.199\n)\n\n => Array\n(\n => 51.79.19.31\n)\n\n => Array\n(\n => 58.65.222.20\n)\n\n => Array\n(\n => 103.27.203.97\n)\n\n => Array\n(\n => 111.88.7.242\n)\n\n => Array\n(\n => 14.171.232.77\n)\n\n => Array\n(\n => 46.101.22.182\n)\n\n => Array\n(\n => 103.94.219.19\n)\n\n => Array\n(\n => 139.190.83.30\n)\n\n => Array\n(\n => 223.190.27.184\n)\n\n => Array\n(\n => 182.185.183.34\n)\n\n => Array\n(\n => 91.74.181.242\n)\n\n => Array\n(\n => 222.252.107.14\n)\n\n => Array\n(\n => 137.97.8.28\n)\n\n => Array\n(\n => 46.101.16.229\n)\n\n => Array\n(\n => 122.53.254.229\n)\n\n => Array\n(\n => 106.201.17.180\n)\n\n => Array\n(\n => 123.24.170.129\n)\n\n => Array\n(\n => 182.185.180.79\n)\n\n => Array\n(\n => 223.190.17.4\n)\n\n => Array\n(\n => 213.108.105.1\n)\n\n => Array\n(\n => 171.22.76.9\n)\n\n => Array\n(\n => 202.66.178.164\n)\n\n => Array\n(\n => 178.62.97.171\n)\n\n => Array\n(\n => 167.179.110.209\n)\n\n => Array\n(\n => 223.230.147.172\n)\n\n => Array\n(\n => 76.218.195.160\n)\n\n => Array\n(\n => 14.189.186.178\n)\n\n => Array\n(\n => 157.41.45.143\n)\n\n => Array\n(\n => 223.238.22.53\n)\n\n => Array\n(\n => 111.88.7.244\n)\n\n => Array\n(\n => 5.62.57.19\n)\n\n => Array\n(\n => 106.201.25.216\n)\n\n => Array\n(\n => 117.217.205.33\n)\n\n => Array\n(\n => 111.88.7.215\n)\n\n => Array\n(\n => 106.201.13.77\n)\n\n => Array\n(\n => 50.7.93.29\n)\n\n => Array\n(\n => 123.201.70.112\n)\n\n => Array\n(\n => 39.42.108.226\n)\n\n => Array\n(\n => 27.5.198.29\n)\n\n => Array\n(\n => 223.238.85.187\n)\n\n => Array\n(\n => 171.49.176.32\n)\n\n => Array\n(\n => 14.248.79.242\n)\n\n => Array\n(\n => 46.219.211.183\n)\n\n => Array\n(\n => 185.244.212.251\n)\n\n => Array\n(\n => 14.102.84.126\n)\n\n => Array\n(\n => 106.212.191.52\n)\n\n => Array\n(\n => 154.72.153.203\n)\n\n => Array\n(\n => 14.175.82.64\n)\n\n => Array\n(\n => 141.105.139.131\n)\n\n => Array\n(\n => 182.156.103.98\n)\n\n => Array\n(\n => 117.217.204.75\n)\n\n => Array\n(\n => 104.140.83.115\n)\n\n => Array\n(\n => 119.152.62.8\n)\n\n => Array\n(\n => 45.125.247.94\n)\n\n => Array\n(\n => 137.97.37.252\n)\n\n => Array\n(\n => 117.217.204.73\n)\n\n => Array\n(\n => 14.248.79.133\n)\n\n => Array\n(\n => 39.37.152.52\n)\n\n => Array\n(\n => 103.55.60.54\n)\n\n => Array\n(\n => 102.166.183.88\n)\n\n => Array\n(\n => 5.62.60.162\n)\n\n => Array\n(\n => 5.62.60.163\n)\n\n)\n```\nEnergy Savings: mjrube94's Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Energy Savings",
null,
"",
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"",
null,
"# Energy Savings\n\nApril 8th, 2006 at 01:23 am\n\nA few weeks ago I wrote about an energy saving kit being offered to NJ gas customers for \\$5. I got it today. It included weather stripping, low flow shower heads and faucet aerators, and outlet cover insulators. It also came with two cards with energy and water savings info. Some of the tips on the cards are basic, but others I didn't know, so I figured I'd pass them along. Check out www.niagaraconservation.com for more info.\n\nOne of the things I liked was the listing of kwh used by many appliances, which I didn't know and I'm going to start paying attention to:\n\nDishwasher- 30kWh w/drying cycle, 20 w/o\nOven 350 degrees for 1 hour- 60 kWh\nMicrowave Oven - 16-40 kWh\nToaster- 4-12 kWh\nCoffee Maker- 12kWh\nWashing Machine (5 loads/wk) - 7 kWh\nDryer (5 loads/wk) - 100 kWh\nTelevision 23-25\" (4 hrs/day) - 20 kWh\nVCR/DVD - 5 kWh\nPC Desktop/Monitor - 10 kWh\nCeiling fan - 18 kWh\nCentral Air (750 hrs/yr) - 2,250 kWh/yr\nWater Heater (family of 4) - 250-400 kWh\nRefrig (17cu ft,frost free,10 hrs/day) - 150\n\nMy goal for the weekend is to check my electric bill to see what I'm being charged for a kWh to get a sense for home much \\$\\$ each of these things is costing me. Then I can figure out where best to direct my efforts. (Line Drying clothes looks like a frontrunner already!).\n\nA few other little known (to me) facts:\n*You can cut your water heating bills in half by using low-flow aerating showerheads and faucets.\n*Use task lighting. Instead of lighting an entire room, focus light where you need it.\n*An outdoor central air conditioner unit operating in the shade can use 10% less electricity than if in the sun. (Ours are...not sure how you'd move them if they weren't)\n\nInteresting stuff...I'm looking forward to looking into this further to try and cut down my bill...\n\n### 3 Responses to “Energy Savings”\n\n1. contrary1 Says:\n\nI've tackled my energy bills here, treating it just like it was a challenge game of some sort. I \"win\" if it is lower then the same month last year! And, you're right, dry those clothes any way BUT in the dryer.\n\n2. Ima saver Says:\n\nLots like the hot water heater turned off is where I am saving. thanks for those figures!!\n\n3. Champion Cheapskate Says:\n\nUseful power usage breakdown. I try not to overlook the obvious like unplugging unused appliances and purchasing highly rated energy star gadgets. Doors open or closed? Curtain drawn and open? Vents open or closed? It depends.\n\n(Note: If you were logged in, we could automatically fill in these fields for you.)\n Name: * Email: Will not be published. Subscribe: Notify me of additional comments to this entry. URL: Verification: * Please spell out the number 4. [ Why? ]\n\nvB Code: You can use these tags: [b] [i] [u] [url] [email]"
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"https://www.savingadvice.com/blogs/images/search/top_left.php",
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"https://www.savingadvice.com/blogs/images/search/top_right.php",
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"https://www.savingadvice.com/blogs/images/search/bottom_left.php",
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http://www.newthinktank.com/2013/04/solving-programming-problems/ | [
"",
null,
"# Solving Programming Problems",
null,
"To finish off my Java Algorithm tutorial, I thought it would be interesting to cover solving programming problems in general. So, in this tutorial I’ll answer the question I’ve been getting, which is how to print a tree data structure.\n\nOn our journey to better understand how to solve problems I will first solve the basic problem. Then in the next part of the tutorial I will perfect printing any type of tree. The code below will better explain the process of solving this problem.\n\nCode From the Video\n\n```4 ROW TREE\n\n_______1\n___1_______1\n_1___1___1___1\n1_1_1_1_1_1_1_1\n\n1st Row Indent 7 Spaces 0\n2nd Row Indent 3 Spaces 7\n3rd Row Indent 1 Spaces 3\n4th Row Indent 0 Spaces 1\n\nSpaces : 0 and then whatever Indent was\n\nFirst Index Per Row\n0\n1 2\n3 4 5 6\n7 8 9 10 11 12 13 14\n.5 * (-2 + (Math.pow(2, iteration)))\n\nItems Per Row\n1, 2, 4, 8\nMath.pow(2, iteration - 1)\n\nMax Index Per Row\nindexToPrint + itemsPerRow\n\nIndent Number\nIndent Number Space Number\nIndent Number Space Number Space...\n\nI need 1 index followed by multiple numbers & spaces every time```\n\n```public void printTree(int rows) {\n\nint spaces = 0;\n\nint iteration = 1;\n\nwhile (iteration <= rows) {\n\nint indent = (int) Math\n.abs(((Math.pow(-2, -iteration)) * (-16 + (Math.pow(2,\niteration)))));\n\nint indexToPrint = (int) (.5 * (-2 + (Math.pow(2, iteration))));\n\nint itemsPerRow = (int) (Math.pow(2, iteration - 1));\n\nint maxIndexToPrint = indexToPrint + itemsPerRow;\n\nfor (int j = 0; j < indent; j++)\nSystem.out.print(\" \");\n\nfor (int l = indexToPrint; l < maxIndexToPrint; l++) {\n\nSystem.out.print(theHeap[l].key);\n\nfor (int k = 0; k < spaces; k++)\nSystem.out.print(\" \");\n\n}\n\nspaces = indent;\n\niteration++;\n\nSystem.out.println();\n\n}\n\n}\n```"
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"http://www.newthinktank.com/wp-content/themes/WhosWho/images/share.gif",
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"http://www.newthinktank.com/wp-content/uploads/2013/04/Solving-Programming-Problems.png",
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https://inspirated.com/2009/09/10/howto-use-latex-mathematical-expressions-in-pygtk | [
"# Inspirated",
null,
"",
null,
"",
null,
"September 10, 2009\n\n## HOWTO: Use LaTeX mathematical expressions in PyGTK\n\nFiled under: Blog — krkhan @ 10:04 pm\n\nI had never really laid my hands on LaTeX until I required it in one of the helper applications for my graduation project. Unfortunately, the requirement wasn’t as simple as producing some documents as I had to embed mathematical expressions on the fly in my PyGTK apps. Googling around for the solution, I found GtkMathView which accomplished something similar to this albeit using MathML. However, my luck ran out on me again as the widget lacked Python bindings. The other solution was to generate transparent PNGs on the fly and include them as GtkImages. This worked rather well, as the final code allowed easy modifications to the generated expressions.\n\nRequirements for the code were:\n\nFinal results:",
null,
"And the simple code behind it:\n\n1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 #!/usr/bin/env python \"\"\"An example demonstrating usage of latexmath2png module for embedding math equations in PyGTK Author: Kamran Riaz Khan <[email protected]> \"\"\" import gtk import os import latexmath2png pre = 'gtktex_' eqs = [ r'$\\alpha_i > \\beta_i$', r'$\\sum_{i=0}^\\infty x_i$', r'$\\left(\\frac{5 - \\frac{1}{x}}{4}\\right)$', r'$s(t) = \\mathcal{A}\\sin(2 \\omega t)$', r'$\\sum_{n=1}^\\infty\\frac{-e^{i\\pi}}{2^n}$' ] latexmath2png.math2png(eqs, os.getcwd(), prefix = pre) def window_destroy(widget): for i in range(0, len(eqs)): os.unlink(os.path.join(os.getcwd(), '%s%d.png' % (pre, i + 1))) gtk.main_quit() window = gtk.Window() window.set_border_width(10) window.set_title('LaTeX Equations in GTK') window.connect('destroy', window_destroy) vbox = gtk.VBox(spacing = 10) window.add(vbox) images = [None] * len(eqs) for i in range(len(eqs)): images[i] = gtk.image_new_from_file('%s%d.png' % (pre, i + 1)) vbox.pack_start(images[i]) window.show_all() gtk.main()\nTags: , , , , , , , , , ,\n\n1. […] HOWTO: Use LaTeX mathematical expressions in PyGTK […]\n\nPingback by (GUI, Mathematical Equations, Scientific Plotting) = (GTK+, LaTeX, Matplotlib) | Inspirated — January 6, 2010 @ 10:16 am\n\n2. at first, it didn’t work on my pc. then i noticed that the directory name where i extracted file had non-ascii chars in it and it stalls because of os.path etc. :)) anyway, good work\n\nComment by Mustafa Yılmaz — April 14, 2010 @ 5:18 am"
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"https://inspirated.com/wordpress/wp-content/themes/inspirated/header/header_banner_3.png",
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"https://inspirated.com/uploads/gtktex.png",
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https://metanumbers.com/464699 | [
"# 464699 (number)\n\n464,699 (four hundred sixty-four thousand six hundred ninety-nine) is an odd six-digits prime number following 464698 and preceding 464700. In scientific notation, it is written as 4.64699 × 105. The sum of its digits is 38. It has a total of 1 prime factor and 2 positive divisors. There are 464,698 positive integers (up to 464699) that are relatively prime to 464699.\n\n## Basic properties\n\n• Is Prime? Yes\n• Number parity Odd\n• Number length 6\n• Sum of Digits 38\n• Digital Root 2\n\n## Name\n\nShort name 464 thousand 699 four hundred sixty-four thousand six hundred ninety-nine\n\n## Notation\n\nScientific notation 4.64699 × 105 464.699 × 103\n\n## Prime Factorization of 464699\n\nPrime Factorization 464699\n\nPrime number\nDistinct Factors Total Factors Radical ω(n) 1 Total number of distinct prime factors Ω(n) 1 Total number of prime factors rad(n) 464699 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 13.0491 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0\n\nThe prime factorization of 464,699 is 464699. Since it has a total of 1 prime factor, 464,699 is a prime number.\n\n## Divisors of 464699\n\n2 divisors\n\n Even divisors 0 2 1 1\nTotal Divisors Sum of Divisors Aliquot Sum τ(n) 2 Total number of the positive divisors of n σ(n) 464700 Sum of all the positive divisors of n s(n) 1 Sum of the proper positive divisors of n A(n) 232350 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 681.688 Returns the nth root of the product of n divisors H(n) 2 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors\n\nThe number 464,699 can be divided by 2 positive divisors (out of which 0 are even, and 2 are odd). The sum of these divisors (counting 464,699) is 464,700, the average is 232,350.\n\n## Other Arithmetic Functions (n = 464699)\n\n1 φ(n) n\nEuler Totient Carmichael Lambda Prime Pi φ(n) 464698 Total number of positive integers not greater than n that are coprime to n λ(n) 464698 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 38738 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares\n\nThere are 464,698 positive integers (less than 464,699) that are coprime with 464,699. And there are approximately 38,738 prime numbers less than or equal to 464,699.\n\n## Divisibility of 464699\n\n m n mod m 2 3 4 5 6 7 8 9 1 2 3 4 5 4 3 2\n\n464,699 is not divisible by any number less than or equal to 9.\n\n## Classification of 464699\n\n• Arithmetic\n• Prime\n• Deficient\n\n### Expressible via specific sums\n\n• Polite\n• Non-hypotenuse\n\n• Prime Power\n• Square Free\n\n## Base conversion (464699)\n\nBase System Value\n2 Binary 1110001011100111011\n3 Ternary 212121110002\n4 Quaternary 1301130323\n5 Quinary 104332244\n6 Senary 13543215\n8 Octal 1613473\n10 Decimal 464699\n12 Duodecimal 1a4b0b\n20 Vigesimal 2i1ej\n36 Base36 9ykb\n\n## Basic calculations (n = 464699)\n\n### Multiplication\n\nn×y\n n×2 929398 1394097 1858796 2323495\n\n### Division\n\nn÷y\n n÷2 232350 154900 116175 92939.8\n\n### Exponentiation\n\nny\n n2 215945160601 100349500186124099 46632312386991682681201 21669988933922647950271423499\n\n### Nth Root\n\ny√n\n 2√n 681.688 77.4564 26.1092 13.5967\n\n## 464699 as geometric shapes\n\n### Circle\n\n Diameter 929398 2.91979e+06 6.78412e+11\n\n### Sphere\n\n Volume 4.20343e+17 2.71365e+12 2.91979e+06\n\n### Square\n\nLength = n\n Perimeter 1.8588e+06 2.15945e+11 657184\n\n### Cube\n\nLength = n\n Surface area 1.29567e+12 1.0035e+17 804882\n\n### Equilateral Triangle\n\nLength = n\n Perimeter 1.3941e+06 9.3507e+10 402441\n\n### Triangular Pyramid\n\nLength = n\n Surface area 3.74028e+11 1.18263e+16 379425\n\n## Cryptographic Hash Functions\n\nmd5 0a6f453429669ec39f4ba1bc5b76749c 1ebb35f04cd186e30eb21684a9e079d223611979 c703bea753331123de6773781bb6efb236e252d9a7eccaf34ad72e3a176e8c35 510967341c317bcaf4d040e22cb2d0fdd4694b716fe3e4c18becd94f8b44a7002a7d059136a99730163e62dca9c68489c03dbfddc92996b153cda8cb2ff5b188 d79a4beeb7316de46e5e3863753c662ffde3020d"
]
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https://mathoverflow.net/questions/402310/lemoine-lozada-circles | [
"I made some rookie attempt to define the 4th Lemoine circle recently. The alternative name for this circle was suggested yesterday. Further investigation revealed a family of circles associated with the Brocard axis. Supposedly the name of Ehrmann-Lozada circles (or Lemoine-Lozada circles) can be safely attached to them.\n\nSpecifically, it appears that for each Lozada circle it is possible to construct at least one circle of the Lemoine type, with its center on the line that is going through the Symmedian point and the circumcenter of the triangle.\n\nstep 1: Let A'B'C' be the midheight triangle of ABC and let BA and CA be the orthogonal projections of B' and C' on BC, respectively. Build AB, CB, AC and BC similarly. These last six points lie on a circle with center X(9729). The circle is called the 1st Lozada circle. (César Lozada, March 17, 2016)\n\nstep 2: Circumcircles of the triangles BcAcX(6), AbCbX(6), BaCaX(6) cut the sides of the original triangle at six points, that define the 1st Ehrmann-Lozada circle. Its center O_1 is not in the ETC.",
null,
"Ehrmann-Lozada circles can be constructed in the similar manner for each(?) of the ten remaining Lozada circles:\n\n1. https://www.geogebra.org/geometry/pg9mrmzx 2 new circles of the Lemoine type\n2. https://geogebra.org/geometry/uqnpqjeh 2 new circles of the Lemoine type\n3. https://www.geogebra.org/geometry/t4jpn9dp only 1 new circle of the Lemoine type\n\n...\n11. https://www.geogebra.org/m/kppjdcny gives us only 1 new circle of the Lemoine type...\n\nMore links will be added soon. None of these points O_i is included in Clark Kimberling's encyclopaedia.\n\nFinding a strict proof of existence of these new circles is the main question here. Bit I am also interested in what makes Lozada points/circles so special? (Taylor point belongs to the Brocards Axis as well, but for some reason it is impossible to construct an Ehrmann-Taylor circle in the same way!) Are these Ehrmann-Lozada circles also Tucker circles?\n\nOnce you have worked through even a handful of examples of a construction, it can be worthwhile to abstract things a bit in hopes of glimpsing a general principle.\n\nA cyclic sextuple of points —two on each side-line of a triangle— is determined by a triple of of those points (one on each side-line). We can ask when such a triple gives rise to a secondary cyclic sextuple by the OP's construction, recapped below. (In what follows, \"$$X_Y$$\" and \"$$X_Y'$$\" indicate points on the side-line opposite triangle vertex $$X$$.)",
null,
"• Given $$A_B$$, $$B_C$$, $$C_A$$, define $$A_C$$, $$B_A$$, $$C_B$$ as the \"other\" points where $$\\bigcirc A_BB_CC_A$$ meets the (side-lines of) the triangle.\n• Further, given a specified point $$S$$, let $$B_C'$$, $$C_B'$$ be the \"other\" points where $$\\bigcirc SB_AC_A$$ meets the triangle; likewise, we have $$C_A'$$ and $$A_C'$$ via $$\\bigcirc SC_BA_B$$, and $$A_B'$$ and $$B_A'$$ via $$\\bigcirc SA_CB_C$$.\n\nDefinition. Triple $$A_B$$, $$B_C$$, $$C_A$$ is magical with respect to $$S$$ if the six derived points $$A_B'$$, $$A_C'$$, $$B_C'$$, $$B_A'$$, $$C_A'$$, $$C_B'$$ lie on a circle (which I'll call the triple's (First) Magic Circle).\n\n(Note that the definition says nothing about the centers of the circumcircles involved. I'll get to that.)\n\nFor general $$S$$, the condition of magicality is complicated to express, involving a quartic relation. The condition happens to be replete with side-squares $$a^2$$, $$b^2$$, $$c^2$$, and the relation factors nicely when $$S$$ is the symmedian point, with barycentric coordinates $$(a^2:b^2:c^2)$$. For this case, we can state\n\nTheorem. Triple $$A_B$$, $$B_C$$, $$C_A$$ is magical with respect to symmedian point $$S$$ if and only if $$\\frac{\\alpha}{c^2}=\\frac{\\beta}{a^2}=\\frac{\\gamma}{b^2} \\qquad\\text{or}\\qquad \\frac{1-\\alpha}{b^2}=\\frac{1-\\beta}{c^2}=\\frac{1-\\gamma}{a^2} \\tag{1}$$ for the signed ratios $$\\alpha := \\frac{|BA_B|}{|BC|}\\qquad \\beta :=\\frac{|CB_C|}{|CA|} \\qquad \\gamma := \\frac{|AC_A|}{|AB|}$$\n\nProof is by Mathematica-assisted symbol crunching (that I won't reproduce here), fueled mostly by power-of-a-point relations among all the points on the side-lines; the relations encoding the role of $$S$$ are what bog the calculations down. For the symmedian case, I suspect there's a not-unreasonable synthetic proof of $$(1)$$. (I should acknowledge: I don't doubt $$(1)$$, and much (most? all?) of what follows, already appears in the literature.)\n\nIf the first proportion in $$(1)$$ holds for $$A_B$$, $$B_C$$, $$C_A$$, then the second holds for $$A_C$$, $$B_A$$, $$C_B$$ (redefining $$\\alpha$$, $$\\beta$$, $$\\gamma$$ appropriately). Consequently, the latter triple is also magical, which allows us to say\n\nIf triple $$A_B$$, $$B_C$$, $$C_A$$ is magical with respect to symmedian point $$S$$, then the six \"other\" points where $$\\bigcirc SA_BB_A$$, $$\\bigcirc SB_CC_A$$, $$\\bigcirc SC_AA_C$$ meet the triangle are also concyclic, giving a Second Magic Circle.",
null,
"Moreover, the construction is symmetric in the given points $$A_B$$, $$B_C$$, $$C_A$$ and derived points $$A_B'$$, $$B_C'$$, $$C_A'$$; thus, the derived triple is necessarily also magical (as is derived triple $$A_C'$$, $$B_A'$$, $$C_B'$$). We don't get any new circles, however; rather, we get the circumcircle of our original points (their \"Zeroth Magic Circle\"), and Second Magic Circle noted above. We effectively have a magical triple of circles: any one gives rise to the other two.\n\nFor a bit of metric specificity ... Define $$\\kappa_0$$ as the (normalized) constant of proportionality in $$(1)$$. $$\\frac{\\kappa_0}{a^2+b^2+c^2} \\;:=\\; \\frac{\\alpha}{c^2}=\\frac{\\beta}{a^2}=\\frac{\\gamma}{b^2}$$ Likewise defining $$\\kappa_1$$ and $$\\kappa_2$$ relative to the First and Second Magic Circles, we find $$\\kappa_{i+1} = \\frac{3 - \\kappa_i}{2 - \\kappa_i}$$ Also, defining $$k_i$$ as the radius of $$i$$-th Magic Circle, we have $$k_i^2 = \\frac{r^2}{(a^2 + b^2 + c^2)^2} \\left(\\;(a^2 + b^2 + c^2)^2 (1-\\kappa_i) \\;+\\; (a^2 b^2+b^2c^2+c^2a^2)\\kappa_i^2\\;\\right)$$ where $$r$$ is the circumradius of $$\\triangle ABC$$.\n\nRecall that the definition of a magical triple of points made no assumptions about the centers of the original or derived circles; in particular, it seems to have ignored what seems to be the key property motivating OP's question. It happens that the symmedian case gives us that property for free:\n\nIf triple $$A_B$$, $$B_C$$, $$C_A$$ is magical with respect to symmedian point $$S$$, then its circumcenter lies on the Brocard axis (joining $$S$$ to the circumcenter $$O$$ of $$\\triangle ABC$$).\n\nSince the construction transfers magic to the derived points, we can say the Zeroth, First, and Second Magic Circles have centers $$K_i$$ on the Brocard axis. And, specifically, $$\\frac{|OK_i|}{|OS|}=\\frac{\\kappa_i}{2}$$\n\nThis, and the radius calculation, tell us how to construct a Magic Circle about any point on the Brocard axis. (Sanity check: $$\\kappa=0$$ corresponds to $$k=r$$. The circle is the triangle's circumcircle. Points $$A_B$$, $$B_C$$, $$C_A$$ coincide with the vertices, so $$\\alpha=\\beta=\\gamma=0$$.)\n\nAlso, one can confirm that each of the three Magic Circles is a Tucker circle. Converting the above results into Tucker/Brocard parameters is left as an exercise to the reader.\n\nOkay, that's enough of that.\n\nI don't know if there's a unifying principle in that family of circles, or if \"the $$n$$-th Lozada Circle\" is simply \"the $$n$$-th circle that Lozada happened to notice had a neat construction and a center on the Brocard axis\". So, I can't say whether it's surprising —or maybe it's obvious— that members of the family are magical. Nevertheless, I'll explicitly confirm the magical nature of three cases illustrated by OP. (I'm afraid I don't have the patience to work through all eleven.)\n\nTo avoid a little clutter, define $$\\lambda := \\kappa/(a^2+b^2+c^2)$$ as the un-normalized proportionality factor in $$(1)$$. Revealing magic is simply a matter of showing that the value of $$\\lambda$$ is symmetric in triangle metrics.\n\n• First Lozada Circle: $$A_B$$, $$B_C$$, $$C_A$$ (and $$A_C$$, $$B_A$$, $$C_B$$) are orthogonal projections of mid-height points. Straightforward right-triangle trig yields $$|BA_B|=r\\sin^2C\\sin A, \\; |CB_C|=r\\sin^2A\\sin B, \\; |AC_A|=r\\sin^2B \\sin C\\;\\to\\; \\lambda=\\frac{1}{8r^2}$$\n• Second Lozada Circle: $$A_B$$, $$B_C$$, $$C_A$$ (and $$A_C$$, $$B_A$$, $$C_B$$) are orthogonal projections of points where altitudes meet the circle having the orthocenter and centroid as a diameter. A coordinate bash gives $$|BA_B| = \\frac{8|\\triangle ABC|^2}{3ab^2}, \\quad |CB_C|=\\cdots, \\quad |AC_A|=\\cdots \\quad\\to\\quad \\lambda = \\frac{8|\\triangle ABC|^2}{3a^2b^2c^2}$$\n• Fourth Lozada Circle: $$A_B$$, $$B_C$$, $$C_A$$ (and $$A_C$$, $$B_A$$, $$C_B$$) are points where the side-lines meet lines through \"opposite\" sides of Pythagoras-esque squares erected on the triangle's sides. Here again, simple trig allows us to write $$|BA_B|=-\\frac{c}{\\sin B}, \\quad |CB_C|=\\cdots, \\quad |AC_A|=\\cdots \\quad\\to\\quad \\lambda=-\\frac{1}{2|\\triangle ABC|}$$ (I don't know why OP asserts the existence of \"only one\" circle derived from the Fourth Lozada Circle. There are two.)\n\nOP seems to be suggesting that the Taylor circle doesn't quite fit with this investigation. In fact, the Taylor Circle (rather, each of two triples of points it determines on the triangle) is magical:\n\n• Taylor Circle: $$A_B$$, $$B_C$$, $$C_A$$ (and $$A_C$$, $$B_A$$, $$C_B$$) are orthogonal projections of the feet of triangle's altitudes onto the side-lines. Specifically, with $$X_Y$$ the projection of the foot from vertex $$Y$$ projected onto the side-line opposite vertex $$X$$, we find $$|BA_B|= a\\sin^2C \\qquad |CB_C|=b\\sin^2A \\qquad |AC_A|=c\\sin^2B \\quad\\to\\quad \\lambda=\\frac{1}{4r^2}$$ Here's a figure showing a Taylor Circle (blue) and associated First and Second Magic Circles:",
null,
"I'll stop typing now.\n\n• I appreciate the amount of labour you have put into this! Believe it or not, there is one more way to construct 6 point circles that I am pinning my last hopes on. 'The 'inscribed-circumscribed' circle with its centre at X1385 and the 'ninepoint-circumscribed' one with its centre at X140 can be both obtained with the help of the same new(?) 'method'. The circles themselves are totally unknown (hopefully) and might have some arguably good properties. They actually greatly remind me of the first two Droz Farny circles... Sep 2, 2021 at 7:39\n• As concerns this part: \"I should acknowledge: I don't doubt (1) and much of what follows, already appears in the literature\" it is a pity that as always in geometry there is no royal way to instantly verify it. Perhaps some advanced Blockchain(?) solution will be invented in the near future to address this issue. Sep 2, 2021 at 7:44\n• \"[I]t is a pity that [...] there is no royal way to instantly verify [that a result exists in the literature]\" That's why it's important to read the literature. :) I say that mostly to admonish myself: If I'd paid more attention to the most-basic lore of Tucker circles in MathWorld, I would've realized sooner that \"magic\" & \"Tucker-ness\" are equivalent properties of circles. (I should update the answer to make this clear.) Thus, it actually is \"obvious\" that Lozada/Taylor circles listed are \"magic\", because they're \"known\" to be Tucker. (Even so, confirming by (1) is a nice exercise.)\n– Blue\nSep 2, 2021 at 10:18\n• @Blue, I presume that in your comment you mean \"magic with respect to X(6)\" rather than \"magic\" in general? Sep 2, 2021 at 12:04\n• @PeterTaylor: Yes, sorry. Lack of specificity was a casualty of the character limit; since I have space to burn here, I'll be explicit: \"Tucker-ness\" and \"magic with respect to the symmedian point\" are equivalent properties of circles.\n– Blue\nSep 2, 2021 at 12:12"
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"https://i.stack.imgur.com/fegWQ.png",
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"https://i.stack.imgur.com/kUsqS.png",
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.8371547,"math_prob":0.99960726,"size":7688,"snap":"2023-40-2023-50","text_gpt3_token_len":2313,"char_repetition_ratio":0.13079126,"word_repetition_ratio":0.06205674,"special_character_ratio":0.30593133,"punctuation_ratio":0.13163197,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99992526,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,1,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-11-30T11:12:07Z\",\"WARC-Record-ID\":\"<urn:uuid:2472d379-12ea-4539-af64-bd043994195b>\",\"Content-Length\":\"128650\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:47fa4502-7676-4bbc-8188-0c6d9f495039>\",\"WARC-Concurrent-To\":\"<urn:uuid:82b63aba-7c3c-4199-96a7-5c8f41efdfda>\",\"WARC-IP-Address\":\"172.64.150.182\",\"WARC-Target-URI\":\"https://mathoverflow.net/questions/402310/lemoine-lozada-circles\",\"WARC-Payload-Digest\":\"sha1:ZLTJXKNBREI5WEDFQ7ZCPMBTPUEUBJ5F\",\"WARC-Block-Digest\":\"sha1:P3TS6RHXEEM2ODVZ5A5WLLFERHKWV6SA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-50/CC-MAIN-2023-50_segments_1700679100184.3_warc_CC-MAIN-20231130094531-20231130124531-00152.warc.gz\"}"} |
https://physics.stackexchange.com/questions/404307/why-is-the-standard-model-gauge-group-a-simple-product | [
"# Why is the Standard Model gauge group a simple product?\n\nThe Standard Model gauge group is always given as $\\text{SU}(3)\\times\\text{SU}(2)\\times\\text{U}(1)$ (possibly quotiented by some subgroup that acts trivially on all konwn fields). However, at the level of (perturbative) Yang-Mills theory, only the local structure of this group is needed, that is, the algebra $\\mathfrak{su}(3)\\oplus\\mathfrak{su}(2)\\oplus\\mathfrak{u}(1)$. However, since this is only a local structure, it would still be the Lie algebra of a more complicated bundle of the three gauge groups. So, why do we say that the Standard Model gauge group can be written as a trivial product of three gauge groups, instead of something more topologically interesting? Are there theoretical or experimental considerations that force the trivialization?\n\nNote: I am not asking why the QCD term is $\\text{SU}(3)$ instead of some other group with the same Lie algebra (and similarly for the $\\text{SU}(2)$ and $U(1)$ parts), since that question was answered here.\n\n• \"However, since this is only a local structure, it would still be the Lie algebra of a more complicated bundle of the three gauge groups.\" What do you mean by this? Why \"bundle\" (bundle over what base?)? Can you give an example of a different group that you think has the same Lie algebra? Note that twisting the product structure, e.g. as a semi-direct product, would also twist the Lie algebra structure, i.e. the sum of the algebras would not be a direct sum anymore. – ACuriousMind May 6 '18 at 19:07\n• This article by David Tong might be relevant to your question (he argues that the gauge group of the standard model is actually an undetermined quotient of the direct product) : arxiv.org/abs/1705.01853 – Antoine May 6 '18 at 19:26\n• @ACuriousMind Construct an arbitrary $\\text{SU}(3)$ bundle over $\\text{SU}(2)$, then construct a $\\text{U}(1)$ bundle over that resulting bundle, as manifolds. This will locally look like the typical standard model gauge group, and so the local structure will be the same, no? – Bob Knighton May 6 '18 at 19:43\n• How is this question different from the linked one? I don't think you're characterizing the linked question correctly, it seems to be asking the same thing as this one. – tparker May 6 '18 at 19:59\n• The local structure as a manifold will be the same, but you don't have any group structure on such bundles - while you could locally in trivial patchs use the group structure from the trivial product, it is not evident that this lifts to a well-defined group structure on the whole bundle. – ACuriousMind May 6 '18 at 20:06\n\nActually while the Lie algebra $$su(3) \\oplus su(2) \\oplus u(1)$$ of the standard model is a direct sum of three simple Lie algebras, the gauge group itself appears to be the group $$S(U(3) \\times U(2))$$."
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https://2021.help.altair.com/2021/embedse/buffer.htm | [
"## buffer\n\nBlock Category: Matrix Operation\n\nInputs: Scalar\n\nDescription: The buffer block places a sequence of values in a buffer based on the buffer length, the time between successive samples, and the duration of the simulation. The buffer produces a single vector output.\n\nWhen the buffer block is added to a feedback loop, the buffer updates on its dt value.\n\nThis block is useful for performing basic digital signal processing operations.",
null,
"Add “Buffer Full” Output Pin: Adds an output pin that goes high when the buffer is full.\n\nAdd Input Trigger: Runs if trigger input is true (1); or does nothing if false (0).\n\nBuffer Length: Determines the number of samples; that is, the size of the buffer.\n\ndT: Determines the time between successive samples; that is, the rate at which samples of the incoming signal are collected and placed in the buffer. When a buffer block is added to a feedback loop, the buffer updates on its dt value.\n\nReverse Buffer: Reverses the order of elements in the buffer. Instead of newest to oldest they are ordered oldest to newest.\n\nUse Circular Buffer: When activated, the buffer block performs full buffer chunks at a time. It starts writing at the far end of the buffer and finishes at the first element. If you also activated Add “Buffer Full” Output Pin, a flag is set indicating the buffer is ready.\n\n#### Examples\n\n1. Basic buffer operation\n\nConsider the following buffer block, with a buffer length of 4 and time between successive samples of 0.01.",
null,
"For simplicity, let the simulation step size be equal to 0.01. If the input to the buffer is an arbitrary non-zero signal, such as a ramp signal, then after two simulation time steps, the output of buffer is a vector of length 4, with the first two elements containing non-zero values and the remaining two still at zero. At the very next time step, the simulation appears as:",
null,
"The previous values are pushed down the vector by one cell, and the top cell is occupied by the latest sample. Once the simulation goes beyond four time steps, the output will be a full vector.\n\nObviously, if the input signal itself is zero for some points, those values will be reflected accurately in the output.\n\n2. Computation of FFT and inverse FFT\n\nConsider a simple example, where a sinusoidal signal is converted to frequency domain via FFT, and then reconstructed using the IFFT.",
null,
"A sinusoid block generates a sinusoid signal with a frequency of 1 rad/sec. The signal is passed through a buffer block of length 128 samples and a time between successive samples of 0.01. The output of the buffer block is connected to an fft block, which computes a 128-sample FFT of the original sinusoid at a sampling rate of 0.01.\n\nThe output of the fft block is Fourier coefficients. The individual coefficients are accessed using a vecToScalar block. The first four coefficients are plotted to show their variation with time.\n\nSignal reconstruction is performed by feeding the output of the fft block to an ifft block to compute the IFFT. The output of the ifft block is a vector of length 128 samples. The contents of this vector are just 128 sinusoid reconstructions, with each sinusoid trailing the preceding sinusoid by an amount equal to the sampling rate.\n\nThe first element in the ifft output vector does not have any delay because zero time has elapsed between the FFT and IFFT phases. In most real-world situations, however, there is a small, non-zero delay between the input signal and its reconstruction that is introduced by the processor performing the numerical computations of FFT and IFFT algorithms."
]
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"https://2021.help.altair.com/2021/embedse/ImagesExt/image23_89.png",
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"https://2021.help.altair.com/2021/embedse/ImagesExt/image23_90.jpg",
null,
"https://2021.help.altair.com/2021/embedse/ImagesExt/image23_91.jpg",
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"https://2021.help.altair.com/2021/embedse/ImagesExt/image23_92.jpg",
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https://jjthetutor.com/vector-calculus-application-of-line-integral-scalar-potential-work-done-by-force/ | [
"# Vector Calculus- Application of Line Integral |Scalar Potential | Work Done By Force |\n\nFollow me on Unacademy for More Videos on IIT JAM|GATE|CSIR NET\n–~–\nThis video lecture of Vector Calculus- Application of Line Integral | Work Done By Force | Scalar Potential | Example & Solution will help Engineering and Basic Science students to understand following topic of Mathematics:\n1. What is Application of line Integral ?\n2. How to find Work done by force ?\n3. What scalar potential & how to find integration of independent of path line integral ?\n4. What is conservative vector field and how to find scalar potential ?\n#VectorCalculus #LineIntegral #Application #GATE #JAM\nThis Concept is very important in Engineering & Basic Science Students. This video is very useful for B.Sc./B.Tech students also preparing NET, GATE and JAM Aspirants.\nFind Online Engineering Math 2018 Online Solutions Of Vector Calculus- Application of Line Integral | Work Done By Force | Scalar Potential | Example & Solution By GP Sir (Gajendra Purohit)\nDo Like & Share this Video with your Friends. If you are watching for the first time then Subscribe to our Channel and stay updated for more videos around Mathematics.\nSubscribe our channel: https://bit.ly/2tNoIMh\nKeep Watching OUR latest video.\nTopic:- Asymptotes\nTopic:- Laplace Transform\nTopic:- Vector Calculus\nTopic:- Partial Differential Equation\nTopic:- Second Order Differential Equation with Variable\nTopic:- Higher Order Differential Equation With Constant Coefficient\nTopic:- Differential Equation (First Order & Degree)\nTopic:- Fourier Series\nTopic:- Matrices\nTopic:-Fourier Transform\nTopic:-Curve Tracing\nTopic:-Volume of Solids of Revolution\nTopic:-Surface Area of Solids of Revolutions\nTopic:- Double Integration & Area By Double Integration\nTopic: Change of Order of Integration\nTopic: Integral Calculus\n\nTopic: Differential Calculus\nTopic: Integral Transform\nTopic:-Curve Tracing\nTopic:-Volume of Solids of Revolution\nTopic:-Surface Area of Solids of Revolutions\nTopic:- Double Integration & Area By Double Integration\nTopic: Change of Order of Integration"
]
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https://math.stackexchange.com/questions/1060064/a-sequence-converges-if-and-only-if-every-subsequence-converges | [
"A sequence converges if and only if every subsequence converges?\n\nI want to prove this and intuitively it makes sense. But I'm having a hard time coming up with a proof. So if a sequence converges, then we have a natural number for which the distance between all terms after it and the limit point get arbitrarily small. So how can I show that this also holds for every subsequence (which is like a subset of a sequence)?\n\n(are subsequences always infinite?)\n\nCould I suppose that there is a subsequence that doesn't converge to that limit, and find a contradiction? (and do the same for the other direction?)\n\n• It seems like no effort has been made by the OP to try to answer the question. – Lost1 Dec 10 '14 at 0:18\n• What is the OP? – user42 Dec 10 '14 at 0:19\n• OP = original poster; the one who asked the original question. BTW, subsequences are always infinite, yes. – Henno Brandsma Dec 10 '14 at 8:04\n\nLet $x_n \\to x$. Then given $\\varepsilon> 0$, there exists an $N \\in \\mathbb N$ such that $|x_n - x| < \\varepsilon$ for all $n \\geq N$. In words, it means that if we go out far enough, $N$ terms, we can talk about the rest of the terms of the sequence as being close enough, within $\\varepsilon$, to the limit, $x$.\nIf you take any subsequence, say $(x_{n_k})_{k\\in\\mathbb N}$, then we can say that the $N^{th}$ term of the subsequence is at least, or beyond, the $N^{th}$ term of the actual sequence. Thus, it shares the same property that the terms of the sequence are within a desired distance from the limit of the main sequence."
]
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https://digitaloptionswina.web.app/rehrer44621bu/effects-of-interest-rates-on-inflation-1332.html | [
"Effects of interest rates on inflation\n\nBecause of inflation's impact, the interest rate on a fixed income security can be expressed in two ways: The nominal, or stated, interest rate is the rate of interest An important indirect effect of interest rates on inflation may have occurred through high deposit rates and the reduced demand for domestic currency deposits. Thus, a change in certain inflation has a predictable effect on nominal interest rates, and no effect on real interest rates. Effects of Inflation Uncertainty: With\n\nThe basic puzzle about the so-called Fisher effect, in which movements in short- term interest rates primarily reflect fluctuations in expected inflation, is why a For the computation of the effective tax rates, assumptions on economic parameters have to be made - in particular on the values of the inflation and interest rate. In contrast, the loanable funds theory implies a negative impact of inflation instability and interest rates since high uncertainty leads consumers to protect Interest rate stance and inflation objective 15. 2.2. Implicit econometric estimates of studies assessing the effect of money growth on inflation and complicates\n\nCutting interest rates didn't boost inflation. macroeconomic theories, is the Fisher effect—a positive relationship between the nominal interest rate and inflation.\n\nThe real interest rate is nominal interest rates minus inflation. Thus if interest rates rose from 5% to 6% but inflation increased from 2% to 5.5 %. This actually represents a cut in real interest rates from 3% (5-2) to 0.5% (6-5.5) Thus in this circumstance the rise in nominal interest rates actually represents expansionary monetary policy. Inflation will also affect interest rate levels. The higher the inflation rate, the more interest rates are likely to rise. This occurs because lenders will demand higher interest rates as Interest rates, bond yields (prices) and inflation expectations correlate with one another. Movements in short-term interest rates, as dictated by a nation's central bank, will affect different bonds with different terms to maturity differently, depending on the market's expectations of future levels of inflation. How do interest rates affect inflation? Interest rate is simply the rate at which interest gets paid by borrowers for money gotten from borrowers. It, therefore, suffices to say that lower interest rates imply more money in circulation or borrowing thus making the customers spend more. Inflation and interest rates are in close relation to each other, and frequently referenced together in economics. Inflation refers to the rate at which prices for goods and services rise. Interest rate means the amount of interest paid by a borrower to a lender, and is set by central banks.\n\n6 Dec 2019 When interest rates are low, individuals and businesses tend to demand more loans. Each bank loan increases the money supply in a fractional\n\n31 Jul 2019 The Federal Reserve is expected to cut its benchmark interest rate on perhaps because they have a real-time effect on how much it costs to borrow. The Fed often adjusts rates in response to inflation — the increase in 1 Oct 2015 This column illustrates how the effects of an interest rate lift-off, from the (fixed or adjustable rate), the speed of the lift-off, and the inflation rate 11 Jul 2019 Will cryptocurrency have a bigger impact on the economy than we realize? Not Interest Rates -- May Have The Biggest Impact On Inflation 16 Jan 2020 Speculation about a cut in interest rates has intensified after the UK's inflation rate sank to its lowest level for more than three years. See related. 18 Mar 2016 Additionally, the impact of unanticipated real and nominal interest rate changes can vary by sector depending upon the characteristic leverage 20 Jan 2018 The quantitative effect of a change in the repo rate on other nominal interest rates with longer maturities depends on how expected the change is.\n\n11 Jul 2019 Will cryptocurrency have a bigger impact on the economy than we realize? Not Interest Rates -- May Have The Biggest Impact On Inflation\n\nThe Central Bank usually increase interest rates when inflation is predicted to rise above their inflation target. Higher interest rates tend to moderate economic average relationships among interest rates, inflation rates, and money growth effect” occurs in reality (though it is hard to see it in the data) and may regard it\n\naverage relationships among interest rates, inflation rates, and money growth effect” occurs in reality (though it is hard to see it in the data) and may regard it\n\nInflation will also affect interest rate levels. The higher the inflation rate, the more interest rates are likely to rise. This occurs because lenders will demand higher interest rates as\n\nIn contrast, the loanable funds theory implies a negative impact of inflation instability and interest rates since high uncertainty leads consumers to protect Interest rate stance and inflation objective 15. 2.2. Implicit econometric estimates of studies assessing the effect of money growth on inflation and complicates"
]
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https://percent.info/of/69/how-to-calculate-69-percent-of-674000.html | [
"69 percent of 674000",
null,
"",
null,
"Here we will show you how to calculate sixty-nine percent of six hundred seventy-four thousand. Before we continue, note that 69 percent of 674000 is the same as 69% of 674000. We will write it both ways throughout this tutorial to remind you that it is the same.\n\n69 percent means that for each 100, there are 69 of something. This page will teach you three different methods you can use to calculate 69 percent of 674000.\n\nWe think that illustrating multiple ways of calculating 69 percent of 674000 will give you a comprehensive understanding of what 69% of 674000 means, and provide you with percent knowledge that you can use to calculate any percentage in the future.\n\nTo solidify your understanding of 69 percent of 674000 even further, we have also created a pie chart showing 69% of 674000. On top of that, we will explain and calculate \"What is not 69 percent of 674000?\"\n\nCalculate 69 percent of 674000 using a formula\nThis is the most common method to calculate 69% of 674000. 674000 is the Whole, 69 is the Percent, and the Part is what we are calculating. Below is the math and answer to \"What is 69% of 674000?\" using the percent formula.\n\n(Whole × Percent)/100 = Part\n(674000 × 69)/100 = 465060\n69% of 674000 = 465060\n\nGet 69 percent of 674000 with a percent decimal number\nYou can convert any percent, such as 69.00%, to 69 percent as a decimal by dividing the percent by one hundred. Therefore, 69% as a decimal is 0.69. Here is how to calculate 69 percent of 674000 with percent as a decimal.\n\nWhole × Percent as a Decimal = Part\n674000 × 0.69 = 465060\n69% of 674000 = 465060\n\nGet 69 percent of 674000 with a fraction function\nThis is our favorite method of calculating 69% of 674000 because it best illustrates what 69 percent of 674000 really means. The facts are that it is 69 per 100 and we want to find parts per 674000. Here is how to illustrate and show you the answer using a function with fractions.\n\n Part 674000\n=\n 69 100\n\nPart = 465060\n\n69% of 674000 = 465060\n\nNote: To solve the equation above, we first multiplied both sides by 674000 and then divided the left side to get the answer.\n\n69 percent of 674000 illustrated\nBelow is a pie chart illustrating 69 percent of 674000. The pie contains 674000 parts, and the blue part of the pie is 465060 parts or 69 percent of 674000.",
null,
"Note that it does not matter what the parts are. It could be 69 percent of 674000 dollars, 69 percent of 674000 people, and so on. The pie chart of 69% of 674000 will look the same regardless what it is.\n\nWhat is not 69 percent of 674000?\nWhat is not 69 percent of 674000? In other words, what is the red part of our pie above? We know that the total is 100 percent, so to calculate \"What is not 69%?\" you deduct 69% from 100% and then take that percent from 674000:\n\n100% - 69% = 31%\n(674000 × 31)/100 = 208940\n\nAnother way of calculating the red part is to subtract 465060 from 674000.\n\n674000 - 465060 = 208940\n\nThat is the end of our tutorial folks. We hope we accomplished our goal of making you a percent expert - at least when it comes to calculating 69 percent of 674000.\n\nPercent of a Number\nGo here if you need to calculate the percent of a different number.\n\n69 percent of 675000\nHere is the next percent tutorial on our list that may be of interest.\n\nCopyright | Privacy Policy | Disclaimer | Contact"
]
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"https://percent.info/images/percent-of/69-percent.jpg",
null,
"https://percent.info/images/percent-of/of-674000.jpg",
null,
"https://percent.info/images/pie-charts/pie-chart-showing-69-percent.png",
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https://www.jneea.com/editors | [
"# Board of Editors\n\n### Editors-in-Chief\n\nGaston M. N'Guérékata (Morgan State University, U. S. A.)\nEvolution equations; abstract harmonic analysis; almost periodic and almost automorphic functions; fractional differential equations.\nGisèle Mophou (Université des Antilles, Guadeloupe, France)\nEvolution equations; fractional differential equations; almost periodic and almost automorphic functions; control theory.\nDynamical systems; evolution equations; partial functional differential equations; functional differential equations.\n\n#### Managing Editors\n\nMaximilian Hasler (Université des Antilles, Martinique, France)\nNonlinear algebraic analysis; function spaces; partial differential equations; mathematical physics; field theory; computational mathematics; number theory.\nPiotr Kasprzak (Adam Mickiewicz University, Poland)\nFunctions of bounded variation, almost periodic functions, integral and differential equations, fractional differential equations.\nLi Fang (Yunnan Normal University, P. R. of China)\nFractional differential equations; operator families; functional differential equations.\nMahamadi Warma (George Mason University, USA)\nFractional partial differential equations, nonlocal partial differential equations, evolution equations, linear and non linear partial differential equations, potential theory, dirichlet forms, feller semigroups, control theory of PDEs, optimal control of PDEs.\n\n### Editors\n\nNasir Uddin Ahmed (University of Ottawa, Canada)\nSystems: deterministic, stochastic, distributed; differential equations and inclusions, optimal control; linear and nonlinear filtering, identification; hemodynamics of artificial heart; modeling of space station, stability and control; suspension bridges and their stability; dynamics of computer communication network and optimization.\nGeorge A. Anastassiou (University of Memphis, U. S. A.)\nApproximation theory; inequalities; computational analysis; operators theory; ordinary differential equations.\nAbdon Atangana (University of the Free State, Republic of South Africa)\nFractional calculus and applications to real world problems; perturbation and asymptotic methods; iterative, numerical and analytical methods for differential equations; integral transforms.\nMathematical ecology, epidemiology and evolution of infectious diseases, host-pathogen interactions; optimal control of biological resources, in-host hiv models, delay-differential equations, nonlinear dynamics, nonstandard finite difference method, reaction-diffusion equations.\nMouffak Benchohra (University of Sidi Bel Abbes, Algeria)\nInitial and boundary value problems (ivps, bvps) for differential and functional differential equations and inclusions; impulsive differential and functional differential equations and inclusions; controllability of differential and functional differential equations and inclusions; fuzzy differential equations and inclusions; abstract evolution equations and inclusions; dynamic equations and inclusions on time scales; fractional differential equations and inclusions.\nMartin Bohner (Missouri University of Science and Technology , U. S. A.)\nDifference equations; control theory; oscillation; variational systems.\nClaudio Cuevas (Universidade Federal de Pernambuco, Brazil)\nDifference equations; periodicity and ergodicity; dispersive estimates; fractional differential equations; functional differential equations; integral and integro-differential operators.\nHui-Sheng Ding (Jiangxi Normal University, P. R. of China)\nAsymptotic behavior of evolution equations including periodicity, almost periodicity, almost automorphy, boundedness and stability; abstract evolution equation, functional differential equation, integral equation, difference equation, dynamical systems.\nCiprian G. S. Gal (Florida International University, U. S. A.)\nNonlinear partial differential equations: Cahn-Hilliard equations, phase-field systems and related models; Hyperbolic equations and related models; Fluid dynamics: the study of incompressible multiphase flows, Navier-Stokes equations; Elliptic and parabolic boundary value problems. Dynamic boundary conditions for PDEs. Nonlocal mathematical models; Attractor's theory of dissipative systems; Perturbation theory, convergence rates to steady states, stability issues.\nJerome A. Goldstein (University of Memphis, U. S. A.)\nEvolution equations; calculus of variations; electron densities in quantum theory; nonlinear PDE.\nDinh Nho Hào (Hanoi Institute of Mathematics , Vietnam)\nPartial differential equations (PDE); inverse and ill-posed problems in PDE; nonlinear analysis; optimal control theory; numerical analysis.\nOlaniyi Samuel Iyiola (Clarkson University, USA)\nNumerical analysis; algorithms & analysis for partial differential equations, integer and non-integer order; computational mathematics; simulations in application, PDE models, optimization; nonlinear operator theory involving fixed point algorithms; variational inequality, optimization and variational analysis, equilibrium problems, and nonlinear integral equations.\nXinzhi Liu (University of Waterloo, Canada)\nOrdinary differential equations; functional differential equations; dynamical systems; stability and control theory.\nOperator theory; ordinary differential equations; integral equations; abstract harmonic analysis.\nMehran Mahdavi (Bowie State University, USA)\nTheory of equations involving abstract volterra operators (causal operators); ordinary differential equations; partial differential equations; complex analysis; dynamical systems.\nOusseynou Nakoulima (University Ouaga 3S, Burkina Faso)\nEvolution equations; nonlinear analysis; control theory.\nMinh Van Nguyen (Columbus State University, U. S. A.)\nAsymptotic behavior of differential equations; semigroup theory and applications.\nStanislas Ouaro (Université Ouaga I, Burkina Faso)\nPartial differential equations (PDE); difference equations; functional equations.\nMichael Ruzhansky (Imperial College London, United Kingdom)\nHyperbolic partial differential equations; Schrödinger equations; Fourier analysis.\nStefan Siegmund (TU Dresden, Germany)\nNonautonomous dynamical systems; bifurcation theory; transient dynamics; coherent structures.\nTi-Jun Xiao (School of Mathematical Sciences, Fudan University, P. R. of China)\nEvolution equations; nonlinear analysis; operator families; control theory.\nYong Zhou (Xiangtan University, P. R. of China)\nFractional differential equations, functional differential equations.\nEnrique Zuazua (Basque Center for Applied Mathematics, Spain)\nPartial differential equations; systems control and numerical analysis; modelling, analysis, computer simulation and control and design of natural phenomena and other problems arising in r+d+i."
]
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| {"ft_lang_label":"__label__en","ft_lang_prob":0.7581366,"math_prob":0.9783589,"size":7000,"snap":"2022-40-2023-06","text_gpt3_token_len":1416,"char_repetition_ratio":0.2528588,"word_repetition_ratio":0.05620609,"special_character_ratio":0.17685714,"punctuation_ratio":0.22318007,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9985588,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-02-01T12:15:12Z\",\"WARC-Record-ID\":\"<urn:uuid:e42bc63e-e39b-4529-8078-608885faa986>\",\"Content-Length\":\"22556\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aaff6e24-5b8b-4a42-91cc-68e09ec951d7>\",\"WARC-Concurrent-To\":\"<urn:uuid:1c7ef510-d9bf-484f-ae68-d1cce5f917af>\",\"WARC-IP-Address\":\"188.130.25.202\",\"WARC-Target-URI\":\"https://www.jneea.com/editors\",\"WARC-Payload-Digest\":\"sha1:RB4PN3TTKZ3RMQAU5OMBY7WM4RHVJRGB\",\"WARC-Block-Digest\":\"sha1:EEVIEMXINFG5YI5272ED55UM26W3OCNE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-06/CC-MAIN-2023-06_segments_1674764499934.48_warc_CC-MAIN-20230201112816-20230201142816-00480.warc.gz\"}"} |
https://www.physicsforums.com/search/4141047/ | [
"# Search results\n\n1. ### Physics Regarding postdoc\n\nthank you, but i didnt get the specified results\n2. ### Specific activity of radionuclides\n\nwhat is the formula to determine the specific activity of radionulcide in sediments\n3. ### Physics Regarding postdoc\n\nI have worked on environmental radiation levels n determining the radioactivity in the environmental samples. I would like to do post doc in the field. I request to suggest me the posts available all over the world.\n4. ### Correlation between macronutirients and radionuclides\n\ni got the point regarding. thank you. since clay particles have high cation exchange capacity, why cant the radinuclides can be embbeded in clays??\n5. ### Granulometric analysis\n\nthank you for the answer sir, the link i went through is also useful in clarifying the doubts.\n6. ### Correlation between macronutirients and radionuclides\n\nthank you sir.\n7. ### Correlation between macronutirients and radionuclides\n\nAs both exist in clay particles, that have high exchange capacity?\n8. ### Correlation between macronutirients and radionuclides\n\ndo the presence of macronutrients (N,P,K) increase the amount of radionuclides (U,Th,K) in the sediments. how are they related?\n9. ### Granulometric analysis\n\nthanks for the reply, do there exists any relation between the minerals (Zn, Cu, Mn, Fe) that is found in the sediments with that of radionuclides (U, Th, K)?\n10. ### Granulometric analysis\n\nwhy do we need to determine the particle size while estimating the activity of radionuclides? is there any relation between the particle size (sand, silt and clay) and primordial radionuclide? will the content of uranium, thorium and potassium rely on particle sizes?\n11. ### Regarding minerological analysis\n\nthank you for the clarification.\n12. ### Regarding minerological analysis\n\nwhich oxide is responsible for the increase of uranium -238 in the sediments of rivers. is it U(IV) oxide is soluble or U(VI) is soluble in water?\n13. ### Regarding minerological analysis\n\nUranium can exist in five oxidation states: +2, +3, +4, +5 and +6. However, only the +4 and +6 states are stable enough to be of practical importance. Tetravalent uranium is reasonably stable and forms hydroxides, hydrated fluorides and phosphates of low solubility. Hexavalent uranium is the...\n14. ### Regarding minerological analysis\n\nwhy we need to find the minerals in the sediments while estimating the amount of radionuclides. although there exist a cation exchange capacity in the minerals, do they actually concentrate the primordial radionuclides in them. further on granulometric analysis, do the clay particles retain the...\n\nthank you sir\n\nwhat is the difference between Bq/m3 and Bq/Kg/h. can the exhaltion rate of radon and thoron for the sediments can can be in the units of Bq/Kg/h.\n\nthank you\n\nthank you\n19. ### B How to calculate the uncertainity errors\n\nwhether poissons's distribution give the exact value of the errors associated at same instant of time.\n20. ### B How to calculate the uncertainity errors\n\nfurther i have been working on environmental radiation, to estimate the amounts. as per the Bharathidasan Univesity , we need some reviwers who could evaluate the thesis and send back the report. i request to the persons who are willing to do me a a favour regarding this.\n21. ### B How to calculate the uncertainity errors\n\nfurther on as Th-232 has to be estimated from the soil/sediments. as it is not a gamma ray emitter, the daughter product of Tl-208 is considered. how safe is this approximation, since Tl-208 is of third branch of the decay series, do we need to multiply the amounts by 3. or the counts directly...\n22. ### B How to calculate the uncertainity errors\n\nthank you, i had understood the concept behind it. i whole heartedly thank you as i would derive the equation with the uncertainity errors. including the activity, background counts and the time interval for each count.\n23. ### B How to calculate the uncertainity errors\n\nwe have meausred the primordial radionuclides in the sediments using NaI gamma ray detector. we measured the counts of the radionucldies and the activity of the radionuclides such as U, Th and K. we would be adding some uncertainity errors along with the specific activity with + and - sign. i...\n24. ### B What is the self absorption parameter\n\ni would like to know the self absorption parameter of 3x3 inches NaI crystal. which is used in estim,ating the radionuclides\n25. ### B How to calculate the uncertainity errors\n\ndo any one explain me pl. i would like to know the uncertainty error in measuring the specific activity of the primordial radionuclides"
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https://asone.ai/polymath/index.php?title=Hyper-optimistic_conjecture&direction=next&oldid=765 | [
"Hyper-optimistic conjecture\n\nGil Kalai and Tim Gowers have proposed a “hyper-optimistic” conjecture.\n\nLet $c^\\mu_n$ be the maximum equal-slices measure of a line-free set. For instance, $c^\\mu_0 = 1$, $c^\\mu_1 = 2$, and $c^\\mu_2 = 4$.\n\nAs in the unweighted case, every time we find a subset $B$ of the grid $\\Delta_n := \\{ (a,b,c): a+b+c=n\\}$ without equilateral triangles, it gives a line-free set $\\Gamma_B := \\bigcup_{(a,b,c) \\in B} \\Gamma_{a,b,c}$. The equal-slices measure of this set is precisely the cardinality of B. Thus we have the lower bound $c^\\mu_n \\geq \\overline{c}^\\mu_n$, where $\\overline{c}^\\mu_n$ is the largest size of equilateral triangles in $\\Delta_n$. The computation of the $\\overline{c}^\\mu_n$ is Fujimura's problem.\n\nHyper-optimistic conjecture: We in fact have $c^\\mu_n = \\overline{c}^\\mu_n$. In other words, to get the optimal equal-slices measure for a line-free set, one should take a set which is a union of slices $\\Gamma_{a,b,c}$.\n\nThis conjecture, if true, will imply the DHJ theorem. Note also that all our best lower bounds for the unweighted problem to date have been unions of slices. Also, the k=2 analogue of the conjecture is true, and is known as the LYM inequality (in fact, for k=2 we have $c^\\mu_n = \\overline{c}^\\mu_n = 1$ for all n).\n\nSmall values of $c^\\mu_n$\n\nI have now found the extremal solutions for the weighted problem in the hyper-optimistic conjecture, again using integer programming.\n\nThe first few values are\n\n• $c^\\mu_0=1$ (trivial)\n• $c^\\mu_1=2$ (trivial)\n• $c^{\\mu}_2=4$ with 3 solutions\n• $c^{\\mu}_3=6$ with 9 solutions\n• $c^{\\mu}_4=9$ with 1 solution\n• $c^{\\mu}_5=12$ with 1 solution\n\nComparing this with the known bounds for $\\overline{c}^\\mu_n$ we see that the hyper-optimistic conjecture is true for $n \\leq 5$.\n\nn=2 by hand\n\nOne should in fact be able to get the Pareto-optimal and extremal statistics for the slice densities $\\alpha_{a,b,c} := |A \\cap \\Gamma_{a,b,c}|/|\\Gamma_{a,b,c}|$ in this case.\n\nn=3 by hand\n\n$c^{\\mu}_3=6$:\n\nIf we have all Three points of the form xxx removed Then the remaining points have value 7 and We have covered all lines any set of moving coordinates And all constant points equal to one value this leaves The lines xab a,b not equal. Each point of the set abc covers three of these lines the entire set covers each of these lines there is no duplication the only alternative is to remove a point abc and cover the lines with points of the form aab which have a higher weight and only cover one line each this would lower the weight so the maximum weight occurs when all of abc is omitted along with the three points xxx and the weight is 6\n\nIf we have only two points removed of the form xxx then The weight is at most 8 say the point not removed is 222 Then we must cover the lines xx2 and x22 we have three six such Lines and all the xx2 must be covered one at a time by either 112 Or 332 the x22 must be covered one at a time by 322 or 122 These points must be removed and the that lowers the weight To 8 - 3*(2/6) – 3*(2/6) = 6 again we have c^{\\mu} must be less than 6\n\nIf we have one point removed say 111 then we must cover all lines of The form xx2 xx3 and x22 and x33 the best we can do is to cover these lines Two of at once as no point can cover x22 and x33 or xx2 and xx3 And we have all points of the form aab when a = 2 and b =3 Or b=2 and a=3 Only points of the form xxy can cover these lines and as we have 12 lines Cover 2 at a times we must have 6 such lines which will have weight 2/6 So we will have removed one point with weight one and six more With weight 1/3 giving remaining weight 7 and we will have Lines of the form x11 and xx1 to cover as well as x12 and x13 The lines of the form x11 will have to be covered by 3 lines with Weight 1/3 however we are not done as we can split a point of the Form 223 into 113 and 221 then we will cover the lines 22x and xx3 With two lines instead of one but we will cover one line of the Form 11x however we due this we will either have to split 3 pairs and we will have weight 6 since 7-1=6 or split two and use one point of the form 11x for the remaining line of the form 11x or again the total is 6 or split one and use 2 and again the remaining points total 6.\n\nFinally we have no points of the form xxx removed but then we will have a line of the form xxx."
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https://byjus.com/maths/2090-in-words/ | [
"",
null,
"# 2090 in Words\n\nThe number 2090 in words is written as Two thousand ninety. In both the International and Indian System of Numerals, 2090 is written as 2,090 and read as Two thousand ninety. We may use 2090 as “that gadget costs 2090 rupees” in this context, 2090 represents an amount; that is why 2090 is a Cardinal Number since it represents some quantity.\n\n 2090 in Words Two Thousand Ninety Two Thousand Ninety in Numbers 2090\n\n## 2090 in English Words\n\nIn English words, 2090 is written as “Two thousand ninety”.",
null,
"## How to Write 2090 in Words?\n\nTo write 2090 in words, we shall use the place value. In the place value chart, write 2 in thousands, 0 in hundreds, 9 in tens and 0 in ones. Let us make a place value chart to write the number 2090 in words.\n\n Thousands Hundreds Tens Ones 2 0 9 0\n\nThus, we can use the expanded form as\n\n2 × Thousand + 0 × Hundred + 9 × Ten + 0 × One\n\n= 2 × 1000 + 0 × 100 + 9 × 10 + 0 × 1\n\n= 2000 + 0 + 90 + 0\n\n= 2090\n\n= Two thousand ninety.\n\n2090 is a Natural Number, the successor of 2089 and the predecessor of 2091\n\n2090 in words – Two thousand ninety\n\n• Is 2090 an odd number? – No\n• Is 2090 an even number? – Yes\n• Is 2090 a perfect square number? – No\n• Is 2090 a perfect cube number? – No\n• Is 2090 a prime number? – No\n• Is 2090 a composite number? – Yes\n\n## Related Articles\n\n2090 in words is written as Two thousand ninety.\n\n## Frequently Asked Questions on 2090 in Words\n\nQ1\n\n### How to write 2090 in words?\n\n2090 in words is written as Two thousand ninety.\nQ2\n\n### How to write 2090 in the International and Indian System of Numerals?\n\nIn both, the system of numerals, 2090 in words, is written as Two thousand ninety.\nQ3\n\n### How to write 2090 in a place value chart?\n\nIn the place value chart, write 2 in thousands, 0 in hundreds, 9 in tens and 0 in ones, respectively."
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https://laceshawl.savingadvice.com/2009/04/ | [
"User Real IP - 34.204.193.85\n```Array\n(\n => Array\n(\n => 182.68.68.92\n)\n\n => Array\n(\n => 101.0.41.201\n)\n\n => Array\n(\n => 43.225.98.123\n)\n\n => Array\n(\n => 2.58.194.139\n)\n\n => Array\n(\n => 46.119.197.104\n)\n\n => Array\n(\n => 45.249.8.93\n)\n\n => Array\n(\n => 103.12.135.72\n)\n\n => Array\n(\n => 157.35.243.216\n)\n\n => Array\n(\n => 209.107.214.176\n)\n\n => Array\n(\n => 5.181.233.166\n)\n\n => Array\n(\n => 106.201.10.100\n)\n\n => Array\n(\n => 36.90.55.39\n)\n\n => Array\n(\n => 119.154.138.47\n)\n\n => Array\n(\n => 51.91.31.157\n)\n\n => Array\n(\n => 182.182.65.216\n)\n\n => Array\n(\n => 157.35.252.63\n)\n\n => Array\n(\n => 14.142.34.163\n)\n\n => Array\n(\n => 178.62.43.135\n)\n\n => Array\n(\n => 43.248.152.148\n)\n\n => Array\n(\n => 222.252.104.114\n)\n\n => Array\n(\n => 209.107.214.168\n)\n\n => Array\n(\n => 103.99.199.250\n)\n\n => Array\n(\n => 178.62.72.160\n)\n\n => Array\n(\n => 27.6.1.170\n)\n\n => Array\n(\n => 182.69.249.219\n)\n\n => Array\n(\n => 110.93.228.86\n)\n\n => Array\n(\n => 72.255.1.98\n)\n\n => Array\n(\n => 182.73.111.98\n)\n\n => Array\n(\n => 45.116.117.11\n)\n\n => Array\n(\n => 122.15.78.189\n)\n\n => Array\n(\n => 14.167.188.234\n)\n\n => Array\n(\n => 223.190.4.202\n)\n\n => Array\n(\n => 202.173.125.19\n)\n\n => Array\n(\n => 103.255.5.32\n)\n\n => Array\n(\n => 39.37.145.103\n)\n\n => Array\n(\n => 140.213.26.249\n)\n\n => Array\n(\n => 45.118.166.85\n)\n\n => Array\n(\n => 102.166.138.255\n)\n\n => Array\n(\n => 77.111.246.234\n)\n\n => Array\n(\n => 45.63.6.196\n)\n\n => Array\n(\n => 103.250.147.115\n)\n\n => Array\n(\n => 223.185.30.99\n)\n\n => Array\n(\n => 103.122.168.108\n)\n\n => Array\n(\n => 123.136.203.21\n)\n\n => Array\n(\n => 171.229.243.63\n)\n\n => Array\n(\n => 153.149.98.149\n)\n\n => Array\n(\n => 223.238.93.15\n)\n\n => Array\n(\n => 178.62.113.166\n)\n\n => Array\n(\n => 101.162.0.153\n)\n\n => Array\n(\n => 121.200.62.114\n)\n\n => Array\n(\n => 14.248.77.252\n)\n\n => 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119.153.40.69\n)\n\n => Array\n(\n => 49.36.133.64\n)\n\n => Array\n(\n => 103.255.4.249\n)\n\n => Array\n(\n => 198.144.154.15\n)\n\n => Array\n(\n => 1.22.46.172\n)\n\n => Array\n(\n => 103.255.5.46\n)\n\n => Array\n(\n => 27.56.195.188\n)\n\n => Array\n(\n => 203.101.167.53\n)\n\n => Array\n(\n => 117.230.62.195\n)\n\n => Array\n(\n => 103.240.194.186\n)\n\n => Array\n(\n => 107.170.166.118\n)\n\n => Array\n(\n => 101.53.245.80\n)\n\n => Array\n(\n => 157.43.13.208\n)\n\n => Array\n(\n => 137.97.100.77\n)\n\n => Array\n(\n => 47.31.150.208\n)\n\n => Array\n(\n => 137.59.222.65\n)\n\n => Array\n(\n => 103.85.127.250\n)\n\n => Array\n(\n => 103.214.119.32\n)\n\n => Array\n(\n => 182.255.49.52\n)\n\n => Array\n(\n => 103.75.247.72\n)\n\n => Array\n(\n => 103.85.125.250\n)\n\n => Array\n(\n => 183.83.253.167\n)\n\n => Array\n(\n => 1.39.222.111\n)\n\n => Array\n(\n => 111.119.185.9\n)\n\n => Array\n(\n => 111.119.187.10\n)\n\n => Array\n(\n => 39.37.147.144\n)\n\n => Array\n(\n => 103.200.198.183\n)\n\n => Array\n(\n => 1.39.222.18\n)\n\n => Array\n(\n => 198.8.80.103\n)\n\n => Array\n(\n => 42.108.1.243\n)\n\n => Array\n(\n => 111.119.187.16\n)\n\n => Array\n(\n => 39.40.241.8\n)\n\n => Array\n(\n => 122.169.150.158\n)\n\n => Array\n(\n => 39.40.215.119\n)\n\n => Array\n(\n => 103.255.5.77\n)\n\n => Array\n(\n => 157.38.108.196\n)\n\n => Array\n(\n => 103.255.4.67\n)\n\n => Array\n(\n => 5.62.60.62\n)\n\n => Array\n(\n => 39.37.146.202\n)\n\n => Array\n(\n => 110.138.6.221\n)\n\n => Array\n(\n => 49.36.143.88\n)\n\n => Array\n(\n => 37.1.215.39\n)\n\n => Array\n(\n => 27.106.59.190\n)\n\n => Array\n(\n => 139.167.139.41\n)\n\n => Array\n(\n => 114.142.166.179\n)\n\n => Array\n(\n => 223.225.240.112\n)\n\n => Array\n(\n => 103.255.5.36\n)\n\n => Array\n(\n => 175.136.1.48\n)\n\n => Array\n(\n => 103.82.80.166\n)\n\n => Array\n(\n => 182.185.196.126\n)\n\n => Array\n(\n => 157.43.45.76\n)\n\n => Array\n(\n => 119.152.132.49\n)\n\n => Array\n(\n => 5.62.62.162\n)\n\n => Array\n(\n => 103.255.4.39\n)\n\n => Array\n(\n => 202.5.144.153\n)\n\n => Array\n(\n => 1.39.223.210\n)\n\n => Array\n(\n => 92.38.176.154\n)\n\n => Array\n(\n => 117.230.186.142\n)\n\n => Array\n(\n => 183.83.39.123\n)\n\n => Array\n(\n => 182.185.156.76\n)\n\n => Array\n(\n => 104.236.74.212\n)\n\n => Array\n(\n => 107.170.145.187\n)\n\n => Array\n(\n => 117.102.7.98\n)\n\n => Array\n(\n => 137.59.220.0\n)\n\n => Array\n(\n => 157.47.222.14\n)\n\n => Array\n(\n => 47.15.206.82\n)\n\n => Array\n(\n => 117.230.159.99\n)\n\n => Array\n(\n => 117.230.175.151\n)\n\n => Array\n(\n => 157.50.97.18\n)\n\n => Array\n(\n => 117.230.47.164\n)\n\n => Array\n(\n => 77.111.244.34\n)\n\n => Array\n(\n => 139.167.189.131\n)\n\n => Array\n(\n => 1.39.204.103\n)\n\n => Array\n(\n => 117.230.58.0\n)\n\n => Array\n(\n => 182.185.226.66\n)\n\n => Array\n(\n => 115.42.70.119\n)\n\n => Array\n(\n => 171.48.114.134\n)\n\n => Array\n(\n => 144.34.218.75\n)\n\n => Array\n(\n => 199.58.164.135\n)\n\n => Array\n(\n => 101.53.228.151\n)\n\n => Array\n(\n => 117.230.50.57\n)\n\n => Array\n(\n => 223.225.138.84\n)\n\n => Array\n(\n => 110.225.67.65\n)\n\n => Array\n(\n => 47.15.200.39\n)\n\n => Array\n(\n => 39.42.20.127\n)\n\n => Array\n(\n => 117.97.241.81\n)\n\n => Array\n(\n => 111.119.185.11\n)\n\n => Array\n(\n => 103.100.5.94\n)\n\n => Array\n(\n => 103.25.137.69\n)\n\n => Array\n(\n => 47.15.197.159\n)\n\n => Array\n(\n => 223.188.176.122\n)\n\n => Array\n(\n => 27.4.175.80\n)\n\n => Array\n(\n => 181.215.43.82\n)\n\n => Array\n(\n => 27.56.228.157\n)\n\n => Array\n(\n => 117.230.19.19\n)\n\n => Array\n(\n => 47.15.208.71\n)\n\n => Array\n(\n => 119.155.21.176\n)\n\n => Array\n(\n => 47.15.234.202\n)\n\n => Array\n(\n => 117.230.144.135\n)\n\n => Array\n(\n => 112.79.139.199\n)\n\n => Array\n(\n => 116.75.246.41\n)\n\n => Array\n(\n => 117.230.177.126\n)\n\n => Array\n(\n => 212.103.48.134\n)\n\n => Array\n(\n => 102.69.228.78\n)\n\n => Array\n(\n => 117.230.37.118\n)\n\n => Array\n(\n => 175.143.61.75\n)\n\n => Array\n(\n => 139.167.56.138\n)\n\n => Array\n(\n => 58.145.189.250\n)\n\n => Array\n(\n => 103.255.5.65\n)\n\n => Array\n(\n => 39.37.153.182\n)\n\n => Array\n(\n => 157.43.85.106\n)\n\n => Array\n(\n => 185.209.178.77\n)\n\n => Array\n(\n => 1.39.212.45\n)\n\n => Array\n(\n => 103.72.7.16\n)\n\n => Array\n(\n => 117.97.185.244\n)\n\n => Array\n(\n => 117.230.59.106\n)\n\n => Array\n(\n => 137.97.121.103\n)\n\n => Array\n(\n => 103.82.123.215\n)\n\n => Array\n(\n => 103.68.217.248\n)\n\n => Array\n(\n => 157.39.27.175\n)\n\n => Array\n(\n => 47.31.100.249\n)\n\n => Array\n(\n => 14.171.232.139\n)\n\n => Array\n(\n => 103.31.93.208\n)\n\n => Array\n(\n => 117.230.56.77\n)\n\n => Array\n(\n => 124.182.25.124\n)\n\n => Array\n(\n => 106.66.191.242\n)\n\n => Array\n(\n => 175.107.237.25\n)\n\n => Array\n(\n => 119.155.1.27\n)\n\n => Array\n(\n => 72.255.6.24\n)\n\n => Array\n(\n => 192.140.152.223\n)\n\n => Array\n(\n => 212.103.48.136\n)\n\n => Array\n(\n => 39.45.134.56\n)\n\n => Array\n(\n => 139.167.173.30\n)\n\n => Array\n(\n => 117.230.63.87\n)\n\n => Array\n(\n => 182.189.95.203\n)\n\n => Array\n(\n => 49.204.183.248\n)\n\n => Array\n(\n => 47.31.125.188\n)\n\n => Array\n(\n => 103.252.171.13\n)\n\n => Array\n(\n => 112.198.74.36\n)\n\n => Array\n(\n => 27.109.113.152\n)\n\n => Array\n(\n => 42.112.233.44\n)\n\n => Array\n(\n => 47.31.68.193\n)\n\n => Array\n(\n => 103.252.171.134\n)\n\n => Array\n(\n => 77.123.32.114\n)\n\n => Array\n(\n => 1.38.189.66\n)\n\n => Array\n(\n => 39.37.181.108\n)\n\n => Array\n(\n => 42.106.44.61\n)\n\n => Array\n(\n => 157.36.8.39\n)\n\n => Array\n(\n => 223.238.41.53\n)\n\n => Array\n(\n => 202.89.77.10\n)\n\n => Array\n(\n => 117.230.150.68\n)\n\n => Array\n(\n => 175.176.87.60\n)\n\n => Array\n(\n => 137.97.117.87\n)\n\n => Array\n(\n => 132.154.123.11\n)\n\n => Array\n(\n => 45.113.124.141\n)\n\n => Array\n(\n => 103.87.56.203\n)\n\n => Array\n(\n => 159.89.171.156\n)\n\n => Array\n(\n => 119.155.53.88\n)\n\n => Array\n(\n => 222.252.107.215\n)\n\n => Array\n(\n => 132.154.75.238\n)\n\n => Array\n(\n => 122.183.41.168\n)\n\n => Array\n(\n => 42.106.254.158\n)\n\n => Array\n(\n => 103.252.171.37\n)\n\n => Array\n(\n => 202.59.13.180\n)\n\n => Array\n(\n => 37.111.139.137\n)\n\n => Array\n(\n => 39.42.93.25\n)\n\n => Array\n(\n => 118.70.177.156\n)\n\n => Array\n(\n => 117.230.148.64\n)\n\n => Array\n(\n => 39.42.15.194\n)\n\n => Array\n(\n => 137.97.176.86\n)\n\n => Array\n(\n => 106.210.102.113\n)\n\n => Array\n(\n => 39.59.84.236\n)\n\n => Array\n(\n => 49.206.187.177\n)\n\n => Array\n(\n => 117.230.133.11\n)\n\n => Array\n(\n => 42.106.253.173\n)\n\n => Array\n(\n => 178.62.102.23\n)\n\n => Array\n(\n => 111.92.76.175\n)\n\n => Array\n(\n => 132.154.86.45\n)\n\n => Array\n(\n => 117.230.128.39\n)\n\n => Array\n(\n => 117.230.53.165\n)\n\n => Array\n(\n => 49.37.200.171\n)\n\n => Array\n(\n => 104.236.213.230\n)\n\n => Array\n(\n => 103.140.30.81\n)\n\n => Array\n(\n => 59.103.104.117\n)\n\n => Array\n(\n => 65.49.126.79\n)\n\n => Array\n(\n => 202.59.12.251\n)\n\n => Array\n(\n => 37.111.136.17\n)\n\n => Array\n(\n => 163.53.85.67\n)\n\n => Array\n(\n => 123.16.240.73\n)\n\n => Array\n(\n => 103.211.14.183\n)\n\n => Array\n(\n => 103.248.93.211\n)\n\n => Array\n(\n => 116.74.59.127\n)\n\n => Array\n(\n => 137.97.169.254\n)\n\n => Array\n(\n => 113.177.79.100\n)\n\n => Array\n(\n => 74.82.60.187\n)\n\n => Array\n(\n => 117.230.157.66\n)\n\n => Array\n(\n => 169.149.194.241\n)\n\n => Array\n(\n => 117.230.156.11\n)\n\n => Array\n(\n => 202.59.12.157\n)\n\n => Array\n(\n => 42.106.181.25\n)\n\n => Array\n(\n => 202.59.13.78\n)\n\n => Array\n(\n => 39.37.153.32\n)\n\n => Array\n(\n => 177.188.216.175\n)\n\n => Array\n(\n => 222.252.53.165\n)\n\n => Array\n(\n => 37.139.23.89\n)\n\n => Array\n(\n => 117.230.139.150\n)\n\n => Array\n(\n => 104.131.176.234\n)\n\n => Array\n(\n => 42.106.181.117\n)\n\n => Array\n(\n => 117.230.180.94\n)\n\n => Array\n(\n => 180.190.171.5\n)\n\n => Array\n(\n => 150.129.165.185\n)\n\n => Array\n(\n => 51.15.0.150\n)\n\n => Array\n(\n => 42.111.4.84\n)\n\n => Array\n(\n => 74.82.60.116\n)\n\n => Array\n(\n => 137.97.121.165\n)\n\n => Array\n(\n => 64.62.187.194\n)\n\n => Array\n(\n => 137.97.106.162\n)\n\n => Array\n(\n => 137.97.92.46\n)\n\n => Array\n(\n => 137.97.170.25\n)\n\n => Array\n(\n => 103.104.192.100\n)\n\n => Array\n(\n => 185.246.211.34\n)\n\n => Array\n(\n => 119.160.96.78\n)\n\n => Array\n(\n => 212.103.48.152\n)\n\n => Array\n(\n => 183.83.153.90\n)\n\n => Array\n(\n => 117.248.150.41\n)\n\n => Array\n(\n => 185.240.246.180\n)\n\n => Array\n(\n => 162.253.131.125\n)\n\n => Array\n(\n => 117.230.153.217\n)\n\n => Array\n(\n => 117.230.169.1\n)\n\n => Array\n(\n => 49.15.138.247\n)\n\n => Array\n(\n => 117.230.37.110\n)\n\n => Array\n(\n => 14.167.188.75\n)\n\n => Array\n(\n => 169.149.239.93\n)\n\n => Array\n(\n => 103.216.176.91\n)\n\n => Array\n(\n => 117.230.12.126\n)\n\n => Array\n(\n => 184.75.209.110\n)\n\n => Array\n(\n => 117.230.6.60\n)\n\n => Array\n(\n => 117.230.135.132\n)\n\n => Array\n(\n => 31.179.29.109\n)\n\n => Array\n(\n => 74.121.188.186\n)\n\n => Array\n(\n => 117.230.35.5\n)\n\n => Array\n(\n => 111.92.74.239\n)\n\n => Array\n(\n => 104.245.144.236\n)\n\n => Array\n(\n => 39.50.22.100\n)\n\n => Array\n(\n => 47.31.190.23\n)\n\n => Array\n(\n => 157.44.73.187\n)\n\n => Array\n(\n => 117.230.8.91\n)\n\n => Array\n(\n => 157.32.18.2\n)\n\n => Array\n(\n => 111.119.187.43\n)\n\n => Array\n(\n => 203.101.185.246\n)\n\n => Array\n(\n => 5.62.34.22\n)\n\n)\n```\nArchive for April, 2009: Laceshawl's Personal Finance Blog\n << Back to all Blogs Login or Create your own free blog Layout: Blue and Brown (Default) Author's Creation\nHome > Archive: April, 2009",
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"# Archive for April, 2009\n\n## Got a job!\n\nApril 28th, 2009 at 06:54 am\n\nOnly 5 hours cleaning a week, but for nice folks and the amount earned doesn't interfere with my benefit. Now I can start contributing to Kiwisaver again. My paln is to keep living on just the amount of the benefit and save all extra money into my EF (which currently stands at nz2,358)\n\n## All spick and dandy\n\nApril 21st, 2009 at 05:47 am\n\nMy church is looking for someone to clean. Just a few hours a week, but one must start somewhere, so I volunteered to spend the next week doing the cleaning. If it works out I should get the job.\n\n## Mend and make do\n\nApril 20th, 2009 at 07:47 am\n\nA friend has this old hand knitted jersey he's very fond of. He complained to me the cuffs and neckband were wearing out and asked if I could do something to mend it. They were quite ragged, so I unravelled them and knitted up new ones using some old wool I had. Result, one smart looking jersey. He was very pleased.\n\nI think poeple who know these skills are going to be frequently called on in the coming months.\n\n## Night out\n\nApril 19th, 2009 at 12:44 am\n\nLast night a friend rang up to say she had tickets to the National Youth Choir and did I want to go. So I jumped at the chance and hunted in my wardrobe for some glad rags to wear. Concert going ladies always seem to go for shawls in a big way, so I wore one of my hand knitted ones. I got several compliments, which is pretty good as the yarn only cost me a dollar, and the rest of my clothes also came from the thrift store - total considerably less than the cost of the tickets. Anyway, had a wonderful time, the choir was terrific, but there was a sour end to the evening. While they were performing someone broke into the changing rooms and stole their wallets and other property. Hope the cops catch them soon.\n\n## Up and down\n\nApril 15th, 2009 at 06:02 am\n\nAbout 15 years ago, New Zealand deicided to get rid of the 1 cent coin, as you couldn't buy anything with it. From then on, everything was rounded to the nearest 5 cents. However, in their attempts to make us think we were getting a bargain, the shops continued to price in nines. So you'd see a can of beans at .99 and think, less than a dollar, but when you got to the checkout, it cost you a dollar anyway. But if you bought 3 cans, \\$2.97 rounded down to \\$2.95. Two cents saved!\n\nNow we have got rid of the 5 cent as well and everything is rounded to the nearest 10 cents. Meanwhile the shops continue to sell mushrooms at nz9.99 a kilogram, so your purchase may come to nz4.76, which rounds up. Add the other things you're buying ending in 9 cents, and it may go up or down. Personally, I make it a game to get a total which will round down.\n\nThis morning I lucked out. I bought fruit and veg for nz9.78, so paid two cents extra. However, if I'd paid by debit card, they would have charged me the exact amount. Go figure.\n\n## You win some, you lose some\n\nApril 13th, 2009 at 06:50 am\n\nI went on a retreat over Easter, which I've been doing for several years. Lovely peaceful setting, good meals provided, and all for only nz\\$100. Definitely not giving that up. i took lunch up with me, but on the way back down we stopped at a very expensive cafe, and I spent \\$13 on lunch, because I didn't want to say no to the others. Oh well, if it's the only time I eat out this month I'll be doing well.\n\n## Reward for good behaviour\n\nApril 9th, 2009 at 12:10 am\n\nEven though it was really cold outside I decided to go to Bible study at church. On my way I found a dollar. Wouldn't it be nice if I got money every time I did. guess a lot more people would go to church.\n\n## Two in a row\n\nApril 8th, 2009 at 03:48 am\n\nNSDs that is, if you don't count a direct debit to pay the credit union loan. The weather this week is horrible, yucky sleet and hail. I am not at all tempted to go out, so it looks like tomorrow will make it a hat trick. Of course I haven't received the electricity bill yet...\n\n## A new jacket\n\nApril 6th, 2009 at 08:02 am\n\nis what I'm looking for, since we're approaching winter in the Southern Hemisphere. I'm hoping to get one in the thrift stores for no more than nz\\$30, but no luck so far Ideally, I'd like it to be waterproof, but good enough to wear to work. I hope I find one soon, as I keep being tempted to impulse buy all sorts of pretty tops and skirts which I don't need, and all those two dollars an item keep adding up. The only other clothes I need are new socks and bra. Will wait for sale at The Warehouse.\n\n## and thereby hangs a yarn\n\nApril 3rd, 2009 at 06:30 am\n\nI put in another p/t job application today, working at a yarn store. I would love that job, although it doesn't pay very much. But I would have to watch I don't spend all my money on knitting supplies.\n\nAt the moment I'm knitting baby clothes for Pregnancy Help, but the wool was all donated.\n\n## Ran out of coffee\n\nApril 2nd, 2009 at 06:32 am\n\nI briefly considered giving it up and drinking water, then I went and bought more coffee. nz2.50. I'll save on something else. No point having piles of money if I can't have what I like with some of the excess.\n\nI have these periodic savings binges where I stop spending on virtually everything to save money, but that's not sustainable long term. I actually enjoy spending money on something I want, even if Depression-raised family voices keep telling me I should stick it under the mattress, in case I need it more later. I have no intention of being a multimillionaire centenarian full of regrets for how I deprived myself. So I'm planning to go on a retreat at Easter after my Lenten frugality."
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https://gitlab.xiph.org/xiph/aom-rav1e/-/commit/5f5c132fef4ffedf9e81ec9a4e1db9e73bcf0af6 | [
"### Jointly optimizing deringing and clpf\n\n```We now signal joint strengths and use a greedy algorithm for the search.\n\nlow-latency, cpu-used=4:\n\nPSNR | PSNR Cb | PSNR Cr | PSNR HVS | SSIM | MS SSIM | CIEDE 2000\n-0.0792 | 0.3551 | 0.4393 | -0.0108 | -0.1338 | -0.0141 | 0.1452\n\nChange-Id: I619ae1c7c7d7ec04fe993cabc5773b07c3f5b201```\nparent 800df032\n ... ... @@ -366,8 +366,7 @@ typedef struct { int send_dq_bit; #endif // CONFIG_NEW_QUANT /* deringing gain *per-superblock* */ int8_t dering_gain; int8_t clpf_strength; int8_t cdef_strength; #if CONFIG_DELTA_Q int current_q_index; #endif ... ...\n ... ... @@ -24,74 +24,6 @@ int dering_level_table[DERING_STRENGTHS] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 17, 20, 24, 28, 33, 39, 46, 54, 63 }; #ifndef NDEBUG static int is_sorted(const int *arr, int num) { int sorted = 1; while (sorted && num-- > 1) sorted &= arr[num] >= arr[num - 1]; return sorted; } #endif uint32_t levels_to_id(const int lev[DERING_REFINEMENT_LEVELS], const int str[CLPF_REFINEMENT_LEVELS]) { uint32_t id = 0; int i; assert(is_sorted(lev, DERING_REFINEMENT_LEVELS)); assert(is_sorted(str, CLPF_REFINEMENT_LEVELS)); for (i = 0; i < DERING_REFINEMENT_LEVELS; i++) id = id * DERING_STRENGTHS + lev[i]; for (i = 0; i < CLPF_REFINEMENT_LEVELS; i++) id = id * CLPF_STRENGTHS + str[i]; return id; } void id_to_levels(int lev[DERING_REFINEMENT_LEVELS], int str[CLPF_REFINEMENT_LEVELS], uint32_t id) { int i; for (i = CLPF_REFINEMENT_LEVELS - 1; i >= 0; i--) { str[i] = id % CLPF_STRENGTHS; id /= CLPF_STRENGTHS; } for (i = DERING_REFINEMENT_LEVELS - 1; i >= 0; i--) { lev[i] = id % DERING_STRENGTHS; id /= DERING_STRENGTHS; } // Pack tables int j; for (i = j = 1; i < DERING_REFINEMENT_LEVELS && j < DERING_REFINEMENT_LEVELS; i++) if (lev[j - 1] == lev[j]) memmove(&lev[j - 1], &lev[j], (DERING_REFINEMENT_LEVELS - j) * sizeof(*lev)); else j++; for (i = j = 1; i < CLPF_REFINEMENT_LEVELS && j < DERING_REFINEMENT_LEVELS; i++) if (str[j - 1] == str[j]) memmove(&str[j - 1], &str[j], (CLPF_REFINEMENT_LEVELS - i) * sizeof(*str)); else j++; assert(is_sorted(lev, DERING_REFINEMENT_LEVELS)); assert(is_sorted(str, CLPF_REFINEMENT_LEVELS)); } void cdef_get_bits(const int *lev, const int *str, int *dering_bits, int *clpf_bits) { int i; *dering_bits = *clpf_bits = 1; for (i = 1; i < DERING_REFINEMENT_LEVELS; i++) (*dering_bits) += lev[i] != lev[i - 1]; for (i = 1; i < CLPF_REFINEMENT_LEVELS; i++) (*clpf_bits) += str[i] != str[i - 1]; *dering_bits = get_msb(*dering_bits); *clpf_bits = get_msb(*clpf_bits); } int sb_all_skip(const AV1_COMMON *const cm, int mi_row, int mi_col) { int r, c; int maxc, maxr; ... ... @@ -212,8 +144,7 @@ static void copy_sb8_16(UNUSED AV1_COMMON *cm, uint16_t *dst, int dstride, } void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, MACROBLOCKD *xd, uint32_t global_level, int clpf_strength_u, int clpf_strength_v) { int clpf_strength_u, int clpf_strength_v) { int r, c; int sbr, sbc; int nhsb, nvsb; ... ... @@ -231,12 +162,11 @@ void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, MACROBLOCKD *xd, int dering_left; int coeff_shift = AOMMAX(cm->bit_depth - 8, 0); int nplanes = 3; int lev[DERING_REFINEMENT_LEVELS]; int str[CLPF_REFINEMENT_LEVELS]; int *lev; int chroma_dering = xd->plane.subsampling_x == xd->plane.subsampling_y && xd->plane.subsampling_x == xd->plane.subsampling_y; id_to_levels(lev, str, global_level); lev = cm->cdef_strengths; nvsb = (cm->mi_rows + MAX_MIB_SIZE - 1) / MAX_MIB_SIZE; nhsb = (cm->mi_cols + MAX_MIB_SIZE - 1) / MAX_MIB_SIZE; av1_setup_dst_planes(xd->plane, frame, 0, 0); ... ... @@ -277,11 +207,13 @@ void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, MACROBLOCKD *xd, level = dering_level_table [lev[cm->mi_grid_visible[MAX_MIB_SIZE * sbr * cm->mi_stride + MAX_MIB_SIZE * sbc] ->mbmi.dering_gain]]; ->mbmi.cdef_strength] / CLPF_STRENGTHS]; clpf_strength = str[cm->mi_grid_visible[MAX_MIB_SIZE * sbr * cm->mi_stride + lev[cm->mi_grid_visible[MAX_MIB_SIZE * sbr * cm->mi_stride + MAX_MIB_SIZE * sbc] ->mbmi.clpf_strength]; ->mbmi.cdef_strength] % CLPF_STRENGTHS; clpf_strength += clpf_strength == 3; curr_row_dering[sbc] = 0; if ((level == 0 && clpf_strength == 0) || ... ...\n ... ... @@ -11,15 +11,8 @@ #ifndef AV1_COMMON_DERING_H_ #define AV1_COMMON_DERING_H_ // ceil(log2(DERING_STRENGTHS^DERING_REFINEMENT_LEVELS * // CLPF_STRENGTHS^CLPF_REFINEMENT_LEVELS)) #define DERING_LEVEL_BITS (22) #define MAX_DERING_LEVEL (1LL << DERING_LEVEL_BITS) #define DERING_REFINEMENT_BITS 2 #define DERING_REFINEMENT_LEVELS 4 #define CLPF_REFINEMENT_BITS 1 #define CLPF_REFINEMENT_LEVELS 2 #define CDEF_MAX_STRENGTHS 16 #define CDEF_STRENGTH_BITS 7 #define DERING_STRENGTHS 21 #define CLPF_STRENGTHS 4 ... ... @@ -37,19 +30,11 @@ extern \"C\" { extern int dering_level_table[DERING_STRENGTHS]; uint32_t levels_to_id(const int lev[DERING_REFINEMENT_LEVELS], const int str[CLPF_REFINEMENT_LEVELS]); void id_to_levels(int lev[DERING_REFINEMENT_LEVELS], int str[CLPF_REFINEMENT_LEVELS], uint32_t id); void cdef_get_bits(const int *lev, const int *str, int *dering_bits, int *clpf_bits); int sb_all_skip(const AV1_COMMON *const cm, int mi_row, int mi_col); int sb_compute_dering_list(const AV1_COMMON *const cm, int mi_row, int mi_col, dering_list *dlist); void av1_cdef_frame(YV12_BUFFER_CONFIG *frame, AV1_COMMON *cm, MACROBLOCKD *xd, uint32_t global_level, int clpf_strength_u, int clpf_strength_v); int clpf_strength_u, int clpf_strength_v); void av1_cdef_search(YV12_BUFFER_CONFIG *frame, const YV12_BUFFER_CONFIG *ref, AV1_COMMON *cm, MACROBLOCKD *xd); ... ...\n ... ... @@ -397,11 +397,9 @@ typedef struct AV1Common { int mib_size; // Size of the superblock in units of MI blocks int mib_size_log2; // Log 2 of above. #if CONFIG_CDEF uint32_t dering_level; int dering_lev[DERING_REFINEMENT_LEVELS]; int clpf_str[CLPF_REFINEMENT_LEVELS]; int dering_bits; int clpf_bits; int nb_cdef_strengths; int cdef_strengths[CDEF_MAX_STRENGTHS]; int cdef_bits; int clpf_strength_u; int clpf_strength_v; #endif ... ...\n ... ... @@ -2402,14 +2402,11 @@ static void decode_partition(AV1Decoder *const pbi, MACROBLOCKD *const xd, if (bsize == BLOCK_64X64) { #endif if (!sb_all_skip(cm, mi_row, mi_col)) { cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col]->mbmi.dering_gain = aom_read_literal(r, cm->dering_bits, ACCT_STR); cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col]->mbmi.clpf_strength = aom_read_literal(r, cm->clpf_bits, ACCT_STR); cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col]->mbmi.cdef_strength = aom_read_literal(r, cm->cdef_bits, ACCT_STR); } else { cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col]->mbmi.dering_gain = cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col] ->mbmi.clpf_strength = 0; cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col]->mbmi.cdef_strength = 0; } } #endif // CONFIG_CDEF ... ... @@ -2673,11 +2670,14 @@ static void setup_loopfilter(AV1_COMMON *cm, struct aom_read_bit_buffer *rb) { #if CONFIG_CDEF static void setup_cdef(AV1_COMMON *cm, struct aom_read_bit_buffer *rb) { cm->dering_level = aom_rb_read_literal(rb, DERING_LEVEL_BITS); int i; cm->cdef_bits = aom_rb_read_literal(rb, 2); cm->nb_cdef_strengths = 1 << cm->cdef_bits; for (i = 0; i < cm->nb_cdef_strengths; i++) { cm->cdef_strengths[i] = aom_rb_read_literal(rb, CDEF_STRENGTH_BITS); } cm->clpf_strength_u = aom_rb_read_literal(rb, 2); cm->clpf_strength_v = aom_rb_read_literal(rb, 2); id_to_levels(cm->dering_lev, cm->clpf_str, cm->dering_level); cdef_get_bits(cm->dering_lev, cm->clpf_str, &cm->dering_bits, &cm->clpf_bits); } #endif // CONFIG_CDEF ... ... @@ -4950,10 +4950,9 @@ void av1_decode_frame(AV1Decoder *pbi, const uint8_t *data, } #if CONFIG_CDEF if ((cm->dering_level || cm->clpf_strength_u || cm->clpf_strength_v) && !cm->skip_loop_filter) { av1_cdef_frame(&pbi->cur_buf->buf, cm, &pbi->mb, cm->dering_level, cm->clpf_strength_u, cm->clpf_strength_v); if (!cm->skip_loop_filter) { av1_cdef_frame(&pbi->cur_buf->buf, cm, &pbi->mb, cm->clpf_strength_u, cm->clpf_strength_v); } #endif // CONFIG_CDEF ... ...\n ... ... @@ -2785,14 +2785,10 @@ static void write_modes_sb(AV1_COMP *const cpi, const TileInfo *const tile, if (bsize == BLOCK_64X64 && #endif // CONFIG_EXT_PARTITION !sb_all_skip(cm, mi_row, mi_col)) { if (cm->dering_bits) if (cm->cdef_bits != 0) aom_write_literal(w, cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col] ->mbmi.dering_gain, cm->dering_bits); if (cm->clpf_bits) aom_write_literal(w, cm->mi_grid_visible[mi_row * cm->mi_stride + mi_col] ->mbmi.clpf_strength, cm->clpf_bits); ->mbmi.cdef_strength, cm->cdef_bits); } #endif } ... ... @@ -3496,7 +3492,11 @@ static void encode_loopfilter(AV1_COMMON *cm, struct aom_write_bit_buffer *wb) { #if CONFIG_CDEF static void encode_cdef(const AV1_COMMON *cm, struct aom_write_bit_buffer *wb) { aom_wb_write_literal(wb, cm->dering_level, DERING_LEVEL_BITS); int i; aom_wb_write_literal(wb, cm->cdef_bits, 2); for (i = 0; i < cm->nb_cdef_strengths; i++) { aom_wb_write_literal(wb, cm->cdef_strengths[i], CDEF_STRENGTH_BITS); } aom_wb_write_literal(wb, cm->clpf_strength_u, 2); aom_wb_write_literal(wb, cm->clpf_strength_v, 2); } ... ...\n ... ... @@ -3522,14 +3522,17 @@ static void loopfilter_frame(AV1_COMP *cpi, AV1_COMMON *cm) { } #if CONFIG_CDEF if (is_lossless_requested(&cpi->oxcf)) { cm->dering_level = cm->clpf_strength_u = cm->clpf_strength_v = 0; cm->clpf_strength_u = cm->clpf_strength_v = 0; cm->cdef_bits = 0; cm->cdef_strengths = 0; cm->nb_cdef_strengths = 1; } else { // Find cm->dering_level, cm->clpf_strength_u and cm->clpf_strength_v av1_cdef_search(cm->frame_to_show, cpi->Source, cm, xd); // Apply the filter av1_cdef_frame(cm->frame_to_show, cm, xd, cm->dering_level, cm->clpf_strength_u, cm->clpf_strength_v); av1_cdef_frame(cm->frame_to_show, cm, xd, cm->clpf_strength_u, cm->clpf_strength_v); // Pack the clpf chroma strengths into two bits each cm->clpf_strength_u -= cm->clpf_strength_u == 4; ... ...\n ... ... @@ -20,6 +20,64 @@ #include \"av1/encoder/clpf_rdo.h\" #include \"av1/encoder/encoder.h\" #define TOTAL_STRENGTHS (DERING_STRENGTHS * CLPF_STRENGTHS) /* Search for the best strength to add as an option, knowing we already selected nb_strengths options. */ static uint64_t search_one(int *lev, int nb_strengths, uint64_t mse[][TOTAL_STRENGTHS], int sb_count) { uint64_t tot_mse[TOTAL_STRENGTHS]; int i, j; uint64_t best_tot_mse = (uint64_t)1 << 63; int best_id = 0; memset(tot_mse, 0, sizeof(tot_mse)); for (i = 0; i < sb_count; i++) { int gi; uint64_t best_mse = (uint64_t)1 << 63; /* Find best mse among already selected options. */ for (gi = 0; gi < nb_strengths; gi++) { if (mse[i][lev[gi]] < best_mse) { best_mse = mse[i][lev[gi]]; } } /* Find best mse when adding each possible new option. */ for (j = 0; j < TOTAL_STRENGTHS; j++) { uint64_t best = best_mse; if (mse[i][j] < best) best = mse[i][j]; tot_mse[j] += best; } } for (j = 0; j < TOTAL_STRENGTHS; j++) { if (tot_mse[j] < best_tot_mse) { best_tot_mse = tot_mse[j]; best_id = j; } } lev[nb_strengths] = best_id; return best_tot_mse; } /* Search for the set of strengths that minimizes mse. */ static uint64_t joint_strength_search(int *best_lev, int nb_strengths, uint64_t mse[][TOTAL_STRENGTHS], int sb_count) { uint64_t best_tot_mse; int i; best_tot_mse = (uint64_t)1 << 63; /* Greedy search: add one strength options at a time. */ for (i = 0; i < nb_strengths; i++) { best_tot_mse = search_one(best_lev, i, mse, sb_count); } /* Trying to refine the greedy search by reconsidering each already-selected option. */ for (i = 0; i < 4 * nb_strengths; i++) { int j; for (j = 0; j < nb_strengths - 1; j++) best_lev[j] = best_lev[j + 1]; best_tot_mse = search_one(best_lev, nb_strengths - 1, mse, sb_count); } return best_tot_mse; } static double compute_dist(uint16_t *x, int xstride, uint16_t *y, int ystride, int nhb, int nvb, int coeff_shift) { int i, j; ... ... @@ -50,21 +108,24 @@ void av1_cdef_search(YV12_BUFFER_CONFIG *frame, const YV12_BUFFER_CONFIG *ref, int level; int dering_count; int coeff_shift = AOMMAX(cm->bit_depth - 8, 0); uint64_t best_tot_mse = 0; uint64_t best_tot_mse = (uint64_t)1 << 63; uint64_t tot_mse; int sb_count; int nvsb = (cm->mi_rows + MAX_MIB_SIZE - 1) / MAX_MIB_SIZE; int nhsb = (cm->mi_cols + MAX_MIB_SIZE - 1) / MAX_MIB_SIZE; int *sb_index = aom_malloc(nvsb * nhsb * sizeof(*sb_index)); uint64_t(*mse)[DERING_STRENGTHS][CLPF_STRENGTHS] = uint64_t(*mse)[DERING_STRENGTHS * CLPF_STRENGTHS] = aom_malloc(sizeof(*mse) * nvsb * nhsb); int clpf_damping = 3 + (cm->base_qindex >> 6); int i; int lev[DERING_REFINEMENT_LEVELS]; int best_lev[DERING_REFINEMENT_LEVELS]; int str[CLPF_REFINEMENT_LEVELS]; int best_str[CLPF_REFINEMENT_LEVELS]; double lambda = exp(cm->base_qindex / 36.0); static int log2[] = { 0, 1, 2, 2 }; int best_lev[CDEF_MAX_STRENGTHS]; int nb_strengths; int nb_strength_bits; int quantizer; double lambda; quantizer = av1_ac_quant(cm->base_qindex, 0, cm->bit_depth) >> (cm->bit_depth - 8); lambda = .12 * quantizer * quantizer / 256.; src = aom_memalign(32, sizeof(*src) * cm->mi_rows * cm->mi_cols * 64); ref_coeff = ... ... @@ -143,7 +204,7 @@ void av1_cdef_search(YV12_BUFFER_CONFIG *frame, const YV12_BUFFER_CONFIG *ref, i + (i == 3), clpf_damping, coeff_shift); copy_dering_16bit_to_16bit(dst, MAX_MIB_SIZE << bsize, tmp_dst, dlist, dering_count, bsize); mse[sb_count][gi][i] = (int)compute_dist( mse[sb_count][gi * CLPF_STRENGTHS + i] = (int)compute_dist( dst, MAX_MIB_SIZE << bsize, &ref_coeff[(sbr * stride * MAX_MIB_SIZE << bsize) + (sbc * MAX_MIB_SIZE << bsize)], ... ... @@ -155,85 +216,38 @@ void av1_cdef_search(YV12_BUFFER_CONFIG *frame, const YV12_BUFFER_CONFIG *ref, sb_count++; } } best_tot_mse = (uint64_t)1 << 63; int l0; for (l0 = 0; l0 < DERING_STRENGTHS; l0++) { int l1; lev = l0; for (l1 = l0; l1 < DERING_STRENGTHS; l1++) { int l2; lev = l1; for (l2 = l1; l2 < DERING_STRENGTHS; l2++) { int l3; lev = l2; for (l3 = l2; l3 < DERING_STRENGTHS; l3++) { int cs0; lev = l3; for (cs0 = 0; cs0 < CLPF_STRENGTHS; cs0++) { int cs1; str = cs0; for (cs1 = cs0; cs1 < CLPF_STRENGTHS; cs1++) { uint64_t tot_mse = 0; str = cs1; for (i = 0; i < sb_count; i++) { int gi; int cs; uint64_t best_mse = (uint64_t)1 << 63; for (gi = 0; gi < DERING_REFINEMENT_LEVELS; gi++) { for (cs = 0; cs < CLPF_REFINEMENT_LEVELS; cs++) { if (mse[i][lev[gi]][str[cs]] < best_mse) { best_mse = mse[i][lev[gi]][str[cs]]; } } } tot_mse += best_mse; } // Add the bit cost int dering_diffs = 0, clpf_diffs = 0; for (i = 1; i < DERING_REFINEMENT_LEVELS; i++) dering_diffs += lev[i] != lev[i - 1]; for (i = 1; i < CLPF_REFINEMENT_LEVELS; i++) clpf_diffs += str[i] != str[i - 1]; tot_mse += (uint64_t)(sb_count * lambda * (log2[dering_diffs] + log2[clpf_diffs])); if (tot_mse < best_tot_mse) { for (i = 0; i < DERING_REFINEMENT_LEVELS; i++) best_lev[i] = lev[i]; for (i = 0; i < CLPF_REFINEMENT_LEVELS; i++) best_str[i] = str[i]; best_tot_mse = tot_mse; } } } } } nb_strength_bits = 0; /* Search for different number of signalling bits. */ for (i = 0; i <= 3; i++) { nb_strengths = 1 << i; tot_mse = joint_strength_search(best_lev, nb_strengths, mse, sb_count); /* Count superblock signalling cost. */ tot_mse += (uint64_t)(sb_count * lambda * i); /* Count header signalling cost. */ tot_mse += (uint64_t)(nb_strengths * lambda * CDEF_STRENGTH_BITS); if (tot_mse < best_tot_mse) { best_tot_mse = tot_mse; nb_strength_bits = i; } } for (i = 0; i < DERING_REFINEMENT_LEVELS; i++) lev[i] = best_lev[i]; for (i = 0; i < CLPF_REFINEMENT_LEVELS; i++) str[i] = best_str[i]; id_to_levels(lev, str, levels_to_id(lev, str)); // Pack tables cdef_get_bits(lev, str, &cm->dering_bits, &cm->clpf_bits); nb_strengths = 1 << nb_strength_bits; cm->cdef_bits = nb_strength_bits; cm->nb_cdef_strengths = nb_strengths; for (i = 0; i < nb_strengths; i++) cm->cdef_strengths[i] = best_lev[i]; for (i = 0; i < sb_count; i++) { int gi, cs; int best_gi, best_clpf; int gi; int best_gi; uint64_t best_mse = (uint64_t)1 << 63; best_gi = best_clpf = 0; for (gi = 0; gi < (1 << cm->dering_bits); gi++) { for (cs = 0; cs < (1 << cm->clpf_bits); cs++) { if (mse[i][lev[gi]][str[cs]] < best_mse) { best_gi = gi; best_clpf = cs; best_mse = mse[i][lev[gi]][str[cs]]; } best_gi = 0; for (gi = 0; gi < cm->nb_cdef_strengths; gi++) { if (mse[i][best_lev[gi]] < best_mse) { best_gi = gi; best_mse = mse[i][best_lev[gi]]; } } cm->mi_grid_visible[sb_index[i]]->mbmi.dering_gain = best_gi; cm->mi_grid_visible[sb_index[i]]->mbmi.clpf_strength = best_clpf; cm->mi_grid_visible[sb_index[i]]->mbmi.cdef_strength = best_gi; } aom_free(src); ... ... @@ -245,5 +259,4 @@ void av1_cdef_search(YV12_BUFFER_CONFIG *frame, const YV12_BUFFER_CONFIG *ref, AOM_PLANE_U); av1_clpf_test_plane(cm->frame_to_show, ref, cm, &cm->clpf_strength_v, AOM_PLANE_V); cm->dering_level = levels_to_id(best_lev, best_str); }\nMarkdown is supported\n0% or .\nYou are about to add 0 people to the discussion. 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http://www.ijpsonline.com/articles/solubility-prediction-of-satranidazole-in-propylene-glycolwater-mixtures-using-extended-hildebrand-solubility-approach.html?view=mobile | [
"*Corresponding Author:\nRathi P.B\nShri Bhagwan College of Pharmacy, N-6, CIDCO, Auranagabad-431 003, India\nE-mail: [email protected]\n Date of Submission 19 January 2011 Date of Revision 21 October 2011 Date of Acceptance 31 October 2011 Indian J Pharm Sci, 2011, 73 (6): 670-674\n\n## Abstract\n\nExtended Hildebrand solubility approach is used to estimate the solubility of satranidazole in binary solvent systems. The solubility of satranidazole in various propylene glycol-water mixtures was analyzed in terms of solute-solvent interactions using a modified version of Hildebrand-Scatchard treatment for regular solutions. The solubility equation employs term interaction energy (W) to replace the geometric mean (δ1δ2), where δand δ2 are the cohesive energy densities for the solvent and solute, respectively. The new equation provides an accurate prediction of solubility once the interaction energy, W, is obtained. In this case, the energy term is regressed against a polynomial in δ1 of the binary mixture. A quartic expression of W in terms of solvent solubility parameter was found for predicting the solubility of satranidazole in propylene glycol-water mixtures. The expression yields an error in mole fraction solubility of ~3.74%, a value approximating that of the experimentally determined solubility. The method has potential usefulness in preformulation and formulation studies during which solubility prediction is important for drug design.\n\n## Keywords\n\nExtended Hildebrand solubility approach, propylene glycol, regular solution theory, satranidazole, solubility parameter\n\nSolubility data on drugs and pharmaceutical adjuncts in mixed solvents have wide applications in the drug sciences. Knowledge of interaction forces between solutes and solvents are of considerable theoretical and practical interest throughout the physical and biological sciences . The theory of solution is one of the most challenging branches of physical chemistry. The Hildebrand-Scatchard theory of regular solution is the pioneer approach in this field, used to estimate solubility only for relatively non-polar drugs in non-polar solvents . An irregular solution is one in which self-association of solute or solvent, solvation of the solute by the solvent molecules, or complexation of two or more solute species are involved . Polar systems exhibit irregular solution behaviour and are commonly encountered in pharmacy. Extended Hildebrand solubility approach (EHSA), modification of the Hildebrand-Scatchard equation, permits calculation of the solubility of polar and non-polar solutes in solvents ranging from non-polar hydrocarbons to highly polar solvents such as water, ethanol, and glycols . The solubility parameters of solute and solvent were introduced to explain the behaviour of regular and irregular solutions . The EHSA has been developed to reproduce the solubility of drugs and other solids in the binary solvent systems .\n\nThe Hildebrand-Scatchard equation for the solubility of crystalline solids in a regular solution may be written as ,",
null,
"(1a)",
null,
"(1b)\n\nThe Extended Hildebrand equation for the solubility of solids in an irregular solution may be written as ,",
null,
"(2)\n\nFrom the geometric mean:",
null,
"(3a)\n\nIn pharmaceutical solutions, the geometric mean of δ1 and δ2 is too restrictive and ordinarily provides a poor fit to experimental data in irregular solutions. The assumption that the geometric mean of two geometric parameters δ1δ2 (Eqn. 1) can be replaced by a less restrictive term W (Eqn. 2), interaction energy parameter, which is allowed to take on values as required to yield correct mole fraction solubilities, X2 as ,",
null,
"(3b)\n\nwhere, K is the proportionality factor relating W to the geometric mean of solubility parameter.\n\nIn Eqn. 1 and Eqn. 2, X2 and X2 i are the mole fraction solubility and ideal mole fraction solubility of the solute respectively. The terms δ1 and δ2 are the solubility parameters for the solvent and solute respectively. The geometric mean, δ1 δ2, provides a reasonable estimate of solvent-solute interaction in regular (ordinarily non-polar) mixtures, whereas W or K δ1 δ2 is required to express solubility’s in non-regular systems (irregular solutions) of drugs in associating mixed solvents.\n\nThe term negative logarithm of the ideal solubility (–log X2 i) can be taken as ,",
null,
"(4)\n\nWhere, ΔHf is heat of fusion of the crystalline drug molecule, T0 is the melting point of solute in absolute degrees.\n\nThe term A in equations 1 and 2 is defined as,",
null,
"(5)\n\nWhere, V2 is the molar volume of the solute as a hypothetical supercooled liquid at solution temperature, R is the universal gas constant, T is the absolute temperature, 298.2°K, of the experiment and Φ1, the volume fraction of the solvent, is ,",
null,
"(6)\n\nWhere, V1 is the molar volume of the solvent at 25°. The term logarithmic solute activity coefficient (log γ2) from Eqn. 2 and Eqn. 5 can be written as ,",
null,
"(7)\n\nA better approach is not to restrict the interaction term W to a geometric mean but evaluate it experimentally from the solubility of the solute in various solvent concentrations in a binary mixture employing Eqn. 2. An empirical equation for W as a function of solubility parameters of the solvent mixture remains to be discovered. Then, backcalculating W and substituting into Eqn. 2 permit the mole fraction solubility of a drug (solute) to be predicted in essentially any solvent mixture. Therefore, the present investigation pertains to the utility of EHSA in relation to the satranidazole solubility in propylene glycol (PG)-water binary solvent mixtures.\n\nSatranidazole, obtained as gift sample from Alkem Laboratories Ltd., Baddi, India, was purified by recrystallization from acetone. Propylene glycol and acetone were purchased from ICPA Laboratories; Ankleshwar, India and Qualigens Fine Chemicals, Mumbai, India respectively. Throughout the study freshly prepared double distilled water was used for experimental purpose. All chemicals and reagents used in the study were of analytical grade and used as such. Double beam UV/Vis spectrophotometer, Shimadzu model 1601 with spectral bandwidth of 2 nm, wavelength accuracy ±0.5 nm and a pair of 10 mm matched quartz cells was used to measure absorbance of the resulting solutions. Citizen balance, CX-100, was used for weighing of Satranidazole. Differential Scanning Calorimeter, Shimadzu TA-60 WS, was used for determination of melting point and heat of fusion of satranidazole.\n\nThe solubility of satranidazole was determined in binary solvent mixtures of PG and water. Double distilled water was used to prepare mixtures with PG in concentrations of 0-100% by volume of PG. About 10 ml of PG, water, or binary solvent blends were introduced into screw-capped vials containing an excess amount of satranidazole. After being sealed with several turns of electrical tape, the vials were submerged in water at 25±0.4° and were shaken at 150 rpm for 24 h in a constanttemperature bath. Preliminary studies showed that this time period was sufficient to ensure saturation at 25°.\n\nAfter equilibration, the solutions were microfiltered (0.45 μm) and the filtrate was then diluted with double distilled water to carry out the spectrophotometric determination at the maximum wavelength of absorption of the satranidazole (λmax-319.80 nm). The solubility of the satranidazole was determined at least three times for this solvent mixture, and the average value was taken. The densities of the solvent mixtures and the filtrates of saturated solutions of satranidazole were determined in triplicate at 25±0.4° using 10-ml specific gravity bottle. Once the densities of solutions are known, the solubilities can be expressed in mole fraction scale.\n\nThe solubility parameters of the solvents were obtained from the literature [15,16]. The solubility parameter of satranidazole was calculated previously by Fedor’s group contribution method [17,18], which was confirmed by solubility analysis in dioxane-water blend.\n\nThe thermogram of satranidazole was obtained with a differential scanning calorimeter. The melting point and heat of fusion were measured. Sample of 8.8 mg in perforated pan was heated at a rate of 15°/min under nitrogen purge. The temperature range studied was 25-225°. The molar enthalpy of fusion of satranidazole was 112.30 J/g (7763.838 cal/mol) and the temperature of fusion is 461.83°K. Neither decomposition nor polymorphic change was observed at the experimental temperature range. The ideal mole fraction solubility of satranidazole was calculated from these values (–logX2i = 1.61). The mole fraction solubilities of satranidazole at 25±0.4° in PG-water binary mixtures which cover a large range of the solubility parameter scale, from 14.80 to 23.40 (Cal/cm3)0.5, are listed in Table 1. The experimental mole fraction solubility of satranidazole at 25±0.4° in PG-water mixtures is plotted in fig. 1 versus the solubility parameter, δ1, of the various mixed solvent systems. The mole fraction solubility of satranidazole (δ2=14.80) in PG (δ1=14.80), water (δ1=23.4), and in the mixture of the two solvents is represented by the solid circles in fig. 1. The maximum solubility of satranidazole in the mixture is X2=0.0002545 mol/l and occurs at δ1=14.80. This value is well below the ideal solubility, X2i =0.0245614 mol/l, as predicted from regular solution theory. The discrepancy between the results using the original Hildebrand-Scatchard equation and experimental points demonstrates that Eqn. 1a and Eqn. 1b cannot be used to predict drug solubility in PG-water binary solvent systems. This behavior has been dealt with the theoretical replacement of mean geometric solubility parameters (δ1δ2) term with the interaction energy term (W).\n\nPG-Water (%v/v) Solubility d1 V1 Density of Mol. Wt X2(obs) W(obs)\n(g/ml) (Cal/cm3)0.5 blend of blend\n0:100 0.0005025 23.40 18.00 0.9980 18.00 3.1350E-05 330.26\n10:90 0.0005350 22.54 23.56 1.0018 23.81 4.3980E-05 310.93\n20:80 0.0005350 21.68 29.12 1.0056 29.62 5.4505E-05 292.19\n30:70 0.0005495 20.82 34.68 1.0094 35.43 6.6711E-05 274.17\n40:60 0.0006073 19.96 40.24 1.0132 41.24 8.5499E-05 256.95\n50:50 0.0006435 19.10 45.80 1.0170 47.05 1.0298E-04 240.38\n60:40 0.0007990 18.24 51.36 1.0208 52.86 1.4313E-04 224.74\n70:30 0.0009111 17.38 56.92 1.0246 58.67 1.8048E-04 209.71\n80:20 0.0009870 16.52 62.48 1.0284 64.48 2.1410E-04 195.35\n90:10 0.0010593 15.66 68.04 1.0322 70.29 2.4957E-04 181.71\n100:0 0.0010015 14.80 73.60 1.0360 76.10 2.5450E-04 168.63\n\nδ1= Solubility parameter of solvent blend, V1= molar volume of the solvent blend, and W is calculated from Eqn. 2\n\nTable 1: Mole Fraction Solubility Of Satranidazole\n\nFigure 1: Solubility parameter versus mole fraction solubility profile. ♦ Experimental solubilities and back-calculated solubilities from Eq. 2. Highest mole fraction solubility obtained is, X2 = 2.545*10-4 when δ1 = 14.80 (Cal/cm3)0.5 in PG-water mixtures\n\nEqn. 2, differs from Eqn. 1, in that the geometric mean is not used, hence provides an accurate prediction of solubility once W is obtained. Although W presently cannot be estimated based on fundamental physicochemical properties of the solute and solvent, W may be regressed against a polynomial in δ1 of the PG-water binary solvent mixtures (fig. 2). The following quadratic, cubic, and quartic equations were obtained using the experimental solubility data for satranidazole in PG-water mixtures: Wcal = 59.67364 + 0.15209 δ1 + 0.48767 δ1 2 (n = 11, R2 = 0.99999816)--- (8), Wcal = 39.74407 + 3.36052 δ1 + 0.31765 δ12 + 0.00297 δ13 (n = 11, R2 = 0.99999892)- - -(9), Wcal = -132.17742 + 40.31464 δ1 – 2.63481 δ1 2 + 0.10690 δ1 3 - 0.00136 δ1 4 (n = 11, R2 = 0.99999971)---(10).\n\nFigure 2: Solubility parameter versus interaction energy profile. W(cal) obtained from quartic regression Eqn. 10, for satranidazole in PG-water mixtures at 25±0.4° and correlation coefficient, r2, is 0.9999 for n = 11\n\nThe W values calculated using these expressions compared favorably with the original W values computed using Eqn. 2. The solid line plotted in fig. 2 was obtained employing the quartic expression (Eqn. 10). The calculated solubility curve fits the experimental data points quite well (figs. 1 and 3), predicting the solubility of satranidazole in PG-water mixtures at most points within an error of ~3.74%, approximating the error in experimentally determined solubility values. These polynomials are used successfully for the calculation of W, at any value of solubility parameter (δ1), which was then subsequently employed to calculate mole fraction solubility of solute (X2cal) in a solvent blend using backward regression. Representative data along with validation parameters are summarized in Table 1. Wcal values are indicating the significant interaction of satranidazole and solvent molecules at the peak of solubility profile.\n\nFigure 3: Comparison of observed and calculated mole fraction solubility. Comparison of 11 observed satranidazole solubilities in PG Water systems at 25 ± 0.40 with solubilities predicted by extended Hildebrand approach. The intercept of the line is 0.0000009, and the slope is 0.997. The correlation coefficient, r2, is 0.998 for n = 11\n\nValidation of Eqn. 10 was done by comparing experimentally obtained and calculated values of mole fraction solubility by estimating residuals and percent difference (Table 2). The predictive capability of the model for satranidazole is represented in fig. 3, which indicates a very high degree of correlation coefficient (R2) 0.998 and negligible intercept equal to zero.\n\nW(obs) W(cal) X2(obs) X2(cal) logγ2/A(obs) logγ2/A(cal) Residual Percent difference\n330.263794 330.277402 3.1350E-05 3.1691E-05 16.764513 16.737296 -1.0877E-02 -1.1\n310.934967 310.905039 4.3980E-05 4.2946E-05 15.913766 15.973622 2.3510E-02 2.4\n292.190240 292.192184 5.4505E-05 5.4589E-05 15.374020 15.370132 -1.5463E-03 -0.2\n274.169268 274.190019 6.6711E-05 6.7820E-05 14.865963 14.824463 -1.6631E-02 -1.7\n256.945357 256.931866 8.5499E-05 8.4587E-05 14.242986 14.269967 1.0666E-02 1.1\n240.383226 240.433191 1.0298E-04 1.0715E-04 13.775647 13.675717 -4.0511E-02 -4.1\n224.739543 224.691600 1.4313E-04 1.3778E-04 12.950614 13.046501 3.7379E-02 3.7\n209.713663 209.686837 1.8048E-04 1.7667E-04 12.369174 12.422825 2.1086E-02 2.1\n195.350960 195.380792 2.1410E-04 2.1923E-04 11.940579 11.880916 -2.3980E-02 -2.4\n181.705894 181.717492 2.4957E-04 2.5188E-04 11.555913 11.532715 -9.2549E-03 -0.9\n168.633292 168.623108 2.5450E-04 2.5245E-04 11.505517 11.525884 8.0566E-03 0.8\n\nTable 2: Experimental And Calculated Mole Fraction Solubilities.\n\nEHSA employs a power series (quartic) equation in δ1 to back-calculate W, which reproduces the solubility of satranidazole in PG-water mixtures within the accuracy ordinarily achieved in such experimental solubility results. On the basis of validation parameters, it can be expressed that the behavior of irregular solution can be quantified more precisely using EHSA. The procedure can be explored further to predict the solubility of satranidazole in any other binary solvent mixtures. Simultaneously, this tool may become useful in optimization problems of clear solution formulations. Thus the method has potential usefulness in preformulation and formulation studies during which solubility prediction is important for drug design.\n\n## Acknowledgements\n\nAuthor wishes to express his gratitude to M/S Alkem Laboratories Limited, Baddi, India for providing gift sample of Satranidazole."
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https://techtutorialsx.com/2020/12/13/python-opencv-add-slider-to-window/ | [
"# Python OpenCV: How to add a slider to a window\n\nContents\n\n## Introduction\n\nIn this tutorial, we will learn how to add a slider to a OpenCV window where we are displaying an image, using Python.\n\nA slider might be useful in scenarios where we want to test different values for a function and we want to change those values manually in a simple user interface. For example, if we are converting an image to black and white (like covered here) and we want to test different thresholds, we can use a slider to change the threshold value.\n\n## A simple example\n\nAs usual, we will start by importing the cv2 module. This will make available the functions we need to display an image and add a slider to it.\n\n```import cv2\n```\n\nAfter that, we will read an image from the file system. To do so, we simply need to call the imread function, passing as input the path to the image. You can check in more detail how to read and display an image on this previous tutorial.\n\n```img = cv2.imread('C:/Users/N/Desktop/monument.png')\n```\n\nAfter that, we will display the image in a window, with a call to the imshow function. As first input we need to pass the name of the window and as second the image. Note that we will define a variable to hold the name of the window since we will need to also use it when creating the slider.\n\n```windowName = 'image'\n\ncv2.imshow(windowName, img)\n```\n\nTo create a slider, we need to call the createTrackbar function from the cv2 module (note that OpenCV refers to this widget as trackbar). The function receives the following parameters:\n\n• Name of the slider. We will call it “slider”;\n• Name of the window where the slider will be created. We will pass the previously created variable that holds the image window name;\n• Initial value of the slider in the scale. We will start at zero (the range for this widget always starts at 0);\n• Maximum value of the slider scale. We will use a value of 100;\n• Callback function that will be called every time the position of the slider changes. We will check its definition below, but it will be called on_change.\n```cv2.createTrackbar('slider', windowName, 0, 100, on_change)\n```\n\nWe will then wait for the user to press a key to destroy the created window and end the program.\n\n```cv2.waitKey(0)\ncv2.destroyAllWindows()\n```\n\nTo finalize, we will check the implementation of the on_change function. It will receive as input a variable with the current value of the slider. In the implementation of the function, we will just print the value received.\n\n```def on_change(value):\nprint(value)\n```\n\nThe complete code can be seen below.\n\n```import cv2\n\ndef on_change(value):\nprint(value)\n\nwindowName = 'image'\n\ncv2.imshow(windowName, img)\ncv2.createTrackbar('slider', windowName, 0, 100, on_change)\n\ncv2.waitKey(0)\ncv2.destroyAllWindows()\n```\n\nTo test the previous code, simply run it in a tool of your choice. I’ll be using PyCharm, a Python IDE.\n\nYou should get a result similar to figure 1. As can be seen, a slider was added to the image window, as expected. If you change its position, you should then see its current value getting printed to the console.\n\n## A more complex example\n\nNow that we already covered the basics in the previous section, we will create a slightly more complex application. Instead of simply printing the value of the slider to the console, we will now draw this value in the image.\n\nAll the code will be the same except for the callback function. So, we will focus our analysis on the on_change function.\n\nThe first thing we will do is creating a copy of the original image. This is needed because the function to write text changes the image. So, if we always use the original image, the numbers will overlap over each other as we are changing the slider. To avoid this, every time the callback executes, we copy the original image (unaltered) and write the slider value on the image copy.\n\nThe procedure to copy an image was covered in this post. In short, we simply need to call the copy method on our original image.\n\n```imageCopy = img.copy()\n```\n\nTo write text to an image, we need to call the putText function from the cv2 module. It receives the following parameters:\n\n• Image to write the text. We will pass our image copy;\n• Text to be drawn, as a string. We will pass the value from the slider, after converting it to a string;\n• A tuple with the coordinates (x and y) of the bottom left corner where to put the text. We will put the text on the bottom left of the image. We will use x equal to 0 and y equal to the height of the image minus 10 pixels (more details on how to get the image dimensions here);\n• The font, from the cv2 module. You can check the possible values here. We are going to use FONT_HERSHEY_SIMPLEX;\n• The scale of the text. We will use a value of 1.0 to maintain the font-specific base size;\n• A tuple with the BGR values for the color of the font. We will set it to black (B=0, G=0, R=0);\n• The thickness of the lines to draw the text. We will use a value of 4.\n```cv2.putText(imageCopy, str(val), (0, imageCopy.shape - 10), cv2.FONT_HERSHEY_SIMPLEX, 1.0, (0, 0, 0), 4)\n```\n\nNow that we have drawn the slider value in the image, we will display it in a window.\n\n```cv2.imshow(windowName, imageCopy)\n```\n\nThe whole callback function can be seen in the code snippet below.\n\n```def on_change(val):\n\nimageCopy = img.copy()\n\ncv2.putText(imageCopy, str(val), (0, imageCopy.shape - 10), cv2.FONT_HERSHEY_SIMPLEX, 1.0, (0, 0, 0), 4)\ncv2.imshow(windowName, imageCopy)\n```\n\nThe complete code can be seen below.\n\n```import cv2\n\ndef on_change(val):\n\nimageCopy = img.copy()\n\ncv2.putText(imageCopy, str(val), (0, imageCopy.shape - 10), cv2.FONT_HERSHEY_SIMPLEX, 1.0, (0, 0, 0), 4)\ncv2.imshow(windowName, imageCopy)\n\nwindowName = 'image'\n\ncv2.imshow(windowName, img)\ncv2.createTrackbar('slider', windowName, 0, 100, on_change)\n\ncv2.waitKey(0)\ncv2.destroyAllWindows()\n```\n\nAfter running the previous code, you should get a result like shown on figure 2. As can be seen, when moving the slider, the value will be drawn in the bottom left corner of the image."
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https://emacs-china.org/t/topic/4830?u=jjpandari | [
"# 复习 Lisp： Syntax and Semantics\n\n2 个赞\n\nThe simplest Lisp forms, atoms, can be divided into two categories: symbols and everything else. A symbol, evaluated as a form, is considered the name of a variable and evaluates to the current value of the variable.\n\nAll other atoms–numbers and strings are the kinds you’ve seen so far–are self-evaluating objects. This means when such an expression is passed to the notional evaluation function, it’s simply returned.\n\nThings get more interesting when we consider how lists are evaluated. All legal list forms start with a symbol, but three kinds of list forms are evaluated in three quite different ways. To determine what kind of form a given list is, the evaluator must determine whether the symbol that starts the list is the name of a function, a macro, or a special operator. If the symbol hasn’t been defined yet–as may be the case if you’re compiling code that contains references to functions that will be defined later–it’s assumed to be a function name.12 I’ll refer to the three kinds of forms as function call forms, macro forms, and special forms.\n\nThe evaluation rule for function call forms is simple: evaluate the remaining elements of the list as Lisp forms and pass the resulting values to the named function. This rule obviously places some additional syntactic constraints on a function call form: all the elements of the list after the first must themselves be well-formed Lisp forms.\n\nWhen it seems like overkill to define a new function with DEFUN, you can create an “anonymous” function using a LAMBDA expression. As discussed in Chapter 3, a LAMBDA expression looks like this:\n\n``````(lambda (parameters) body)\n``````\n\nOne way to think of LAMBDA expressions is as a special kind of function name where the name itself directly describes what the function does. This explains why you can use a LAMBDA expression in the place of a function name with #’.\n\n``````(funcall #'(lambda (x y) (+ x y)) 2 3) ==> 5\n``````\n\nYou can even use a LAMBDA expression as the “name” of a function in a function call expression. If you wanted, you could write the previous FUNCALL expression more concisely.\n\n``````((lambda (x y) (+ x y)) 2 3) ==> 5\n``````\n\nBut this is almost never done;\n\n``````((lambda () \"fff\"))\n``````\n\n`````` cl 规定 lambda 表达式就是一种 “特殊的” 函数名，\n``````\n\n`````` (setq a \"111\")\n(a 1 1)``````\n\n(+ 1 2)\n\n1 个赞\n\n``````(setq foo (list #'bar #'baz)) ;; `bar' `baz' 均为合法函数定义，且 `bar' 是无参数的函数。\n``````\n\n``````(funcall (car foo))\n``````\n\n1 个赞\n\n(symbolp (lambda () “”)) => nil\n\nlambda 返回的似乎就是一个 第一个元素是 lambda 的列表…\n\n``````Welcome to GNU CLISP 2.49.60 (2017-06-25) <http://clisp.org/>\n\nCopyright (c) Bruno Haible, Michael Stoll 1992, 1993\nCopyright (c) Bruno Haible, Marcus Daniels 1994-1997\nCopyright (c) Bruno Haible, Pierpaolo Bernardi, Sam Steingold 1998\nCopyright (c) Bruno Haible, Sam Steingold 1999-2000\nCopyright (c) Sam Steingold, Bruno Haible 2001-2010\n\nType :h and hit Enter for context help.\n\n> (defun test () \"f\")\nTEST\n> (setq a (list #'test))\n(#<FUNCTION TEST NIL (DECLARE (SYSTEM::IN-DEFUN TEST)) (BLOCK TEST \"f\")>)\n> (funcall (car a))\n\"f\"\n> (defun test () \"g\")\nTEST\n> (funcall (car a))\n\"f\"\n>\n``````\n\nelisp:\n\n``````;; This is *scratch* buffer.\n\n(defun test () \"fff\") => test\n\n(setq a (list #'test)) => (test)\n\n(funcall (car a)) => \"fff\"\n\n(defun test () \"ggg\") => test\n\n(funcall (car a)) => \"ggg\"``````\n2 个赞\n\n2 个赞\n\nEmacs 里面 `lambda` 是个宏，展开来是 `#'(lambda ...)`，后者就是对 `funtion` 这个 special form 的调用。你说的没错。\n\n``````(setq foo (lambda () 1))\n\nfoo ;; => (lambda nil 1)\n\n(car foo) ;; => lambda\n``````\n\n``````(setq lexical-binding t)\n\n(lambda () 1) ;; => (closure (t) nil 1)\n\n(car (lambda nil 1)) ;; => closure\n\n;; 最有趣的要来了\n\n(fset 'bar '(clozure (t) nil 1))\n\n(bar) ;; => 1\n\n;; 还不够刺激？\n\n(funcall '(closure (t) nil 1)) ;; => 1\n\n((closure (t) (x) x) 2) ;; => 2\n``````\n\n2 个赞\n\n``````(setq foo (byte-compile (lambda nil \"foo\")))\n;; => #[nil \"\\300\\207\" [\"foo\"] 1 \"foo\"]\n(funcall foo)\n;; => \"foo\"\n(car foo) ;; => Err\n;; 做个对比\n(listp foo) ;; => nil\n(listp (lambda nil \"foo\")) ;; => t\n``````\n\nlambda 作为函数编译以后很明显就不是列表了。\n\n2 个赞"
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https://tolstoy.newcastle.edu.au/R/help/05/12/17367.html | [
"# [R] lattice legend colors recycling sooner than expected\n\nFrom: alejandro munoz <guerinche_at_gmail.com>\nDate: Sat 10 Dec 2005 - 04:03:50 EST\n\ndear r-helpers,\n\nit seems the colors in an automatically generated lattice legend recycle after the 8th color, even when the user has set e.g. superpose.symbol\\$col to be longer than 8. the following example will illustrate what i mean:\n\nz <- data.frame(x=rep(letters[1:15], each=4), y=rnorm(60),\n\n``` groups=rep(LETTERS[1:3], 20))\n```\nlibrary(nlme)\nlibrary(lattice)\nplot(groupedData(y ~ x | groups, data=z))\n# symbol colors recycle after every 8 for plot and legend symbols;\n# e.g. a, h, and o are cyan.\n\ntrellis.par.set(superpose.symbol = list(col=rainbow(15))) plot(groupedData(y ~ x | groups, data=z))\n# each dot in the plot has a different color, but colors in legend\n# still recycle every 8 points; e.g. cyan, violet, and blue aren't in legend.\n\nin case this is an nlme issue, i examined the three plot.XXXGroupedData methods, but could not find any obvious fixes.\n\nis there a simple way of having auto.key \"know\" how many color entries it should have?\n\nalejandro\n\[email protected] mailing list"
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https://web2.0calc.com/questions/geometry_26021 | [
"+0\n\n# geometry\n\n0\n191\n1\n\nThe width, length, and height of a rectangular prism are each increased by 10%. What is the percent increase in the surface area of the prism? Express your answer to the nearest whole number.\n\nJun 8, 2021\n\n#1\n+2\n\nThe scale factor = 1.10\n\nThe % increase = (1.10)^2 = 1.21 = 21% increase",
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https://calculatorsonline.org/mix-number-to-as-a-decimal/14_12/59 | [
"# 14 12/59 as a decimal\n\nHere you will see step by step solution to convert 14 12/59 mix number fraction into decimal. 14 12/59 as a decimal is 14.20339. Please check the explanation that how to write 14 12/59 as a decimal.\n\n## Answer: 14 12/59 as a decimal is\n\n= 14.20339\n\n### How to convert 14 12/59 mix number in a decimal form?\n\nTo convert the 14 12/59 as a decimal form simply divide the numerator by denominator then add result with whole part 14, in this case 12 is called the numerator and 59 is called a denominator, 14 is called whole number and the fraction bar is just called 'divided by'.\n\n#### How to write mix fraction 14 12/59 as a decimal?\n\nFollow these easy steps to convert 14 12/59 mixed fraction into decimal-\n\n• Divide 12 [numerator] by 59 [denominator] like this:\n• = 12/59\n= 12 ÷ 59 = 0.20339\n• Now add proper fraction result with whole number - 14\n• = 14 + 0.20339 = 14.20339\n\nTherefore, the 14 12/59 mixed fraction as a decimal final answer is 14.20339. The 14 12/59 mixed fraction is simplified as much as possible."
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https://www.mathcelebrity.com/search.php?searchInput=side | [
"",
null,
"Your Search returned 554 results for side\n\nNot the results you were looking for? Suggest this term be built on our contact us page\n\n%60 of the freshman ate pizza at lunch today. If 180 freshman ate pizza, how many freshman are enro\n60% of the freshman ate pizza [URL='http://www.mathcelebrity.com/community/x-apple-data-detectors://3']at lunch today[/URL]. If 180 freshman ate pizza, how many freshman are enrolled at our school? 60% of x = 180 We write this as 0.6x = 180 Divide each side by 0.6 to isolate x. We get x = 300 freshman\n\n(4x - 20)/8 = 9y for x\n(4x - 20)/8 = 9y for x Cross multiply: 4x - 20 = 8 * 9y 4x - 20 = 72y Add 20 to each side to isolate x: 4x - 20 + 20 = 72y + 20 Cancel the 20 on the left side, we get: 4x = 72y + 20 Divide each side by 4: 4x/4 = (72y + 20)/4 Cancel the 4 on the left side: x = [B](72y + 20)/4[/B]\n\n-2 <= x +4 < 9\n-2 <= x +4 < 9 Subtract 4 from each piece: -2 - 4 <= x < 5 Simplify: [B]-6 <= x < 5 [/B] To find the interval notation, we set up our notation: [LIST] [*]The left side has a solid bracket, since we have an equal sign: [*]The right side has an open parentheses, since we have no equal sign [*][B][-6, 5)[/B] [/LIST]\n\n-3x<= -9 or 5+x<6\n-3x<= -9 or 5+x<6 Take each piece: -3x<= -9 Divide each side by -3: x>=3 Now take 5 + x < 6 5 + x < 6 Subtract 5 from each side: x < 1 Joining together the two inequalities, we have: x<1 or x>=3 Use our [URL='http://www.mathcelebrity.com/interval-notation-calculator.php?num=x%3C1orx%3E%3D3&pl=Show+Interval+Notation']interval notion calculator[/URL] to find the interval notation of this compound inequality\n\n1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers.\n1 integer is 7 times another. If the product of the 2 integers is 448, then find the integers. Let the first integer be x and the second integer be y. We have the following two equations: [LIST=1] [*]x = 7y [*]xy = 448 [/LIST] Substitute (1) into (2), we have: (7y)y = 448 7y^2 = 448 Divide each side by 7 y^2 = 64 y = -8, 8 We use 8, since 8*7 = 56, and 56*8 =448. So the answer is [B](x, y) = (8, 56)[/B]\n\n1/2a-10b=c solve for a\n1/2a-10b=c solve for a Multiply each side of the equation by 2: 2/2a - 2(10)b = 2c Simplify: a - 20b = 2c Add 20b to each side: a - 20b + 20b = 2c + 20b Cancel the 20b on the left side: [B]a = 2c + 20b [/B] You can also factor out a 2 on the left side for another version of this answer: [B]a = 2(c + 10b)[/B]\n\n1/a + 1/b = 1/2 for a\n1/a + 1/b = 1/2 for a Subtract 1/b from each side to solve this literal equation: 1/a + 1/b - 1/b = 1/2 - 1/b Cancel the 1/b on the left side, we get: 1/a = 1/2 - 1/b Rewrite the right side, using 2b as a common denominator: 1/a = (b - 2)/2b Cross multiply: a(b - 2) = 2b Divide each side by (b - 2) a = [B]2b/(b - 2)[/B]\n\n10ac-x/11=3 for a\n10ac-x/11=3 for a Add x/11 to each side of the equation to isolate a: 10ac - x/11 + x/11 = 3 + x/11 Cancelling the x/11 on the left side, we get: 10ac = 3 + x/11 Divide each side by 10c to isolate a: 10ac/10c = 3 + x/11 Cancelling the 10c on the left side, we get: a = [B]3/10c + x/110c[/B]\n\n15 cats, 10 have stripes, 7 have stripes and green eyes, how many cats have just green eyes\nLet G be green eyes and S be Stripes, and SG be Stripes and Green Eyes. [U]Set up an equation[/U] Total Cats = Green Eyes + Stripes - Green Eyes and Stripes 15 = G + 10 - 7 15 = G + 3 [U]Subtract 3 from each side:[/U] [B]G = 12[/B]\n\n15y + 13/c = m for y\n15y + 13/c = m for y Subtract 13/c from each side to isolate the y term: 15y + 13/c - 13/c = m - 13/c Cancel the 13/c on the left side and we get 15y = m - 13/c Now, divide each side by 15 to isolate y: 15y/15 = (m - 13/c)/15 Cancel the 15 on the left side, and we get: y = [B](m - 13/c)/15[/B]\n\n175 students separated into n classes is 25\n175/n = 25 25n = 175 Divide each side by 25 [B]n = 7 classes[/B]\n\n175 students separated into n classes is 25\n175 students separated into n classes is 25 [U]Divide 175 by n[/U] 175/n [U]The word [I]is[/I] means equal to, so set this expression equal to 25[/U] 175/n = 25 [U]Cross multiply[/U] 25n = 175 [U]Divide each side by 25[/U] [B]n = 7[/B]\n\n2 consecutive even integers that equal 118\nLet x be the first even integer. That means the next consecutive even integer must be x + 2. Set up our equation: x + (x + 2) = 118 Group x terms 2x + 2 = 118 Subtract 2 from each side 2x = 116 Divide each side by 2 x = 58 Which means the next consecutive even integer is 58 + 2 = 60 So our two consecutive even integers are [B]58, 60[/B] Check our work: 58 + 60 = 118\n\n2 consecutive odd integers such that their product is 15 more than 3 times their sum\n2 consecutive odd integers such that their product is 15 more than 3 times their sum. Let the first integer be n. The next odd, consecutive integer is n + 2. We are given the product is 15 more than 3 times their sum: n(n + 2) = 3(n + n + 2) + 15 Simplify each side: n^2 + 2n = 6n + 6 + 15 n^2 + 2n = 6n + 21 Subtract 6n from each side: n^2 - 4n - 21 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-4n-21%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: n = (-3, 7) If we use -3, then the next consecutive odd integer is -3 + 2 = -1. So we have [B](-3, -1)[/B] If we use 7, then the next consecutive odd integer is 7 + 2 = 9. So we have [B](7, 9)[/B]\n\n2 fair sided die are rolled. How many ways can the dice be rolled to sum exactly 6?\n2 fair sided die are rolled. How many ways can the dice be rolled to sum exactly 6? [URL='https://www.mathcelebrity.com/2dice.php?gl=1&pl=6&opdice=1&rolist=+2%2C3%2C9%2C10&dby=2%2C3%2C5&ndby=4%2C5&montect=+100']Using our 2 dice calculator[/URL], we get the following options: [LIST] [*]2,4 [*]3,3 [*]4,2 [*]1,5 [*]5,1 [/LIST] The probability of rolling a sum of 6 is [B]5/36[/B]\n\n2 numbers that add up makes 5 but multiplied makes -36\n2 numbers that add up makes 5 but multiplied makes -36 Let the first number be x and the second number be y. We're given two equations: [LIST=1] [*]x + y = 5 [*]xy = -36 [/LIST] Rearrange equation (1) by subtracting y from each side: [LIST=1] [*]x = 5 - y [*]xy = -36 [/LIST] Substitute equation (1) for x into equation (2): (5 - y)y = -36 5y - y^2 = -36 Add 36 to each side: -y^2 + 5y + 36 = 0 We have a quadratic equation. To solve this, we [URL='https://www.mathcelebrity.com/quadratic.php?num=-y%5E2%2B5y%2B36%3D0&pl=Solve+Quadratic+Equation&hintnum=0']type it in our search engine and solve[/URL] to get: y = ([B]-4, 9[/B]) We check our work for each equation: [LIST=1] [*]-4 + 9 = -5 [*]-4(9) = -36 [/LIST] They both check out\n\n2 numbers that are equal have a sum of 60\n2 numbers that are equal have a sum of 60 Let's choose 2 arbitrary variables for the 2 numbers x, y Were given 2 equations: [LIST=1] [*]x = y <-- Because we have the phrase [I]that are equal[/I] [*]x + y = 60 [/LIST] Because x = y in equation (1), we can substitute equation (1) into equation (2) for x: y + y = 60 Add like terms to get: 2y = 60 Divide each side by 2: 2y/2 = 60/2 Cancel the 2's and we get: y = [B]30 [/B] Since x = y, x = y = 30 x = [B]30[/B]\n\n20% of a number is x. What is 100% of the number? Assume x>0.\n20% of a number is x. What is 100% of the number? Assume x>0. Let the number be n. We're given: 0.2n = x <-- Since 20% = 0.2 To find n, we multiply each side of the equation by 5: 5(0.2)n = 5x n = [B]5x[/B]\n\n26 students 15 like vanilla 16 like chocolate. 3 do not like either flavour. How many like both vani\n26 students 15 like vanilla 16 like chocolate. 3 do not like either flavour. How many like both vanilla and chocolate Define our people: [LIST=1] [*]We have Vanilla Only [*]Chocolate Only [*]Both Vanilla and Chocolate [*]Neither Vanilla Nor Chocolate [*]Add up 1-4 to get our total [/LIST] Total = Vanilla Only + Chocolate Only - Vanilla and Chocolate + Neither 26 = 15 + 16 - V&C + 3 26 = 34 - V&C Subtract 34 from each side -V&C = -8 Multiply each side by -1 [B]V&C = 8[/B]\n\n2d = (a - b)/(b - c) for d\n2d = (a - b)/(b - c) for d Divide each side by 2 to isolate d: 2d/2 = (a - b)/2(b - c) Cancel the 2's on the left side, we get: d = [B](a - b)/2(b - c)[/B]\n\n2m - n/3 = 5m for n\n2m - n/3 = 5m for n Subtract 2m from each side of the equation: 2m-n/3 - 2m = 5m - 2m -n/3 = 3m Multiply each side of the equation by -3 to isolate n: -3 * -n/3 = -3 * 3m n = [B]-9m[/B]\n\n2x^2+4x < 3x+6\n2x^2+4x < 3x+6 Subtract 3x from both sides: 2x^2 + x < 6 Subtract 6 from both sides 2x^2 + x - 6 < 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=2x%5E2%2Bx-6&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x < 1.5 and x < -2 When we take the intersection of these, it's [B]x < 1.5[/B]\n\n36 PAGES AND IT 3/8CM THICK, WHAT IS THE THICKNESS OF 1 PAGE?\n36 PAGES AND IT 3/8CM THICK, WHAT IS THE THICKNESS OF 1 PAGE? Set up a proportion in pages to cm: 36 pages /3/8cm = 1 page/x cm Cross multiply: 36x = 3/8 Divide each side by 36 x = 3/(8 * 36) x = 1/(8*12) x = [B]1/96 cm[/B]\n\n3k^3 = rt for t\n3k^3 = rt for t This is a literal equation. Let's divide each side of the equation by r, to isolate t: 3k^3/r = rt/r Cancel the r's on the right side, and we get: t = [B]3k^3/r[/B]\n\n4 rectangular strips of wood, each 30 cm long and 3 cm wide, are arranged to form the outer section\n4 rectangular strips of wood, each 30 cm long and 3 cm wide, are arranged to form the outer section of a picture frame. Determine the area inside the wooden frame. Area inside forms a square, with a length of 30 - 3 - 3 = 24. We subtract 3 twice, because we account for 2 rectangular strips with a width of 3. Area of a square is side * side. So we have 24 * 24 = [B]576cm^2[/B]\n\n5 -8| -2n|=-75\nSubtract 5 from each side: -8|-2n| = -80 Divide each side by -8 |-2n| = 10 Since this is an absolute value equation, we need to setup two equations: -2n = 10 -2n = -10 Solving for the first one by dividing each side by -2, we get: n = -5 Solving for the second one by dividing each side by -2, we get: n = 5\n\n5 books and 5 bags cost \\$175. What is the cost of 2 books and 2 bags\n5 books and 5 bags cost \\$175. What is the cost of 2 books and 2 bags Let the cost of each book be b and the cost of each bag be c. We're given 5b + 5c = 175 We can factor this as: 5(b + c) = 175 Divide each side of the equation by 5, we get: (b + c) = 35 The problem asks for 2b + 2c Factor out 2: 2(b + c) we know from above that (b + c) = 35, so we substitute: 2(35) [B]70[/B]\n\n5/9v+w=z,for v\n5/9v+w=z,for v Subtract w from each side: 5/9v = z - w Multiply each side by 9/5 [B]v = 9(z - w)/5[/B]\n\n508 people are there, the daily price is \\$1.25 for kids and \\$2.00 for adults. The receipts totaled \\$\n508 people are there, the daily price is \\$1.25 for kids and \\$2.00 for adults. The receipts totaled \\$885.50. How many kids and how many adults were there? Assumptions: [LIST] [*]Let the number of adults be a [*]Let the number of kids be k [/LIST] Given with assumptions: [LIST=1] [*]a + k = 508 [*]2a + 1.25k = 885.50 (since cost = price * quantity) [/LIST] Rearrange equation (1) by subtracting c from each side to isolate a: [LIST=1] [*]a = 508 - k [*]2a + 1.25k = 885.50 [/LIST] Substitute equation (1) into equation (2): 2(508 - k) + 1.25k = 885.50 Multiply through: 1016 - 2k + 1.25k = 885.50 1016 - 0.75k = 885.50 To solve for k, we [URL='https://www.mathcelebrity.com/1unk.php?num=1016-0.75k%3D885.50&pl=Solve']type this equation into our search engine[/URL] and we get: k = [B]174[/B] Now, to solve for a, we substitute k = 174 into equation 1 above: a = 508 - 174 a = [B]334[/B]\n\n6 sided die probability to roll a odd number or a number less than 6\n6 sided die probability to roll a odd number or a number less than 6 First, we'll find the set of rolling an odd number. [URL='https://www.mathcelebrity.com/1dice.php?gl=1&opdice=1&pl=Odds&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get: Odd = {1, 3, 5} Next, we'll find the set of rolling less than a 6. [URL='https://www.mathcelebrity.com/1dice.php?gl=4&pl=6&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']From this dice calculator[/URL], we get: Less than a 6 = {1, 2, 3, 4, 5} The question asks for [B]or[/B]. Which means a Union: {1, 3, 5} U {1, 2, 3, 4, 5} = {1, 2, 3, 4, 5} This probability is [B]5/6[/B]\n\n63 is the sum of 24 and helenas age\nSet up an equation where h is Helena's age. h + 24 = 63 [URL='http://www.mathcelebrity.com/1unk.php?num=h%2B24%3D63&pl=Solve']Subtract 24 from each side[/URL] h = 39\n\n9 rulers cost the same as 11 erasers. One eraser cost 0.09 cents. What is the cost of 1 ruler\n9 rulers cost the same as 11 erasers. One eraser cost 0.09 cents. What is the cost of 1 ruler Let the cost of a ruler be r. We're given: 9r = 11(0.09) 9r = 0.99 Divide each side by 9 and we get: r = [B]0.11[/B]\n\n963 animals on a farm, 159 sheep and 406 cows and pigs. How many are pigs?\n963 animals on a farm, 159 sheep and 406 cows and pigs. How many are pigs? Set up equation to represent the total animals on the farm Total Animals = Cows + Pigs + Sheep Now plug in what is given 963 = 406 + Pigs + 159 Simplify: Pigs + 565 = 963 Subtract 565 from each side [B]Pigs = 398[/B]\n\n993 cold drinks bottles are to be placed in crates. Each crate can hold 9 bottles. How many crates w\n993 cold drinks bottles are to be placed in crates. Each crate can hold 9 bottles. How many crates would be needed and how many bottles will remain? Let c equal the number of crates 9 bottles per crate * c = 993 9c = 993 Solve for [I]c[/I] in the equation 9c = 993 [SIZE=5][B]Step 1: Divide each side of the equation by 9[/B][/SIZE] 9c /9 = 993/9 c = 110.33333333333 Since we can't have fractional crates, we round up 1 to the next full crate c = [B]111[/B]\n\na +?b +?c =?180 for b\na +?b +?c =?180 for b We have a literal equation. Subtract (a + c) from each side of the equation to isolate b: a + b + c - (a + c) = 180 - (a + c) The (a + c) cancels on the left side, so we have: [B]b = 180 - (a + c)[/B] or, distributing the negative sign: [B]b = 180 - a - c[/B]\n\nA 12 feet ladder leans against the side of a house. The bottom of the ladder is 9 feet from the side\nA 12 feet ladder leans against the side of a house. The bottom of the ladder is 9 feet from the side of the house. How high is the top of the ladder from the ground? If necessary, round your answer to the nearest tenth. We have a right triangle, where 12 is the hypotenuse. [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=9&hypinput=12&pl=Solve+Missing+Side']Using our right triangle calculator[/URL], we get: side = [B]7.9[/B]\n\na 12 sided die is rolled find the probability of rolling a number greater than 7\na 12 sided die is rolled find the probability of rolling a number greater than 7 We assume this is a fair die, not loaded. This means each side 1-12 has an equal probability of 1/12 of being rolled. The problem asks, P(Roll > 7) Greater than 7 means our sample space is {8, 9, 10, 11, 12} If each of these 5 faces have an equal probability of being rolled, then we have: P(Roll > 7) = P(Roll = 8) + P(Roll = 9) + P(Roll = 10) + P(Roll = 11) + P(Roll = 12) P(Roll > 7) = 1/12 + 1/12 + 1/12 + 1/12 + 1/12 P(Roll > 7) =[B] 5/12[/B]\n\nA 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find t\nA 12-sided die is rolled. The set of equally likely outcomes is {1,2,3,4,5,6,7,8,9,10,11,12}. Find the probability of rolling a number less than 6. We have 12 outcomes. Less than 6 means 1, 2, 3, 4, 5. Our probability P(x < 6) is: P(x < 6) = [B]5/12[/B]\n\nA 13 ft. ladder is leaning against a building 12 ft. up from the ground. How far is the base of the\nA 13 ft. ladder is leaning against a building 12 ft. up from the ground. How far is the base of the ladder from the building? This is a classic 5-12-13 pythagorean triple, where the hypotenuse is 13, and the 2 sides are 5 and 12. The building and the ground form a right triangle. [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=12&hypinput=13&pl=Solve+Missing+Side']You can see the proof here[/URL]...\n\nA 13ft ladder leans against the side of a house. The bottom of the ladder is 10ft from the side of t\nA 13ft ladder leans against the side of a house. The bottom of the ladder is 10ft from the side of the house. How high is the top of the ladder from the ground? If necessary, round your answer to the nearest tenth. We have a right triangle. Hypotenuse = 13, one leg = 10. We use our [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=10&hypinput=13&pl=Solve+Missing+Side']Pythagorean theorem Calculator to solve for the other leg[/URL]: s = [B]8.3066[/B]\n\nA 15 feet piece of string is cut into two pieces so that the longer piece is 3 feet longer than twic\nA 15 feet piece of string is cut into two pieces so that the longer piece is 3 feet longer than twice the shorter piece. If the shorter piece is x feet long, find the lengths of both pieces. If the shorter piece is x, the longer piece is 20 - x We also are given 15 - x = 2x + 3 Add x to each side: 3x + 3 = 15 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B3%3D15&pl=Solve']equation calculator[/URL], we get a shorter piece of: [B]x = 4[/B] The longer piece is: 15 - x 15 - 4 [B]11[/B]\n\nA 20 feet piece of string is cut into two pieces so that the longer piece is 5 feet longer than twic\nA 20 feet piece of string is cut into two pieces so that the longer piece is 5 feet longer than twice the shorter piece. If the shorter piece is x feet long, find the lengths of both pieces. If the shorter piece is x, the longer piece is 20 - x We also are given 20 - x = 2x + 5 Add x to each side: 3x + 5 = 20 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3x%2B5%3D20&pl=Solve']equation calculator[/URL], we get a shorter piece of: [B]x = 5 [/B] The longer piece is: 20 - x 20 - 5 [B]15[/B]\n\nA 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an ho\nA 3 hour river cruise goes 15 km upstream and then back again. The river has a current of 2 km an hour. What is the boat's speed and how long was the upstream journey? [U]Set up the relationship of still water speed and downstream speed[/U] Speed down stream = Speed in still water + speed of the current Speed down stream = x+2 Therefore: Speed upstream =x - 2 Since distance = rate * time, we rearrange to get time = Distance/rate: 15/(x+ 2) + 15 /(x- 2) = 3 Multiply each side by 1/3 and we get: 5/(x + 2) + 5/(x - 2) = 1 Using a common denominator of (x + 2)(x - 2), we get: 5(x - 2)/(x + 2)(x - 2) + 5(x + 2)/(x + 2)(x - 2) (5x - 10 + 5x + 10)/5(x - 2)/(x + 2)(x - 2) 10x = (x+2)(x-2) We multiply through on the right side to get: 10x = x^2 - 4 Subtract 10x from each side: x^2 - 10x - 4 = 0 This is a quadratic equation. To solve it, [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2-10x-4%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type it in our search engine[/URL] and we get: Speed of the boat in still water =X=5 +- sq. Root of 29 kmph We only want the positive solution: x = 5 + sqrt(29) x = 10.38 [U]Calculate time for upstream journey:[/U] Time for upstream journey = 15/(10.38 - 2) Time for upstream journey = 15/(8.38) Time for upstream journey = [B]1.79[/B] [U]Calculate time for downstream journey:[/U] Time for downstream journey = 15/(10.38 + 2) Time for downstream journey = 15/(12.38) Time for downstream journey = [B]1.21[/B]\n\nA 50-foot pole and a 70-foot pole are 30 feet apart. If you were to run a line between the tops of t\nA 50-foot pole and a 70-foot pole are 30 feet apart. If you were to run a line between the tops of the two poles, what is the minimum length of cord you would need? The difference between the 70 foot and 50 foot pole is: 70 - 50 = 20 foot height difference. So we have a right triangle, with a height of 20, base of 30. We want to know the hypotenuse. Using our [URL='https://www.mathcelebrity.com/pythag.php?side1input=20&side2input=30&hypinput=&pl=Solve+Missing+Side']Pythagorean theorem calculator to solve for hypotenuse[/URL], we get: hypotenuse = [B]36.06 feet[/B]\n\nA 6-sided die is rolled once. What is the probability of rolling a number less than 4?\nA 6-sided die is rolled once. What is the probability of rolling a number less than 4? Using our [URL='https://www.mathcelebrity.com/1dice.php?gl=4&pl=4&opdice=1&rolist=+2%2C3%2C4&dby=+2%2C3%2C5&montect=+100']one dice calculator[/URL], we get: P(N < 4) = [B]1/2[/B]\n\nA 6000 seat theater has tickets for sale at \\$24 and \\$40. How many tickets should be sold at each pri\nA 6000 seat theater has tickets for sale at \\$24 and \\$40. How many tickets should be sold at each price for a sellout performance to generate a total revenue of \\$188,800? Let x be the number of \\$24 tickets, and y be the number of \\$40 tickets. We have: [LIST=1] [*]24x + 40y = 188,800 [*]x + y = 6,000 [*]Rearrange (2) to solve for x: x = 6000 - y [*]Plug in (3) to (1): [/LIST] 24(6000 - y) + 40y = 188800 144,000 - 24y + 40y = 188,800 16y + 144,000 = 188,800 Subtract 144,000 from each side: 16y = 44,800 Divide each side by 16 y = 2,800 (\\$40 tickets) Plug this into (2) x + 2,800 = 6000 Subtract 2,800 from each side: x = 3,200 (\\$24 tickets)\n\nA 74 inch rake is Leaning against a wall. The top of the rake hits the wall 70 inches above the grou\nA 74 inch rake is Leaning against a wall. The top of the rake hits the wall 70 inches above the ground. How far is the bottom of the rake from the base of the wall? We have a right triangle. Hypotenuse is the rake length fo 74 inches. One of the legs is 70. We [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=70&hypinput=74&pl=Solve+Missing+Side']use our right triangle calculator to solve for the other leg[/URL]: [B]24 inches[/B]\n\na 9-foot rope is cut into two pieces one piece is x feet express the length of the other piece in te\na 9-foot rope is cut into two pieces one piece is x feet express the length of the other piece in terms of x Piece 1 + Piece 2 = 9 Piece 1 = x x + Piece 2 = 9 Subtracting x from each side, we get: x - x + Piece 2 = 9 - x Cancel the x's on the left side, we get: Piece 2 = [B]9 - x [/B] Check our work: x + 9 - x ? 9 9 = 9\n\na = v^2/r for r\na = v^2/r for r Start by cross multiplying to get r out of the denominator: ar = v^2 Divide each side of the equation by a to isolate r: ar/a = v^2/a Cancel the a's on the left side, and we get: r = [B]v^2/a[/B]\n\nA bird was sitting 12 meters from the base of an oak tree and flew 15 meters to reach the top of the\nA bird was sitting 12 meters from the base of an oak tree and flew 15 meters to reach the top of the tree. How tall is the tree? So we have a [U]right triangle[/U]. Hypotenuse is 15. Base is 12. We want the length of the leg. The formula for a right triangle relation of sides is a^2 + b^2 = c^2 where c is the hypotenuse and a, b are the sides Rearranging this equation to isolate a, we get a^2 = c^2 - b^2 Taking the square root of both sides, we get a = sqrt(c^2 - b^2) a = sqrt(15^2 - 12^2) a = sqrt(225 - 144) a = sqrt(81) a = [B]9 meters[/B]\n\na boat traveled 336 km downstream with the current. The trip downstream took 12 hours. write an equa\na boat traveled 336 km downstream with the current. The trip downstream took 12 hours. write an equation to describe this relationship We know the distance (d) equation in terms of rate (r) and time (t) as: d = rt We're given d = 336km and t = 12 hours, so we have: [B]336 km = 12t [/B] <-- this is our equation Divide each side by 12 to solve for t: 12t/12 = 336/12 t = [B]28 km / hour[/B]\n\nA book publishing company has fixed costs of \\$180,000 and a variable cost of \\$25 per book. The books\nA book publishing company has fixed costs of \\$180,000 and a variable cost of \\$25 per book. The books they make sell for \\$40 each. [B][U]Set up Cost Function C(b) where b is the number of books:[/U][/B] C(b) = Fixed Cost + Variable Cost x Number of Units C(b) = 180,000 + 25(b) [B]Set up Revenue Function R(b):[/B] R(b) = 40b Set them equal to each other 180,000 + 25b = 40b Subtract 25b from each side: 15b = 180,000 Divide each side by 15 [B]b = 12,000 for break even[/B]\n\nA boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find th\nA boy is 10 years older than his brother. In 4 years he will be twice as old as his brother. Find the present age of each? Let the boy's age be b and his brother's age be c. We're given two equations: [LIST=1] [*]b = c + 10 [*]b + 4 = 2(c + 4) [/LIST] Substitute equation (1) into equation (2): (c + 10) + 4 = 2(c + 4) Simplify by multiplying the right side through and grouping like terms: c + 14 = 2c + 8 [URL='https://www.mathcelebrity.com/1unk.php?num=c%2B14%3D2c%2B8&pl=Solve']Type this equation into our search engine[/URL] and we get: c = [B]6[/B] Now plug c = 6 into equation (1): b = 6 + 10 b = [B]16[/B]\n\nA car is purchased for \\$24,000 . Each year it loses 30% of its value. After how many years will t\nA car is purchased for \\$24,000 . Each year it loses 30% of its value. After how many years will the car be worth \\$7300 or less? (Use the calculator provided if necessary.) Write the smallest possible whole number answer. Set up the depreciation equation D(t) where t is the number of years in the life of the car: D(t) = 24,000/(1.3)^t The problem asks for D(t)<=7300 24,000/(1.3)^t = 7300 Cross multiply: 7300(1.3)^t = 24,000 Divide each side by 7300 1.3^t = 24000/7300 1.3^t = 3.2877 Take the natural log of both sides: LN(1.3^t) = LN(3.2877) Using the natural log identities, we have: t * LN(1.3) = 1.1902 t * 0.2624 = 1.1902 Divide each side by 0.2624 t = 4.5356 [B]Rounding this up, we have t = 5[/B]\n\nA cashier has 44 bills, all of which are \\$10 or \\$20 bills. The total value of the money is \\$730. How\nA cashier has 44 bills, all of which are \\$10 or \\$20 bills. The total value of the money is \\$730. How many of each type of bill does the cashier have? Let a be the amount of \\$10 bills and b be the amount of \\$20 bills. We're given two equations: [LIST=1] [*]a + b = 44 [*]10a + 20b = 730 [/LIST] We rearrange equation 1 in terms of a. We subtract b from each side and we get: [LIST=1] [*]a = 44 - b [*]10a + 20b = 730 [/LIST] Now we substitute equation (1) for a into equation (2): 10(44 - b) + 20b = 730 Multiply through to remove the parentheses: 440 - 10b + 20b = 730 Group like terms: 440 + 10b = 730 Now, to solve for b, we [URL='https://www.mathcelebrity.com/1unk.php?num=440%2B10b%3D730&pl=Solve']type this equation into our search engine[/URL] and we get: b = [B]29 [/B] To get a, we take b = 29 and substitute it into equation (1) above: a = 44 - 29 a = [B]15 [/B] So we have [B]15 ten-dollar bills[/B] and [B]29 twenty-dollar bills[/B]\n\nA certain group of woman has a 0.69% rate of red/green color blindness. If a woman is randomly selec\nA certain group of woman has a 0.69% rate of red/green color blindness. If a woman is randomly selected, what is the probability that she does not have red/green color blindness? 0.69% = 0.0069. There exists a statistics theorem for an event A that states: P(A) + P(A') = 1 where A' is the event not happening In this case, A is the woman having red/green color blindness. So A' is the woman [U][B][I]not[/I][/B][/U][I] having red/green color blindness[/I] So we have: 0.0069 + P(A') = 1 Subtract 0.0069 from each side, we get: P(A') = 1 - 0.0069 P(A') = [B]0.9931[/B]\n\nA certain Illness is spreading at a rate of 10% per hour. How long will it take to spread to 1,200 p\nA certain Illness is spreading at a rate of 10% per hour. How long will it take to spread to 1,200 people if 3 people initially exposes? Round to the nearest hour. Let h be the number of hours. We have the equation: 3 * (1.1)^h = 1,200 Divide each side by 3: 1.1^h = 400 [URL='https://www.mathcelebrity.com/natlog.php?num=1.1%5Eh%3D400&pl=Calculate']Type this equation into our search engine [/URL]to solve for h: h = 62.86 To the nearest hour, we round up and get [B]h = 63[/B]\n\nA certain number added to its square is 30\nLet x be the number. We have: x^2 + x = 30 Subtract 30 from each side: x^2 + x - 30 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get potential solutions of: [B]x = 5 or x = -6[/B] Check 5: 5 + 5^2 = 5 + 25 = [B]30[/B] Check -6 -6 + -6^2 = -6 + 36 = [B]30[/B]\n\nA company now has 4900 employees nationwide. It wishes to reduce the number of employees by 300 per\nA company now has 4900 employees nationwide. It wishes to reduce the number of employees by 300 per year through retirements, until its total employment is 2560. How long will this take? Figure out how many reductions are needed 4900 - 2560 = 2340 We want 300 per year for retirements, so let x equal how many years we need to get 2340 reductions. 300x = 2340 Divide each side by 300 x = 7.8 years. If we want full years, we would do 8 full years\n\nA computer screen has a diagonal dimension of 19 inches and a width of 15 inches. Approximately what\nA computer screen has a diagonal dimension of 19 inches and a width of 15 inches. Approximately what is the height of the screen? We have a right triangle, with hypotenuse of 19, and width of 15. [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=15&hypinput=19&pl=Solve+Missing+Side']Using our right triangle calculator, we get [/URL][B]height = 11.662[/B]\n\nA construction company can remove 2/3 tons of dirt from a construction site each hour. How long wil\nA construction company can remove 2/3 tons of dirt from a construction site each hour. How long will it take them to remove 30 tons of dirt from the site? Let h be the number of hours. We have the following equation: 2/3h = 30 Multiply each side by 3: 2(3)h/3 = 30 * 3 Cancel the 3 on the left side: 2h = 90 [URL='https://www.mathcelebrity.com/1unk.php?num=2h%3D90&pl=Solve']Type 2h = 90 into the search engine[/URL], we get [B]h = 45[/B].\n\nA construction company needs to remove 24 tons of dirt from a construction site. They can remove 3/8\nA construction company needs to remove 24 tons of dirt from a construction site. They can remove 3/8 ton s of dirt each hour. How long will I it take to remove the dirt? Let h be the number of hours it takes, we have: 3/8h = 24 Multiply each side by 8/3 h = 24(8)/3 24/3 = 8, so we have: h = 8(8) h = [B]64 hours[/B]\n\nA construction crew has just built a new road. They built 43.75 kilometers of road at a rate of 7 ki\nA construction crew has just built a new road. They built 43.75 kilometers of road at a rate of 7 kilometers per week. How many weeks did it take them? Let w = weeks 7 kilometers per week * w = 43.75 To solve for w, we divide each side of the equation by 7: 7w/7 = 43.75/7 Cancel the 7's, we get: w = [B]6.25 [/B]\n\nA cookie recipe uses 10 times as much flour as sugar. If the total amount of these ingredients is 8\nA cookie recipe uses 10 times as much flour as sugar. If the total amount of these ingredients is 8 1/4 cups, how much flour and how much sugar would it be? Let f be the number of cups of sugar. And let f be the number of cups of flour. We're given two equations: [LIST=1] [*]f = 10s [*]s + f = 8 & 1/4 [/LIST] Substitute (1) into (2): s + 10s = 8 & 1/4 11fs= 33/4 <-- 8 & 1/4 = 33/4 Cross multiply: 44s = 33 Divide each side by 44: s= 33/44 Divide top and bottom by 11 and we get s [B]= 3/4 or 0.75[/B] Now substitute this into (1): f = 10(33/44) [B]f = 330/44 or 7 & 22/44 or 7.5[/B]\n\nA cube has an edge that is x cm long. What is the capacity of C(x)?\nA cube has an edge that is x cm long. What is the capacity of C(x)? Capacity is another word for volume, or the amount an object will hold. Given a side x, the capacity (volume) of a cube is: C(x) = [B]x^3[/B]\n\nA cubical storage box has edges that are 2 feet 4 inches long. What is the volume of the storage box\nA cubical storage box has edges that are 2 feet 4 inches long. What is the volume of the storage box? Since 1 foot = 12 inches, we have: 2 feet 4 inches = 2(12) + 4 2 feet 4 inches = 24 + 4 2 feet 4 inches = 28 inches We type [URL='https://www.mathcelebrity.com/cube.php?num=28&pl=Side&type=side&show_All=1']cube side = 28[/URL] into our search engine to get: V = [B]21952 cubic inches[/B]\n\nA dad gave his 3 sons each the same amount of money in an envelope. He took \\$20 from one son for get\nA dad gave his 3 sons each the same amount of money in an envelope. He took \\$20 from one son for getting a D on a math test and he gave another son an extra \\$35 for doing extra chores. Combined, the sons had \\$81. Figure out how much each son had. Let x, y, and z be the money each son received. To begin, x = y = z. But then, Dad took 20 from son X and gave it to son Y. So now, x = y - 20 Next, he gave Son Z an extra \\$35 for chores So z is now y + 35 since y and z used to be equal Combined, they all have 81. x + y + z = 181 But with the changes, it is: (y - 20) + y + (y + 35) Combine like terms: 3y - 20 + 35 = 81 3y + 15 = 81 Subtract 15 from each side: 3y = 66 Divide each side by 3 to isolate y y = 22 Since x = y - 20, x = 2 Since z = y + 35, we have z = 57 Checking our work, 2 + 22 + 57 = 81.\n\nA dice has six sides. The dice is rolled once. What is the probability that a six will be the result\nA dice has six sides. The dice is rolled once. What is the probability that a six will be the result. P(6) = [B]1/6[/B]\n\nA fair six-sided die is rolled. Describe the sample space.\nA fair six-sided die is rolled. Describe the sample space. [B]{1, 2, 3, 4, 5, 6}[/B]\n\nA fake coin has heads on both sides, if the coin tossed once, what is the probability of getting a h\nA fake coin has heads on both sides, if the coin tossed once, what is the probability of getting a head? Since you always flip a head, we have: P(Head) = [B]1 or 100%[/B]\n\nA father is K years old and his son is M years younger. The sum of their ages is 53.\nA father is K years old and his son is M years younger. The sum of their ages is 53. Father's Age = K Son's Age = K - M and we know K + (K - M) = 53 Combine like terms: 2K - M = 53 Add M to each side: 2K - M + M = 53 + M Cancel the M's on the left side, we get: 2K = 53+ M Divide each side by 2: 2K/2 = (53 + M)/2 Cancel the 2's on the left side: K = [B](53 + M)/2[/B]\n\na fever is generally considered to be a body temperature greater than 100 F. You friend has a temper\na fever is generally considered to be a body temperature greater than 100 F. You friend has a temperature of 37 C. Does you friend have a fever? 37 Celsius equals 98.6 F. Since this is less than 100F, your friend does not have a fever.\n\nA first number plus twice a second number is 6. Twice the first number plus the second totals 15. Fi\nA first number plus twice a second number is 6. Twice the first number plus the second totals 15. Find the numbers. Let the first number be x. Let the second number be y. We're given two equations: [LIST=1] [*]x + 2y = 6 [*]2x + y = 15 [/LIST] Multiply the first equation by -2: [LIST=1] [*]-2x - 4y = -12 [*]2x + y = 15 [/LIST] Now add them -2x + 2x - 4y + y = -12 + 15 -3y = 3 Divide each side by -3: y = 3/-3 y =[B] -1[/B] Plug this back into equation 1: x + 2(-1) = 6 x - 2 = 6 To solve for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x-2%3D6&pl=Solve']type this equation into our search engine[/URL] and we get: x = [B]8[/B]\n\nA first number plus twice a second number is 7. Twice the first number plus the second totals 23. Fi\nA first number plus twice a second number is 7. Twice the first number plus the second totals 23. Find the numbers Let the first number be a and the second number be b. We have: [LIST=1] [*]a + 2b = 7 [*]2a + b = 23 [/LIST] Rearrange (1) into (3) (3) a = 7 - 2b Substitute (3) into (2): 2(7 - 2b) + b = 23 Multiply through: 14 - 4b + b = 23 Combine like terms: 14 - 3b = 23 Subtract 14 from each side: -3b = 9 Divide each side by -3 [B]b = -3[/B] Substitute this into (3) a = 7 - 2b a = 7 - 2(-3) a = 7 + 6 [B]a = 13[/B] [B](a, b) = (13, -3)[/B]\n\nA flower bed is to be 3 m longer than it is wide. The flower bed will an area of 108 m2 . What will\nA flower bed is to be 3 m longer than it is wide. The flower bed will an area of 108 m2 . What will its dimensions be? A flower bed has a rectangle shape, so the area is: A = lw We are given l = w + 3 Plugging in our numbers given to us, we have: 108 = w(w + 3) w^2 + 3w = 108 Subtract 108 from each side: w^2 + 3w - 108 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B3w-108%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get: w = (9, -12) Since length cannot be negative, w = 9. And l = 9 + 3 --> l = 12 So we have [B](l, w) = (12, 9)[/B] Checking our work, we have: A = (12)9 A = 108 <-- Match!\n\nA food truck sells salads for \\$6.50 each and drinks for \\$2.00 each. The food trucks revenue from sel\nA food truck sells salads for \\$6.50 each and drinks for \\$2.00 each. The food trucks revenue from selling a total of 209 salads and drinks in one day was \\$836.50. How many salads were sold that day? Let the number of drinks be d. Let the number of salads be s. We're given two equations: [LIST=1] [*]2d + 6.50s = 836.50 [*]d + s = 209 [/LIST] We can use substitution to solve this system of equations quickly. The question asks for the number of salads (s). Therefore, we want all expressions in terms of s. Rearrange Equation 2 by subtracting s from both sides: d + s - s = 209 - s Cancel the s's, we get: d = 209 - s So we have the following system of equations: [LIST=1] [*]2d + 6.50s = 836.50 [*]d = 209 - s [/LIST] Substitute equation (2) into equation (1) for d: 2(209 - s) + 6.50s = 836.50 Multiply through to remove the parentheses: 418 - 2s + 6.50s = 836.50 To solve this equation for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=418-2s%2B6.50s%3D836.50&pl=Solve']type it into our search engine and we get[/URL]: s = [B]93[/B]\n\na group of students and teachers are going on a field trip. one ninth of the group will fit on 1/3 o\na group of students and teachers are going on a field trip. one ninth of the group will fit on 1/3 of a school bus how many buses are needed to transport the entire group 1/9g = 1/3b We want to find g, so we multiply each side through by 9 g = 9/3b Simplify: g = 3b, so we need [B]3 buses[/B]\n\nA helicopter rose vertically 300 m and then flew west 400 m how far was the helicopter from it�s sta\nA helicopter rose vertically 300 m and then flew west 400 m how far was the helicopter from it�s starting point? The distance forms a right triangle. We want the distance of the hypotenuse. Using our [URL='http://www.mathcelebrity.com/pythag.php?side1input=300&side2input=400&hypinput=&pl=Solve+Missing+Side']right triangle calculator[/URL], we get a distance of [B]500[/B]. We also could use a shortcut on this problem. If you divide 300 and 400 by 100, you get 3 and 4. Since we want the hypotenuse, you get the famous 3-4-5 triangle ratio. So the answer is 5 * 100 = 500.\n\nA hexagon has a total 126 dots and a equal number of dots on each side. how many dots on each side?\nA hexagon has a total 126 dots and a equal number of dots on each side. how many dots on each side? Since it has an equal number of dots on each side, each side has: Number of dots on each side = 126 dots / 6 sides Number of dots on each side = [B]21 dots per side[/B]\n\nA home is to be built on a rectangular plot of land with a perimeter of 800 feet. If the length is 2\nA home is to be built on a rectangular plot of land with a perimeter of 800 feet. If the length is 20 feet less than 3 times the width, what are the dimensions of the rectangular plot? [U]Set up equations:[/U] (1) 2l + 2w = 800 (2) l = 3w - 20 [U]Substitute (2) into (1)[/U] 2(3w - 20) + 2w = 800 6w - 40 + 2w = 800 [U]Group the w terms[/U] 8w - 40 = 800 [U]Add 40 to each side[/U] 8w = 840 [U]Divide each side by 8[/U] [B]w = 105 [/B] [U]Substitute w = 105 into (2)[/U] l = 3(105) - 20 l = 315 - 20 [B]l = 295[/B]\n\nA house costs 3.5 times as much as the lot. Together they sold for \\$135,000. Find the cost of each\nA house costs 3.5 times as much as the lot. Together they sold for \\$135,000. Find the cost of each. Let the house cost be h, and the lot cost be l. We have the following equations: [LIST=1] [*]h = 3.5l [*]h + l = 135,000 [/LIST] Substitute (1) into (2) 3.5l + l = 135,000 Combine like terms: 4.5l = 135,000 Divide each side by 4.5 to isolate l [B]l = 30,000[/B] Substitute this back into equation (1) h = 3.5(30,000) [B]h = 105,000[/B]\n\nA jet left Nairobi and flew east at an average speed of 231 mph. A passenger plane left four hours l\nA jet left Nairobi and flew east at an average speed of 231 mph. A passenger plane left four hours later and flew in the same direction but with an average speed of 385 mph. How long did the jet fly before the passenger plane caught up? Jet distance = 231t Passenger plane distance = 385(t - 4) 385(t - 4) = 231t 385t - 1540 = 231t Subtract 231t from each side 154t = 1540 [URL='https://www.mathcelebrity.com/1unk.php?num=154t%3D1540&pl=Solve']Type 154t = 1540[/URL] into the search engine, we get [B]t = 10. [/B] Check our work: Jet distance = 231(10) = 2,310 Passenger plane distance = 385(10 - 4) = 385 * 6 = 2,310\n\nA jet plane traveling at 550 mph over takes a propeller plane traveling at 150 mph that had a 3 hour\nA jet plane traveling at 550 mph over takes a propeller plane traveling at 150 mph that had a 3 hours head start. How far from the starting point are the planes? Use the formula D = rt where [LIST] [*]D = distance [*]r = rate [*]t = time [/LIST] The plan traveling 150 mph for 3 hours: Time 1 = 150 Time 2 = 300 Time 3 = 450 Now at Time 3, the other plane starts Time 4 = 600 Time 5 = 750 Time 6 = 450 + 150t = 550t Subtract 150t 400t = 450 Divide each side by 400 t = 1.125 Plug this into either distance equation, and we get: 550(1.125) = [B]618.75 miles[/B]\n\nA ladder 25 feet long is leaning against a wall. If the base of the ladder is 7 feet from the wall,\nA ladder 25 feet long is leaning against a wall. If the base of the ladder is 7 feet from the wall, how high up the wall does the ladder reach? We have a right triangle, where the ladder is the hypotenuse, and we want the measurement of one leg. Set up the pythagorean theorem with these given items using our P[URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=7&hypinput=25&pl=Solve+Missing+Side']ythagorean Theorem Calculator[/URL]. We get Side 1 = [B]24 feet.[/B]\n\nA ladder is 25 ft long. The ladder needs to reach to a window that is 24 ft above the ground. How fa\nA ladder is 25 ft long. The ladder needs to reach to a window that is 24 ft above the ground. How far away from the building should the bottom of the ladder be placed? We have a right triangle, where the ladder is the hypotenuse, and the window side is one side. Using our right triangle and the [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=24&hypinput=25&pl=Solve+Missing+Side']pythagorean theorem calculator[/URL], we get a length of [B]7 ft [/B]for the ladder bottom from the wall.\n\nA ladder rests 2.5 m from the base of a house. If the ladder is 4 m long, how far up the side of the\nA ladder rests 2.5 m from the base of a house. If the ladder is 4 m long, how far up the side of the house will the ladder reach? We have a right triangle with the hypotenuse as 4, the one leg as 2.5 We want to solve for the other leg length. We use our [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=2.5&hypinput=4&pl=Solve+Missing+Side']right triangle solver[/URL] to get [B]3.122[/B]\n\nA local Dunkin� Donuts shop reported that its sales have increased exactly 16% per year for the last\nA local Dunkin� Donuts shop reported that its sales have increased exactly 16% per year for the last 2 years. This year�s sales were \\$80,642. What were Dunkin' Donuts' sales 2 years ago? Declare variable and convert numbers: [LIST] [*]16% = 0.16 [*]let the sales 2 years ago be s. [/LIST] s(1 + 0.16)(1 + 0.16) = 80,642 s(1.16)(1.16) = 80,642 1.3456s = 80642 Solve for [I]s[/I] in the equation 1.3456s = 80642 [SIZE=5][B]Step 1: Divide each side of the equation by 1.3456[/B][/SIZE] 1.3456s/1.3456 = 80642/1.3456 s = 59930.142687277 s = [B]59,930.14[/B]\n\nA local shop sold 499 hamburgers and cheese burgers. There were 51 fewer cheese burgers sold. How ma\nA local shop sold 499 hamburgers and cheese burgers. There were 51 fewer cheese burgers sold. How many hamburgers were sold? Let h = number of hamburgers sold and c be the number of cheeseburgers sold. We have two equations: (1) c = h - 51 (2) c + h = 499 Substitute (1) into (2) h - 51 + h = 499 Combine like terms 2h - 51 = 499 Add 51 to both sides 2h = 550 Divide each side by 2 to isolate h [B]h = 275[/B]\n\nA man is 5 years older than his wife, and the daughter age is half of the mother, and if you add the\nA man is 5 years older than his wife, and the daughter age is half of the mother, and if you add their ages is equal 100 Let the man's age be m. Let the wife's age be w. Let the daughter's age be d. We're given: [LIST=1] [*]m = w + 5 [*]d = 0.5m [*]d + m + w = 100 [/LIST] Rearrange equation 1 in terms of w my subtracting 5 from each side: [LIST=1] [*]w = m - 5 [*]d = 0.5m [*]d + m + w = 100 [/LIST] Substitute equation (1) and equation (2) into equation (3) 0.5m + m + m - 5 = 100 We [URL='https://www.mathcelebrity.com/1unk.php?num=0.5m%2Bm%2Bm-5%3D100&pl=Solve']type this equation into our search engine[/URL] to solve for m and we get: m = [B]42 [/B] Now, substitute m = 42 into equation 2 to solve for d: d = 0.5(42) d = [B]21 [/B] Now substitute m = 42 into equation 1 to solve for w: w = 42 - 5 w = [B]37 [/B] To summarize our ages: [LIST] [*]Man (m) = 42 years old [*]Daughter (d) = 21 years old [*]Wife (w) = 37 years old [/LIST]\n\nA man purchased 20 tickets for a total of \\$225. The tickets cost \\$15 for adults and \\$10 for children\nA man purchased 20 tickets for a total of \\$225. The tickets cost \\$15 for adults and \\$10 for children. What was the cost of each ticket? Declare variables: [LIST] [*]Let a be the number of adult's tickets [*]Let c be the number of children's tickets [/LIST] Cost = Price * Quantity We're given two equations: [LIST=1] [*]a + c = 20 [*]15a + 10c = 225 [/LIST] Rearrange equation (1) in terms of a: [LIST=1] [*]a = 20 - c [*]15a + 10c = 225 [/LIST] Now that I have equation (1) in terms of a, we can substitute into equation (2) for a: 15(20 - c) + 10c = 225 Solve for [I]c[/I] in the equation 15(20 - c) + 10c = 225 We first need to simplify the expression removing parentheses Simplify 15(20 - c): Distribute the 15 to each term in (20-c) 15 * 20 = (15 * 20) = 300 15 * -c = (15 * -1)c = -15c Our Total expanded term is 300-15c Our updated term to work with is 300 - 15c + 10c = 225 We first need to simplify the expression removing parentheses Our updated term to work with is 300 - 15c + 10c = 225 [SIZE=5][B]Step 1: Group the c terms on the left hand side:[/B][/SIZE] (-15 + 10)c = -5c [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] -5c + 300 = + 225 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 300 and 225. To do that, we subtract 300 from both sides -5c + 300 - 300 = 225 - 300 [SIZE=5][B]Step 4: Cancel 300 on the left side:[/B][/SIZE] -5c = -75 [SIZE=5][B]Step 5: Divide each side of the equation by -5[/B][/SIZE] -5c/-5 = -75/-5 c = [B]15[/B] Recall from equation (1) that a = 20 - c. So we substitute c = 15 into this equation to solve for a: a = 20 - 15 a = [B]5[/B]\n\nA man stands at point p, 45 metres from the base of a building that is 20 metres high. Find the angl\nA man stands at point p, 45 metres from the base of a building that is 20 metres high. Find the angle of elevation of the top of the building from the man. Draw a right triangle ABC where Side A is from the bottom of the building to the man and Side B is the bottom of the building to the top of the building. Using right triangle calculations, we want Angle A which is the angle of elevation. [URL='http://www.mathcelebrity.com/righttriangle.php?angle_a=&a=20&angle_b=&b=45&c=&pl=Calculate+Right+Triangle']Angle of Elevation[/URL] which is [B]23.9625�[/B]\n\nA man's age (a) 10 years ago is 43.\nA man's age (a) 10 years ago is 43. Years ago means we subtract [B]a - 10 = 43 [/B] If the problem asks you to solve for a, we type this equation into our math engine and we get: Solve for [I]a[/I] in the equation a - 10 = 43 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants -10 and 43. To do that, we add 10 to both sides a - 10 + 10 = 43 + 10 [SIZE=5][B]Step 2: Cancel 10 on the left side:[/B][/SIZE] a = [B]53[/B]\n\nA math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 poin\nA math test is worth 100 points and has 38 problems. Each problem is worth either 5 points or 2 points. How many problems of each point value are on the test? Let's call the 5 point questions m for multiple choice. Let's call the 2 point questions t for true-false. We have two equations: [LIST=1] [*]m + t = 38 [*]5m + 2t = 100 [/LIST] Rearrange (1) to solve for m - subtract t from each side: 3. m = 38 - t Now, substitute (3) into (2) 5(38 - t) + 2t = 100 190 - 5t + 2t = 100 Combine like terms: 190 - 3t = 100 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=190-3t%3D100&pl=Solve']equation solver[/URL], we get [B]t = 30[/B]. Plugging t = 30 into (1), we get: 30 + t = 38 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=m%2B30%3D38&pl=Solve']equation solver[/URL] again, we get [B]m = 8[/B]. Check our work for (1) 8 + 30 = 38 <-- Check Check our work for (2) 5(8) + 2(30) ? 100 40 + 60 ? 100 100 = 100 <-- Check You could also use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=m+%2B+t+%3D+38&term2=5m+%2B+2t+%3D+100&pl=Cramers+Method']simultaneous equations calculator[/URL]\n\nA medium orange has 70 calories. This is 10 calories less then 1/4 of the calories in a sugar krunch\nA medium orange has 70 calories. This is 10 calories less then 1/4 of the calories in a sugar krunchy. How many calories are in a sugar crunchy? Let s = calories in a sugar crunch. Let o = 70 be the calories in a medium orange. Set up the equation: o = 1/4s - 10 70 = 1/4s - 10 Add 10 to each side 1/4s = 80 Multiply each side by 4 [B]s = 320[/B]\n\nA Middleweight UFC fighter weighs between 170 lbs and 185 lbs.\nA Middleweight UFC fighter weighs between 170 lbs and 185 lbs. Let w be the UFC fighter's weight: We have a compound inequality. Right side includes 185 lbs. because between means includes 185lbs. Left side includes 170 lbs. because between means includes 17lb0s [B]170 <= w <= 185[/B]\n\nA motorboat travels 408 kilometers in 8 hours going upstream and 546 kilometers in 6 hours going dow\nA motorboat travels 408 kilometers in 8 hours going upstream and 546 kilometers in 6 hours going downstream. What is the rate of the boat in still water and what is the rate of the current? [U]Assumptions:[/U] [LIST] [*]B = the speed of the boat in still water. [*]S = the speed of the stream [/LIST] Relative to the bank, the speeds are: [LIST] [*]Upstream is B - S. [*]Downstream is B + S. [/LIST] [U]Use the Distance equation: Rate * Time = Distance[/U] [LIST] [*]Upstream: (B-S)6 = 258 [*]Downstream: (B+S)6 = 330 [/LIST] Simplify first by dividing each equation by 6: [LIST] [*]B - S = 43 [*]B + S = 55 [/LIST] Solve this system of equations by elimination. Add the two equations together: (B + B) + (S - S) = 43 + 55 Cancelling the S's, we get: 2B = 98 Divide each side by 2: [B]B = 49 mi/hr[/B] Substitute this into either equation and solve for S. B + S = 55 49 + S = 55 To solve this, we [URL='https://www.mathcelebrity.com/1unk.php?num=49%2Bs%3D55&pl=Solve']type it in our search engine[/URL] and we get: S = [B]6 mi/hr[/B]\n\nA movie theater has a seating capacity of 143. The theater charges \\$5.00 for children, \\$7.00 for stu\nA movie theater has a seating capacity of 143. The theater charges \\$5.00 for children, \\$7.00 for students, and \\$12.00 of adults. There are half as many adults as there are children. If the total ticket sales was \\$ 1030, How many children, students, and adults attended? Let c be the number of children's tickets, s be the number of student's tickets, and a be the number of adult's tickets. We have 3 equations: [LIST=1] [*]a + c + s = 143 [*]a = 0.5c [*]12a + 5c + 7s =1030 [/LIST] Substitute (2) into (1) 0.5c + c + s = 143 1.5c + s = 143 Subtract 1.5c from each side 4. s = 143 - 1.5c Now, take (4) and (2), and plug it into (3) 12(0.5c) + 5c + 7(143 - 1.5c) = 1030 6c + 5c + 1001 - 10.5c = 1030 Combine like terms: 0.5c + 1001 = 1030 Use our [URL='http://www.mathcelebrity.com/1unk.php?num=0.5c%2B1001%3D1030&pl=Solve']equation calculator[/URL] to get [B]c = 58[/B]. Plug this back into (2) a = 0.5(58) [B]a = 29 [/B] Now take the a and c values, and plug it into (1) 29 + 58 + s = 143 s + 87 = 143 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=s%2B87%3D143&pl=Solve']equation calculator[/URL] again, we get [B]s = 56[/B]. To summarize, we have: [LIST] [*]29 adults [*]58 children [*]56 students [/LIST]\n\nA national political party has a budget of \\$30,000,000 to spend on the inauguration of the new presi\nA national political party has a budget of \\$30,000,000 to spend on the inauguration of the new president. 16% of the costs will be paid to personnel, 12% of the costs will go toward food, and 10% will go to decorations. How much money will go for personnel, food, and decorations? [LIST] [*]Personnel Costs = 0.16 * 30,000,000 = \\$4,800,000 [*]Food Costs = 0.12 * 30,000,000 = \\$3,600,000 [*]Decoration Costs = 0.10 * 30,000,000 = \\$3,000,000 [/LIST]\n\nA new company president is said to have caused the company \"to do a 180.\" Before the new president,\nA new company president is said to have caused the company \"to do a 180.\" Before the new president, the company was losing money. What is the company most likely doing under the new president? A 180 is a completely different direction. Since 180 degrees means the other way, a half-circle, a switch in direction. This means if the company was losing money, after doing a \"180\", they're making money.\n\nA non-profit organization is having a couple�s banquet for a fundraiser. The banquet hall will only\nA non-profit organization is having a couple�s banquet for a fundraiser. The banquet hall will only hold 250 people. The President, Vice-President, two volunteers, and a guest speaker will be working the event. How many couples will be able to attend the banquet? We subtract the 5 people working the event to get: 250 - 5 = 245 A couple is 2 people, so we have 245/2 = 122.5 We round down to [B]122 couples[/B].\n\nA number multiplied by 6 and divided by 5 give four more than a number?\nA number multiplied by 6 and divided by 5 give four more than a number? A number is represented by an arbitrary variable, let's call it x. Multiply by 6: 6x Divide by 5 6x/5 The word \"gives\" means equals, so we set this equal to 4 more than a number, which is x + 4. 6x/5 = x + 4 Now, multiply each side of the equation by 5, to eliminate the fraction on the left hand side: 6x(5)/5 = 5(x + 4) The 5's cancel on the left side, giving us: 6x = 5x + 20 Subtract 5x from each side [B]x = 20[/B] Check our work from our original equation: 6x/5 = x + 4 6(20)/5 ? 20 + 4 120/5 ?24 24 = 24 <-- Yes, we verified our answer\n\nA parallelogram has a perimeter of 48 millimeters. Two of the sides are each 20 millimeters long. Wh\nA parallelogram has a perimeter of 48 millimeters. Two of the sides are each 20 millimeters long. What is the length of each of the other two sides? 2 sides * 20 mm each is 40 mm subtract this from the perimeter of 48: 48 - 40 = 8 Since the remaining two sides equal each other, their length is: 8/2 = [B]4mm[/B]\n\nA parallelogram has a perimeter of 54 centimeters. Two of the sides are each 17 centimeters long. Wh\nA parallelogram has a perimeter of 54 centimeters. Two of the sides are each 17 centimeters long. What is the length of each of the other two sides? A parallelogram is a rectangle bent on it's side. So we have the perimeter formula P below: P = 2l + 2w We're given w = 17 and P = 54. So we plug this into the formula for perimeter: 2l + 2(17) = 54 2l + 34 = 54 Using our [URL='https://www.mathcelebrity.com/1unk.php?num=2l%2B34%3D54&pl=Solve']equation calculator[/URL], we get [B]l = 10[/B].\n\nA parking meter contains 27.05 in quarters and dimes. All together there are 146 coins. How many of\nA parking meter contains 27.05 in quarters and dimes. All together there are 146 coins. How many of each coin are there? Let d = the number of dimes and q = the number of quarters. We have two equations: (1) d + q = 146 (2) 0.1d + 0.25q = 27.05 Rearrange (1) into (3) solving for d (3) d = 146 - q Substitute (3) into (2) 0.1(146 - q) + 0.25q = 27.05 14.6 - 0.1q + 0.25q = 27.05 Combine q's 0.15q + 14.6 = 27.05 Subtract 14.6 from each side 0.15q = 12.45 Divide each side by 0.15 [B]q = 83[/B] Plugging that into (3), we have: d = 146 - 83 [B]d = 63[/B]\n\nA passenger train left station A at 6:00 P.M. Moving with the average speed 45 mph, it arrived at st\nA passenger train left station A at 6:00 P.M. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A transit train left from station A 1 hour later than the passenger train, but it arrived at the station B at the same time with the passenger train. What was the average speed of the transit train? [U]Passenger Train[/U] [LIST] [*]45 miles per hour and it got there in 4 hours. [/LIST] Using our formula D = rt where: [LIST] [*]D = Distance [*]r = rate [*]t = time [/LIST] [LIST] [*]D = rt [*]D = 45(4) [*]D = 180 miles from Station A to Station B [/LIST] Transit Train [LIST] [*]It has to go the same distance, 180 miles, so D = 180 [*]It made it there in 3 hours. This is r [*]We want to solve for t [/LIST] D = rt 180 = 3r Divide each side by 3 [B]r = 60 miles per hour[/B]\n\nA penny has a diameter of 19 millimeters. What is the radius of the penny.\nA penny has a diameter of 19 millimeters. What is the radius of the penny. D = 2r To solve for r, we divide each side by 2: r = D/2 Plugging in D = 19, we get: r = [B]19/2 or 9.5[/B]\n\nA person invested 30,000 in stocks and bonds. Her investment in bonds is 2000 more than 1-third her\nA person invested 30,000 in stocks and bonds. Her investment in bonds is 2000 more than 1-third her investments in stocks. How much did she invest in stocks? How much did she invest in bonds? Let the stock investment be s, and the bond investment be b. We're given: [LIST=1] [*]b + s = 30000 [*]b = 1/3s + 2000 [/LIST] Plug in (2) to (1): 1/3s + 2000 + s = 30000 Group like terms: (1/3 + 1)s + 2000 = 30000 Since 1 = 3/3, we have: 4/3s + 2000 = 30000 Subtract 2000 from each side: 4/3s + 2000 - 2000 = 30000 - 2000 Cancel the 2000's on the left side, we get: 4/3s = 28000 [URL='https://www.mathcelebrity.com/1unk.php?num=4%2F3s%3D28000&pl=Solve']Typing this equation into our calculator[/URL], we get: s = [B]21,000[/B]\n\nA pet supply chain called pet city has 15 hamsters and 12 gerbils for sale at its seaside location.\nA pet supply chain called pet city has 15 hamsters and 12 gerbils for sale at its seaside location. At its livingston location there are 19 hamsters and 10 gerbils. Which location has a lower ratio of hamsters to gerbils? Seaside ratio 15/12 = 1.25 Livingston ratio 19/10 = 1.9 Since 1.25 < 1.9, Seaside has the lower ratio of hamsters to gerbils\n\nA pile of coins, consisting of quarters and half dollars, is worth 11.75. If there are 2 more quarte\nA pile of coins, consisting of quarters and half dollars, is worth 11.75. If there are 2 more quarters than half dollars, how many of each are there? Let h be the number of half-dollars and q be the number of quarters. Set up two equations: (1) q = h + 2 (2) 0.25q + 0.5h = 11.75 [U]Substitute (1) into (2)[/U] 0.25(h + 2) + 0.5h = 11.75 0.25h + 0.5 + 0.5h = 11.75 [U]Group h terms[/U] 0.75h + 0.5 = 11.75 [U]Subtract 0.5 from each side[/U] 0.75h = 11.25 [U]Divide each side by h[/U] [B]h = 15[/B] [U]Substitute h = 15 into (1)[/U] q = 15 + 2 [B]q = 17[/B]\n\nA population grows at 6% per year. How many years does it take to triple in size?\nA population grows at 6% per year. How many years does it take to triple in size? With a starting population of P, and triple in size means 3 times the original, we want to know t for: P(1.06)^t = 3P Divide each side by P, and we have: 1.06^t = 3 Typing this equation into our search engine to solve for t, we get: t = [B]18.85 years[/B] Note: if you need an integer answer, we round up to 19 years\n\nA private jet flies the same distance in 4 hours that a commercial jet flies in 2 hours. If the spee\nA private jet flies the same distance in 4 hours that a commercial jet flies in 2 hours. If the speed of the commercial jet was 154 mph less than 3 times the speed of the private jet, find the speed of each jet. Let p = private jet speed and c = commercial jet speed. We have two equations: (1) c = 3p - 154 (2) 4p =2c Plug (1) into (2): 4p = 2(3p - 154) 4p = 6p - 308 Subtract 4p from each side: 2p - 308 = 0 Add 308 to each side: 2p = 308 Divide each side by 2: [B]p = 154[/B] Substitute this into (1) c = 3(154) - 154 c = 462 - 154 [B]c = 308[/B]\n\nA professor assumed there was a correlation between the amount of hours people were expose to sunlig\nA professor assumed there was a correlation between the amount of hours people were expose to sunlight and their blood vitamin D level. The null hypothesis was that the population correlation was__ a. Positive 1.0 b. Negative 1.0 c. Zero d. Positive 0.50 [B]c. Zero[/B] Reason: Since the professor wanted to assume a correlation (either positive = 1.0 or negative = -1.0), then we take the other side of that assumption for our null hypothesis and say that there is no correlation (Zero)\n\nA quarter of the learners in a class have blond hair and two thirds have brown hair. The rest of the\nA quarter of the learners in a class have blond hair and two thirds have brown hair. The rest of the learners in the class have black hair. How many learners in the class if 9 of them have blonde hair? Total learners = Blond + Brown + Black Total Learners = 1/4 + 2/3 + Black Total Learners will be 1, the sum of all fractions 1/4 + 2/3 + Black = 1 Using common denominators of 12, we have: 3/12 + 8/12 + Black = 12/12 11/12 + Black = 12/12 Subtract 11/12 from each side: Black = 1/12 Let t be the total number of people in class. We are given for blondes: 1/4t = 9 Multiply each side by 4 [B]t = 36[/B] Brown Hair 2/3(36) = 24 Black Hair 1/12(36) = 3\n\nA realtor makes an annual salary of \\$25000 plus a 3% commission on sales. If a realtor's salary is \\$\nA realtor makes an annual salary of \\$25000 plus a 3% commission on sales. If a realtor's salary is \\$67000, what was the amount of her sales? Total post-salary pay = \\$67,000 - \\$25,000 = \\$42,000 Let Sales be s. So 0.03s = \\$42,000 Divide each side by 0.03 s = \\$1,400,000\n\na recipe of 20 bread rolls requires 5 tablespoons of butter. How many tablespoons of butter are need\na recipe of 20 bread rolls requires 5 tablespoons of butter. How many tablespoons of butter are needed for 30 bread rolls? Set up a proportion of bread rolls per tablespoons of butter where t is the amount of tablespoons of butter needed for 30 bread rolls: 20/5 = 30/t Cross multiply our proportion: Numerator 1 * Denominator 2 = Denominator 1 * Numerator 2 20t = 30 * 5 20t = 150 Divide each side of the equation by 20: 20t/20 = 150/20 Cancel the 20's on the left side and we get: t = [B]7.5[/B]\n\na rectangle has an area of 238 cm 2 and a perimeter of 62 cm. What are its dimensions?\na rectangle has an area of 238 cm 2 and a perimeter of 62 cm. What are its dimensions? We know the rectangle has the following formulas: Area = lw Perimeter = 2l + 2w Given an area of 238 and a perimeter of 62, we have: [LIST=1] [*]lw = 238 [*]2(l + w) = 62 [/LIST] Divide each side of (1) by w: l = 238/w Substitute this into (2): 2(238/w + w) = 62 Divide each side by 2: 238/w + w = 31 Multiply each side by w: 238w/w + w^2 = 31w Simplify: 238 + w^2 = 31w Subtract 31w from each side: w^2 - 31w + 238 = 0 We have a quadratic. So we run this through our [URL='https://www.mathcelebrity.com/quadratic.php?num=w%5E2-31w%2B238%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL] and we get: w = (14, 17) We take the lower amount as our width and the higher amount as our length: [B]w = 14 l = 17 [/B] Check our work for Area: 14(17) = 238 <-- Check Check our work for Perimeter: 2(17 + 14) ? 62 2(31) ? 62 62 = 62 <-- Check\n\nA rectangular football pitch has its length equal to twice its width and a perimeter of 360m. Find i\nA rectangular football pitch has its length equal to twice its width and a perimeter of 360m. Find its length and width. The area of a rectangle (A) is: A = lw --> where l is the length and w is the width We're given l = 2w, so we substitute this into the Area equation: A = (2w)w A = 2w^2 We're given the area of the pitch is 360, so we set: 2w^2 = 360 We [URL='https://www.mathcelebrity.com/1unk.php?num=2w%5E2%3D360&pl=Solve']type this equation into our search engine[/URL], follow the links, and get: w = [B]6*sqrt(5) [/B] Now we take this, and substitute it into this equation: 6*sqrt(5)l = 360 Dividing each side by 6*sqrt(5), we get: l = [B]60/sqrt(5)[/B]\n\nA rectangular garden is 5 ft longer than it is wide. Its area is 546 ft2. What are its dimensions?\nA rectangular garden is 5 ft longer than it is wide. Its area is 546 ft2. What are its dimensions? [LIST=1] [*]Area of a rectangle is lw. lw = 546ft^2 [*]We know that l = w + 5. [/LIST] Substitute (2) into (1) (w + 5)w = 546 w^2 + 5w = 546 Subtract 546 from each side w^2 + 5w - 546 = 0 Using the positive root in our [URL='http://www.mathcelebrity.com/quadratic.php?num=w%5E2%2B5w-546%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get [B]w = 21[/B]. This means l = 21 + 5. [B]l = 26[/B]\n\nA retired couple invested \\$8000 in bonds. At the end of one year, they received an interest payment\nA retired couple invested \\$8000 in bonds. At the end of one year, they received an interest payment of \\$584. What was the simple interest rate of the bonds? For simple interest, we have: Balance * interest rate = Interest payment 8000i = 584 Divide each side of the equation by 8000 to isolate i: 8000i/8000 = 584/8000 Cancelling the 8000's on the left side, we get: i = 0.073 Most times, interest rates are expressed as a percentage. Percentage interest = Decimal interest * 100% Percentage interest = 0.073 * 100% Multiplying by 100 is the same as moving the decimal point two places right: Percentage interest = [B]7.3%[/B]\n\nA Salesperson receives a weekly salary of \\$100 plus a 5.5% commission on sales. Her salary last week\nA Salesperson receives a weekly salary of \\$100 plus a 5.5% commission on sales. Her salary last week was \\$1090. What were her sales that week? \\$1,090 - 100 = \\$990. This is her commission. Let s = Sales. So 0.055s = \\$990 Divide each side by 0.055. s = \\$18,000\n\nA skier is trying to decide whether or not to buy a season ski pass. A daily pass costs \\$75. A seas\nA skier is trying to decide whether or not to buy a season ski pass. A daily pass costs \\$75. A season ski pass costs \\$350. The skier would have to rent skis with either pass for \\$20 per day. How many days would the skier have to go skiing in order to make the season pass less expensive than the daily passes? Let d be the number of days: Daily Plan cost: 75d + 20d = 95d Season Pass: 350 + 20d We want to find d such that 350 + 20d < 95d Subtract 20d from each side 75d > 350 Divide each side by 75 d > 4.66667 [B]d = 5[/B]\n\na son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age.\na son is 1/4 of his fathers age. the difference in their ages is 30. what is the fathers age. Declare variables: [LIST] [*]Let f be the father's age [*]Let s be the son's age [/LIST] We're given two equations: [LIST=1] [*]s = f/4 [*]f - s = 30. [I]The reason why we subtract s from f is the father is older[/I] [/LIST] Using substitution, we substitute equaiton (1) into equation (2) for s: f - f/4 = 30 To remove the denominator/fraction, we multiply both sides of the equation by 4: 4f - 4f/4 = 30 *4 4f - f = 120 3f = 120 To solve for f, we divide each side of the equation by 3: 3f/3 = 120/3 Cancel the 3's on the left side and we get: f = [B]40[/B]\n\nA spherical water tank holds 11,500ft^3 of water. What is the diameter?\nA spherical water tank holds 11,500ft^3 of water. What is the diameter? The tank holding amount is volume. And the volume of a sphere is: V = (4pir^3)/3 We know that radius is 1/2 of diameter: r =d/2 So we rewrite our volume function: V = 4/3(pi(d/2)^3) We're given V = 11,500 so we have: 4/3(pi(d/2)^3) = 11500 Multiply each side by 3/4 4/3(3/4)(pi(d/2)^3) = 11,500*3/4 Simplify: pi(d/2)^3 = 8625 Since pi = 3.1415926359, we divide each side by pi: (d/2)^3 = 8625/3.1415926359 (d/2)^3 = 2745.42 Take the cube root of each side: d/2 = 14.0224 Multiply through by 2: [B]d = 28.005[/B]\n\nA square has a perimeter of 24 inches. What is the area of the square?\nA square has a perimeter of 24 inches. What is the area of the square? Perimeter of a square = 4s where s = the length of a side. Therefore, we have: 4s = P 4s = 24 Using our equation solver, [URL='https://www.mathcelebrity.com/1unk.php?num=4s%3D24&pl=Solve']we type in 4s = 24[/URL] and get: s = 6 The problems asks for area of a square. It's given by A = s^2 Plugging in s = 6, we get: A = 6^2 A = 6 * 6 A = [B]36 [/B] Now if you want a shortcut in the future, type in the shape and measurement you know. Such as: [I][URL='https://www.mathcelebrity.com/square.php?num=24&pl=Perimeter&type=perimeter&show_All=1']square perimeter = 24[/URL][/I] From the link, you'll learn every other measurement about the square.\n\nA square of an integer is the integer. Find the integer.\nA square of an integer is the integer. Find the integer. Let the integer be n. The square means we raise n to the power of 2, so we have: n^2 = n Subtract n from each side: n^2 - n = n - n n^2 - n = 0 Factoring this, we get: n(n - 1) = 0 So n is either [B]0 or 1[/B].\n\na stone mason builds 7 houses in 3 days. How many days does it take to build 11 houses?\na stone mason builds 7 houses in 3 days. How many days does it take to build 11 houses? The build rate of houses per days is proportional. Set up a proportion of [I]houses to days[/I] where d is the number of days it takes to build 11 houses: 7/3 = 11/d Cross multiply: Numerator 1 * Denominator 2 = Denominator 1 * Numerator 2 7d = 11 * 3 7d = 33 Divide each side of the equation by 7: 7d/7 = 33/7 d = [B]4.7142857142857[/B]\n\nA store owner bought 240 cartons of eggs. The owner sold 5/8 of the eggs and set aside 5 cartons. Ho\nA store owner bought 240 cartons of eggs. The owner sold 5/8 of the eggs and set aside 5 cartons. How many cartons of eggs did the owner have left to sale? If the owner sold 5/8 of the eggs, they have 1 - 5/8 left. 1 = 8/8, so we have 8/8 - 5/8 = 3/8 left 3/8 (240 cartons) = 90 cartons remaining The owner set aside 5 cartons. We're left with 90 cartons - 5 cartons = [B]85 cartons[/B]\n\nA sum of money doubles in 20 years on simple interest. It will get triple at the same rate in: a.\nA sum of money doubles in 20 years on simple interest. It will get triple at the same rate in: a. 40 years b. 50 years c. 30 years d. 60 years e. 80 years Simple interest formula if we start with 1 dollar and double to 2 dollars: 1(1 + i(20)) = 2 1 + 20i = 2 Subtract 1 from each side: 20i = 1 Divide each side by 20 i = 0.05 Now setup the same simple interest equation, but instead of 2, we use 3: 1(1 + 0.05(t)) = 3 1 + 0.05t = 3 Subtract 1 from each side: 0.05t = 2 Divide each side by 0.05 [B]t = 40 years[/B]\n\nA super deadly strain of bacteria is causing the zombie population to double every day. Currently, t\nA super deadly strain of bacteria is causing the zombie population to double every day. Currently, there are 25 zombies. After how many days will there be over 25,000 zombies? We set up our exponential function where n is the number of days after today: Z(n) = 25 * 2^n We want to know n where Z(n) = 25,000. 25 * 2^n = 25,000 Divide each side of the equation by 25, to isolate 2^n: 25 * 2^n / 25 = 25,000 / 25 The 25's cancel on the left side, so we have: 2^n = 1,000 Take the natural log of each side to isolate n: Ln(2^n) = Ln(1000) There exists a logarithmic identity which states: Ln(a^n) = n * Ln(a). In this case, a = 2, so we have: n * Ln(2) = Ln(1,000) 0.69315n = 6.9077 [URL='https://www.mathcelebrity.com/1unk.php?num=0.69315n%3D6.9077&pl=Solve']Type this equation into our search engine[/URL], we get: [B]n = 9.9657 days ~ 10 days[/B]\n\nA taxi charges a flat rate of \\$1.50 with an additional charge of \\$0.80 per mile. Samantha wants to s\nA taxi charges a flat rate of \\$1.50 with an additional charge of \\$0.80 per mile. Samantha wants to spend less than \\$12 on a ride. Which inequality can be used to find the distance Samantha can travel? [LIST] [*]Each ride will cost 1.50 + 0.8x where x is the number of miles per trip. [*]This expression must be less than 12. [/LIST] [U]Setup the inequality:[/U] 1.5 + 0.8x < 12 [U]Subtracting 1.5 from each side of the inequality[/U] 0.8x < 10.5 [U]Simplifying even more by dividing each side of the inequality by 0.8, we have:[/U] [B]x < 13.125[/B]\n\nA taxi charges a flat rate of 1.75, plus an additional 0.65 per mile. If Erica has at most 10 to spe\nA taxi charges a flat rate of 1.75, plus an additional 0.65 per mile. If Erica has at most 10 to spend on the cab ride, how far could she travel? Setup an equation where x is the number of miles traveled: 0.65x + 1.75 = 10 Subtract 1.75 from each side: 0.65x = 8.25 Divide each side by 0.65 [B]x = 12.69 miles[/B] If we do full miles, we round down to 12.\n\na textbook store sold a combined total of 296 sociology and history text books in a week. the number\na textbook store sold a combined total of 296 sociology and history text books in a week. the number of history textbooks sold was 42 less than the number of sociology textbooks sold. how many text books of each type were sold? Let h = history book and s = sociology books. We have 2 equations: (1) h = s - 42 (2) h + s = 296 Substitute (1) to (2) s - 42 + s = 296 Combine variables 2s - 42 = 296 Add 42 to each side 2s = 338 Divide each side by 2 s = 169 So h = 169 - 42 = 127\n\nA tow truck charges a service fee of \\$50 and an additional fee of \\$1.75 per mile. What distance was\nA tow truck charges a service fee of \\$50 and an additional fee of \\$1.75 per mile. What distance was Marcos car towed if he received a bill for \\$71 Set up a cost equation C(m) where m is the number of miles: C(m) = Cost per mile * m + Service Fee Plugging in the service fee of 50 and cost per mile of 1.75, we get: C(m) = 1.75m + 50 The question asks for what m is C(m) = 71. So we set C(m) = 71 and solve for m: 1.75m + 50 = 71 Solve for [I]m[/I] in the equation 1.75m + 50 = 71 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants 50 and 71. To do that, we subtract 50 from both sides 1.75m + 50 - 50 = 71 - 50 [SIZE=5][B]Step 2: Cancel 50 on the left side:[/B][/SIZE] 1.75m = 21 [SIZE=5][B]Step 3: Divide each side of the equation by 1.75[/B][/SIZE] 1.75m/1.75 = 21/1.75 m = [B]12[/B]\n\nA traveler is walking on a moving walkway in an airport. the traveler must walk back on the walkway\nA traveler is walking on a moving walkway in an airport. the traveler must walk back on the walkway to get a bag he forgot. the traveler's ground speed is 2 ft/s against the walkway and 6 ft/s with the walkway. what is the traveler's speed off the walkway? What is the speed of the moving walkway. We have two equations, where w is the speed of the walkway and t is the speed of the traveler. [LIST=1] [*]t - w = 2 [*]t + w = 6 [*]Rearrange (1) to solve for t: t = w + 2 [/LIST] Now plug (3) into (2) (w + 2) + w = 6 Combine like terms: 2w + 2 = 6 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2w%2B2%3D6&pl=Solve']equation solver[/URL], we get [B]w = 2[/B] Plug this into (1) t - 2 = 2 Add 2 to each side [B]t = 4[/B]\n\nA trench is 40 feet long and the trencher goes 2 feet per minute. How long does it take to dig the\nA trench is 40 feet long and the trencher goes 2 feet per minute. How long does it take to dig the trench? 2 feet per minute * x minutes = 40 feet Divide each side by 2 [B]x = 20 minutes[/B]\n\na triangle has side lengths of 12,16, and 20 centimeters. is it a right triangle?\na triangle has side lengths of 12,16, and 20 centimeters. is it a right triangle? First, we see if we can simplify. So we [URL='https://www.mathcelebrity.com/gcflcm.php?num1=12&num2=16&num3=20&pl=GCF']type GCF(12,16,20) [/URL]and we get 4. We divide the 3 side lengths by 4: 12/4 = 3 16/4 = 4 20/4 = 5 And lo and behold, we get a Pythagorean Triple of 3, 4, 5. So [B]yes, this is a right triangle[/B].\n\nA vendor sells h hot dogs and s sodas. If a hot dog costs twice as much as a soda, and if the vendor\nA vendor sells h hot dogs and s sodas. If a hot dog costs twice as much as a soda, and if the vendor takes in a total of d dollars, how many cents does a soda cost? Let the cost of the soda be p. So the cost of a hot dog is 2p. The total cost of hot dogs: 2hp The total cost of sodas: ps The total cost of both equals d. So we set the total cost of hots dogs plus sodas equal to d: 2hp + ps = d We want to know the cost of a soda (p). So we have a literal equation. We factor out p from the left side: p(2h + s) = d Divide each side of the equation by (2h + s) p(2h + s)/(2h + s) = d/(2h + s) Cancel the (2h + s) on the left side, we get: p = [B]d/(2h + s[/B])\n\nA woman walked for 5 hours, first along a level road, then up a hill, and then she turned around and\nA woman walked for 5 hours, first along a level road, then up a hill, and then she turned around and walked back to the starting point along the same path. She walks 4mph on level ground, 3 mph uphill, and 6 mph downhill. Find the distance she walked. Hint: Think about d = rt, which means that t = d/r. Think about each section of her walk, what is the distance and the rate. You know that the total time is 5 hours, so you know the sum of the times from each section must be 5. Let Level distance = L and hill distance = H. Add the times it took for each section of the walk: L/4 + H /3 + H/6 + L/4 = 5 The LCD of this is 12 from our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=4&num2=3&num3=6&pl=LCM']LCD Calculator[/URL] [U]Multiply each side through by our LCD of 12[/U] 3L + 4H + 2H + 3L = 60 [U]Combine like terms:[/U] 6L + 6H = 60 [U]Divide each side by 3:[/U] 2L + 2H = 20 The woman walked [B]20 miles[/B]\n\na writer can write a novel at a rate of 3 pages per 5 hour work. if he wants to finish the novel in\na writer can write a novel at a rate of 3 pages per 5 hour work. if he wants to finish the novel in x number of pages, determine a function model that will represent the accumulated writing hours to finish his novel if 3 pages = 5 hours, then we divide each side by 3 to get: 1 page = 5/3 hours per page So x pages takes: 5x/3 hours Our function for number of pages x is: [B]f(x) = 5x/3[/B]\n\nA young dad, who was a star football player in college, set up a miniature football field for his fi\nA young dad, who was a star football player in college, set up a miniature football field for his five-year-old young daughter, who was already displaying an unusual talent for place-kicking. At each end of the mini-field, he set up goal posts so she could practice kicking extra points and field goals. He was very careful to ensure the goalposts were each straight up and down and that the crossbars were level. On each set, the crossbar was six feet long, and a string from the top of each goalpost to the midpoint between them on the ground measured five feet. How tall were the goalposts? How do you know this to be true? The center of each crossbar is 3 feet from each goalpost. We get this by taking half of 6, since midpoint means halfway. Imagine a third post midway between the two goal posts. It has the same height as the two goalposts. From the center post, the string from the top of a goalpost to the base of the center post, and half the crossbar form and right triangle with hypotenuse 5 feet and one leg 3 feet. [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=3&hypinput=5&pl=Solve+Missing+Side']Using the Pythagorean Theorem[/URL], the other leg -- the height of each post -- is 4 feet.\n\nA+B+D=255 B+15=A D+12=B A=\nA+B+D=255 B+15=A D+12=B A= [LIST=1] [*]A + B + D = 255 [*]B + 15 = A [*]D + 12 = B [*]A = ? [*]Rearrange (3) to solve for D by subtracting 12 from each side: D = B - 12 [/LIST] Substitute (2) and (5) into 1 (B + 15) + B + (B - 12) = 255 Combine like terms: 3B + 3 = 255 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=3b%2B3%3D255&pl=Solve']equation solver[/URL], b = 84 Substitute b = 84 into equation (2): A = 84 + 15 [B]A = 99[/B]\n\nA=0.5(bh), for h\nA=0.5(bh), for h Divide each side by 0.5b [B]h = A/0.5b[/B]\n\nA=2(l+w) for l\nMultiply through: A = 2l + 2w To solve for l, subtract 2w from each side: 2l = A - 2w Divide each side by 2 l = (A - 2w)/2\n\nA=2(l+w) for w\nMultiply through using the distributive property, so we have: A = 2l + 2w Subtract 2l from each side 2w = A - 2l Divide each side by w w = (A - 2l)/2\n\nA=a+b+c+d�4 for c\nA=a+b+c+d�4 for c Assume A and a are different variables: Cross multiply: a + b + c + d = 4A Subtract a, b, and d from each side: a + b + c + d - (a + b + d) = 4A - (a + b + d) Cancel the a + b + d on the left side [B]c = 4A - a - b - d[/B]\n\nAaron bought a bagel and 3 muffins for \\$7.25. Bea bought a bagel and 2 muffins for \\$6. How much is b\nAaron bought a bagel and 3 muffins for \\$7.25. Bea bought a bagel and 2 muffins for \\$6. How much is bagel and how much is a muffin? Let b be the number of bagels and m be the number of muffins. We have two equations: [LIST=1] [*]b + 3m = 7.25 [*]b + 2m = 6 [/LIST] Subtract (2) from (1) [B]m = 1.25 [/B] Plug this into (2), we have: b + 2(1.25) = 6 b + 2.5 = 6 Subtract 2.5 from each side [B]b = 3.5[/B]\n\nacw+cz=y for a\nacw+cz=y for a Solve this literal equation: Subtract cz from each side: acw + cz - cz = y - cz Cancel the cz on the left side: acw = y - cz Divide each side by cw to isolate a: acw/cw = (y - cz)/cw Cancel cw on the left side: [B]a = (y - cz)/cw[/B]\n\nadmission to the school fair is \\$2.50 for students and \\$3.75 for others. if 2848 admissions were col\nadmission to the school fair is \\$2.50 for students and \\$3.75 for others. if 2848 admissions were collected for a total of 10,078.75, how many students attended the fair Let the number of students be s and the others be o. We're given two equations: [LIST=1] [*]o + s = 2848 [*]3.75o + 2.50s = 10078.75 [/LIST] Since we have no coefficients for equation 1, let's solve this the fast way using substitution. Rearrange equation 1 by subtracting o from each side to isolate s [LIST=1] [*]o = 2848 - s [*]3.75o + 2.50s = 10078.75 [/LIST] Now substitute equation 1 into equation 2: 3.75(2848 - s) + 2.50s =10078.75 To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=3.75%282848-s%29%2B2.50s%3D10078.75&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]481[/B]\n\nAge now problems\nLet f be the age of the father and d be the age of the daughter and s be the age of the son. We have: [LIST=1] [*]f = 3s [*]d = s - 3 [*]d - 3 + f - 3 + s - 3 = 63 [/LIST] Simplify (3) d + f + s - 9 = 63 d + f + s = 72 Now, substitute (1) and (2) into the modified (3) (s - 3) + 3s + s = 72 Combine like terms: 5s - 3 = 72 Add 3 to each side 5s = 75 Divide each side by 5 s = 15 We want f, so we substitute s = 15 into (1) f = 3(15) [B]f = 45[/B]\n\nAlgebra Master (Polynomials)\nGiven 2 polynomials this does the following:\n2) Polynomial Subtraction\n\nAlso generates binomial theorem expansions and polynomial expansions with or without an outside constant multiplier.\n\nAlvin is 12 years younger than Elga. The sum of their ages is 60 . What is Elgas age?\nAlvin is 12 years younger than Elga. The sum of their ages is 60 . What is Elgas age? Let a be Alvin's age and e be Elga's age. We have the following equations: [LIST=1] [*]a = e - 12 [*]a + e = 60 [/LIST] Plugging in (1) to (2), we get: (e - 12) + e = 60 Grouping like terms: 2e - 12 = 60 Add 12 to each side: 2e = 72 Divide each side by 2 [B]e = 36[/B]\n\nAmy and ryan operate a car dealing and repair service. For a car detailing (full wash outside and in\nAmy and ryan operate a car dealing and repair service. For a car detailing (full wash outside and inside. Amy charges 40\\$ and Ryan charges 50\\$ . In addition they charge a hourly rate. Amy charges \\$35/h and ryan charges \\$30/h. How many hours does amy and ryan have to work to make the same amount of money? Set up the cost functions C(h) where h is the number of hours. [U]Amy:[/U] C(h) = 35h + 40 [U]Ryan:[/U] C(h) = 30h + 50 To make the same amount of money, we set both C(h) functions equal to each other: 35h + 40 = 30h + 50 To solve for h, we [URL='https://www.mathcelebrity.com/1unk.php?num=35h%2B40%3D30h%2B50&pl=Solve']type this equation into our search engine[/URL] and we get: h = [B]2[/B]\n\nAn ancient Greek was said to have lived 1/4 of his live as a boy, 1/5 as a youth, 1/3 as a man, and\nAn ancient Greek was said to have lived 1/4 of his live as a boy, 1/5 as a youth, 1/3 as a man, and spent the last 13 years as an old man. How old was he when he died? Set up his life equation per time lived as a boy, youth, man, and old man 1/4 + 1/5 + 1/3 + x = 1. Using our [URL='http://www.mathcelebrity.com/gcflcm.php?num1=4&num2=3&num3=5&pl=LCM']LCM Calculator[/URL], we see the LCM of 3,4,5 is 60. This is our common denominator. So we have 15/60 + 12/60 + 20/60 + x/60 = 60/60 [U]Combine like terms[/U] x + 47/60 = 60/60 [U]Subtract 47/60 from each side:[/U] x/60 = 13/60 x = 13 out of the 60 possible years, so he was [B]60 when he died[/B].\n\nAn equilateral triangle has three sides of equal length. What is the equation for the perimeter of a\nAn equilateral triangle has three sides of equal length. What is the equation for the perimeter of an equilateral triangle if P = perimeter and S = length of a side? P = s + s + s [B]P = 3s[/B]\n\nAn experienced accountant can balance the books twice as fast as a new accountant. Working together\nAn experienced accountant can balance the books twice as fast as a new accountant. Working together it takes the accountants 10 hours. How long would it take the experienced accountant working alone? Person A: x/2 job per hour Person B: 1/x job per hour Set up our equation: 1/x + 1/(2x) = 1/10 Multiply the first fraction by 2/2 to get common denominators; 2/(2x) + 1/(2x) = 1/10 Combine like terms 3/2x = 1/10 Cross multiply: 30 = 2x Divide each side by 2: [B]x = 15[/B]\n\nAn irregular pentagon is a five sided figure. The two longest sides of a pentagon are each three tim\nAn irregular pentagon is a five sided figure. The two longest sides of a pentagon are each three times as long as the shortest side. The remaining two sides are each 8m longer than the shortest side. The perimeter of the Pentagon is 79m. Find the length of each side of the Pentagon. Let long sides be l. Let short sides be s. Let medium sides be m. We have 3 equations: [LIST=1] [*]2l + 2m + s = 79 [*]m = s + 8 [*]l = 3s [/LIST] Substitute (2) and (3) into (1): 2(3s) + 2(s + 8) + s = 79 Multiply through and simplify: 6s + 2s + 16 + s = 79 9s + 16 = 79 [URL='https://www.mathcelebrity.com/1unk.php?num=9s%2B16%3D79&pl=Solve']Using our equation calculator[/URL], we get [B]s = 7[/B]. This means from Equation (2): m = 7 + 8 [B]m = 15 [/B] And from equation (3): l = 3(7) [B]l = 21[/B]\n\nAn orchard has 816 apple trees. The number of rows exceeds the number of trees per row by 10. How ma\nAn orchard has 816 apple trees. The number of rows exceeds the number of trees per row by 10. How many trees are there in each row? Let the rows be r and the trees per row be t. We're given two equations: [LIST=1] [*]rt = 816 [*]r = t + 10 [/LIST] Substitute equation (2) into equation (1) for r: (t + 10)t = 816 t^2 + 10t = 816 Subtract 816 from each side of the equation: t^2 + 10t - 816 = 816 - 816 t^2 + 10t - 816 = 0 We have a quadratic equation. To solve this, we [URL='https://www.mathcelebrity.com/quadratic.php?num=t%5E2%2B10t-816%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type it in our search engine [/URL]and we get: t = (24, -34) Since the number of trees per row can't be negative, we choose [B]24[/B] as our answer\n\nAndrea has one hour to spend training for an upcoming race she completes her training by running ful\nAndrea has one hour to spend training for an upcoming race she completes her training by running full speed in the distance of the race and walking back the same distance to cool down if she runs at a speed of 9 mph and walks back at a speed of 3 mph how long should she plan on spending to walk back Let r = running time. Let w = walking time We're given two equations [LIST=1] [*]r + w = 1 [*]9r = 3w [/LIST] Rearrange equation (1) by subtract r from each side: [LIST=1] [*]w = 1 - r [*]9r = 3w [/LIST] Now substitute equation (1) into equation (2): 9r = 3(1 - r) 9r = 3 - 3r To solve for r, [URL='https://www.mathcelebrity.com/1unk.php?num=9r%3D3-3r&pl=Solve']we type this equation into our search engine[/URL] and we get: r = 0.25 Plug this into modified equation (1) to solve for w, and we get: w = 1. 0.25 [B]w = 0.75[/B]\n\nApril, May and June have 90 sweets between them. May has three-quarters of the number of sweets that\nApril, May and June have 90 sweets between them. May has three-quarters of the number of sweets that June has. April has two-thirds of the number of sweets that May has. How many sweets does June have? Let the April sweets be a. Let the May sweets be m. Let the June sweets be j. We're given the following equations: [LIST=1] [*]m = 3j/4 [*]a = 2m/3 [*]a + j + m = 90 [/LIST] Cross multiply #2; 3a = 2m Dividing each side by 2, we get; m = 3a/2 Since m = 3j/4 from equation #1, we have: 3j/4 = 3a/2 Cross multiply: 6j = 12a Divide each side by 12: a = j/2 So we have: [LIST=1] [*]m = 3j/4 [*]a = j/2 [*]a + j + m = 90 [/LIST] Now substitute equation 1 and 2 into equation 3: j/2 + j + 3j/4 = 90 Multiply each side by 4 to eliminate fractions: 2j + 4j + 3j = 360 To solve this equation for j, we [URL='https://www.mathcelebrity.com/1unk.php?num=2j%2B4j%2B3j%3D360&pl=Solve']type it in our search engine[/URL] and we get: j = [B]40[/B]\n\nArnie bought some bagels at 20 cents each. He ate 4, and sold the rest at 30 cents each. His profit\nArnie bought some bagels at 20 cents each. He ate 4, and sold the rest at 30 cents each. His profit was \\$2.40. How many bagels did he buy? Let x be the number of bagels Arnie sold. We have the following equation: 0.30(x - 4) - 0.20(4) = 2.40 Distribute and simplify: 0.30x - 1.20 - 0.8 = 2.40 Combine like terms: 0.30x - 2 = 2.40 Add 2 to each side: 0.30x = 4.40 Divide each side by 0.3 [B]x = 14.67 ~ 15[/B]\n\nAs a salesperson, you are paid \\$50 per week plus \\$2 per sale. This week you want your pay to be at l\nAs a salesperson, you are paid \\$50 per week plus \\$2 per sale. This week you want your pay to be at least \\$100. What is the minimum number of sales you must make to earn at least \\$100? Set up the inequality where s is the amount of sales you make: 50 + 2s >= 100 We use >= because the phrase [I]at least[/I] 100 means 100 or more Subtract 50 from each side: 2s >= 50 Divide each side by 2 [B]s >= 25[/B]\n\nAt 4am the outside temperature was -28 By 4pm it rose 38� What was the temperature at 4pm\nAt 4am the outside temperature was -28 By 4pm it rose 38� What was the temperature at 4pm -28 + 38 = [B]10 degrees (above zero)[/B]\n\nAt a local fitness center, members pay a \\$10 membership fee and \\$3 for each aerobics class. Nonme\nAt a local fitness center, members pay a \\$10 membership fee and \\$3 for each aerobics class. Nonmembers pay \\$5 for each aerobics class. For what number of aerobics classes will the cost for members and nonmembers be the same? Set up the cost functions where x is the number of aerobics classes: [LIST] [*]Members: C(x) = 10 + 3x [*]Non-members: C(x) = 5x [/LIST] Set them equal to each other 10 + 3x = 5x Subtract 3x from both sides: 2x = 10 Divide each side by 2 [B]x = 5 classes[/B]\n\nAt a local fitness center, members pay an \\$8 membership fee and \\$3 for each aerobics class. Nonmembe\nAt a local fitness center, members pay an \\$8 membership fee and \\$3 for each aerobics class. Nonmembers pay \\$5 for each aerobics class. For what number of aerobics classes will the cost for members be equal to nonmembers? Set up two cost equations C(x): [LIST=1] [*]Members: C(x) = 8 + 3x [*]Nonmembers: C(x) = 5x [/LIST] Set the two cost equations equal to each other: 8 + 3x = 5x Subtract 3x from each side 2x = 8 Divide each side by 2 [B]x = 4[/B]\n\nAt the end of the week, Francesca had a third of her babysitting money left after spending \\$14.65 on\nAt the end of the week, Francesca had a third of her babysitting money left after spending \\$14.65 on a movie and popcorn and another \\$1.35 on a pen. How much did she earn babysitting? Let the original amount of money earned for babysitting be b. We're given: [LIST=1] [*]Start with b [*]Spending 14.65 for a movie means we subtract 14.65 from b: b - 14.65 [*]Spending 1.35 on a pen means we subtract another 1.35 from step 2: b - 14.65 - 1.35 [*]Francesca has a third of her money left. So we set step 3 equal to 1/3 of b [/LIST] b - 14.65 - 1.35 = b/3 Multiply each side of the equation by 3 to remove the fraction 3(b - 14.65 - 1.35) = 3b/3 3b - 43.95 - 4.05 = b To solve this equation for b, [URL='https://www.mathcelebrity.com/1unk.php?num=3b-43.95-4.05%3Db&pl=Solve']we type it in our search engine[/URL] and we get: b =[B] 24[/B]\n\nAt what simple interest rate will 4500\\$ amount to 8000\\$ in 5 years?\nAt what simple interest rate will 4500\\$ amount to 8000\\$ in 5 years? Simple Interest is written as 1 + it. With t = 5, we have: 4500(1 + 5i) = 8000 Divide each side by 4500 1 + 5i = 1.77777778 Subtract 1 from each side: 5i = 0.77777778 Divide each side by 5 i = 0.15555 As a percentage we multiply by 100 to get [B]15.5%[/B]\n\nAustin has 15 CDs, which is 3 less than his sister has. How many CDs does his sister have?\nAustin has 15 CDs, which is 3 less than his sister has. How many CDs does his sister have? Let s be the number of CD's his sister has and a be the number Austin has [LIST=1] [*]a = 15 [*]a = s - 3 [/LIST] Substitute (1) into (2) 15 = s - 3 Add 3 to each side [B]s = 18[/B]\n\nAva is 4 times as old as Peter. What equation can be used to find Peter�s age?\nAva is 4 times as old as Peter. What equation can be used to find Peter�s age? [U]Assumptions[/U] Let a be Ava's age Let p be Peter's age We're given: a = 4p To find Peter's age, we divide each side of the equation by 4 to get: a/4 = 4p/4 p = [B]a/4[/B]\n\nAva set her watch 2 seconds behind, every day it sets back 1 second. How many days has it been since\nAva set her watch 2 seconds behind, every day it sets back 1 second. How many days has it been since she last set her watch if it is 41 seconds behind? Right now: Watch is 2 seconds behind [U]Let d be the day after right now[/U] (1)d + 2 = 41 d + 2 = 41 [U]Subtract 2 from each side[/U] [B]d = 39[/B]\n\nax + b = cx - d\nWe are solving for x: Subtract b from each side: ax = cx - d - b Subtract cx from each side: ax - cx = -d - b Factor out x from the left side: x(a - c) = -d - b Divide each side by (a - c) x = (-d - b)/(a - c)\n\nB+c =10/a for a\nB+c =10/a for a Cross multiply: a(B + c) = 10 Divide each side by a [B]a = 10/(B + c)[/B]\n\nBarney has \\$450 and spends \\$3 each week. Betty has \\$120 and saves \\$8 each week. How many weeks will\nBarney has \\$450 and spends \\$3 each week. Betty has \\$120 and saves \\$8 each week. How many weeks will it take for them to have the same amount of money? Let w be the number of weeks that go by for saving/spending. Set up Barney's balance equation, B(w). Spending means we [U]subtract[/U] B(w) = Initial Amount - spend per week * w weeks B(w) = 450 - 3w Set up Betty's balance equation, B(w). Saving means we [U]add[/U] B(w) = Initial Amount + savings per week * w weeks B(w) = 120 + 8w The same amount of money means both of their balance equations B(w) are equal. So we set Barney's balance equal to Betty's balance and solve for w: 450 - 3w = 120 + 8w Add 3w to each side to isolate w: 450 - 3w + 3w = 120 + 8w + 3w Cancelling the 3w on the left side, we get: 450 = 120 + 11w Rewrite to have constant on the right side: 11w + 120 = 450 Subtract 120 from each side: 11w + 120 - 120 = 450 - 120 Cancelling the 120's on the left side, we get: 11w = 330 To solve for w, we divide each side by 11 11w/11 = 330/11 Cancelling the 11's on the left side, we get: w = [B]30 [MEDIA=youtube]ifG_q-utgJI[/MEDIA][/B]\n\nBawi solves a problem that has an answer of x = -4. He first added 7 to both sides of the equal sign\nBawi solves a problem that has an answer of x = -4. He first added 7 to both sides of the equal sign, then divided by 3. What was the original equation [LIST=1] [*]If we added 7 to both sides, that means we had a minus 7 (-7) to start with as a constant. Since subtraction undoes addition. [*]If we divided by 3, this means we multiplied x by 3 to begin with. Since division undoes multiplication [/LIST] So we have the start equation: 3x - 7 If the answer was x = -4, then we plug this in to get our number on the right side of the equation: 3(-4) - 7 -12 - 7 -19 This means our original equation was: [B]3x - 7 = -19[/B] And if we want to solve this to prove our answer, we [URL='https://www.mathcelebrity.com/1unk.php?num=3x-7%3D-19&pl=Solve']type the equation into our search engine [/URL]and we get: x = -4\n\nBen has \\$4.50 in quarters(Q) and dimes(D). a)Write an equation expressing the total amount of money\nBen has \\$4.50 in quarters(Q) and dimes(D). a)Write an equation expressing the total amount of money in terms of the number of quarters and dimes. b)Rearrange the equation to isolate for the number of dimes (D) a) The equation is: [B]0.1d + 0.25q = 4.5[/B] b) Isolate the equation for d. We subtract 0.25q from each side of the equation: 0.1d + 0.25q - 0.25q = 4.5 - 0.25q Cancel the 0.25q on the left side, and we get: 0.1d = 4.5 - 0.25q Divide each side of the equation by 0.1 to isolate d: 0.1d/0.1 = (4.5 - 0.25q)/0.1 d = [B]45 - 2.5q[/B]\n\nBesides 8 and 1, what is one factor of 8\nBesides 8 and 1, what is one factor of 8. Using our [URL='http://www.mathcelebrity.com/factoriz.php?num=8&pl=Show+Factorization']factor calculator[/URL], or entering the shortcut [B]Factor 8[/B], we get the following factors: 1, 2, 4, 8 Excluding 1 and 8, we have [B]2, 4[/B]\n\nBeth made a trip to the train station and back. On the trip there she traveled 45 km/h and on the re\nBeth made a trip to the train station and back. On the trip there she traveled 45 km/h and on the return trip she went 30 km/h. How long did the trip there take if the return trip took six hours? We use the distance formula: D = rt where D = distance, r = rate, and t = time. Start with the return trip: D = 45(6) D = 270 The initial trip is: 270= 30t Divide each side by 30 [B]t = 9 hours[/B]\n\nBinomial Multiplication (FOIL)\nMultiplies out the product of 2 binomials in the form (a + b)(c + d) with 1 unknown variable.\nThis utilizes the First-Outside-Inside-Last (F.O.I.L.) method.\n\nBlake writes 4 pages per hour. How many hours will Blake have to spend writing this week in order to\nBlake writes 4 pages per hour. How many hours will Blake have to spend writing this week in order to have written a total of 16 pages? [U]Let x = the number of hours Blake needs to write[/U] 4 pages per hour * x hours = 16 [U]Divide each side by 4[/U] [B]x = 4 hours[/B]\n\nBrice has 1200 in the bank. He wants to save a total of 3000 by depositing 40 per week from his payc\nBrice has 1200 in the bank. He wants to save a total of 3000 by depositing 40 per week from his paycheck. How many weeks will it take until he saves 3000? Remaining Savings = 3,000 - 1,200 = 1,800 40 per week * x weeks = 1,800 40x = 1800 Divide each side of the equation by 40 [B]x = 45 weeks[/B]\n\nBrighthouse charges \\$120 a month for their basic plan, plus \\$2.99 for each on demand movie you buy.\nBrighthouse charges \\$120 a month for their basic plan, plus \\$2.99 for each on demand movie you buy. Write and solve and inequality to find how many on demand movies could you buy if you want your bill to be less than \\$150 for the month. Let x equal to the number room movie rentals per month. Our inequality is: 120 + 2.99x < 150 To solve for the number of movies, Add 120 to each side 2.99x < 30 Divide each side by 2.99 x < 10.03, which means 10 since you cannot buy a fraction of a movie\n\nby + 2/3 = c for y\nby + 2/3 = c for y Subtract 2/3 from each side of the literal equation: by + 2/3 - 2/3 = c - 2/3 Cancel the 2/3 on the left side to get: by = c - 2/3 Divide each side by b to isolate y: by/b = (c - 2/3)/b Cancel the b's on the left side to get: y = [B](c - 2/3)/b[/B]\n\nby + 2/3 = c, for y\nby + 2/3 = c, for y Subtract 2/3 from each side: by = c - 2/3 Divide each side by b y = [B](c - 2/3)/b[/B]\n\nc=59f-288 for f\nc=59f-288 for f Add 288 to each side: c + 288 = 59f - 288 + 288 Cancel the 288 on the right side, we get: 59f = c + 288 Divide each side by 59 to isolate f: 59f/59 = (c + 288)/59 Cancel the 59 on the left side, we get: f = [B](c + 288)/59[/B]\n\ncalculate cos(x) given tan(x)=8/15\ncalculate cos(x) given tan(x)=8/15 tan(x) = sin(x)/cos(x) sin(x)/cos(x) = 8/15 Cross multiply: 15sin(x) = 8cos(x) Divide each side by 8 [B]cos(x) = 15sin(x)/8[/B]\n\nCaleb earns points on his credit card that he can use towards future purchases.\nLet f = dollars spent on flights, h dollars spent on hotels, and p dollars spent on all other purchases. [U]Set up our equations:[/U] (1) 4f + 2h + p = 14660 (2) f + h + p = 9480 (3) f = 2h + 140 [U]First, substitute (3) into (2)[/U] (2h + 140) + h + p = 9480 3h + p + 140 = 9480 3h + p = 9340 [U]Subtract 3h to isolate p to form equation (4)[/U] (4) p = 9340 - 3h [U]Take (3) and (4), and substitute into (1)[/U] 4(2h + 140) + 2h + (9340 - h) = 14660 [U]Multiply through[/U] 8h + 560 + 2h + 9340 - 3h = 14660 [U]Combine h terms and constants[/U] (8 + 2 - 3)h + (560 + 9340) = 14660 7h + 9900 = 14660 [U]Subtract 9900 from both sides:[/U] 7h = 4760 [U]Divide each side by 7[/U] [B]h = 680[/B] [U]Substitute h = 680 into equation (3)[/U] f = 2(680) + 140 f = 1360 + 140 [B]f = 1,500[/B] [U] Substitute h = 680 and f = 1500 into equation (2)[/U] 1500 + 680 + p = 9480 p + 2180 = 9480 [U]Subtract 2180 from each side:[/U] [B]p = 7,300[/B]\n\ncan someone help me with how to work out this word problem?\nConsider a paper cone, pointing down, with the height 6 cm and the radius 3 cm; there is currently 9/4 (pie) cubic cm of water in the cone, and the cone is leaking at a rate of 2 cubic centimeters of water per second. How fast is the water level changing, in cm per second?\n\nCevian Triangle Relations\nGiven a triangle with a cevian, this will solve for the cevian or segments or sides based on inputs\n\nChi-Square χ2 Test\nThis calculator determines a χ2 chi-square test on a test statistic and determines if it is outside an accepted range with critical value test and conclusion.\n\nChris, Alex and Jesse are all siblings in the same family. Alex is 5 years older than chris. Jesse i\nChris, Alex and Jesse are all siblings in the same family. Alex is 5 years older than chris. Jesse is 6 years older than Alex. The sum of their ages is 31 years. How old is each one of them? Set up the relational equations where a = Alex's age, c = Chris's aged and j = Jesse's age [LIST=1] [*]a = c + 5 [*]j = a + 6 [*]a + c + j = 31 [*]Rearrange (1) in terms of c: c = a - 5 [/LIST] [U]Plug in (4) and (2) into (3)[/U] a + (a - 5) + (a + 6) = 31 [U]Combine like terms:[/U] 3a + 1 = 31 [U]Subtract 1 from each side[/U] 3a = 30 [U]Divide each side by 3[/U] [B]a = 10[/B] [U]Plug in 1 = 10 into Equation (4)[/U] c = 10 - 5 [B]c = 5[/B] Now plug 1 = 10 into equation (2) j = 10 + 6 [B]j = 16[/B]\n\nClass A has 8 pupils and class B has 10 pupils. Both classes sit the same maths test. The mean sco\nClass A has 8 pupils and class B has 10 pupils. Both classes sit the same maths test. The mean score for class A is 55. The mean score for both classes is 76. What is the mean score (rounded to 1 DP) in the maths test for class B Mean of the sum equals the sum of the means. U(A + B) = U(A) + U(B) 76 = 55 + U(B) Subtract 55 from each side, we get: [B]U(B) = 21[/B]\n\nColin was thinking of a number. Colin divides by 8, then adds 1 to get an answer of 2. What was the\nColin was thinking of a number. Colin divides by 8, then adds 1 to get an answer of 2. What was the original number? Let the number be n. Divide by 8: n/8 Then add 1: n/8 + 1 The phrase [I]get an answer[/I] of means an equation, so we set n/8 + 1 equal to 2: n/8 + 1 = 2 To solve for n, we subtract 1 from each side to isolate the n term: n/8 + 1 - 1 = 2 - 1 Cancel the 1's on the left side, we get: n/8 = 1 Cross multiply: n = 8*1 n = [B]8[/B]\n\nCompany a charges \\$25 plus \\$0.10 a mile. Company b charges \\$20 plus \\$0.15 per mile. How far would yo\nCompany a charges \\$25 plus \\$0.10 a mile. Company b charges \\$20 plus \\$0.15 per mile. How far would you need to travel to get each charge to be the same? Let x be the number of miles traveled Company A charge: C = 25 + 0.10x Company B charge: C = 20 + 0.15x Set up an equation find out when the charges are the same. 25 + 0.10x = 20 + 0.15x Combine terms and simplify 0.05x = 5 Divide each side of the equation by 0.05 to isolate x x = [B]100[/B]\n\nConnor runs 2 mi more each day than David. The sum of the distances they run each week is 56 mi. How\nConnor runs 2 mi more each day than David. The sum of the distances they run each week is 56 mi. How far does David run each day? Let Connor's distance be c Let David's distance be d We're given two equations: [LIST=1] [*]c = d + 2 [*]7(c + d) = 56 [/LIST] Simplifying equation 2 by dividing each side by 7, we get: [LIST=1] [*]c = d + 2 [*]c + d = 8 [/LIST] Substitute equation (1) into equation (2) for c d + 2 + d = 8 To solve for d, we [URL='https://www.mathcelebrity.com/1unk.php?num=d%2B2%2Bd%3D8&pl=Solve']type this equation into our calculation engine[/URL] and we get: d = [B]3[/B]\n\nConsider a firm that has two assembly lines, 1 and 2, both producing calculator. Assume that you hav\nConsider a firm that has two assembly lines, 1 and 2, both producing calculator. Assume that you have purchased a calculator and it turns out to be defective. And the line 1 produces 60% of all calculators produced. L1: event that the calculator is produced on line 1 L2: event that the calculator is produced on line 2 Suppose that your are given the conditional information: 10% of the calculator produced on line 1 is defective 20% of the calculator produced on line 2 is defective Q: If we choose one defective, what is the probability that the defective calculator comes from Line 1 and Line2? L1 = event that the calculator is produced on line 1 = 0.6 L2 = event that the calculator is produced on line 2 = 1 - 0.6 = 0.4 D = Defective D|L1 Defective from Line 1 = 0.1 D|L2 = Defective from Line 2 = 0.20 [U]Defective from Line 1[/U] P(L1|D) = P(L1)P(D/L1) / [ P(L1)P(D/L1) + P(L2)P(D/L2)] P(L1|D) = (.60)(.10) /[(.60)(.10)+ (.40)(.20)] [B]P(L1|D) = 0.4286[/B] [U]Defective from Line 2[/U] P(L2|D) = P(L2)P(D/L2) / [ P(L1)P(D/L1) + P(L2)P(D/L2)] P(L2|D) = (.40)(.20) /[(.60)(.10)+ (.40)(.20)] [B]P(L2|D) = 0.5714[/B]\n\nConsider a probability model consisting of randomly drawing two colored balls from a jar containing\nConsider a probability model consisting of randomly drawing two colored balls from a jar containing 2 red and 1 blue balls. What is the Sample Space of this experiment? (assume B= blue and R=red) The sample space is the list of all possible events [LIST] [*]RRB [*]RBR [*]BRR [/LIST]\n\nConsider the case of a manufacturer who has an automatic machine that produces an important part. Pa\nConsider the case of a manufacturer who has an automatic machine that produces an important part. Past records indicate that at the beginning of the data the machine is set up correctly 70 percent of the time. Past experience also shows that if the machine is set up correctly it will produce good parts 90 percent of the time. If it is set up incorrectly, it will produce good parts 40 percent of the time. Since the machine will produce 60 percent bad parts, the manufacturer is considering using a testing procedure. If the machine is set up and produces a good part, what is the revised probability that it is set up correctly? [U]Determine our events:[/U] [LIST] [*]C = Correctly Set Machine = 0.7 [*]C|G = Correctly Set Machine And Good Part = 0.9 [*]I = Incorrectly Set Machine = 1 - 0.7 = 0.3 [*]I|G = Incorrectly Set Machine And Good Part = 0.4 [*]B< = BAD PARTS = 0.60 [/LIST] P[correctly set & part ok] = P(C) * P(C|G) P[correctly set & part ok] = 70% * 90% = 63% P[correctly set & part ok] = P(I) * P(I|G) P[incorrectly set & part ok] = 30% *40% = 12% P[correctly set | part ok] = P[correctly set & part ok]/(P[correctly set & part ok] + P[incorrectly set & part ok]) P[correctly set | part ok] = 63/(63+12) = [B]0.84 or 84%[/B]\n\nConsider the following 15 numbers 1, 2, y - 4, 4, 5, x, 6, 7, 8, y, 9, 10, 12, 3x, 20 - The mean o\nConsider the following 15 numbers 1, 2, y - 4, 4, 5, x, 6, 7, 8, y, 9, 10, 12, 3x, 20 - The mean of the last 10 numbers is TWICE the mean of the first 10 numbers - The sum of the last 2 numbers is FIVE times the sum of the first 3 numbers (i) Calculate the values of x and y We're given two equations: [LIST=1] [*](x + 6 + 7 + 8 + y + 9 + 10 + 12 + 3x + 20)/10 = 2(1 + 2 + y - 4 + 4 + 5 + x + 6 + 7 + 8 + y)/10 [*]3x - 20 = 5(1 + 2 + y - 4) [/LIST] Let's evaluate and simplify: [LIST=1] [*](x + 6 + 7 + 8 + y + 9 + 10 + 12 + 3x + 20)/10 = (1 + 2 + y - 4 + 4 + 5 + x + 6 + 7 + 8 + y)/5 [*]3x - 20 = 5(y - 1) [/LIST] Simplify some more: [URL='https://www.mathcelebrity.com/polynomial.php?num=x%2B6%2B7%2B8%2By%2B9%2B10%2B12%2B3x%2B20&pl=Evaluate'](x + 6 + 7 + 8 + y + 9 + 10 + 12 + 3x + 20)/10[/URL] = (4x + y + 72)/10 [URL='https://www.mathcelebrity.com/polynomial.php?num=1%2B2%2By-4%2B4%2B5%2Bx%2B6%2B7%2B8%2By&pl=Evaluate'](1 + 2 + y - 4 + 4 + 5 + x + 6 + 7 + 8 + y)/5[/URL] = (2y + x + 29)/5 5(y - 1) = 5y - 5 So we're left with: [LIST=1] [*](4x + y + 72)/10 = (2y + x + 29)/5 [*]3x - 20 = 5y - 5 [/LIST] Cross multiply equations in 1, we have: 5(4x + y + 72) = 10(2y + x + 29) 20x + 5y + 360 = 20y + 10x + 290 We have: [LIST=1] [*]20x + 5y + 360 = 20y + 10x + 290 [*]3x - 20 = 5y - 5 [/LIST] Combining like terms: [LIST=1] [*]10x - 15y = -70 [*]3x - 5y = 15 [/LIST] Now we have a system of equations which we can solve any of three ways: [LIST] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10x+-+15y+%3D+-70&term2=3x+-+5y+%3D+15&pl=Substitution']Substitution Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10x+-+15y+%3D+-70&term2=3x+-+5y+%3D+15&pl=Elimination']Elimination Method[/URL] [*][URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=10x+-+15y+%3D+-70&term2=3x+-+5y+%3D+15&pl=Cramers+Method']Cramer's Rule[/URL] [/LIST] No matter which method we choose, we get the same answer: (x, y) = [B](-115, -72)[/B]\n\nConsider the formula for the area of a trapezoid: A=12h(a+b) . Is it mathematically simpler to solve\nConsider the formula for the area of a trapezoid: A=12h(a+b) . Is it mathematically simpler to solve for a, b, or h? Why? Solve for each of these variables to demonstrate. The variable \"h\" is the easiest to solve for. Because you only have one step. Let's review: Divide each side of the equation by 12(a + b) h = 12(a + b)/A Solving for \"a\", we two steps. Divide each side by 12h: A/12h = a + b Subtract b from each side a = A/12h - b Solving for \"b\" takes two steps as well. Divide each side by 12h: A/12h = a + b Subtract a from each side b = A/12h - a\n\nContainer Arrangements\nGiven a set of items inside a container, this calculates the probability that you draw certain items in the following fashion:\nDraw all the items\nDraw any of the items\nHow many ways can you choose m items of a, n items of b, o items of c, etc.\n\ncot(?)=12 and ? is in Quadrant I, what is sin(?)?\ncot(?)=12 and ? is in Quadrant I, what is sin(?)? cot(?) = cos(?)/sin(?) 12 = cos(?)/sin(?) Cross multiply: 12sin(?) = cos(?) Divide each side by 12: sin(?) = [B]12cos(?)[/B]\n\nCountry A produces about 7 times the amount of diamonds in carats produce in Country B. If the total\nCountry A produces about 7 times the amount of diamonds in carats produce in Country B. If the total produced in both countries is 40,000,000 carats, find the amount produced in each country. Set up our two given equations: [LIST=1] [*]A = 7B [*]A + B = 40,000,000 [/LIST] Substitute (1) into (2) (7B) + B = 40,000,000 Combine like terms 8B = 40,000,000 Divide each side by 8 [B]B = 5,000,000[/B] Substitute this into (1) A = 7(5,000,000) [B]A = 35,000,000[/B]\n\nCube\nSolves for Volume (Capacity), Lateral Area,Surface Area, and the value of a side for a cube.\n\ncx+b/d=y for b\ncx+b/d=y for b Subtract cx from each side to isolate b/d: cx - cx + b/d = y - cx Cancel the cx terms on each side: b/d = y - cx Cross multiply: b = [B]d(y - cx)[/B]\n\nDan makes 11 an hour working at the local grocery store. Over the past year he has saved 137.50 towa\nDan makes 11 an hour working at the local grocery store. Over the past year he has saved 137.50 toward a new pair of retro sneakers. If sneakers cost 240, how many hours will he need to be able to buy the sneakers? Figure out his remaining savings target: 240 - 137.50 = 102.50 Let x equal the number of remaining hours Dan needs to work 11x = 102.50 Divide each side by 11 x = 9.318 We round up for a half-hour to 9.5, or a full hour to 10.\n\nDecagon\nSolves for the side, perimeter, and area of a decagon.\n\nDeon opened his account starting with \\$650 and he is going to take out \\$40 per month. Mai opened up\nDeon opened his account starting with \\$650 and he is going to take out \\$40 per month. Mai opened up her account with a starting amount of \\$850 and is going to take out \\$65 per month. When would the two accounts have the same amount of money? We set up a balance equation B(m) where m is the number of months. [U]Set up Deon's Balance equation:[/U] Withdrawals mean we subtract from our current balance B(m) = Starting Balance - Withdrawal Amount * m B(m) = 650 - 40m [U]Set up Mai's Balance equation:[/U] Withdrawals mean we subtract from our current balance B(m) = Starting Balance - Withdrawal Amount * m B(m) = 850 - 65m When the two accounts have the same amount of money, we can set both balance equations equal to each other and solve for m: 650 - 40m = 850 - 65m Solve for [I]m[/I] in the equation 650 - 40m = 850 - 65m [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables -40m and -65m. To do that, we add 65m to both sides -40m + 650 + 65m = -65m + 850 + 65m [SIZE=5][B]Step 2: Cancel -65m on the right side:[/B][/SIZE] 25m + 650 = 850 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 650 and 850. To do that, we subtract 650 from both sides 25m + 650 - 650 = 850 - 650 [SIZE=5][B]Step 4: Cancel 650 on the left side:[/B][/SIZE] 25m = 200 [SIZE=5][B]Step 5: Divide each side of the equation by 25[/B][/SIZE] 25m/25 = 200/25 m = [B]8[/B]\n\nDiana invested \\$3000 in a savings account for 3 years. She earned \\$450 in interest over that time pe\nDiana invested \\$3000 in a savings account for 3 years. She earned \\$450 in interest over that time period. What interest rate did she earn? Use the formula I=Prt to find your answer, where I is interest, P is principal, r is rate and t is time. Enter your solution in decimal form rounded to the nearest hundredth. For example, if your solution is 12%, you would enter 0.12. Our givens are: [LIST] [*]I = 450 [*]P = 3000 [*]t = 3 [*]We want r [/LIST] 450 = 3000(r)(3) 450 = 9000r Divide each side by 9000 [B]r = 0.05[/B]\n\nDivide 73 into two parts whose product is 402\nDivide 73 into two parts whose product is 40 Our first part is x Our second part is 73 - x The product of the two parts is: x(73 - x) = 40 Multiplying through, we get: -x^2 + 73x = 402 Subtract 40 from each side, we get: -x^2 + 73x - 402 = 0 This is a quadratic equation. To solve this, we type it in our search engine, choose \"solve Quadratic\", and we get: [LIST=1] [*]x = [B]6[/B] [*]x = [B]67[/B] [/LIST]\n\nDylan is playing darts. He hit the bullseye on 5 out of his last 20 tosses. Considering this data, h\nDylan is playing darts. He hit the bullseye on 5 out of his last 20 tosses. Considering this data, how many bullseyes would you expect Dylan to get during his next 16 tosses? We have a proportion of bullseyes to tosses where b is the number of bullseyes for 16 tosses: 5/20 = b/16 [URL='https://www.mathcelebrity.com/prop.php?num1=5&num2=b&den1=20&den2=16&propsign=%3D&pl=Calculate+missing+proportion+value']Type this proportion into our search engine[/URL] and we get: b = [B]4[/B]\n\nEach side of a square is lengthened by 3 inches . The area of this new, larger square is 25 square\nEach side of a square is lengthened by 3 inches . The area of this new, larger square is 25 square inches. Find the length of a side of the original square. area of a square is s^2 New square has sides s + 3, so the area of 25 is: (s + 3)^2 = 25 [URL='https://www.mathcelebrity.com/1unk.php?num=%28s%2B3%29%5E2%3D25&pl=Solve']Solving for s[/URL], we get: s = [B]2[/B]\n\nEmily is three years older than twice her sister Mary�s age. The sum of their ages is less than 30.\nEmily is three years older than twice her sister Mary�s age. The sum of their ages is less than 30. What is the greatest age Mary could be? Let e = Emily's age and m = Mary's age. We have the equation e = 2m + 3 and the inequality e + m < 30 Substitute the equation for e into the inequality: 2m + 3 + m < 30 Add the m terms 3m + 3 < 30 Subtract 3 from each side of the inequality 3m < 27 Divide each side of the inequality by 3 to isolate m m < 9 Therefore, the [B]greatest age[/B] Mary could be is 8, since less than 9 [U]does not include[/U] 9.\n\nEquation 2y+5x=40. Interprt the intercepts\nEquation 2y+5x=40. Interprt the intercepts Y intercept is when X = 0 2y + 5(0) = 40 2y = 40 Divide each side by 2 [B]y = 20 [/B] X intercept is when Y = 0 2(0) + 5x = 40 5x = 40 Divide each side by 5 [B]x = 8[/B]\n\nEquilateral Triangle\nGiven a side (a), this calculates the following items of the equilateral triangle:\n* Perimeter (P)\n* Semi-Perimeter (s)\n* Area (A)\n* altitudes (ha,hb,hc)\n* medians (ma,mb,mc)\n* angle bisectors (ta,tb,tc)\n\nErik is rolling two regular six-sided number cubes. What is the probability that he will roll an eve\nErik is rolling two regular six-sided number cubes. What is the probability that he will roll an even number on one cube and a prime number on the other? P(Even on first cube) = (2,4,6) / 6 total choices P(Even on first cube) = 3/6 P(Even on first cube) = 1/2 <-- [URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F6&frac2=3%2F8&pl=Simplify']Using our fraction simplify calculator[/URL] P(Prime on second cube) = (2,3,5) / 6 total choices P(Prime on second cube) = 3/6 P(Prime on second cube) = 1/2 <-- [URL='https://www.mathcelebrity.com/fraction.php?frac1=3%2F6&frac2=3%2F8&pl=Simplify']Using our fraction simplify calculator[/URL] Since each event is independent, we have: P(Even on the first cube, Prime on the second cube) = P(Even on the first cube) * P(Prime on the second cube) P(Even on the first cube, Prime on the second cube) = 1/2 * 1/2 P(Even on the first cube, Prime on the second cube) = [B]1/4[/B]\n\nExpand Master and Build Polynomial Equations\nThis calculator is the ultimate expansion tool to multiply polynomials. It expands algebraic expressions listed below using all 26 variables (a-z) as well as negative powers to handle polynomial multiplication. Includes multiple variable expressions as well as outside multipliers.\nAlso produces a polynomial equation from a given set of roots (polynomial zeros). * Binomial Expansions c(a + b)x\n* Polynomial Expansions c(d + e + f)x\n* FOIL Expansions (a + b)(c + d)\n* Multiple Parentheses Multiplications c(a + b)(d + e)(f + g)(h + i)\n\nExplain the steps you would take to find an equation for the line perpendicular to 4x - 5y = 20 and\nExplain the steps you would take to find an equation for the line perpendicular to 4x - 5y = 20 and sharing the same y-intercept Get this in slope-intercept form by adding 5y to each side: 4x - 5y + 5y = 5y + 20 Cancel the 5y's on the left side and we get: 5y + 20 = 4x Subtract 20 from each side 5y + 20 - 20 = 4x - 20 Cancel the 20's on the left side and we get: 5y = 4x - 20 Divide each side by 5: 5y/5 = 4x/5 - 4 y = 4x/5 - 4 So we have a slope of 4/5 to find our y-intercept, we set x = 0: y = 4(0)/5 - 4 y = 0 - 4 y = -4 If we want a line perpendicular to the line above, our slope will be the negative reciprocal: The reciprocal of 4/5 is found by flipping the fraction making the numerator the denominator and the denominator the numerator: m = 5/4 Next, we multiply this by -1: -5/4 So our slope-intercept of the perpendicular line with the same y-intercept is: [B]y = -5x/4 - 4[/B]\n\ney/n + k = t for y\ney/n + k = t for y Let's take this literal equation in pieces: Subtract k from each side: ey/n + k - k = t - k Cancel the k's on the left side, we have: ey/n = t - k Now multiply each side by n: ney/n = n(t - k) Cancel the n's on the left side, we have: ey = n(t - k) Divide each side by e: ey/e = n(t - k)/e Cancel the e's on the left side, we have: [B]y = n(t - k)/e[/B]\n\nf - g = 1/4b for b\nf - g = 1/4b for b Multiply each side of the equation by 4 to remove the 1/4 and isolate b: 4(f - g) = 4/4b 4/4 = 1, so we have: b = [B]4(f - g)[/B] [I]the key to this problem was multiplying by the reciprocal of the constant[/I]\n\nf(x)=a(b)^x and we know that f(3)=17 and f(7)=3156. what is the value of b\nf(x)=a(b)^x and we know that f(3)=17 and f(7)=3156. what is the value of b Set up both equations with values When x = 3, f(3) = 17, so we have a(b)^3 = 17 When x = 7, f(7) = 3156, so we have a(b)^7 = 3156 Isolate a in each equation a = 17/(b)^3 a = 3156/(b)^7 Now set them equal to each other 17/(b)^3 = 3156/(b)^7 Cross Multiply 17b^7 = 3156b^3 Divide each side by b^3 17b^4 = 3156 Divide each side by 17 b^4 = 185.6471 [B]b = 3.6912[/B]\n\nf+g/e=r for g\nf+g/e=r for g Subtract f from each side g/e = r - f Multiply each side by e [B]g = e(r - f)[/B]\n\nF/B=(M-N*L)/D for L\nF/B=(M-N*L)/D for L Cross multiply: DF/B = M - N*L Subtract M from each side: DF/B - M = -N*L Divide each side by -N [B]L = -DF/BN[/B]\n\nFarmer Yumi has too many plants in her garden. If she starts out with 150 plants and is losing them\nFarmer Yumi has too many plants in her garden. If she starts out with 150 plants and is losing them at a rate of 4% each day, how long will it take for her to have 20 plants left? Round UP to the nearest day. We set up the function P(d) where d is the number of days sine she started losing plants: P(d) = Initial plants * (1 - Loss percent / 100)^d Plugging in our numbers, we get: 20 = 150 * (1 - 4/100)^d 20 = 150 * (1 - 0.04)^d Read left to right so it's easier to read: 150 * 0.96^d = 20 Divide each side by 150, and we get: 0.96^d = 0.13333333333 To solve this logarithmic equation for d, we [URL='https://www.mathcelebrity.com/natlog.php?num=0.96%5Ed%3D0.13333333333&pl=Calculate']type it in our search engine[/URL] and we get: d = 49.35 The problem tells us to round up, so we round up to [B]50 days[/B]\n\nFind a linear function f, given f(16)=-2 and f(-12)=-9. Then find f(0)\nFind a linear function f, given f(16)=-2 and f(-12)=-9. Then find f(0). We've got 2 points: (16, -2) and (-12, -9) Calculate the slope (m) of this line using: m = (y2 - y1)/(x2 - x1) m = (-9 - -2)/(-12 - 16) m = -7/-28 m = 1/4 The line equation is denoted as: y = mx + b Let's use the first point (x, y) = (16, -2) -2 = 1/4(16) + b -2 = 4 + b Subtract 4 from each side, and we get: b = -6 So our equation of the line is: y = 1/4x - 6 The questions asks for f(0): y = 1/4(0) - 6 y = 0 - 6 [B]y = -6[/B]\n\nfind all sets of two consecutive positive odd integers whose sum is no greater than 18\nSo x + y <=18 y = x + 1 x + x + 1 <=18 2x + 1 <= 18 Subtract 1 from both sides 2x <= 17 x<=8.5 --> 8 So we have {(0,1),(1,2),(2,3),(3,4),(4,5),(5,6),(6,7),(7,8),(8,9)}\n\nfind the difference between a mountain with an altitude of 1,684 feet above sea level and a valley\nfind the difference between a mountain with an altitude of 1,684 feet above sea level and a valley 216 feet below sea level. Below sea level is the same as being on the opposite side of zero on the number line. To get the difference, we do the following: 1,684 - (-216) Since subtracting a negative is a positive, we have: 1,684 + 216 [B]1,900 feet[/B]\n\nFind the largest of three consecutive even integers when six times the first integers is equal to fi\nFind the largest of three consecutive even integers when six times the first integers is equal to five times the middle integer. Let the first of the 3 consecutive even integers be n. The second consecutive even integer is n + 2. The third (largest) consecutive even integer is n + 4. We are given 6n = 5(n + 2). Multiply through on the right side, and we get: 6n = 5n + 10 [URL='https://www.mathcelebrity.com/1unk.php?num=6n%3D5n%2B10&pl=Solve']Typing 6n = 5n + 10 into the search engine[/URL], we get n = 10. Remember, n was our smallest of 3 consecutive even integers. So the largest is: n + 4 10 + 4 [B]14[/B]\n\nFind two consecutive integers if the sum of their squares is 1513\nFind two consecutive integers if the sum of their squares is 1513 Let the first integer be n. The next consecutive integer is (n + 1). The sum of their squares is: n^2 + (n + 1)^2 = 1513 n^2 + n^2 + 2n + 1 = 1513 2n^2 + 2n + 1 = 1513 Subtract 1513 from each side: 2n^2 + 2n - 1512 = 0 We have a quadratic equation. We [URL='https://www.mathcelebrity.com/quadratic.php?num=2n%5E2%2B2n-1512%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this into our search engine[/URL] and get: n = (-27, 28) Let's take the positive solution. The second integer is: n + 1 28 + 1 = 29\n\nFind two consecutive positive integers such that the sum of their squares is 25\nFind two consecutive positive integers such that the sum of their squares is 25. Let the first integer be x. The next consecutive positive integer is x + 1. The sum of their squares equals 25. We write this as:: x^2 + (x + 1)^2 Expanding, we get: x^2 + x^2 + 2x + 1 = 25 Group like terms: 2x^2 + 2x + 1 = 25 Subtract 25 from each side: 2x^2 + 2x - 24 = 0 Simplify by dividing each side by 2: x^2 + x - 12 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get x = 3 or x = -4. The problem asks for positive integers, so we discard -4, and use 3. This means, our next positive integer is 3 + 1 = 4. So we have [B](3, 4) [/B]as our answers. Let's check our work: 3^2 + 4^2 = 9 + 16 = 25\n\nFind y if the line through (1,y) and (4,5) has a slope of 3\nFind y if the line through (1,y) and (4,5) has a slope of 3. Slope formula is: m = (y2 - y1)/(x2 - x1) With m = 3, we have: 3 = (5 - y)/(4 - 1) 3 = (5 - y)/3 Cross multiply: 5 - y = 9 Subtract 5 from each side -y = 4 Multiply each side by -1 [B]y = -4[/B]\n\nFixed cost 500 marginal cost 8 item sells for 30\nfixed cost 500 marginal cost 8 item sells for 30. Set up Cost Function C(x) where x is the number of items sold: C(x) = Marginal Cost * x + Fixed Cost C(x) = 8x + 500 Set up Revenue Function R(x) where x is the number of items sold: R(x) = Revenue per item * items sold R(x) = 30x Set up break even function (Cost Equals Revenue) C(x) = R(x) 8x + 500 = 30x Subtract 8x from each side: 22x = 500 Divide each side by 22: x = 22.727272 ~ 23 units for breakeven\n\nFoster is centering a photo that is 9/1/2 inches wide on a scrapbook pages that is 10 inches wide. H\nFoster is centering a photo that is 9/1/2 inches wide on a scrapbook pages that is 10 inches wide. How far from each side of the pages should he put the picture? Enter your answer as a mixed number. First, determine your margins, which is the difference between the width and photo width, divided by 2. 10 - 9 & 1/2 = 1/2 1/2 / 2 = [B]1/4[/B]\n\nFV-O/T=A for o\nFV-O/T=A for o Add O/T to each side: FV-O/T + O/T = A + O/T We have: A + O/T = FV Subtract A from each side: A - A + O/T = FV + A Cancelling the A's, e have: O/T = FV - A Cross multiply the T: [B]O = T(FV - A)[/B]\n\nf^2+5g=3md for d\nf^2+5g = 3md for d Divide each side by 3m to isolate d: (f^2+5g)/3m = 3md/3md Cancel the 3m on the right side and we get: d = [B](f^2+5g)/3m[/B]\n\nGary has three less pets than Abe. If together they own 15 pets, how many pets does Gary own?\nLet g = Gary's pets and a = Abe's pets. We are given two equations: (1) g = a - 3 (2) a + g = 15 Substitute (1) into (2) a + (a - 3) = 15 Combine Like Terms: 2a - 3 = 15 Add 3 to each side: 2a = 18 Divide each side by 2 to isolate a: a = 9 --> Abe has 9 pets Substitute a = 9 into Equation (1) g = 9 - 3 g = 6 --> Gary has 6 pets\n\nGeocache puzzle help\nLet x equal the number of sticks he started with. We have: The equation is 4/5 * (3/4 * (2/3 * (0.5x - 0.5) -1/3) - 0.75) - 0.2 = 19 Add 0.2 to each side: 4/5 * (3/4 * (2/3 * (0.5x - 0.5) -1/3) - 0.75) = 19.2 Multiply each side by 5/4 (3/4 * (2/3 * (0.5x - 0.5) - 1/3) - 0.75) = 24 Multiply the inside piece first: 2/6x - 2/6 - 1/3 2/6x - 4/6 Now subtract 0.75 which is 3/4 2/6x - 4/6 - 3/4 4/6 is 8/12 and 3/4 is 9/12, so we have: 2/6x - 17/12 Now multiply by 3/4 6/24x - 51/48 = 24 Simplify: 1/4x - 17/16 = 24 Multiply through by 4 x - 17/4 = 96 Since 17/4 = 4.25, add 4.25 to each side x = 100.25 Since he did not cut up any sticks, he has a full stick to start with: So x = [B]101[/B]\n\nGeocache puzzle help\nLet x equal the number of sticks he started with. We have: The equation is 4/5 * (3/4 * (2/3 * (0.5x - 0.5) -1/3) - 0.75) - 0.2 = 19 Add 0.2 to each side: 4/5 * (3/4 * (2/3 * (0.5x - 0.5) -1/3) - 0.75) = 19.2 Multiply each side by 5/4 (3/4 * (2/3 * (0.5x - 0.5) - 1/3) - 0.75) = 24 Multiply the inside piece first: 2/6x - 2/6 - 1/3 2/6x - 4/6 Now subtract 0.75 which is 3/4 2/6x - 4/6 - 3/4 4/6 is 8/12 and 3/4 is 9/12, so we have: 2/6x - 17/12 Now multiply by 3/4 6/24x - 51/48 = 24 Simplify: 1/4x - 17/16 = 24 Multiply through by 4 x - 17/4 = 96 Since 17/4 = 4.25, add 4.25 to each side x = 100.25 Since he did not cut up any sticks, he has a full stick to start with: So x = [B]101[/B]\n\nGiven the rectangular prism below, if AB = 6 in., AD = 8 in. and BF = 24, find the length of FD.\nGiven the rectangular prism below, if AB = 6 in., AD = 8 in. and BF = 24, find the length of FD. [IMG]http://www.mathcelebrity.com/images/math_problem_library_129.png[/IMG] If AB = 6 and AD = 8, by the Pythagorean theorem, we have BD = 10 from our [URL='http://www.mathcelebrity.com/pythag.php?side1input=6&side2input=8&hypinput=&pl=Solve+Missing+Side']Pythagorean Theorem[/URL] Calculator Using that, we have another right triangle which we can use the [URL='http://www.mathcelebrity.com/pythag.php?side1input=10&side2input=24&hypinput=&pl=Solve+Missing+Side']pythagorean theorem[/URL] calculator to get [B]FD = 26[/B]\n\nHappy Paws charges \\$16.00 plus \\$1.50 per hour to keep a dog during the day. Woof Watchers charges \\$1\nHappy Paws charges \\$16.00 plus \\$1.50 per hour to keep a dog during the day. Woof Watchers charges \\$11.00 plus \\$2.75 per hour. Complete the equation and solve it to find for how many hours the total cost of the services is equal. Use the variable h to represent the number of hours. Happy Paws Cost: C = 16 + 1.5h Woof Watchers: C = 11 + 2.75h Setup the equation where there costs are equal 16 + 1.5h = 11 + 2.75h Subtract 11 from each side: 5 + 1.5h = 2.75h Subtract 1.5h from each side 1.25h = 5 Divide each side by 1.25 [B]h = 4[/B]\n\nHari planted 324 plants in such a way that there were as many rows of plants as there were number of\nHari planted 324 plants in such a way that there were as many rows of plants as there were number of columns. Find the number of rows and columns. Let r be the number of rows and c be the number of columns. We have the area: rc = 324 Since rows equal columns, we have a square, and we can set r = c. c^2 = 324 Take the square root of each side: [B]c = 18[/B] Which means [B]r = 18[/B] as well. What we have is a garden of 18 x 18.\n\nheat loss of a glass window varies jointly as the window's area and the difference between the outsi\nheat loss of a glass window varies jointly as the window's area and the difference between the outside and the inside temperature. a window 6 feet wide by 3 feet long loses 1,320 btu per hour when the temperature outside is 22 degree colder than the temperature inside. Find the heat loss through a glass window that is 3 feet wide by 5 feet long when the temperature outside is 9 degree cooler than the temperature inside. Find k of the equation: 6*3*22*k = 1320 396k = 1,320 k = 3.33333 [URL='https://www.mathcelebrity.com/1unk.php?num=396k%3D1320&pl=Solve']per our equation solver[/URL] Now, find the heat loss for a 3x5 window when the temperature is 9 degrees cooler than the temperature inside: 3*5*9*3.333333 = [B]450 btu per hour[/B]\n\nHeptagon\nSolves for side length, perimeter, and area of a heptagon.\n\nHexagon\nThis calculator solves for side length (s), Area (A), and Perimeter (P) of a hexagon given one of the 3 entries.\n\nHow much do 10 pieces of candy cost if 1000 pieces cost 100.00?\nHow much do 10 pieces of candy cost if 1000 pieces cost 100.00? Set up a proportion of pieces to cost 10/x = 1000/100 Divide the right side by 100 on top and bottom 10/x = 10/1 [B]x = 1[/B]\n\nHow much would you need to deposit in an account now in order to have \\$6000 in the account in 10 yea\nHow much would you need to deposit in an account now in order to have \\$6000 in the account in 10 years? Assume the account earns 6% interest compounded monthly. We start with a balance of B. We want to know: B(1.06)^10 = 6000 B(1.79084769654) = 6000 Divide each side of the equation by 1.79084769654 to solve for B B = [B]3,350.37[/B]\n\nHow much would you need to deposit in an account now in order to have \\$6000 in the account in 15 yea\nHow much would you need to deposit in an account now in order to have \\$6000 in the account in 15 years? Assume the account earns 8% interest compounded monthly. 8% compounded monthly = 8/12 = 0.6667% per month. 15 years = 15*12 = 180 months We want to know an initial balance B such that: B(1.00667)^180 = \\$6,000 3.306921B = \\$6,000 Divide each side by 3.306921 [B]B = \\$1,814.38[/B]\n\nI only own blue blankets and red blankets. 8 out of every 15 blankets I have are red.\nI only own blue blankets and red blankets. 8 out of every 15 blankets I have are red. If have i 45 blankets, how many are blue? If 8 out of 15 blankets are red, then 15 - 8 = 7 are blue So 7 out of every 15 blankets are blue. Set up a proportion of blue blankets to total blankets where b is the number of blue blankets in 45 blankets 7/15 = b/45 Cross multiply: If 2 proportions are equal, then we can do the following: Numerator 1 * Denominator 2 = Denominator 1 * Numerator 2 15b = 45 * 7 15b = 315 To solve for b, divide each side of the equation by 15: 15b/15 = 315/15 Cancel the 15's on the left side and we get: b = [B]21[/B]\n\nIf \\$9000 grows to \\$9720 in 2 years find the simple interest rate.\nIf \\$9000 grows to \\$9720 in 2 years find the simple interest rate. Simple interest formula is Initial Balance * (1 + tn) = Current Balance We have [LIST] [*]Initial Balance = 9000 [*]Current Balance = 9720 [*]n = 2 [/LIST] Plugging in these values, we get: 9000 * (1 + 2t) = 9720 Divide each side by 9000 1 + 2t = 1.08 Subtract 1 from each sdie 2t = 0.08 Divide each side by 2 t = [B]0.04 or 4%[/B]\n\nIf (a - b)/b = 3/7, which of the following must also be true?\nIf (a - b)/b = 3/7, which of the following must also be true? A) a/b = -4/7 B) a/b = 10/7 C) (a + b)/b = 10/7 D) (a - 2b)/b = -11/7 We can rewrite (a - b)/b as: a/b - b/b = 3/7 Since b/b = 1, we have: a/b - 1 = 3/7 Since -1 = -7/7, we have: a/b - 7/7 = 3/7 Add 7/7 to each side: a/b - 7/7 + 7/7 = 3/7 + 7/7 Cancel the 7/7 on the left side, we get: [B]a/b = 10/7 or Answer B[/B]\n\nIf 100 runners went with 4 bicyclists and 5 walkers, how many bicyclists would go with 20 runners an\nIf 100 runners went with 4 bicyclists and 5 walkers, how many bicyclists would go with 20 runners and 2 walkers? [U]Set up a joint variation equation, for the 100 runners, 4 bicyclists, and 5 walkers:[/U] 100 = 4 * 5 * k 100 = 20k [U]Divide each side by 20[/U] k = 5 <-- Coefficient of Variation [U]Now, take scenario 2 to determine the bicyclists with 20 runners and 2 walkers[/U] 20 = 2 * 5 * b 20 = 10b [U]Divide each side by 10[/U] [B]b = 2[/B]\n\nIf 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numer\nIf 2 is added to the numerator and denominator it becomes 9/10 and if 3 is subtracted from the numerator and denominator it become 4/5. Find the fractions. Convert 2 to a fraction with a denominator of 10: 20/2 = 10, so we multiply 2 by 10/10: 2*10/10 = 20/10 Add 2 to the numerator and denominator: (n + 2)/(d + 2) = 9/10 Cross multiply and simplify: 10(n + 2) = 9(d + 2) 10n + 20 = 9d + 18 Move constants to right side by subtracting 20 from each side and subtracting 9d: 10n - 9d = -2 Subtract 3 from the numerator and denominator: (n - 3)/(d - 3) = 4/5 Cross multiply and simplify: 5(n - 3) = 4(d - 3) 5n - 15 = 4d - 12 Move constants to right side by adding 15 to each side and subtracting 4d: 5n - 4d = 3 Build our system of equations: [LIST=1] [*]10n - 9d = -2 [*]5n - 4d = 3 [/LIST] Multiply equation (2) by -2: [LIST=1] [*]10n - 9d = -2 [*]-10n + 8d = -6 [/LIST] Now add equation (1) to equation (2) (10 -10)n (-9 + 8)d = -2 - 6 The n's cancel, so we have: -d = -8 Multiply through by -1: d = 8 Now bring back our first equation from before, and plug in d = 8 into it to solve for n: 10n - 9d = -2 10n - 9(8) = -2 10n - 72 = -2 To solve for n, we [URL='https://www.mathcelebrity.com/1unk.php?num=10n-72%3D-2&pl=Solve']plug this equation into our search engine[/URL] and we get: n = 7 So our fraction, n/d = [B]7/8[/B]\n\nIf 3(c + d) = 5, what is the value of c + d?\nIf 3(c + d) = 5, what is the value of c + d? A) 3/5 B) 5/3 C) 3 D) 5 Divide each side of the equation by 3 to [U]isolate[/U] c + d 3(c + d)/3 = 5/3 Cancel the 3's on the left side, we get: c + d = [B]5/3, or answer B[/B]\n\nIf 4x+7=xy-6, then what is the value of x, in terms of y\nIf 4x+7=xy-6, then what is the value of x, in terms of y Subtract xy from each side: 4x + 7 - xy = -6 Add 7 to each side: 4x - xy = -6 - 7 4x - xy = -13 Factor out x: x(4 - y) = -13 Divide each side of the equation by (4 - y) [B]x = -13/(4 - y)[/B]\n\nIf 7 times the square of an integer is added to 5 times the integer, the result is 2. Find the integ\nIf 7 times the square of an integer is added to 5 times the integer, the result is 2. Find the integer. [LIST] [*]Let the integer be \"x\". [*]Square the integer: x^2 [*]7 times the square: 7x^2 [*]5 times the integer: 5x [*]Add them together: 7x^2 + 5x [*][I]The result is[/I] means an equation, so we set 7x^2 + 5x equal to 2 [/LIST] 7x^2 + 5x = 2 [U]This is a quadratic equation. To get it into standard form, we subtract 2 from each side:[/U] 7x^2 + 5x - 2 = 2 - 2 7x^2 + 5x - 2 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=7x%5E2%2B5x-2%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Type this problem into our search engine[/URL], and we get two solutions: [LIST=1] [*]x = 2/7 [*]x= -1 [/LIST] The problem asks for an integer, so our answer is x[B] = -1[/B]. [U]Let's check our work by plugging x = -1 into the quadratic:[/U] 7x^2 + 5x - 2 = 0 7(-1)^2 + 5(-1) - 2 ? 0 7(1) - 5 - 2 ? 0 0 = 0 So we verified our answer, [B]x = -1[/B].\n\nIf 800 feet of fencing is available, find the maximum area that can be enclosed.\nIf 800 feet of fencing is available, find the maximum area that can be enclosed. Perimeter of a rectangle is: 2l + 2w = P However, we're given one side (length) is bordered by the river and the fence length is 800, so we have: So we have l + 2w = 800 Rearranging in terms of l, we have: l = 800 - 2w The Area of a rectangle is: A = lw Plug in the value for l in the perimeter into this: A = (800 - 2w)w A = 800w - 2w^2 Take the [URL='https://www.mathcelebrity.com/dfii.php?term1=800w+-+2w%5E2&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']first derivative[/URL]: A' = 800 - 4w Now set this equal to 0 for maximum points: 4w = 800 [URL='https://www.mathcelebrity.com/1unk.php?num=4w%3D800&pl=Solve']Typing this equation into the search engine[/URL], we get: w = 200 Now plug this into our perimeter equation: l = 800 - 2(200) l = 800 - 400 l = 400 The maximum area to be enclosed is; A = lw A = 400(200) A = [B]80,000 square feet[/B]\n\nif a + b = 2 and a2 - b2 = -4, what is the value of a - b?\nif a+b=2 and a2-b2=-4, what is the value of a-b? a^2 - b^2 = -4 Factor this: (a + b)(a - b) = -4 We know from above, (a +b) = 2, so substitute: 2(a - b) = -4 Divide each side by 2 [B](a - b) = -2[/B]\n\nif a number is added to its square, the result is 72. find the number\nif a number is added to its square, the result is 72. find the number. Let the number be n. We're given: n + n^2 = 72 Subtract 72 from each side, we get: n^2 + n - 72 = 0 This is a quadratic equation. [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn-72%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']We type this equation into our search engine[/URL], and we get: [B]n = 8 and n = -9[/B]\n\nIf EF = 9x - 17, FG = 17x - 14, and EG = 20x + 17, what is FG?\nIf EF = 9x - 17, FG = 17x - 14, and EG = 20x + 17, what is FG? By segment addition, we know that: EF + FG = EG Substituting in our values for the 3 segments, we get: 9x - 17 + 17x - 14 = 20x + 17 Group like terms and simplify: (9 + 17)x + (-17 - 14) = 20x - 17 26x - 31 = 20x - 17 Solve for [I]x[/I] in the equation 26x - 31 = 20x - 17 [SIZE=5][B]Step 1: Group variables:[/B][/SIZE] We need to group our variables 26x and 20x. To do that, we subtract 20x from both sides 26x - 31 - 20x = 20x - 17 - 20x [SIZE=5][B]Step 2: Cancel 20x on the right side:[/B][/SIZE] 6x - 31 = -17 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants -31 and -17. To do that, we add 31 to both sides 6x - 31 + 31 = -17 + 31 [SIZE=5][B]Step 4: Cancel 31 on the left side:[/B][/SIZE] 6x = 14 [SIZE=5][B]Step 5: Divide each side of the equation by 6[/B][/SIZE] 6x/6 = 14/6 x = [B]2.3333333333333[/B]\n\nIf f(x) = 3x + 1 and g(x) = x^2 + 2x, find x when f(g(x)) = 10\nIf f(x) = 3x + 1 and g(x) = x^2 + 2x, find x when f(g(x)) = 10 [U]Evaluate f(g(x))[/U] f(g(x)) = 3(x^2 + 2x) + 1 f(g(x)) = 3x^2 + 6x + 1 [U]When f(g(x)) = 10, we have[/U] 10 = 3x^2 + 6x + 1 [U]Subtract 10 from each side:[/U] 3x^2 + 6x - 9 = 0 Divide each side of the equation by 3 x^2 + 2x - 3 = 0 Factor, we have: (x + 3)(x - 1) = 0 So x is either [B]1 or -3[/B]\n\nIf f(x) = ax^2 + bx + c and f(0) = 1 and f(-1) = 3, what is a - b\nIf f(x) = ax^2 + bx + c and f(0) = 1 and f(-1) = 3, what is a - b Evaluate f(0) f(0) = a(0^2) + b(0) + c f(0) = a(0) + b(0) + c f(0) = c Since f(0) = 1, we have c = 1 Evaluate f(-1) f(-1) = a(-1^2) + b(-1) + c f(-1) = a(1) - b + c f(-1) = a - b + c Since f(-1) = 3, we have: a - b + c = -3 We learned above that f(0) = 1, so c = 1. Plug c = 1 into f(-1) a - b + 1 = -3 Subtract 1 from each side: a - b + 1 - 1 = -3 - 1 Cancel the 1's on the left side and we get: a - b = [B]-4[/B]\n\nIf Mr hernandez has 5 times as many students as Mr daniels and together they have 150 students how m\nIf Mr hernandez has 5 times as many students as Mr daniels and together they have 150 students how many students do each have? Let h = Mr. Hernandez's students and d = Mr. Daniels students. We are given two equations: (1) h = 5d (2) d + h = 150 Substitute equation (1) into equation (2) d + (5d) = 150 Combine like terms: 6d = 150 Divide each side of the equation by 6 to isolate d d = 25 <-- Mr. Daniels Students Now, plug the value for d into equation (1) h = 5(25) h = 125 <-- Mr. Hernandez students\n\nif n(A) = 6, n(A intersect B) = 2 and n(A union B) = 11, find n(B)\nif n(A) = 6, n(A intersect B) = 2 and n(A union B) = 11, find n(B). n(A union B) = n(A) + n(B) - n(A intersect B) Plugging in our given values, we have: 11 = 6 + n(B) - 2 11 = 4 + n(B) Subtract 4 from each side: [B]n(B) = 7[/B]\n\nIf p+4=2 and q-3=2, what is the value of qp?\nIf p+4=2 and q-3=2, what is the value of qp? Isolate p by subtracting 4 from each side using our [URL='http://www.mathcelebrity.com/1unk.php?num=p%2B4%3D2&pl=Solve']equation calculator[/URL] p = -2 Isolate q by adding 3 to each side using our [URL='http://www.mathcelebrity.com/1unk.php?num=q-3%3D2&pl=Solve']equation calculator[/URL]: q = 5 pq = (-2)(5) [B]pq = -10[/B]\n\nIf tanx = 3/4 ,what is cosx?\nIf tanx = 3/4 ,what is cosx? tan(x) = sin(x)/cos(x), so we have: sin(x)/cos(x) = 3/4 cross multiply: 4sin(x) = 3cos(x) Divide each side by 3 to isolate cos(x): cos(x) = [B]4sin(x)/3 [/B]\n\nIf the circumference of a circular rug is 16? feet, then what is the area of the rug in terms of pi\nIf the circumference of a circular rug is 16? feet, then what is the area of the rug in terms of pi C = 2pir, so we have: C = 16? 16? = 2?r Divide each side by 2?: r = 16?/2? r = 8 Now, the area of a circle A is denoted below: A = ?r^2 Given r = 8 from above, we have: A = ?(8)^2 A = [B]64?[/B]\n\nIf the equation of a line passes through the points (1, 3) and (0, 0), which form would be used to w\nIf the equation of a line passes through the points (1, 3) and (0, 0), which form would be used to write the equation of the line? [URL='https://www.mathcelebrity.com/slope.php?xone=1&yone=3&slope=+&xtwo=0&ytwo=0&bvalue=+&pl=You+entered+2+points']Typing (1,3),(0,0) into the search engine[/URL], we get a point-slope form: [B]y - 3 = 3(x - 1)[/B] If we want mx + b form, we have: y - 3 = 3x - 3 Add 3 to each side: [B]y = 3x[/B]\n\nIf the perimeter of a rectangular field is 120 feet and the length of one side is 25 feet, how wide\nIf the perimeter of a rectangular field is 120 feet and the length of one side is 25 feet, how wide must the field be? The perimeter of a rectangle P, is denoted as: P = 2l + 2w We're given l = 25, and P = 120, so we have 2(25) + 2w = 120 Simplify: 2w + 50 = 120 [URL='https://www.mathcelebrity.com/1unk.php?num=2w%2B50%3D120&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]w = 35[/B]\n\nif the point (.53,y) is on the unit circle in quadrant 1, what is the value of y?\nif the point (.53,y) is on the unit circle in quadrant 1, what is the value of y? Unit circle equation: x^2 + y^2 = 1 Plugging in x = 0.53, we get (0.53)^2 + y^2 = 1 0.2809 + y^2 = 1 Subtract 0.2809 from each side: y^2 = 0.7191 y = [B]0.848[/B]\n\nif the ratio of 2x to 5y is 3 to 4, what is the ratio of x to y?\nif the ratio of 2x to 5y is 3 to 4, what is the ratio of x to y? Set up our given ratio: 2x/5y = 3/4 Cross multiply: 2x * 4 = 5y * 3 8x = 15y Divide each side by 8: 8x/8 = 15y/8 x = 15y/8 Now divide each side by y to find x/y: x/y = 15y/8y x/y =[B] 15/8[/B]\n\nIf the speed of an aeroplane is reduced by 40km/hr, it takes 20 minutes more to cover 1200m. Find th\nIf the speed of an aeroplane is reduced by 40km/hr, it takes 20 minutes more to cover 1200m. Find the time taken by aeroplane to cover 1200m initially. We know from the distance formula (d) using rate (r) and time (t) that: d = rt Regular speed: 1200 = rt Divide each side by t, we get: r = 1200/t Reduced speed. 20 minutes = 60/20 = 1/3 of an hour. So we multiply 1,200 by 3 3600 = (r - 40)(t + 1/3) If we multiply 3 by (t + 1/3), we get: 3t + 1 So we have: 3600 = (r - 40)(3t + 1) Substitute r = 1200/t into the reduced speed equation: 3600 = (1200/t - 40)(3t + 1) Multiply through and we get: 3600 = 3600 - 120t + 1200/t - 40 Subtract 3,600 from each side 3600 - 3600 = 3600 - 3600 - 120t + 1200/t - 40 The 3600's cancel, so we get: - 120t + 1200/t - 40 = 0 Multiply each side by t: -120t^2 - 40t + 1200 = 0 We've got a quadratic equation. To solve for t, [URL='https://www.mathcelebrity.com/quadratic.php?num=-120t%5E2-40t%2B1200%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this in our search engine[/URL] and we get: t = -10/3 or t = 3. Since time [I]cannot[/I] be negative, our final answer is: [B]t = 3[/B]\n\nIf x/2y = 3/4, what is the value of y/x?\nIf x/2y = 3/4, what is the value of y/x? Cross multiply this proportion: 4x = 3(2y) 4x = 6y Divide each side by x: 4x/x = 6y/x The x's cancel, and we have: 6y/x = 4 Divide each side by 6: 6y/6x = 4/6 The 6's on the left cancel, we have: y/x = 4/6 We can simplify this. [URL='https://www.mathcelebrity.com/fraction.php?frac1=4%2F6&frac2=3%2F8&pl=Simplify']Type in Simplify 4/6 into the search engine[/URL], and we get 2/3. y/x = [B]2/3[/B]\n\nif x2 is added to x, the sum is 42\nIf x2 is added to x, the sum is 42. x^2 + x = 42 Subtract 42 from both sides: x^2 + x - 42 = 0 We have a quadratic equation. Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-42%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation solver[/URL], we get: [B]x = 6 and x = -7 [/B] Since the problem does not state positive number solutions, they are both answers.\n\nIf you can buy 1?3 of a box of chocolates for 6 dollars, how much can you purchase for 4 dollars? Wr\nIf you can buy 1?3 of a box of chocolates for 6 dollars, how much can you purchase for 4 dollars? Write your answer as a fraction of a box. Set up a proportion of dollars to boxes where b is the number of boxes for \\$4: 6/1/3 = 4/b Cross multiply: 6b = 4/3 Multiply each side by 1/6 to isolate b: b = 4/18 [URL='https://www.mathcelebrity.com/gcflcm.php?num1=4&num2=18&num3=&pl=GCF+and+LCM']Type in GCF(4,18) into the search engine[/URL]. We get a greatest common factor of 2. Divide 4 and 18 in the fraction by 2. We get the reduced fraction of: [B]b = 2/9[/B]\n\nIf you divide my brother's age by 3 and then add 20, you will get my age, which is 31. What is my br\nIf you divide my brother's age by 3 and then add 20, you will get my age, which is 31. What is my brothers age? Let b be the brother's age. We're given the following relationship for the brother's age and my age: b/3 + 20 = 31 Subtract 20 from each side: b/3 + 20 - 20 = 31 - 20 Cancel the 20's on the left side and we get: b/3 = 11 Cross multiply, and we get: b = 3 * 11 b = [B]33 [/B] Check our work using b = 33 for b/3 + 20 = 31: 33/3 + 20 ? 31 11 + 20 ? 31 31 = 31\n\nIf you triple a number and then add 10, you get one-half of the original number. What is the number\nIf you triple a number and then add 10, you get one-half of the original number. What is the number? Let the number be n. We have: 3n + 10 = 0.5n Subtract 0.5n from each side 2.5n + 10 = 0 Subtract 10 from each side: 2.5n = -10 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2.5n%3D-10&pl=Solve']equation calculator,[/URL] we get: [B]n = -4[/B]\n\nIn 16 years, Ben will be 3 times as old as he is right now.\nIn 16 years, Ben will be 3 times as old as he is right now. Let Ben's age right now be b. We have, in 16 years, Ben's age will be 3 times what his age is now: b + 16 = 3b Subtract b from each side: 2b = 16 Divide each side by 2 [B]b = 8[/B] Check our work: 16 years from now, Ben's age is 8 + 16 = 24 And, 8 x 3 = 24\n\nIn 45 years, Gabriela will be 4 times as old as she is right now.\nIn 45 years, Gabriela will be 4 times as old as she is right now. Let a be Gabriela's age. we have: a + 45 = 4a Subtract a from each side: 3a = 45 Divide each side by a [B]a = 15[/B]\n\nIn a class there are 5 more boys than girls. There are 13 students in all. How many boys are there i\nIn a class there are 5 more boys than girls. There are 13 students in all. How many boys are there in the class? We start by declaring variables for boys and girls: [LIST] [*]Let b be the number of boys [*]Let g be the number of girls [/LIST] We're given two equations: [LIST=1] [*]b = g + 5 [*]b + g = 13 [/LIST] Substitute equation (1) for b into equation (2): g + 5 + g = 13 Grouping like terms, we get: 2g + 5 = 13 Subtract 5 from each side: 2g + 5 - 5 = 13 - 5 Cancel the 5's on the left side and we get: 2g = 8 Divide each side of the equation by 2 to isolate g: 2g/2 = 8/2 Cancel the 2's on the left side and we get: g = 4 Substitute g = 4 into equation (1) to solve for b: b = 4 + 5 b = [B]9[/B]\n\nIn a given year, Houston has good air quality 48% of the days, moderate air quality 41% of the days,\nIn a given year, Houston has good air quality 48% of the days, moderate air quality 41% of the days, and unhealthy air quality 4% of the days. How many days per year do residents have unhealthy air quality? 4% of 365 days in a year = [B]14.6 days. If we are talking full days, we have 14.[/B]\n\nIn a newspaper, it was reported that yearly robberies in Springfield were up 40% to 77 in 2012 from\nIn a newspaper, it was reported that yearly robberies in Springfield were up 40% to 77 in 2012 from 2011. How many robberies were there in Springfield in 2011? Let r be the number of robberies in 2011. We have: Robberies in 2012 = Robberies in 2011 * 1.4 77 = r * 1.4 Divide each side by 1.4 [B]r = 55[/B]\n\nin a presidential election ohio had 20 electoral votes. this is 14 less than texas had. how many ele\nIn a presidential election ohio had 20 electoral votes. This is 14 less than Texas had. How many electoral votes did Texas have? Let 0 = Ohio votes and t = Texas votes. We have: [LIST=1] [*]o = 20 [*]0 = t - 14 [/LIST] [U]Substitute (1) into (2)[/U] 20 = t - 14 [U]Add 14 to each side[/U] [B]t = 34[/B]\n\nIn Super Bowl XXXV, the total number of points scored was 41. The winning team outscored the losing\nIn Super Bowl XXXV, the total number of points scored was 41. The winning team outscored the losing team by 27 points. What was the final score of the game? In Super Bowl XXXV, the total number of points scored was 41. The winning team outscored the losing team by 27 points. What was the final score of the game? Let w be the winning team's points, and l be the losing team's points. We have two equations: [LIST=1] [*]w + l = 41 [*]w - l = 27 [/LIST] Add the two equations: 2w = 68 Divide each side by 2 [B]w = 34[/B] Substitute this into (1) 34 + l = 41 Subtract 34 from each side [B]l = 7[/B] Check your work: [LIST=1] [*]34 + 7 = 41 <-- check [*]34 - 7 = 27 <-- check [/LIST] The final score of the game was [B]34 to 7[/B]. You could also use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=w+%2B+l+%3D+41&term2=w+-+l+%3D+27&pl=Cramers+Method']simultaneous equation solver[/URL].\n\nIn this class of 4/5 students are right handed. if there are 20 right handed students, what is the t\nIn this class of 4/5 students are right handed. if there are 20 right handed students, what is the total number of students in this class? Let x be the total number of students in the class. We have: 4/5x = 20 Cross multiplying or using our [URL='http://www.mathcelebrity.com/1unk.php?num=4x%3D100&pl=Solve']equation calculator[/URL], we get: 4x = 100 Divide each side by 4 [B]x = 25[/B]\n\nIn x years time, Peter will be 23 years old. How old is he now?\nIn x years time, Peter will be 23 years old. How old is he now? Let Peter's current age be a. In x years time means we add x to a, so we're given: a + x = 23 We want to find a, s we subtract x from each side to get: a + x - x = 23 - x Cancel the x terms on the left side and we get: a = [B]23 - x[/B]\n\nIs 3 standard deviations above the means considered an outlier?\nIs 3 standard deviations above the means considered an outlier? [B]Yes.[/B] Using the empirical rule, we know that: [LIST] [*]68% of the values lie within one standard deviation of the mean [*]95% of the values lie within two standard deviations of the mean [/LIST] Anything out side of two standard deviations is considered an outlier.\n\nIsabel will run less than 36 minutes today. So far, she has run 22 minutes. What are the possible nu\nIsabel will run less than 36 minutes today. So far, she has run 22 minutes. What are the possible numbers of additional minutes she will run? Set up our inequality. If she ran 22 minutes, we need to find an expression to find out the remaining minutes x + 22 < 36 Subtract 22 from each side: x < 14 Remember, she cannot run negative minutes, so our lower bound is 0, so we have: [B]0 < x < 14 [/B]\n\nIsosceles Triangle\nGiven a long side (a) and a short side (b), this determines the following items of the isosceles triangle:\n* Area (A)\n* Semi-Perimeter (s)\n* Altitude a (ha)\n* Altitude b (hb)\n* Altitude c (hc)\n\nIt takes 60 minutes for 7 people to paint 5 walls. How many minutes does it take 10 people to paint\nIt takes 60 minutes for 7 people to paint 5 walls. How many minutes does it take 10 people to paint 10 walls? Rate * Time = Output Let \"Rate\" (r) be the rate at which [B]one person[/B] works. So we have: 7r * 60 = 5 Multiply through and simplify: 420r = 5 Divide each side by 5 to isolate r: r = 1/84 So now we want to find out how many minutes it takes 10 people to paint 10 walls using this rate: 10rt = 10 With r = 1/84, we have: 10t/84 = 10 Cross multiply: 10t = 840 To solve for t, we t[URL='https://www.mathcelebrity.com/1unk.php?num=10t%3D840&pl=Solve']ype this equation into our search engine[/URL] and we get: t = [B]84 minutes[/B]\n\nj - m/4 = 4k for m\nj - m/4 = 4k for m Multiply each side by 4: 4j - 4m/4 = 4(4k) Simplify: 4j - m = 16k Add m to each side: 4j - m + m = 16k + m The m's cancel on the left side, so we have: 4j = 16k + m Subtract 16k from each side: 4j - 16k = 16k - 16k + m The 16k cancels on the right side, so we're left with: [B]m = 4j - 16k or 4(j - 4k)[/B]\n\nJack bought 7 tickets for a movie. He paid \\$7 for each adult ticket and \\$4 for each child ticket. Ja\nJack bought 7 tickets for a movie. He paid \\$7 for each adult ticket and \\$4 for each child ticket. Jack spent \\$40 for the tickets Let a = Number of adult tickets and c be the number of child tickets. [LIST=1] [*]7a + 4c = 40 [*]a + c = 7 [*]Rearrange (2): a = 7 - c [/LIST] Now substitute a in (3) into (1): 7(7 - c) + 4c = 40 49 - 7c + 4c = 40 49 - 3c = 40 Add 3c to each side and subtract 40: 3c = 9 Divide each side by 3: [B]c = 3 [/B] Substitute c = 3 into Equation (2) a + 3 = 7 Subtract 3 from each side: [B]a = 4[/B]\n\nJake used 5 boxes to pack 43.5 kg of books. If the boxes each weighed the same and held 8 books, wh\nJake used 5 boxes to pack 43.5 kg of books. If the boxes each weighed the same and held 8 books, what did each book weigh? [U]Set up equations were w is the weight of each book:[/U] [LIST=1] [*]5 boxes * 8 books * w = 43.5 [*]40w = 43.5 [/LIST] [U]Divide each side by 40[/U] [B]w = 1.0875 kg[/B]\n\nJamie spent \\$15.36 on several items at the store. he spent an equal amount on each item. if jamie sp\nJamie spent \\$15.36 on several items at the store. he spent an equal amount on each item. if jamie spent \\$1.92 on each item, how many items did he buy? Let x equal the number of items Jamie bought. We have: 1.92x = 15.36 Divide each side by 1.92 [B]x = 8[/B]\n\nJane received 183 more votes than jack. If jack received n votes, how many votes did Jane receive? Let j = jane's votes. We have, j + 183 = n Subtract 183 from each side: j = n - 183\n\nJanet drove 395 kilometers and the trip took 5 hours. How fast was Janet traveling?\nJanet drove 395 kilometers and the trip took 5 hours. How fast was Janet traveling? Distance = Rate * Time We're given D = 395 and t = 5 We want Rate. We divide each side of the equation by time: Distance / Time = Rate * Time / Time Cancel the Time's on each side and we get: Rate = Distance / Time Plugging our numbers in, we get: Rate = 395/5 Rate = [B]79 kilometers[/B]\n\nJay purchased tickets for a concert. To place the order, a handling charge of \\$7 per ticket was cha\nJay purchased tickets for a concert. To place the order, a handling charge of \\$7 per ticket was charged. A sales tax of 4% was also charged on the ticket price and the handling charges. If the total charge for four tickets was \\$407.68, what was the ticket price? Round to the nearest dollar. with a ticket price of t, we have the total cost written as: 1.04 * (7*4 + 4t)= 407.68 Divide each side by 1.04 1.04 * (7*4 + 4t)/1.04= 407.68/1.04 Cancel the 1.04 on the left side and we get: 7*4 + 4t = 392 28 + 4t = 392 To solve this equation for t, we [URL='https://www.mathcelebrity.com/1unk.php?num=28%2B4t%3D392&pl=Solve']type it in our math engine[/URL] and we get: t = [B]91[/B]\n\nJenny went shoe shopping. Now she has 5 more pairs than her brother. Together they have 25 pairs. Ho\nJenny went shoe shopping. Now she has 5 more pairs than her brother. Together they have 25 pairs. How many pairs does Jenny have and how many pairs does her brother have? [U]Let j be the number of shoes Jenny has and b be the number of s hoes her brother has. Set up 2 equations:[/U] (1) b + j = 25 (2) j = b + 5 [U]Substitute (2) into (1)[/U] b + (b + 5) = 25 [U]Group the b terms[/U] 2b + 5 = 25 [U]Subtract 5 from each side[/U] 2b = 20 [U]Divide each side by b[/U] [B]b = 10 [/B] [U]Substitute b = 10 into (2)[/U] j = 10 + 5 [B]j = 15[/B]\n\nJill made 122 muffins. She put them into 3 boxes and has two muffins left. How many are in each box\nJill made 122 muffins. She put them into 3 boxes and has two muffins left. How many are in each box if they all contain the same amount of muffins? Let m equal the number of muffins per box. We're told that we have 3 boxes and 2 muffins left after filling up all 3 boxes. 3m + 2 = 122 To solve for m, we subtract 2 from each side: 3m + 2 - 2 = 122 - 2 Cancel the 2's on the left side and we get: 3m = 120 Divide each side by 3 to isolate m: 3m/3 = 120/3 Cancel the 3's on the left side and we get: m = [B]40[/B]\n\nJim was thinking of a number. Jim adds 20 to it, then doubles it and gets an answer of 99.2. What wa\nJim was thinking of a number. Jim adds 20 to it, then doubles it and gets an answer of 99.2. What was the original number? Start with x. Add 20 to it x + 20 Double it 2(x + 20) Set this equal to 99.2 2(x + 20) = 99.2 Divide each side by 2: x + 20 = 49.6 Subtract 20 from each side: x = [B]29.6[/B]\n\nJohn Adams was born in 1732 and became president in 1797. Harry S. Truman was born in 1884 and becam\nJohn Adams was born in 1732 and became president in 1797. Harry S. Truman was born in 1884 and became President in 1945. Who was older when he became president? Adams: [URL='http://www.mathcelebrity.com/longdiv.php?num1=1797&num2=1732&pl=Subtract']1797 - 1732[/URL] = 65 Truman: [URL='http://www.mathcelebrity.com/longdiv.php?num1=1945&num2=1884&pl=Subtract']1945 - 1884[/URL] = 61 Adams was older.\n\nJohn read the first 114 pages of a novel, which was 3 pages less than 1/3\nJohn read the first 114 pages of a novel, which was 3 pages less than 1/3 Set up the equation for the number of pages (p) in the novel 1/3p - 3 = 114 Add 3 to each side 1/3p = 117 Multiply each side by 3 [B]p = 351[/B]\n\nJohn read the first 114 pages of a novel, which was 3 pages less than 1/3 of the novel.\nJohn read the first 114 pages of a novel, which was 3 pages less than 1/3 of the novel. Let n be the number of pages in the novel. We have: 1/3n - 3 = 114 Multiply each side by 3: n - 9 = 342 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=n-9%3D342&pl=Solve']equation solver[/URL], we get [B]n = 351[/B].\n\nJonathan earns a base salary of \\$1500 plus 10% of his sales each month. Raymond earns \\$1200 plus 15%\nJonathan earns a base salary of \\$1500 plus 10% of his sales each month. Raymond earns \\$1200 plus 15% of his sales each month. How much will Jonathan and Raymond have to sell in order to earn the same amount each month? [U]Step 1: Set up Jonathan's sales equation S(m) where m is the amount of sales made each month:[/U] S(m) = Commission percentage * m + Base Salary 10% written as a decimal is 0.1. We want decimals to solve equations easier. S(m) = 0.1m + 1500 [U]Step 2: Set up Raymond's sales equation S(m) where m is the amount of sales made each month:[/U] S(m) = Commission percentage * m + Base Salary 15% written as a decimal is 0.15. We want decimals to solve equations easier. S(m) = 0.15m + 1200 [U]The question asks what is m when both S(m)'s equal each other[/U]: The phrase [I]earn the same amount [/I]means we set Jonathan's and Raymond's sales equations equal to each other 0.1m + 1500 = 0.15m + 1200 We want to isolate m terms on one side of the equation. Subtract 1200 from each side: 0.1m + 1500 - 1200 = 0.15m + 1200 - 1200 Cancel the 1200's on the right side and we get: 0.1m - 300 = 0.15m Next, we subtract 0.1m from each side of the equation to isolate m 0.1m - 0.1m + 300 = 0.15m - 0.1m Cancel the 0.1m terms on the left side and we get: 300 = 0.05m Flip the statement since it's an equal sign to get the variable on the left side: 0.05m = 300 To solve for m, we divide each side of the equation by 0.05: 0.05m/0.05 = 300/0.05 Cancelling the 0.05 on the left side, we get: m = [B]6000[/B]\n\nJuan runs out of gas in a city. He walks 30yards west and then 16 yards south looking for a gas stat\nJuan runs out of gas in a city. He walks 30yards west and then 16 yards south looking for a gas station. How far is he from his starting point? Juan is located on a right triangle. We calculate the hypotenuse: 30^2 + 16^2 = Hypotenuse^2 900 + 256 = Hypotenuse^2 Hypotenuse^2 = 1156 Take the square root of each side: [B]Hypotenuse = 34 yards[/B]\n\nJuan spent at most \\$2.50 on apples and oranges. He bought 5 apples at \\$0.36 each. What is the most h\nJuan spent at most \\$2.50 on apples and oranges. He bought 5 apples at \\$0.36 each. What is the most he spent on oranges? Let a be spending apples and o be spending on oranges, we have: [LIST=1] [*]a + o <= 2.36 <-- At most means less than or equal to [*]a = 5 * 0.36 = 1.8 [/LIST] Substitute (2) into (1) 1.8 + o <= 2.36 Subtract 1.8 from each side [B]o <= 0.56[/B]\n\nJulia has a bucket of water that weighs 10lbs. The total weight is 99% water. She leaves the bucke\nJulia has a bucket of water that weighs 10lbs. The total weight is 99% water. She leaves the bucket outside overnight and some of the water evaporates, in the morning the water is only 98% of the total weight. What is the new weight? Setup the proportion: 0.99/10 = 0.98/w Using our [URL='http://www.mathcelebrity.com/prop.php?num1=0.99&num2=0.98&den1=10&den2=w&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL], we get [B]w = 9.899 lbs[/B].\n\nJulia owes 18.20 for the month of November. Her plan costs 9.00 for the first 600 text messages and\nJulia owes 18.20 for the month of November. Her plan costs 9.00 for the first 600 text messages and .10 cents for additional texts. How many texts did she send out? Let m be the number of messages. We have a cost function of: C(m) = 9 + 0.1(m - 600) We are given C(m) = 18.20 18.20 = 9 + 0.1(m - 600) 18.20 = 9 + 0.1m - 60 Combine like terms: 18.20 = 0.1m - 51 Add 51 to each side 0.1m = 69.20 Divide each side by 0.1 [B]m = 692[/B]\n\nJulie is making a documentary about how Boxerville residents see their town. If she talks to 7 peopl\nJulie is making a documentary about how Boxerville residents see their town. If she talks to 7 people, and each interview lasts 4 minutes, how long will the film be? 7 people * 4 minutes each = [B]28 minutes[/B]\n\nk+m-7=n for k\nk+m/-7=n for k Add m/7 to both sides [B]k = n + m/7[/B]\n\nk=g-a/5 for g\nk=g-a/5 for g Add a/5 to each side; k + a/5 = g - a/5 + a/5 Cancel the a/5 terms on the right side, and we get: g = [B]k + a/5[/B]\n\nKarin has 3 to spend in the arcade. The game she likes costs 50c per play. What are the possible num\nKarin has 3 to spend in the arcade. The game she likes costs 50c per play. What are the possible numbers of times that she can play? [U]Let x = the number of games Karin can play with her money[/U] 0.5x = 3 [U]Divide each side by 0.5[/U] [B]x = 6[/B]\n\nKatie is twice as old as her sister Mara. The sum of their age is 24.\nLet k = Katie's age and m = Mara's age. We have 2 equations: (1) k = 2m (2) k + m = 24 Substitute (1) into (2) (2m) + m = 24 Combine like terms: 3m = 24 Divide each side of the equation by 3 to isolate m m = 8 If m = 8, substituting into (1) or (2), we get k = 16.\n\nKelly took clothes to the cleaners 3 times last month. First, she brought 4 shirts and 1 pair of sla\nKelly took clothes to the cleaners 3 times last month. First, she brought 4 shirts and 1 pair of slacks and paid11.45. Then she brought 5 shirts, 3 pairs of slacks, and 1 sports coat and paid 27.41. Finally, she brought 5 shirts and 1 sports coat and paid 16.94. How much was she charged for each shirt, each pair of slacks, and each sports coat? Let s be the cost of shirts, p be the cost of slacks, and c be the cost of sports coats. We're given: [LIST=1] [*]4s + p = 11.45 [*]5s + 3p + c = 27.41 [*]5s + c = 16.94 [/LIST] Rearrange (1) by subtracting 4s from each side: p = 11.45 - 4s Rearrange (3)by subtracting 5s from each side: c = 16.94 - 5s Take those rearranged equations, and plug them into (2): 5s + 3(11.45 - 4s) + (16.94 - 5s) = 27.41 Multiply through: 5s + 34.35 - 12s + 16.94 - 5s = 27.41 [URL='https://www.mathcelebrity.com/1unk.php?num=5s%2B34.35-12s%2B16.94-5s%3D27.41&pl=Solve']Group like terms using our equation calculator [/URL]and we get: [B]s = 1.99 [/B] <-- Shirt Cost Plug s = 1.99 into modified equation (1): p = 11.45 - 4(1.99) p = 11.45 - 7.96 [B]p = 3.49[/B] <-- Slacks Cost Plug s = 1.99 into modified equation (3): c = 16.94 - 5(1.99) c = 16.94 - 9.95 [B]c = 6.99[/B] <-- Sports Coat cost\n\nKendra is half as old as Morgan and 3 years younger than Lizzie. The total of their ages is 39. How\nKendra is half as old as Morgan and 3 years younger than Lizzie. The total of their ages is 39. How old are they? Let k be Kendra's age, m be Morgan's age, and l be Lizzie's age. We're given: [LIST=1] [*]k = 0.5m [*]k = l - 3 [*]k + l + m = 39 [/LIST] Rearranging (1) by multiplying each side by 2, we have: m = 2k Rearranging (2) by adding 3 to each side, we have: l = k + 3 Substituting these new values into (3), we have: k + (k + 3) + (2k) = 39 Group like terms: (k + k + 2k) + 3 = 39 4k + 3 = 39 [URL='https://www.mathcelebrity.com/1unk.php?num=4k%2B3%3D39&pl=Solve']Type this equation into the search engine[/URL], and we get: [B]k = 9 [/B] Substitute this back into (1), we have: m = 2(9) [B]m = 18 [/B] Substitute this back into (2), we have: l = (9) + 3 [B][B]l = 12[/B][/B]\n\nKerry asked a bank teller to cash 390 check using 20 bills and 50 bills. If the teller gave her a to\nKerry asked a bank teller to cash 390 check using 20 bills and 50 bills. If the teller gave her a total of 15 bills, how many of each type of bill did she receive? Let t = number of 20 bills and f = number of 50 bills. We have two equations. (1) 20t + 50f = 390 (2) t + f = 15 [U]Rearrange (2) into (3) for t, by subtracting f from each side:[/U] (3) t = 15 - f [U]Now substitute (3) into (1)[/U] 20(15 - f) + 50f = 390 300 - 20f + 50f = 390 [U]Combine f terms[/U] 300 + 30f = 390 [U]Subtract 300 from each side[/U] 30f = 90 [U]Divide each side by 30[/U] [B]f = 3[/B] [U]Substitute f = 3 into (3)[/U] t = 15 - 3 [B]t = 12[/B]\n\nKevin and randy have a jar containing 41 coins, all of which are either quarters or nickels. The tot\nKevin and randy have a jar containing 41 coins, all of which are either quarters or nickels. The total value of the jar is \\$7.85. How many of each type? Let d be dimes and q be quarters. Set up two equations from our givens: [LIST=1] [*]d + q = 41 [*]0.1d + 0.25q = 7.85 [/LIST] [U]Rearrange (1) by subtracting q from each side:[/U] (3) d = 41 - q [U]Now, substitute (3) into (2)[/U] 0.1(41 - q) + 0.25q = 7.85 4.1 - 0.1q + 0.25q = 7.85 [U]Combine q terms[/U] 0.15q + 4.1 = 7.85 [U]Using our [URL='http://www.mathcelebrity.com/1unk.php?num=0.15q%2B4.1%3D7.85&pl=Solve']equation calculator[/URL], we get:[/U] [B]q = 25[/B] [U]Substitute q = 25 into (3)[/U] d = 41 - 25 [B]d = 16[/B]\n\nKevin is 4 times old as Daniel and is also 6 years older than Daniel\nKevin is 4 times old as Daniel and is also 6 years older than Daniel. Let k be Kevin's age and d be Daniel's age. We have 2 equations: [LIST=1] [*]k = 4d [*]k = d + 6 [/LIST] Plug (1) into (2): 4d = d + 6 Subtract d from each side: 4d - d = d - d + 6 Cancel the d terms on the right side and simplify: 3d = 6 Divide each side by 3: 3d/3 = 6/3 Cancel the 3 on the left side: d = 2 Plug this back into equation (1): k = 4(2) k = 8 So Daniel is 2 years old and Kevin is 8 years old\n\nkira will spend less than 27 on gifts. so far, she has spent 12\\$. what are the possible additional a\nkira will spend less than 27 on gifts. so far, she has spent 12\\$. what are the possible additional amounts she will spend? The key word in this problem is [I]less than[/I]. So we know this is an inequality. Let s be Kira's possible spend. We have: s + 12 < 27 To solve for s in this inequality, we subtract 12 from each side: s + 12 - 12 < 27 - 12 Cancel the 12's on the left side, and we get: [B]s < 15 [/B] [I]The summary here is Kira can spend anything up to [U]but not including[/U] 15[/I]\n\nKites\nThis calculates perimeter and/or area of a kite given certain inputs such as short and long side, short and long diagonal, or angle between short and long side\n\nlarger of 2 numbers is 12 more than the smaller number. if the sum of the 2 numbers is 74 find the 2\nlarger of 2 numbers is 12 more than the smaller number. if the sum of the 2 numbers is 74 find the 2 numbers Declare Variables for each number: [LIST] [*]Let l be the larger number [*]Let s be the smaller number [/LIST] We're given two equations: [LIST=1] [*]l = s + 12 [*]l + s = 74 [/LIST] Equation (1) already has l solved for. Substitute equation (1) into equation (2) for l: s + 12 + s = 74 Solve for [I]s[/I] in the equation s + 12 + s = 74 [SIZE=5][B]Step 1: Group the s terms on the left hand side:[/B][/SIZE] (1 + 1)s = 2s [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 2s + 12 = + 74 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 12 and 74. To do that, we subtract 12 from both sides 2s + 12 - 12 = 74 - 12 [SIZE=5][B]Step 4: Cancel 12 on the left side:[/B][/SIZE] 2s = 62 [SIZE=5][B]Step 5: Divide each side of the equation by 2[/B][/SIZE] 2s/2 = 62/2 s = [B]31[/B] To solve for l, we substitute in s = 31 into equation (1): l = 31 + 12 l = [B]43[/B]\n\nlarger of 2 numbers is 4 more than the smaller. the sum of the 2 is 40. what is the larger number?\nlarger of 2 numbers is 4 more than the smaller. the sum of the 2 is 40. what is the larger number? Declare variables for the 2 numbers: [LIST] [*]Let l be the larger number [*]Let s be the smaller number [/LIST] We're given two equations: [LIST=1] [*]l = s + 4 [*]l + s = 40 [/LIST] To get this problem in terms of the larger number l, we rearrange equation (1) in terms of l. Subtract 4 from each side in equation (1) l - 4 = s + 4 - 4 Cancel the 4's and we get: s = l - 4 Our given equations are now: [LIST=1] [*]s = l - 4 [*]l + s = 40 [/LIST] Substitute equation (1) into equation (2) for s: l + l - 4 = 40 Grouping like terms for l, we get: 2l - 4 = 40 Add 4 to each side: 2l - 4 + 4 = 40 + 4 Cancelling the 4's on the left side, we get 2l = 44 Divide each side of the equation by 2 to isolate l: 2l/2 = 44/2 Cancel the 2's on the left side and we get: l = [B]22[/B]\n\nLarry Mitchell invested part of his \\$31,000 advance at 6% annual simple interest and the rest at 7%\nLarry Mitchell invested part of his \\$31,000 advance at 6% annual simple interest and the rest at 7% annual simple interest. If the total yearly interest from both accounts was \\$2,090, find the amount invested at each rate. Let x be the amount invested at 6%. Then 31000 - x is invested at 7%. We have the following equation: 0.06x + (31000 - x)0.07 = 2090 Simplify: 0.06x + 2170 - 0.07x = 2090 Combine like Terms -0.01x + 2170 = 2090 Subtract 2170 from each side -0.01x = -80 Divide each side by -0.01 x = [B]8000 [/B]at 6% Which means at 7%, we have: 31000 - 8000 = [B]23,000[/B]\n\nLaura spent half of her money on a necklace. She spent 14.60 of what was left having dinner with car\nLaura spent half of her money on a necklace. She spent 14.60 of what was left having dinner with carolyn. if she had 3.90 left, how much money did she start out with? Let x equal Laura's starting money 1/2x = 14.60 + 3.90 1/2x = 18.5 Divide each side by 1/2 [B]x = \\$37[/B]\n\nLauren wrote a total of 6 pages over 2 hours. How many hours will Lauren have to spend writing this\nLauren wrote a total of 6 pages over 2 hours. How many hours will Lauren have to spend writing this week in order to have written a total of 9 pages? Solve using unit rates. 6 pages per 2 hours equals 6/2 = 3 pages per hour as a unit rate Set up equation using h hours: 3h = 9 Divide each side by 3 [B]h = 3[/B]\n\nLeslie has 8 pencils. She has 9 fewer pencils than Michelle. How many pencils does Michelle have?\nLet m = the number of pencils Michelle has. So, Leslie has m - 9 = 8. Add 9 to both sides: m = 17. So Michelle has 17 pencils, and Leslie has 8, which is 9 fewer than 17\n\nLet P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For\nLet P(n) and S(n) denote the product and the sum, respectively, of the digits of the integer n. For example, P(23) = 6 and S(23) = 5. Suppose N is a two-digit number such that N = P(N) + S(N). What could N be? Is there more than one answer? For example, for 23 P(23) = 6 and S(23) = 5, but 23 could not be the N that we want since 23 <> 5 + 6 Let t = tens digit and o = ones digit P(n) = to S(n) = t + o P(n) + S(n) = to + t + o N = 10t + o Set them equal to each other N = P(N) + S(N) 10t + o = to + t + o o's cancel, so we have 10t = to + t Subtract t from each side, we have 9t = to Divide each side by t o = 9 So any two-digit number with 9 as the ones digit will work: [B]{19,29,39,49,59,69,79,89,99}[/B]\n\nLinda takes classes at both Westside Community College and Pinewood Community College. At Westside,\nLinda takes classes at both Westside Community College and Pinewood Community College. At Westside, class fees are \\$98 per credit hour, and at Pinewood, class fees are \\$115 per credit hour. Linda is taking a combined total of 18 credit hours at the two schools. Suppose that she is taking w credit hours at Westside. Write an expression for the combined total dollar amount she paid for her class fees. Let p be the number of credit hours at Pinewood. We have two equations: [LIST] [*]98w for Westside [*]115p at Pinewood [*]w + p = 18 [*]Total fees: [B]98w + 115p[/B] [/LIST]\n\nLiz harold has a jar in her office that contains 47 coins. Some are pennies and the rest are dimes.\nLiz harold has a jar in her office that contains 47 coins. Some are pennies and the rest are dimes. If the total value of the coins is 2.18, how many of each denomination does she have? [U]Set up two equations where p is the number of pennies and d is the number of dimes:[/U] (1) d + p = 47 (2) 0.1d + 0.01p = 2.18 [U]Rearrange (1) into (3) by solving for d[/U] (3) d = 47 - p [U]Substitute (3) into (2)[/U] 0.1(47 - p) + 0.01p = 2.18 4.7 - 0.1p + 0.01p = 2.18 [U]Group p terms[/U] 4.7 - 0.09p = 2.18 [U]Add 0.09p to both sides[/U] 0.09p + 2.18 = 4.7 [U]Subtract 2.18 from both sides[/U] 0.09p = 2.52 [U]Divide each side by 0.09[/U] [B]p = 28[/B] [U]Now substitute that back into (3)[/U] d =47 - 28 [B]d = 19[/B]\n\nLucas has nickels,dimes,and quarters in the ratio 1:3:2. If 10 of Lucas coins are quarters, how many\nLucas has nickels,dimes,and quarters in the ratio 1:3:2. If 10 of Lucas coins are quarters, how many nickels and dimes does Lucas have? 1 + 3 + 2 = 6. Quarters account for 2/6 which is 1/3 of the total coin count. Let x be the total number of coins. We have: 1/3x = 10 Multiply each side by 3 x = 30 We have the following ratios and totals: [LIST] [*]Nickels: 1/6 * 30 = [B]5 nickels[/B] [*]Dimes: 3/6 * 30 = [B]15 dimes[/B] [*]Quarters: 2/6 * 30 = [B]10 quarters[/B] [/LIST]\n\nM/n = p-6 for m\nM/n = p-6 for m Solve this literal equation by multiplying each side by n to isolate M: Mn/n = n(p - 6) Cancelling the n terms on the left side, we get: [B]M = n(p - 6)[/B]\n\nm/x = k-6 for m\nm/x = k-6 for m To solve this literal equation, multiply each side by x: x(m/x) = x(k - 6) The x's cancel on the left side, so we get: m = [B]x(k - 6)[/B]\n\nm=u/k-r/k for k\nm=u/k-r/k for k Multiply both sides by k to eliminate the k denominator: km = uk/k - rk/k Cancel the k's on the right side and we get km = u - r Divide each side by m: km/m = (u - r)/m Cancel the m on the left side: [B]k = (u - r)/m[/B]\n\nManuel can pay for his car insurance on a monthly basis, but if he pays an entire year's insurance i\nManuel can pay for his car insurance on a monthly basis, but if he pays an entire year's insurance in advance, he'll receive a \\$40 discount. His discounted bill for the year would then be \\$632. What is the monthly fee for his insurance? His full bill F, is denoted as: F - 40 = 632 [URL='https://www.mathcelebrity.com/1unk.php?num=f-40%3D632&pl=Solve']If we add 40 to each side[/URL], we get: F = [B]\\$672[/B]\n\nMarcela is having a presidential debate watching party with all of her friends, She will be making c\nMarcela is having a presidential debate watching party with all of her friends, She will be making chicken wings and hot dogs. Each chicken wing costs \\$2 to make and each hot dog costs \\$3. She needs to spend at least \\$500. Marcela knows that she will make more than 50 chicken wings and hot dogs combined. She also knows that she will make less than 120 chicken wings and less that 100 hot dogs. What are her inequalities? Let c be the number of chicken wings and h be the number of hot dogs. Set up the given inequalities: [LIST=1] [*]c + h > 50 [I]Marcela knows that she will make more than 50 chicken wings and hot dogs combined.[/I] [*]2c + 3h >= 500 [I]She needs to spend at least \\$500[/I] [*]c < 120 [I]She also knows that she will make less than 120 chicken wings[/I] [*]h < 100 [I]and less that 100 hot dogs[/I] [/LIST]\n\nMaria bought 7 boxes. A week later half of all her boxes were destroyed in a fire. There are now onl\nMaria bought 7 boxes. A week later half of all her boxes were destroyed in a fire. There are now only 22 boxes left. With how many did she start? [U]Let x be the starting box number. We have:[/U] (x + 7)/2 = 22 [U]Cross multiply[/U] x + 7 = 44 [U]Subtract 7 from each side[/U] [B]x = 37[/B]\n\nMarissa has 24 coins in quarters and nickels. She has 3 dollars. How many of the coins are quarters?\nLet n be the number of nickels and q be the number of quarters. We have two equations: (1) n + q = 24 (2) 0.05n + 0.25q = 3 Rearrange (1) to solve for n in terms of q for another equation (3) (3) n = 24 - q Plug (3) into (2) 0.05(24 - q) + 0.25q = 3 Multiply through: 1.2 - 0.05q + 0.25q = 3 Combine q terms 0.2q + 1.2 = 3 Subtract 1.2 from each side: 0.2q = 1.8 Divide each side by 0.2 [B]q = 9[/B]\n\nMarty is 3 years younger than 6 times his friend Warrens age. The sum of their ages is greater than\nMarty is 3 years younger than 6 times his friend Warrens age. The sum of their ages is greater than 11. What is the youngest age Warren can be? Let m be Marty's age and w be Warren's age. We have two equations: (1) m = 6w - 3 (2) m + w > 11 Plug (1) into (2) 6w - 3 + w > 11 Combine w terms 7w - 3 > 11 Add 3 to each side 7w > 14 Divide each side by 7 w > 2 which means [B]w = 3[/B] as the youngest age.\n\nVolume of rectangular prism is: V = lwh Plugging in the numbers you gave: 195 = (6)(5)h 195 = 30h Divide each side by 30 h = 6.5 6.5 feet is 6 feet, 6 inches. This is 2 inches more than your actor, so [B]yes[/B], he will fit in the box standing up.\n\nMatilda needs at least \\$112 to buy an new dress. She has already saved \\$40. She earns \\$9 an hour bab\nMatilda needs at least \\$112 to buy an new dress. She has already saved \\$40. She earns \\$9 an hour babysitting. Write and solve and inequality to find how many hours she will need to babysit to buy the dress. Subtract remaining amount needed after savings: 112 - 40 = 72 Let h be her hourly wages for babysitting. We have the equation: [B]9h = 72[/B] Divide each side by 9 [B]h = 8[/B]\n\nMax and Bob went to the store. Max bought 2 burgers and 2 drinks for \\$5.00 bob bought 3 burgers and\nMax and Bob went to the store. Max bought 2 burgers and 2 drinks for \\$5.00. Bob bought 3 burgers and 1 drink for \\$5.50. How much is each burger and drink? [U]Set up the givens where b is the cost of a burger and d is the cost of a drink:[/U] Max: 2b + 2d = 5 Bob: 3b + d = 5.50 [U]Rearrange Bob's equation by subtracting 3b from each side[/U] (3) d = 5.50 - 3b [U]Now substitute that d equation back into Max's Equation[/U] 2b + 2(5.50 - 3b) = 5 2b + 11 - 6b = 5 [U]Combine b terms:[/U] -4b + 11 = 5 [U]Subtract 11 from each side[/U] -4b = -6 [U]Divide each side by -4[/U] b = 3/2 [B]b = \\$1.50[/B] [U]Now plug that back into equation (3):[/U] d = 5.50 - 3(1.50) d = 5.50 - 4.50 [B]d = \\$1.00[/B]\n\nMichelle and Julie sold 65 cupcakes. If Julie sold 9 more cupcakes than Michelle, how many cupcakes\nMichelle and Julie sold 65 cupcakes. If Julie sold 9 more cupcakes than Michelle, how many cupcakes did each of them sell? Let m = Michelle's cupcakes and j = Julie's cupcakes. We have two equations: m + j = 65 j = m + 9 Substituting, we get: m + (m + 9) = 65 Combine like terms, we get: 2m + 9 = 65 Subtract 9 from each side: 2m = 56 Divide each side by 2 to isolate m m = 28 If m = 28, then j = 28 + 9 = 37 So (m, j) = (28, 37)\n\nmike went to canalside with \\$40 to spend. he rented skates for \\$10 and paid \\$3 per hour to skate.wha\nmike went to canalside with \\$40 to spend. he rented skates for \\$10 and paid \\$3 per hour to skate.what is the greatest number of hours Mike could have skated? Let h be the number of hours of skating. We have the cost function C(h): C(h) = Hourly skating rate * h + rental fee C(h) = 3h + 10 The problem asks for h when C(h) = 40: 3h + 10 = 40 To solve this equation for h, we [URL='https://www.mathcelebrity.com/1unk.php?num=3h%2B10%3D40&pl=Solve']type it in our search engine[/URL] and we get: h = [B]10[/B]\n\nMindy and troy combined ate 9 pieces of the wedding cake. Mindy ate 3 pieces of cake troy had 1/4 of\nMindy and troy combined ate 9 pieces of the wedding cake. Mindy ate 3 pieces of cake troy had 1/4 of the total cake. Write an equation to determine how many pieces of cake (c) that were in total Let c be the total number of pieces of cake. Let m be the number of pieces Mindy ate. Let t be the number of pieces Troy ate. We have the following given equations: [LIST] [*]m + t = 9 [*]m = 3 [*]t = 1/4c [/LIST] Combining (2) and (3) into (1), we have: 3 + 1/4c = 9 Subtract 3 from each side: 1/4c = 6 Cross multiply: [B]c = 24[/B]\n\nMr. Jimenez has a pool behind his house that needs to be fenced in. The backyard is an odd quadrilat\nMr. Jimenez has a pool behind his house that needs to be fenced in. The backyard is an odd quadrilateral shape and the pool encompasses the entire backyard. The four sides are 1818a, 77b, 1111a, and 1919b in length. How much fencing? (the length of the perimeter) would he need to enclose the pool? The perimeter P is found by adding all 4 sides: P = 1818a + 77b + 1111a + 1919b Group the a and b terms P = (1818 + 1111)a + (77 + 1919b) [B]P = 2929a + 1996b[/B]\n\nMr. Smith wants to spend less than \\$125 at a zoo. A ticket cost \\$7 he is taking 2 kids with him. Use\nMr. Smith wants to spend less than \\$125 at a zoo. A ticket cost \\$7 he is taking 2 kids with him. Use p to represent the other money he can spend there. 2 kids and Mr. Smith = 3 people. Total Ticket Cost is 3 people * 7 per ticket = 21 If he has 125 to spend, we have the following inequality using less than or equal to (<=) since he can spend up to or less than 125: p + 21 <= 125 Subtract 21 from each side: [B]p <= 104[/B]\n\nMr. Winkle downloaded 34 more songs than Mrs. Winkle downloaded. Together they downloaded 220 songs. How many songs did each download? Let x = Mr. Winkle downloads and y = Mrs. Winkle downloads. We then have x = y + 34 and x + y = 220. Substitute equation 1 into equation 2, we have: (y + 34) + y = 220 2y + 34 = 220 Subtract 34 from each side: 2y = 186 Divide each side by 2: y = 93 (Mrs. Winkle) x = 93 + 34 x = 127 (Mr. Winkle)\n\nn = b + d^2a for a\nn = b + d^2a for a Let's start by isolating the one term with the a variable. Subtract b from each side: n - b = b - b + d^2a Cancel the b terms on the right side and we get: n - b = d^2a With the a term isolated, let's divide each side of the equation by d^2: (n - b)/d^2 = d^2a/d^2 Cancel the d^2 on the right side, and we'll display this with the variable to solve on the left side: a = [B](n - b)/d^2 [MEDIA=youtube]BCEVsZmoKoQ[/MEDIA][/B]\n\nn=i*x+y for i\nn=i*x+y for i This is a literal equation. Subtract y from each side of the equation: n - y = i*x + y - y The y's cancel on the right side, so we have: n - y = ix Divide each side of the equation by x, to isolate i (n - y)/x = ix/x The x's cancel on the right side, so we have: i = [B](n - y)/x[/B]\n\nNate jars 2 liters of jam everyday. How many days did Nate spend making jam if he jarred 8 liters of\nNate jars 2 liters of jam everyday. How many days did Nate spend making jam if he jarred 8 liters of jam? 2 liters per 1 day and 8 liters per x days. Set up a proportion: 2/1 = 8/x Cross multiply: 2x = 8 Divide each side by 2 x = [B]4 days[/B].\n\nNeed help on this question\nConsider the recurrence relation T(n) =2 if n = 1, T(n?1) + 4n?2 if n > 1 (i) Derive the closed form expression f(n) for this recurrence relation. (ii) Prove that T(n) = f(n),?n ?N\n\nNonagon\nCalculates the side, perimeter, and area of a nonagon\n\nNotebooks cost \\$1.39 each. What are the possible numbers of notebooks that can be purchased with \\$10\nNotebooks cost \\$1.39 each. What are the possible numbers of notebooks that can be purchased with \\$10? Let n be the number of notebooks you can purchase. We have the following inequality: 1.39n <= 10 Divide each side by 1.39 n <= 7.194 We want whole notebooks, we cannot buy fractions of notebooks, so we have: n <= 7 The question asks for the possible numbers of notebooks we can buy. This implies we buy at least 1, but our inequality says not more than 7. So our number set is: [B]N = {1, 2, 3, 4, 5, 6, 7}[/B]\n\nN^2=5qd for d\nN^2=5qd for d Divide each side by 5q to isolate d: N^2/5q = 5qd/5q Cancel 5q on the right side and we get: d = [B]N^2/5q[/B]\n\nOceanside Bike Rental Shop charges \\$15.00 plus \\$9.00 per hour for renting a bike. Dan paid \\$51.00 to\nOceanside Bike Rental Shop charges \\$15.00 plus \\$9.00 per hour for renting a bike. Dan paid \\$51.00 to rent a bike. How many hours was he hiking for? Set up the cost equation C(h) where h is the number of hours needed to rent the bike: C(h) = Cost per hour * h + rental charge Using our given numbers in the problem, we have: C(h) = 9h + 15 The problem asks for h, when C(h) = 51. 9h + 15 = 51 To solve for h, we [URL='https://www.mathcelebrity.com/1unk.php?num=9h%2B15%3D51&pl=Solve']plug this equation into our search engine[/URL] and we get: h = [B]4[/B]\n\nOceanside Bike Rental Shop charges 16 dollars plus 6 dollars an hour for renting a bike. Mary paid 5\nOceanside Bike Rental Shop charges 16 dollars plus 6 dollars an hour for renting a bike. Mary paid 58 dollars to rent a bike. How many hours did she pay to have the bike checked out ? Set up the cost function C(h) where h is the number of hours you rent the bike: C(h) = Hourly rental cost * h + initial rental charge C(h) = 6h + 16 Now the problem asks for h when C(h) = 58, so we set C(h) = 58: 6h + 16 = 58 To solve this equation for h, we [URL='https://www.mathcelebrity.com/1unk.php?num=6h%2B16%3D58&pl=Solve']type it in our math engine[/URL] and we get: h = [B]7 hours[/B]\n\nOctagon\nCalculate side, area, and perimeter of an octagon based on inputs\n\nohn read the first 114 pages of a novel, which was 3 pages less than1/3 of the novel\nLet p be the novel pages. We have 1/3p - 3 = 114 Add 3 to each side 1/3p = 117 Multiply each side by 3 p = 351\n\nOliver invests \\$1,000 at a fixed rate of 7% compounded monthly, when will his account reach \\$10,000?\nOliver invests \\$1,000 at a fixed rate of 7% compounded monthly, when will his account reach \\$10,000? 7% monthly is: 0.07/12 = .00583 So we have: 1000(1 + .00583)^m = 10000 divide each side by 1000; (1.00583)^m = 10 Take the natural log of both sides; LN (1.00583)^m = LN(10) Use the identity for natural logs and exponents: m * LN (1.00583) = 2.30258509299 0.00252458479m = 2.30258509299 m = 912.064867899 Round up to [B]913 months[/B]\n\nOne number is equal to the square of another. Find the numbers if both are positive and their sum is\nOne number is equal to the square of another. Find the numbers if both are positive and their sum is 650 Let the number be n. Then the square is n^2. We're given: n^2 + n = 650 Subtract 650 from each side: n^2 + n - 650 = 0 We have a quadratic equation. [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn-650%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']We type this into our search engine[/URL] and we get: n = 25 and n = -26 Since the equation asks for a positive solution, we use [B]n = 25[/B] as our first solution. the second solution is 25^2 = [B]625[/B]\n\nOne third of the bagels in a bakery are sesame bagels. There are 72 sesame bagels.\nOne third of the bagels in a bakery are sesame bagels. There are 72 sesame bagels. Set up our equation where b is the number of total bagels 72 = b/3 Multiply each side by 3 [B]b = 216[/B]\n\np/q = f/q- f for f\np/q = f/q- f for f Isolate f in this literal equation. Factor out f on the right side: p/q = f(1/q - 1) Rewriting the term in parentheses, we get: p/q = f(1 - q)/q Cross multiply: f = pq/q(1 - q) Cancelling the q/q on the right side, we get: f = [B]p/(1 - q)[/B]\n\np/q=f/q-f for f\np/q=f/q-f for f To solve this literal equation for f, let's factor out f on the right side: p/q=f(1/q-1) Divide each side by (1/q - 1) p/(q(1/q - 1)) = f(1/q-1)/(1/q - 1) Cancelling the (1/q - 1) on the right side, we get: f = p/(1/q - 1) Rewriting this since (1/q -1) = (1 - q)/q since q/q = 1 we have: f = [B]pq/(1 - q)[/B]\n\nP/v=nr/t for r\nP/v=nr/t for r Cross multiply to solve this literal equation: Pt = nrv Divide each side of the equation by nv: Pt/nv = nrv/nv Cancel the nv's on the right side, we get: r = [B]Pt/nv[/B]\n\nP=15+5d/11 for d\nSubtract 15 from each side: 5d/11 = P - 15 Multiply each side by 11 5d = 11p - 165 Divide each side of the equation by d: d = (11p - 165) ------------ 5\n\nP=ab/c, for c\nP=ab/c, for c Cross multiply: cP = ab Divide each side by P [B]c = (ab)/P[/B]\n\nPentagons\nGiven a side length and an apothem, this calculates the perimeter and area of the pentagon.\n\nPerimeter of a rectangle is 372 yards. If the length is 99 yards, what is the width?\nPerimeter of a rectangle is 372 yards. If the length is 99 yards, what is the width? The perimeter P of a rectangle with length l and width w is: 2l + 2w = P We're given P = 372 and l = 99, so we have: 2(99) + 2w = 372 2w + 198 = 372 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants 198 and 372. To do that, we subtract 198 from both sides 2w + 198 - 198 = 372 - 198 [SIZE=5][B]Step 2: Cancel 198 on the left side:[/B][/SIZE] 2w = 174 [SIZE=5][B]Step 3: Divide each side of the equation by 2[/B][/SIZE] 2w/2 = 174/2 w = [B]87[/B]\n\nPet supplies makes a profit of \\$5.50 per bag, if the store wants to make a profit of no less than \\$5\nPet supplies makes a profit of \\$5.50 per bag, if the store wants to make a profit of no less than \\$5225, how many bags does it need to sell? 5.5ob >= \\$5,225 Divide each side of the inequality by \\$5.50 b >=9.5 bags, so round up to a whole number of 10 bags.\n\nPleasantburg has a population growth model of P(t)=at2+bt+P0 where P0 is the initial population. Sup\nPleasantburg has a population growth model of P(t)=at^2+bt+P0 where P0 is the initial population. Suppose that the future population of Pleasantburg t years after January 1, 2012, is described by the quadratic model P(t)=0.7t^2+6t+15,000. In what month and year will the population reach 19,200? Set P(t) = 19,200 0.7t^2+6t+15,000 = 19,200 Subtract 19,200 from each side: 0.7t^2+6t+4200 = 0 The Quadratic has irrational roots. So I set up a table below to run through the values. At t = 74, we pass 19,200. Which means we add 74 years to 2012: 2012 + 74 = [B]2086[/B] t 0.7t^2 6t Add 15000 Total 1 0.7 6 15000 15006.7 2 2.8 12 15000 15014.8 3 6.3 18 15000 15024.3 4 11.2 24 15000 15035.2 5 17.5 30 15000 15047.5 6 25.2 36 15000 15061.2 7 34.3 42 15000 15076.3 8 44.8 48 15000 15092.8 9 56.7 54 15000 15110.7 10 70 60 15000 15130 11 84.7 66 15000 15150.7 12 100.8 72 15000 15172.8 13 118.3 78 15000 15196.3 14 137.2 84 15000 15221.2 15 157.5 90 15000 15247.5 16 179.2 96 15000 15275.2 17 202.3 102 15000 15304.3 18 226.8 108 15000 15334.8 19 252.7 114 15000 15366.7 20 280 120 15000 15400 21 308.7 126 15000 15434.7 22 338.8 132 15000 15470.8 23 370.3 138 15000 15508.3 24 403.2 144 15000 15547.2 25 437.5 150 15000 15587.5 26 473.2 156 15000 15629.2 27 510.3 162 15000 15672.3 28 548.8 168 15000 15716.8 29 588.7 174 15000 15762.7 30 630 180 15000 15810 31 672.7 186 15000 15858.7 32 716.8 192 15000 15908.8 33 762.3 198 15000 15960.3 34 809.2 204 15000 16013.2 35 857.5 210 15000 16067.5 36 907.2 216 15000 16123.2 37 958.3 222 15000 16180.3 38 1010.8 228 15000 16238.8 39 1064.7 234 15000 16298.7 40 1120 240 15000 16360 41 1176.7 246 15000 16422.7 42 1234.8 252 15000 16486.8 43 1294.3 258 15000 16552.3 44 1355.2 264 15000 16619.2 45 1417.5 270 15000 16687.5 46 1481.2 276 15000 16757.2 47 1546.3 282 15000 16828.3 48 1612.8 288 15000 16900.8 49 1680.7 294 15000 16974.7 50 1750 300 15000 17050 51 1820.7 306 15000 17126.7 52 1892.8 312 15000 17204.8 53 1966.3 318 15000 17284.3 54 2041.2 324 15000 17365.2 55 2117.5 330 15000 17447.5 56 2195.2 336 15000 17531.2 57 2274.3 342 15000 17616.3 58 2354.8 348 15000 17702.8 59 2436.7 354 15000 17790.7 60 2520 360 15000 17880 61 2604.7 366 15000 17970.7 62 2690.8 372 15000 18062.8 63 2778.3 378 15000 18156.3 64 2867.2 384 15000 18251.2 65 2957.5 390 15000 18347.5 66 3049.2 396 15000 18445.2 67 3142.3 402 15000 18544.3 68 3236.8 408 15000 18644.8 69 3332.7 414 15000 18746.7 70 3430 420 15000 18850 71 3528.7 426 15000 18954.7 72 3628.8 432 15000 19060.8 73 3730.3 438 15000 19168.3 74 3833.2 444 15000 19277.2\n\nDistance = Rate x Time 6.4 meters = 4 meters/minute * t Divide each side by 4 [B]t = 1.6 minutes[/B]\n\nTime 1, distance apart is 105 + 85 = 190 So every hour, the distance between them is 190 * t where t is the number of hours. Set up our distance function: D(t) = 190t We want D(t) = 494 190t = 494 Divide each side by 190 [B]t = 2.6 hours[/B]\n\nFigure 1, we have a cone, cylinder, and cube. Let's get the volume of each Cone volume = pir^2h/3 radius = s/2 h = s Cone Volume = pi(s/2)^2(s)/3 Cone Volume = pis^3/12 Volume of cube = s^3 Volume of cylinder = pir^2h Volume of cylinder = pi(s/2)^2s Volume of cylinder = pis^3/2 But Figure 2 has no sizes? For sides, height, etc. So I cannot answer the question until I have that.\n\nplease solve the fifth word problem\nFind what was used: Used Money = Prepaid original cost - Remaining Credit Used Money = 20 - 17.47 Used Money = 2.53 Using (m) as the number of minutes, we have the following cost equation: C(m) = 0.11m C(m) = 2.53, so we have: 0.11m = 2.53 Divide each side by 0.11 [B]m = 23[/B]\n\nplease solve the third word problem\nA Web music store offers two versions of a popular song. The size of the standard version is 2.7 megabytes (MB). The size of the high-quality version is 4.7 MB. Yesterday, the high-quality version was downloaded three times as often as the standard version. The total size downloaded for the two versions was 4200 MB. How many downloads of the standard version were there? Let s be the standard version downloads and h be the high quality downloads. We have two equations: [LIST=1] [*]h = 3s [*]2.7s + 4.7h = 4200 [/LIST] Substitute (1) into (2) 2.7s + 4.7(3s) = 4200 2.7s + 14.1s = 4200 Combine like terms: 16.8s = 4200 Divide each side by 16.8 [B]s = 250[/B]\n\nPolygon Side\nDetermines the sides of a polygon given an interior angle sum.\n\nPolygons\nUsing various input scenarios of a polygon such as side length, number of sides, apothem, and radius, this calculator determines Perimeter or a polygon and Area of the polygon. This also determines interior angles of a polygon and diagonals of a polygon as well as the total number of 1 vertex diagonals.\n\npr=xf/y for r\npr=xf/y for r So for this literal equation, we divide each side of the equation by p to isolate r. pr/p = xf/yp Cancel the p's on the left side and we get: r = [B]xf/yp [MEDIA=youtube]6ekuN4H3mM4[/MEDIA][/B]\n\nPut the number 123456789 exactly ones in the bubble so that each edge adds up to say number\nPut the number 123456789 exactly ones in the bubble so that each edge adds up to say number [B] Each side adds up to 17 [IMG]https://www.mathcelebrity.com/images/triangle_sum_17.png[/IMG] [/B]\n\nPythagorean Theorem\nFigures out based on user entry the missing side or missing hypotenuse of a right triangle. In addition, the calculator shows the proof of the Pythagorean Theorem and then determines by numerical evaluation if the 2 sides and hypotenuse you entered are a right triangle using the Pythagorean Theorem\n\nQ is a point on segment PR. If PQ = 2.7 and PR = 6.1, what is QR?\nQ is a point on segment PR. If PQ = 2.7 and PR = 6.1, what is QR? From segment addition, we know that: PQ + QR = PR Plugging our given numbers in, we get: 2.7 + QR = 6.1 Subtract 2.7 from each side, and we get: 2.7 - 2.7 + QR = 6.1 - 2.7 Cancelling the 2.7 on the left side, we get: QR = [B]3.4[/B]\n\nq=c+d/5 for d\nq=c+d/5 for d Subtract c from each side to solve this literal equation: q - c = c - c + d/5 Cancel the c's on the right side, we get d/5 = q - c Multiply each side by 5: 5d/5 = 5(q - c) Cancel the 5's on the left side, we get: [B]d = 5(q - c)[/B]\n\nq=rs/2-p;p\nq=rs/2-p;p Add p to each side: q + p = rs/2 Subtract q from each side: [B]p = rs/2 - q[/B]\n\nr=l^2w/2 for w\nr=l^2w/2 for w Solve this literal equation by isolating w. Cross multiply: 2r = l^2w Divide each side by l^2 w = [B]2r/l^2[/B]\n\nRachel buys some scarves that cost \\$10 each and 2 purses that cost \\$16 each. The cost of Rachel's to\nRachel buys some scarves that cost \\$10 each and 2 purses that cost \\$16 each. The cost of Rachel's total purchase is \\$62. What equation can be used to find n, the number of scarves that Rebecca buys Scarves Cost + Purses Cost = Total Cost [U]Calculate Scarves Cost[/U] Scarves cost = Cost per scarf * number of scarves Scarves cost = 10n [U]Calculate Purses Cost[/U] Purses cost = Cost per purse * number of purses Purses cost = 16 * 2 Purses cost = 32 Total Cost = 62. Plug in our numbers and values to the Total Cost equation : 10n + 32 = 62 Solve for [I]n[/I] in the equation 10n + 32 = 62 [SIZE=5][B]Step 1: Group constants:[/B][/SIZE] We need to group our constants 32 and 62. To do that, we subtract 32 from both sides 10n + 32 - 32 = 62 - 32 [SIZE=5][B]Step 2: Cancel 32 on the left side:[/B][/SIZE] 10n = 30 [SIZE=5][B]Step 3: Divide each side of the equation by 10[/B][/SIZE] 10n/10 = 30/10 n = [B]3[/B]\n\nRearrange the following equation to make x the subject, and select the correct rearrangement from th\nRearrange the following equation to make x the subject, and select the correct rearrangement from the list below 3x + 2y 1 -------- = --- 4x + y 3 [LIST] [*]x = 7y/13 [*]x = 7y/5 [*]x = -7y [*]x = -3y [*]x = 3y/5 [*]x = -5y/13 [*]x = -y [/LIST] Cross multiply: 3(3x - 2y) = 4x + y Multiply the left side through 9x - 6y = 4x + y Subtract 4x from each side and add 6y to each side 5x = 7y Divide each side by 5 to isolate x, the subject of an equation is the variable to the left [B]x = 7y/5[/B]\n\nrectangle abcd prove: triangle adc is congruent to triangle bcd\nrectangle abcd prove: triangle adc is congruent to triangle bcd 1. Given: ABCD is a rectangle 2. AB = CD since opposite sides of rectangle are congruent 3. BC = AD since opposite sides of rectangle are congruent 4. AC = AC by the Reflexive Property of Equality 5. triangle ADC = triangle CBA by the Side-Side-Side (SSS) Property\n\nRectangle Word Problem\nSolves word problems based on area or perimeter and variable side lengths\n\nRhombus\nGiven inputs of a rhombus, this calculates the following:\nPerimeter of a Rhombus\nArea of a Rhombus\nSide of a Rhombus\n\nRicks age increased by 24 is 69\nLet a be Rick's age We have a + 24 = 69 Subtract 24 from each side [B]a = 45[/B]\n\nRobert has 45 dollars. He buys 6 tshirts and has 7 dollars left over. How much did each tshirt cost?\nLet x be the price of one t-shirt. Set up an equation: 6 times the number of t-shirts plus 7 dollars left over get him to a total of 45 6x = 45 - 7 6x = 38 Divide each side by 6 [B]x = 6.33[/B]\n\nrs+h^2=1 for h\nrs+h^2=1 for h Subtract rs from each side to isolate h: rs - rs + h^2 = 1 - rs Cancel the rs on the left side: h^2 = 1 - rs Take the square root of each side: sqrt(h^2) = sqrt(1 - rs) [B]h = +- sqrt(1 -rs)[/B]\n\nRunning from the top of a flagpole to a hook in the ground there is a rope that is 9 meters long. If\nRunning from the top of a flagpole to a hook in the ground there is a rope that is 9 meters long. If the hook is 4 meters from the base of the flagpole, how tall is the flagpole? We have a right triangle, with hypotenuse of 9 and side of 4. [URL='https://www.mathcelebrity.com/pythag.php?side1input=&side2input=4&hypinput=9&pl=Solve+Missing+Side']Using our Pythagorean Theorem calculator[/URL], we get a flagpole height of [B]8.063[/B].\n\ns = tu^2 for u\ns = tu^2 for u Divide each side by t u^2 = s/t Take the square root of each side [LIST] [*]u = sqrt(s/t) [*]u = -sqrt(s/t) [/LIST] We have two answers due to negative number squared is positive\n\ns=u^2t for t\ns=u^2t for t Divide each side by u^2 to isolate t: u^2t/u^2 = s/u^2 Cancel the u^2 on the left side, we get: t = [B]s/u^2[/B]\n\ns=w-10e/m for w\ns=w-10e/m for w Add 10e/m to each side to isolate w: s + 10e/m = w - 10e/m + 10e/m Cancel the 10e/m on the right side, and we get: w = [B]s + 10e/m[/B]\n\nSally and Adam works a different job. Sally makes \\$5 per hour and Adam makes \\$4 per hour. They each\nSally and Adam works a different job. Sally makes \\$5 per hour and Adam makes \\$4 per hour. They each earn the same amount per week but Adam works 2 more hours. How many hours a week does Adam work? [LIST] [*]Let [I]s[/I] be the number of hours Sally works every week. [*]Let [I]a[/I] be the number of hours Adam works every week. [*]We are given: a = s + 2 [/LIST] Sally's weekly earnings: 5s Adam's weekly earnings: 4a Since they both earn the same amount each week, we set Sally's earnings equal to Adam's earnings: 5s = 4a But remember, we're given a = s + 2, so we substitute this into Adam's earnings: 5s = 4(s + 2) Multiply through on the right side: 5s = 4s + 8 <-- [URL='https://www.mathcelebrity.com/expand.php?term1=4%28s%2B2%29&pl=Expand']multiplying 4(s + 2)[/URL] [URL='https://www.mathcelebrity.com/1unk.php?num=5s%3D4s%2B8&pl=Solve']Typing this equation into the search engine[/URL], we get s = 8. The problem asks for Adam's earnings (a). We plug s = 8 into Adam's weekly hours: a = s + 2 a = 8 + 2 [B]a = 10[/B]\n\nSalma purchased a prepaid phone card for 30. Long distance calls cost 9 cents a minute using this ca\nSalma purchased a prepaid phone card for 30. Long distance calls cost 9 cents a minute using this card. Salma used her card only once to make a long distance call. If the remaining credit on her card is 28.38, how many minutes did her call last? [U]Set up the equation where m is the number of minutes used:[/U] 0.09m = 30 - 28.38 0.09m = 1.62 [U]Divide each side by 0.09[/U] [B]m = 18[/B]\n\nSam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which\nSam and Jeremy have ages that are consecutive odd integers. The product of their ages is 783. Which equation could be used to find Jeremy's age, j, if he is the younger man. Let Sam's age be s. Let' Jeremy's age be j. We're given: [LIST=1] [*]s = j + 2 <-- consecutive odd integers [*]sj = 783 [/LIST] Substitute (1) into (2): (j + 2)j = 783 j^2 + 2j = 783 Subtract 783 from each side: j^2 + 2j - 783 = 0 <-- This is the equation to find Jeremy's age. To solve this, [URL='https://www.mathcelebrity.com/quadratic.php?num=j%5E2%2B2j-783%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']we type this quadratic equation into the search engine[/URL] and get: j = 27, j = -29. Since ages cannot be negative, we have: [B]j = 27[/B]\n\nSam leaves school to go home. He walks 10 blocks North and then 8 blocks west. How far is John from\nSam leaves school to go home. He walks 10 blocks North and then 8 blocks west. How far is John from the school? Sam walked at a right angle. His distance from home to school is the hypotenuse. Using our [URL='https://www.mathcelebrity.com/pythag.php?side1input=8&side2input=10&hypinput=&pl=Solve+Missing+Side']Pythagorean theorem calculator[/URL], we get: [B]12.806 blocks[/B]\n\nShalini gave 0.4 of her plums to her brother and 20% to her sister. She kept 16 for herself how many\nShalini gave 0.4 of her plums to her brother and 20% to her sister. She kept 16 for herself how many plums did she have at first? Let p be the number of plums Shalini started with. We have: [LIST] [*]0.4 given to her brother [*]20% which is 0.2 given away to her sister [*]What this means is she kept 1 - (0.4 + 0.2) = 1 - 0.6 = 0.4 for herself [/LIST] 0.4p = 16 Divide each side by 0.4 [B]p = 40[/B]\n\nShalini gave 0.4 of her plums to her brother and 20% to her sister. She kept 16 for herself. How man\nShalini gave 0.4 of her plums to her brother and 20% to her sister. She kept 16 for herself. How many plums did she have first? Let's convert everything to decimals. 20% = 0.2 So Shalini gave 0.4 + 0.2 = 0.6 of the plums away. Which means she has 1 = 0.6 = 0.4 or 40% left over. 40% represents 16 plums So our equation is 0.4p = 16 where p is the number of plums to start with Divide each side by 0.4 [B]p = 40[/B]\n\nShe earns \\$20 per hour as a carpenter and \\$25 per hour as a blacksmith, last week Giselle worked bot\nShe earns \\$20 per hour as a carpenter and \\$25 per hour as a blacksmith, last week Giselle worked both jobs for a total of 30 hours, and a total of \\$690. How long did Giselle work as a carpenter and how long did she work as a blacksmith? Assumptions: [LIST] [*]Let b be the number of hours Giselle worked as a blacksmith [*]Let c be the number of hours Giselle worked as a carpenter [/LIST] Givens: [LIST=1] [*]b + c = 30 [*]25b + 20c = 690 [/LIST] Rearrange equation (1) to solve for b by subtracting c from each side: [LIST=1] [*]b = 30 - c [*]25b + 20c = 690 [/LIST] Substitute equation (1) into equation (2) for b 25(30 - c) + 20c = 690 Multiply through: 750 - 25c + 20c = 690 To solve for c, we [URL='https://www.mathcelebrity.com/1unk.php?num=750-25c%2B20c%3D690&pl=Solve']type this equation into our search engine[/URL] and we get: c = [B]12 [/B] Now, we plug in c = 12 into modified equation (1) to solve for b: b = 30 - 12 b = [B]18[/B]\n\nShe ordered 6 large pizzas. Luckily, she had a coupon for 3 off each pizza. If the bill came to 38.9\nShe ordered 6 large pizzas. Luckily, she had a coupon for 3 off each pizza. If the bill came to 38.94, what was the price for a large pizza? [U]Determine additional amount the pizzas would have cost without the coupon[/U] 6 pizzas * 3 per pizza = 18 [U]Add 18 to our discount price of 38.94[/U] Full price for 6 large pizzas = 38.94 + 18 Full price for 6 large pizzas = 56.94 Let x = full price per pizza before the discount. Set up our equation: 6x = 56.94 Divide each side by 6 [B]x = \\$9.49[/B]\n\nSheila wants build a rectangular play space for her dog. She has 100 feet of fencing and she wants i\nSheila wants build a rectangular play space for her dog. She has 100 feet of fencing and she wants it to be 5 times as long as it is wide. What dimensions should the play area be? Sheila wants: [LIST=1] [*]l =5w [*]2l + 2w = 100 <-- Perimeter [/LIST] Substitute (1) into (2) 2(5w) + 2w = 100 10w + 2w = 100 12w = 100 Divide each side by 12 [B]w = 8.3333[/B] Which means l = 5(8.3333) -->[B] l = 41.6667[/B]\n\nSolve 11 - 1/2y = 3 + 6x for y\nSolve 11 - 1/2y = 3 + 6x for y Subtract 11 from each side so we can isolate the y term: 11 -11 - 1/2y = 3 + 6x - 11 Cancelling the 11's on the left side, we get: -1/2y = 6x - 8 <-- Since 3 - 11 = -8 Multiply both sides of the equation by -2 to remove the -1/2 on the left side: -2(-1/2)y = -2(6x - 8) Simplifying, we get: y = [B]-12x + 16[/B]\n\nSolve a= (a + b + c + d)/4 for c\nSolve a= (a + b + c + d)/4 for c Cross multiply: 4a = a + b + c + d Subtract a + b+ d from each side to isolate c: 4a - a - b - d = a + b + c + d - a - b - d Canceling the a, b, and d from the right side, we get: c = [B]3a - b - d [/B]\n\nSolve for h. rs + h^2 = l\nSolve for h. rs + h^2 = l [U]Subtract rs from each side to isolate h:[/U] rs - rs + h^2 = l - rs [U]Cancel the rs terms on the left side, and we get:[/U] h^2 = l - rs [U]Take the square root of each side:[/U] h = [B]sqrt(l - rs)[/B]\n\nSolve for x\nExpand the right side: 1/3x + 1/2 = 6/4x - 10 Simplify as 6/4 is 3/2 x/3 + 1/2 = 3x/2 - 10 Common denominator of 2 and 3 is 6. So we have: 2x/6 + 1/2 = 9x/6 -10 Subtract 2x/6 from each side 7x/6 - 10 = 1/2 Add 10 to each side. 10 is 20/2 7x/6 = 21/2 Using our [URL='http://www.mathcelebrity.com/prop.php?num1=7x&num2=21&den1=6&den2=2&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL], we get: [B]x = 9[/B]\n\nSolve mgh=1/2mv^2+1/2(2/5)mr^2(v^2/r^2) for v\nSolve mgh=1/2mv^2+1/2(2/5)mr^2(v^2/r^2) for v 1/2(2/5) = 1/5 since the 2's cancel r^2/r^2 = 1 So we simplify, and get: mgh=1/2mv^2+1/5(mv^2) for v Divide each side by m, so m's cancel in each term on the left and right side: gh = 1/2v^2 + 1/5(v^2) Combine like terms for v^2 on the right side: 1/2 + 1/5 = 7/10 from our [URL='https://www.mathcelebrity.com/fraction.php?frac1=1%2F2&frac2=1%2F5&pl=Add']fraction calculator[/URL] So we have: gh = 7v^2/10 Multiply each side by 10: 10gh = 7v^2 Now divide each side by 7 10gh/7 = v^2 Take the square root of each side: [B]v = sqrt(10gh/7)[/B]\n\nSophie and Claire are having a foot race. Claire is given a 100-foot head-start. If Sophie is runn\nSophie and Claire are having a foot race. Claire is given a 100-foot head-start. If Sophie is running at 5 feet per second and Claire is running at 3 feet per second. i. After how many seconds will Sophie catch Claire? ii. If the race is 500 feet, who wins? i. Sophie's distance formula is given as D = 5s Claire's distance formula is given as D = 3s + 100 Set them equal to each other 5s = 3s + 100 Subtract 3s from both sides: 2s = 100 Divide each side by 2 [B]s = 50[/B] ii. [B]Sophie since after 50 seconds, she takes the lead and never gives it back.[/B]\n\nSpecial Triangles: Isosceles and 30-60-90\nGiven an Isosceles triangle (45-45-90) or 30-60-90 right triangle, the calculator will solve the 2 remaining sides of the triangle given one side entered.\n\nSportStation.Store - #1 Sports Equipment Online Store!\n\nSquares\nSolve for Area of a square, Perimeter of a square, side of a square, diagonal of a square.\n\nSquaring a number equals 5 times that number\nSquaring a number equals 5 times that number. The phrase [I]a number[/I] means an arbitrary variable, let's call it x. Squaring this number: x^2 5 times this number means we multiply by 5: 5x The phrase [I]equals[/I] means we set both expressions equal to each other: [B]x^2 = 5x [/B] <-- This is our algebraic expression If you want to solve for x, then we subtract 5x from each side: x^2 - 5x = 5x - 5x Cancel the 5x on the right side, leaving us with 0: x^2 - 5x = 0 Factor out x: x(x - 5) So we get x = 0 or [B]x = 5[/B]\n\nStacy sells art prints for \\$12 each. Her expenses are \\$2.50 per print, plus \\$38 for equipment. How m\nStacy sells art prints for \\$12 each. Her expenses are \\$2.50 per print, plus \\$38 for equipment. How many prints must she sell for her revenue to equal her expenses? Let the art prints be p Cost function is 38 + 2p Revenue function is 12p Set cost equal to revenue 12p = 38 + 2p Subtract 2p from each side 10p = 38 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=10p%3D38&pl=Solve']equation calculator[/URL] gives us [B]p = 3.8[/B]\n\nstandard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and\nStandard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: [B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]\n\nSteve woke up and it was -12 Fahrenheit outside the weatherman said it was supposed to warm up to 20\nSteve woke up and it was -12 Fahrenheit outside the weatherman said it was supposed to warm up to 20 degrees. how many degrees will the temperature increase We start with a temperature of -12F Warming up means we [U][B]add[/B][/U] degrees to the original temperature. -12 + 20 = [B]+8F[/B]\n\nSteven has some money. If he spends \\$9, then he will have 3/5 of the amount he started with.\nSteven has some money. If he spends \\$9, then he will have 3/5 of the amount he started with. Let the amount Steven started with be s. We're given: s - 9 = 3s/5 Multiply each side through by 5 to eliminate the fraction: 5(s - 9) = 5(3s/5) Cancel the 5's on the right side and we get: 5s - 45 = 3s To solve for s, we [URL='https://www.mathcelebrity.com/1unk.php?num=5s-45%3D3s&pl=Solve']type this equation into our search engine[/URL] and we get: s = [B]22.5[/B]\n\nSum of a number and it's reciprocal is 6. What is the number?\nSum of a number and it's reciprocal is 6. What is the number? Let the number be n. The reciprocal is 1/n. The word [I]is[/I] means an equation, so we set n + 1/n equal to 6 n + 1/n = 6 Multiply each side by n to remove the fraction: n^2 + 1 = 6n Subtract 6n from each side: [B]n^2 - 6n + 1 = 0 [/B]<-- This is our algebraic expression If the problem asks you to solve for n, then you [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-6n%2B1%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']type this quadratic equation into our search engine[/URL].\n\nSuper Snack, a convenience store, charges \\$4.35 for a large chicken sandwich and two large colas. Fo\nSuper Snack, a convenience store, charges \\$4.35 for a large chicken sandwich and two large colas. For a large chicken sandwich and a large cola, they charge \\$4.00. How much are the Super Snack large chicken sandwiches? The difference between the orders is \\$0.35 and 1 large cola. Therefore, 1 large cola = \\$0.35. And if we use the first order of one large chicken sandwich and one large cola, we get: Large Chicken Sandwich + 0.35 = 4.35 Subtract 0.35 from each side, and we get: Large Chicken Sandwich = \\$[B]4.00[/B]\n\nSuppose Briley has 10 coins in quarters and dimes and has a total of 1.45. How many of each coin doe\nSuppose Briley has 10 coins in quarters and dimes and has a total of 1.45. How many of each coin does she have? Set up two equations where d is the number of dimes and q is the number of quarters: (1) d + q = 10 (2) 0.1d + 0.25q = 1.45 Rearrange (1) into (3) to solve for d (3) d = 10 - q Now plug (3) into (2) 0.1(10 - q) + 0.25q = 1.45 Multiply through: 1 - 0.1q + 0.25q = 1.45 Combine q terms 0.15q + 1 = 1.45 Subtract 1 from each side 0.15q = 0.45 Divide each side by 0.15 [B]q = 3[/B] Plug our q = 3 value into (3) d = 10 - 3 [B]d = 7[/B]\n\nSuppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in ga\nSuppose that the weight (in pounds) of an airplane is a linear function of the amount of fuel (in gallons) in its tank. When carrying 20 gallons of fuel, the airplane weighs 2012 pounds. When carrying 55 gallons of fuel, it weighs 2208 pounds. How much does the airplane weigh if it is carrying 65 gallons of fuel? Linear functions are written in the form of one dependent variable and one independent variable. Using g as the number of gallons and W(g) as the weight, we have: W(g) = gx + c where c is a constant We are given: [LIST] [*]W(20) = 2012 [*]W(55) = 2208 [/LIST] We want to know W(65) Using our givens, we have: W(20) = 20x + c = 2012 W(55) = 55x + c = 2208 Rearranging both equations, we have: c = 2012 - 20x c = 2208 - 55x Set them both equal to each other: 2012 - 20x = 2208 - 55x Add 55x to each side: 35x + 2012 = 2208 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=35x%2B2012%3D2208&pl=Solve']equation solver[/URL], we see that x is 5.6 Plugging x = 5.6 back into the first equation, we get: c = 2012 - 20(5.6) c = 2012 - 112 c = 2900 Now that we have all our pieces, find W(65) W(65) = 65(5.6) + 2900 W(65) = 264 + 2900 W(65) = [B]3264[/B]\n\nT = mg - mf for f\nT = mg - mf for f Subtract mg from each side: T - mg = mg - mg - mf Cancel the mg on the right side and we get: T - mg = -mf Multiply each side by -1: -(T - mg) = -(-mf) mg - T = mf Now Divide each side by m to isolate f: (mg - T)/m = mf/m Cancel the m on the right side and we get: f = [B](mg - T)/m[/B]\n\ntammy earns \\$18000 salary with 4% comission on sales. How much should she sell to earn \\$55,000 total\ntammy earns \\$18000 salary with 4% comission on sales. How much should she sell to earn \\$55,000 total We have a commission equation below: Sales * Commission percent = Salary We're given 4% commission percent and 55,000 salary. With 4% as 0.04, we have: Sales * 0.04 = 55,000 Divide each side of the equation by 0.04, and we get: Sales = [B]1,375,000[/B]\n\nThe average of a number and double the number is 25.5\nLet x equal \"a number\". Double the number is 2x. The average is (x + 2x)/2 Combine the terms in the numerator: 3x/2 The word [I]is[/I] means equal to, so we set 3x/2 equal to 25.5 3x/2 = 25.5 Cross multiply the 2: 3x = 51 Divide each side by 3 [B]x = 17[/B]\n\nThe base of a triangle with a height of 7 units is represented by the formula b=2/7A. The base of th\nThe base of a triangle with a height of 7 units is represented by the formula b=2/7A. The base of the triangle is less than 10 units. Write and solve an inequality that represents the possible area A of the triangle We're given: b=2/7A We're also told that b is less than 10. So we have: 2/7A < 10 2A/7 < 10 Cross multiply: 2A < 7 * 10 2A < 70 Divide each side of the inequality by 2 to isolate A 2A/2 < 70/2 Cancel the 2's on the left side and we get: A < [B]35[/B]\n\nThe club uses the function S(t) = -4,500t + 54,000 to determine the salvage S(t) of a fertilizer ble\nThe club uses the function S(t) = -4,500t + 54,000 to determine the salvage S(t) of a fertilizer blender t years after its purchase. How long will it take the blender to depreciate completely? Complete depreciation means the salvage value is 0. So S(t) = 0. We need to find t to make S(t) = 0 -4,500t + 54,000 = 0 Subtract 54,000 from each side -4,500t = -54,000 Divide each side by -4,500 [B]t = 12[/B]\n\nThe cost of renting a rototiller is \\$19.50 for the first hour and \\$7.95 for each additional hour. Ho\nThe cost of renting a rototiller is \\$19.50 for the first hour and \\$7.95 for each additional hour. How long can a person have the rototiller if the cost must be less than \\$95? Setup the inequality: \\$19.50 + \\$7.95x < \\$95 Subtract 19.50 from both sides: 7.95x < 75.50 Divide each side of the inequality by 7.95 to isolate x x < 9.5 The next lowest integer is 9. So we take 9 + the first hour of renting to get [B]10 total hours[/B]. Check our work: \\$7.95 * 9.5 + \\$19.50 \\$71.55 + \\$19.50 = \\$91.05\n\nThe difference between 2 numbers is 108. 6 times the smaller is equal to 2 more than the larger. Wh?\nThe difference between 2 numbers is 108. 6 times the smaller is equal to 2 more than the larger. What are the numbers? Let the smaller number be x. Let the larger number be y. We're given: [LIST=1] [*]y - x = 108 [*]6x = y + 2 [/LIST] Rearrange (1) by adding x to each side: [LIST=1] [*]y = x + 108 [/LIST] Substitute this into (2): 6x = x + 108 + 2 Combine like terms 6x = x +110 Subtract x from each side: 5x = 110 [URL='https://www.mathcelebrity.com/1unk.php?num=5x%3D110&pl=Solve']Plugging this equation into our search engine[/URL], we get: x = [B]22[/B]\n\nThe difference between two numbers is 25. The smaller number is 1/6th of the larger number. What is\nThe difference between two numbers is 25. The smaller number is 1/6th of the larger number. What is the value of the smaller number Let the smaller number be s. Let the larger number be l. We're given two equations: [LIST=1] [*]l - s = 25 [*]s = l/6 [/LIST] Plug in equation (2) into equation (1): l - l/6 = 25 Multiply each side of the equation by 6 to remove the fraction: 6l - l = 150 To solve for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=6l-l%3D150&pl=Solve']type this equation into our search engine[/URL] and we get: l = 30 To solve for s, we plug in l = 30 into equation (2) above: s = 30/6 [B]s = 5[/B]\n\nThe difference between two positive numbers is 5 and the square of their sum is 169\nThe difference between two positive numbers is 5 and the square of their sum is 169. Let the two positive numbers be a and b. We have the following equations: [LIST=1] [*]a - b = 5 [*](a + b)^2 = 169 [*]Rearrange (1) by adding b to each side. We have a = b + 5 [/LIST] Now substitute (3) into (2): (b + 5 + b)^2 = 169 (2b + 5)^2 = 169 [URL='https://www.mathcelebrity.com/community/forums/calculator-requests.7/create-thread']Run (2b + 5)^2 through our search engine[/URL], and you get: 4b^2 + 20b + 25 Set this equal to 169 above: 4b^2 + 20b + 25 = 169 [URL='https://www.mathcelebrity.com/quadratic.php?num=4b%5E2%2B20b%2B25%3D169&pl=Solve+Quadratic+Equation&hintnum=+0']Run that quadratic equation in our search engine[/URL], and you get: b = (-9, 4) But the problem asks for [I]positive[/I] numbers. So [B]b = 4[/B] is one of our solutions. Substitute b = 4 into equation (1) above, and we get: a - [I]b[/I] = 5 [URL='https://www.mathcelebrity.com/1unk.php?num=a-4%3D5&pl=Solve']a - 4 = 5[/URL] [B]a = 9 [/B] Therefore, we have [B](a, b) = (9, 4)[/B]\n\nThe difference of 2 positive numbers is 54. The quotient obtained on dividing the 1 by the other is\nThe difference of 2 positive numbers is 54. The quotient obtained on dividing the 1 by the other is 4. Find the numbers. Let the numbers be x and y. We have: [LIST] [*]x - y = 54 [*]x/y = 4 [*]Cross multiply x/y = 4 to get x = 4y [*]Now substitute x = 4y into the first equation [*](4y) - y = 54 [*]3y = 54 [*]Divide each side by 3 [*][B]y = 18[/B] [*]If x = 4y, then x = 4(18) [*][B]x = 72[/B] [/LIST]\n\nThe difference of two numbers is 12 and their mean is 15. Find the two numbers\nThe difference of two numbers is 12 and their mean is 15. Find the two numbers. Let the two numbers be x and y. We're given: [LIST=1] [*]x - y = 12 [*](x + y)/2 = 15. <-- Mean is an average [/LIST] Rearrange equation 1 by adding y to each side: x - y + y = y + 12 Cancelling the y's on the left side, we get: x = y + 12 Now substitute this into equation 2: (y + 12 + y)/2 = 15 Cross multiply: y + 12 + y = 30 Group like terms for y: 2y + 12 = 30 [URL='https://www.mathcelebrity.com/1unk.php?num=2y%2B12%3D30&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]y = 9[/B] Now substitute this into modified equation 1: x = y + 12 x = 9 + 12 [B]x = 21[/B]\n\nThe distance between consecutive bases is 90 feet. An outfielder catches the ball on the third base\nThe distance between consecutive bases is 90 feet. An outfielder catches the ball on the third base line about 40 feet behind third base. How far would the outfielder have to throw the ball to first base? We have a right triangle. From home base to third base is 90 feet. We add another 40 feet to the outfielder behind third base to get: 90 + 40 = 130 The distance from home to first is 90 feet. Our hypotenuse is the distance from the outfielder to first base. [URL='https://www.mathcelebrity.com/pythag.php?side1input=130&side2input=90&hypinput=&pl=Solve+Missing+Side']Using our Pythagorean theorem calculator[/URL], we get: d = [B]158.11 feet[/B]\n\nThe fraction has a value of 3/5. The sum of the numerator and the denominator was 40. What was the f\nThe fraction has a value of 3/5. The sum of the numerator and the denominator was 40. What was the fraction? We're given two equations with a fraction with numerator (n) and denominator (d): [LIST=1] [*]n + d = 40 [*]n/d = 3/5 [/LIST] Cross multiply equation 2, we get: 5n = 3d Divide each side by 5: 5n/5 = 3d/5 n = 3d/5 Substitute this into equation 1: 3d/5 + d = 40 Multiply through both sides of the equation by 5: 5(3d/5) = 5d = 40 * 5 3d + 5d =200 To solve this equation for d, we [URL='https://www.mathcelebrity.com/1unk.php?num=3d%2B5d%3D200&pl=Solve']type it in our search engine and we get[/URL]: d = [B]25 [/B] Now substitute that back into equation 1: n + 25 = 40 Using [URL='https://www.mathcelebrity.com/1unk.php?num=n%2B25%3D40&pl=Solve']our equation solver again[/URL], we get: n = [B]15[/B]\n\nThe function f(x) = x^3 - 48x has a local minimum at x = and a local maximum at x = ?\nThe function f(x) = x^3 - 48x has a local minimum at x = and a local maximum at x = ? f'(x) = 3x^2 - 48 Set this equal to 0: 3x^2 - 48 = 0 Add 48 to each side: 3x^2 = 48 Divide each side by 3: x^2 = 16 Therefore, x = -4, 4 Test f(4) f(4) = 4^3 - 48(4) f(4) = 64 - 192 f(4) = [B]-128 <-- Local minimum[/B] Test f(-4) f(-4) = -4^3 - 48(-4) f(-4) = -64 + 192 f(-4) = [B]128 <-- Local maximum[/B]\n\nThe function P(x) = -30x^2 + 360x + 785 models the profit, P(x), earned by a theatre owner on the ba\nThe function P(x) = -30x^2 + 360x + 785 models the profit, P(x), earned by a theatre owner on the basis of a ticket price, x. Both the profit and the ticket price are in dollars. What is the maximum profit, and how much should the tickets cost? Take the [URL='http://www.mathcelebrity.com/dfii.php?term1=-30x%5E2+%2B+360x+%2B+785&fpt=0&ptarget1=0&ptarget2=0&itarget=0%2C1&starget=0%2C1&nsimp=8&pl=1st+Derivative']derivative of the profit function[/URL]: P'(x) = -60x + 360 We find the maximum when we set the profit derivative equal to 0 -60x + 360 = 0 Subtract 360 from both sides: -60x = -360 Divide each side by -60 [B]x = 6 <-- This is the ticket price to maximize profit[/B] Substitute x = 6 into the profit equation: P(6) = -30(6)^2 + 360(6) + 785 P(6) = -1080 + 2160 + 785 [B]P(6) = 1865[/B]\n\nThe graph of a polynomial f(x) = (2x - 3)(x - 4)(x + 3) has x-intercepts at 3 values. What are they?\nThe graph of a polynomial f(x) = (2x - 3)(x - 4)(x + 3) has x-intercepts at 3 values. What are they? A few things to note: [LIST] [*]X-intercepts are found when y (or f(x)) is 0. [*]On the right side, we have 3 monomials. [*]Therefore, y or f(x) could be 0 when [U]any[/U] of these monomials is 0 [/LIST] The 3 monomials are: [LIST=1] [*]2x - 3 = 0 [*]x - 4 = 0 [*]x + 3 = 0 [/LIST] Find all 3 x-intercepts: [LIST=1] [*]2x - 3 = 0. [URL='https://www.mathcelebrity.com/1unk.php?num=2x-3%3D0&pl=Solve']Using our equation calculator[/URL], we see that x = [B]3/2 or 1.5[/B] [*]x - 4 = 0 [URL='https://www.mathcelebrity.com/1unk.php?num=x-4%3D0&pl=Solve']Using our equation calculator[/URL], we see that x = [B]4[/B] [*]x + 3 = 0 [URL='https://www.mathcelebrity.com/1unk.php?num=x%2B3%3D0&pl=Solve']Using our equation calculator[/URL], we see that x = [B]-3[/B] [/LIST] So our 3 x-intercepts are: x = [B]{-3, 3/2, 4}[/B]\n\nThe height of an object t seconds after it is dropped from a height of 300 meters is s(t)=-4.9t^2 +3\nThe height of an object t seconds after it is dropped from a height of 300 meters is s(t)=-4.9t^2 +300. Find the average velocity of the object during the first 3 seconds? (b) Use the Mean value Theorem to verify that at some time during the first 3 seconds of the fall the instantaneous velocity equals the average velocity. Find that time. Average Velocity: [ f(3) - f(0) ] / ( 3 - 0 ) Calculate f(3): f(3) = -4.9(3^2) + 300 f(3) = -4.9(9) + 300 f(3) = -44.1 + 300 f(3) = 255.9 Calculate f(0): f(0) = -4.9(0^2) + 300 f(0) = 0 + 300 f(0) = 300 So we have average velocity: Average velocity = (255.9 - 300)/(3 - 0) Average velocity = -44.1/3 Average velocity = -[B]14.7 [/B] Velocity is the first derivative of position s(t)=-4.9t^2 +300 s'(t) = -9.8t So we set velocity equal to average velocity: -9.8t = -14.7 Divide each side by -9.8 to solve for t, we get [B]t = 1.5[/B]\n\nThe length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 fee\nThe length of a rectangular building is 6 feet less than 3 times the width. The perimeter is 120 feet. Find the width and length of the building. P = 2l + 2w Since P = 120, we have: (1) 2l + 2w = 120 We are also given: (2) l = 3w - 6 Substitute equation (2) into equation (1) 2(3w - 6) + 2w = 120 Multiply through: 6w - 12 + 2w = 120 Combine like terms: 8w - 12 = 120 Add 12 to each side: 8w = 132 Divide each side by 8 to isolate w: w =16.5 Now substitute w into equation (2) l = 3(16.5) - 6 l = 49.5 - 6 l = 43.5 So (l, w) = (43.5, 16.5)\n\nThe length of the flag is 2 cm less than 7 times the width. The perimeter is 60cm. Find the length a\nThe length of the flag is 2 cm less than 7 times the width. The perimeter is 60cm. Find the length and width. A flag is a rectangle shape. So we have the following equations Since P = 2l + 2w, we have 2l + 2w = 60 l = 7w - 2 Substitute Equation 1 into Equation 2: 2(7w -2) + 2w = 60 14w - 4 + 2w = 60 16w - 4 = 60 Add 4 to each side 16w = 64 Divide each side by 16 to isolate w w = 4 Which means l = 7(4) - 2 = 28 - 2 = 26\n\nThe perimeter of a rectangle is 400 meters. The length is 15 meters less than 4 times the width. Fin\nThe perimeter of a rectangle is 400 meters. The length is 15 meters less than 4 times the width. Find the length and the width of the rectangle. l = 4w - 15 Perimeter = 2l + 2w Substitute, we get: 400 = 2(4w - 15) + 2w 400 = 8w - 30 + 2w 10w - 30 = 400 Add 30 to each side 10w = 370 Divide each side by 10 to isolate w w = 37 Plug that back into our original equation to find l l = 4(37) - 15 l = 148 - 15 l = 133 So we have (l, w) = (37, 133)\n\nThe perimeter of a square with side a\nThe perimeter of a square with side a Perimeter of a square is 4s where s is the side length. With s = a, we have: P = [B]4a[/B]\n\nThe perpendicular height of a right-angled triangle is 70 mm longer than the base. Find the perimete\nThe perpendicular height of a right-angled triangle is 70 mm longer than the base. Find the perimeter of the triangle if its area is 3000. [LIST] [*]h = b + 70 [*]A = 1/2bh = 3000 [/LIST] Substitute the height equation into the area equation 1/2b(b + 70) = 3000 Multiply each side by 2 b^2 + 70b = 6000 Subtract 6000 from each side: b^2 + 70b - 6000 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=b%5E2%2B70b-6000%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: b = 50 and b = -120 Since the base cannot be negative, we use b = 50. If b = 50, then h = 50 + 70 = 120 The perimeter is b + h + hypotenuse Using the [URL='http://www.mathcelebrity.com/righttriangle.php?angle_a=&a=70&angle_b=&b=50&c=&pl=Calculate+Right+Triangle']right-triangle calculator[/URL], we get hypotenuse = 86.02 Adding up all 3 for the perimeter: 50 + 70 + 86.02 = [B]206.02[/B]\n\nThe points -5, -24 and 5,r lie on a line with slope 4. Find the missing coordinate r. Slope = (y2 -\nThe points -5, -24 and 5,r lie on a line with slope 4. Find the missing coordinate r. Slope = (y2 - y1)/(x2 - x1) Plugging in our numbers, we get: 4 = (r - -24)/(5 - -5) 4 = (r +24)/10 Cross multiply: r + 24 = 40 Subtract 24 from each side: [B]r = 16[/B]\n\nThe points 6,4 and 9,r lie on a line with slope 3. Find the missing coordinate r.\nThe points 6,4 and 9,r lie on a line with slope 3. Find the missing coordinate r. Slope = (y2 - y1)/(x2 - x1) Plugging in our numbers, we get: 3 = (r - 4)/(9 - 6) 3 = (r - 4)/3 Cross multiply: r - 4 = 9 Add 4 to each side: [B]r = 13[/B]\n\nThe product of two positive numbers is 96. Determine the two numbers if one is 4 more than the other\nThe product of two positive numbers is 96. Determine the two numbers if one is 4 more than the other. Let the 2 numbers be x and y. We have: [LIST=1] [*]xy = 96 [*]x = y - 4 [/LIST] [U]Substitute (2) into (1)[/U] (y - 4)y = 96 y^2 - 4y = 96 [U]Subtract 96 from both sides:[/U] y^2 - 4y - 96 = 0 [U]Factoring using our quadratic calculator, we get:[/U] (y - 12)(y + 8) So y = 12 and y = -8. Since the problem states positive numbers, we use [B]y = 12[/B]. Substituting y = 12 into (2), we get: x = 12 - 4 [B]x = 8[/B] [B]We have (x, y) = (8, 12)[/B]\n\nThe residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes\nThe residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes was 6 to 5. If there were 4570 no votes, what was the total number of votes? Set up a proportion where y is the number of yes votes to 4570 no votes 6/5 = y/4570 Using our [URL='http://www.mathcelebrity.com/prop.php?num1=6&num2=y&den1=5&den2=4570&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator,[/URL] we get: [B]y = 5484[/B]\n\nThe residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes\nThe residents of a city voted on whether to raise property taxes. The ratio of yes votes to no votes was 4 to 3 . If there were 2958 no votes, what was the total number of votes? Set up a ratio of yes to no votes 4/3 = x/2958 Using our [URL='http://www.mathcelebrity.com/prop.php?num1=4&num2=x&den1=3&den2=2958&propsign=%3D&pl=Calculate+missing+proportion+value']proportion calculator[/URL], we get x = 3,944 for yes votes. Adding yes votes and no votes together to get total votes, we get: Total Votes = Yes Votes + No Votes Total Votes = 3,944 + 2,958 Total Votes = [B]6,902[/B]\n\nThe sales price of a new compact disc player is \\$210 at a local discount store. At the store where t\nThe sales price of a new compact disc player is \\$210 at a local discount store. At the store where the sale is going on, each new cd is on sale for \\$11. If Kyle purchases a player and some cds for \\$243 how many cds did he purchase? If Kyle bought the player, he has 243 - 210 = 33 left over. Each cd is 11, so set up an equation to see how many CDs he bought: 11x = 33 Divide each side by 11 [B]x = 3[/B]\n\nThe science club charges 4.50 per car at their car wash. Write and solve and inequality to find how\nThe science club charges 4.50 per car at their car wash. Write and solve and inequality to find how many cars they have to wash to earn at least 300 Let x be the number of cars they wash. Set up our inequality. Note, at least 300 means 300 or greater, so we use greater than or equal to. [U]Inequality:[/U] [B]4.50x >= 300 [/B] [U]So solve for x, divide each side by 4[/U] [B]x >= 66.67[/B]\n\nThe sides of a triangle are consecutive numbers. If the perimeter of the triangle is 240 m, find the\nThe sides of a triangle are consecutive numbers. If the perimeter of the triangle is 240 m, find the length of each side Let the first side be n. Next side which is consecutive is n + 1 Next side which is consecutive is n + 1 + 1 = n + 2 So we have the sum of 3 consecutive numbers is 240. We type in [I][URL='https://www.mathcelebrity.com/sum-of-consecutive-numbers.php?num=sumof3consecutivenumbersis240&pl=Calculate']sum of 3 consecutive numbers is 240[/URL][/I] into our search engine and we get: [B]79, 80, 81[/B]\n\nThe singular form of the word \"dice\" is \"die\". Tom was throwing a six-sided die. The first time he t\nThe singular form of the word \"dice\" is \"die\". Tom was throwing a six-sided die. The first time he threw, he got a three; the second time he threw, he got a three again. What's the probability of getting a three at the third time? Since all trials are independent: 1/6 * 1/6 * 1/6 = [B]1/216[/B]\n\nThe Square of a positive integer is equal to the sum of the integer and 12. Find the integer\nThe Square of a positive integer is equal to the sum of the integer and 12. Find the integer Let the integer be x. [LIST] [*]The sum of the integer and 12 is written as x + 12. [*]The square of a positive integer is written as x^2. [/LIST] We set these equal to each other: x^2 = x + 12 Subtract x + 12 from each side: x^2 - x - 12 = 0 We have a quadratic function. [URL='https://www.mathcelebrity.com/quadratic.php?num=x%5E2-x-12%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Run it through our search engine[/URL] and we get x = 3 and x = -4. The problem asks for a positive integer, so we have [B]x = 3[/B]\n\nThe square of a positive integer minus twice its consecutive integer is equal to 22. find the intege\nThe square of a positive integer minus twice its consecutive integer is equal to 22. Find the integers. Let x = the original positive integer. We have: [LIST] [*]Consecutive integer is x + 1 [*]x^2 - 2(x + 1) = 22 [/LIST] Multiply through: x^2 - 2x - 2 = 22 Subtract 22 from each side: x^2 - 2x - 24 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2-2x-24%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we get: x = 6 and x = -4 Since the problem states [U]positive integers[/U], we use: x = 6 and x + 1 = 7 [B](6, 7)[/B]\n\nthe square root of twice a number is 4 less than the number\nWrite this out, let the number be x. sqrt(2x) = x - 4 since 4 less means subtract Square each side: sqrt(2x)^2 = (x - 4)^2 2x = x^2 - 8x + 16 Subtract 2x from both sides x^2 - 10x + 16 = 0 Using the [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2+-+10x+%2B+16+%3D+0&pl=Solve+Quadratic+Equation&hintnum=0']quadratic calculator[/URL], we get two potential solutions x = (2, 8) Well, 2 does not work, since sqrt(2*2) = 2 which is not 4 less than 2 However, 8 does work: sqrt(2*8) = sqrt(16) = 4, which is 4 less than the number 8.\n\nThe sum of 2 numbers is 18. 3 times the greater number exceeds 4 times the smaller number by 5. Find\nThe sum of 2 numbers is 18. 3 times the greater number exceeds 4 times the smaller number by 5. Find the numbers. Let the first number be x. The second number is y. We have: [LIST=1] [*]x + y = 18 [*]3x = 4y + 5 [/LIST] Rearrange (2), by subtracting 4y from each side: 3x - 4y = 5 So we have a system of equations: [LIST=1] [*]x + y = 18 [*]3x - 4y = 5 [/LIST] Using our [URL='https://www.mathcelebrity.com/simultaneous-equations.php?term1=x+%2B+y+%3D+18&term2=3x+-+4y+%3D+5&pl=Cramers+Method']simultaneous equations calculator[/URL], we get: [B]x = 11 y = 7[/B]\n\nThe sum of 2 numbers is 70. The difference of these numbers is 24. Write and solve a system of equat\nThe sum of 2 numbers is 70. The difference of these numbers is 24. Write and solve a system of equations to determine the numbers. Let the two numbers be x and y. We have the following equations: [LIST=1] [*]x + y = 70 [*]x - y = 24 [/LIST] Add (1) to (2): 2x = 94 Divide each side by 2 [B]x = 47[/B] Plug this into (1) 47 + y = 70 Subtract 47 from each side, we have: [B]y = 23[/B]\n\nThe sum of 5x and 2x is at least 70\n[I]Is at least [/I]means greater than or equal to: 5x + 2x >= 70 If we combine like terms, we have: 7x >=70 We can further simplify by dividing each side of the inequality by 7 x >=10 If you want the interval notation for that, use the [URL='http://www.mathcelebrity.com/interval-notation-calculator.php?num=x%3E%3D10&pl=Show+Interval+Notation']interval notation calculator[/URL].\n\nThe sum of a number and its square is 72. find the numbers?\nThe sum of a number and its square is 72. find the numbers? Let the number be n. We have: n^2 + n = 72 Subtract 72 from each side: n^2 + n - 72 = 0 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=n%5E2%2Bn-72%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic calculator[/URL], we have: [B]n = 8 or n = -9 [/B] Since the numbers do not state positive or negative, these are the two solutions.\n\nthe sum of a number and itself is 8\nA number means an arbitrary variable, let's call it x. The sum of a number and itself means adding the number to itself x + x Simplified, we have 2x The word is means equal to, so we have an algebraic expression of: [B]2x= 8 [/B] IF you need to solve this equation, divide each side by 2 [B]x = 4[/B]\n\nthe sum of n and twice n is 12\nTwice n means we multiply n by 2 2n The sum of n and twice n means we add n + 2n The word [I]is[/I] means equal to, so we set that expression above equal to 12 n + 2n = 12 Combine like terms: 3n = 12 Divide each side of the equation by 3 to isolate n n = 4 Check our work Twice n is 2*4 = 8 Add that to n = 4 8 + 4 12\n\nThe sum of the ages of levi and renee is 89 years. 7 years ago levi's age was 4 times renees age. Ho\nThe sum of the ages of levi and renee is 89 years. 7 years ago levi's age was 4 times renees age. How old is Levi now? Let Levi's current age be l. Let Renee's current age be r. Were given two equations: [LIST=1] [*]l + r = 89 [*]l - 7 = 4(r - 7) [/LIST] Simplify equation 2 by multiplying through: [LIST=1] [*]l + r = 89 [*]l - 7 = 4r - 28 [/LIST] Rearrange equation 1 to solve for r and isolate l by subtracting l from each side: [LIST=1] [*]r = 89 - l [*]l - 7 = 4r - 28 [/LIST] Now substitute equation (1) into equation (2): l - 7 = 4(89 - l) - 28 l - 7 = 356 - 4l - 28 l - 7 = 328 - 4l To solve for l, we [URL='https://www.mathcelebrity.com/1unk.php?num=l-7%3D328-4l&pl=Solve']type the equation into our search engine[/URL] and we get: l = [B]67[/B]\n\nThe sum of the digits of a 2 digit number is 10. The value of the number is four more than 15 times\nThe sum of the digits of a 2 digit number is 10. The value of the number is four more than 15 times the unit digit. Find the number. Let the digits be (x)(y) where t is the tens digit, and o is the ones digit. We're given: [LIST=1] [*]x + y = 10 [*]10x+ y = 15y + 4 [/LIST] Simplifying Equation (2) by subtracting y from each side, we get: 10x = 14y + 4 Rearranging equation (1), we get: x = 10 - y Substitute this into simplified equation (2): 10(10 - y) = 14y + 4 100 - 10y = 14y + 4 [URL='https://www.mathcelebrity.com/1unk.php?num=100-10y%3D14y%2B4&pl=Solve']Typing this equation into our search engine[/URL], we get: y = 4 Plug this into rearranged equation (1), we get: x = 10 - 4 x = 6 So our number xy is [B]64[/B]. Let's check our work against equation (1): 6 + 4 ? 10 10 = 10 Let's check our work against equation (2): 10(6)+ 4 ? 15(4) + 4 60 + 4 ? 60 + 4 64 = 64\n\nThe sum of the digits of a certain two-digit number is 16. Reversing its digits increases the number\nThe sum of the digits of a certain two-digit number is 16. Reversing its digits increases the number by 18. What is the number? Let x and (16-x) represent the original ten and units digits respectively Reversing its digits increases the number by 18 Set up the relational equation [10x + (16-x)] + 18 = 10(16 - x) + x Solving for x 9x + 34 = 160 - 9x Combine like terms 18x = 126 Divide each side of the equation by 18 18x/18 = 126/18 x = 7 So 16 - x = 16 - 7 = 9 The first number is 79, the other number is 97. and 97 - 79 = 18 so we match up. The number in our answer is [B]79[/B]\n\nThe sum of the squares of two consecutive positive integers is 61. Find these two numbers.\nThe sum of the squares of two consecutive positive integers is 61. Find these two numbers. Let the 2 consecutive integers be x and x + 1. We have: x^2 + (x + 1)^2 = 61 Simplify: x^2 + x^2 + 2x + 1 = 61 2x^2 + 2x + 1 = 61 Subtract 61 from each side: 2x^2 + 2x - 60 = 0 Divide each side by 2 x^2 + x - 30 Using our [URL='http://www.mathcelebrity.com/quadratic.php?num=x%5E2%2Bx-30&pl=Solve+Quadratic+Equation&hintnum=+0']quadratic equation calculator[/URL], we get: x = 5 and x = -6 The question asks for [I]positive integers[/I], so we use [B]x = 5. [/B]This means the other number is [B]6[/B].\n\nThe sum of three consecutive integers is 42\nLet the 3 integers be x, y, and z. y = x + 1 z = y + 1, or x + 2. Set up our equation: x + (x + 1) + (x + 2) = 42 Group our variables and constants: (x + x + x) + (1 + 2) = 42 3x + 3 = 42 Subtract 3 from each side: 3x = 39 Divide each side of the equation by 3: [B]x = 13 So y = x + 1 = 14 z = x + 2 = 15 (x,y,z) = (13,14,15)[/B]\n\nThe sum of twice an integer and 3 times the next consecutive integer is 48\nThe sum of twice an integer and 3 times the next consecutive integer is 48 Let the first integer be n This means the next consecutive integer is n + 1 Twice an integer means we multiply n by 2: 2n 3 times the next consecutive integer means we multiply (n + 1) by 3 3(n + 1) The sum of these is: 2n + 3(n + 1) The word [I]is[/I] means equal to, so we set 2n + 3(n + 1) equal to 48: 2n + 3(n + 1) = 48 Solve for [I]n[/I] in the equation 2n + 3(n + 1) = 48 We first need to simplify the expression removing parentheses Simplify 3(n + 1): Distribute the 3 to each term in (n+1) 3 * n = (3 * 1)n = 3n 3 * 1 = (3 * 1) = 3 Our Total expanded term is 3n + 3 Our updated term to work with is 2n + 3n + 3 = 48 We first need to simplify the expression removing parentheses Our updated term to work with is 2n + 3n + 3 = 48 [SIZE=5][B]Step 1: Group the n terms on the left hand side:[/B][/SIZE] (2 + 3)n = 5n [SIZE=5][B]Step 2: Form modified equation[/B][/SIZE] 5n + 3 = + 48 [SIZE=5][B]Step 3: Group constants:[/B][/SIZE] We need to group our constants 3 and 48. To do that, we subtract 3 from both sides 5n + 3 - 3 = 48 - 3 [SIZE=5][B]Step 4: Cancel 3 on the left side:[/B][/SIZE] 5n = 45 [SIZE=5][B]Step 5: Divide each side of the equation by 5[/B][/SIZE] 5n/5 = 45/5 Cancel the 5's on the left side and we get: n = [B]9[/B]\n\nThe total age of three cousins is 48. Suresh is half as old as Hakima and 4 years older than Saad. h\nThe total age of three cousins is 48. Suresh is half as old as Hakima and 4 years older than Saad. How old are the cousins? Let a be Suresh's age, h be Hakima's age, and c be Saad's age. We're given: [LIST=1] [*]a + h + c = 48 [*]a = 0.5h [*]a = c + 4 [/LIST] To isolate equations in terms of Suresh's age (a), let's do the following: [LIST] [*]Rewriting (3) by subtracting 4 from each side, we get c = a - 4 [*]Rewriting (2) by multiply each side by 2, we have h = 2a [/LIST] We have a new system of equations: [LIST=1] [*]a + h + c = 48 [*]h = 2a [*]c = a - 4 [/LIST] Plug (2) and (3) into (1) a + (2a) + (a - 4) = 48 Group like terms: (1 + 2 + 1)a - 4 = 48 4a - 4 = 48 [URL='https://www.mathcelebrity.com/1unk.php?num=4a-4%3D48&pl=Solve']Typing this equation into our search engine[/URL], we get: [B]a = 13 [/B]<-- Suresh's age This means that Hakima's age, from modified equation (2) above is: h = 2(13) [B]h = 26[/B] <-- Hakima's age This means that Saad's age, from modified equation (3) above is: c = 13 - 4 [B]c = 9[/B] <-- Saad's age [B] [/B]\n\nthere are 120 calories in 3/4 cup serving of cereal. How many Calories are there in 6 cups of cereal\n120/3/4 = x/6 Cross multiply: 0.75x = 720 Divide each side of the equation by 0.75 [B]x = 960[/B]\n\nThere are 13 animals in the barn. some are chickens and some are pigs. there are 40 legs in all. How\nThere are 13 animals in the barn. some are chickens and some are pigs. there are 40 legs in all. How many of each animal are there? Chickens have 2 legs, pigs have 4 legs. Let c be the number of chickens and p be the number of pigs. Set up our givens: (1) c + p = 13 (2) 2c + 4p = 40 [U]Rearrange (1) to solve for c by subtracting p from both sides:[/U] (3) c = 13 - p [U]Substitute (3) into (2)[/U] 2(13 - p) + 4p = 40 26 - 2p + 4p = 40 [U]Combine p terms[/U] 2p + 26 = 40 [U]Subtract 26 from each side:[/U] 2p = 14 [U]Divide each side by 2[/U] [B]p = 7[/B] [U]Substitute p = 7 into (3)[/U] c = 13 - 7 [B]c = 6[/B] For a shortcut, you could use our [URL='http://www.mathcelebrity.com/simultaneous-equations.php?term1=c+%2B+p+%3D+13&term2=2c+%2B+4p+%3D+40&pl=Cramers+Method']simultaneous equations calculator[/URL]\n\nThere are 2 consecutive integers. Twice the first increased by the second yields 16. What are the nu\nThere are 2 consecutive integers. Twice the first increased by the second yields 16. What are the numbers? Let x be the first integer. y = x + 1 is the next integer. We have the following givens: [LIST=1] [*]2x + y = 16 [*]y = x + 1 [/LIST] Substitute (2) into (1) 2x + (x + 1) = 16 Combine x terms 3x + 1 = 16 Subtract 1 from each side 3x = 15 Divide each side by 3 [B]x = 5[/B] So the other integer is 5 + 1 = [B]6[/B]\n\nThere are 32 female performers in a dance recital. The ratio of men to women is 3:8. How many men ar\nThere are 32 female performers in a dance recital. The ratio of men to women is 3:8. How many men are in the dance recital? 3:8 = x:32 3/8 = x/32 Cross multiply: 8x = 96 Divide each side by 8 x = 12 Check our work: 12:32 Divide each part by 4 12/4 = 3 and 32/4 = 8 so we have 3:8 :)\n\nThere are 85 students in a class, 40 of them like math,31 of them like science, 12 of them like both\nThere are 85 students in a class, 40 of them like math,31 of them like science, 12 of them like both, how many don't like either. We have the following equation: Total Students = Students who like math + students who like science - students who like both + students who don't like neither. Plug in our knowns, we get: 85 = 40 + 31 - 12 + Students who don't like neither 85 = 59 + Students who don't like neither Subtract 59 from each side, we get: Students who don't like neither = 85 - 59 Students who don't like neither = [B]26[/B]\n\nThere are two numbers. The sum of 4 times the first number and 3 times the second number is 24. The\nThere are two numbers. The sum of 4 times the first number and 3 times the second number is 24. The difference between 2 times the first number and 3 times the second number is 24. Find the two numbers. Let the first number be x and the second number be y. We have 2 equations: [LIST=1] [*]4x + 3y = 24 [*]2x - 3y = 24 [/LIST] Without doing anything else, we can add the 2 equations together to eliminate the y term: (4x + 2x) + (3y - 3y) = (24 + 24) 6x = 48 Divide each side by 6: [B]x = 8 [/B] Substitute this into equation (1) 4(8) + 3y = 24 32 + 3y = 24 [URL='https://www.mathcelebrity.com/1unk.php?num=32%2B3y%3D24&pl=Solve']Type 32 + 3y = 24 into our search engine[/URL] and we get [B]y = 2.6667[/B].\n\nThere is an escalator that is 1090.3 feet long and drops a vertical distance of 193.4 feet. What is\nThere is an escalator that is 1090.3 feet long and drops a vertical distance of 193.4 feet. What is its angle of depression? The sin of the angle A is the length of the opposite side / hypotenuse. sin(A) = Opposite / Hypotenuse sin(A) = 193.4 / 1090/3 sin(A) = 0.1774 [URL='https://www.mathcelebrity.com/anglebasic.php?entry=0.1774&pl=arcsin']We want the arcsin(0.1774)[/URL]. [B]A = 10.1284[/B]\n\nThere were 150 students at a dance. There were 16 more boys than girls. How many boys were there?\nSet up two equations: (1) b = g + 16 (2) b + g = 150 Substitute equation (1) into (2) (g + 16) + g = 150 Combine like terms 2g + 16 = 150 Subtract 16 from each side 2g = 134 Divide each side by 2 to isolate g g = 67 Substitute this into equation (1) b = 67 + 16 [B]b = 83[/B]\n\nTiffany is 59 years old. The sum of the ages of Tiffany and Maria is 91. How old is Maria?\nTiffany is 59 years old. The sum of the ages of Tiffany and Maria is 91. How old is Maria? Tiffany + Maria = 91 59 + Maria = 91 Subtract 59 from each side Maria = 91 - 59 [B]Maria = 32[/B]\n\nTime and Distance\nLet h be the number of hours that pass when Charlie starts. We have the following equations: [LIST] [*]Charlie: D = 40h + 9 [*]Danny: D = 55h [/LIST] Set them equal to each other: 40h + 9 = 55h Subtract 40h from both sides: 15h = 9 h = 3/5 [B]3/5 of an hour = 3(60)/5 = 36 minutes[/B]\n\nTime and Distance\nThank you so much [QUOTE=\"math_celebrity, post: 1003, member: 1\"]Let h be the number of hours that pass when Charlie starts. We have the following equations: [LIST] [*]Charlie: D = 40h + 9 [*]Danny: D = 55h [/LIST] Set them equal to each other: 40h + 9 = 55h Subtract 40h from both sides: 15h = 9 h = 3/5 [B]3/5 of an hour = 3(60)/5 = 36 minutes[/B][/QUOTE]\n\nTina's mom made brownies. When tinas friend came over they ate 1/3 of the brownies. Her sister ate 2\nTina's mom made brownies. When tinas friend came over they ate 1/3 of the brownies. Her sister ate 2 and her dad ate 4. If there are 26 brownies left. How many did her mom make Let b denote the number of brownies Tina's mom made. We're given: b - 1/3b - 2 - 4 = 26 Combining like terms, we have: 2b/3 - 6 = 26 Add 6 to each side, we get: 2b/3 = 32 To solve this equation for b, we [URL='https://www.mathcelebrity.com/prop.php?num1=2b&num2=32&den1=3&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']type it in our math engine[/URL] and we get: b = [B]48[/B]\n\nToday is my birthday! Four-fifths of my current age is greater than three-quarters of my age one yea\nToday is my birthday! Four-fifths of my current age is greater than three-quarters of my age one year from now. Given that my age is an integer number of years, what is the smallest my age could be? Let my current age be a. We're given: 4/5a > 3/4(a + 1) Multiply through on the right side: 4a/5 > 3a/4 + 3/4 Let's remove fractions by multiply through by 5: 5(4a/5) > 5(3a/4) + 5(3/4) 4a > 15a/4 + 15/4 Now let's remove the other fractions by multiply through by 4: 4(4a) > 4(15a/4) + 4(15/4) 16a > 15a + 15 [URL='https://www.mathcelebrity.com/1unk.php?num=16a%3E15a%2B15&pl=Solve']Typing this inequality into our search engine[/URL], we get: a > 15 This means the smallest [I]integer age[/I] which the problem asks for is: 15 + 1 = [B]16[/B]\n\nTom has 6 fewer pencils than ari. Tom has 7 pencils. How many pencils does Ari have?\nTom has 6 fewer pencils than ari. Tom has 7 pencils. How many pencils does Ari have? 6 fewer means less. Let a = Ari's pencils. We have: a - 6 = 7 Add 6 to each side [B]a = 13[/B]\n\nTriangle Inequality\nThis calculator displays 2 scenarios\n1) Enter 3 sides of a triangle, and it will determine if the side lengths satisfy the properties of the triangle inequality and form a triangle\n2) Enter 2 sides of a triangle, and this will determine an acceptable range for the length of the 3rd side of a triangle so that the 3rd side respects the Triangle Inequality.\n\nTriangle KLM has vertices at . k(-2,-2), l(10,-2), m(4,4) What type of triangle is KLM?\nTriangle KLM has vertices at . k(-2,-2), l(10,-2), m(4,4) What type of triangle is KLM? [URL='https://www.mathcelebrity.com/slope.php?xone=-2&yone=-2&slope=+2%2F5&xtwo=10&ytwo=-2&pl=You+entered+2+points']Side 1: KL[/URL] = 12 [URL='https://www.mathcelebrity.com/slope.php?xone=10&yone=-2&slope=+2%2F5&xtwo=4&ytwo=4&pl=You+entered+2+points']Side 2: LM[/URL] = 8.4853 [URL='https://www.mathcelebrity.com/slope.php?xone=-2&yone=2&slope=+2%2F5&xtwo=4&ytwo=4&pl=You+entered+2+points']Side 3: KM[/URL] = 6.3246 Then, we want to find the type of triangle. Using our [URL='https://www.mathcelebrity.com/tribasic.php?side1input=12&side2input=8.4853&side3input=6.3246&angle1input=&angle2input=&angle3input=&pl=Solve+Triangle']triangle solver with our 3 sides[/URL], we get: [B]Obtuse, Scalene[/B]\n\nTriangle Solver and Classify Triangles\nSolves a triangle including area using the following solving methods\nSide-Angle-Side (SAS) Side Angle Side\nAngle-Side-Angle (ASA) Angle Side Angle\nSide-Side-Angle (SSA) Side Angle Side\nSide-Side-Side (SSS) Side Side Side\nArea (A) is solved using Herons Formula\nLaw of Sines\nLaw of Cosines\n\nAlso classifies triangles based on sides and angles entered.\n\nTriangle with perimeter\nA triangle with a perimeter of 120. What degree are the three sides?\n\nTriangle with perimeter\nWhat kind of triangle? Do you have side lengths? I need more information.\n\nTrue False Equations\nDetermines if a set of addition and subtraction of numbers on each side of an equation are equivalent. Also known as true or false equations\n\ntwice the difference of a number and 3 is equal to 3 times the sum of a number and 2\ntwice the difference of a number and 3 is equal to 3 times the sum of a number and 2. We've got 2 algebraic expressions here. Let's take them in parts. Left side algebraic expression: twice the difference of a number and 3 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]The word [I]difference[/I] means we subtract 3 from the variable x [*]x - 3 [*]Twice this difference means we multiply (x - 3) by 2 [*]2(x - 3) [/LIST] Right side algebraic expression: 3 times the sum of a number and 2 [LIST] [*]The phrase [I]a number[/I] means an arbitrary variable, let's call it x. [*]The word [I]sum[/I] means we add 2 to the variable x [*]x + 2 [*]3 times the sum means we multiply (x + 2) by 3 [*]3(x + 2) [/LIST] Now, we have both algebraic expressions, the problem says [I]is equal to[/I] This means we have an equation, where we set the left side algebraic expression equal to the right side algebraic expression using the equal sign (=) to get our answer [B]2(x - 3) = 3(x + 2)[/B]\n\ntwice the difference of a number and 55 is equal to 3 times the sum of a number and 8\ntwice the difference of a number and 55 is equal to 3 times the sum of a number and 8 Take this algebraic expression in pieces. Left side: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The difference of this number and 55 means we subtract 55 from x x - 55 Twice the difference means we multiply x - 55 by 2 2(x - 55) Right side: The phrase [I]a number[/I] means an arbitrary variable, let's call it x. x The sum of a number and 8 means we add 8 to x x + 8 3 times the sum means we multiply x + 8 by 3 3(x + 8) Now that we have the left and right side of the expressions, we see the phrase [I]is equal to[/I]. This means an equation, so we set the left side equal to the right side: [B]2(x - 55) = 3(x + 8)[/B]\n\ntwo mechanics worked on a car. the first mechanic worked for 10 hours, and the second mechanic worke\ntwo mechanics worked on a car. the first mechanic worked for 10 hours, and the second mechanic worked for 5 hours. together they charged a total of 1225. what was the rate charged per hour by each mechanic if the sum of the two rates was 170 per hour? Set up two equations: (1) 10x + 5y = 1225 (2) x + y = 170 Rearrange (2) x = 170 - y Substitute that into (1) 10(170 - y) + 5y = 1225 1700 - 10y + 5y = 1225 1700 - 5y = 1225 Move 5y to the other side 5y + 1225 = 1700 Subtract 1225 from each side 5y =475 Divide each side by 5 [B]y = 95[/B] Which means x = 170 - 95, [B]x = 75[/B]\n\nTwo numbers have a sum of 20. If one number is p, express the other in terms of p.\nTwo numbers have a sum of 20. If one number is p, express the other in terms of p. If the sum is 20 and one number is p, then let the other number be q. We have: p + q = 20 We want q, so we subtract p from each side: [B]q = 20 - p[/B]\n\nTwo numbers total 50 and have a difference of 28. Find the two numbers.\nTwo numbers total 50 and have a difference of 28. Find the two numbers. Using x and y as our two numbers, we write the following 2 equations: [LIST=1] [*]x + y = 50 [*]x - y = 28 [/LIST] Add the 2 rows: 2x = 78 Divide each side by 2: [B]x = 39[/B] If x = 39, then from (1), we have y = 50 - 39 [B]y = 11[/B]\n\nTwo numbers total 83 and have a difference of 17 find the two numbers\nLet the numbers be x and y. Set up our givens: [LIST=1] [*]x + y = 83 [*]x - y = 17 [*]Rearrange (2), by adding y to each side, we have: x = 17 + y [/LIST] [U]Substitute (3) into (1):[/U] (17 + y) + y = 83 [U]Group y terms[/U] 2y + 17 = 83 [U]Using our [URL='http://www.mathcelebrity.com/1unk.php?num=2y%2B17%3D83&pl=Solve']equation solver[/URL], we get:[/U] [B]y = 33 [/B] [U]Substitute that into (3)[/U] x = 17 + 33 [B]x = 50 [/B] So our two numbers (x, y) = (33, 50)\n\nu=ak/b for a\nCross multiply: ub = ak Divide each side of the equation by k to isolate a: a = ub/k\n\nU=ak/b, for a\nU=ak/b, for a [U]Cross multiply:[/U] Ub = ak [U]Divide each side by k[/U] [B]a = Ub/k[/B]\n\nV ? E + F = 2 for e\nV ? E + F = 2 for e To solve this literal equation, we want to isolate e. Add E to both sides: V ? E + F + E = 2 + E The E's cancel on the left side, so we have: V + F = 2 + E Subtract 2 from each side: V + F - 2 = 2 + E + 2 The 2's cancel on the right side, so we have: E = [B]V + F - 2[/B]\n\nVideo store movie rental plans. Plan A 25 membership fee plus 1.25 for movie. Plan B 40 for unlimite\nVideo store movie rental plans. Plan A 25 membership fee plus 1.25 for movie. Plan B 40 for unlimited rentals. What number of movies rentals is plan B less than plan A? Let x equal the number of movies rented and C the cost for rentals Plan A: C = 1.25x + 25 Plan B: C = 40 Set up the inequality: 1.25x + 25 > 40 Subtract 25 from each side: 1.25x > 15 Divide each side of the inequality by 1.25 x > 12 So [B]13[/B] rentals or more make Plan B less than Plan A.\n\nvw^2+y=x for w\nvw^2+y=x for w This is an algebraic expression. Subtract y from each side: vw^2 + y - y = x - y The y's cancel on the left side, so we're left with: vw^2 = x - y Divide each side by v w^2 = (x - y)/v Take the square root of each side: w = [B]Sqrt((x - y)/v)[/B]\n\nWater flows from tank A to tank B at the rate of 2 litres per minute.\n[QUOTE=\"Jahn, post: 78, member: 5\"]Water flows from tank A to tank B at the rate of 2 litres per minute. Initially tank A has 200 litres in it and tank B has 100 Litres in it. Water drains from tank B at 0.5 litres per minute. After how many minutes are there equal volumes of water in the 2 tanks? Write an equation and solve it.[/QUOTE] Tank A: V = 200 - 2x Tank B: V = 100 - 0.5x Where x is the number of minutes passed. Set them equal to each other 200 - 2x = 100 - 0.5x Subtract 100 from each side: 100 - 2x = -0.5x Add 2x to each side: 1.5x = 100 Divide each side of the equation by x: x = 66.66666667\n\nwhat integer is tripled when 9 is added to 3 fourths of it?\nwhat integer is tripled when 9 is added to 3 fourths of it? Let the integer be n. Tripling an integer means multiplying it by 3. We're given: 3n = 3n/4 + 9 Since 3 = 12/4, we have: 12n/4 = 3n/4 + 9 Subtract 3n/4 from each side: 9n/4 = 9 [URL='https://www.mathcelebrity.com/prop.php?num1=9n&num2=9&den1=4&den2=1&propsign=%3D&pl=Calculate+missing+proportion+value']Typing this equation into the search engine[/URL], we get: [B]n = 4[/B]\n\nWhat is the annual nominal rate compounded daily for a bond that has an annual yield of 5.4%? Round\nWhat is the annual nominal rate compounded daily for a bond that has an annual yield of 5.4%? Round to three decimal places. Use a 365 day year. [U]Set up the accumulation equation:[/U] (1+i)^365 = 1.054 [U]Take the natural log of each side[/U] 365 * Ln(1 + i) = 1.054 Ln(1 + i) = 0.000144089 [U]Use each side as a exponent to eulers constant e[/U] (1 + i) = e^0.000144089 1 + i = 1.000144099 i = 0.000144099 or [B].0144099%[/B]\n\nWhat is the sample space for a 10 sided die?\nWhat is the sample space for a 10 sided die? Sample space means the set of all possible outcomes. For a 10-sided die, we have: [B]{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}[/B]\n\nWhat number when multiplied by four exceeds itself by 42?\nWhat number when multiplied by four exceeds itself by 42? Let the number be n. We have: 4n = n + 42 Subtract n from each side: 3n = 42 Divide each side by 3 [B]n = 14[/B]\n\nWhat pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the\nWhat pair of consecutive integers gives the following: 7 times the smaller is less than 6 times the larger? Let x and y be consecutive integers, where y = x + 1 We have 7x < 6y as our inequality. Substituting x, y = x + 1, we have: 7x < 6(x + 1) 7x < 6x + 6 Subtracting x from each side, we have: x < 6, so y = 6 + 1 = 7 (x, y) = (6, 7)\n\nWhen 28 is subtracted from the square of a number, the result is 3 times the number. Find the negati\nWhen 28 is subtracted from the square of a number, the result is 3 times the number. Find the negative solution. Let the number be n. Square of a number: n^2 28 is subtracted from the square of a number: n^2 - 28 3 times the number: 3n [I]The result is[/I] mean an equation, so we set n^2 - 28 = 3n n^2 - 28 = 3n Subtract 3n from each side: n^2 - 3n - 28 = 3n - 3n The right side cancels to 0, so we have: n^2 - 3n - 28 = 0 This is a quadratic equation in standard form, so we [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-3n-28%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']use our quadratic calculator[/URL] to solve: We get two solutions for n: n = (-4, 7) The question asks for the negative solution, so our answer is: [B]n = -4[/B]\n\nWhen 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 intege\nWhen 3 consecutive positive integers are multiplied, the product is 16 times the sum of the 3 integers. What is the difference of the product minus the sum? Let the 3 consecutive positive integers be: [LIST=1] [*]x [*]x + 1 [*]x + 2 [/LIST] The product is: x(x + 1)(x + 2) The sum is: x + x + 1 + x + 2 = 3x + 3 We're told the product is equivalent to: x(x + 1)(x + 2) = 16(3x + 3) x(x + 1)(x + 2) = 16 * 3(x + 1) Divide each side by (x + 1) x(x + 2) = 48 x^2 + 2x = 48 x^2 + 2x - 48 = 0 Now subtract the sum from the product: x^2 + 2x - 48 - (3x + 3) [B]x^2 - x - 51[/B]\n\nWhen 4 is subtracted from the square of a number, the result is 3 times the number. Find the positiv\nWhen 4 is subtracted from the square of a number, the result is 3 times the number. Find the positive solution. Let the number be n. We have: n^2 - 4 = 3n Subtract 3n from each side: n^2 - 3n - 4 = 0 [URL='https://www.mathcelebrity.com/quadratic.php?num=n%5E2-3n-4%3D0&pl=Solve+Quadratic+Equation&hintnum=+0']Typing this quadratic equation into the search engine[/URL], we get: n = (-1, 4) The problem asks for the positive solution, so we get [B]n = 4[/B].\n\nWhen 4 times a number is increased by 40, the answer is the same as when 100 is decreased by the num\nWhen 4 times a number is increased by 40, the answer is the same as when 100 is decreased by the number. Find the number The phrase [I]a number, [/I]means an arbitrary variable, let's call it \"x\". 4 times a number, increased by 40, means we multiply 4 times x, and then add 40 4x + 40 100 decreased by the number means we subtract x from 100 100 - x The problem tells us both of these expressions are the same, so we set them equal to each other: 4x + 40 = 100 - x Add x to each side: 4x + x + 40 = 100 - x + x The x's cancel on the right side, so we have: 5x + 40 = 100 [URL='https://www.mathcelebrity.com/1unk.php?num=5x%2B40%3D100&pl=Solve']Typing this equation into the search engine[/URL], we get [B]x = 12[/B].\n\nWhen Ms. Thelma turned on her oven, the temperature inside was 70 degrees F. The temperature began t\nWhen Ms. Thelma turned on her oven, the temperature inside was 70 degrees F. The temperature began to rise at a rate of 20 degrees per minute. How Long did it take for the oven to reach 350 degrees F? Figure out how many degrees we have left: 350 - 70 = 280 Let m = minutes 20m = 280 Divide each side by m [B]m = 14[/B]\n\nWhich of the following equations represents a line that is parallel to the line with equation y = -3\nWhich of the following equations represents a line that is parallel to the line with equation y = -3x + 4? A) 6x + 2y = 15 B) 3x - y = 7 C) 2x - 3y = 6 D) x + 3y = 1 Parallel lines have the same slope, so we're looking for a line with a slope of 3, in the form y = mx + b. For this case, we want a line with a slope of -3, like our given line. If we rearrange A) by subtracting 6x from each side, we get: 2y = -6x + 15 Divide each side by 2, we get: y = -3x + 15/2 This line is in the form y = mx + b, where m = -3. So our answer is [B]A[/B].\n\nwriting and solving equations\nYour answer is correct. Here is how I set up the profit equation where h is the hours worked and x is the supply cost: P(h) = 15.35h + x We know P(4) = 141.73 P(4) = 15.35(4) + x 141.73 = 15.35(4) + x Simplify 141.73 = 61.4 + x Subtract 61.4 from each side: [B]x = 80.33[/B]\n\nwy - ma = ay/n for y\nwy - ma = ay/n for y Subtract ay/n from each side: wy - ma - ay/n = ay/n - ay/n wy - ma - ay/n = 0 Now add ma to each side: wy - ay/n = ma Factor out y: y(w - a/n) = ma Divide each side by (w - a/n) y = [B]ma/(w - a/n)[/B]\n\nx + 8y/4 = 9y for x\nx + 8y/4 = 9y for x Step 1: Isolate x by subtracting 8y/4 from each side: x + 8y/4 - 8y/4 = 9y - 8y/4 Cancel 8y/4 on the left side: x =[B] 9y - 8y/4 [MEDIA=youtube]5NLDNw_T8GU[/MEDIA][/B]\n\nx - 5y = 6 for x\nx - 5y = 6 for x Add 5y to each side to solve this literal equation for x. x - 5y + 5y = 6 + 5y Cancel the 5y on each side, we get: x = [B]6 + 5y[/B]\n\nx is a multiple of 6 and 1 ? x ? 16\nx is a multiple of 6 and 1 ? x ? 16. We want multiples of 6 between 1 and 16. We start with 6. Another multiple of 6 is 12 The next multiple of 6 is 18, which is out side the range of 1 ? x ? 16. So our number set is [B]x = {6, 12}[/B]\n\nx is a woman who served as US president before 2000\nx is a woman who served as US president before 2000 No woman US presidents before 2000, so we have the empty set. [B]x = {}[/B]\n\nX varies directly with the cube root of y when x=1 y=27. Calculate y when x=4\nX varies directly with the cube root of y when x=1 y=27. Calculate y when x=4 Varies directly means there is a constant k such that: x = ky^(1/3) When x = 1 and y = 27, we have: 27^1/3(k) = 1 3k = 1 To solve for k, we[URL='https://www.mathcelebrity.com/1unk.php?num=3k%3D1&pl=Solve'] type in our equation into our search engine[/URL] and we get: k = 1/3 Now, the problem asks for y when x = 4. We use our variation equation above with k = 1/3 and x = 4: 4 = y^(1/3)/3 Cross multiply: y^(1/3) = 4 * 3 y^(1/3) =12 Cube each side: y^(1/3)^3 = 12^3 y = [B]1728[/B]\n\nX+y/3=5 for x\n(X+y)/3=5 for x Cross multiply: x + y = 15 Subtract y from each side: [B]x = 15 - y[/B]\n\nx-m=n+p, for x\nx-m=n+p, for x Add m to each side: x - m + m = n + p + m Cancelling the m's on each side: x = [B]n + p + m[/B]\n\nx/3 - g = a for x\nx/3 - g = a for x Add g to each side so we can isolate the x term: x/3 - g + g = a + g Cancel the g terms on the left side and we get: x/3 = a + g Multiply each side by 3 to isolate x: 3(x/3) = 3(a + g) Cancelling the 3's on the left side, we get: x = [B]3(a + g)[/B]\n\nx/5-7=2q for x\nx/5-7=2q for x Add 7 to each side: x/5 -7 + 7 = 2q + 7 Cancel the 7's on the left side, we get: x/5 = 2q + 7 Cross multiply the 5: x = 5(2q + 7) x = [B]10q + 35[/B]\n\nx/r - h = 4 for x\nx/r - h = 4 for x Add h to each side: x/r - h + h = h + 4 Cancel the h's on the left side, we get: x/r = h + 4 Multiply each side by r to isolate x: xr/r = r(h + 4) Cancel the r's on the left side, we get: x = [B]r(h + 4)[/B]\n\nx/y + 9 = n for y\nx/y + 9 = n for y First, subtract 9 from each side to isolate the y term: x/y + 9 - 9 = n - 9 Cancel the 9's on the left side, and we get: x/y = n - 9 Cross multiply: x = y(n - 9) Divide each side by (n - 9): x/(n - 9) = y(n - 9)/(n - 9) Cancel the (n - 9) on the right side, and we get: y = [B]x/(n - 9)[/B]\n\nx/y = z - 8 for x\nx/y = z - 8 for x Multiply each side by y to isolate x: y*(x/y) = y(z - 8) The y's cancel out on the left side, so we have: x = [B]y(z - 8)[/B]\n\nx/y = z - 8 for x\nx/y = z - 8 for x We start by seeing that x is isolated. To remove y from the left side, we multiply each side of the equation by y: xy/y = y(z - 8) Cancelling y on the left side, we get our answer of: x = [B]y(z - 8) [MEDIA=youtube]_HNyGlnnQdQ[/MEDIA][/B]\n\nX= p-w/m for p\nAdd w/m to each side: p = X + w/m\n\nXavier has \\$132 to buy a video game. Each game costs \\$12. Write an equation to find the number of ga\nXavier has \\$132 to buy a video game. Each game costs \\$12. Write an equation to find the number of games Xavier can purchase. Let g be the number of games, we have a cost function C(g) C(g) = 12g We want to find g such that C(g) = 132 12g = 132 Divide each side by 12 [B]g = 11[/B]\n\ny/2+c=d for y\nMultiply each side by 2 to isolate y. y +2c = 2d Subtract 2c from each side of the equation: y = 2d - 2c This can also be written y = 2(d - c)\n\nYesterday a car rental agency rented 4 convertibles and 30 other vehicles. Considering this data, ho\nYesterday a car rental agency rented 4 convertibles and 30 other vehicles. Considering this data, how many of the first 68 vehicles rented today should you expect to be convertibles? 30 other vehicles + 4 convertibles = 34 cars 34 * 2 = 68 30 * 2 other vehicles + 4 * 2 convertibles = 68 cars 60 other vehicles and [B]8 convertibles[/B]\n\nYolanda runs each lap in 7 minutes. She will run less than 35 minutes today. What are the possible n\nYolanda runs each lap in 7 minutes. She will run less than 35 minutes today. What are the possible numbers of laps she will run today? 7 minutes per lap must be less than 35 minutes. Let l be the number of laps 7l < 35 Divide each side by 7 [B]l < 5[/B]\n\nYou and a friend want to start a business and design t-shirts. You decide to sell your shirts for \\$1\nYou and a friend want to start a business and design t-shirts. You decide to sell your shirts for \\$15 each and you paid \\$6.50 a piece plus a \\$50 set-up fee and \\$25 for shipping. How many shirts do you have to sell to break even? Round to the nearest whole number. [U]Step 1: Calculate Your Cost Function C(s) where s is the number of t-shirts[/U] C(s) = Cost per Shirt * (s) Shirts + Set-up Fee + Shipping C(s) = \\$6.50s + \\$50 + \\$25 C(s) = \\$6.50s + 75 [U]Step 2: Calculate Your Revenue Function R(s) where s is the number of t-shirts[/U] R(s) = Price Per Shirt * (s) Shirts R(s) = \\$15s [U]Step 3: Calculate Break-Even Point[/U] Break Even is where Cost = Revenue. Set C(s) = R(s) \\$6.50s + 75 = \\$15s [U]Step 4: Subtract 6.5s from each side[/U] 8.50s = 75 [U]Step 5: Solve for s[/U] [URL='https://www.mathcelebrity.com/1unk.php?num=8.50s%3D75&pl=Solve']Run this through our equation calculator[/URL] to get s = 8.824. We round up to the next integer to get [B]s = 9[/B]. [B][URL='https://www.facebook.com/MathCelebrity/videos/10156751976078291/']FB Live Session[/URL][/B]\n\nYou and your friend are playing a number-guessing game. You ask your friend to think of a positive n\nYou and your friend are playing a number-guessing game. You ask your friend to think of a positive number, square the number, multiply the result by 2, and then add three. If your friend's final answer is 53, what was the original number chosen? Let n be our original number. Square the number means we raise n to the power of 2: n^2 Multiply the result by 2: 2n^2 And then add three: 2n^2 + 3 If the friend's final answer is 53, this means we set 2n^2 + 3 equal to 53: 2n^2 + 3 = 53 To solve for n, we subtract 3 from each side, to isolate the n term: 2n^2 + 3 - 3 = 53 - 3 Cancel the 3's on the left side, and we get: 2n^2 = 50 Divide each side of the equation by 2: 2n^2/2 = 50/2 Cancel the 2's, we get: n^2 = 25 Take the square root of 25 n = +-sqrt(25) n = +-5 We are told the number is positive, so we discard the negative square root and get: n = [B]5[/B]\n\nYou and your friend are saving for a vacation. You start with the same amount and save for the same\nYou and your friend are saving for a vacation. You start with the same amount and save for the same number of weeks. You save 75 per week, and your friend saves 50 per week. When vacation time comes, you have 950, and your friend has 800. How much did you start with, and for how many weeks did you save? [U]Let w be the number of weeks. Set up two equations where s is the starting amount:[/U] (1) s + 75w =950 (2) s + 50w = 800 [U]Rearrange (1) into (3) to solve for s by subtracting 75w[/U] (3) s = 950 - 75w [U]Rearrange (2) into (4) to solve for s by subtracting 50w[/U] (4) s = 800 - 50w [U]Set (3) and (4) equal to each other so solve for w[/U] 950 - 75w = 800 - 50w [U]Add 75w to each side, and subtract 950 from each side:[/U] 25w = 150 [U]Divide each side by w[/U] [B]w = 6[/B] Now plug w = 6 into (3) s = 950 - 75(6) s = 950 - 450 [B]s = 500[/B]\n\nYou are baking muffins for your class. There are 17 total students in your class and you have baked\nYou are baking muffins for your class. There are 17 total students in your class and you have baked 5 muffins. Write and solve an equation to find the additional number x of muffins you need to bake in order to have 2 muffins for each student. Write your equation so that the units on each side of the equation are muffins per student. 2 muffins per student = 17*2 = 34 muffins. We have an equation with a given 5 muffins, how much do we need (x) to get to 34 muffins (2 per student): x + 5 = 34 To solve for x, we type this equation into our search engine and we get: x = [B]29[/B]\n\nYou are baking muffins for your class. There are 17 total students in your class and you have baked\nYou are baking muffins for your class. There are 17 total students in your class and you have baked 5 muffins. Write and solve an equation to find the additional number x of muffins you need to bake in order to have 2 muffins for each student. Write your equation so that the units on each side of the equation are muffins per student. [U]Calculate total muffins:[/U] Total muffins = 2 muffins per student * 17 students Total muffins = 34 [U]Set up the equation using x for muffins:[/U] [B]x + 5 = 34 [/B] [U]To Solve this equation for x, we [URL='https://www.mathcelebrity.com/1unk.php?num=x%2B5%3D34&pl=Solve']type it in our search engine[/URL] and we get:[/U] x = [B]29 [/B]\n\nYou are comparing the costs of producing shoes at two different manufacturers. Company 1 charges \\$5\nYou are comparing the costs of producing shoes at two different manufacturers. Company 1 charges \\$5 per pair of shoes plus a \\$650 flat fee. Company 2 charges \\$4 per pair of shoes plus a \\$700 flat fee. How many pairs of shoes are produced when the total costs for both companies are equal? Let s be the number of shoes. We have two equations: (1) C = 5s + 650 (2) C = 4s + 700 Set the costs equal to each other 5s + 650 = 4s + 700 Subtract 4s from each side s + 650 = 700 Subtract 650 from each side [B]s =50[/B]\n\nYou are offered two different sales jobs. The first company offers a straight commission of 6% of th\nYou are offered two different sales jobs. The first company offers a straight commission of 6% of the sales. The second company offers a salary of \\$330 per week plus 2% of the sales. How much would you have to sell in a week in order for the straight commission offer to be at least as good? Let s be the sales and C be the weekly commission for each sales job. We have the following equations: [LIST=1] [*]C = 0.06s [*]C = 330 + 0.02s [/LIST] Set them equal to each other: 0.06s = 330 + 0.02s Subtract 0.02s from each side: 0.04s = 330 Using our [URL='http://www.mathcelebrity.com/1unk.php?num=0.04s%3D330&pl=Solve']equation solver[/URL], we get [B]s = 8,250[/B]\n\nYou borrowed \\$25 from your friend. You paid him back in full after 6 months. He charged \\$2 for inter\nYou borrowed \\$25 from your friend. You paid him back in full after 6 months. He charged \\$2 for interest. What was the annual simple interest rate that he charged you? Use the formula: I = Prt. We have I = 2, P = 25, t = 0.5 2 = 25(r)0.5 Divide each side by 0.5 4 = 25r Divide each side by 25 r = 4/25 [B]r = 0.16[/B] As a percentage, this is [B]16%[/B]\n\nYou can afford monthly deposits of \\$270 into an account that pays 3.0% compounded monthly. How long\nYou can afford monthly deposits of \\$270 into an account that pays 3.0% compounded monthly. How long will it be until you have \\$11,100 to buy a boat. Round to the next higher month. [U]Set up our accumulation expression:[/U] 270(1.03)^n = 11100 1.03^n = 41.1111111 [U]Take the natural log of both sides[/U] n * Ln(1.03) = 41.1111111 n = 3.71627843/0.0295588 n = 125.72 so round up to [B]126[/B]\n\nYou can pay a daily entrance fee of \\$3 or purchase a membership for the 12 week summer season for \\$8\nYou can pay a daily entrance fee of \\$3 or purchase a membership for the 12 week summer season for \\$82 and pay only \\$1 per day to swim. How many days would you have to swim to make the membership worthwhile? Set up cost equations: Daily entrance fee: 3d where d is the number of days of membership Membership fee 82 + 1d Set them equal to each other 82 + 1d = 3d Subtract d from each side: 2d = 82 Divide each side by 2 [B]d = 41[/B]\n\nYou deposit \\$2000 in an account that earns simple interest at an annual rate of 4%. How long must yo\nYou deposit \\$2000 in an account that earns simple interest at an annual rate of 4%. How long must you leave the money in the account to earn \\$500 in interest? The simple interest formula for the accumulated balance is: I = Prt We are given P = 2,000, r = 0.04, and I = 500. 500 = 2000(0.04)t 80t = 500 Divide each side by 80 t = 6.25 years.\n\nIf you have \\$6 left over, then 8 notebooks cost \\$22 - \\$6 = \\$16. 8 notebooks = \\$16 Divide each side of the equation by 8 Each notebook is \\$2\n\nYou have \\$10.00 to spend on tacos. Each taco costs \\$0.50. Write and solve an inequality that explain\nYou have \\$10.00 to spend on tacos. Each taco costs \\$0.50. Write and solve an inequality that explains how many tacos you can buy. Let's start with t as the number of tacos. We know that cost = price * quantity, so we have the following inequality for our taco spend: [B]0.5t <= 10 [/B] Divide each side of the inequality by 0.5 to isolate t: 0.5t/0.5 <= 10/0.5 Cancel the 0.5 on the left side and we get: t <= [B]20 [MEDIA=youtube]yy51EsGi1nM[/MEDIA][/B]\n\nYou have \\$37 to plant garden. If you spend \\$12.25 on seeds, how many packs of vegetable plants can\nYou have \\$37 to plant garden. If you spend \\$12.25 on seeds, how many packs of vegetable plants can you buy for 2.75 each? [U]How much do we have to spend on plants?[/U] \\$37 - 12.25 = \\$24.75 [U]Calculate how many vegetable plants we can buy. Set up an equation where x = vegetable plants[/U] 2.75x = 24.75 Divide each side by 2.75 [B]x = 9[/B]\n\nYou have a total of 42 math and science problems for homework. You have 10 more math problems than s\nYou have a total of 42 math and science problems for homework. You have 10 more math problems than science problems. How many problems do you have in each subject? Let m be the math problems and s be the science problems. We have two equations: (1) m + s = 42 (2) m = s + 10 Substitute (2) into (1) (s + 10) + s = 42 Combine like terms 2s + 10 = 42 Subtract 10 from each side 2s = 32 Divide each side by 2 [B]s = 16[/B] So that means m = 16 + 10 --> [B]m = 26 (m, s) = (26, 16)[/B]\n\nYou roll a standard, fair, 5-sided die and see what number you get. Find the sample space of this ex\nYou roll a standard, fair, 5-sided die and see what number you get. Find the sample space of this experiment. Write your answer using { } symbols, and write your values in order with a comma but no spaces between Sample Space: [B]{1,2,3,4,5}[/B]\n\nYou roll two six-sided dice. What is the probability that the sum is less than 13?\nYou roll two six-sided dice. What is the probability that the sum is less than 13? The probability is [B]1, or 100%[/B], since the maximum sum of two six-sided dice is 12.\n\nYour friends in class want you to make a run to the vending machine for the whole group. Everyone pi\nYour friends in class want you to make a run to the vending machine for the whole group. Everyone pitched in to make a total of \\$12.50 to buy snacks. The fruit drinks are \\$1.50 and the chips are \\$1.00. Your friends want you to buy a total of 10 items. How many drinks and how many chips were you able to purchase? Let c be the number of chips. Let f be the number of fruit drinks. We're given two equations: [LIST=1] [*]c + f = 10 [*]c + 1.5f = 12.50 [/LIST] Rearrange equation 1 by subtracting f from both sides: [LIST=1] [*]c = 10 - f [*]c + 1.5f = 12.50 [/LIST] Substitute equation (1) into equation (2): 10 - f + 1.5f = 12.50 To solve for f, we [URL='https://www.mathcelebrity.com/1unk.php?num=10-f%2B1.5f%3D12.50&pl=Solve']type this equation into our search engine[/URL] and we get: [B]f = 5[/B] Now, substitute this f = 5 value back into modified equation (1) above: c = 10 - 5 [B]c = 5[/B]\n\nyour starting salary at a new company is 45000. Each year you receive a 2% raise. How long will it t\nyour starting salary at a new company is 45000. Each year you receive a 2% raise. How long will it take you to make \\$80000? Let y be the number of years of compounding the 2% raise. Since 2% as a decimal is 0.02, we have the following equation for compounding the salary: 45000 * (1.02)^y = 80000 Divide each side by 45000: (1.02)^y = 1.77777777778 To solve this equation for y, we [URL='https://www.mathcelebrity.com/natlog.php?num=1.02%5Ey%3D1.77777777778&pl=Calculate']type it in our search engine[/URL] and we get: y = [B]29.05[/B] [B]Or just over 29 years[/B]\n\nz = (x + y)/mx; Solve for x\nz = (x + y)/mx; Solve for x Cross multiply: zmx = x + y Subtract x from each side zmx - x = y Factor out x x(zm - 1) = y Divide each sdie by zm - 1 x = y/(zm - 1)\n\nz/w=x+z/x+y for z\nz/w=x+z/x+y for z This is a literal equation. Let's isolate z on one side. Subtract z/x from each side. z/w - z/x = x + y Factor our z on the left side: z(1/w - 1/x) = x + y Divide each side by (1/w - 1/x) z = x + y/(1/w - 1/x) To remove reciprocals in the denominator, we rewrite 1/w - 1/x with a common denominator xw (x - w)/xw Then multiply x + y by the reciprocal z = [B](x + y)xw/(x - w)[/B]\n\nz=m-x+y, for x\nz=m-x+y, for x This is a literal equation. Let's add subtract (m + y) from each side: z - (m + y) = m - x + y - (m + y) The m + y terms cancel on the right side, so we have: z - m - y = -x Multiply each side by -1 to isolate x: -1(z - m - y) = -(-x) x = [B]m + y - z[/B]\n\nzy-dm=ky/t for y\nzy-dm=ky/t for y Isolate terms with y to solve this literal equation. Subtract zy from each side: zy - dm - zy = ky/t - zy Cancel the zy terms on the left side, we get: -dm = ky/t - zy Factor out y: y(k/t - z) = -dm Divide each side by (k/t - z) y = -dm/(k/t - z) (k/t - z) can be rewritten as (k - tz)/t We multiply -dm by the reciprocal of this quotient to get our simplified literal equation: y = [B]-dmt/(k - tz)[/B]\n\n|(x-7)/5|<=4\n|(x-7)/5|<=4 Set up two equations: [LIST=1] [*](x-7)/5 <= 4 [*](x-7)/5 > -4 [/LIST] Cross Multiply (1): x - 7 <= 20 Add 7 to each side: x <= 27 Cross Multiply (2): x - 7 > -20 Add 7 to each side: x > -13"
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http://www.iust.ac.ir/ijoce/search.php?sid=1&slc_lang=en&author=Khatibinia | [
"## Search published articles\n\nShowing 11 results for Khatibinia\n\nS. Shojaee, M. Arjomand, M. Khatibinia,\nVolume 3, Issue 1 (3-2013)\nAbstract\n\nAn efficient method for size and layout optimization of the truss structures is presented in this paper. In order to this, an efficient method by combining an improved discrete particle swarm optimization (IDPSO) and method of moving asymptotes (MMA) is proposed. In the hybrid of IDPSO and MMA, the nodal coordinates defining the layout of the structure are optimized with MMA, and afterwards the results of MMA are used in IDPSO to optimize the cross-section areas. The results show that the hybrid of IDPSO and MMA can effectively accelerate the convergence rate and can quickly reach the optimum design.\nM. Khatibinia, H. Chiti, A. Akbarpour , H. R. Naseri,\nVolume 6, Issue 1 (1-2016)\nAbstract\n\nThis study focuses on the shape optimization of concrete gravity dams considering dam–water–foundation interaction and nonlinear effects subject to earthquake. The concrete gravity dam is considered as a two–dimensional structure involving the geometry and material nonlinearity effects. For the description of the nonlinear behavior of concrete material under earthquake loads, the Drucker–Prager model based on the associated flow rule is adopted in this study. The optimum design of concrete gravity dams is achieved by the hybrid of an improved gravitational search algorithm (IGSA) and the orthogonal crossover (OC), called IGSA–OC. In order to reduce the computational cost of optimization process, the support vector machine approach is employed to approximate the dam response instead of directly evaluating it by a time–consuming finite element analysis. To demonstrate the nonlinear behavior of concrete material in the optimum design of concrete gravity dams, the shape optimization of a real dam is presented and compared with that of dam considering linear effect.\nH. Chiti, M. Khatibinia, A. Akbarpour , H. R. Naseri,\nVolume 6, Issue 3 (9-2016)\nAbstract\n\nThe paper deals with the reliability–based design optimization (RBDO) of concrete gravity dams subjected to earthquake load using subset simulation. The optimization problem is formulated such that the optimal shape of concrete gravity dam described by a number of variables is found by minimizing the total cost of concrete gravity dam for the given target reliability. In order to achieve this purpose, a framework is presented whereby subset simulation is integrated with a hybrid optimization method to solve the RBDO approach of concrete gravity dam. Subset simulation with Markov Chain Monte Carlo (MCMC) sampling is utilized to estimate accurately the failure probability of dams with a minimum number of samples. In this study, the concrete gravity dam is treated as a two–dimensional structure involving the material nonlinearity effects and dam–reservoir–foundation interaction. An efficient metamodel in conjunction with subset simulation–MCMC is provided to reduce the computational cost of dynamic analysis of dam–reservoir–foundation system. The results demonstrate that the RBDO approach is more appropriate than the deterministic optimum approach for the optimal shape design of concrete gravity dams.\n\nM. Roodsarabi, M. Khatibinia , S. R. Sarafrazi,\nVolume 6, Issue 3 (9-2016)\nAbstract\n\nThis study focuses on the topology optimization of structures using a hybrid of level set method (LSM) incorporating sensitivity analysis and isogeometric analysis (IGA). First, the topology optimization problem is formulated using the LSM based on the shape gradient. The shape gradient easily handles boundary propagation with topological changes. In the LSM, the topological gradient method as sensitivity analysis is also utilized to precisely design new holes in the interior domain. The hybrid of these gradients can yield an efficient algorithm which has more flexibility in changing topology of structure and escape from local optimal in the optimization process. Finally, instead of the conventional finite element method (FEM) a Non–Uniform Rational B–Splines (NURBS)–based IGA is applied to describe the field variables as the geometry of the domain. In IGA approach, control points play the same role with nodes in FEM, and B–Spline functions are utilized as shape functions of FEM for analysis of structure. To demonstrate the performance of the proposed method, three benchmark examples widely used in topology optimization are presented. Numerical results show that the proposed method outperform other LSMs.\n\nM. Khatibinia, H. Gholami, S. F. Labbafi,\nVolume 6, Issue 4 (10-2016)\nAbstract\n\nTuned mass dampers (TMDs) are as a efficient control tool in order to reduce undesired vibrations of tall buildings and large–span bridges against lateral loads such as wind and earthquake. Although many researchers has been widely investigated TMD systems due to its simplicity and application, the optimization of parameters and placement of TMD are challenging tasks. Furthermore, ignoring the effects of soil–structure interaction (SSI) may lead to unrealistic desig of structure and its dampers. Hence, the effects of SSI should be considered in the design of TMD. Therefore, the main aim of this study is to optimize parameters of TMD subjected to earthquake and considering the effects of SSI. In this regard, the parameters of TMD including mass, stiffness and damping optimization are considered as the variables of optimization. The maximum absolute displacement and acceleration of structure are also simultaneously selected as objective functions. The multi –objective particle swarm optimization (MOPSO) algorithm is adopted to find the optimal parameters of TMD. In this study, the Lagrangian method is utilized for obtaining the equations of motion for SSI system, and the time domain analysis is implemented based on Newmark method. In order to investigate the effects of SSI in the optimal design of TMD, a 40 storey shear building with a TMD subjected to the El–Centro earthquake is considered. The numerical results show that the SSI effects have the significant influence on the optimum parameters of TMD.\n\nM. Feizbakhsh , M. Khatibinia,\nVolume 7, Issue 3 (7-2017)\nAbstract\n\nThis study investigates the prediction model of compressive strength of self–compacting concrete (SCC) by utilizing soft computing techniques. The techniques consist of adaptive neuro–based fuzzy inference system (ANFIS), artificial neural network (ANN) and the hybrid of particle swarm optimization with passive congregation (PSOPC) and ANFIS called PSOPC–ANFIS. Their performances are comparatively evaluated in order to find the best prediction model. In this study, SCC mixtures containing different percentage of nano SiO2 (NS), nano–TiO2 (NT), nano–Al2O3 (NA), also binary and ternary combining of these nanoparticles are selected. The results indicate that the PSOPC–ANFIS approach in comparison with the ANFIS and ANN techniques obtains an improvement in term of generalization and predictive accuracy. Although, the ANFIS and ANN techniques are a suitable model for this purpose, PSO integrated with the ANFIS is a flexible and accurate method due tothe stronger global search ability of the PSOPC algorithm.\n\nM. Khatibinia, M. Roodsarabi, M. Barati,\nVolume 8, Issue 2 (8-2018)\nAbstract\n\nThis paper presents the topology optimization of plane structures using a binary level set (BLS) approach and isogeometric analysis (IGA). In the standard level set method, the domain boundary is descripted as an isocountour of a scalar function of a higher dimensionality. The evolution of this boundary is governed by Hamilton–Jacobi equation. In the BLS method, the interfaces of subdomains are implicitly represented by the discontinuities of BLS functions taking two values 1 or −1. The subdomains interfaces are represented by discontinuities of these functions. Using a two–phase approximation and the BLS approach the original structural optimization problem is reformulated as an equivalent constrained optimization problem in terms of this level set function. For solving drawbacks of the conventional finite element method (FEM), IGA based on a Non–Uniform Rational B–Splines (NURBS) is adopted to describe the field variables as the geometry of the domain. For this purpose, the B–Spline functions are utilized as the shape functions of FEM for analysis of structure and the control points are considered the same role with nodes in FEM. Three benchmark examples are presented to investigate the performance the topology optimization based on the proposed method. Numerical results demonstrate that the BLS method with IGA can be utilized in this field.\nR. Kamgar, M. Khatibinia, M. Khatibinia,\nVolume 9, Issue 2 (4-2019)\nAbstract\n\nMany researches have focused on the optimal design of tuned mass damper (TMD) system without the effect of soil–structure interaction (SSI), so that ignoring the effect of SSI may lead to an undesirable and unrealistic design of TMD. Furthermore, many optimization criteria have been proposed for the optinal design of the TMD system. Hence, the main aim of this study is to compare different optimization criteria for the optimal design of the TMD system considering the effects of SSI in a high–rise building. To acheive this purpose, the optimal TMD for a 40–storey shear building is firstly evaluated by expressing the objective functions in terms of the reduction of structural responses (including the displacement and acceleration) and the limitation of the scaled stroke of TMD. Then, the best optimization criterion is selected, which leads to the best performance for the vibration control of the structure. In this study, the whale optimization algorithm (WOA) is employed to optimize the parameters of the TMD system. The numerical results show that the soil type and selected objective function efficiently affect the optimal design of the TMD system.\nM. Araghi, M. Khatibinia,\nVolume 9, Issue 2 (4-2019)\nAbstract\n\nFlow number of asphalt–aggregate mixtures as an explanatory factor has been proposed in order to assess the rutting potential of asphalt mixtures. This study proposes a multiple–kernel based support vector machine (MK–SVM) approach for modeling of flow number of asphalt mixtures. The MK–SVM approach consists of weighted least squares–support vector machine (WLS–SVM) integrating two kernel functions in order to improve the learning and generalization ability of WLS–SVM. In the proposed method, a linear convex combination of the radial basis function (RBF) and Morlet wavelet kernel functions is adopted, which are considered as the most popular kernel functions. To validate the efficiency of the proposed method, experiments are conducted on a database including 118 uniaxial dynamic creep test results. The results of the statistical criteria show a good agreement between the predicted and measured flow number values. Further, the simulation results demonstrate that the proposed MK–SVM approach has more superior performance than the single kernel based WLS–SVM and other methods found in the literature.\nM. Khatibinia, M. Mahmoudi, H. Eliasi,\nVolume 10, Issue 1 (1-2020)\nAbstract\n\nActive tuned mass damper (ATMD) systems have attracted the considerable attention of researchers for protecting buildings subjected to earthquake loading. This paper presentes the development of an optimal sliding mode control (OSMC) system for a building equipped with ATMD. In the OSMC technique, a linear sliding surface is used with the slope of this surface designed such that a given (or desired) cost function is minimized. The design is obtained by transforming the system into the regular form. In the regular form, the system is divided into two subsystems inclding: a control term explicitly appears, and other control terms do not appear. In order to demonstrate the capability of the OSMC system, an 11–story realistic building with a TMD installed on the top story of the structure is considered. For achieving this purpose, four well–known earthquake records are selected to evaluate the performance of the OSMC system. Results show that the OSMC technique performs better than other control techniques in the reduction of seismic responses of the structure.\nM. Khatibinia, M. Roodsarabi,\nVolume 10, Issue 3 (6-2020)\nAbstract\n\nThe present study proposes a hybrid of the piecewise constant level set (PCLS) method and isogeometric analysis (IGA) approach for structural topology optimization. In the proposed hybrid method, the discontinuities of PCLS functions is used in order to present the geometrical boundary of structure. Additive Operator Splitting (AOS) scheme is also considered for solving the Lagrange equations in the optimization problem subjected to some constraints. For reducing the computational cost of the PCLS method, the Merriman–Bence–Osher (MBO) type of projection scheme is applied. In the optimization process, the geometry of structures is described using the Non–Uniform Rational B–Splines (NURBS)–based IGA instead of the conventional finite element method (FEM). The numerical examples illustrate the efficiency of the PCLS method with IGA in the efficiency, convergence and accuracy compared with the other level set methods (LSMs) in the framework of 2–D structural topology optimization. The results of the topology optimization reveal that the proposed method can obtain the same topology in lower number of convergence iteration.\n\n Page 1 from 1\n\n© 2023 CC BY-NC 4.0 | Iran University of Science & Technology\n\nDesigned & Developed by : Yektaweb"
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https://www.geeksforgeeks.org/gre-geometry-quadrilaterals/?ref=lbp | [
"Related Articles\n• Last Updated : 24 May, 2019\n\nA quadrilateral is a 2 – D shape which contain 4 sides. It consist of four sides, four angles and four vertices.",
null,
"Different type of quadrilateral and their properties:\n\n1. Rectangle:\nThe rectangle, like the square, is one of the most commonly known quadrilaterals. It is defined as having all four interior angles 90° (right angles).",
null,
"• All four angles in a rectangle are 90°.\n• Opposite side are parallel and of same length(or congruent).\n• Diagonals of rectangle bisect each other and divide a the rectangle into two congruent triangles.\n• Area of rectangle = length * breadth.\n• Perimeter of rectangle = 2*(length + breadth)\n\n2. Square:\nIn geometry, a square is a regular quadrilateral, which means that it has four equal sides and four equal angles (90-degree angles, or (100-gradian angles or right angles). It can also be defined as a rectangle in which two adjacent sides have equal length.",
null,
"• All four angles in a rectangle are 90°.\n• All sides are equal and opposite sides are parallel(or congruent).\n• Diagonals of square bisect each other at 90° and divide the square into two congruent triangle.\n• Area of square = side * side\n• Perimeter of square = 4 * side\n\n3. Rhombus:\nA Rhombus is a flat shape with 4 equal straight sides. A rhombus looks like a diamond. All sides have equal length. Opposite sides are parallel, and opposite angles are equal (it is a Parallelogram).",
null,
"• All sides are equal in a rhombus.\n• Opposite angles are equal in rhombus.\n• The diagonals of rhombus intersect each other at equal angles.\n• Area of rhombus = (diagonal1 * diagonal2) / 2\n• Perimeter of rhombus = 4 * side\n\n4. Trapezium:\nIn Euclidean geometry, a convex quadrilateral with at least one pair of parallel sides is referred to as a trapezoid.",
null,
"• Atleast one pair of opposite side is parallel to each other.\n• Area of trapezium = (sum of length of parallel side(or bases)) * height / 2.\n• Perimeter of trapezium = sum of all sides.\n\n5. Parallelogram:\nIn Euclidean geometry, a parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite or facing sides of a parallelogram are of equal length and the opposite angles of a parallelogram are of equal measure.",
null,
"• Opposite sides are parallel and congurent to each other.\n• Opposite angles are congurent to each other.\n• Sum of adjacent angles is 180°\n• Diagonals of parallelogram bisect each other and divide the parallelogram into two congurent triangles.\n• Area of parallelogram = length * breadth\n• Perimeter of parallelogram = 2(length + breadth)\n\nWhat will be the area and perimeter of given rectangle if length of rectangle is 5cm and breadth of rectangle is 7cm?",
null,
"Solution:\n\n```Area of rectangle = length * breadth.\nArea of rectangle = 5 * 7 = 35 cm2\nPerimeter of rectangle = 2*(length + breadth)\nPerimeter of rectangle = 2*(5 + 7) = 24cm\n```\n\nWhat will be the are of given rhombus if diagonal of rhombus are 4cm and 6cm?",
null,
"```Area of rhombus = (diagonal1 * diagonal2) / 2\nArea of rhombus = (4 * 6) / 2 = 12 cm2\n```\n\nWhat will be area and perimeter of given trapezium if AB is parallel to CD?",
null,
"```Area of trapezium = (sum of length of parallel side(or bases)) * height / 2.\nArea of trapezium = (3 + 4. 5) * 3 / 2\n= 11.25 cm2\nPerimeter of trapezium = Sum of all sides\n= 3 + 3 + 3.5 + 4.5 = 14 cm\n```\nMy Personal Notes arrow_drop_up"
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412170845/quadli.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412173606/quadli2.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412173423/quadli1.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412174653/quadli6.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412174418/quadli5.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412173805/quadli4.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412173606/quadli2.png",
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"https://media.geeksforgeeks.org/wp-content/uploads/20190412174653/quadli6.png",
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https://ai.stackexchange.com/questions/6102/can-a-neural-network-work-out-the-concept-of-distance | [
"# Can a neural network work out the concept of distance?\n\nImagine a game where it is a black screen apart from a red pixel and a blue pixel. Given this game to a human, they will first see that pressing the arrow keys will move the red pixel. The next thing they will try is to move the red pixel onto the blue pixel.\n\nGive this game to an AI, it will randomly move the red pixel until a million tries later it accidentally moves onto the blue pixel to get a reward. If the AI had some concept of distance between the red and blue pixel, it might try to minimize this distance.\n\nWithout actually programming in the concept of distance, if we take the pixels of the game can we calculate a number(s), such as \"entropy\", that would be lower when pixels are far apart than when close together? It should work with other configurations of pixels. Such as a game with three pixels where one is good and one is bad. Just to give the neural network more of a sense of how the screen looks? Then give the NN a goal, such as \"try to minimize the entropy of the board as well as try to get rewards\".\n\nIs there anything akin to this in current research?\n\n• I don't think you would directly feed it the image and have it find the distance... Rather you would feed it a set of numbers that represent distance, vertical distance, horizontal distance, etc. Apr 18, 2018 at 14:19\n• @Pheo yes, but you would have to feed it different values for every type of \"game\". Whereas what I'm saying is, could we have some global type of value that is high when pixels are grouped together and low when pixels are spaced apart? Apr 21, 2018 at 17:46\n• \"The next thing they will try is to move the red pixel onto the blue pixel.\" might to do not will \"red\" and \"blue\" are most times are enemys so you will start to increase distance before the blue pixel notice you.\n– Lee\nOct 23, 2019 at 6:27\n\nI'm going to take your question at face value, and go really deep into this topic.\n\nYes, they can. The typical human mind can. But consider the human mind. Millions, if not billions, of neurons. In fact, one can consider distance as a human concept, simply a theory developed from interactions with the world.\n\nTherefore, given a year or two, with a ton of neurons on your hand, you could replicate this scenario. That is if your computer is as parallel as the human mind. The short explanation is that the human mind is very parallel.\n\nHowever, it would be simpler to calculate the distance with a program, not an AI, and simply feed the result to the AI that would make the decisions.\n\nConsider the amount of time you have spent looking at a screen. If you can tell the (approximate) distance between two pixels, so can a Neural Network, as you are one. However, add the amount of time you have spent alive and learning into the equation, and it becomes a disaster.\n\n### The human brain is parallel\n\nThis is a result of the fact that all of the neurons in the human brain are independent of each other. They can run true simultaneous actions, thus making the action of interpreting images and such much easier, as blocks of neurons can \"think\" independent of the operations of the others, limiting what would be \"lag\" to a minuscule amount.\n\nYou can create AI to \"see\" as a human. As you said, giving the human the keys, he will click randomly. He just needs to know which keys he presses that brings him closer to other objects on the screen. I think the basics of an AI is object recognition. I would try to create a script to map the screen objects of the game. There are legal examples in Python.\n\nI would try to follow a path like this:\n\n• Make the AI understand that by clicking the arrows or the WASD and it is in the context GAME, the object that move pixels according to the direction, represents the main author (the player).\n\n• In parallel: map all boundaries of the region and index different objects within that region to automatically have the coordinate domain and object distance. AI needs to SEE (stream) the game and through images to categorize objects. Do you understand what I mean?\n\n• In parallel: The AI needs to be aware of all texts and information that is on the screen (all mapped, remember?). You need to understand when a text changes or something different happens. For example: whenever he returns to the initial position of each phase, whenever he has a count, what happens when the cout reaches zero or a common number that generates another type of change.\n\n• He needs to understand what is repeated at every \"respawn\". You also need to understand what \"respawn\" is. Maybe a certain map position on every map it returns whenever a count on the screen ends. Or when it comes up against a certain type of object (mapped object)\n\nTo be honest, if you want to create a super intelligent robot, you can go following all the steps that go through the heads of different humans, or the best humans, or the rules of each game. But sometimes it's easier to build specific bots to perform specific tasks. It depends on what you want to do\n\n• He was not asking how you would do it, but rather can you do it. Apr 18, 2018 at 17:53\n• It is possible to do it in several ways. I passed the way I would take to create the template. It is not a theory, it is a process that can encompass other processes according to the evolution of AI.\n– GIA\nApr 18, 2018 at 18:12\n\nWhat you mention there is the perfect example for path-planning, which is extensively researched in AI.\n\nPlease look for A-star algorithm and how to enhance it with neural networks :)\n\nWe can break down the problem as follows:\n\nFirst, if you have two points on a plane and feed the coordinates of those points to a neural network (e.g., a vector $$< x_0, y_0, x_1, y_1 >$$) and and train it on a label thats the actual distance (e.g., $$\\sqrt{(x_0 - y_0)^2 + (x_1-y_1)^2}$$), it should be able to learn this relationship with arbitrarily close accuracy.\n\nNext, if you have an image similar to what you describe, and feed that through a different neural network (e.g., a CNN), and as labels you used the the points of the two dots (once again $$< x_0, y_0, x_1, y_1 >$$), then it should be able to learn that relationship with arbitrarily close accuracy once again.\n\nOf course, there's no reason to do this in two separate neural network, so we can just combine the two end-to-end have a model that takes the image as input and the distance as output.\n\nThis model would need to be trained on labeled data, however, so you'd either need to generate the data yourself or label images.\n\nBut if you wanted it to learn the notion of closing a distance in a less supervised way, you'd need to use reinforcement learning. In this case, you'd have to setup an environment that incentivises the agent to reduce the distance. This could be as simple as gaining reward if an action reduces the distance.\n\nAnother approach would be to incentivise the agent using future reward. That is, it's reward doesn't just come from the results of the next immediet state, but there's also contributions from the next possible state, and the one after that, and so on. This is the idea behind Deep Q-Learning, and I implement a simple example (very similar to what you're describing) in this notebook.\n\nSo, now the question is: has this implementation done something other than randomly moving around until it follows a path to success?\n\nIn your example, you talk about rewarding the agent when it lands on the goal. But in what I described, it gained reward by moving closer to the goal (either through the Q-Function or directly from the environment). It is able to do so by learning some abstract idea of distance (which can be illustrated in the supervised version).\n\nWhen a human learns this, it's for the same exact reason: the human is gaining a reward for moving in that direction through a sense of future rewards.\n\nI'd say that, given enough training and data, reinforcement learning could learn this concept with ease. As far as other rewards being present on the board (e.g., \"minimise the entropy of the board as well as try to get rewards\"), you need to think about what it is you're asking. Would you rather the agent minimize distance or maximize reward? Cause, in general, it can't do both. If you're looking for some balance between the two, then really you're just redefining the reward to also consider the distance."
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https://simplystatistics.org/posts/2016-02-17-non-tidy-data/ | [
"# Non-tidy data\n\nJeff Leek\n2016-02-17\n\nDuring the discussion that followed the ggplot2 posts from David and I last week we started talking about tidy data and the man himself noted that matrices are often useful instead of “tidy data” and I mentioned there might be other data that are usefully “non tidy”. Here I will be using tidy/non-tidy according to Hadley’s definition. So tidy data have:\n\n• One variable per column\n• One observation per row\n• Each type of observational unit forms a table\n\nI push this approach in my guide to data sharing and in a lot of my personal work. But note that non-tidy data can definitely be already processed, cleaned, organized and ready to use.\n\n@hadleywickham @drob @mark_scheuerell I’m saying that not all data are usefully tidy (and not just matrices) so I care more abt flexibility\n\n— Jeff Leek (@jtleek) February 12, 2016\n\nThis led to a very specific blog request:\n\n@jtleek @drob I want a blog post on non-tidy data!\n\nSo I thought I’d talk about a couple of reasons why data are usefully non-tidy. The basic reason is that I usually take a problem first, not solution backward approach to my scientific research. In other words, the goal is to solve a particular problem and the format I chose is the one that makes it most direct/easy to solve that problem, rather than one that is theoretically optimal. To illustrate these points I’ll use an example from my area.\n\nExample data\n\nOften you want data in a matrix format. One good example is gene expression data or data from another high-dimensional experiment. David talks about one such example in his post here. He makes the (valid) point that for students who aren’t going to do genomics professionally, it may be more useful to learn an abstract tool such as tidy data/dplyr. But for those working in genomics, this can make you do unnecessary work in the name of theory/abstraction.\n\nHe analyzes the data in that post by first tidying the data.\n\n``` library(dplyr) library(tidyr) library(stringr) library(readr) library(broom) original_data % separate(NAME, c(\"name\", \"BP\", \"MF\", \"systematic_name\", \"number\"), sep = \"\\\\|\\\\|\") %>% mutate_each(funs(trimws), name:systematic_name) %>% select(-number, -GID, -YORF, -GWEIGHT) %>% gather(sample, expression, G0.05:U0.3) %>% separate(sample, c(\"nutrient\", \"rate\"), sep = 1, convert = TRUE) ```\n\nIt isn’t 100% tidy as data of different types are in the same data frame (gene expression and metadata/phenotype data belong in different tables). But its close enough for our purposes. Now suppose that you wanted to fit a model and test for association between the “rate” variable and gene expression for each gene. You can do this with David’s tidy data set, dplyr, and the broom package like so:\n\n``` rate_coeffs = cleaned_data %>% group_by(name) %>% do(fit = lm(expression ~ rate + nutrient, data = .)) %>% tidy(fit) %>% dplyr::filter(term==\"rate\") ```\n\nOn my computer we get something like:\n\n``` system.time( cleaned_data %>% group_by(name) %>% + do(fit = lm(expression ~ rate + nutrient, data = .)) %>% + tidy(fit) %>% + dplyr::filter(term==\"rate\")) |==========================================================|100% ~0 s remaining user system elapsed 12.431 0.258 12.364 ```\n\nLet’s now try that analysis a little bit differently. As a first step, lets store the data in two separate tables. A table of “phenotype information” and a matrix of “expression levels”. This is the more common format used for these type of data. Here is the code to do that:\n\n``` expr = original_data %>% select(grep(\"[0:9]\",names(original_data))) rownames(expr) = original_data %>% separate(NAME, c(\"name\", \"BP\", \"MF\", \"systematic_name\", \"number\"), sep = \"\\\\|\\\\|\") %>% select(systematic_name) %>% mutate_each(funs(trimws),systematic_name) %>% as.matrix() vals = data.frame(vals=names(expr)) pdata = separate(vals,vals,c(\"nutrient\", \"rate\"), sep = 1, convert = TRUE) expr = as.matrix(expr) ```\n\nIf we leave the data in this format we can get the model fits and the p-values using some simple linear algebra\n\n``` expr = as.matrix(expr) mod = model.matrix(~ rate + as.factor(nutrient),data=pdata) rate_betas = expr %*% mod %*% solve(t(mod) %*% mod) ```\n\nThis gives the same answer after re-ordering\n\n``` all(abs(rate_betas[,2]- rate_coeffs\\$estimate[ind]) < 1e-5,na.rm=T) TRUE ```\n\nBut this approach is much faster.\n\n``` system.time(expr %*% mod %*% solve(t(mod) %*% mod)) user system elapsed 0.015 0.000 0.015 ```\n\nThis requires some knowledge of linear algebra and isn’t pretty. But it brings us to the first general point: you might not use tidy data because some computations are more efficient if the data is in a different format.\n\nMany examples from graphical models, to genomics, to neuroimaging, to social sciences rely on some kind of linear algebra based computations (matrix multiplication, singular value decompositions, eigen decompositions, etc.) which are all optimized to work on matrices, not tidy data frames. There are ways to improve performance with tidy data for sure, but they would require an equal amount of custom code to take advantage of say C, or vectorization properties in R.\n\nOk now the linear regressions here are all treated independently, but it is very well known that you get much better performance in terms of the false positive/true positive tradeoff if you use an empirical Bayes approach for this calculation where you pool variances.\n\nIf the data are in this matrix format you can do it with R like so:\n\n``` library(limma) fit_limma = lmFit(expr,mod) ebayes_limma = eBayes(fit_limma) topTable(ebayes_limma) ```\n\nThis approach is again very fast, optimized for the calculations being performed and performs much better than the one-by-one regression approach. But it requires the data in matrix or expression set format. Which brings us to the second general point: ****you might not use tidy data because many functions require a different, also very clean and useful data format, and you don’t want to have to constantly be switching back and forth. ****Again, this requires you to be more specific to your application, but the potential payoffs can be really big as in the case of limma.\n\nI’m showing an example here with expression sets and matrices, but in NLP the data are often input in the form of lists, in graphical analyses as matrices, in genomic analyses as GRanges lists, etc. etc. etc. One option would be to rewrite all infrastructure in your area of interest to accept tidy data formats but that would be going against conventions of a community and would ultimately cost you a lot of work when most of that work has already been done for you.\n\nThe final point, which I won’t discuss here is that data are often usefully represented in a non-tidy way. Examples include the aforementioned GRanges list which consists of (potentially) ragged arrays of intervals and quantitative measurements about them. You could force these data to be tidy by the definition above, but again most of the infrastructure is built around a different format that is much more intuitive for that type of data. Similarly data from other applications may be more suited to application specific formats.\n\nIn summary, tidy data is a useful conceptual idea and is often the right way to go for general, small data sets, but may not be appropriate for all problems. Here are some examples of data formats (biased toward my area, but there are others) that have been widely adopted, have a ton of useful software, but don’t meet the tidy data definition above. I will define these as “processed data” as opposed to “tidy data”.\n\nI’m sure there are a ton more I’m missing and would be happy to get some suggestions on Twitter too."
]
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https://www.nuclear-power.net/nuclear-engineering/radiation-protection/equivalent-dose/sievert-unit-of-equivalent-dose/calculation-of-shielded-dose-rate-in-sieverts/ | [
"Calculation of Shielded Dose Rate in Sieverts",
null,
"In radiation protection, the sievert is a derived unit of equivalent dose and effective dose. The sievert represents the equivalent biological effect of the deposit of a joule of gamma rays energy in a kilogram of human tissue. Unit of sievert is of importance in radiation protection and was named after the Swedish scientist Rolf Sievert, who did a lot of the early work on dosimetry in radiation therapy.\n\nCalculation of Shielded Dose Rate in Sieverts\n\nAssume the point isotropic source which contains 1.0 Ci of 137Cs, which has a half-life of 30.2 years. Note that the relationship between half-life and the amount of a radionuclide required to give an activity of one curie is shown below. This amount of material can be calculated using λ, which is the decay constant of certain nuclide:",
null,
"About 94.6 percent decays by beta emission to a metastable nuclear isomer of barium: barium-137m. The main photon peak of Ba-137m is 662 keV. For this calculation, assume that all decays go through this channel.\n\nCalculate the primary photon dose rate, in gray per hour (Gy.h-1), at the outer surface of a 5 cm thick lead shield. Then calculate the equivalent dose rate. Assume that this external radiation field penetrate uniformly through the whole body. Primary photon dose rate neglects all secondary particles. Assume that the effective distance of the source from the dose point is 10 cm. We shall also assume that the dose point is soft tissue and it can reasonably be simulated by water and we use the mass energy absorption coefficient for water.\n\nSolution:\n\nThe primary photon dose rate is attenuated exponentially, and the dose rate from primary photons, taking account of the shield, is given by:",
null,
"As can be seen, we do not account for the buildup of secondary radiation. If secondary particles are produced or if the primary radiation changes its energy or direction, then the effective attenuation will be much less. This assumption generally underestimates the true dose rate, especially for thick shields and when the dose point is close to the shield surface, but this assumption simplifies all calculations. For this case the true dose rate (with the buildup of secondary radiation) will be more than two times higher.\n\nTo calculate the absorbed dose rate, we have to use in the formula:\n\n• k = 5.76 x 10-7\n• S = 3.7 x 1010 s-1\n• E = 0.662 MeV\n• μt/ρ = 0.0326 cm2/g (values are available at NIST)\n• μ = 1.289 cm-1 (values are available at NIST)\n• D = 5 cm\n• r = 10 cm\n\nResult:\n\nThe resulting absorbed dose rate in grays per hour is then:",
null,
"Since the radiation weighting factor for gamma rays is equal to one and we have assumed the uniform radiation field, we can directly calculate the equivalent dose rate from the absorbed dose rate as:",
null,
"If we want to account for the buildup of secondary radiation, then we have to include the buildup factor. The extended formula for the dose rate is then:",
null,
"——–\n\nReferences:\n\n1. Knoll, Glenn F., Radiation Detection and Measurement 4th Edition, Wiley, 8/2010. ISBN-13: 978-0470131480.\n2. Stabin, Michael G., Radiation Protection and Dosimetry: An Introduction to Health Physics, Springer, 10/2010. ISBN-13: 978-1441923912.\n3. Martin, James E., Physics for Radiation Protection 3rd Edition, Wiley-VCH, 4/2013. ISBN-13: 978-3527411764.\n4. U.S.NRC, NUCLEAR REACTOR CONCEPTS\n5. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.\n\nNuclear and Reactor Physics:\n\n1. J. R. Lamarsh, Introduction to Nuclear Reactor Theory, 2nd ed., Addison-Wesley, Reading, MA (1983).\n2. J. R. Lamarsh, A. J. Baratta, Introduction to Nuclear Engineering, 3d ed., Prentice-Hall, 2001, ISBN: 0-201-82498-1.\n3. W. M. Stacey, Nuclear Reactor Physics, John Wiley & Sons, 2001, ISBN: 0- 471-39127-1.\n4. Glasstone, Sesonske. Nuclear Reactor Engineering: Reactor Systems Engineering, Springer; 4th edition, 1994, ISBN: 978-0412985317\n5. W.S.C. Williams. Nuclear and Particle Physics. Clarendon Press; 1 edition, 1991, ISBN: 978-0198520467\n6. G.R.Keepin. Physics of Nuclear Kinetics. Addison-Wesley Pub. Co; 1st edition, 1965\n7. Robert Reed Burn, Introduction to Nuclear Reactor Operation, 1988.\n8. U.S. Department of Energy, Nuclear Physics and Reactor Theory. DOE Fundamentals Handbook, Volume 1 and 2. January 1993.\n9. Paul Reuss, Neutron Physics. EDP Sciences, 2008. ISBN: 978-2759800414.\n\nSievert"
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