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http://www.ccsenet.org/journal/index.php/ijsp/article/view/0/40417	[ "### A Bivariate Index for Visually Measuring Marginal Inhomogeneity in Square Tables\n\n• Shuji Ando\n\n#### Abstract\n\nFor square tables, the marginal homogeneity model which has a structure that the row marginal distribution is equal to the column marginal distribution was proposed. Thereafter, various extended models of marginal homogeneity have been proposed, these models can be classified into two types marginal inhomogeneity. On the other hand, various indexes which measure the degree of deviation from marginal homogeneity have been proposed. However these indexes cannot concurrently define degrees of deviation from marginal homogeneity with respect to two types marginal inhomogeneity. This paper proposes a bivariate index that can concurrently define degrees of deviation from those. The proposed bivariate index would also be utility for visually comparing degrees of deviation from marginal homogeneity in several tables using confidence regions.", null, "This work is licensed under a Creative Commons Attribution 4.0 License.\n• ISSN(Print): 1927-7032\n• ISSN(Online): 1927-7040\n• Started: 2012\n• Frequency: bimonthly\n\n### Journal Metrics\n\n• h-index (December 2019): 15\n• i10-index (December 2019): 24\n• h5-index (December 2019): N/A\n• h5-median(December 2019): N/A" ]	[ null, "data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAAFAAAAAPCAMAAABEF7i9AAAAllBMVEUAAAD///+rsapERER3d3eIiIjMzMzu7u4iIiKUmZO6v7rKzsoODg4RERFVVVUNDQ0NDg0PEA8zMzNLTEtbXltmZmZydnF9gn2AgICPkI+ZmZmqqqq7u7vFxsXIzMgNDQwZGRkgICAhISEkJSMnKCcuMC4xMzE5Ozk7PTtBQkFCQkJDQ0Nna2eGhoaHh4ezuLLGysbd3d1wVGpAAAAAv0lEQVR42q2T2RKCMAxFk3QDBUEUF3Df9+X/f87WqTiMLwK906e2c3pv0gA4FzqVAZLRBYTdYoFP1O7WdPcB+oEoHmEbotaoIfCw+rqW0Cea9ZoBY4HSA6YYeBLZTmce2/raSxWBQ5ig10HOQ5XlyGOdeWrP3wurOhyAVKYrncRk5gHR/VE4/N/fj0OWhopnyKOywxpAigTmIaS6hskTz9tSDWsB90vHXfbntwJ4Wjv4h7otV8s7LppOiutZdqwXrQAI6Vlm1kYAAAAASUVORK5CYII=", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.83040124,"math_prob":0.47154158,"size":1704,"snap":"2020-24-2020-29","text_gpt3_token_len":425,"char_repetition_ratio":0.13058823,"word_repetition_ratio":0.036885247,"special_character_ratio":0.2112676,"punctuation_ratio":0.08300395,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9635285,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-03T15:17:59Z\",\"WARC-Record-ID\":\"<urn:uuid:c8d24417-48dc-4b0a-97ab-f24f21ea4b44>\",\"Content-Length\":\"15437\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:5ee0a847-1a6a-4542-914d-db65033dacca>\",\"WARC-Concurrent-To\":\"<urn:uuid:158156ca-012e-4d52-9e86-c897133641d5>\",\"WARC-IP-Address\":\"66.155.58.71\",\"WARC-Target-URI\":\"http://www.ccsenet.org/journal/index.php/ijsp/article/view/0/40417\",\"WARC-Payload-Digest\":\"sha1:RNMVQRSIYCOEMNNGZ6LBT6FPSMSBIGNG\",\"WARC-Block-Digest\":\"sha1:MTDSU3MLUFXX64H2NAR5ZMIBHD6SZKXU\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347435238.60_warc_CC-MAIN-20200603144014-20200603174014-00030.warc.gz\"}"}
https://www.teachoo.com/4444/713/Example-6---Total-revenue-received-from-sale-of-x-units/category/Examples/	[ "", null, "", null, "", null, "1. Chapter 6 Class 12 Application of Derivatives (Term 1)\n2. Serial order wise\n3. Examples\n\nTranscript\n\nExample 6 The total revenue in Rupees received from the sale of x units of a product is given by R(x) = 3x2 + 36x + 5. Find the marginal revenue, when x = 5, where by marginal revenue we mean the rate of change total revenue with respect to the number of items sold at an instant.Marginal revenue is rate of change of total revenue w. r. t the number of unit sold Let MR be marginal revenue So, MR = 𝒅𝑹/𝒅𝒙 Given, Total revenue = R (𝑥) = 3𝑥2 + 36𝑥 + 5 We need to find marginal revenue when 𝑥 = 5 i.e. MR when 𝑥 = 5 MR = 𝑑(𝑅(𝑥))/𝑑𝑥 MR = (𝑑 (3𝑥2 + 36𝑥 + 5) )/𝑑𝑥 MR = (𝑑(3𝑥2))/𝑑𝑥 + (𝑑(36𝑥))/𝑑𝑥 + (𝑑(5))/𝑑𝑥 MR = 3 (𝑑(𝑥2))/𝑑𝑥 + 36 (𝑑(𝑥))/𝑑𝑥 + 0 MR = 3 × 2𝑥 + 36 MR = 6𝑥 + 36 MR when 𝑥 = 5 MR = 6 (5) + 36 MR = 30 + 36 MR = 66 Hence, the required marginal revenue is Rs. 66", null, "" ]	[ null, "https://d1avenlh0i1xmr.cloudfront.net/bc49d502-1926-43fa-a322-85718968cf40/slide21.jpg", null, "https://d1avenlh0i1xmr.cloudfront.net/5bc0bf66-b3a9-4721-98fb-cfa5e2a1d8eb/slide22.jpg", null, "https://d1avenlh0i1xmr.cloudfront.net/faef0de7-d010-40b7-be43-681b40010d29/slide23.jpg", null, "https://delan5sxrj8jj.cloudfront.net/misc/Davneet+Singh.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7646306,"math_prob":0.99996924,"size":2168,"snap":"2021-31-2021-39","text_gpt3_token_len":773,"char_repetition_ratio":0.31423292,"word_repetition_ratio":0.13580246,"special_character_ratio":0.32380074,"punctuation_ratio":0.03194103,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999902,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,4,null,4,null,4,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T04:06:07Z\",\"WARC-Record-ID\":\"<urn:uuid:ce631707-17e1-4c2e-ad4d-441a264e91d6>\",\"Content-Length\":\"85862\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:b9e31f0e-f2a3-4c15-b48b-fe43194b0825>\",\"WARC-Concurrent-To\":\"<urn:uuid:3ca9cf57-ea04-4b42-b08f-9dfb46648f0e>\",\"WARC-IP-Address\":\"23.22.5.68\",\"WARC-Target-URI\":\"https://www.teachoo.com/4444/713/Example-6---Total-revenue-received-from-sale-of-x-units/category/Examples/\",\"WARC-Payload-Digest\":\"sha1:TRSGD43TLELC7HHNQU4VZCREHXTLVO26\",\"WARC-Block-Digest\":\"sha1:V3RWH5CR6A3XZSFIPMRRZDH7WSPBTOWW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057496.18_warc_CC-MAIN-20210924020020-20210924050020-00044.warc.gz\"}"}
https://www.geeksforgeeks.org/gate-gate-cs-2001-question-44/	[ "# GATE \| GATE-CS-2001 \| Question 44\n\nConsider the following program\n\n `Program` `P2 ` ` ``var` `n: int: ` ` ``procedure` `W(``var` `x: int) ` ` ``begin` ` ``x=x+``1``; ` ` ``print x; ` ` ``end` ` ` ` ``procedure` `D ` ` ``begin` ` ``var` `n: int; ` ` ``n=``3``; ` ` ``W(n); ` ` ``end` `begin` `//beginP2 ` ` ``n=``10``; ` ` ``D; ` `end`\n\nIf the language has dynamic scoping and parameters are passed by reference, what will be printed by the program?\n(A) 10\n(B) 11\n(C) 3\n(D) None of the above\n\nExplanation:\n\nIn static scoping or compile-time scoping the free variables (variables used in a function that are neither local variables nor parameters of that function) are referred as global variables because at compile only global variables are available.\nIn dynamic scoping or run-time scoping the free variables are referred as the variables in the most recent frame of function call stack. In the given code in the function call of procedure W the local variable x is printed i.e 4. Under dynamic scoping if x would have not been there in procedure W then we would refer to x of the function in function call stack i.e procedure D and the main function but since x is a local variable not a free variable we referred to the local variable hence 4 will be printed.\n\nThis solution is contributed by Parul Sharma.\n\nQuiz of this Question\n\nMy Personal Notes arrow_drop_up\n\nArticle Tags :\n\nBe the First to upvote.\n\nPlease write to us at [email protected] to report any issue with the above content." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.7348879,"math_prob":0.8743665,"size":2050,"snap":"2019-43-2019-47","text_gpt3_token_len":620,"char_repetition_ratio":0.2453568,"word_repetition_ratio":0.25138122,"special_character_ratio":0.31268293,"punctuation_ratio":0.067010306,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96037054,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-12T23:41:25Z\",\"WARC-Record-ID\":\"<urn:uuid:a6387883-0a67-48e7-9eeb-cf8b2afd1510>\",\"Content-Length\":\"90738\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:771af67a-aa98-4062-8359-a9fa3bb86b3d>\",\"WARC-Concurrent-To\":\"<urn:uuid:9cf6bbe9-6eeb-49c7-8a8e-a317bd023454>\",\"WARC-IP-Address\":\"104.96.221.51\",\"WARC-Target-URI\":\"https://www.geeksforgeeks.org/gate-gate-cs-2001-question-44/\",\"WARC-Payload-Digest\":\"sha1:G26NB32Q6TMQTKRMDJ23YSXNXQDTSD5M\",\"WARC-Block-Digest\":\"sha1:XKR2YUCF4VL3EUQ5RIWACHLBZGFYTGLG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496665809.73_warc_CC-MAIN-20191112230002-20191113014002-00372.warc.gz\"}"}
https://stage.geogebra.org/m/kmjxvgjf	[ "# im.g.2.7.2 Proving the Angle-Side-Angle (ASQA\n\n1. Two triangles have 2 pairs of corresponding angles congruent, and the corresponding sides between those angles are given as congruent. Modify the sketch to fit that description. 2. The triangles are labeled WXY and DEF, such that angle W is congruent to angle D, angle X is congruent to angle E. You should modify the segments such that WX = DE. 3. Use a sequence of rigid motions to take triangle WXY onto triangle DEF." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8422035,"math_prob":0.91969526,"size":730,"snap":"2023-40-2023-50","text_gpt3_token_len":188,"char_repetition_ratio":0.13636364,"word_repetition_ratio":0.0,"special_character_ratio":0.23013699,"punctuation_ratio":0.18954249,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9878034,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-03T03:15:26Z\",\"WARC-Record-ID\":\"<urn:uuid:ee1776f0-a6a9-4186-b946-35fc5181c307>\",\"Content-Length\":\"46070\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a9f460bc-b3ad-41fc-8075-97823f618390>\",\"WARC-Concurrent-To\":\"<urn:uuid:da07675f-9f37-4508-b6ad-4118a5f4c078>\",\"WARC-IP-Address\":\"3.162.103.11\",\"WARC-Target-URI\":\"https://stage.geogebra.org/m/kmjxvgjf\",\"WARC-Payload-Digest\":\"sha1:7TRH7WPM36ATQC5YGKDHBO4JXNKWUUI5\",\"WARC-Block-Digest\":\"sha1:YHJKOAB4UQB7T7PNEX7INMFNKOMQBT6P\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233511053.67_warc_CC-MAIN-20231003024646-20231003054646-00154.warc.gz\"}"}
https://redmine.ruby-lang.org/issues/1408	[ "## Feature #1408", null, "", null, "### 0.1.to_r not equal to (1/10)\n\nStatus:\nClosed\nPriority:\nNormal\nTarget version:\n[ruby-core:23318]\n\nDescription\n\n=begin\n\\$ ruby -e 'p 0.1.to_r'\n(3602879701896397/36028797018963968)\n\nwhereas\n\n\\$ ruby -e 'p \"0.1\".to_r'\n(1/10)\n=end\n\nRelated issues\n\n Has duplicate Ruby master - Bug #5309: 0.6.to_r != \"0.6\".to_r Rejected 09/13/2011 Actions\n\n####", null, "Updated by phasis68 (Heesob Park)over 11 years ago\n\n=begin\n2009/4/27 Martin DeMello [email protected]:\n\nOn Sun, Apr 26, 2009 at 2:51 PM, Heesob Park [email protected] wrote:\n\n\\$ ruby -e 'p 0.1.to_r'\n(3602879701896397/36028797018963968)\n\nwhereas\n\n\\$ ruby -e 'p \"0.1\".to_r'\n(1/10)\n\ninvoked, 0.1 is already a binary float, with the implicit rounding\noff.\n\nIn theory, Float#to_r could be done through Float#to_s#to_r.\n\nRegards,\n\nPark Heesob\n\n=end\n\n####", null, "Updated by shyouhei (Shyouhei Urabe)over 11 years ago\n\n=begin\nHeesob Park wrote:\n\n2009/4/27 Martin DeMello [email protected]:\n\nOn Sun, Apr 26, 2009 at 2:51 PM, Heesob Park [email protected] wrote:\n\n\\$ ruby -e 'p 0.1.to_r'\n(3602879701896397/36028797018963968)\n\nwhereas\n\n\\$ ruby -e 'p \"0.1\".to_r'\n(1/10)\ninvoked, 0.1 is already a binary float, with the implicit rounding\noff.\n\nIn theory, Float#to_r could be done through Float#to_s#to_r.\n\n-1. That loses data.\n\n=end\n\n####", null, "Updated by rogerdpack (Roger Pack)over 11 years ago\n\n=begin\n\n-1 that loses data.\n\nTrue--however the (current) code for String#to_s attempts to determine whether the floating point number \"is the equivalent default for the rounded value\" (i.e. if it round trips).\nDo you think that using a comparisong like this (similar to what Park suggested) would be good enough for deducing the true original value? (I've thought of proposing a similar thing for BigDecimal,\n\nex: BigDecimal(0.1) => #\n\n-=r\n=end\n\n####", null, "Updated by tadf (tadayoshi funaba)over 11 years ago\n\n=begin\nto_r should provide exact conversion.\nI think ruby may provide \"rationalize\" on common lisp or scheme.\nbut not yet.\n\n=end\n\n####", null, "Updated by nobu (Nobuyoshi Nakada)over 11 years ago\n\n=begin\nHi,\n\nAt Fri, 1 May 2009 21:12:52 +0900,\nRoger Pack wrote in [ruby-core:23345]:\n\nTrue--however the (current) code for String#to_s attempts to\ndetermine whether the floating point number \"is the\nequivalent default for the rounded value\" (i.e. if it round\ntrips).\n\nIndex: rational.c\n===================================================================\n--- rational.c (revision 23433)\n+++ rational.c (working copy)\n@@ -1286,4 +1286,5 @@ integer_to_r(VALUE self)\n}\n\n+#if 0\nstatic void\nfloat_decode_internal(VALUE self, VALUE rf, VALUE rn)\n@@ -1299,5 +1300,4 @@ float_decode_internal(VALUE self, VALUE\n}\n\n-#if 0\nstatic VALUE\nfloat_decode(VALUE self)\n@@ -1310,11 +1310,82 @@ float_decode(VALUE self)\n#endif\n\n+#if FLT_RADIX == 2 && SIZEOF_BDIGITS * 2 * CHAR_BIT > DBL_MANT_DIG\n+# ifdef HAVE_LONG_LONG\n+# define BDIGITDBL2NUM(x) ULL2NUM(x)\n+# else\n+# define BDIGITDBL2NUM(x) ULONG2NUM(x)\n+# endif\n+#else\n+# define NEEDS_FDIV\n+static ID id_fdiv;\n+fun2(fdiv)\n+#endif\n+\n+static VALUE\n+float_r_round(double a, double f, int n)\n+{\n\n• int i, r; +#ifdef BDIGITDBL2NUM\n• BDIGIT_DBL fn = (BDIGIT_DBL)fabs(f);\n• BDIGIT_DBL d1 = (BDIGIT_DBL)1 << -n, d2 = d1;\n• BDIGIT_DBL rv = d1 % fn;\n• VALUE b, d;\n• if (rv < 10) {\n• for (i = 1, r = (int)rv; i <= r; ++i) {\n• if ((double)fn / --d2 != a) break;\n• if (fn % (d1 = d2) == 0) break;\n• }\n• }\n• else if ((rv = fn - rv) && rv < 10) {\n• for (i = 1, r = (int)rv; i <= r; ++i) {\n• if ((double)fn / ++d2 != a) break;\n• if (fn % (d1 = d2) == 0) break;\n• }\n• }\n• b = BDIGITDBL2NUM(fn);\n• d = BDIGITDBL2NUM(d1);\n• if (f < 0) b = f_negate(b); +#else\n• VALUE d2, fn, rv;\n• VALUE b = rb_dbl2big(f);\n• VALUE d = rb_big_pow(rb_uint2big(FLT_RADIX), INT2FIX(-n));\n• if (FIXNUM_P(d)) {\n• d = rb_uint2big(FIX2LONG(d));\n• }\n• d2 = d;\n• fn = f_abs(b);\n• rv = rb_big_modulo(d, fn);\n• if (FIXNUM_P(rv) && (r = FIX2LONG(rv)) < 10) {\n• for (i = 1; i <= r; ++i) {\n• d2 = f_sub(d2, INT2FIX(1));\n• if (RFLOAT_VALUE(f_fdiv(fn, d2)) != a) break;\n• if (f_mod(fn, d = d2) == INT2FIX(0)) break;\n• }\n• }\n• else if (FIXNUM_P(rv = f_sub(fn, rv)) && (r = FIX2LONG(rv)) < 10) {\n• for (i = 1; i <= r; ++i) {\n• if (RFLOAT_VALUE(f_fdiv(fn, d2)) != a) break;\n• if (f_mod(fn, d = d2) == INT2FIX(0)) break;\n• }\n• } +#endif\n• return rb_rational_new(b, d); +} + static VALUE float_to_r(VALUE self) {\n• VALUE f, n;\n• double a, f;\n• int n;\n\n• float_decode_internal(self, &f, &n);\n\n• a = RFLOAT_VALUE(self);\n\n• f = frexp(a, &n);\n\n• f = ldexp(f, DBL_MANT_DIG);\n\n• n -= DBL_MANT_DIG;\n\n• if (n <= DBL_MANT_DIG && f != 0) {\n\n• return float_r_round(a, f, n);\n\n• }\n\n}\n\n@@ -1569,4 +1640,7 @@ Init_Rational(void)\nid_to_s = rb_intern(\"to_s\");\nid_truncate = rb_intern(\"truncate\");\n+#ifdef NEEDS_FDIV\n\n• id_fdiv = rb_intern(\"fdiv\");\n+#endif\n\nml = (long)(log(DBL_MAX) / log(2.0) - 1);\n\n--\n\n=end\n\n####", null, "Updated by matz (Yukihiro Matsumoto)over 11 years ago\n\n=begin\nHi,\n\nIn message \"Re: [ruby-core:23465] Re: [Feature #1408] 0.1.to_r not equal to (1/10)\"\n\nCould you explain how this patch differs from the original?\n\n``` matz.\n```\n\n=end\n\n####", null, "Updated by nobu (Nobuyoshi Nakada)over 11 years ago\n\n=begin\nHi,\n\nAt Mon, 18 May 2009 11:15:16 +0900,\nYukihiro Matsumoto wrote in [ruby-core:23487]:\n\nCould you explain how this patch differs from the original?\n\nSearches more reduceable numerator which can round trip. Since\nit just tries the numerator only in very restricted condtion,\nbetter result may be achieved by trying also the denominator,\nin other cases. In fact, the patch works for very simple\ncases, e.g. 0.1 and (1.0/3.0), but doesn't for 0.24.\n\n--\n\n=end\n\n####", null, "Updated by yugui (Yuki Sonoda)over 11 years ago\n\n• Target version changed from 1.9.1 to 1.9.2\n\n=begin\n\n=end\n\n####", null, "Updated by marcandre (Marc-Andre Lafortune)about 11 years ago\n\n• Category set to core\n• Assignee set to matz (Yukihiro Matsumoto)\n\n=begin\n\n=end\n\n####", null, "Updated by marcandre (Marc-Andre Lafortune)about 11 years ago\n\n=begin\nSorry to be late to the party on this one.\n\nIt is important to remember that a Float is always an approximation.\n\n1.0 has to be understood as 1.0 +/- EPSILON, where the EPSILON is platform dependent. 1.0 is not more equal to 1 than to 1 + EPSILON/2. Indeed, there is no way to distinguish either when they are stored as floats.\n\nTo believe that Float#to_s loses data is wrong. If r.to_s returns \"1.2\", it implies that 1.2 is one of the values in the range of possible values for that floating number. It could have been 1.2000...0006. Or something else. There is no way to know, so #to_s chooses, wisely, to return the simplest value in the range.\n\nThere are many rationals that would be encoded as floats the same way. There is no magic way to know that the \"exact\" value was exactly 12/10 or 5404319552844595/4503599627370496, or anything in between. All have the same representation as a float. There is no reason to believe that the missing (binary) decimals that couldn't be written in space allowed where all 0. Actually, there is reason to believe that they were probably non zero, because fractions that can not be expressed with a finite number of terms in their expansion in a given base all have a recurring expansion. I.e. if the significand does not end with a whole bunch of zeros (rational has finite expansion) then it probably ends with an infinite pattern (say 011011011 in binary, or 333333 in decimal).\n\nFor any given float, there is one and only one rational with the smallest denominator that falls in the range of its possible values. It is currently given by Number#rationalize, and I really do not understand why #to_r would return anything else.\n\nI cannot see any purpose to any other fraction. Moreover, the current algorithm, which returns the middle of the range of possibilities, is platform dependent since the range of possibilities is platform dependent. That makes it even less helpful.\n\nIs there an example where one would want 0.1.to_r to be 3602879701896397/36028797018963968 ?\n\nDo we really think that 0.1.to_r to be 3602879701896397/36028797018963968 corresponds to the principle of least surprise?\nNote that I'm writing that fraction but with a different native double encoding, the fraction would be different.\n\n=end\n\n####", null, "Updated by znz (Kazuhiro NISHIYAMA)over 10 years ago\n\n• Status changed from Open to Assigned\n• Target version changed from 1.9.2 to 2.0.0\n\n=begin\n\n=end\n\n####", null, "Updated by marcandre (Marc-Andre Lafortune)over 10 years ago\n\n=begin\nWhy isn't Float#to_r simply calling Float#rationalize ?\n\n=end\n\n####", null, "Updated by mwaechter (Matthias Wächter)over 10 years ago\n\n=begin\nAm 20.09.2009 06:17, schrieb Marc-Andre Lafortune:\n\nSorry to be late to the party on this one.\n\nI’m late as well ;)\n\nIt is important to remember that a Float is always an approximation.\n\nNo. It is an approximation only for:\n\n• conversion from most decimal numbers, especially floats, and\n• calculations that drop digits.\n\nYou can do exact math in a limited range of operations, and the question\nshould be whether the approximation approach should overrule this exact\nmath range of use, especially considering that conversion back to\ndecimal could be done precisely, however, sometimes requiring a bunch\nof digits.\n\n1.0 has to be understood as 1.0 +/- EPSILON, where the EPSILON is platform\ndependent. 1.0 is not more equal to 1 than to 1 + EPSILON/2. Indeed, there\nis no way to distinguish either when they are stored as floats.\n\nIf what’s stored in the Float is your precise result, you certainly\nwould not ask for precision reduction just because it could have been\nthe result of an imprecise calculation.\n\nTo believe that Float#to_s loses data is wrong.\n\nI think there should be both a Float#to_s and Float#to_nearest_s. The\nfirst would be precise, the second would output the “shortest” decimal\nrepresentation within ±EPSILON/2.\n\nIf r.to_s returns \"1.2\", it implies that 1.2 is one of the values in the\nrange of possible values for that floating number. It could have been\n1.2000...0006. Or something else. There is no way to know, so #to_s chooses,\nwisely, to return the simplest value in the range.\n\nThis is based on the assumption that no-one would ever care about\nFloat’s precision.\n\nThere are many rationals that would be encoded as floats the same way. There\nis no magic way to know that the \"exact\" value was exactly 12/10 or\n5404319552844595/4503599627370496, or anything in between. All have the same\nrepresentation as a float. There is no reason to believe that the missing\n(binary) decimals that couldn't be written in space allowed where all 0.\nActually, there is reason to believe that they were probably non zero,\nbecause fractions that can not be expressed with a finite number of terms in\ntheir expansion in a given base all have a recurring expansion. I.e. if the\nsignificand does not end with a whole bunch of zeros (rational has finite\nexpansion) then it probably ends with an infinite pattern (say 011011011 in\nbinary, or 333333 in decimal).\n\nFor any given float, there is one and only one rational with the smallest\ndenominator that falls in the range of its possible values. It is currently\ngiven by Number#rationalize, and I really do not understand why #to_r would\nreturn anything else.\n\nI cannot see any purpose to any other fraction. Moreover, the current algorithm,\nwhich returns the middle of the range of possibilities, is platform dependent\nsince the range of possibilities is platform dependent. That makes it even less\n\nIs there an example where one would want 0.1.to_r to be\n3602879701896397/36028797018963968 ?\n\nIf the binary/Float’s representation of\n3602879701896397/36028797018963968 is the real result of the\ncalculation? How do you know?\n\nDo we really think that 0.1.to_r to be 3602879701896397/36028797018963968\ncorresponds to the principle of least surprise?\n\nFalse assumption here. Using floats for exact decimal math already\nviolates POLS. Don’t blame the messenger, i.e. the converter back to\ndecimal, the only part of the game that could always be precise.\n\nNote that I'm writing that fraction but with a different native double\nencoding, the fraction would be different.\n\nSure. Great to have different levels of precision/imprecision from the\ncomputers.\n\nAnd portability is not always the issue, otherwise there would have\nnever been different native floating point precisions.\n\n– Matthias\n\n=end\n\n####", null, "Updated by mwaechter (Matthias Wächter)over 10 years ago\n\n=begin\nHello Marc-Andre,\n\nOn 19.04.2010 00:14, Marc-Andre Lafortune wrote:\n\nI hope my dissent will not sound too harsh.\n\nNot at all.\n\nArguing that 0.1.to_r should be 3602879701896397/36028797018963968 is\nthe same as arguing that 0.1.to_s should outputs these 55 decimals.\n\nRight, that’s my point. 0.1 as a Float has a precise meaning in binary as in decimal, so Float#to_s should keep those 55 decimals. That’s why I said\nthat Float#to_nearest_s – choose a better name or an option to Fload#to_s – should be created that does »what everyone expects« to_s to do.\n\nThe same applies to Float#to_r. It should be as precise as possible, which it is currently. The function that does »what everyone expects« should be\nFloat#to_nearest_r in the same way as for the string representation.\n\nFor these reasons, the set S is of little interest to anybody.\n\nThe problem is that most people think that Floating point arithmetic is precise, which it is only for the the cases I described in my last mail.\n\nWhat is interesting is the set of real numbers. Floating numbers are\nused to represent them approximately. To add to my voice, here are a\ncouple of excerpts from the first links that come up on google\n(highlight mine):\n\n\"In computing, floating point describes a system for representing\nnumbers that would be too large or too small to be represented as\nintegers. Numbers are in general represented approximately to a\nfixed number of significant digits and scaled using an exponent.\"\nhttp://en.wikipedia.org/wiki/Floating_point\n\n\"Squeezing infinitely many real numbers into a finite number of bits\nrequires an approximate representation.... Therefore the result of a\nfloating-point calculation must often be rounded in order to fit back\ninto its finite representation. This rounding error is the\ncharacteristic feature of floating-point computation.\" source:\nhttp://docs.sun.com/source/806-3568/ncg_goldberg.html\n\nThat’s where the problem starts. Everyone thinks he can do exact math on a computer, and the only problem was the approximation of the binary\nrepresentation of a real number, characterized by ±EPSILON/2. No, the real issue is the approximation of calculations which not only accumulates\nEPSILON with each calculation, but it can shift EPSILON to any order. Think of something trivial like (1E-40+0.1-0.1) returning 0.0 vs.\n(1E-40+0.3-0.2-0.1) returning -2.7E-17. There is no real math in floats.\n\nOne can go as far as saying that availability of math-like operators and math-like precedence in a programming language supports the expectations of\nreal-number-like behavior and precision. But this is slightly off-topic, and in fact method calls for simple math are not doing any good to\nreadability. Math-like operator precedence is different and something completely unnecessary in a programming language, IMHO.\n\nNote that typing 0.1 in Ruby is a \"calculation\" which consists in\nfinding the member of S closest to 1/10.\n\nYour final question was: how do I know that the value someone is\ntalking about is 0.1 and not\n0.1000000000000000055511151231257827021181583404541015625 (or\nequivalently 3602879701896397/36028797018963968) ?\n\nI call it common sense.\n\nIt looks so obvious when we are talking about 0.1. If we talk about any other number with 80 digits, my point may become clearer.\n\nWhat do you do if it’s not 0.1 a.k.a. 0.1000000000000000055511151231257827021181583404541015625 but\n0.09999999999999997779553950749686919152736663818359375 (the result of (0.3-0.2)? What’s the difference for your argument? Now we will not get back\nthe expected nearest 0.1 anyway without applying the actually required/expected rounding constraints.\n\nIf it’s just about 0.1.to_r, i.e. converting from a decimal constant number to rational, use String#to_r.\n\nBottom line: Floats are not exact in terms of math, but they are exact in terms of computer-level implementation, implementing IEEE 754. We should\nrespect the latter and help people deal with the former.\n\n– Matthias\n\n=end\n\n####", null, "Updated by tadf (tadayoshi funaba)over 10 years ago\n\n=begin\n\nWhy isn't Float#to_r simply calling Float#rationalize ?\n\na = 0.5337486539516013\nb = 0.5337486539516012\n\na == b #=> false\n\na.to_r == a #=> true\na.rationalize == a #=> false\n\na.to_r == b #=> false\na.rationalize == b #=> true\n\nactually, flonum is restricted rational number.\nhowever, rationalize bends the value.\n\nto_r is the simplest and the cheapest way, rationalize is not so.\n\nmoreover, various languages support exact conversion (e.g. CL, Scheme, Haskell, Squeak, Python).\n\n=end\n\n####", null, "Updated by mrkn (Kenta Murata)over 10 years ago\n\n=begin\nFloat#rationalize is added again at r27503.\n\nOn 2010/05/06, at 7:23, Marc-Andre Lafortune wrote:\n\nMaybe a kind Japanese reader can provide the gist of [ruby-dev:41061]\nto explain why was Float#rationalize removed?\n\nI would also appreciate opinions as to why it wouldn't be a net\nimprovement if to_r used the rationalize algorithm and some other\nmethods were provided for anyone wanting the value of the\nrepresentation (e.g. Float#representation which would return [sign,\nmantissa, significand] and/or Float#representation_to_r would give the\nrational corresponding to the internal representation of that float)\n\n--\nKenta Murata\nOpenPGP FP = FA26 35D7 4F98 3498 0810 E0D5 F213 966F E9EB 0BCC\n\nE-mail: [email protected]" ]	[ null, "https://www.gravatar.com/avatar/666b4e17b4bb0e2d999037a25f65a7cb", null, "https://www.gravatar.com/avatar/6738588a11b852833edf6aec90ef6fa3", null, "https://www.gravatar.com/avatar/666b4e17b4bb0e2d999037a25f65a7cb", null, "https://www.gravatar.com/avatar/9d2f78236e45a335301ba1195026105d", null, "https://www.gravatar.com/avatar/bec38d63650c8912b6ba9b557fb953b9", null, "https://www.gravatar.com/avatar/6d5ca2abd3ce607577fe509ef4af1614", null, "https://www.gravatar.com/avatar/f1d6cc2b735bfd82c8773172da2aeab9", null, "https://www.gravatar.com/avatar/6738588a11b852833edf6aec90ef6fa3", null, "https://www.gravatar.com/avatar/f1d6cc2b735bfd82c8773172da2aeab9", null, "https://www.gravatar.com/avatar/1e7e007723cfe59320d994a7db4ab4cf", null, "https://www.gravatar.com/avatar/243d1a845bc4318650e6c505916182f2", null, "https://www.gravatar.com/avatar/243d1a845bc4318650e6c505916182f2", null, "https://www.gravatar.com/avatar/8e497efbe99e2fa051316337d03624d9", null, "https://www.gravatar.com/avatar/243d1a845bc4318650e6c505916182f2", null, "https://www.gravatar.com/avatar/150c2e7bb8354b13dee87fc5e5de09b0", null, "https://www.gravatar.com/avatar/150c2e7bb8354b13dee87fc5e5de09b0", null, "https://www.gravatar.com/avatar/6d5ca2abd3ce607577fe509ef4af1614", null, "https://www.gravatar.com/avatar/7cca11c5257fda526eeb4b1ada28f904", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84228015,"math_prob":0.92782927,"size":17953,"snap":"2020-45-2020-50","text_gpt3_token_len":5137,"char_repetition_ratio":0.10752688,"word_repetition_ratio":0.36541143,"special_character_ratio":0.3118142,"punctuation_ratio":0.14685908,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98099333,"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],"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,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T06:36:44Z\",\"WARC-Record-ID\":\"<urn:uuid:017b6d3a-841c-4d2d-b121-6dc518baaef3>\",\"Content-Length\":\"72444\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a5a90d2f-c165-4fdd-8890-4b6f9b9fb346>\",\"WARC-Concurrent-To\":\"<urn:uuid:b96c7093-0656-4568-af4f-b7bc030fe44c>\",\"WARC-IP-Address\":\"52.1.26.21\",\"WARC-Target-URI\":\"https://redmine.ruby-lang.org/issues/1408\",\"WARC-Payload-Digest\":\"sha1:KFIUNWLGKWEUGZ4XECHHQ3J7WWPBLIKZ\",\"WARC-Block-Digest\":\"sha1:LTVBWZKQCCNYZBAC67EKYUKKXZKFAVQV\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107893402.83_warc_CC-MAIN-20201027052750-20201027082750-00606.warc.gz\"}"}
https://global-sci.org/intro/article_detail/getBib?article_id=12168	[ "@Article{AAMM-9-651, author = {}, title = {A Combined Discontinuous Galerkin Method for Saltwater Intrusion Problem with Splitting Mixed Procedure}, journal = {Advances in Applied Mathematics and Mechanics}, year = {2018}, volume = {9}, number = {3}, pages = {651--666}, abstract = {\n\nIn this paper, a new combined method is presented to simulate saltwater intrusion problem. A splitting positive definite mixed element method is used to solve the water head equation, and a symmetric discontinuous Galerkin (DG) finite element method is used to solve the concentration equation. The introduction of these two numerical methods not only makes the coefficient matrixes symmetric positive definite, but also does well with the discontinuous problem. The convergence of this method is considered and the optimal L 2 -norm error estimate is also derived.\n\n}, issn = {2075-1354}, doi = {https://doi.org/10.4208/aamm.2015.m1026}, url = {http://global-sci.org/intro/article_detail/aamm/12168.html} }" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87469333,"math_prob":0.9011057,"size":990,"snap":"2020-24-2020-29","text_gpt3_token_len":231,"char_repetition_ratio":0.09939148,"word_repetition_ratio":0.04411765,"special_character_ratio":0.26767677,"punctuation_ratio":0.15204678,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.97149354,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-05-29T17:56:09Z\",\"WARC-Record-ID\":\"<urn:uuid:6e694823-2687-4ccb-85d3-a053584a23f8>\",\"Content-Length\":\"1363\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:33ad955d-557b-45f3-876a-7506aad837d3>\",\"WARC-Concurrent-To\":\"<urn:uuid:082bd300-26a2-462b-b30c-f327f3329315>\",\"WARC-IP-Address\":\"47.52.67.10\",\"WARC-Target-URI\":\"https://global-sci.org/intro/article_detail/getBib?article_id=12168\",\"WARC-Payload-Digest\":\"sha1:XDMUP6OFNXOIRHPTEMPC6ETR4ED2OCGM\",\"WARC-Block-Digest\":\"sha1:W5OE3EHYRZF75Z6WEZ7TFMSOKM5VASX4\",\"WARC-Identified-Payload-Type\":\"application/x-bibtex-text-file\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347405558.19_warc_CC-MAIN-20200529152159-20200529182159-00227.warc.gz\"}"}
https://unix.stackexchange.com/questions/264058/merge-multiple-lines-in-same-file-based-on-column-1	[ "# Merge multiple lines in same file based on column 1\n\nI'm still learning programming and I've tried many things but just cannot get the correct format. I have a tab delimited file with 17 columns and many (around 50.000) rows. The file is already sorted by first column. I want to merge rows that have the same first column (A), but all other 16 columns are different and I want to keep all the information in one row, preferably in the same column with semicolon ; as a delimiter between them. I want to keep tab as a delimiter in the output file. Thank you so much for the answers and if you could also explain the answer where I went wrong that would be even better :).\n\nI've tried so far:\n\n``````awk -F'\\t' 'NF>1{a[\\$1] = a[\\$1]\";\"\\$2}END{for(i in a){print i\"\"a[i]}}' filename.txt\n\nperl -F',' -anle 'next if /^\\$/;\\$h{\\$F} = \\$h{\\$F}.\", \".\\$F;\nEND{print \\$_,\\$h{\\$_},\"\\n\" for sort keys %h}' filename.txt\n``````\n\nFILE FORMAT (other 15 columns have the same format as column B)\n\n``````A B C\n123 fvv ggg\n123 kjf ggg\n123 ccd att\n567 abc gst\n567 abc hgt\n879 ttt tyt\n``````\n\nThe output I want (I need all 17 columns and for columns 2-16 I need the same output as in column B and C). All cases of B should be under B and all cases of C should be under C and all cases of D should be under D etc. So the output has 17 columns just like the input and instead of 50.000 rows, it now should have around 20.000, because there are many repeats for column 1 (for this particular file):\n\n``````A B C\n123 fvv;kjf;ccd ggg;ggg;att\n567 abc;abc gst;hgt\n879 ttt lll\n``````\n• `sed ':1;\\$!N;s/^\\(\\(\\S\\+\\s\\+\\).\\)\\n\\2/\\1;/;t1;P;D' filename.txt` – Costas Feb 18 '16 at 12:44\n• This doesn't do exactly what I want. It gives output in such a way that it repeats all 16 columns in this way (but for 16): 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 and since i have for some cases 50 rows, with the same 1st row, this makes it hard to read as well as give an error due to so many columns (16 x 50 = 800 !!) . – Fluorine Feb 18 '16 at 12:55\n• Hang on, what output do you want for the multiple columns? Should all different cases of \"B\" be under \"B\"? Or should they be concatenated? Please edit your question and add a more representative example of your input and desired output so we can see how you want us to deal with the multiple columns. – terdon Feb 18 '16 at 13:23\n• Yes all cases of B should be under B and all cases of C should be under C and all cases of D should be under D etc. So the output has 17 columns just like the input and instead of 50.000 rows, it now should have around 20.000 (for this particular file). – Fluorine Feb 18 '16 at 14:48\n• If you don't mind having the values separated by commas instead of semicolons you could try with `gnu datamash` e.g. assuming `infile` is sorted by 1st column you could run `{ head -n 1; datamash -g 1 \\$(printf 'collapse %s ' {2..15}); } <infile \| column -t` – don_crissti Feb 18 '16 at 15:32\n\n``````awk '{\nif(NR!=1){a[\\$1]=\\$2\";\"a[\\$1]}\nelse print \\$0}\nEND{\nn = asorti(a, b);\nfor (n in b) {\nprint b[n],a[b[n]]\n}\n}'\n``````\n\nA perl solution:\n\n``````\\$ perl -F\"\\t\" -anle 'if(\\$.==1){print; next} push @{\\$k{\\$F}},@F[1..\\$#F];\nEND{print \"\\$_\\t\" . join(\";\",@{\\$k{\\$_}}) for sort keys(%k)}' file\nA B\n123 fvv;kjf;ccd\n567 abc;abc\n879 ttt\n``````\n\nThis can work on an arbitrary number of fields. It does, however, require loading quite a few things into memory and that might be a problem if your file is large.\n\nAs for where you went wrong, we can't tell you unless you explain what actually happened but, off the top of my head, you perl attempt would fail because:\n\n• You are using `-F,` which sets the field separator to a comma when your input has tabs.\n• You are using `-l` and `print \"foo\\n\"`. The `-l` already adds a newline to each print call, so you'll have multiple, blank lines.\n• You are using `\\$h{\\$F}.\", \".\\$F;` to append, so the first time that is run and `\\$h{\\$F}` is not defined, you will add an extra `,` at the beginning of your stored value.\n• You are only looking at the second field, ignoring all others.\n\nSimilarly, your `awk` will fail because:\n\n• You are printing `foo\"\"bar` which will concatenate the output with no space between each field. You want `print foo,bar` and you also want `OFS=\"\\t\"` for tab-separated output.\n• You are only looking at the second field, ignoring all others.\n\napologies for this one-liner, but here it goes --\n\n``````awk 'BEGIN{FS=\"\\t\"} {for(i=2; i<=NF; i++) { if (!a[\\$1]) a[\\$1]=\\$1FS\\$i ;else a[\\$1]=a[\\$1]\";\"\\$i};if (\\$1 != old) b[j++] = a[old];old=\\$1 } END{for (i=0; i<j; i++) print b[i] }' 1\n\n123 fvv ;kjf;ccd\n567 abc;abc\n879 ttt\n``````\n\nPerl can do this handily, using a hash:\n\n``````#!/usr/bin/env perl\n\nuse strict;\nuse warnings;\n\nmy %stuff;\nmy @header = split ' ', <>;\n\n#read in the data to \"stuff\"\nwhile ( <> ) {\nmy ( \\$key, \\$value ) = split;\npush ( @{\\$stuff{\\$key}}, \\$value );\n}\n\nforeach my \\$key ( sort keys %stuff ) {\nprint \\$key, \"\\t\", join \";\", @{\\$stuff{\\$key}},\"\\n\";\n}\n``````\n\nOutput:\n\n``````A B\n123 fvv;kjf;ccd;\n567 abc;abc;\n879 ttt;\n``````\n\nWhere you went wrong? Honestly, I would suggest it's in trying to compress it all into a one liner. That is - in my opinion - really bad practice. At best, it promotes inscrutable code that's hard to follow.\n\nThe above could be condensed down, but it really pays to do it longhand first.\n\nIn order to support multiple columns, then you start to hit a slight bit of a nuisance with column widths.\n\nThis works, but produces output that indentation isn't neatly aligned:\n\n``````#!/usr/bin/env perl\n\nuse strict;\nuse warnings;\n\nmy %stuff;\n\nmy ( \\$id, @header ) = split ' ', <>;\n\nwhile ( <> ) {\nmy ( \\$key, @values ) = split;\nmy %row;\npush ( @{\\$stuff{\\$key}{\\$_}}, \\$row{\\$_} ) for keys %row;\n}\n\nprint join ( \"\\t\", \\$id, @header),\"\\n\";\nforeach my \\$key ( sort keys %stuff ) {\nprint join (\"\\t\", \\$key, map { join \";\", @{\\$stuff{\\$key}{\\$_}}} @header), \"\\n\";\n}\n``````\n\nOutput of:\n\n``````A B C\n123 fvv;kjf;ccd ggg;ggg;att\n567 abc;abc gst;hgt\n879 ttt tyt\n``````\n\nIf tab separation isn't suitable for your needs, you can use `sprintf` to do the formatting:\n\n``````my \\$format = '%12s';\nprint map { sprintf(\\$format, \\$_) } ( \\$id, @header),\"\\n\";\nforeach my \\$key ( sort keys %stuff ) {\nprint map { sprintf(\\$format, \\$_) } ( \\$key, map { join \";\", @{\\$stuff{\\$key}{\\$_}}} @header),\"\\n\";\n}\n``````\n\nWe make some use of `map` here, which I appreciate isn't exactly an obvious thing.\n\nWhat it does is take a list, and apply a transformation to each element. So - in the above example:\n\n``````print join (\"\\y\", map { join \";\", @\\$_ } ([1,2,3],[4,5,6],[7,8,9]) )\n``````\n\nWill generate:\n\n``````1;2;3 4;5;6 7;8;9\n``````\n\nThe `map` operation is saying 'join each subarray on ;' and then return that as a list... that we can then join with a tab. That's basically what the above is doing.\n\n• I rolled back my edit (which, ironically enough, added a one-liner and was written before your warning against them) because I noticed that your solution only works for 2 fields. The OP has multiple fields and has only included 2 as an example. – terdon Feb 18 '16 at 13:17\n• Good point - I'll amend my example. (And might try and one-liner-it, but might not :)) – Sobrique Feb 18 '16 at 13:20\n• This script gives me only column 1 and 2 and for columns 3-17 only column name, rest is empty. – Fluorine Feb 18 '16 at 14:43\n• Yes, sorry, still rejigging it with the new sample input. – Sobrique Feb 18 '16 at 14:43\n• Amended. Works with sample data now. – Sobrique Feb 18 '16 at 14:52\n``````awk '\nfunction p(n,A){\ns = n\nfor(i=2;i<=NF;i++){\ns = s \"\\t\" A[i]\nA[i] = \\$i\n}\nif(n)\nprint s\n}\nNR==1{\nprint\nnext\n}\n\\$1==n{\nfor(i=2;i<=NR;i++)\nA[i] = A[i] \";\" \\$i\nnext\n}\n{\np(n,A)\nn = \\$1\n}\nEND{\np(n,A)\n}\n' file\n``````\n\nI want to merge rows that have the same first column (A), but all other 16 columns are different and I want to keep all the information in one row\n\nYou got some fine answers, but I want to point out to you that what you've described in those 30 words is known in SQL as a join. If you import your 50,000 lines into two tables in SQLite, you get the effect you want with\n\n``````select from R join S on r1 = s1\n``````\n\nwhere r1 and s1 are column names that you determine.\n\nThere are a lot of advantages to using SQL for something like this, especially as the join and selection criteria grow more complex. It's one of the reasons SQL was invented.\n\n• I'm fairly certain that you misunderstood the problem. In SQL parlance, the problem most resembles a GROUP BY where the aggregating functions concatenate strings. – Barefoot IO Feb 18 '16 at 23:22\n• Not a misunderstanding, just a simplification. The JOIN is the hard part (and it is a join). The concatenation is easy. It could be done in the caller, or in SQL with `SELECT r1, r2 \|\| ';' \|\| s2, r3 \|\| ';' s3`, etc. – James K. Lowden Feb 18 '16 at 23:37" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86212534,"math_prob":0.8475043,"size":1731,"snap":"2019-26-2019-30","text_gpt3_token_len":548,"char_repetition_ratio":0.11580776,"word_repetition_ratio":0.0,"special_character_ratio":0.34199885,"punctuation_ratio":0.1291866,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95885414,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-07-22T22:44:14Z\",\"WARC-Record-ID\":\"<urn:uuid:0339670a-1b52-4060-a926-cfa08591bc5b>\",\"Content-Length\":\"182659\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d72913f4-0f4b-40d0-85e2-60c77bdcc5fd>\",\"WARC-Concurrent-To\":\"<urn:uuid:459d5d59-b6c1-4c27-b62a-35d31a58aa84>\",\"WARC-IP-Address\":\"151.101.193.69\",\"WARC-Target-URI\":\"https://unix.stackexchange.com/questions/264058/merge-multiple-lines-in-same-file-based-on-column-1\",\"WARC-Payload-Digest\":\"sha1:PSSGBJL42NEEZ2H5DQSBSKULKUT2MBQT\",\"WARC-Block-Digest\":\"sha1:BXFNB4YXHO3L42WSWF3A2ESF5PHCCUVN\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-30/CC-MAIN-2019-30_segments_1563195528290.72_warc_CC-MAIN-20190722221756-20190723003756-00143.warc.gz\"}"}
https://exactlyhowlong.com/how-long-does-it-take-to-count-to-a-million-and-why/	[ "# How Long Does It Take To Count To A Million (And Why)?\n\n## Exact Answer: 11 Days, 13 Hours And 46 MInutes\n\nThere are only a few things that have no end meaning that they are infinite. One is the time, and the second is the number system. Time also gets denoted with the help of a number that has no possible end.\n\nIf a person were to count to a million, it would take a lot of time to reach there. The process is not only time-consuming but also quite complicated. Besides that, the longest-living person’s life would feel insufficient if the entire number system is to get counted. That includes both positive and negative numbers.\n\n## How Long Does It Take To Count To A Million?\n\nCounting might sound like an easy task only when it is about counting a hundred or maybe a thousand numbers in the correct order. But when it comes to numbers beyond a specific limit, it becomes hard to focus and complete the counting correctly.\n\nThe number has been the most creative discovery that man could ever make. The reason is that everything is either counted or measured in numbers only, whether it is time or distance. Hence knowing how to count the number becomes the foundation for a person’s learning.\n\nWhen it is counting numbers, the starting numbers are easy as they are one word. It then increases to multiple words. Considering the concept of numbers, if a person starts counting, it will take around 11 days, 13 hours, and 46 minutes to count till a million. That is keeping a standard time of one number getting counted per second.\n\nHowever, counting numbers without any break is impossible for a human. Hence, it is possible to count till a million in 11 days, 13 hours, and 46 minutes only for an artificial intelligence operated machine or computerized system,\n\nIt goes on and on if the counting gets increased substantially. This rule applies to only English numbers. Another regional way of counting numbers would take different times as the pronunciation time would be different.\n\n## Why Does It Take That Long To Count A Million?\n\nNumbers are infinite, as the number system got created in a way, where there is no end to numbers. The reason is that even the counting of all existing numbers is a task that would go on for billions of years. Maybe even more than that.\n\nThere are many reasons behind so much time getting invested in the count to a million. A few of the reasons are as follows:\n\n• The primary reason is that the number increases from one digit to a seven-digit number when a million starts. The time to pronounce the words per number also increases to seven digits. Collectively the time goes high.\n• Besides that, if a human counts the number, they would mandatorily need some essential breaks. So if all the time gets invested into counting, that would take about 89 to 90 days to count till a million. That is considering all the essential breaks get excluded.\n\nA person also needs to be focused while counting to a large number like a million. Since the number are so repetitive, it is easy to get confused. There are many studies with regards to how much it would take to count till a million.\n\nSome studies consider one number per second, while some may take the number of syllables that come in each word between one and one million into consideration. There are also a lot of other methods to know how much goes into counting to a million.\n\n## Conclusion\n\nBesides the standard timing to count numbers, if a person can count fast, the counting time would even reduce. The faster a person or a machine will count, the lesser time it will take to count to a million.\n\nConsidering the advancement of AI technologies, a human does not have to count on its own. The machine can count to a million much faster than humans without making any errors.\n\n## References\n\nSubscribe\nNotify of", null, "" ]	[ null, "data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%2056%2056'%3E%3C/svg%3E", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9494036,"math_prob":0.9362337,"size":4353,"snap":"2022-40-2023-06","text_gpt3_token_len":956,"char_repetition_ratio":0.15267877,"word_repetition_ratio":0.020242915,"special_character_ratio":0.21525384,"punctuation_ratio":0.09490741,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96417785,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-09-28T17:01:15Z\",\"WARC-Record-ID\":\"<urn:uuid:242f802c-5c47-463a-ae01-157d861a1fbe>\",\"Content-Length\":\"177127\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:bf0e6681-1f01-4597-995b-b7414032f3b6>\",\"WARC-Concurrent-To\":\"<urn:uuid:ba561a42-333a-4c0c-8ab4-a678957757e7>\",\"WARC-IP-Address\":\"172.67.200.218\",\"WARC-Target-URI\":\"https://exactlyhowlong.com/how-long-does-it-take-to-count-to-a-million-and-why/\",\"WARC-Payload-Digest\":\"sha1:JGXNSEDTJ7C4WF4AAFD3IGBJVL7H3AT6\",\"WARC-Block-Digest\":\"sha1:5IRZ2752ARSS36BRRNXJQTURJ7LDFSJK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-40/CC-MAIN-2022-40_segments_1664030335257.60_warc_CC-MAIN-20220928145118-20220928175118-00476.warc.gz\"}"}
https://dl.bukkit.org/threads/my-explosion-regen-code-has-strange-behaviour.274751/	[ "# my explosion regen code has strange behaviour\n\nDiscussion in 'Plugin Development' started by xize, Jun 1, 2014.\n\nNot open for further replies.\n1. Offline\n\n### xize\n\nHello,\n\nso I'm trying to update a very old code I made in the past in order to have more schedulers instead of one globally called scheduler.\n\nhowever while it tries to generate blocks it gives false/positives because its just skipping lots of blocks for example when ive a normal tnt explosion the scheduler runs 3 times while it should do more than 3 because more than 3 blocks where exploded. I highly suspect the Comperator is doing something weird.\n\nmy event so far:\n\nCode:java\n` @EventHandler public void onExplosion(EntityExplodeEvent e) { System.out.print(\"event has been runned.\"); for(Block block : e.blockList()) { if(block.getType() == Material.CHEST \|\| block.getType() == Material.FURNACE \|\| block.getType() == Material.ANVIL \|\| block.getType() == Material.BED \|\| block.getType() == Material.BREWING_STAND \|\| block.getType() == Material.WOOD_DOOR \|\| block.getType() == Material.IRON_DOOR \|\| block.getType() == Material.TORCH \|\| block.getType() == Material.REDSTONE_WIRE \|\| block.getType() == Material.REDSTONE_COMPARATOR \|\| block.getType() == Material.REDSTONE_TORCH_ON \|\| block.getType() == Material.REDSTONE_TORCH_OFF) { e.setCancelled(true); } else if(block.getType() == Material.TNT) { //leave empty ignore tnt } else { blocks.put(block.getLocation(), block.getState().getData()); block.setType(Material.AIR); } } System.out.print(\"regen object has been created.\"); RegenObject regen = new RegenObject(blocks); xEssentials.getRegenList().getList.add(regen); } `\n\nmy RegenObject class where my scheduler is in:\nCode:java\n` public class RegenObject implements Runnable { //note ive also just removed the comperator here but still the same behaviour. private TreeMap<Location, MaterialData> blocks = new TreeMap<Location, MaterialData>(new Comparator<Location>() { @Override public int compare(Location o1, Location o2) { return Integer.valueOf(o1.getBlockY()).compareTo(o2.getBlockY()); } }); private BukkitTask task; public RegenObject(TreeMap<Location, MaterialData> blocks) { this.blocks = blocks; startRegen(); } /** * @author xize * @param starts the block regen task. / public void startRegen() { if(!(task instanceof BukkitTask)) { this.task = Bukkit.getScheduler().runTaskTimer(xEssentials.getPlugin(), this, 0L, 1L); } } /* * @author xize * @param stops the regen task. / public void stopRegen() { if(task instanceof BukkitTask) { this.task.cancel(); this.task = null; } if(xEssentials.getRegenList().getList.contains(this)) { xEssentials.getRegenList().getList.remove(this); } } @SuppressWarnings(\"deprecation\") @Override public void run() { System.out.print(\"scheduler is still running.\"); Iterator<Entry<Location, MaterialData>> it = blocks.entrySet().iterator(); if(it.hasNext()) { Map.Entry<Location, MaterialData> entry = (Map.Entry<Location, MaterialData>) it.next(); Location loc = entry.getKey(); MaterialData data = entry.getValue(); System.out.print(\"setting block!\"); loc.getBlock().setTypeIdAndData(data.getItemTypeId(), data.getData(), true); loc.getWorld().playEffect(loc, Effect.STEP_SOUND, data.getItemTypeId()); it.remove(); blocks.remove(loc); } else { System.out.print(\"its stopped regen\"); stopRegen(); } } @Override public String toString() { return \"RegenObject [blocks=\" + blocks + \", task=\" + task + \", getClass()=\" + getClass() + \", hashCode()=\" + hashCode() + \", toString()=\" + super.toString() + \"]\"; } @Override public int hashCode() { final int prime = 31; int result = 1; result = prime result + ((blocks == null) ? 0 : blocks.hashCode()); result = prime * result + ((task == null) ? 0 : task.hashCode()); return result; } @Override public boolean equals(Object obj) { if (this == obj) return true; if (obj == null) return false; if (getClass() != obj.getClass()) return false; RegenObject other = (RegenObject) obj; if (blocks == null) { if (other.blocks != null) return false; } else if (!blocks.equals(other.blocks)) return false; if (task == null) { if (other.task != null) return false; } else if (!task.equals(other.task)) return false; return true; }} `\n\nif someone could explain me what is 'corrupting' the resulting behaviour I'm very happy thanks", null, "#1" ]	[ null, "https://dl.bukkit.org/threads/my-explosion-regen-code-has-strange-behaviour.274751/smilies/happy.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.86977303,"math_prob":0.7649444,"size":273,"snap":"2020-24-2020-29","text_gpt3_token_len":69,"char_repetition_ratio":0.09293681,"word_repetition_ratio":0.5365854,"special_character_ratio":0.26373628,"punctuation_ratio":0.19607843,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95228237,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-06-04T17:27:57Z\",\"WARC-Record-ID\":\"<urn:uuid:e0ee1152-0e3e-4097-abea-3993375991e2>\",\"Content-Length\":\"54954\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:96125501-e532-4b3a-8ea2-fc98e3bda454>\",\"WARC-Concurrent-To\":\"<urn:uuid:9e3efbcd-1f29-4417-8d03-266c6ebb08a9>\",\"WARC-IP-Address\":\"104.19.162.33\",\"WARC-Target-URI\":\"https://dl.bukkit.org/threads/my-explosion-regen-code-has-strange-behaviour.274751/\",\"WARC-Payload-Digest\":\"sha1:KVFO6LOITIGOODSIUHBU64HD4PKZ35HJ\",\"WARC-Block-Digest\":\"sha1:N7Y4B6WOSNHBFXWMBPKBOXJS2DUA6DIZ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-24/CC-MAIN-2020-24_segments_1590347445880.79_warc_CC-MAIN-20200604161214-20200604191214-00132.warc.gz\"}"}
https://gis.stackexchange.com/questions/122567/generate-random-points-in-polygon-algorithm-question?noredirect=1	[ "# Generate random points in polygon. Algorithm question\n\ni created a Random Point Generator in python using the algorithm described here by @whuber. The algorithm is implemented using Shapely, PostGIS and psycopg2. To boost performance pythons multiprocessing and threading libraries are used. The source code can be found in my GitHub Repository. The code generates a total of 57.990.970 Points within 6388 Polygons in ~3517 sec on my Xeon-1231v3 with 16GB Ram.\n\n``````Procedure SimpleRandomSample(P:Polygon, N:Integer) {\nU = Sorted list of N independent uniform values between 0 and 1\nReturn SRS(P, BoundingBox(P), U)\n}\n\nProcedure SRS(P:Polygon, B:Rectangle, U:List) {\nN = Length(U)\nIf (N == 0) {Return empty list}\naP = Area(P)\nIf (aP <= 0) {\nError(\"Cannot sample degenerate polygons.\")\nReturn empty list\n}\nt = 2\nIf (aPt < Area(B)) {\n# Cut P into pieces\nIf (Width(B) > Height(B)) {\nB1 = Left(B); B2 = Right(B)\n} Else {\nB1 = Bottom(B); B2 = Top(B)\n}\nP1 = Clip(P, B1); P2 = Clip(P, B2)\nK = Search(U, Area(P1) / aP)\nV = Concatenate( SRS(P1, B1, U[1::K]), SRS(P2, B2, U[K+1::N]) )\n} Else {\n# Sample P\nV = empty list\nmaxIter = tN + 5Sqrt(tN)\nWhile(Length(V) < N and maxIter > 0) {\nDecrement maxIter\nQ = RandomPoint(B)\nIf (Q In P) {Append Q to V}\n}\nIf (Length(V) < N) {\nError(\"Too many iterations.\")\n}\n}\nReturn V\n}\n``````\n\nBut concerning the algorithm I have an open question. It mentions a `U = Sorted list of N independent uniform values between 0 and 1` which i could build using numpy.uniform(0, 1, numberOfValues). But then the algorithm will generate weird results, like one part of the polygon will not have any points in it. I work around this issue by weighting the points to be created by the recursive call using `k = int(round(n * (p1.area / polygon.area)))`.\n\nSo my question is why this List of uniform values between 0 and 1 should be used in the first place.\n\n• Bill could answer more completely, but I'd think the discrete set of intervals (randomly chosen) would force the distribution to be more complete in less time. Can you quantify these \"weird results\" you'd expect? Nov 18 '14 at 21:25\n• Well the weird result appeared once t was set so something smaller than 2. It started splitting the Polygon and basicly one part of it had no points in it. The problem was that K becomes a value between 0 and 1, but after splitting U like `U[1::K]` the max value in U might become smaller than K which forces `U[K+1::N]` to have a length of 0, thus no points are generated in this recursive call. Nov 18 '14 at 21:32\n• K is an integer between 0 and N. Nov 18 '14 at 22:09\n• Ehm stupid me, I wanted to say the area of the sub-polygon p1 divided by the whole area is something between 0-1. So Search does look in U where the index of the element is which is smaller than the result. But again if U was split before, lets say at 0.5 and the result is 0.7, K will be the last index of U, thus in p2 no points will be created. It only would make sense to search U if you recreate it every time SRS is run. But then also you would not need to pass it to the function. Just the number of points to be generated would be sufficient. Nov 18 '14 at 22:24" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8175435,"math_prob":0.9853266,"size":1949,"snap":"2021-31-2021-39","text_gpt3_token_len":575,"char_repetition_ratio":0.08534704,"word_repetition_ratio":0.05730659,"special_character_ratio":0.31246793,"punctuation_ratio":0.12713936,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9978833,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T18:00:37Z\",\"WARC-Record-ID\":\"<urn:uuid:60d547d5-35a0-48a4-a656-5c3b0a3c715c>\",\"Content-Length\":\"162998\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:603b6c9c-719e-49db-9e8f-5ca2e8c22a0b>\",\"WARC-Concurrent-To\":\"<urn:uuid:65f80ecd-b6dd-4ac2-bd26-ac227c2effba>\",\"WARC-IP-Address\":\"151.101.1.69\",\"WARC-Target-URI\":\"https://gis.stackexchange.com/questions/122567/generate-random-points-in-polygon-algorithm-question?noredirect=1\",\"WARC-Payload-Digest\":\"sha1:D7FNR7USX5MT7Z3OHYNYJR53TEQUNQRF\",\"WARC-Block-Digest\":\"sha1:JJ6FA26ULF2LCXHRMBPOMOPUTIRYASNH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057427.71_warc_CC-MAIN-20210923165408-20210923195408-00607.warc.gz\"}"}
https://rdrr.io/cran/did/f/vignettes/pre-testing.Rmd	[ "Pre-Testing in a DiD Setup using the did Package In did: Treatment Effects with Multiple Periods and Groups\n\nknitr::opts_chunk$set( collapse = TRUE, comment = \"#>\" ) library(did) # Source the currently version of the did package (based on our Dropbox) # fldr <- here::here(\"R/\") # sapply(paste0(fldr,list.files(fldr)), source) # Source simulation designs source(here::here(\"vignettes/setup_sims.R\")) Introduction This vignette provides a discussion of how to conduct pre-tests in DiD setups using the did package • One appealing feature of many DiD applications with multiple periods is that the researcher can pre-test the parallel trends assumptions. • The idea here is simple: although one can not always test whether parallel trends itself holds, one can check if it holds in periods before treated units actually become treated. • Importantly, this is just a pre-test; it is different from an actual test. Whether or not the parallel trends assumption holds in pre-treatment periods does not actually tell you if it holds in the current period (and this is when you need it to hold!). It is certainly possible for the identifying assumptions to hold in previous periods but not hold in current periods; it is also possible for identifying assumptions to be violated in previous periods but for them to hold in current periods. That being said, we view the pre-test as a piece of evidence on the credibility of the DiD design in a particular application. • In this vignette, we demonstrate that the approach used in the did package for pre-testing may work substantially better than the more common \"event study regression\". Common Approaches to Pre-Testing in Applications By far the most common approach to pre-testing in applications is to run an event-study regression. Here, the idea is to run a regression that includes leads and lags of the treatment dummy variable such as $$Y_{it} = \\theta_t + \\eta_i + \\sum_{l=-\\mathcal{T}}^{\\mathcal{T}-1} D_{it}^l \\mu_l + v_{it}$$ where$D_{it}^l = 1$if individual$i$has been exposed to the treatment for$l$periods in period$t$, and$D_{it}^l = 0$otherwise. To be clear here, it is helpful to give some examples. Suppose individual$i$becomes treated in period 3. Then, •$D_{it}^0 = 1$when$t=3$and is equal to 0 in other time periods •$D_{it}^2 = 1$when$t=5$and is equal to 0 in other time periods •$D_{it}^{-2} = 1$when$t=1$and is equal to 0 in other time periods. And$\\mu_l$is interpreted as the effect of treatment for different lengths of exposure to the treatment. Typically,$\\mu_{-1}$is normalized to be equal to 0, and we follow that convention here. It is common to interpret estimated$\\mu_l$'s with$l < 0$as a way to pre-test the parallel trends assumption. Pitfalls with Event Study Regressions time.periods <- 4 reset.sim() bett <- betu <- rep(0, time.periods) te <- 0 set.seed(1814) Best Case Scenario for Pre-Testing First, let's start with a case where an event study regression is going to work well for pre-testing the parallel trends assumption # generate dataset with 4 time periods time.periods <- 4 # generate dynamic effects te.e <- time.periods:1 # generate data set with these parameters # (main thing: it generates a dataset that satisfies # parallel trends in all periods...including pre-treatment) data <- build_sim_dataset() head(data) The main thing to notice here: • The dynamics are common across all groups. This is the case where an event-study regression will work. Next, a bit more code #----------------------------------------------------------------------------- # modify the dataset a bit so that we can run an event study #----------------------------------------------------------------------------- # generate leads and lags of the treatment Dtl <- sapply(-(time.periods-1):(time.periods-2), function(l) { dtl <- 1( (data$period == data$G + l) & (data$G > 0) )\ndtl\n})\nDtl <- as.data.frame(Dtl)\ncnames1 <- paste0(\"Dtmin\",(time.periods-1):1)\ncolnames(Dtl) <- c(cnames1, paste0(\"Dt\",0:(time.periods-2)))\ndata <- cbind.data.frame(data, Dtl)\nrow.names(data) <- NULL\n\n#-----------------------------------------------------------------------------\n# run the event study regression\n#-----------------------------------------------------------------------------\n\nlibrary(plm)\n\n# run event study regression\n# normalize effect to be 0 in pre-treatment period\nes <- plm(Y ~ Dtmin3 + Dtmin2 + Dt0 + Dt1 + Dt2,\ndata=data, model=\"within\", effect=\"twoways\",\nindex=c(\"id\",\"period\"))\n\nsummary(es)\n\n#-----------------------------------------------------------------------------\n# make an event study plot\n#-----------------------------------------------------------------------------\n\n# some housekeeping for making the plot\n# add 0 at event time -1\ncoefs1 <- coef(es)\nses1 <- sqrt(diag(summary(es)$vcov)) idx.pre <- 1:(time.periods-2) idx.post <- (time.periods-1):length(coefs1) coefs <- c(coefs1[idx.pre], 0, coefs1[idx.post]) ses <- c(ses1[idx.pre], 0, ses1[idx.post]) exposure <- -(time.periods-1):(time.periods-2) cmat <- data.frame(coefs=coefs, ses=ses, exposure=exposure) library(ggplot2) ggplot(data=cmat, mapping=aes(y=coefs, x=exposure)) + geom_line(linetype=\"dashed\") + geom_point() + geom_errorbar(aes(ymin=(coefs-1.96ses), ymax=(coefs+1.96ses)), width=0.2) + ylim(c(-2,5)) + theme_bw() You will notice that everything looks good here. The pre-test performs well (the caveat to this is that the standard errors are \"pointwise\" and would be better to have uniform confidence bands though this does not seem to be standard practice in applications). We can compare this to what happens using the did package: # estimate group-group time average treatment effects did_att_gt <- att_gt(yname=\"Y\", tname=\"period\", idname=\"id\", gname=\"G\", data=data, bstrap=FALSE, cband=FALSE) summary(did_att_gt) # plot them ggdid(did_att_gt) # aggregate them into event study plot did_es <- aggte(did_att_gt, type=\"dynamic\") # plot the event study ggdid(did_es) Overall, everything looks good using either approach. (Just to keep things fair, we report pointwise confidence intervals for group-time average treatment effects, but it is easy to get uniform confidence bands by setting the options bstrap=TRUE, cband=TRUE to the call to att_gt.) Pitfall: Selective Treatment Timing Sun and Abraham (2020) point out a major limitation of event study regressions: when there is selective treatment timing the$\\mu_l$end up being weighted averages of treatment effects across different lengths of exposures. Selective treatment timing means that individuals in different groups experience systematically different effects of participating in the treatment from individuals in other groups. For example, there would be selective treatment timing if individuals choose to be treated in earlier periods if they tend to experience larger benefits from participating in the treatment. This sort of selective treatment timing is likely to be present in many applications in economics / policy evaluation. Contrary to event study regressions, pre-tests based on group-time average treatment effects (or based on group-time average treatment effects that are aggregated into an event study plot) are still valid even in the presence of selective treatment timing. To see this in action, let's keep the same example as before, but add selective treatment timing. reset.sim() bett <- betu <- rep(0, time.periods) te <- 0 set.seed(1814) # generate dataset with 4 time periods time.periods <- 4 # generate dynamic effects te.e <- time.periods:1 # generate selective treatment timing # ( this is what is different here ) te.bet.ind <- time.periods:1 / (time.periods/2) # generate data set with these parameters # (main thing: it generates a dataset that satisfies # parallel trends in all periods...including pre-treatment) data <- build_sim_dataset() # run through same code as in earlier example...omitted #----------------------------------------------------------------------------- # modify the dataset a bit so that we can run an event study #----------------------------------------------------------------------------- # generate leads and lags of the treatment Dtl <- sapply(-(time.periods-1):(time.periods-2), function(l) { dtl <- 1( (data$period == data$G + l) & (data$G > 0) )\ndtl\n})\nDtl <- as.data.frame(Dtl)\ncnames1 <- paste0(\"Dtmin\",(time.periods-1):1)\ncolnames(Dtl) <- c(cnames1, paste0(\"Dt\",0:(time.periods-2)))\ndata <- cbind.data.frame(data, Dtl)\nrow.names(data) <- NULL\n\n#-----------------------------------------------------------------------------\n# run the event study regression\n#-----------------------------------------------------------------------------\n\nlibrary(plm)\n# run event study regression\n# normalize effect to be 0 in pre-treatment period\nes <- plm(Y ~ Dtmin3 + Dtmin2 + Dt0 + Dt1 + Dt2,\ndata=data, model=\"within\", effect=\"twoways\",\nindex=c(\"id\",\"period\"))\n\nsummary(es)\n#-----------------------------------------------------------------------------\n# make an event study plot\n#-----------------------------------------------------------------------------\n\n# some housekeeping for making the plot\n# add 0 at event time -1\ncoefs1 <- coef(es)\nses1 <- sqrt(diag(summary(es)\\$vcov))\nidx.pre <- 1:(time.periods-2)\nidx.post <- (time.periods-1):length(coefs1)\ncoefs <- c(coefs1[idx.pre], 0, coefs1[idx.post])\nses <- c(ses1[idx.pre], 0, ses1[idx.post])\nexposure <- -(time.periods-1):(time.periods-2)\n\ncmat <- data.frame(coefs=coefs, ses=ses, exposure=exposure)\n# run through same code as before...omitted\n\n# new event study plot\nggplot(data=cmat, mapping=aes(y=coefs, x=exposure)) +\ngeom_line(linetype=\"dashed\") +\ngeom_point() +\ngeom_errorbar(aes(ymin=(coefs-1.96ses), ymax=(coefs+1.96ses)), width=0.2) +\nylim(c(-2,5)) +\ntheme_bw()\n\nIn contrast to the last case, it is clear that things have gone wrong here. Parallel trends holds in all time periods and for all groups here, but the event study regression incorrectly rejects that parallel trends holds -- this is due to the selective treatment timing.\n\nWe can compare this to what happens using the did package:\n\n# estimate group-group time average treatment effects\ndid.att.gt <- att_gt(yname=\"Y\",\ntname=\"period\",\nidnam=\"id\",\ngname=\"G\",\ndata=data\n)\nsummary(did.att.gt)\n\n# plot them\nggdid(did.att.gt)\n# aggregate them into event study plot\ndid.es <- aggte(did.att.gt, type=\"dynamic\")\n\n# plot the event study\nggdid(did.es)\n\nThis is the correct performance (up to aforementioned caveats about multiple hypothesis testing).\n\nConditional Moment Tests\n\nAnother main use case for the did package is when the parallel trends assumptions holds after conditioning on some covariates. This is likely to be important in many applications. For example, to evaluate the effect of participating in a job training program on earnings, it is likely to be important to condition on an individual's education. This would be true if (i) the distribution of education is different for individuals that participate in job training relative to those that don't (this is very likely to hold as people that participate in job training tend to have less education than those who do not), and (ii) if the path of earnings (absent participating in job training) depends on an individual's education. See Heckman, Ichimura, and Todd (1998) and Abadie (2005) for more discussion.\n\nEven when one includes covariates to estimate group-time average treatment effects, pre-tests based only on group-time average treatment effects can fail to detect some violations of the parallel trends assumption. To give an example, suppose that the only covariate is binary variable for an individual's sex. Pre-tests based on group-time average treatment effects could fail to detect violations of the conditional parallel trends assumption in cases where it is violated in one direction for men and in the other direction for women.\n\nThe did package contains an additional pre-test for the conditional parallel trends assumption in the conditional_did_pretest function.\n\n# not run (this code can be substantially slower)\nreset.sim()\nset.seed(1814)\nnt <- 1000\nnu <- 1000\ncdp <- conditional_did_pretest(\"Y\", \"period\", \"id\", \"G\", xformla=~X, data=data)\ncdp\n\nTry the did package in your browser\n\nAny scripts or data that you put into this service are public.\n\ndid documentation built on Jan. 13, 2021, 4:12 p.m." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78267026,"math_prob":0.87015516,"size":12551,"snap":"2022-05-2022-21","text_gpt3_token_len":2961,"char_repetition_ratio":0.18610026,"word_repetition_ratio":0.28263402,"special_character_ratio":0.31750458,"punctuation_ratio":0.12971482,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99111015,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2022-01-24T06:22:34Z\",\"WARC-Record-ID\":\"<urn:uuid:d3be7103-6d6a-4fbe-a0dd-551efa6cc637>\",\"Content-Length\":\"59385\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e06f0228-7d73-444f-a416-285e24fc28bc>\",\"WARC-Concurrent-To\":\"<urn:uuid:819a7f73-9951-467a-8a90-3aeed6c62531>\",\"WARC-IP-Address\":\"51.81.83.12\",\"WARC-Target-URI\":\"https://rdrr.io/cran/did/f/vignettes/pre-testing.Rmd\",\"WARC-Payload-Digest\":\"sha1:RTHGFZD5EUR3XHBQDQWLPGW5JQEBANQN\",\"WARC-Block-Digest\":\"sha1:TZN4SPF3GAKIIKIS35KJZTEXZ7IC4P6B\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2022/CC-MAIN-2022-05/CC-MAIN-2022-05_segments_1642320304515.74_warc_CC-MAIN-20220124054039-20220124084039-00188.warc.gz\"}"}
https://www.vadepares.cat/triangle-proofs-worksheet/	[ "# Triangle Proofs Worksheet", null, "Some of the worksheets for this concept are congruent triangles 2 column proofs, geometric proofs, , geometryh work proofs in two column form, triangle proofs s sas asa aas, using cpctc with triangle congruence, proving triangles are congruent by sas asa, assignment date period. ∠ ≅ ∠y c 1.", null, "Pythagorean Theorem Maze Worksheet Pythagorean theorem\n\n### Geometry worksheet triangle congruence proofs name:", null, "Triangle proofs worksheet. 2) why is an altitude? Triangle proofs worksheet part 1. The point that divides a segment into two congruent segments.\n\nAnswer key for 4 4 practice worksheet. Therefore, they have the same length. 1) why is the triangle isosceles?\n\nTriangle congruence proofs worksheet answers together with practical themes. Worksheet by kuta software llc 3 answers to naming angles and angle addition postulate naaaap 1 g 3. 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Having the exact same size and shape and there by having the exact same measures.\n\nMy geometry students need practice with proving triangles congruent using sss sas congruent triangles worksheet proving triangles congruent triangle worksheet a collection of congruent triangles worksheets on key concepts like congruent parts of congruent triangles … The ray that divides an angle into two congruent angles. Make sure you understand the main areas of study, like a basic definition of a.\n\nHaving the exact same size and shape and there by having the exact same measures. Belum ada komentar untuk 30 congruent triangles proofs worksheet answers A triangle with 2 sides of the same length is isosceles.\n\nBecause you should supply programs in a reputable along with reputable origin, all of us present very helpful info on different subject matter plus topics. You will receive your score and answers at the end. About this quiz & worksheet.", null, "Geometry Puzzle Worksheets Bundle Geometry Resources and", null, "Triangle Congruence Worksheet Answer Key Beautiful", null, "Isosceles and Equilateral Triangles Puzzle Worksheet", null, "Proving Triangles Congruent Worksheet Answers Congruent", null, "Classified Triangle Congruence Proofs Worksheet Triangle", null, "20 Geometric Proofs Worksheet with Answers in 2020", null, "50 Triangle Inequality theorem Worksheet in 2020", null, "Hypotenuse Leg theorem Worksheet 30 Triangle Congruence", null, "Recent Parallel Lines and Transversals Worksheet Answers", null, "Pin on Customize Design Worksheet Online", null, "Geometric Shapes Worksheet 2nd Grade 2nd Grade Geometry", null, "congruent is = like circumference is perimeter in geometry", null, "Pin on Printable Blank Worksheet Template", null, "Proving Triangles Congruent Proof Activity Geometry", null, "Due Thursday October 23 Due Thursday October 23 San Marino", null, "Pin on Professionally Designed Worksheets", null, "Relationships in Triangles INB Pages Triangle inequality", null, "Segments in Triangles Foldable Teaching geometry", null, "Readable Proving Triangles Congruent Worksheet" ]	[ null, "https://i2.wp.com/www.vadepares.cat/wp-content/uploads/2021/02/images-18045.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/13/80/1b/13801b4612d058472a7b581932b4e750.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/81/4e/26/814e26deade7b961bf5f231a905b88c4.jpg", null, "https://i1.wp.com/i.pinimg.com/originals/db/37/1e/db371e20dffa6393b2a1423263d1cb6a.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/0a/ce/dd/0aceddf0a068c7b724ff72a8303d08a3.jpg", null, "https://i1.wp.com/i.pinimg.com/736x/ec/16/f6/ec16f643ccf0f07e3dfe14d7b24c0b68--geometry-worksheets-secondary-math.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/77/35/6c/77356c70d5620fab2f0f0c27dc103af5.jpg", null, "https://i2.wp.com/i.pinimg.com/originals/e6/bb/34/e6bb34b799f1e1786fa98762829596d5.jpg", null, "https://i0.wp.com/i.pinimg.com/736x/10/58/e4/1058e46a38e7a8af74620ff56bfebb5b.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/d7/19/69/d719699b59657d612bd64c21091dbdd4.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/24/ea/12/24ea12c026a92f0f155a2a6519a09cb3.jpg", null, "https://i1.wp.com/i.pinimg.com/originals/e1/d2/82/e1d2827632ed876f5d14ad0b2f4f168e.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/27/56/63/275663ba76aad64b4ea8ff73ade9cddb.jpg", null, "https://i1.wp.com/i.pinimg.com/originals/2c/dd/ea/2cddea78d0d046c71401b0accf458361.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/f6/cd/26/f6cd26b2807cd9a7fcb8ed3407809aba.png", null, "https://i0.wp.com/i.pinimg.com/originals/2d/bb/2b/2dbb2bc554ab5ab707ea2c58110410d1.jpg", null, "https://i2.wp.com/i.pinimg.com/originals/a9/18/0f/a9180fff79728a249319bc8d9b466fbe.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/b0/1f/bf/b01fbf26766334e800aa622d595896ec.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/96/9e/17/969e17ddd35420b2d639e9be901221d4.jpg", null, "https://i0.wp.com/i.pinimg.com/originals/81/4e/26/814e26deade7b961bf5f231a905b88c4.jpg", null, "https://i2.wp.com/i.pinimg.com/originals/d4/80/1f/d4801f90224b4fd8f10feccbad88408e.jpg", null, "https://i2.wp.com/i.pinimg.com/originals/41/1e/0e/411e0eba23fe9d87ba15384c213e97eb.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8336257,"math_prob":0.9053974,"size":7941,"snap":"2021-31-2021-39","text_gpt3_token_len":1736,"char_repetition_ratio":0.22565201,"word_repetition_ratio":0.06586345,"special_character_ratio":0.19153759,"punctuation_ratio":0.09604105,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99746877,"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],"im_url_duplicate_count":[null,1,null,3,null,2,null,1,null,3,null,1,null,2,null,2,null,1,null,1,null,3,null,7,null,1,null,1,null,2,null,2,null,1,null,5,null,2,null,2,null,1,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-08-01T17:15:49Z\",\"WARC-Record-ID\":\"<urn:uuid:0a2218a0-8c99-4b39-97ef-748eec9f997f>\",\"Content-Length\":\"68605\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1d0b7160-8b46-4d47-86ee-168272539234>\",\"WARC-Concurrent-To\":\"<urn:uuid:9cdc5214-cd4f-455b-902d-2d791771f2d6>\",\"WARC-IP-Address\":\"104.21.76.249\",\"WARC-Target-URI\":\"https://www.vadepares.cat/triangle-proofs-worksheet/\",\"WARC-Payload-Digest\":\"sha1:NX5KFOGO7K3DFFTTCGD3KR4EUV76EFDU\",\"WARC-Block-Digest\":\"sha1:DU2YS77WKG2QMX22HUI53USM6LUE4ITW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046154214.63_warc_CC-MAIN-20210801154943-20210801184943-00565.warc.gz\"}"}
http://mural.maynoothuniversity.ie/5003/	[ "Emergent Phenomena in Matrix Models\n\nKaltenbrunner, Thomas (2014) Emergent Phenomena in Matrix Models. PhD thesis, National University of Ireland Maynooth.", null, "", null, "Preview", null, "", null, "", null, "", null, "", null, "more...", null, "Add this article to your Mendeley library\n\nAbstract\n\nIn this work we perform a careful study of different matrix models and particularly the property of emergent phenomena in them. We start discussing a 2-matrix model of Yang-Mills type that exhibits an emergent topology in the strong coupling limit. We use Monte-Carlo simulations to obtain various observables that allow us to get more insight in the transition from the non-commutative regime towards the commutative, strong coupling, limit. We will continue to discuss higher dimensional Yang-Mills matrix models, focusing on the lowest dimensional case that is well defned, D = 3, and on the large-D limit. While we discuss the possibility of an emergent topology in 3 dimensions, we find that the behaviour of this type of models changes towards random matrices for large D. In the second part of the thesis we will add a Myers term to the Yang-Mills type models which extends the possible solutions to the model by fuzzy spaces. We carry out a 1-loop calculation for a general SU(d) symmetric solution to this class of models. We will then turn to a numerical study of a model that incorporates the simplest case of a fuzzy manifold, the fuzzy sphere. We will further study fuzzy CP2 which appears as a solution to the 8-dimensional Yang-Mills-Myers model. Numerical results from a Monte-Carlo simulation will be used to compare with the analytical results obtained earlier. We will further construct a slightly modified 8-dimensional model that has a fuzzy complex projective plane as the ground state in phase space. In total, we find four different phases in this model, which we will describe in detail after numerically mapping the phase diagram.\n\nItem Type: Thesis (PhD) Emergent Phenomena; Matrix Models; Faculty of Science and Engineering > Mathematical Physics 5003 IR eTheses 04 Jun 2014 11:44", null, "Item control page" ]	[ null, "http://mural.maynoothuniversity.ie/5003/1.hassmallThumbnailVersion/thesis.pdf", null, "http://mural.maynoothuniversity.ie/5003/1.haspreviewThumbnailVersion/thesis.pdf", null, "http://mural.maynoothuniversity.ie/images/icons8-Twitter-26.png", null, "http://mural.maynoothuniversity.ie/images/facebook.png", null, "http://mural.maynoothuniversity.ie/images/linkedin.png", null, "http://mural.maynoothuniversity.ie/images/googleplus.png", null, "http://mural.maynoothuniversity.ie/images/sendmail.png", null, "http://www.mendeley.com/graphics/mendeley.png", null, "http://mural.maynoothuniversity.ie/style/images/action_view.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8620342,"math_prob":0.8444881,"size":2178,"snap":"2019-26-2019-30","text_gpt3_token_len":476,"char_repetition_ratio":0.10579577,"word_repetition_ratio":0.0,"special_character_ratio":0.20615244,"punctuation_ratio":0.0964467,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9648945,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18],"im_url_duplicate_count":[null,1,null,1,null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-06-18T15:32:49Z\",\"WARC-Record-ID\":\"<urn:uuid:2645e91e-df62-43f4-a91b-89cfbe32efcf>\",\"Content-Length\":\"30865\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6d45bd1f-37ea-4951-b706-11e8ada0f693>\",\"WARC-Concurrent-To\":\"<urn:uuid:143eade5-2d97-4819-9696-ac7dae1cb52a>\",\"WARC-IP-Address\":\"13.69.156.231\",\"WARC-Target-URI\":\"http://mural.maynoothuniversity.ie/5003/\",\"WARC-Payload-Digest\":\"sha1:NBF2EMZ22EXOKQJWRO24CGJL2IGUUZHL\",\"WARC-Block-Digest\":\"sha1:V6WLCQ22I3RY7H7ZS2HTJHF3O3RINIKD\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-26/CC-MAIN-2019-26_segments_1560627998755.95_warc_CC-MAIN-20190618143417-20190618165417-00311.warc.gz\"}"}
http://oeis.org/A163778	[ "The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.", null, "Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)\n A163778 Odd terms in A054639. 9\n 3, 5, 9, 11, 23, 29, 33, 35, 39, 41, 51, 53, 65, 69, 81, 83, 89, 95, 99, 105, 113, 119, 131, 135, 155, 173, 179, 183, 189, 191, 209, 221, 231, 233, 239, 243, 245, 251, 261, 273, 281, 293, 299, 303, 309, 323, 329, 359, 371, 375, 393, 411, 413, 419, 429 (list; graph; refs; listen; history; text; internal format)\n OFFSET 1,1 COMMENTS Previous name was: The A_1-primes (Archimedes_1 primes). We have: (1) N is an A_1-prime iff N is odd, p=2N+1 is a prime number and only one of +2 and -2 generates Z_p^* (the multiplicative group of Z_p); (2) N is an A_1-prime iff p=2N+1 is a prime number and exactly one of the following holds: (a) N == 1 (mod 4) and +2 generates Z_p^* but -2 does not, (b) N == 3 (mod 4) and -2 generates Z_p^* but +2 does not. The A_1-primes are the odd T- or Twist-primes (the T-primes are the same as the Queneau-numbers, A054639). For the related A_0-, A^+_1- and A^-_1-primes, see A163777, A163779 and A163780. Considered as a set, the present sequence is the union of the A^+_1-primes (A163779) and the A^-_1-primes (A163780). It is also equal to the difference of A054639 and the A_0-primes (A163777). LINKS P. R. J. Asveld, Table of n, a(n) for n=1..6706. P. R. J. Asveld, Permuting operations on strings and their relation to prime numbers, Discrete Applied Mathematics 159 (2011) 1915-1932. P. R. J. Asveld, Permuting operations on strings and the distribution of their prime numbers (2011), TR-CTIT-11-24, Dept. of CS, Twente University of Technology, Enschede, The Netherlands. P. R. J. Asveld, Some Families of Permutations and Their Primes (2009), TR-CTIT-09-27, Dept. of CS, Twente University of Technology, Enschede, The Netherlands. P. R. J. Asveld, Permuting Operations on Strings-Their Permutations and Their Primes, Twente University of Technology, 2014. MATHEMATICA follow[s_, f_] := Module[{t, k}, t = f[s]; k = 1; While[t>s, k++; t = f[t]]; If[s == t, k, 0]]; okQ[n_] := n>1 && n == follow[1, Function[j, Ceiling[n/2] + (-1)^jCeiling[ (j-1)/2]]]; A163778 = Select[Range, okQ] ( Jean-François Alcover, Jun 07 2018, after Andrew Howroyd ) PROG (PARI) Follow(s, f)={my(t=f(s), k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)} ok(n)={n>1 && n==Follow(1, j->ceil(n/2) + (-1)^jceil((j-1)/2))} select(ok, [1..1000]) \\\\ Andrew Howroyd, Nov 11 2017 (PARI) ok(n)={n>1 && n%2==1 && isprime(2n+1) && znorder(Mod(2, 2n+1)) == if(n%4==3, n, 2n)} select(ok, [1..1000]) \\\\ Andrew Howroyd, Nov 11 2017 CROSSREFS Cf. A054639, A163777, A163779, A163780, A294434, A294673. Sequence in context: A113488 A092917 A256220 A328643 A160358 A319084 Adjacent sequences: A163775 A163776 A163777 * A163779 A163780 A163781 KEYWORD nonn AUTHOR Peter R. J. Asveld, Aug 11 2009 EXTENSIONS a(33)-a(55) from Andrew Howroyd, Nov 11 2017 New name from Joerg Arndt, Mar 23 2018 STATUS approved\n\nLookup \| Welcome \| Wiki \| Register \| Music \| Plot 2 \| Demos \| Index \| Browse \| More \| WebCam\nContribute new seq. or comment \| Format \| Style Sheet \| Transforms \| Superseeker \| Recent\nThe OEIS Community \| Maintained by The OEIS Foundation Inc.\n\nLast modified August 12 17:17 EDT 2020. Contains 336439 sequences. (Running on oeis4.)" ]	[ null, "http://oeis.org/banner2021.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.66515356,"math_prob":0.94873077,"size":2885,"snap":"2020-34-2020-40","text_gpt3_token_len":1157,"char_repetition_ratio":0.10898993,"word_repetition_ratio":0.09368636,"special_character_ratio":0.4967071,"punctuation_ratio":0.24680851,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9862174,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-12T21:46:25Z\",\"WARC-Record-ID\":\"<urn:uuid:7cae849f-18be-4a0c-b781-742aa77a61b9>\",\"Content-Length\":\"21079\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:110bb15e-79bf-4db7-b829-e65722163548>\",\"WARC-Concurrent-To\":\"<urn:uuid:d2eb41f8-7123-4198-a362-bc58cb6b52ef>\",\"WARC-IP-Address\":\"104.239.138.29\",\"WARC-Target-URI\":\"http://oeis.org/A163778\",\"WARC-Payload-Digest\":\"sha1:VNNP4LEJ7Y6H6VAEZRNQP3MXF6OGPLO6\",\"WARC-Block-Digest\":\"sha1:SNHRNLORINQF2S2OMKDMNFVMHICIF2YR\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439738944.95_warc_CC-MAIN-20200812200445-20200812230445-00267.warc.gz\"}"}
https://www.readchemistry.com/2019/02/IR-spectroscopy-in-organic-compounds.html	[ "## Breaking News\n\n#### Infrared Spectroscopy\n\n Infrared (IR) spectroscopy is a simple, rapid, and nondestructive instrumental technique that can give evidence for the presence of various functional groups. If you had a sample of unknown identity, among the first things you would do is obtain an infrared spectrum, along with determining its solubility in common solvents and its melting and/ or boiling point.\n\n Infrared spectroscopy, as with all forms of spectroscopy, depends on the interaction of molecules or atoms with electromagnetic radiation.\n\n Infrared radiation causes atoms and groups of atoms of organic compounds to vibrate with increased amplitude about the covalent bonds that connect them. (Infrared radiation is not of sufficient energy to excite electrons, as is the case when some molecules interact with visible, ultraviolet, or higher energy forms of light.)\n\n Since the functional groups of organic molecules include specific arrangements of bonded atoms, absorption of IR radiation by an organic molecule will occur at specific frequencies characteristic of the types of bonds and atoms present in the specific functional groups of that molecule. These vibrations are quantized, and as they occur, the compounds absorb IR energy in particular regions of the IR portion of the spectrum.\n\n#### Fourier transform infrared (FTIR) spectrometer\n\n The figure shows diagram of a Fourier transform infrared (FTIR) spectrometer.\n\n FTIR spectrometers employ a Michelson interferometer, which splits the radiation beam from the IR source so that it reflects simultaneously from a moving mirror and a fixed mirror, leading to interference.\n\n After the beams recombine, they pass through the sample to the detector and are recorded as a plot of time versus signal intensity, called an interferogram.\n\n The overlapping wavelengths and the intensities of their respective absorptions are then converted to a spectrum by applying a mathematical operation called a Fourier transform.\n\n The FTIR method eliminates the need to scan slowly over a range of wavelengths, as was the case with older types of instruments called dispersive IR spectrometers, and therefore FTIR spectra can be acquired very quickly.\n\n The FTIR method also allows greater throughput of IR energy. The combination of these factors gives FTIR spectra strong signals as compared to background noise (i.e., a high signal to noise ratio) because radiation throughput is high and rapid scanning allows multiple spectra to be averaged in a short period of time. The result is enhancement of real signals and cancellation of random noise.\n\n An infrared spectrometer (Fig. 1) operates by passing a beam of IR radiation through a sample and comparing the radiation transmitted through the sample with that transmitted in the absence of the sample. Any frequencies absorbed by the sample will be apparent by the difference. The spectrometer plots the results as a graph showing absorbance versus frequency or wavelength.\n\n#### Wavenumbers\n\n The position of an absorption band (peak) in an IR spectrum is specified in units of wavenumbers (ῡ).\n\n Wavenumbers are the reciprocal of wavelength when wavelength is expressed in centimeters (the unit is cm-1), and therefore give the number of wave cycles per centimeter. The larger the wavenumber, the higher is the frequency of the wave, and correspondingly the higher is the frequency of the bond absorption. IR absorptions are sometimes, though less commonly, reported in terms of wavelength (λ), in which case the units are micrometers (μm; old name micron, μ). Wavelength is the distance from crest to crest of a wave.\n\n In their vibrations covalent bonds behave as if they were tiny springs connecting the atoms. When the atoms vibrate, they can do so only at certain frequencies, as if the bonds were “tuned.” Because of this, covalently bonded atoms have only particular vibrational energy levels; that is, the levels are quantized.\n\n The excitation of a molecule from one vibrational energy level to another occurs only when the compound absorbs IR radiation of a particular energy, meaning a particular wavelength or frequency. Note that the energy (E ) of absorption is directly proportional to the frequency of radiation (ʋ) because ΔE = hʋ, and inversely proportional to the wavelength (λ) because c/λ and therefore ΔE = hc / λ\n\n Molecules can vibrate in a variety of ways. Two atoms joined by a covalent bond can undergo a stretching vibration where the atoms move back and forth as if joined by a spring. Three atoms can also undergo a variety of stretching and bending vibrations.\n\n#### Characteristic Infrared Absorptions of Groups\n\n The frequency of a given stretching vibration in an IR spectrum can be related to two factors. These are the masses of the bonded atoms—light atoms vibrate at higher frequencies than heavier ones—and the relative stiffness of the bond. (These factors are accounted for in Hooke’s law, a relationship you may study in introductory physics.) Triple bonds are stiffer (and vibrate at higher frequencies) than double bonds, and double bonds are stiffer (and vibrate at higher frequencies) than single bonds. We can see some of these effects in this Table:\n\n Notice that stretching frequencies of groups involving hydrogen (a light atom) such as C-H, N-H, and O-H all occur at relatively high frequencies:\n\n Notice, too, that triple bonds vibrate at higher frequencies than double bonds:\n\n Not all molecular vibrations result in the absorption of IR energy. In order for a vibration to occur with the absorption of IR energy, the dipole moment of the molecule must change as the vibration occurs.\n\n** Thus, methane does not absorb IR energy for symmetric streching of the four C-H bonds; asymmetric stretching, on the other hand, does lead to an IR absorption.Symmetrical vibrations of the carbon–carbon double and triple bonds of ethene and ethyne do not result in the absorption of IR radiation, either\n\n### Solved problem\n\nThe infrared spectrum of l-hexyne shows a sharp absorption peak near 2100 cm-1 due to stretching of its triple bond. However, 3-hexyne shows no absorption in that region. Explain" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90911424,"math_prob":0.93052745,"size":7503,"snap":"2021-31-2021-39","text_gpt3_token_len":1626,"char_repetition_ratio":0.13401787,"word_repetition_ratio":0.021648627,"special_character_ratio":0.19099027,"punctuation_ratio":0.08582375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.957968,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-23T23:49:48Z\",\"WARC-Record-ID\":\"<urn:uuid:3014b11a-5279-4e93-aa42-9efbea6db3ec>\",\"Content-Length\":\"315926\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:6b966893-a2ba-44f9-b579-cd9ca3a5b825>\",\"WARC-Concurrent-To\":\"<urn:uuid:c7081492-f2d8-4e8b-ba9d-55f112b59610>\",\"WARC-IP-Address\":\"172.217.1.211\",\"WARC-Target-URI\":\"https://www.readchemistry.com/2019/02/IR-spectroscopy-in-organic-compounds.html\",\"WARC-Payload-Digest\":\"sha1:SWMHXMTGTB5GWQFBISRK4PBNM4IYF63E\",\"WARC-Block-Digest\":\"sha1:TF6GHK6G4YNIIZTGJULXJCSEC6BBJHG7\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057479.26_warc_CC-MAIN-20210923225758-20210924015758-00635.warc.gz\"}"}
https://www.lessonplanet.com/teachers/multiplying-decimals-and-mixed-numbers	[ "", null, "# Multiplying Decimals and Mixed Numbers\n\n##### This Multiplying Decimals and Mixed Numbers lesson plan also includes:\n\nStudents practice multiplying decimals and/or mixed numbers, explore the effects of multiplying decimals and mixed numbers, and practice predicting the effects of multiplying a number by a decimal or mixed number." ]	[ null, "data:image/png;base64,R0lGODlhAQABAAD/ACwAAAAAAQABAAACADs", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.80119914,"math_prob":0.99237627,"size":2753,"snap":"2021-31-2021-39","text_gpt3_token_len":607,"char_repetition_ratio":0.13604948,"word_repetition_ratio":0.0046189376,"special_character_ratio":0.19723937,"punctuation_ratio":0.077283375,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.987845,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-27T23:29:31Z\",\"WARC-Record-ID\":\"<urn:uuid:0525b41c-be2e-4d71-bc0b-9f5e7b294e18>\",\"Content-Length\":\"101467\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8710822a-f82c-4592-9bbe-1155120724f7>\",\"WARC-Concurrent-To\":\"<urn:uuid:87677104-5829-4546-aa09-db4130c02a6c>\",\"WARC-IP-Address\":\"3.142.116.236\",\"WARC-Target-URI\":\"https://www.lessonplanet.com/teachers/multiplying-decimals-and-mixed-numbers\",\"WARC-Payload-Digest\":\"sha1:44CJQG5NRVD2MNR26ITCXZ4FAQPWQDP5\",\"WARC-Block-Digest\":\"sha1:YZJIB2U5MBJGJ7AIAIT4HEIL36R5MQCH\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780058552.54_warc_CC-MAIN-20210927211955-20210928001955-00351.warc.gz\"}"}
https://docs.opencv.org/3.4.2/df/d0c/tutorial_py_fast.html	[ "", null, "OpenCV 3.4.2 Open Source Computer Vision\nFAST Algorithm for Corner Detection\n\n## Goal\n\nIn this chapter,\n\n• We will understand the basics of FAST algorithm\n• We will find corners using OpenCV functionalities for FAST algorithm.\n\n## Theory\n\nWe saw several feature detectors and many of them are really good. But when looking from a real-time application point of view, they are not fast enough. One best example would be SLAM (Simultaneous Localization and Mapping) mobile robot which have limited computational resources.\n\nAs a solution to this, FAST (Features from Accelerated Segment Test) algorithm was proposed by Edward Rosten and Tom Drummond in their paper \"Machine learning for high-speed corner detection\" in 2006 (Later revised it in 2010). A basic summary of the algorithm is presented below. Refer original paper for more details (All the images are taken from original paper).\n\n### Feature Detection using FAST\n\n1. Select a pixel $$p$$ in the image which is to be identified as an interest point or not. Let its intensity be $$I_p$$.\n2. Select appropriate threshold value $$t$$.\n3. Consider a circle of 16 pixels around the pixel under test. (See the image below)", null, "image\n4. Now the pixel $$p$$ is a corner if there exists a set of $$n$$ contiguous pixels in the circle (of 16 pixels) which are all brighter than $$I_p + t$$, or all darker than $$I_p − t$$. (Shown as white dash lines in the above image). $$n$$ was chosen to be 12.\n5. A high-speed test was proposed to exclude a large number of non-corners. This test examines only the four pixels at 1, 9, 5 and 13 (First 1 and 9 are tested if they are too brighter or darker. If so, then checks 5 and 13). If $$p$$ is a corner, then at least three of these must all be brighter than $$I_p + t$$ or darker than $$I_p − t$$. If neither of these is the case, then $$p$$ cannot be a corner. The full segment test criterion can then be applied to the passed candidates by examining all pixels in the circle. This detector in itself exhibits high performance, but there are several weaknesses:\n• It does not reject as many candidates for n < 12.\n• The choice of pixels is not optimal because its efficiency depends on ordering of the questions and distribution of corner appearances.\n• Results of high-speed tests are thrown away.\n• Multiple features are detected adjacent to one another.\n\nFirst 3 points are addressed with a machine learning approach. Last one is addressed using non-maximal suppression.\n\n### Machine Learning a Corner Detector\n\n1. Select a set of images for training (preferably from the target application domain)\n2. Run FAST algorithm in every images to find feature points.\n3. For every feature point, store the 16 pixels around it as a vector. Do it for all the images to get feature vector $$P$$.\n4. Each pixel (say $$x$$) in these 16 pixels can have one of the following three states:", null, "image\n5. Depending on these states, the feature vector $$P$$ is subdivided into 3 subsets, $$P_d$$, $$P_s$$, $$P_b$$.\n6. Define a new boolean variable, $$K_p$$, which is true if $$p$$ is a corner and false otherwise.\n7. Use the ID3 algorithm (decision tree classifier) to query each subset using the variable $$K_p$$ for the knowledge about the true class. It selects the $$x$$ which yields the most information about whether the candidate pixel is a corner, measured by the entropy of $$K_p$$.\n8. This is recursively applied to all the subsets until its entropy is zero.\n9. The decision tree so created is used for fast detection in other images.\n\n### Non-maximal Suppression\n\nDetecting multiple interest points in adjacent locations is another problem. It is solved by using Non-maximum Suppression.\n\n1. Compute a score function, $$V$$ for all the detected feature points. $$V$$ is the sum of absolute difference between $$p$$ and 16 surrounding pixels values.\n2. Consider two adjacent keypoints and compute their $$V$$ values.\n3. Discard the one with lower $$V$$ value.\n\n### Summary\n\nIt is several times faster than other existing corner detectors.\n\nBut it is not robust to high levels of noise. It is dependent on a threshold.\n\n## FAST Feature Detector in OpenCV\n\nIt is called as any other feature detector in OpenCV. If you want, you can specify the threshold, whether non-maximum suppression to be applied or not, the neighborhood to be used etc.\n\nFor the neighborhood, three flags are defined, cv.FAST_FEATURE_DETECTOR_TYPE_5_8, cv.FAST_FEATURE_DETECTOR_TYPE_7_12 and cv.FAST_FEATURE_DETECTOR_TYPE_9_16. Below is a simple code on how to detect and draw the FAST feature points.\n\nimport numpy as np\nimport cv2 as cv\nfrom matplotlib import pyplot as plt\n# Initiate FAST object with default values\nfast = cv.FastFeatureDetector_create()\n# find and draw the keypoints\nkp = fast.detect(img,None)\nimg2 = cv.drawKeypoints(img, kp, None, color=(255,0,0))\n# Print all default params\nprint( \"Threshold: {}\".format(fast.getThreshold()) )\nprint( \"nonmaxSuppression:{}\".format(fast.getNonmaxSuppression()) )\nprint( \"neighborhood: {}\".format(fast.getType()) )\nprint( \"Total Keypoints with nonmaxSuppression: {}\".format(len(kp)) )\ncv.imwrite('fast_true.png',img2)\n# Disable nonmaxSuppression\nfast.setNonmaxSuppression(0)\nkp = fast.detect(img,None)\nprint( \"Total Keypoints without nonmaxSuppression: {}\".format(len(kp)) )\nimg3 = cv.drawKeypoints(img, kp, None, color=(255,0,0))\ncv.imwrite('fast_false.png',img3)\n\nSee the results. First image shows FAST with nonmaxSuppression and second one without nonmaxSuppression:", null, "image" ]	[ null, "https://docs.opencv.org/3.4.2/opencv-logo-small.png", null, "https://docs.opencv.org/3.4.2/fast_speedtest.jpg", null, "https://docs.opencv.org/3.4.2/fast_eqns.jpg", null, "https://docs.opencv.org/3.4.2/fast_kp.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87486374,"math_prob":0.99317545,"size":5435,"snap":"2023-40-2023-50","text_gpt3_token_len":1277,"char_repetition_ratio":0.10329589,"word_repetition_ratio":0.0045454544,"special_character_ratio":0.24599816,"punctuation_ratio":0.11764706,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9956041,"pos_list":[0,1,2,3,4,5,6,7,8],"im_url_duplicate_count":[null,null,null,1,null,1,null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-02T01:33:40Z\",\"WARC-Record-ID\":\"<urn:uuid:b0ac9f7b-00c4-4ac0-85e5-a4ec3bee675b>\",\"Content-Length\":\"14650\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:138cd8e0-bd04-4559-acfe-660d5c30e382>\",\"WARC-Concurrent-To\":\"<urn:uuid:999933de-637d-4987-9494-a2b289c51752>\",\"WARC-IP-Address\":\"64.91.246.43\",\"WARC-Target-URI\":\"https://docs.opencv.org/3.4.2/df/d0c/tutorial_py_fast.html\",\"WARC-Payload-Digest\":\"sha1:5M2ET53SU3YMDYRD2F3IZAJPJZBJPDGJ\",\"WARC-Block-Digest\":\"sha1:5V6LDLVZ5LVHXXV7Y2GSIOT37X2SFSVZ\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510942.97_warc_CC-MAIN-20231002001302-20231002031302-00374.warc.gz\"}"}
http://www.indiabix.com/aptitude/problems-on-hcf-and-lcm/	[ "# Aptitude - Problems on H.C.F and L.C.M\n\n## Why Aptitude Problems on H.C.F and L.C.M?\n\nIn this section you can learn and practice Aptitude Questions based on \"Problems on H.C.F and L.C.M\" and improve your skills in order to face the interview, competitive examination and various entrance test (CAT, GATE, GRE, MAT, Bank Exam, Railway Exam etc.) with full confidence.\n\n## Where can I get Aptitude Problems on H.C.F and L.C.M questions and answers with explanation?\n\nIndiaBIX provides you lots of fully solved Aptitude (Problems on H.C.F and L.C.M) questions and answers with Explanation. Solved examples with detailed answer description, explanation are given and it would be easy to understand. All students, freshers can download Aptitude Problems on H.C.F and L.C.M quiz questions with answers as PDF files and eBooks.\n\n## Where can I get Aptitude Problems on H.C.F and L.C.M Interview Questions and Answers (objective type, multiple choice)?\n\nHere you can find objective type Aptitude Problems on H.C.F and L.C.M questions and answers for interview and entrance examination. Multiple choice and true or false type questions are also provided.\n\n## How to solve Aptitude Problems on H.C.F and L.C.M problems?\n\nYou can easily solve all kind of Aptitude questions based on Problems on H.C.F and L.C.M by practicing the objective type exercises given below, also get shortcut methods to solve Aptitude Problems on H.C.F and L.C.M problems.\n\n### Exercise :: Problems on H.C.F and L.C.M - General Questions\n\n1.\n\nFind the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.\n\n A. 4 B. 7 C. 9 D. 13\n\nExplanation:\n\nRequired number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)\n\n= H.C.F. of 48, 92 and 140 = 4.\n\n2.\n\nThe H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:\n\n A. 276 B. 299 C. 322 D. 345\n\nExplanation:\n\nClearly, the numbers are (23 x 13) and (23 x 14).", null, "Larger number = (23 x 14) = 322.\n\n3.\n\nSix bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?\n\n A. 4 B. 10 C. 15 D. 16\n\nExplanation:\n\nL.C.M. of 2, 4, 6, 8, 10, 12 is 120.\n\nSo, the bells will toll together after every 120 seconds(2 minutes).\n\n In 30 minutes, they will toll together 30 + 1 = 16 times. 2\n\n4.\n\nLet N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:\n\n A. 4 B. 5 C. 6 D. 8\n\nExplanation:\n\nN = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)\n\n= H.C.F. of 3360, 2240 and 5600 = 1120.\n\nSum of digits in N = ( 1 + 1 + 2 + 0 ) = 4\n\n5.\n\nThe greatest number of four digits which is divisible by 15, 25, 40 and 75 is:\n\n A. 9000 B. 9400 C. 9600 D. 9800\n\nExplanation:\n\nGreatest number of 4-digits is 9999.\n\nL.C.M. of 15, 25, 40 and 75 is 600.\n\nOn dividing 9999 by 600, the remainder is 399.", null, "Required number (9999 - 399) = 9600." ]	[ null, "http://www.indiabix.com/_files/images/aptitude/1-sym-tfr.gif", null, "http://www.indiabix.com/_files/images/aptitude/1-sym-tfr.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.78929347,"math_prob":0.9148939,"size":1542,"snap":"2021-43-2021-49","text_gpt3_token_len":531,"char_repetition_ratio":0.12808843,"word_repetition_ratio":0.082539685,"special_character_ratio":0.42801556,"punctuation_ratio":0.23325062,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99707586,"pos_list":[0,1,2,3,4],"im_url_duplicate_count":[null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-08T09:51:04Z\",\"WARC-Record-ID\":\"<urn:uuid:dd6f68bd-1f54-4b48-b341-938056fa87ef>\",\"Content-Length\":\"35704\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:7428b7a1-5ec6-41a0-8ffc-5a0d2b3bde80>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a4e2ec0-b19a-43ea-9325-5caa82e2cb0a>\",\"WARC-IP-Address\":\"35.244.11.196\",\"WARC-Target-URI\":\"http://www.indiabix.com/aptitude/problems-on-hcf-and-lcm/\",\"WARC-Payload-Digest\":\"sha1:FLFVOUHRL4Q6UUUCKMGCINETLTZFPA3R\",\"WARC-Block-Digest\":\"sha1:IZ2236TVLW4WE2HPILO7ZZROXEXUHE3Q\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964363465.47_warc_CC-MAIN-20211208083545-20211208113545-00213.warc.gz\"}"}
https://forcoder.su/bayesian-analysis-python-introduction-2nd/	[ "# Bayesian Analysis with Python: Introduction to statistical modeling and probabilistic programming using PyMC3 and ArviZ, 2nd Edition", null, "Bayesian Analysis with Python: Introduction to statistical modeling and probabilistic programming using PyMC3 and ArviZ, 2nd Edition by Osvaldo Martin\nEnglish \| 2018 \| ISBN: 1789341652 \| 356 Pages \| True PDF, EPUB \| 75 MB\n\nBayesian modeling with PyMC3 and exploratory analysis of Bayesian models with ArviZ\nThe second edition of Bayesian Analysis with Python is an introduction to the main concepts of applied Bayesian inference and its practical implementation in Python using PyMC3, a state-of-the-art probabilistic programming library, and ArviZ, a new library for exploratory analysis of Bayesian models.\nThe main concepts of Bayesian statistics are covered using a practical and computational approach. Synthetic and real data sets are used to introduce several types of models, such as generalized linear models for regression and classification, mixture models, hierarchical models, and Gaussian processes, among others.\nBy the end of the book, you will have a working knowledge of probabilistic modeling and you will be able to design and implement Bayesian models for your own data science problems. After reading the book you will be better prepared to delve into more advanced material or specialized statistical modeling if you need to.\nWhat you will learn\n\n• Build probabilistic models using the Python library PyMC3\n• Analyze probabilistic models with the help of ArviZ\n• Acquire the skills required to sanity check models and modify them if necessary\n• Understand the advantages and caveats of hierarchical models\n• Find out how different models can be used to answer different data analysis questions\n• Compare models and choose between alternative ones\n• Discover how different models are unified from a probabilistic perspective\n• Think probabilistically and benefit from the flexibility of the Bayesian framework" ]	[ null, "https://forcoder.su/wp-content/cover/bayesian-analysis-python-introduction-2nd.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8789411,"math_prob":0.82432234,"size":1719,"snap":"2020-10-2020-16","text_gpt3_token_len":334,"char_repetition_ratio":0.14635569,"word_repetition_ratio":0.0,"special_character_ratio":0.18789995,"punctuation_ratio":0.061594203,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9931862,"pos_list":[0,1,2],"im_url_duplicate_count":[null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-02-21T04:16:59Z\",\"WARC-Record-ID\":\"<urn:uuid:11ca54d6-1861-4f81-b50e-62cc7b951e77>\",\"Content-Length\":\"36813\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:a8cbbd46-1ad0-43f5-9027-61accfc3079a>\",\"WARC-Concurrent-To\":\"<urn:uuid:51f33e70-2b46-4d82-91d1-9609579576b7>\",\"WARC-IP-Address\":\"104.18.61.39\",\"WARC-Target-URI\":\"https://forcoder.su/bayesian-analysis-python-introduction-2nd/\",\"WARC-Payload-Digest\":\"sha1:RSVX6PUSQIKYBOSUOT7ZFGMSD3MX3THR\",\"WARC-Block-Digest\":\"sha1:XFM2OXAQDS4MMMIUO4SCEJPMGA4TSNCW\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-10/CC-MAIN-2020-10_segments_1581875145438.12_warc_CC-MAIN-20200221014826-20200221044826-00038.warc.gz\"}"}
http://forums.wolfram.com/mathgroup/archive/1995/Oct/msg00376.html	[ "", null, "", null, "", null, "", null, "", null, "", null, "", null, "Re: Simplifying vector expressions\n\n• Subject: [mg2168] Re: [mg2140] Simplifying vector expressions\n• From: danl (Daniel Lichtblau)\n• Date: Wed, 11 Oct 1995 05:59:13 GMT\n• Approved: [email protected]\n• Distribution: local\n• Newsgroups: wri.mathgroup\n• Organization: Wolfram Research, Inc.\n• Sender: daemon at wri.com ( )\n\n```\nThe difficulty lies in the fact that Times is Listable.\n\nIn:= vec1 = { a, b, c }\nOut= {a, b, c}\n\nIn:= vec2 = (1/x) * vec1 // InputForm\nOut//InputForm= {a/x, b/x, c/x}\n\nIn:= ClearAttributes[Times, Listable]\n\nIn:= vec3 = (1/x) * vec1 // InputForm\nOut//InputForm= {a, b, c}/x\n\nYou can change this as above, but if you do not change it back prior\nto invoking any internal functions that rely on this property, be\nprepared for strange results. In particular, I'dn ot be surprised if\nSimplify got flaky, and it is not clear to me how Expand could work\nif one argument involved a List.\nA tactic that might serve your purposes is to (i) use Factor et al\nas they are (that is, with Times left as Listable), (ii) find out the\nPolynomialGCD of all elements in your list, (iii) divide through by\nthat gcd to remove it from each element (might need to apply Together\nto get the job done completely), and (iv) make Times nonListable as\nabove, and now multiply through by that gcd to restore it. Note that\nthis will pull out factors in numerators as well as denominators,\nwhich may or may not be what you want.\n\nDaniel Lichtblau\nWRI\ndanl at wri.com\n\nBegin forwarded message:\n\n>Date: Wed, 4 Oct 1995 01:57:42 -0400\n>From: shenkin at still3.chem.columbia.edu (Peter Shenkin)\n>Subject: [mg2140] Simplifying vector expressions\n>Organization: MacroModel Development Group, Chemistry, Columbia U.,\n>NY, NY\n\nConsider the following vectors:\n\nIn:= vec1 = { a, b, c }\n\nOut= {a, b, c}\n\nIn:= vec2 = (1/x) * vec1\n\na b c\nOut= {-, -, -}\nx x x\n\nI am looking for a mechanism I can apply to vec2 to put it in the\nform:\n\n1\n(-) { a, b, c }\nx\n\nor\n\n{ a, b, c }\n-----------\nx\n\nThis would be the list equivalent of Factor[], as shown by the\n\nfollowing example:\n\nIn:= expr = a/x + b/x + c/x\n\na b c\nOut= - + - + -\nx x x\n\nIn:= Factor[ expr ]\n\na + b + c\nOut= ---------\nx\n\nDoes this facility (for lists) exist in Mathematica, or in a package\nI can obtain easily? If not, can you suggest guidelines for implementing\nof algebraic functionality: Simplify[], Expand[], etc., in addition\nto Factor[].\n\nThanks,\n-P.\n\n--\n\n****** When somebody says, \"It's a matter of principle,\"...\n*****\nPeter S. Shenkin, Box 768 Havemeyer Hall, Chemistry, Columbia\nUniv.,\nNY, NY 10027; shenkin at columbia.edu; (212)854-5143; FAX:\n678-9039\n********* ...it's a sure sign he wants the whole pie.\n***********\n\n```\n\n• Prev by Date: Re: A simple swap function\n• Next by Date: Comparison of overlapping numbers\n• Previous by thread: Simplifying vector expressions\n• Next by thread: Re: Simplifying vector expressions" ]	[ null, "http://forums.wolfram.com/mathgroup/images/head_mathgroup.gif", null, "http://forums.wolfram.com/mathgroup/images/head_archive.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/1.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/9.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/9.gif", null, "http://forums.wolfram.com/mathgroup/images/numbers/5.gif", null, "http://forums.wolfram.com/mathgroup/images/search_archive.gif", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8259493,"math_prob":0.8934587,"size":2755,"snap":"2020-34-2020-40","text_gpt3_token_len":809,"char_repetition_ratio":0.090512544,"word_repetition_ratio":0.033333335,"special_character_ratio":0.338657,"punctuation_ratio":0.19392918,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9801218,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14],"im_url_duplicate_count":[null,null,null,null,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-08-04T23:48:05Z\",\"WARC-Record-ID\":\"<urn:uuid:258cef30-6551-42d8-a033-1b8ad37ca2bf>\",\"Content-Length\":\"46225\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:8cba7f9f-ce1e-4a85-9378-4aaa60ae8fb0>\",\"WARC-Concurrent-To\":\"<urn:uuid:eab256b6-b21c-4b18-90b7-5f556d2e248c>\",\"WARC-IP-Address\":\"140.177.205.73\",\"WARC-Target-URI\":\"http://forums.wolfram.com/mathgroup/archive/1995/Oct/msg00376.html\",\"WARC-Payload-Digest\":\"sha1:67PZC3J46AVJV2STYWZW4ZQVXRYHQU2T\",\"WARC-Block-Digest\":\"sha1:CDYTCEYTC3WWBYXMYKYH4PPGN7E4WUYM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-34/CC-MAIN-2020-34_segments_1596439735885.72_warc_CC-MAIN-20200804220455-20200805010455-00438.warc.gz\"}"}
http://bnyy.milanoporteeserramenti.it/logic-gates-questions-and-answers-pdf.html	[ "# Logic Gates Questions And Answers Pdf\n\n9 BOOLEAN ANALYSIS OF LOGIC CIRCUITS 80 2. HTML MCQ Question with Answer. For the electronics circuits and signals a logic 1 will represent closed switch, a high voltage, or an “on” lamp, and a logic 0 will represent an open switch, low voltage, or an “off” lamp. Research the part numbers and datasheets of the following logic gate integrated. It has syntax and semantics. Question 12 Suppose you needed a two-input AND gate, but happened to have an unused 3-input AND gate in one of the integrated circuits (\"chips\") already in the system you were building. Philosopher(a) Scholar(a) • x, King(x) Greedy (x) Evil (x) –Variables range over individuals (domain of discourse) •Second order logic. common logic gate. I am electrical engineering, so I was asked questions about SoC architecture, how microprocessors work, semiconductor physics, about past experiences, and the brain teasers included questions to see how you approached tricky questions. If you don’t know the answer to a question, make a guess or come back to it later. The sheet presents a series of activities which focus on the AND, OR, and NOT gates. These are useful, even necessary, in some digital circuit applications. Build a register file out of (already built) registers, mutiplexers, decoders, and logic gates. Susan receives $40,000. Write two characteristics of combinational circuits. In this section you can learn and practice logic puzzles, number puzzles, word puzzles, math puzzles etc. We can classify these Logic gates into the following three categories. Please be ready to send us a copy of CAASPP, i-ready, CTY, Math Kangaroo, AMC8 scores upon request. GATE CSE, ECE, EE, ME, Civil PAst years Question with solutions, GATE Computer Science, GATE Electrical Engineering, GATE Electronics and Communication Engineering, GATE Mechanical Engineering, GATE Civil Engineering All Previous Years GATE Exams Questions and Answers with Well Organized GATE Subject/Topic and Chapterwise. Logic Gates Questions And Answers Logic Gates Questions and Answers. The output of the 74 series GATE of TTL gates is taken from a BJT in a. That is, find the four outputs for all possible input conditions. This resource includes a PowerPoint presentation that teaches about logic gates, with a worksheet designed to be used on the computers, and an online link to practically find out about logic gates. Tech, BE, ME students an interview for various positions like. Boolean Algebra specifies the relationship between Boolean variables which is used to design combinational logic circuits using Logic Gates. 13) [EASY] MOSFET – Multiple Choice Questions –Download here. The two characteristics of combinational circuits are: Cus Search powered by Engineering Interview Questions. Use a truth table to show that X = (X AND Y) OR (X AND NOT Y). These are useful, even necessary, in some digital circuit applications. Solved examples with detailed answer description, explanation are given and it would be easy to understand. Mention any one web browser. The person who has the dice and knows the game, rolls five dice and remarks almost instantly on the answer. You can see the explanation for the questions of sensation and a good user interface. Using Boolean algebra, rewrite the simplified SOP expression to use only 2-input AND and OR gates, and NOT gates (NOT gates are also known as inverters) 4. They can also be told the answer for every throw of the dice that are used in the game. For the electronics circuits and signals a logic 1 will represent closed switch, a high voltage, or an “on” lamp, and a logic 0 will represent an open switch, low voltage, or an “off” lamp. The full forms of the abbreviations TTL and CMOS in reference to logic families are a. Section 1: Logic Gates (Introduction) 3 1. Logic gates using the programmable logic controller (PLC) is the basic thing you must learn if you want to enhance your Electrical and Electronics skills. The answer is C. Digital Logic – Application and Design – John M. Answer: Logic Regression can be defined as: This is a statistical method of examining a dataset having one or more variables that are independent defining an outcome. Get Free Logic Gates Questions And Answers Objective Typethe book as a Kindle file (. Analysis and Synthesis of Synchronous Sequential Circuits. Exam Category: GATE Overflow Tests should give @arjun sir please make a separate tab for test @hareesh22 ya I have received the order Math had contained more than 13 marks few years @Mellophi Yeah both books are great and if. First four problems are basic in nature. This document is intended for guidance only. It has one output and one or more inputs. 2 NOR Gate as a Universal Gate 77 2. A NAND gate is in fact a NOT-AND gate. (iv) Underline the names of the two logic gates that should be used inside the control box. 9 BOOLEAN ANALYSIS OF LOGIC CIRCUITS 80 2. The combinational circuit has no memory element. Learn basic electronics Multiple Choice Questions and Answers (MCQs), \"uses of logic gates\" quiz questions and answers for online degree programs. Please note that all the relevant information about the GATE 2021 is available only at the official web site https://gate. The sheet presents a series of activities which focus on the AND, OR, and NOT gates. Making statements based on opinion; back them up with references or personal experience. It also includes a double sided homework sheet. • This is called Boolean logic • In a circuit schematic each logic gate is represented by a different picture, like the ones shown below. Questions (80) Explanation was long and question became somewhat rhetorical. This tests the basics of mathematical logic, and in particular logical connectives (implication, negation, etc. It must be a NAND gate 43) The value of total collector current in a CB circuit is Ic = αααα IE. All cats like fish. If output of a logic circuit should remain constant at High Voltage level (1) but it goes low then we say static1 hazard exist. Logic Gates ,Semiconductor Electronics - Get topics notes, Online test, Video lectures, Doubts and Solutions for CBSE Class 12-science on TopperLearning. Activity Draw each of the diagrams from earlier using the correct symbols Design your own logic gates and create questions for others in the class to attempt. , experimental or observational) science like physics, biology, or psychology. For the same clock situation, if the R input is at high level (logic 1) and S input is at low level (logic 0), then the SR flip – flop is said to be in RESET state and the output of the SR flip – flop is RESET to 0. We are not yet accepting applications for the Fall. Logic Gates (Introduction) The package Truth Tables and Boolean Algebra set out the basic principles of logic. 15) MOSFET & IC Basics – GATE Problems (Part – II) – Download MCQs from here. Explore the latest questions and answers in Mathematical Logic, and find Mathematical Logic experts. Answer the easy questions first, give each question a full effort, come back later to tackle more challenging questions, and when just a few minutes are remaining fill in any unanswered questions. Download Free NNPC Past Questions and Answers 2019/2020 PDF and how to pass NNPC Aptitude Test 2019. We can classify these Logic gates into the following three categories. Making statements based on opinion; back them up with references or personal experience. (a) Complete the truth tables for the two gates. There are 7 types of logic gates which are: AND gate, OR gate, XOR gate, NAND gate, NOR gate, XNOR gate and NOT gate. Name : _____ Summer 2012 P11 (Q12) (a) (i) Complete the truth table for the following logic circuit, which is made up of NAND gates: Computer-2210 (Bit Pattern) (8) abdul. What is a logic. What logic question can save the gringo's life? You probably remember the answer from the very first problem on this page, don't you :-). com GCSE PHYSICS ELECTRONIC CONTROL High Demand Questions QUESTIONSHEET 8 It is compulsory to wear seat belts when a car is moving. ” One way to view the logical conditional is to think of an obligation or contract. The logic diagrams for the full adder implemented in sum-of-products form are the following: It can also be implemented using two half adders and one OR gate (using XOR gates). Inference rules describe correct. We will suggest you adopt this strategy: Collect the meticulous study material from our website; Get acquainted with the quality content. ' Bill gates laughs again and says, 'OK, how about my daughter's name \"Jennifer Katharine gates\". The binary world of 1s and 0s alone doesn’t allow us to re-land rockets in the middle of the ocean, or deliver packages within minutes through the use of drones, or map the known physical universe and all of its wonders. What is the technical reason for the reduction to. pdf file size 205KB 15 Overview of Sensors & Actuators w/quest. Can you build a device that, logically, behaves like an or gate from only and and not gates? If so, do so (just for the case where and and or gates have only two inputs). This page is all about reasoning, reasoning question and answer, reasoning question and answer in Hindi, reasoning question and answer in English, reasoning question and answer in pdf, reasoning question and answer pdf download, reasoning online test, reasoning question and answer with solution, reasoning question and. Draw the entire ALU, adding the fewest possible gates. Each gate implements a Boolean operation, and is depicted schematically by a shape indicating the operation. Questions (80) Explanation was long and question became somewhat rhetorical. We hope the given Karnataka 2nd PUC Class 12 Computer Science Question Bank with Answers Solutions, Notes, Guide Pdf Free Download of 2nd PUC Computer Science Textbook Questions and Answers, Model Question Papers with Answers, Study Material 2020-2021 in English Medium and Kannada Medium will help you. Design different types of logic gates using CMOS inverter and analyze their transfer characteristics. Learn basic electronics Multiple Choice Questions and Answers (MCQs), \"uses of logic gates\" quiz questions and answers for online degree programs. Introduction. Define sorting. See full list on examsdaily. Table 4-1 Basic rules of Boolean algebra. So that is where I needed to start: create relay logic gates of my design. Use the available op code 11 to select the XOR operation. Question 5 Boolean algebra is a strange sort of math. logic design aim: to design digital systems using the rules of boolean algebra (floyd 4-5/4-6). It's okay to guess Since your score will not be negatively affected by getting questions wrong, be sure to answer, or guess, on each question. Make sure you understand how many questions you will have to answer and how long you have to complete the test. Many logic problems try to distract you or lead you down the wrong path. MathJax reference. One half adder and three full adder. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Introduction. Sir please prepare different numbers of objective questions on diploma base competive exams only. Given here are 100 Question of Computer Basic Skills Info Practice Quiz Online Test with Answers of important questions for various General Knowledge exams and interviews preparation. 2 Remember to take your time here to fully understand the statement. pdf file size 287KB. for programmable logic controllers, many worked examples, multi-choice questions and problems are included in the book with answers to all multi-choice questions and problems given at the end of the book. we no need btech level objective based on gate level is depth in concept. Roth, Thomson Publications, 5th Edition, 2004. A are about Top 50 Networking Questions & Internet Questions to help you increase your GK for better preparations. Write any one application of computer network. (8pt) For the following circuit, complete the truth table. Explain what are the universal logic gates? Universal gate is a gate that can perform all the basic logical operations such as NAND and NOR gates. In this electrical worksheet, students answer a series of 27 open-ended questions about logic circuits. 8) What is a Logic gate? The basic gates that make up the digital system are called a logic gate. Answer: There are 3 ways to connect assertion to RTL: inline, Instantiation and Virtual Instantiation (bind). Research the part numbers and datasheets of the following logic gate integrated. Questions and answers - MCQ with explanation on Computer Science subjects like System Architecture, Introduction to Management, Math For Computer Science, DBMS, C Programming, System Analysis and Design, Data Structure and Algorithm Analysis, OOP and Java, Client Server Application Development, Data Communication and Computer Networks, OS, MIS, Software Engineering, AI, Web Technology and many. MathJax reference. Aptitude Set-2: Questions Solutions. We are not yet accepting applications for the Fall. What logic question can save the gringo's life? You probably remember the answer from the very first problem on this page, don't you :-). Programmable Logic Controller (PLC) Questions and Answers – 21. Answer: There are 3 ways to connect assertion to RTL: inline, Instantiation and Virtual Instantiation (bind). All the switches only pass signals from source to drain, incorrect wiring of the devices will result in high impedance outputs. Digital Electronics Interview Questions & Answer 2020 PDF. The simplest logic gate technically isn’t really a gate at all though some refer to it as a NOT gate. More than one answer bubbled against a question will be treated as a wrong answer. old answer completely. There was a mixture of technical Questions, behavioral questions, and brain teasers. In this electrical worksheet, students answer a series of 27 open-ended questions about logic circuits. Make sure you understand how many questions you will have to answer and how long you have to complete the test. Digital Electronics Interview Questions (Basic / Advanced) Ques11: What is a Logic gate? Ans: The basic gates that make up the digital system are called a logic gate. 111 + 111 = 3. Research the part numbers and datasheets of the following logic gate integrated circuits: Triple 3-input NOR gate Dual 4-input AND gate Single 8-input NAND gate. 1 to 9 are based on the logic gates like AND, OR, NOT, NAND & NOR etc. The term was coined by education researcher Mary Budd Rowe in her journal article, \"Wait-time and Rewards as Instructional Variables, Their Influence in Language, Logic, and Fate Control. Answer any five questions. Preparing for different types of interview questions ahead of time will help you stay calm and answer each question with precision. Rather, logic is a non-empirical science like mathematics. \" She noted that on average, teachers paused only one-and-a-half seconds after asking a question; some waited only a tenth of a second. Download PDF Free download in PDF Software Engineering Objective Type Questions and Answers or Software Engineering mcq from chapter Software Engineeing Fundamentals. All cheat sheets, round-ups, quick reference cards, quick reference guides and quick reference sheets in one page. Digital logic multiple choice questions answers can also be used by any candidate who wants to gain credits in digital logic in BS or MS computer science, Digital electronics multiple choice questions with answers can also be used by student who is pursuing degree/diploma in information technology like BSc or MSc in Information Technology or. Programmable Logic Controller (PLC) Questions and Answers – 23. Truth Tables Another way of working out the output from a logic diagram without having to construct the circuit is by using a truth table A truth table create a map of all of the. Please fill in your answer on the answer sheet provided. Let's start with an easy one - answer this. It has appeared in the volume The Examined Life: Readings from Western Philosophy from Plato to Kant, edited by Stanley Rosen, published in 2000 by Random House. Draw a logic. Click here for Webct. UPSC IAS Prelims Answer Key 2019: Here is the answer key of the UPSC Preliminary Exam 2019 GS Paper 1 with explanations. What is a record? 7. Analysis and Synthesis of Synchronous Sequential Circuits. These describe the only two states that exist in digital logic systems and will be used to represent the in and out conditions of logic gates. Electric Circuits GATE (Graduate Aptitude Test in Engineering) Entrance exams EE Electrical Engineering Electric Circuits GATE Exam EE Electrical Engineering - Objective type Online Test Questions and Answers with Solution, Explanation, Solved Problems. Sir please prepare different numbers of objective questions on diploma base competive exams only. 41) When an input electrical signal A = 10100 is applied to a NOT gate, its output signal is 01011 42) In a certain 2 – input logic gate, when A = 0, B= 0 then C = 1 and when A = 0, B = 1 then again C = 1. The Critical Thinking Company publishes PreK-12+ books and software to develop critical thinking in core subject areas. This is the seventh lesson in the 'How a Computer Works' topic. we no need btech level objective based on gate level is depth in concept. In this section you can learn and practice logic puzzles, number puzzles, word puzzles, math puzzles etc. (ii) What single logic gate has the same function as the above logic circuit?. What is a logic. More than one answer bubbled against a question will be treated as a wrong answer. The Karnaugh Map Provides a method for simplifying Boolean expressions It will produce the simplest SOP and POS expressions Works best for less than 6 variables Similar to a truth table => it maps all possibilities A Karnaugh map is an array of cells arranged in a special manner The number of cells is 2n where n = number of variables A 3-Variable Karnaugh Map:. Question 3. – If too much static electricity builds up on the gate, then the MOSFET may be damaged. Tech, BE, ME students an interview for various positions like. Preparing for different types of interview questions ahead of time will help you stay calm and answer each question with precision. Difference between combinational and sequential circuits, A combinational logic circuit is a digital logic circuit wherein the output is capable of being determined with the help of logic functions related to the current state inputs. pdf file size 42KB 14 Computers / Logic Gates with questions. 90 to get the final answer. Boolean logic represents all values as TRUE (1) or FALSE (0). Mention any one web browser. examples with detailed response description, explanation is given and it would be easy to understand. The circuit that can operate on many binary inputs to perform a particular logic function is called an electronic circuit. Go through our link having the previous year solved papers and get benefited. Also, in saying that logic is the science of reasoning, we do not mean. General science 100 Important Questions and Answers. Combinational Circuit Design with Programmable Logic Devices. i) Construct a BCD to Excess -3 code converter using full adders (8) ii) Design an 8421 to gray code converter. You can choose to pay for your order via PayPal, debit/credit cards or net banking. Here you will find a list of common important questions on html programming in MCQ quiz style with answer for competitive exams and interviews. Download Free NNPC Past Questions and Answers 2019/2020 PDF and how to pass NNPC Aptitude Test 2019. These Computer Organisation objective questions answers for online exam preparations include Flip-flop, logic gate etc. What do you think if we exercise the mind for a while with logic riddles? Test your common sense with witty logical questions with answers. with logic gates. Problems 3 & 4 are based on word statement. 8 ALTERNATE LOGIC-GATE REPRESENTATIONS 79 2. Digital logic design study guide with questions and answers about algorithmic state machine, asynchronous sequential logic, binary systems, Boolean algebra and logic gates, combinational logic, digital integrated circuits, MSI and PID components, registers counters and memory units, Boolean functions, standard graphic symbols, synchronous. Logic gates have one or two inputs that can be turned on or off, the output from the gate will vary depending on the type of logic gate. Following the completion of the payment, you will get an email or text that will confirm your order. Make sure you understand how many questions you will have to answer and how long you have to complete the test. What type of logic gate does this symbol represent. Source: Online GATE Exam Mock Test provided by IITM Aptitude Set-1: Questions Solutions. common logic gate. 1 to 9 are based on the logic gates like AND, OR, NOT, NAND & NOR etc. Programmable Logic Controller (PLC) Questions and Answers – 21. (8pt) For the following circuit, complete the truth table. Explain what is the specialty of NAND and NOR gates?. Question 5 Boolean algebra is a strange sort of math. You can also find ManyBooks' free eBooks from the genres page or recommended category. For example:. The output of the 74 series GATE of TTL gates is taken from a BJT in a. Proofs in predicate logic can be carried out in a manner similar to proofs in propositional logic (Sections 14. Please do not believe in the phishing/fake emails which are being circulated saying that IIT Madras faculty team have specially designed tests for the GATE aspirants. Questions 71 through 73 is one set of common data questions, questions 74 and 75 is another pair of common data questions. Use compile option for synthesis. Some manufacturers put an alarm into the car to remind people to fasten their seat belts. The questions asked in this NET practice paper are from various previous year papers. CE/CZ1105 Digital Logic Tutorial 2 Logic gates and Boolean algebra 1. answers to these questions and the reasoning behind the answers. examples with detailed response description, explanation is given and it would be easy to understand. Power Systems GATE (Graduate Aptitude Test in Engineering) Entrance exams EE Electrical Engineering Power Systems GATE Exam EE Electrical Engineering - Objective type Online Test Questions and Answers with Solution, Explanation, Solved Problems. ) and quantifiers. Questions and answers - MCQ with explanation on Computer Science subjects like System Architecture, Introduction to Management, Math For Computer Science, DBMS, C Programming, System Analysis and Design, Data Structure and Algorithm Analysis, OOP and Java, Client Server Application Development, Data Communication and Computer Networks, OS, MIS, Software Engineering, AI, Web Technology and many. Logical reasoning generally does not require verbal or numerical reasoning although variations exist that do. Q: Why is the VGS rating of Trench 6 automotive logic level MOSFETs limited to 10 V and can it be increased beyond 10 V? A: The VGS rating of 10 V given to Trench 6 logic level MOSFETs is driven by our <1 ppm failure. Earlier students use to consider few options after cracking gate exam such as scientist or a professor but nowadays the scenario has changed and scope has widen up. Digital logic is the application of the Boolean algebra of 0 and 1 to electronic hardware consisting of logic gates connected to form a circuit diagram. 10x01=10 1. Problems 3 & 4 are based on word statement. implement the logic diagram using electronic circuitry. write the truth table 3. Logic Gates Questions and Answers. Also, explain how an understanding of this can be helpful in troubleshooting faulted logic gates. (a) Complete the truth tables for the two gates. 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Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. These frequently asked sample questions on HTML are given with correct choice of answer that you can check instantly. In order to understand the theoretical basics you need to know five basic definitions. Logic Gates Questions And Answers Recognizing the artifice ways to acquire this book logic gates questions and answers is additionally useful. Predicate logic formulas without quantifiers can be verified using derivation. Jurisich, S. Modular Sequential Logic. Set theory (12 questions). 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In the practice of electronics, a logic gate is a device which carries out a logical operation on a certain number of binary inputs and produces just one binary output. A logic gate. Question 2 What does ‘Boolean logic’ mean? Boolean logic describes the states of the inputs and outputs to the logic gates. Programmable Logic Controller (PLC) Questions and Answers – 20. These puzzles are designed to test with Numerical ability, Logical thinking, Maths problem solving with sp. The Following Section consists Multiple Choice Questions on Logic Gates. 0201 Fall 2009 ANSWERS Digital Logic Read: Sections 3. logic design aim: to design digital systems using the rules of boolean algebra (floyd 4-5/4-6). In simple terms, logic gates are the electronic circuits in a digital system. and transistors. Then, most of them are unsolved. pdf file size 42KB 14 Computers / Logic Gates with questions. Deﬁnition 4. 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It's okay to guess Since your score will not be negatively affected by getting questions wrong, be sure to answer, or guess, on each question. View Tut 2 questions. The OLSAT is a test for children that measures abstract thinking and reasoning ability. Explain what is a combinational circuit? In a combinational circuit, the output depends upon present input(s) only i. is a series of transistors connected together to give one or more. Aptitude Practice Paper designed by GATECounsellor TEAM Aptitude Set-1: Questions Solutions. Answer the easy questions first, give each question a full effort, come back later to tackle more challenging questions, and when just a few minutes are remaining fill in any unanswered questions. - Use the truth tables method to determine whether the formula ’: p^:q!p^q is a logical consequence of the formula : :p. 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Digital logic design study guide with questions and answers about algorithmic state machine, asynchronous sequential logic, binary systems, Boolean algebra and logic gates, combinational logic, digital integrated circuits, MSI and PID components, registers counters and memory units, Boolean functions, standard graphic symbols, synchronous. Operational Amplifiers Questions & Answers Electronics Q & A 1. Aptitude Set-2: Questions Solutions. CE/CZ1105 Digital Logic Tutorial 2 Logic gates and Boolean algebra 1. The relationship between the input and the output is based on a certain logic. In Section 14. Explain what is a combinational circuit? In a combinational circuit, the output depends upon present input(s) only i. The basic digital electronic circuit that has one or more inputs and single output is known as Logic gate. Every terminal in a logic gate will always be in one of the two binary states (0) or (1). What is a logic gate? 3. Answer: There are 3 ways to connect assertion to RTL: inline, Instantiation and Virtual Instantiation (bind). A + 0 = A A variable ORed with 0 is always equal to the variable. Learning Objectives 1) To be able to describe how data is stored within a computer in binary form. We'll tackle addition of numbers (which requires many gates) in a later quiz, but for now let's consider a simpler problem: We would like a machine that takes three inputs A, B, and C and returns 1 if and only if all the inputs are 1. If so, this section of Logic Puzzles and Riddles has been framed by us just for the people like you. Fundamentals of Logic Design – Charles H. (8pt) For the following circuit, complete the truth table. plc interview questions for experienced, omron plc interview questions, plc programming questions, multiple choice questions on plc pdf, plc interview questions, plc ladder logic interview questions, plc operator interview questions, PLC Ladder Logic Questions and Answers. Use MathJax to format equations. The SR flip – flops can be designed by using logic gates like NOR gates and NAND gates. Yarbrough, Thomson. This document is intended for guidance only. The Karnaugh Map Provides a method for simplifying Boolean expressions It will produce the simplest SOP and POS expressions Works best for less than 6 variables Similar to a truth table => it maps all possibilities A Karnaugh map is an array of cells arranged in a special manner The number of cells is 2n where n = number of variables A 3-Variable Karnaugh Map:. A strategy is a way of organizing available resources to achieve. Although there are extra transistors given that it uses clocks. Puzzles Questions and Answers with explanation for placement, interview preparations. You can choose to pay for your order via PayPal, debit/credit cards or net banking. ﬁle 03834 Question 10 Many types of logic gate circuits are built with more than two inputs. 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Three Logic Levels are used and they are High, Low, High impedance state. 3-6 Implementation of the given Boolean function using logic gates in both sop and pos forms. An inverter is a logic gate whose output is the inverse or complement of its input. Examples are available on the other pages with step-by-step explanations if you need any clarification. Combines verbal and logical elements with good examples and problems to work through. Draw a logic circuit for A + BC + D. Logic gates using the programmable logic controller (PLC) is the basic thing you must learn if you want to enhance your Electrical and Electronics skills. Based on this, logic gates are named as AND gate, OR gate, NOT gate etc. Boolean algebra and logic gates quiz questions and answers pdf, complement of function, algebric manipulation, dld standard forms, boolean functions in dld test for cisco certifications. 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Along with the previous year question papers for JEE Main, answers keys are also available here. The person who has the dice and knows the game, rolls five dice and remarks almost instantly on the answer. Susan receives$40,000. Multiply 30 by 0. For example: // A synchronous D Flip-Flop moduledff (input logic [7:0] D, input RST_, input CLK, output logic [7:0] Q); always. It has appeared in the volume The Examined Life: Readings from Western Philosophy from Plato to Kant, edited by Stanley Rosen, published in 2000 by Random House. As the page includes Logical Problems Reasoning Questions, you need to answer the questions in the meanwhile; you can know the tips and tricks in solving those issues. Whereas a sequential logic circuit usage the logic function related to current state inputs and previous state inputs. Get Free Logic Gates Questions And Answers Objective Typethe book as a Kindle file (. A + 0 = A A variable ORed with 0 is always equal to the variable. Fully solved Multiple choice questions and answers for competitive examinations. 13) [EASY] MOSFET – Multiple Choice Questions –Download here. pdf from CE CZ1105 at Nanyang Technological University. IndianStudyHub offers many fully Coding and Decoding Questions & Answers : Logical Reasoning pdf free download questions and answers with explanations. acquire the logic gates questions and answers belong to that we provide here and check out the link. Logical Problems deals with the consistent ability to solve the problem. The beginnings of computing and logic gates starts with telephone companies in the early-mid 20th century, and telephone companies did this with relays. \"Electrostatic MCQ\" PDF covers quiz questions about electrostatics, electric field lines, electric flux, electric potential, capacitor, Coulomb’s law, Gauss law, electric and gravitational forces. All the switches only pass signals from source to drain, incorrect wiring of the devices will result in high impedance outputs. These are useful, even necessary, in some digital circuit applications. Browse through all study tools. 3 in the textbook. Build a register file out of (already built) registers, mutiplexers, decoders, and logic gates. This course is an introduction to Logic from a computational perspective. In addition to reading the questions and answers on my site, I would suggest you to check the following, on amazon, as well:. Consider each word in the question for hidden easy answers. A logic system has 5 inputs. Take the Quiz and improve your overall Engineering. With hundreds of chapter-wise questions & answers on Basic Electronics, this is the most comprehensive question bank on the entire internet. Research the part numbers and datasheets of the following logic gate integrated. Programmable Logic Controller (PLC) Questions and Answers – 23. Boolean Algebra is the mathematical foundation of digital circuits. Since we are focusing on only one gate and its expression, it is easy. Modular Sequential Logic. A logic gate. Please be sure to answer all questions as accurately and completely as possible. Most mathematicians will understand the term logic as in mathematical logic. Transmission gates are bi-directional and can be resistive or non-resistive. When you are told to begin you will have 30 minutes to do as many questions as you can. Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question. Technical Articles Toyota Series - Electrical (623 Training Course) Elec01 Essential Electrical Concepts. verification of the truth tables of logic gates using TTL ICS. pdf file size 205KB 15 Overview of Sensors & Actuators w/quest. 13) [EASY] MOSFET – Multiple Choice Questions –Download here. This resource includes a PowerPoint presentation that teaches about logic gates, with a worksheet designed to be used on the computers, and an online link to practically find out about logic gates. Practice test for UGC NET Computer Science Paper. These calculations are shown in detail later in this chapter. Making statements based on opinion; back them up with references or personal experience. In the case of the OR gate, if there is no potential (i. For the first step, we write the logic expressions of individual gates. Digital logic is the application of the Boolean algebra of 0 and 1 to electronic hardware consisting of logic gates connected to form a circuit diagram. 6 LOGIC GATES 69 2. The combinational circuit has no memory element. The output of the 74 series GATE of TTL gates is taken from a BJT in a. Don't follow the first train of thought that enters your head; look at each word and see if there's a simple answer that's easy to miss. The simplest logic gate technically isn’t really a gate at all though some refer to it as a NOT gate. What is a record? 7. Logic Gates Questions And Answers Recognizing the artifice ways to acquire this book logic gates questions and answers is additionally useful. 100+ Logical Reasoning Questions & Answers PDF Download Now Part 1. Download link is provided below to ensure for the Students to download the Regulation 2017 Anna University EE8351 Digital Logic Circuits Lecture Notes, Syllabus, Part-A 2 marks with answers & Part-B 16 marks Questions with answers, Question Bank with answers, All the materials are listed below for the students to make use of it and score Good (maximum) marks with our study materials. 2 7 -8 Verification of state tables of RS, JK, T and D flip-flops using 3 NAND & nor gates. Boolean logic represents all values as TRUE (1) or FALSE (0). Introduction. Logic gates are the building blocks of a digital circuit. Logic gates quiz questions and answers pdf: \"Branch of mathematics which deals with relationships of logic variables is\" with answers for online high school courses. What type of logic gate does this symbol represent. 1 to 9 are based on the logic gates like AND, OR, NOT, NAND & NOR etc. With hundreds of chapter-wise questions & answers on Basic Electronics, this is the most comprehensive question bank on the entire internet. Digital logic design study guide with questions and answers about algorithmic state machine, asynchronous sequential logic, binary systems, Boolean algebra and logic gates, combinational logic, digital integrated circuits, MSI and PID components, registers counters and memory units, Boolean functions, standard graphic symbols, synchronous. Example Question 1. Logic Gates 4 OO Software Design and Construction 2-input Logic Gate Hierarchy It is sensible to view each of the 2-input logic gates as a specialized sub-type of a generic logic gate (a base type) which has 2 input wires and transmits its output to a single output wire. Math Logic Questions with Answers: Simple, Fun and Tricky Math Questions for Kids and Adults. What do you think if we exercise the mind for a while with logic riddles? Test your common sense with witty logical questions with answers. and dc circuits, electronic devices (including thyristors), a knowledge of basic logic gates, flip flops, and Boolean algebra, and college algebra and trigonometry. The answer is C. Making statements based on opinion; back them up with references or personal experience. Because XOR is “OR and NOT AND” we can just use the output of the existing AND and OR gates in the ALU. Although these circuits may be. It can be obtained by connecting a NOT gate in the output of an AND gate. For example: // A synchronous D Flip-Flop moduledff (input logic [7:0] D, input RST_, input CLK, output logic [7:0] Q); always. You can also find ManyBooks' free eBooks from the genres page or recommended category. Electronics Questions and Answers (20) Filter Circuits (9) Founder Speaks (3) GATE Questions and Answers (1) Integrated Circuits (7) Interview (8) JK Flip Flop (3) Logic circuits (1) Logic Gates and Boolean Algebra (10) Long Questions and Answers (3) MCQ (3) Multiple Choice Questions and Answers (57) Number System (1) Operational Amplifier (3. This page is all about reasoning, reasoning question and answer, reasoning question and answer in Hindi, reasoning question and answer in English, reasoning question and answer in pdf, reasoning question and answer pdf download, reasoning online test, reasoning question and answer with solution, reasoning question and. This quiz tests the very basics of logic gates. Electronics Questions and Answers (20) Filter Circuits (9) Founder Speaks (3) GATE Questions and Answers (1) Integrated Circuits (7) Interview (8) JK Flip Flop (3) Logic circuits (1) Logic Gates and Boolean Algebra (10) Long Questions and Answers (3) MCQ (3) Multiple Choice Questions and Answers (57) Number System (1) Operational Amplifier (3. The Hardest Logic Puzzle Ever is a logic puzzle so called by American philosopher and logician George Boolos and published in The Harvard Review of Philosophy in 1996. Hence, Socrates is mortal. Once you submit the order form, you will receive a quote for your assignment. Logic gates are the building blocks of a digital circuit. Logic Gates Questions And Answers Logic Gates Questions and Answers. Has applications in electronics (logic gates) programming (combining instructions) as well as philosophy - what a mixture!. Printable Logic Grid Puzzles. It is intended for the general reader. Logic Gates Worksheet – Truth Tables Label these basic logic gates and fill in their truth tables. For the electronics circuits and signals a logic 1 will represent closed switch, a high voltage, or an “on” lamp, and a logic 0 will represent an open switch, low voltage, or an “off” lamp. In order to understand the theoretical basics you need to know five basic definitions. Based on this, logic gates are named as AND gate, OR gate, NOT gate etc. is a series of transistors connected together to give one or more. Output signal appears only for certain combinations of input signals. UPSC IAS Prelims Answer Key 2019: Here is the answer key of the UPSC Preliminary Exam 2019 GS Paper 1 with explanations. They were produced by question setters, primarily for the benefit of the examiners. It's okay to guess Since your score will not be negatively affected by getting questions wrong, be sure to answer, or guess, on each question. Take the Quiz and improve your overall Engineering. 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Rules for Boolean addition: 0+0 = 0 0+1 = 1 1+0 = 1 1+1 = 1 ﬁle 01298 Question 6 Surveying the rules for Boolean multiplication, the 0 and 1 values seem to resemble the truth table of a very common. Simplification of Switching Functions. ﬁle 03834 Question 10 Many types of logic gate circuits are built with more than two inputs. CIMA P1 Past Paper Questions and Answers Below are a number of past paper questions and answers from the 2010 Syllabus that are still relevant to the CIMA P1 exam. Try These Out: https://bit. Logic gates are used to carry out logical operations on single or multiple binary inputs and give one binary output. 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The solution notes for the most recent two year’s worth of examinations are held back by the department and only made available to supervisors and other. Changes from third edition The fourth edition is a complete restructuring and updating of the third. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. The files are grouped by difficulty (very easy, easy and medium) and are a great activity for all ages. The logic diagrams for the full adder implemented in sum-of-products form are the following: It can also be implemented using two half adders and one OR gate (using XOR gates). In order to make a 4-bit parallel adder minimum circuitry required is_____? Ans. Question 12 Suppose you needed a two-input AND gate, but happened to have an unused 3-input AND gate in one of the integrated circuits (\"chips\") already in the system you were building. Our online logic trivia quizzes can be adapted to suit your requirements for taking some of the top logic quizzes. is part of a circuit used to turn on the parking lights of a car automatically when it is dark Explain your answer. If output of a logic circuit should remain constant at High Voltage level (1) but it goes low then we say static1 hazard exist." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9004369,"math_prob":0.7256657,"size":58693,"snap":"2020-45-2020-50","text_gpt3_token_len":12600,"char_repetition_ratio":0.16693076,"word_repetition_ratio":0.45560965,"special_character_ratio":0.20985466,"punctuation_ratio":0.110814296,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.96974206,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-10-27T00:38:49Z\",\"WARC-Record-ID\":\"<urn:uuid:e530d168-36a7-4cc6-9c80-7cc1a33f3114>\",\"Content-Length\":\"69403\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:e73a52c9-8b1e-40df-8f58-f3819c2c75b1>\",\"WARC-Concurrent-To\":\"<urn:uuid:b3bc4e22-d33e-45cc-add0-7683a604e61e>\",\"WARC-IP-Address\":\"104.18.38.199\",\"WARC-Target-URI\":\"http://bnyy.milanoporteeserramenti.it/logic-gates-questions-and-answers-pdf.html\",\"WARC-Payload-Digest\":\"sha1:XJI2JVNALLXTR54U7DOULEPWQG6EYDPB\",\"WARC-Block-Digest\":\"sha1:JZJS5S3DQGDHW64QJUBBCWV64ERSWRKQ\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-45/CC-MAIN-2020-45_segments_1603107892710.59_warc_CC-MAIN-20201026234045-20201027024045-00191.warc.gz\"}"}
https://ericasadun.com/2017/05/15/tuple-assignments/	[ "Do you have any good examples of when it would be useful to have a tuple but be doing complicated enough stuff with them?\n\nHere are some examples I grepped out of a local folder, including some from third parties:\n\n```var (x, y) = (7.5, 7.5)\nlet (controlPoint1θ, controlPoint2θ) = (dθ / 3.0, 2.0 * dθ / 3.0)\nvar (_, sceneWidth) = boundingNode.boundingSphere\nlet (vMin, vMax) = label.boundingBox\nlet (duration, _) = cameraController?.performFlyover(toFace: mainActor.rotation) ?? (0, 0)\nstruct Point { var (x, y) : (Double, Double) }```\n\nHope that helps.\n\n# One Comment\n\n• Here’s a useful one:\n``` var a = 1 var b = 2```\n\n``` // swap (a, b) = (b, a) ```\n\n```print(a) // 2 print(b) // 1 ```" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8035147,"math_prob":0.95859456,"size":649,"snap":"2023-40-2023-50","text_gpt3_token_len":212,"char_repetition_ratio":0.093023255,"word_repetition_ratio":0.0,"special_character_ratio":0.33436054,"punctuation_ratio":0.21678321,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99737316,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-10-01T04:39:53Z\",\"WARC-Record-ID\":\"<urn:uuid:792d6e12-a2f8-4c30-ab66-d1c19a2e44cc>\",\"Content-Length\":\"41121\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1edd1263-0f7f-45db-8084-0a5df2c5e847>\",\"WARC-Concurrent-To\":\"<urn:uuid:e205e83e-b5ba-43b8-ac20-fe1d0bdfa9f3>\",\"WARC-IP-Address\":\"34.174.22.52\",\"WARC-Target-URI\":\"https://ericasadun.com/2017/05/15/tuple-assignments/\",\"WARC-Payload-Digest\":\"sha1:6IX2YZ6T2H245BMWZABEGAR7LRIF34T2\",\"WARC-Block-Digest\":\"sha1:6CVFE745LK7YZMIH4ENLGXCZHVTH3GYL\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-40/CC-MAIN-2023-40_segments_1695233510781.66_warc_CC-MAIN-20231001041719-20231001071719-00815.warc.gz\"}"}
https://stat.ethz.ch/R-manual/R-patched/library/base/html/duplicated.html	[ "duplicated {base} R Documentation\n\n## Determine Duplicate Elements\n\n### Description\n\n`duplicated()` determines which elements of a vector or data frame are duplicates of elements with smaller subscripts, and returns a logical vector indicating which elements (rows) are duplicates.\n\n`anyDuplicated(.)` is a “generalized” more efficient shortcut for `any(duplicated(.))`.\n\n### Usage\n\n```duplicated(x, incomparables = FALSE, ...)\n\n## Default S3 method:\nduplicated(x, incomparables = FALSE,\nfromLast = FALSE, nmax = NA, ...)\n\n## S3 method for class 'array'\nduplicated(x, incomparables = FALSE, MARGIN = 1,\nfromLast = FALSE, ...)\n\nanyDuplicated(x, incomparables = FALSE, ...)\n## Default S3 method:\nanyDuplicated(x, incomparables = FALSE,\nfromLast = FALSE, ...)\n## S3 method for class 'array'\nanyDuplicated(x, incomparables = FALSE,\nMARGIN = 1, fromLast = FALSE, ...)\n```\n\n### Arguments\n\n `x` a vector or a data frame or an array or `NULL`. `incomparables` a vector of values that cannot be compared. `FALSE` is a special value, meaning that all values can be compared, and may be the only value accepted for methods other than the default. It will be coerced internally to the same type as `x`. `fromLast` logical indicating if duplication should be considered from the reverse side, i.e., the last (or rightmost) of identical elements would correspond to `duplicated = FALSE`. `nmax` the maximum number of unique items expected (greater than one). `...` arguments for particular methods. `MARGIN` the array margin to be held fixed: see `apply`, and note that `MARGIN = 0` may be useful.\n\n### Details\n\nThese are generic functions with methods for vectors (including lists), data frames and arrays (including matrices).\n\nFor the default methods, and whenever there are equivalent method definitions for `duplicated` and `anyDuplicated`, `anyDuplicated(x, ...)` is a “generalized” shortcut for `any(duplicated(x, ...))`, in the sense that it returns the index `i` of the first duplicated entry `x[i]` if there is one, and `0` otherwise. Their behaviours may be different when at least one of `duplicated` and `anyDuplicated` has a relevant method.\n\n`duplicated(x, fromLast = TRUE)` is equivalent to but faster than `rev(duplicated(rev(x)))`.\n\nThe array method calculates for each element of the sub-array specified by `MARGIN` if the remaining dimensions are identical to those for an earlier (or later, when `fromLast = TRUE`) element (in row-major order). This would most commonly be used to find duplicated rows (the default) or columns (with `MARGIN = 2`). Note that `MARGIN = 0` returns an array of the same dimensionality attributes as `x`.\n\nMissing values (`\"NA\"`) are regarded as equal, numeric and complex ones differing from `NaN`; character strings will be compared in a “common encoding”; for details, see `match` (and `unique`) which use the same concept.\n\nValues in `incomparables` will never be marked as duplicated. This is intended to be used for a fairly small set of values and will not be efficient for a very large set.\n\nExcept for factors, logical and raw vectors the default `nmax = NA` is equivalent to `nmax = length(x)`. Since a hash table of size `8*nmax` bytes is allocated, setting `nmax` suitably can save large amounts of memory. For factors it is automatically set to the smaller of `length(x)` and the number of levels plus one (for `NA`). If `nmax` is set too small there is liable to be an error: `nmax = 1` is silently ignored.\n\nLong vectors are supported for the default method of `duplicated`, but may only be usable if `nmax` is supplied.\n\n### Value\n\n`duplicated()`: For a vector input, a logical vector of the same length as `x`. For a data frame, a logical vector with one element for each row. For a matrix or array, and when `MARGIN = 0`, a logical array with the same dimensions and dimnames.\n\n`anyDuplicated()`: an integer or real vector of length one with value the 1-based index of the first duplicate if any, otherwise `0`.\n\n### Warning\n\nUsing this for lists is potentially slow, especially if the elements are not atomic vectors (see `vector`) or differ only in their attributes. In the worst case it is O(n^2).\n\n### References\n\nBecker, R. A., Chambers, J. M. and Wilks, A. R. (1988) The New S Language. Wadsworth & Brooks/Cole.\n\n`unique`.\n\n### Examples\n\n```x <- c(9:20, 1:5, 3:7, 0:8)\n## extract unique elements\n(xu <- x[!duplicated(x)])\n## similar, same elements but different order:\n(xu2 <- x[!duplicated(x, fromLast = TRUE)])\n\n## xu == unique(x) but unique(x) is more efficient\nstopifnot(identical(xu, unique(x)),\nidentical(xu2, unique(x, fromLast = TRUE)))\n\nduplicated(iris)[140:143]\n\nduplicated(iris3, MARGIN = c(1, 3))\nanyDuplicated(iris) ## 143\n\nanyDuplicated(x)\nanyDuplicated(x, fromLast = TRUE)\n```\n\n[Package base version 3.6.0 Index]" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.6998128,"math_prob":0.9530757,"size":4243,"snap":"2019-43-2019-47","text_gpt3_token_len":1103,"char_repetition_ratio":0.15923567,"word_repetition_ratio":0.042979944,"special_character_ratio":0.25524393,"punctuation_ratio":0.17220172,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9702561,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-11-20T22:28:51Z\",\"WARC-Record-ID\":\"<urn:uuid:5f14716b-f1c0-46a9-afdc-2250b4f525cc>\",\"Content-Length\":\"7084\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:670a259b-a0b6-4d35-a392-0841f5c955e8>\",\"WARC-Concurrent-To\":\"<urn:uuid:ea78dfa0-4421-40f6-962b-9a4d35fa7d2e>\",\"WARC-IP-Address\":\"129.132.119.195\",\"WARC-Target-URI\":\"https://stat.ethz.ch/R-manual/R-patched/library/base/html/duplicated.html\",\"WARC-Payload-Digest\":\"sha1:GJTYTPQDQQDFNG6JOB2VDTHP4NTNQ6DZ\",\"WARC-Block-Digest\":\"sha1:HTZRW46QI3TO4MX5J2WNPLP6E52IUFHM\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-47/CC-MAIN-2019-47_segments_1573496670635.48_warc_CC-MAIN-20191120213017-20191121001017-00092.warc.gz\"}"}
https://www.mdpi.com/1996-1073/5/9/3701/htm	[ "Next Article in Journal\nEffects of a Green Space Layout on the Outdoor Thermal Environment at the Neighborhood Level\nNext Article in Special Issue\nComprehensive Exergy Analysis of Three IGCC Power Plant Configurations with CO2 Capture\nPrevious Article in Journal\nA Revisit to the Hydrogen Desorption/Absorption Behaviors of LiAlH4/LiBH4: Effects of Catalysts\nPrevious Article in Special Issue\nAn Innovative Use of Renewable Ground Heat for Insulation in Low Exergy Building Systems\n\nEnergies 2012, 5(9), 3701-3722; https://doi.org/10.3390/en5093701\n\nArticle\nEnergy and Exergy Analysis and Optimization of Combined Heat and Power Systems. Comparison of Various Systems\nby", null, "Michel Feidt 1 and", null, "Monica Costea 2,\n1\nLaboratoire d’Energétique et de Mécanique Théorique et Appliquée (LEMTA), Lorraine University, UMR CNRS 7563, 2 Avenue de la Forêt de Haye, 54516 Vandoeuvre-Les-Nancy Cedex, France\n2\nDepartment of Engineering Thermodynamics, Engines, Thermal and Refrigerating Equipment, University “Politehnica” of Bucharest, Splaiul Independentei 313, 060042 Bucharest, Romania\n\nAuthor to whom correspondence should be addressed.\nReceived: 14 July 2012; in revised form: 3 September 2012 / Accepted: 19 September 2012 / Published: 24 September 2012\n\n## Abstract\n\n:\nThe paper presents a comparison of various CHP system configurations, such as Vapour Turbine, Gas Turbine, Internal Combustion Engine, External Combustion Engine (Stirling, Ericsson), when different thermodynamic criteria are considered, namely the first law efficiency and exergy efficiency. Thermodynamic optimization of these systems is performed intending to maximize the exergy, when various practical related constraints (imposed mechanical useful energy, imposed heat demand, imposed heat to power ratio) or main physical limitations (limited heat availability, maximum system temperature allowed, thermo-mechanical constraints) are taken into account. A sensitivity analysis to model parameters is given. The results have shown that the various added constraints were useful for the design allowing to precise the influence of the model main parameters on the system design. Future perspective of the work and recommendations are stated.\nKeywords:\nthermodynamics; optimization; combined heat and power systems; exergy efficiency; First Law efficiency; constraints\n\n## 1. Introduction\n\nA cogeneration plant, also called a CHP system (Combined Heat and Power Production), can operate at efficiencies greater than those achieved when heat and power are produced in separate or distinct processes. For example, efficiency values go from 35%–40% for electrical or mechanical production, to 80%–85% for the cogeneration system efficiency . The environmental issue should be also considered as an important cogeneration system advantage with respect to carbon dioxide emissions, which are mainly responsible for the greenhouse effect.\nIn the recent past, due to environmental impact considerations and energy efficient use purposes, a renewal and development of combined heat and power systems was increasing from large to small scale CHP systems, even μCHP, and for industrial or building applications [1,2,3,4,5,6,7,8,9,10,11,12,13,14]. New configurations of CHP systems were studied and among them photovoltaic/thermal (PV/T) configurations [15,16,17] or fuel cell CHP systems [2,4,14] are close to implementation in the near future. Also the fuel disposal issue was considered, mainly by various biomass availabilities [16,18,19], or gasoline and hydrogen . Analysis of the CO2 mitigation costs of large-scale biomass-fired cogeneration technologies with CO2 capture and storage was performed , showing that biomass-fired cogeneration plants based on integrated gasification combined cycle technology (CHP-BIGCC) is very energy and emission efficient and also cost competitive compared with other conversion systems. A new analytical approach based on the current models of the solid oxide fuel cell and gas turbine was elaborated , in which multiple irreversibilities existing in real hybrid systems are taken into account. The general performance characteristics of the hybrid system (irreversible solid oxide fuel cell-gas turbine) were revealed and the optimum criteria of the main performance parameters were determined. Other hybrid systems were considered , such as bi-energy technologies (gas and electricity), as a path to transfer loads from one system to another, so an absolute peak load reduction by 17% at the small scale was found. A novel conceptualisation considering the steam cycle of a combined heat and power generator thermodynamically equivalent to a conventional steam cycle generator plus an additional virtual steam cycle heat pump leads to the conclusion that the performance of CHP will tend to be significantly higher than that of real heat pumps operating at similar temperatures. It also shows that the thermodynamic performance advantages of CHP are consistent with the goal of deep, long-term decarbonisation of industrialised economies.\nBesides the particular look at specific characteristics of CHP systems, various criteria to evaluate their performances are used. Multicriteria evaluations according to weighting methodologies have been proposed recently [23,24]. Then, First and Second Law analyses of gas engines, fuel cells or hybrid solar systems [1,5,6,7,11,14] have shown that the energy-saving effect increases with the system scale because the heat to power ratio of the system decreases , or that both the main energy and exergy loss take place at the parabolic trough collector , and that the polymer exchange membrane fuel cell (PEMFC)-based CHP system, operating at atmospheric pressure and low temperature, is the most efficient system when compared to a solid oxide fuel cell (SOFC) one .\nExergy-based criteria were found to give much better guidance for system improvement [3,4,10,12], as they account better for use of energy resources. Thus, the comparison of gasoline and hydrogen fuelled spark ignition internal combustion engines yielded that the hydrogen fuelled engine had a greater proportion of its chemical exergy converted into mechanical exergy, as well as a greater exergy due to heat transfer and smaller combustion irreversibility associated with hydrogen combustion . When looking into internal combustion engine (ICE) poly-generation systems , the analysis provides high primary energy savings and low emissions suggesting that for such systems optimization should be done from an economic and environmental point of view. Finally, exergoeconomic analysis of CHP applications (engines, gas turbine) [6,8,9] or evaluation of CO2 capture and management studies [12,19] complete the overview and come to meet users’ main concerns—available energy and CO2 emission price.\nThe proposed thermodynamics approach perspective points out cold and heat cogeneration systems (CCHP), and also extends to polygeneration systems [24,25]. These concepts and methodologies could help better design, manage and integrate these systems in the future, with respect to environmental and economic concerns.\nThe present analysis focuses on the main CHP systems based on Vapour Turbine, Internal and External Combustion Engines and Gas Turbine Engines. They are modelled as thermal machines with two heat reservoirs, heat losses between the heat reservoirs, and external irreversibilities due to the heat transfer at source and sink. The First Law efficiency and exergetic efficiency criteria are used in order to evaluate the CHP performance. The optimization procedure considers several constraints, namely, imposed ratio of the useful heat to power, mechanical power load, useful heat load or energy rate consumption. The results are given in terms of maximum of the exergy rate of the useful energy delivered by the CHP system and the corresponding optimum parameter expression. The specified upper bounds (maximum maximorum) correspond to the CHP system based on External Combustion Engines.\n\n## 2. Modelling of Various CHP Systems\n\nCombined Heat and Power systems with heat delivery as by-product are considered here. Although this represents a particular case of cogeneration, various processes are possible, characterized by systems or cycles referring to different techniques . The first one developed was the CHP with vapour turbines that appears as an externally fired engine (by a boiler or a steam generator). Other kinds of external combustion engines will be considered hereafter, like Ericsson and Stirling ones. Current well developed CHP systems are composed of internal combustion engines of various sizes going from 1 kWelec to more than 1 MW for industrial or urban applications. The third main systems category used for cogeneration is based on gas turbines, and in this case post combustion could be used. Some more recent studied categories using Fuel Cells or Solar Photovoltaic arrays are also added to the previously mentioned ones.\nThe present analysis is deliberately limited to the main CHP categories, namely Vapour Turbines, External Combustion Engines, Internal Combustion Engines, and Gas Turbine Engine. All these systems are thermo-mechanical ones, each having a high temperature heat source. Generally, the heat is obtained by burning a fuel, whatever the fuel is. Mechanical and then electrical power is produced by the engine (the machine). According to the Second Law of Thermodynamics, the machine gives back heat (a part of the useful energy delivered by CHP, UE) to a cold sink, before rejecting the remaining heat to the environment at T0, the reference temperature.\nThe performances of four main engines acting as CHP systems will be compared hereafter. They are represented by the Carnot, Stirling or Ericsson, Otto or Diesel, and finally Brayton-Joule cycles . Firstly, the model is developed by means of equilibrium thermodynamics, which provides more insight into the CHP system operation by considering heat losses and other irreversibilities. Also various possible performance criteria and constraints associated to the objective function that should be optimized are used in the model and will be presented in the next sections.\n\n## 3. Criteria and Optimizations\n\nThe proposed models were developed for steady state operation assumption in order to get the upper bound of obtainable exergy rate of the useful energy delivered by the CHP system corresponding to a nominal design point. This is crucial to determine the corresponding exergetic efficiency at any rate. The two useful effects of CHP systems are the mechanical power, $W ˙$ and the useful heat transfer rate supplied to the consumer, $Q ˙ U$, both being negatives as the proposed sign convention states. In this sign convention each quantity entering the machine is positive, and each leaving quantity is negative.\nIn most cases the machines are non-adiabatic due to thermal losses to the environment. As a first attempt, lumped analysis suggests that these losses can be represented as an equivalent heat loss between the highest and lowest temperature of the system. They are summarized as $Q ˙ L$, heat transfer rate loss between the hot and cold side (positive quantity).\n\n#### 3.1. First Law Efficiency Criterion, ηICHP\n\nWhatever the engine is, the energy balance can be written as follows (see Figure 1a):\n$Q ˙ S H + W ˙ + Q ˙ S C = 0$\nwhere $Q ˙ S H$ is the heat transfer rate input, the so called energy rate consumption (EC), given by:\n$Q ˙ S H = Q ˙ H + Q ˙ L$\nand $Q ˙ S C$ is the heat transfer rate rejected to the sink, expressed as:\n$Q ˙ S C = Q ˙ C − Q ˙ L$\nwith $Q ˙ H$, $Q ˙ C$ the heat transfer rate entering, respectively leaving the converter.\nThe model assumption regarding the useful heat transfer rate provided by the engine in cogeneration mode operation corresponds to the ideal case of maximum heat transfer rate recovery that yields:\n$Q ˙ U = Q ˙ C$\nThe first law efficiency ηICHP is defined as the ratio of the usable energy rate (UE) and the energy rate consumption, EC. By combining Equations (1), (3) and (4) the first law efficiency with cogeneration is given by:", null, "The consequence is that the first law implies only a “non-adiabatic system efficiency” due to the presence of heat transfer rate loss, $Q ˙ L$, whatever the thermo-mechanical CHP system is. If the system is without losses, the limit ηICHP is one because of the ideal case of maximum heat transfer rate recovery that was considered, whatever the system is. However, one notes that if it could happen for external combustion engines (see Figure 1a,b), it is not the case of internal combustion engines (see Figure 1 c,d), due to the fact that corresponding engines are open systems and TU > T0. So, heat losses to ambient appear. These dissimilarities are due to the heat transfer particularities. Hence, useful heat could be delivered at constant temperature (TSC = TU) in cases a and b, but not for cases c and d, where finite heat source effects cannot be neglected.\nFigure 1. Equilibrium Thermodynamics of CHP thermo-mechanical engines: (a) and (b) –ECE; (c) and (d) –ICE.\nFigure 1. Equilibrium Thermodynamics of CHP thermo-mechanical engines: (a) and (b) –ECE; (c) and (d) –ICE.\nTo conclude the temperature level of the useful heat appears important. It is why exergetic criteria have to be considered in the analysis.\n\n#### 3.2. Exergetic Criterion\n\nWhatever the engine is, the exergy transfer rate of the usable energy is given by :", null, "The useful thermal exergy corresponds to the case where it is looked from the external utility side instead of the working fluid side. This remark confirms the difference between ECE and ICE, as mentioned in Section 3.1.\nThen, the difference between the machine (engine) and the system from the exergetic point of view is indicated hereafter by using two entropy balances in the case of endo-irreversible thermodynamic approach. Their expressions are as follows:\nfor the engine:", null, "", null, "where $S ˙ i$, internal entropy generation rate of the engine; $S ˙ T$, total entropy generation rate of the system including source and sink (the hypothesis is that the useful heat is finally delivered to the environment).\nUpon combining Equations (7) and (8) and calculating, the total entropy generation rate expression of the CHP system results as:", null, "The exergy rate consumed by the system is expressed as:", null, "This result confirms the fact that the heat transfer rate loss due to internal irreversibilities, $I ˙ i = T 0 S ˙ i$, have a negative impact on the exergy efficiency of the system. The exergy efficiency of the engine is slightly different from that of the system due to the fact that $E ˙ x C$ moves from the preceding expression to", null, ". The same methodology will be applied to ICE hereafter.\n\n#### 3.3. General Optimization Procedure\n\nBy considering the same hypothesis introduced in Section 2, a two heat conductances model is develop here corresponding to the one proposed by Sahin and Kodal . The model corresponds to the endoreversible case (without internal irreversibilities), and is also the basic configuration of the CHP vapour turbine system without losses. Then, by means of non-equilibrium Thermodynamics more insight in the CHP system operation is achieved by considering external heat transfer irreversibilities at source and engine contact and at engine and sink contact (Figure 2).\nThe engine entropy balance expression results as:\n$Q ˙ H T H + Q ˙ U T C = 0$\nIt is clear that energy efficiency (equal to one in the ideal case without heat loss) does not depend on TU, nor TH, TC, but the system exergy efficiency does. By combining Equations (1) - (3), (6), (7), the exergy efficiency expression becomes:", null, "Figure 2. Schematic representation of the temperature distribution in a non-equilibrium CHP Carnot system.\nFigure 2. Schematic representation of the temperature distribution in a non-equilibrium CHP Carnot system.\nIf $Q ˙ H = Q ˙ S H$ is imposed, the maximum of efficiency gives back the equilibrium case (TC = TSC = TU; TH = TSH). But the necessary finite heat transfer rate imposes through the entropy balance a constraint relating the two degrees of freedom (TC, TH).\nFor the case of linear heat transfer law considered as an example, where:\n$Q ˙ H = K H ( T S H − T H )$\n$Q ˙ C = K C ( T U − T C )$\nThe optimization with respect to TC, TH by using Lagrangian method (for details of variational calculus see [29,30]) gives:", null, "", null, "where the heat transfer conductances KH, KC are parameters associated to a given design.\nConsequently one gets from Equation (6) combined with Equations (1)–(4), and (16):", null, "Equation (17) corresponds to the endoreversible case without heat losses and shows that optimal exergy does not depend on the level of TU. Furthermore, a constrained dimension imposed to the system, namely the total heat transfer conductance, KH + KC = KT, for the case without heat losses, but with internal irreversibilities of the convertor yields for Equations (16) and (17) the following expressions:", null, "", null, "This results in a new optimum regarding the finite heat transfer conductances when the total heat transfer conductance is fixed.\nBy derivation of Equation (19) the best allocation of heat transfer conductances corresponds to:", null, "Equation (20) expresses that a maximum of the useful heat exergy is obtained at equipartition of the heat transfer conductances for the endoreversible case (without internal irreversibilities, $S ˙ i$ = 0):\n$K H * = K C * = K T 2$\nHence, the maximum exergy rate of the useful energy supplied by CHP results as:", null, "One notes that if lim TC = TU is considered (Chambadal-Novikov limit [31,32]), KC tends to infinity, $T H * = T 0 T S H$ whatever KH is. A numerical example considering TSH = 2000 K, T0 = 300 K, will get $T H $ = 100$60$ ≈ 775 K which is closed to the common values used today. The main result is that Equation (22) is a useful upper bound for all the ECE cases.\n\n## 4. Optimization with Constraints\n\nThe use and design of a CHP system is characterized by the ratio of useful heat transfer rate to useful power, $R = Q ˙ U / W ˙$. Two other parameters can be added to this one corresponding to the two possible priorities, the mechanical power load, $W ˙ 0$, or the useful heat load, $Q ˙ U 0$. The last more significant case could be the hot source heat transfer rate limited to $Q ˙ S H 0$. These alternatives correspond to one technical added constraint that suppresses one degree of freedom for the optimization. The four previously cited cases will be considered hereafter, and new corresponding results for the Carnot endoreversible CHP system without heat losses will be given.\n\n#### 4.1. Carnot CHP System with R Imposed\n\nThe optimization of the energy efficiency of CHP from a general point of view is the same as the optimization of the engine efficiency. By considering the ratio of useful heat transfer rate to useful power in the model and after some simple calculations, it comes for the endoreversible system:\n$T C T H = R 0 1 + R 0$\nwhere R0 is the imposed value of the R ratio whatever expressions are used for the heat transfer law, and the available heat transfer rate at hot source. In the frame of equilibrium thermodynamics, Equation (23) becomes:\n$T U T S H = R 0 1 + R 0$\nOne notes that the ratio R0 is fundamentally related to the Carnot efficiency of the engine according to:\n$R 0 = 1 − 1 / η C$\nFor linear heat transfer law (KH, KC -parameters) considered in the model, the corresponding exergy rate of the useful energy supplied by CHP results by combining Equations (6), (13), (14) and (24) as:", null, "with:\n$X 0 = 1 + R 0 R 0$\nIt is easy to demonstrate that optimum optimorum of useful effect is given again by Equation (22), but R0, TU, TSH are interrelated at the optimum by:\n$1 + R 0 R 0 T U = T S H T 0$\nThis important relation means that for TSH parameter, R0 opt satisfies Equation (28) for a chosen TU or TU opt satisfies Equation (28) for an imposed R0.\nNote that Equation (26) could be enlarged when an irreversibility ratio is considered. Its expression becomes:", null, "This expression allows an optimum for a specific value of X0 relating TSH, T0 and TU, as expressed by Equations (27) and (28). This result is a new one. When imposing finite heat transfer conductances, KH + KC = KT, the conductance equipartition is found again for .\n\n#### 4.2. Carnot CHP System with $W ˙$ Imposed\n\nThe same methodology as for engine optimization is used here, but by adding the constraint $W ˙ = W ˙ 0$, an intermediate variable α appears:", null, "The optimum of $E ˙ x U$ corresponds to equipartition of heat transfer conductances previously found for an endoreversible system, and α opt satisfies the following equation that yields by combining Equations (14), (21), and (30) together with the constraint $W ˙ = W ˙ 0$ in Equation (6) and derivation with respect to TU:", null, "The equilibrium thermodynamics limit is straight forward ($W ˙ 0 = 0$), namely, the corresponding limit of the energy and exergy efficiency tends to one, for reversible and no heat losses operating regime.\nIf", null, ", an interesting limit appears:", null, "and the corresponding approximated yields:", null, "Equation (33) shows that the optimized exergy of the useful energy supplied by CHP is proportional to the imposed power $W ˙$, but amplified by the temperature ratio (TSHT0)/(TSHTU).\n\n#### 4.3. Useful Heat Transfer Rate Imposed\n\nThe heat utility is the priority in this case by its imposed value, $Q ˙ U = Q ˙ U 0$. By following the same steps as in Section 4.2, one finds that the optimum of $E ˙ x U$ corresponds to the equipartition of heat transfer conductances, and αopt becomes:", null, "The corresponding results:", null, "By derivation of Equation (35) with respect to TU, one can show that an optimum optimorum of $E ˙ x U$ exists, but it requires:", null, "Again, Equation (22) is valid for .\n\n#### 4.4. Heat Transfer Rate at the Source Imposed\n\nThe consumed heat transfer rate corresponds to the heat transfer rate input, which is now imposed, $Q ˙ S H = Q ˙ S H 0$. The optimum exergy of the useful energy supplied by CHP associated to this constraint appears for conductance equipartition, and its expression is:", null, "Here again a maximum maximorum exists and corresponds to:", null, "It always leads to the Equation (22) for .\n\n#### 4.5. Partial Conclusion\n\nThe models presented in Section 4 point out that in the presence of constraints the optimum of the useful exergy function does exist and it is given by the Equations (26), (34), and (36). The optimum of these optima [Equation (22)], results by derivation of the previously mentioned relations and it corresponds to special values of the constrained parameters. The expression of does not depend on TU, but is proportional to KT that relates it to the size of the system.\nLastly an interesting constraint to discuss is when TSH = TMAX. It was shown that is an increasing function of TSH. Actually TSH is limited at a given value imposed by the thermomechanical strength of materials, which means that one cannot increase it as much as we want. Moreover, it is known that for non-adiabatic system this limitation corresponds to the stagnation temperature. This is a current practice for solar systems and it could be also for ICE, when considering the stagnation temperature to be less or equal to the adiabatic temperature of combustion. However for a non- adiabatic system, a new optimum exists, limited by the stagnation temperature of the system .\nThese extensions will be summarized hereafter (Section 5). It has been shown that it comes out from material limitations and it is designated by TMAX.\n\n## 5. Discussion—Comparisons\n\nSome results of the model will be exemplified in Table 1 with corresponding comments given hereafter. Extensive examination of CHP configurations using thermo-equilibrium thermodynamics does not exhibit the optima for the studied systems. Moreover, first law efficiencies are not appropriate to qualify performance of these systems, so that exergetic efficiencies are recommended. Actually, First Law analysis does not provide an optima of the useful energy rate. It is not the case for exergy analysis, where the heat quality is accounted. The present study has been done for evaluating the useful exergy rate in relation with finite dimensions of the system given by heat transfer conductances to allocate, and with the maximum allowed temperature for the system.\n\n#### 5.1. Thermodynamic Optimization of Carnot CHP Systems\n\nIt was shown that when Finite Dimensions of the systems are considered, and pre-established criteria are used, optimum configurations exist, relative to temperature distribution and design variables of the system (heat transfer conductances, heat transfer rate, heat exchanger effectiveness). The Carnot cycle was chosen to show the optimization results because it is generally the reference cycle regarding steam turbines. Also, the main constraints applied to the cycle appear to move strongly the obtained optimal results. In any cases we have got the same upper bound given by Equation (19). It represents the upper bound (reference) of the exergy rate obtainable for a Carnot CHP system related to finite size through the total heat transfer conductance, KT, and maximum temperature, TSH, with regard to environment temperature, T0.\nThe proposed model includes heat losses from the system to ambiance, and also internal irreversibilities of the converter. The internal irreversibilities can be considered by two ways, (1) through entropy rate created inside the system $S ˙ i$ or (2) through an irreversibility ratio, which is the most used irreversibility representation. The Carnot CHP system extensions were reported in .\nTable 1 gives the results obtained for the Carnot adiabatic CHP system and when the irreversibility ratio I is considered, and under various proposed added constraints. These results convey to the evidence that the optimal allocation of heat transfer conductances does not depend on the added constraints. In every case, the results differ from equipartition because:", null, "The optimal temperature of the working fluid depends on the added constraint. The temperature TH opt is nondependent explicitly of TU, except when the ratio R = R0 is imposed. The objective function of the studied CHP systems is logically the exergy rate, $E ˙ x U$, of the useful energy delivered by CHP. It is composed by the sum of the useful mechanical exergy, $W ˙$, and the useful heat exergy. Its optimal value is depending on the studied case, as given in Table 1.\nThe maximum maximorum of $\| E ˙ x U \|$ corresponds to:", null, "One notes that this value does not depend on TU, the useful heat temperature level. The same conclusion holds, if the entropy rate method is chosen:", null, "Complementary results have been published, when the CHP system is a non adiabatic one. In that case, the maximum attainable temperature of the system, the stagnation temperature TS, is a useful quantity to be introduced. Hence, a new compromise between the heat transfer conductances and the heat losses has been defined (see for some details).\nTable 1. Carnot CHP system optimization, with added constraint, and irreversibility ratio method.\n optimum TH opt TC opt $\| E ˙ x U \| o p t$ Added constraint without", null, "", null, "", null, "$Q ˙ H = Q ˙ H 0$", null, "", null, "", null, "R = R0", null, "", null, "", null, "$W ˙ = W ˙ 0$", null, "", null, "", null, "$Q ˙ U = Q ˙ U 0$", null, "", null, "", null, " Some details regarding the derivation of αopt expression are given in the Appendix.\n\n#### 5.2. Optimization of Other CHP Configurations\n\nAs can be seen in Figure 1, a Stirling or Ericsson engine is an externally fired CHP system. Hence, the results have great similarity with the previous examined ones (Section 4). Complementary results could be found in a recent report . Contrarily Otto (Diesel) CHP systems, as well as Joule CHP systems, are internally fired. Due to this common issue, an insight to CHP system based on gas turbine engine is given hereafter. The optimization corresponds to the constraint of TMAX, the maximum imposed temperature, as it is well known that this condition is actually the most limiting one for gas turbines, in order to preserve material properties.\nDetails of the model can be found in a recently published paper . The heat exchangers model uses the NTU-effectiveness model: the corresponding effectiveness is εR for recuperator and εU for the useful heat exchanger. For the Joule CHP system cycle illustrated in Figure 1d, the First Law has the same expression remains as for all ICE engines:\n$W ˙ + Q ˙ H + Q ˙ U + Q ˙ C = 0$\nwhere:\n$Q ˙ H = C ˙ ( T M A X − T X )$\nwhere $C ˙ = m ˙ · c p$ is the heat capacity rate of the working fluid in the cycle, and:\n$T X = ε R · T 4 + ( 1 − ε R ) · T 2$\nThe rejected heat transfer rate from the turbine is:\n$Q ˙ U + Q ˙ C = C ˙ ( T 0 − T Y )$\nwith:\n$T Y = T 4 · ( 1 − ε R ) + ε R · T 2$\nBy combining Equations (42)−(46) the mechanical power is:", null, "where T2 results from Equation (7) after some calculations:\n$T 2 = T M A X T 0 k T 4$\nand the irreversibility factor is given by:\n$k = exp [ − S ˙ i / C ˙ ]$\nBy taking into account the above expressions, Equation (6) becomes for this case:", null, "The optimum exergy rate of the useful energy supplied corresponds to the derivation of Equation (50) with respect to T2 or T4 that are related by Equation (48). One can easily get:", null, "", null, "with:", null, "", null, "The corresponding $M A X \| E ˙ x U \|$ is obtained from Equations (50)–(52) as:", null, "Equation (55) allows to one to determine the maximum exergy rate of the useful energy supplied depending on system parameters, namely the effectiveness of the recuperator and useful heat HEX, temperatures and the irreversibility factor.\nAll the results detailed in this section are relative to the gas turbine engine configuration where the recuperator and the useful heat HEX are connected in series at the turbine exit. Therefore, the temperatures constraint should be: TU < TY < T4, in order to ensure the optimisation solution.\nSome limit cases are interesting to note for the previously reported Joule CHP systems:\n• perfect heat recuperation given by εR = 1. It involves:", null, "The maximum exergy rate of the useful energy supplied (Equation (55)) results as:", null, "• perfect useful heat HEX provided by εU = 1. It involves:", null, "For this case Equation (55) with a = aR and b = bR provides the optimum result.\n• perfect heat exchangers: εR = 1 and εU = 1. It involves:\nand:", null, "Equation (60) constitutes an upper bound depending on the three temperatures (T0, TU, TMAX) and also considering the cycle irreversibilities by the factor k. The endoreversible operation corresponding to k = 1 is straightforward.\n• no heat recuperation: εR = 0. For this case:", null, "and Equation (55) yields:", null, "Figure 3. Evolution of the non-dimensional optimum temperature at the turbine exit, $t 4 $, as a function of TU for TMAX variable.\nFigure 3. Evolution of the non-dimensional optimum temperature at the turbine exit, $t 4 $, as a function of TU for TMAX variable.\nFigure 4. Evolution of $t 4 $ as a function of tU for irreversibility factor k as variable.\nFigure 4. Evolution of $t 4 $ as a function of tU for irreversibility factor k as variable.\nNote that a0 = bR, so one could say that Equation (57) is identical to Equation (62). Although they are formally identical, the provided optimum is different due to the variation range of TU which is greater for the present case compared to the previous one (εR = 1). Other gas turbine CHP systems were studied and these results are currently in press [36,37].\nFigure 3 and Figure 4 illustrate the influence of TU level on the optimum temperature T4 at the turbine exit, for different values of the non-dimensional maximum allowed temperature, respectively, irreversibility factor, k. Note that the main parameter of the model is the non-dimensional temperature corresponding to the useful heat temperature level, given by:\n$t U = T U T 0$\nand the non-dimensional value of the maximum allowed temperature with respect to the ambient one, is:\n$t M A X = T M A X T 0$\nSo, if the non-dimensional useful heat temperature level, tU, increases, the optimal non-dimensional temperature t4 diminishes. This decrease is more pronounced as tMAX increases (Figure 3), and remains proportional as k increases (Figure 4). Otherwise, t4 opt increases with the allowed tMAX. For example, the non-dimensional optimum temperature at the turbine exit increases twice when the non-dimensional value of the maximum allowed temperature with respect to the ambient one increases from 2 to 8. Also, t4 opt increases significantly with the internal irreversibility of the turbine that corresponds to the decrease of k (Figure 4). Hence, an almost constant growth of 0.35 is registered by t4 opt when k decreases with 0.25. These models are presently under development.\n\n## 6. Conclusions\n\nThe thermodynamics of Combined Heat and Power Systems has been reviewed, with a particular focus on the most common ones which are the thermo-mechanical systems. Two main categories were proposed:\n-\nECE, External Combustion Engine;\n-\nICE, Internal Combustion Engine;\nOptimization criteria for these systems are reviewed. The analysis confirmed that the First Law efficiency criterion is only representative of the system thermal losses (non-adiabatic operation), so it is recommended to use exergy efficiency, that takes into account the heat quality, and also qualifies the irreversibilities of the converter (engine) or of the system depending on the model used (Section 3.2). Hence, the exergy analysis is revealed by the present report to be the main tool for CHP systems study due to simultaneous consideration of work and heat by their exergy, that differs for the heat from its value.\nFurthermore, it was shown by the mathematical approach particularized for Carnot system that optimization could be performed only if the finite size of the system is considered; equilibrium thermodynamics being not able to provide optima. The finite size optima obtained are coherent with the observed ones.\nVarious significant added constraints were proposed, they being useful for the design of specific systems and allowing to precise the influence on the design of the model parameters. Upper bound of the exergy rate of the usable energy were determined by consideration of the main constraints of the CHP systems: (1) imposed heat to power ratio R0; (2) imposed useful power $W ˙ 0$ (electrical or mechanical priority), (3) imposed useful heat transfer rate $Q ˙ U 0$ (heat demand priority), and lastly (4) imposed available source heat transfer rate $Q ˙ S H 0$. The main tendencies for Carnot CHP system were presented in Table 1 proving the optima existence and showing the dependence on the model parameters of the optimum temperatures of the working gas at source and sink, and of the optimum exergy rate of the usable energy.\nThe upper bound of the exergy rate of the useful energy delivered by the Carnot CHP system was obtained [Equation (22)]. It is a very important result, as it is the equivalent of the “nice radical” of Curzon-Ahlborn approach. Also, its expression contains the total heat transfer conductance to be allocated to the system that appears as the size factor of the maximum exergy rate.\nAmong other CHP systems that have been examined, the Gas Turbine Engine-based one is considered the more representative CHP system for industrial applications. The corresponding optimum of the exergy rate of the useful energy delivered by the CHP system was derived and the upper bound was determined in the case where maximum temperature of the working gas, TMAX, is fixed. This case is the most significant for the present state of the art of Gas Turbine Engine. The expressions of the optimum exergy rate of the useful energy delivered to the consumer were presented for several limiting case, and have shown the dependence on the recuperator and useful heat exchanger effectiveness, temperatures TMAX, TU, T0, and internal irreversibilities by the internal entropy generation rate, $S ˙ i$.\nThe Gas Turbine Engine CHP system illustrates the Internal Combustion Engine CHP configuration. Thus, the present methodology could be applied to Otto or Diesel CHP configurations appropriate for small and household applications. Further extension of these models is presently under development in order to allow the comparison of different constrained cases and to offer an overview on the performance of various existing CHP systems.\n\n## Nomenclature\n\n $C ˙$ heat capacity rate [W K−1]; cp mass specific heat at constant pressure [W kg−1 K−1]; $E ˙ X$ exergy rate [W]; $m ˙$ mass flow rate of the working gas in the cycle [kg s−1]; I irreversibility ratio; K heat transfer conductance [W K−1]; NTU number of heat transfer units; $Q ˙$ heat transfer rate [W]; R ratio of useful heat transfer rate to useful power; $S ˙$ entropy rate [W K−1]; T temperature [K]; t no dimensional temperature; X temperature difference [K]; $W ˙$ mechanical power [W];\n\n## Greek symbols\n\n $ε$ heat exchanger effectiveness; $η$ efficiency; $α$ intermediate variable;\n\n## Subscripts and superscripts\n\n C related to the working fluid, at the sink; c consumed or Carnot; CHP combined heat and power system; ECE external combustion engine; ex exergetic; GT gas turbine; H related to the working fluid, at the source; i internal; ICE internal combustion engine; L loss; PV/T photovoltaic/thermal system; R recuperator; SH source; SC sink; T total; U useful; I related to first law; 0 ambient or imposed value; * optimal\n\n## Appendix\n\nThis annex gives the αopt value announced in Table 1:\nIf $W ˙ = W ˙ 0 < 0$\nαopt is the solution of the equation:\n$a α 2 − b α − c = 0$\nwith:", null, "", null, "", null, "If $Q ˙ U = Q ˙ U 0 < 0$\n$α o p t = d 1 + d$\nwith:", null, "## References\n\n1. 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http://simul.iro.umontreal.ca/latbuilder/doc/d5/d98/hightut.html	[ "Lattice Builder Manual Software Package for Constructing Rank-1 Lattices\nHigh-Level API Tutorial\n\nThis tutorial shows how to use high-level features of the Lattice Builder application programming interface (API), in the context of simulation software.\n\nSimulation software written in an arbitrary language can always make use of Lattice Builder by invoking the command-line tool with appropriate options (see the Command-Line Tutorial) and by parsing its text output to extract the generating vector from the results. This is what the Lattice Builder Web Application does. Calling an external command can be done in most programming languages.\n\nFrom software written in C++, it is also possible avoids the overhead of loading and executing an external program by directly calling functions of the Lattice Builder library instead of invoking the command-line tool. In fact, the command-line tool is just an interface to the API described in this tutorial.\n\nThe complete code of the examples in this tutorial can be found under the latbuilder/examples/tutorial/ directory.\n\nThe Parser namespace contains the tools to make calls to Lattice Builder using text arguments similar to those of the command-line tool (see Command-Line Tutorial). The Parser::Search class provides a parse() method that returns a pointer to a Task::Search instance. The arguments to parse() are all text of the same format as the command-line arguments (in the following order) to:\n\n• --construction\n• --lattice-type\n• --size\n• --dimension\n• --figure-of-merit\n• --weights\n• --filters\n• --multilevel-filters\n• --combiner\n\n# Constructing and Using Ordinary Lattice Rules\n\nThe following piece of code calls Lattice Builder to perform a fast CBC construction of an ordinary lattice rule with $$n=2^8$$ points in dimension 10 using the coordinate-uniform implementation (required by fast CBC) of the $$\\mathcal P_2$$ criterion, with product weights with $$\\gamma_j=0.1$$ for all $$j$$, using no filters:\n\nLatticePoints search()\n{\ncmd.construction = \"fast-CBC\";\ncmd.size = \"2^8\";\ncmd.dimension = \"10\";\ncmd.figure = \"CU:P2\";\ncmd.weights = std::vector<std::string>{\"product:0.1\"};\ncmd.weightsPowerScale = 1.0;\ncmd.normType = \"2\";\nauto search = cmd.parse();\nstd::cout << search << std::endl;\nsearch->execute();\nstd::cout << \"BEST LATTICE: \" << search->bestLattice() << std::endl;\nstd::cout << \"MERIT: \" << search->bestMeritValue() << std::endl;\nconst auto& lat = search->bestLattice();\nreturn LatticePoints(lat.sizeParam().numPoints(), lat.gen());\n}\n\nThe combiner argument is set to a valid value, but it is not used. The resulting lattice definition can be retrieved with search->bestLattice(). Our function returns an instance of the virtual container LatticePoints, defined in tutorial/LatticePoints.h as:\n\n// Copyright (c) 2012 David Munger, Pierre L'Ecuyer, Université de Montréal.\n//\n// This file is part of Lattice Builder.\n//\n// Lattice Builder is free software: you can redistribute it and/or modify\n// the Free Software Foundation, either version 3 of the License, or\n// (at your option) any later version.\n//\n// Lattice Builder is distributed in the hope that it will be useful,\n// but WITHOUT ANY WARRANTY; without even the implied warranty of\n// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n// GNU General Public License for more details.\n//\n// You should have received a copy of the GNU General Public License\n// along with Lattice Builder. If not, see <http://www.gnu.org/licenses/>.\n#ifndef LATTICE_POINTS_H\n#define LATTICE_POINTS_H\n// virtual container for lattice points\nclass LatticePoints {\npublic:\n// standard container type definitions\ntypedef std::vector<double> value_type;\ntypedef size_t size_type;\nLatticePoints(size_type numPoints, std::vector<unsigned long> gen):\nm_numPoints(numPoints),\nm_intGen(std::move(gen)),\nm_gen(m_intGen.size())\n{ updateGen(); }\n// returns the number of points of the lattice rule\nsize_type numPoints() const { return m_numPoints; }\n// returns the number of points of the lattice rule\nsize_type size() const { return m_numPoints; }\n// returns the lattice dimension\nsize_type dimension() const { return m_gen.size(); }\n// returns the i-th lattice point\nvalue_type operator[](size_type i) const\n{\nstd::vector<double> point(dimension());\nfor (size_type j = 0; j < point.size(); j++) {\ndouble x = i m_gen[j];\npoint[j] = x - int(x);\n}\nreturn point;\n}\nprivate:\nsize_type m_numPoints;\nstd::vector<unsigned long> m_intGen;\nstd::vector<double> m_gen;\nvoid updateGen()\n{\nfor (size_type j = 0; j < m_gen.size(); j++)\nm_gen[j] = double(m_intGen[j]) / m_numPoints;\n}\n};\n#endif\n\nwhich can be used to enumerate the lattice points. The following piece of code shows how it could be used by simulation software.\n\nvoid simulate(const LatticePoints& lat)\n{\nfor (size_t i = 0; i < lat.size(); i++)\nstd::cout << \"point \" << i << \":\\t\" << lat[i] << std::endl;\n}\n\nIn this example, we just print the points to standard output; in practice we would use them to integrate multidimensional functions. The complete example can be found in tutorial/ParserFastCBC.cc.\n\n# Constructing and Using Embedded Lattice Rules\n\nTo construct embedded lattices, we change the lattice type to embedded and the combiner to sum:\n\ncmd.construction = \"fast-CBC\";\ncmd.size = \"2^8\";\ncmd.dimension = \"10\";\ncmd.figure = \"CU:P2\";\ncmd.weights = std::vector<std::string>{\"product:0.1\"};\ncmd.weightsPowerScale = 1.0;\ncmd.normType = \"2\";\ncmd.multilevelFilters = std::vector<std::string>{\"norm:P2-SL10\", \"low-pass:1.0\"};\ncmd.combiner = \"sum\";\nauto search = cmd.parse();\n\nWe also add normalization and low-pass filters:\n\ncmd.multilevelFilters = std::vector<std::string>{\"norm:P2-SL10\", \"low-pass:1.0\"};\n\nAnd, our search() functions now returns an instance of EmbeddedLatticePoints instead of LatticePoints:\n\nconst auto& lat = search->bestLattice();\nreturn EmbeddedLatticePoints(lat.sizeParam().base(), lat.sizeParam().maxLevel(), lat.gen());\n\nThe virtual container EmbeddedLatticePoints is defined in tutorial/EmbeddedLatticePoints.h as:\n\n// Copyright (c) 2012 David Munger, Pierre L'Ecuyer, Université de Montréal.\n//\n// This file is part of Lattice Builder.\n//\n// Lattice Builder is free software: you can redistribute it and/or modify\n// the Free Software Foundation, either version 3 of the License, or\n// (at your option) any later version.\n//\n// Lattice Builder is distributed in the hope that it will be useful,\n// but WITHOUT ANY WARRANTY; without even the implied warranty of\n// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the\n// GNU General Public License for more details.\n//\n// You should have received a copy of the GNU General Public License\n// along with Lattice Builder. If not, see <http://www.gnu.org/licenses/>.\n#ifndef EMBEDDED_LATTICE_POINTS_H\n#define EMBEDDED_LATTICE_POINTS_H\n// virtual container for lattice points\nclass EmbeddedLatticePoints {\npublic:\n// standard container type definitions\ntypedef std::vector<double> value_type;\ntypedef size_t size_type;\nEmbeddedLatticePoints(size_type base, size_type maxLevel, std::vector<unsigned long> gen):\nm_base(base),\nm_maxLevel(maxLevel),\nm_intGen(std::move(gen)),\nm_gen(m_intGen.size())\n{ reset(); }\n// returns to level 0\nvoid reset() { m_level = 0; m_numPoints = 1; updateGen(); }\n// increases the level\nvoid extend() { m_level++; m_numPoints = m_base; updateGen(); }\n// returns the base for the number of points\nsize_type base() const { return m_base; }\n// returns the maximum level number\nsize_type maxLevel() const { return m_maxLevel; }\n// returns the current level number\nsize_type level() const { return m_level; }\n// returns the total number of points on the current level\nsize_type numPoints() const { return m_numPoints; }\n// returns the number of points on the current level but not on previous\n// levels\nsize_type size() const { return numPoints() <= 1 ? numPoints() : (base() - 1) numPoints() / base(); }\n// returns the lattice dimension\nsize_type dimension() const { return m_gen.size(); }\n// returns the i-th lattice point that lies on the current levels but not on\n// previous levels\nvalue_type operator[](size_type i) const\n{\nstd::vector<double> point(dimension());\nfor (size_type j = 0; j < point.size(); j++) {\ndouble x = map(i) * m_gen[j];\npoint[j] = x - int(x);\n}\nreturn point;\n}\nprivate:\nsize_type m_base;\nsize_type m_maxLevel;\nstd::vector<unsigned long> m_intGen;\nstd::vector<double> m_gen;\nsize_type m_numPoints;\nsize_type m_level;\nsize_type map(size_type i) const { return numPoints() <= 1 ? i : (base() + 1) * (i + 1) / base(); }\nvoid updateGen()\n{\nfor (size_type j = 0; j < m_gen.size(); j++)\nm_gen[j] = double(m_intGen[j]) / m_numPoints;\n}\n};\n#endif\n\nFinally, we modify our simulate() function to use the points level by level:\n\nvoid simulate(EmbeddedLatticePoints lat)\n{\nwhile (lat.level() <= lat.maxLevel()) {\nstd::cout << \"==> level \" << lat.level() << std::endl;\nfor (size_t i = 0; i < lat.size(); i++)\nstd::cout << \"point \" << i << \":\\t\" << lat[i] << std::endl;\nlat.extend();\n}\n}\n\nThe complete example can be found in tutorial/ParserFastCBCEmbedded.cc." ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.5198962,"math_prob":0.943329,"size":6779,"snap":"2023-14-2023-23","text_gpt3_token_len":1749,"char_repetition_ratio":0.14701107,"word_repetition_ratio":0.23336968,"special_character_ratio":0.2839652,"punctuation_ratio":0.22575387,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99176186,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2023-03-30T20:50:52Z\",\"WARC-Record-ID\":\"<urn:uuid:ba3d9b71-c10c-4e53-8be1-9aaca099012e>\",\"Content-Length\":\"27891\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c12432c1-2457-432e-a700-87aa51e0311a>\",\"WARC-Concurrent-To\":\"<urn:uuid:c41ed641-dd6e-49c8-b7d8-e77dff83f95c>\",\"WARC-IP-Address\":\"132.204.26.154\",\"WARC-Target-URI\":\"http://simul.iro.umontreal.ca/latbuilder/doc/d5/d98/hightut.html\",\"WARC-Payload-Digest\":\"sha1:SSRFEIGRM2SPM5FW6MEXK2D2GOFMG43P\",\"WARC-Block-Digest\":\"sha1:IP3HMX7CZ7VZOUO3FOHTHJJ3ISCKIAXU\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2023/CC-MAIN-2023-14/CC-MAIN-2023-14_segments_1679296949387.98_warc_CC-MAIN-20230330194843-20230330224843-00557.warc.gz\"}"}
https://www.slideserve.com/andra/quiz	[ "", null, "Download", null, "Download Presentation", null, "Quiz\n\n# Quiz\n\nDownload Presentation", null, "## Quiz\n\n- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -\n##### Presentation Transcript\n\n1. Quiz • How many significant figures are there in the number 45.07 cm? (5 points) • Express the operations correctly (5 points each) a. 34.56 cm + 3 cm = b. 34.56 cm X 3 cm =\n\n2. Dimensions, Units, Metric System, Dimensional Analysis Homework: Skim Ch. 1. Complete the handout\n\n3. I. Dimensions. Physics is a science founded on observation and measurement. Measurements are used to convey information gained about the world around us. ____________ are measurable physical properties.\n\n4. I. Dimensions. Physics is a science founded on observation and measurement. Measurements are used to convey information gained about the world around us. Dimensions are measurable physical properties.\n\n5. Here are a few examples: Length, Time, Mass, Charge, Temperature, Rate, Speed\n\n6. Dimensions are divided into two types, ____________ and ____________. Fundamental dimensions are just as their name indicates, these are the basic building blocks used to set the framework of measurements.\n\n7. Dimensions are divided into two types, fundamental and derived. Fundamental dimensions are just as their name indicates, these are the basic building blocks used to set the framework of measurements.\n\n8. We will use only three basic building blocks for the entire first semester: ______, ____, and ____. Each of these is defined in section one of the text.\n\n9. We will use only three basic building blocks for the entire first semester: length, mass, and time. Each of these is defined in section one of the text.\n\n10. Derived dimensions are made of combinations of fundamental dimensions. Here are some examples: area = (length)2,volume = (length)3\n\n11. II. Units. Units are ____________ values given to the dimensions above. Standardization is necessary; else measurements can easily become meaningless. These are three main systems used today: _____, _____, and the _______ system\n\n12. II. Units. Units are standardized values given to the dimensions above. Standardization is necessary; else measurements can easily become meaningless. These are three main systems used today: _____, _____, and the _______ system\n\n13. II. Units. Units are standardized values given to the dimensions above. Standardization is necessary; else measurements can easily become meaningless. These are three main systems used today: m k s, c g s, and the british system\n\n14. The first two are part of the metric system, a system built off of powers of 10. SI, or mks, is the current standard.\n\n15. The mks stands for _____, ________, and ______, representing length, mass, and time\n\n16. The mks stands for meter, kilogram, and second, representing length, mass, and time\n\n17. The other, older system is cgs, which stands for __________, ____, and ______.\n\n18. The other, older system is cgs, which stands for centimeter, gram, and second.\n\n19. Each of the units has the same set of multipliers, allowing each unit to be increased or decreased in amount by factors of ten. These are the multipliers that you must memorize:\n\n20. The main benefit of the metric system is that it is ______, ___________, and _____________________. Here are some examples of the usage of the prefixes:\n\n21. The main benefit of the metric system is that it is simple, easy to use, and based on powers of 10. Here are some examples of the usage of the prefixes:\n\n22. The main benefit of the metric system is that it is simple, easy to use, and based on powers of 10. Here are some examples of the usage of the prefixes: kilogram = 1000 grams millimeter = 0.001 meter decibel = 0.1 bel\n\n23. III. Dimensional Analysis: Dimensional analysis refers to the process of converting one set of units into another. Here are a few examples:\n\n24. 1. 125 mL = ? L\n\n25. 2. 155 km = ? m\n\n26. 3. 8.5 mm = ? dm\n\n27. 3. 8.5 mm = ? dm\n\n28. 3. 8.5 mm = ? dm\n\n29. 4. 0.0278 kg = ? mg\n\n30. 4. 0.0278 kg = ? mg\n\n31. 5. 0.000 008 8 mm = ? dam\n\n32. 5. 0.000 008 8 mm = ? dam\n\n33. The conversions between metric and British units are not as easy. The conversions are actually definitions, and are considered exact. Highlight these for memorizing! 1 foot = 12 inches, 1 yard = 3 feet = 36 inches, 1 mile = 5280 feet 1 inch = 2.54 cm, 1 liter = 1000 cc = 1000 cm3, 1 gallon = 4 quarts 1 day = 24 hours, 1 hour = 60 minutes = 3600 seconds 1 micron = 1 micrometer = 1 μm, The density of water = 1.000 g/cm3.\n\n34. Example #1: How many inches are in 0.00435 mile?\n\n35. Example #1: How many inches are in 0.00435 mile?\n\n36. Example #1: How many inches are in 0.00435 mile?\n\n37. Example #1: How many inches are in 0.00435 mile?\n\n38. Example #1: How many inches are in 0.00435 mile?\n\n39. Example #1: How many inches are in 0.00435 mile? Don’t forget proper significant figures!\n\n40. Example #2: How many firkins can be held in one hogshead?\n\n41. Example #2: How many firkins can be held in one hogshead?\n\n42. Example #2: How many firkins can be held in one hogshead? Since all conversions are exact, this is considered an exact computation.\n\n43. Example #3: How many meters are in one mile?\n\n44. Example #3: How many meters are in one mile?\n\n45. 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https://unclecoder.com/Article/c-plus-plus/194/CPP-Program-tostudy-if-control-statement	[ "UncleCoder.com\n\nUncleCoder.com\n\nFree programming examples and instructions\n\n# CPP Program to study if control statement\n\n## CPP Program to study how to use if control statement\n\nby Krishna\n\nPosted on 27 Jun 2018 Category: c-plus-plus Views: 931\n\nCPP Program to study if control statement\n\nIn this article we will discuss about the basic CPP control structures and its implementation. Like C programming language, CPP also supports all basic control structures such as if, if..else, switch, while, do..while, for loop. So here we will discuss in detail about the ‘if control statement’. Syntax is same as the C if statement.\n\nThe if statement is implemented in two forms\n\n• Simple if statement\n``````if(true condition)\n{\nAction;\n}\n``````\n\n• if…else statement\n``````if(true condition)\n{\nAction;\n}\nelse\n{\nAction;\n}\n``````\n\nHere is a program which demonstrates the if control statement. This program calculates the largest number among the three given numbers.\n\n``````#include<iostream>\n#include<conio.h>\nusing namespace std;\nvoid main()\n{\nint num1, num2, num3;\ncout<<\"Enter the first number \";\ncin>>num1;\ncout<<\"Enter the second number: \";\ncin>>num2;\ncout<<\"Enter the third number: \";\ncin>>num3;\nif((a>b)&&(a>c))\n{\ncout<< “ largest number is”<<num1;\n}\nelse if(b>c)\n{\ncout<< “largest number is”<<num2;\n}\nelse\n{\ncout<< “largest number is”<<num3;\n}\ngetch();\n}\n``````\n\nIn the above program, the first if statement checks whether the first number is greater than the second and third number, if it is true action is executed(here, prints the largest number) and if it is false, the control goes to the if else part or else part.\n\nC code for the above program is also available in unclecoder.com\n\nOutput for the above program is shown as:\n\nOUTPUT", null, "" ]	[ null, 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https://www.nyooztrend.com/2020/10/professional-calculators-are-boon-to.html	[ "# PROFESSIONAL CALCULATORS ARE A BOON TO BUSINESS\n\nOctober 27, 2020 RAWAT 0 Comments\n\nA calculator is an electronic device that is used to perform calculations. Arithmetic calculations can be simple or complex. Nowadays there are different types of oT calculators in the market depending on the necessity of a user. There are different varieties of electronic calculator which are available in the market; these calculators are designed for special purposes.", null, "The main features of business calculators are scientific notations, floating-point notation, inbuilt functions such as logarithmic functions, trigonometric functions, exponential functions, quick access to constants such as pie and e, cursor controls and edit equation and view previous calculations, hexadecimal to binary, octal calculations, complex numbers Junction calculations, statistics and probability, Matrix calculation, equation solving, calculus, conversion of units, physical constants, calculation of simple interest, compound interest, profit and loss and much more. These inbuilt functions help to perform calculations gradually. These advanced financial calculators can calculate up to 250 to 400 steps of calculation at a time the data can also be stored in the memory which is one of the most advanced features of these calculators.\n\nThis is one of the most useful and shallow calculators as the main features of this calculator is built in direct keystrokes for swift switching of exclusive modes. It has Your-line displays allowing easy input conformation change and correction for each parameter. Easily fixed percentage method some of their digit’s bonus can be calculated easily sales and levels with the breakeven point can also be calculated with direct function keys." ]	[ null, "https://1.bp.blogspot.com/-bHw_AHfVryg/X5fhksXVvyI/AAAAAAAAXo8/3rKDYjipA-UWg7KuJ3Dly9Nby6NFvbcUwCLcBGAsYHQ/s16000/istockphoto-1168618923-170667a.jpg", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.91156715,"math_prob":0.9956754,"size":2678,"snap":"2020-45-2020-50","text_gpt3_token_len":516,"char_repetition_ratio":0.17988034,"word_repetition_ratio":0.04896907,"special_character_ratio":0.1695295,"punctuation_ratio":0.10204082,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9834331,"pos_list":[0,1,2],"im_url_duplicate_count":[null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-11-24T18:17:01Z\",\"WARC-Record-ID\":\"<urn:uuid:0777ffeb-e5d4-4191-b75a-b73ea36921f5>\",\"Content-Length\":\"385898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9e52f040-7c7f-436b-8d95-f375d9cc4343>\",\"WARC-Concurrent-To\":\"<urn:uuid:99749a23-0a1d-4be9-83b0-3f07ff945785>\",\"WARC-IP-Address\":\"216.239.32.21\",\"WARC-Target-URI\":\"https://www.nyooztrend.com/2020/10/professional-calculators-are-boon-to.html\",\"WARC-Payload-Digest\":\"sha1:NTQZW2LCP4Y6AX7GRDEAV33IFLNO6BIZ\",\"WARC-Block-Digest\":\"sha1:CZ4ZQ4BOXDRV6PWQTR6DCELKHGS57C4X\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-50/CC-MAIN-2020-50_segments_1606141176922.14_warc_CC-MAIN-20201124170142-20201124200142-00308.warc.gz\"}"}
https://www.webqc.org/molecularweightcalculated-190212-155.html	[ "", null, "#### Chemical Equations Balanced on 02/12/19\n\n Molecular weights calculated on 02/11/19 Molecular weights calculated on 02/13/19\nCalculate molecular weight\n 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 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226\nMolar mass of Mn(NO3)2 is 178.947845\nMolar mass of Ba(ClO3)2 is 304,2294\nMolar mass of LaSD4 is 179.0268771112\nMolar mass of MnCl2 is 125,844045\nMolar mass of KClO2 is 106.5501\nMolar mass of Al2O3 is 101.9612772\nMolar mass of C2H2NO2 is 72.04278\nMolar mass of Fe(NO3)3 is 241,8597\nMolar mass of C10H2O5 is 202.11988\nMolar mass of CuSO4 is 159.6086\nMolar mass of no2 is 518.20206\nMolar mass of AgSO4 is 203.9308\nMolar mass of NO2 is 46.0055\nMolar mass of Al(OH)3 is 78.0035586\nMolar mass of KOH is 56.10564\nMolar mass of f2o3 is 85.9950064\nMolar mass of CaO is 56.0774\nMolar mass of C80H68Cu2N7OP4Re is 1580.636268\nMolar mass of AgNO3 is 169.8731\nMolar mass of CO2 is 44,0095\nMolar mass of NO is 30.0061\nMolar mass of H10O5 is 90.0764\nMolar mass of C12H22O11 is 342.29648\nMolar mass of HNO3 is 63.01284\nMolar mass of fe2o3 is 159.6882\nMolar mass of Cr(PO4)2 is 241.938824\nMolar mass of C2H2 is 26,03728\nMolar mass of He(lbmol) is 4,002602\nMolar mass of He(lbmol) is 4,002602\nMolar mass of C7H8 is 92.13842\nMolar mass of H(lbmol) is 1,00794\nMolar mass of K2O is 94,196\nMolar mass of CH4 is 16.04246\nMolar mass of Mg(ClO3)2 is 191.2074\nMolar mass of KBr is 119.0023\nMolar mass of C2H2*2 is 26,03728\nMolar mass of (NH2)2CO is 60.05526\nMolar mass of k is 39.0983\nMolar mass of h2o is 18.01528\nMolar mass of K is 39.0983\nMolar mass of Cr(NO3)3 is 238.0108\nMolar mass of C7H9NO2 is 139.15186\nMolar mass of C6H6 is 78.11184\nMolar mass of KBr is 119.0023\nMolar mass of Ni3(PO4)2 is 366,022924\nMolar mass of Fe(ClO3)2 is 222.7474\nMolar mass of Fe(ClO3)2 is 222.7474\nMolar mass of Fe(ClO3)2 is 222.7474\nMolar mass of Fe(ClO3)2 is 222.7474\nMolar mass of Fe(ClO3)2 is 222.7474\n 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 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226\nCalculate molecular weight\n Molecular weights calculated on 02/11/19 Molecular weights calculated on 02/13/19\nMolecular masses on 02/05/19\nMolecular masses on 01/13/19\nMolecular masses on 02/12/18\nBy using this website, you signify your acceptance of Terms and Conditions and Privacy Policy.\n© 2019 webqc.org All rights reserved\n Periodic table Unit converters Chemistry tools Chemical Forum Chemistry FAQ Constants Symmetry Search Chemistry links Link to us Suggestion? Contact us Choose languageDeutschEnglishEspañolFrançaisItalianoNederlandsPolskiPortuguêsРусский中文日本語한국어 How to cite? WebQC.Org online education free homework help chemistry problems questions and answers" ]	[ null, "https://www.webqc.org/pictures/webqclogo2.png", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.84211266,"math_prob":0.8666804,"size":1860,"snap":"2019-35-2019-39","text_gpt3_token_len":816,"char_repetition_ratio":0.3890086,"word_repetition_ratio":0.15337424,"special_character_ratio":0.4752688,"punctuation_ratio":0.112068966,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95020413,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-21T16:23:49Z\",\"WARC-Record-ID\":\"<urn:uuid:20bc7758-0176-4096-a230-64186920b48a>\",\"Content-Length\":\"50703\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c730d6c9-9431-458b-8491-7ead84219497>\",\"WARC-Concurrent-To\":\"<urn:uuid:45867899-b474-4f8e-9311-bd0a7b827f93>\",\"WARC-IP-Address\":\"104.18.55.146\",\"WARC-Target-URI\":\"https://www.webqc.org/molecularweightcalculated-190212-155.html\",\"WARC-Payload-Digest\":\"sha1:ZSIQ45JMVOFHBKBS5KUTQTCHSG3EKWF5\",\"WARC-Block-Digest\":\"sha1:JRLWTCMBXRZ7KNWDAG5QBIHKJ42P7GFP\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027316075.15_warc_CC-MAIN-20190821152344-20190821174344-00393.warc.gz\"}"}
https://wpblog.wyzant.com/resources/lessons/math/geometry/triangles/angles_of_triangles/	[ "# Angle Properties of Triangles\n\nNow that we are acquainted with the classifications\nof triangles\n, we can begin our extensive study of the angles of triangles.\nIn many cases, we will have to utilize the angle theorems\nwe’ve seen to help us solve problems and proofs. However, there are some triangle\ntheorems that will be just as essential to know. This first theorem tells\nus that if we know the measures of two angles of a triangle, it is possible to determine\nthe measure of the third angle.\n\n## Triangle Angle Sum Theorem\n\nThe sum of the measures of the interior angles of a triangle is 180.", null, "The diagram above illustrates the Triangle Angle Sum Theorem.\n\nLet’s do some examples involving the Triangle Sum Theorem to help us see its utility.\n\n### Examples\n\n(1) Find the measure of ?C.", null, "Solution:\n\nAs with all problems, we must first use the facts that are given to us. Using the\ndiagram, we are given that", null, "", null, "Since our goal is to find the measure of ?C, we can use the Triangle\nAngle Sum Theorem to solve for the missing angle. So we have", null, "Using the angle measures we were given, we can substitute those values into our\nequation to get.", null, "", null, "", null, "Having ?C measure out to 26° satisfies the property\nthat the sum of the interior angles of a triangle is 180°.\n\n(2) Find the value of x in the diagram below.", null, "Solution:\n\nIn this exercise, we are given that", null, "", null, "", null, "Looking at ?RST, we see that two of three angles are given to us.\nThus, we can apply the Triangle Angle Sum Theorem to figure out the measure of the\nthird angle:", null, "", null, "", null, "", null, "Note that ?SRT is the vertical angle opposite ?QRP,\nso we can deduce that", null, "Then, by the definition of congruent angles, we have", null, "", null, "Now, we have one of three angle measures of ?QRP. Since we know that\nm?P = m?Q = x, we can use the Triangle Angle Sum Theorem as follows", null, "", null, "", null, "", null, "", null, "We have found the measure of ?P and ?Q to be 67.\n\nIn order to comprehend the next theorem, we must learn two more terms that describe\nangles. The angle formed by one side of a triangle with the extension of another\nside is called an exterior angle of the triangle.", null, "Exterior angles get their name because they lie on the outsides of triangles.\n\nThe two angles that are not adjacent, or next to, the exterior angle of the triangle\nare called remote interior angles.", null, "Now that we know what these terms mean, we are ready for a theorem that will help\nus tremendously in our proofs.\n\n## Exterior Angle Theorem\n\nThe measure of an exterior angle of a triangle is equal to the sum of the measures\nof the two remote interior angles.", null, "Adding the measures of the two remote interior angles of a triangle gives the measure\nof the exterior angle.\n\nLet’s see how the Exterior Angle Theorem can be utilized to help us find the measures\nof unknown angles in the examples below.\n\n### Examples\n\n(1) Find the measures of ?1 and ?2 in the figure below.", null, "Solution:\n\nFirst, we can solve for m?1 since we are given the measure of two\nangles of that triangle. This part of the problem is similar to the examples we\nhave already done above. Let’s begin with the statements of what we are given, which\nare:", null, "", null, "Now, we can solve for m?1 by using the Triangle Angle Sum Theorem.\nSo we have", null, "", null, "", null, "", null, "In order to solve for the measure of ?2, we will need to apply the\nExterior Angle Theorem. We know that the two remote interior angles in the figure\nare ?S and ?A. Thus, by the Exterior Angle Theorem,\nthe sum of those angles is equal to the measure of the exterior angle. We have", null, "", null, "", null, "While not always necessary, we can check our solution using our previous knowledge\nof lines. We see that ?1 and ?2 make up ray AK.\nAnd since straight lines have 180° measures, we know that the sum\nof ?1 and ?2 must be 180. Let’s check\nto make sure:", null, "", null, "So, we know we have worked this problem out correctly.\n\n(2) Find m?B.", null, "Solution:\n\nLet’s take a look at the information we have been given first. We know that", null, "", null, "", null, "Right off the bat, we can apply the Exterior Angle Theorem to help us solve the\nproblem. We have", null, "", null, "", null, "", null, "", null, "This does not answer the question, however. The question asked for m?B.\nThe variable x alone does not tell us what the measure of the angle\nis. So, we must plug x = 4 into our equation for m?B:", null, "", null, "", null, "", null, ".\n\nNow we have found that the measure of ?B is 39°.\n\nScroll to Top" ]	[ null, "https://wpblog.wyzant.com/embed/editor/256288", null, "https://wpblog.wyzant.com/embed/editor/256289", null, "https://wpblog.wyzant.com/embed/editor/256290", null, "https://wpblog.wyzant.com/embed/editor/256291", null, "https://wpblog.wyzant.com/embed/editor/256292", null, "https://wpblog.wyzant.com/embed/editor/256293", null, "https://wpblog.wyzant.com/embed/editor/256294", null, "https://wpblog.wyzant.com/embed/editor/256295", null, "https://wpblog.wyzant.com/embed/editor/256296", null, "https://wpblog.wyzant.com/embed/editor/256297", null, "https://wpblog.wyzant.com/embed/editor/256298", null, "https://wpblog.wyzant.com/embed/editor/256299", null, "https://wpblog.wyzant.com/embed/editor/256300", null, "https://wpblog.wyzant.com/embed/editor/256301", null, "https://wpblog.wyzant.com/embed/editor/256302", null, "https://wpblog.wyzant.com/embed/editor/256303", null, "https://wpblog.wyzant.com/embed/editor/256304", null, "https://wpblog.wyzant.com/embed/editor/256305", null, "https://wpblog.wyzant.com/embed/editor/256306", null, "https://wpblog.wyzant.com/embed/editor/256307", null, "https://wpblog.wyzant.com/embed/editor/256308", null, "https://wpblog.wyzant.com/embed/editor/256309", null, "https://wpblog.wyzant.com/embed/editor/256310", null, "https://wpblog.wyzant.com/embed/editor/256311", null, "https://wpblog.wyzant.com/embed/editor/256312", null, "https://wpblog.wyzant.com/embed/editor/256313", null, "https://wpblog.wyzant.com/embed/editor/256314", null, "https://wpblog.wyzant.com/embed/editor/256315", null, "https://wpblog.wyzant.com/embed/editor/256316", null, "https://wpblog.wyzant.com/embed/editor/256317", null, "https://wpblog.wyzant.com/embed/editor/256318", null, "https://wpblog.wyzant.com/embed/editor/256319", null, "https://wpblog.wyzant.com/embed/editor/256320", null, "https://wpblog.wyzant.com/embed/editor/256321", null, "https://wpblog.wyzant.com/embed/editor/256322", null, "https://wpblog.wyzant.com/embed/editor/256323", null, "https://wpblog.wyzant.com/embed/editor/256324", null, "https://wpblog.wyzant.com/embed/editor/256325", null, "https://wpblog.wyzant.com/embed/editor/256326", null, "https://wpblog.wyzant.com/embed/editor/256327", null, "https://wpblog.wyzant.com/embed/editor/256328", null, "https://wpblog.wyzant.com/embed/editor/256329", null, "https://wpblog.wyzant.com/embed/editor/256330", null, "https://wpblog.wyzant.com/embed/editor/256331", null, "https://wpblog.wyzant.com/embed/editor/256332", null, "https://wpblog.wyzant.com/embed/editor/256333", null, "https://wpblog.wyzant.com/embed/editor/256334", null, "https://wpblog.wyzant.com/embed/editor/256335", null, "https://wpblog.wyzant.com/embed/editor/256329", null, "https://wpblog.wyzant.com/embed/editor/256336", null, "https://wpblog.wyzant.com/embed/editor/256337", null, "https://wpblog.wyzant.com/embed/editor/256338", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.9232942,"math_prob":0.99170595,"size":4114,"snap":"2021-43-2021-49","text_gpt3_token_len":992,"char_repetition_ratio":0.20024331,"word_repetition_ratio":0.05419355,"special_character_ratio":0.22848809,"punctuation_ratio":0.12151617,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9999428,"pos_list":[0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100,101,102,103,104],"im_url_duplicate_count":[null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,2,null,2,null,2,null,2,null,2,null,2,null,4,null,2,null,2,null,2,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-12-04T12:52:44Z\",\"WARC-Record-ID\":\"<urn:uuid:ff0f2d0e-db91-4bb9-a8e3-997a9422dd44>\",\"Content-Length\":\"356005\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3afd3b4-7541-41e2-8447-a6c6841e37e5>\",\"WARC-Concurrent-To\":\"<urn:uuid:41d76388-2498-446a-bcc7-6a72c2e67b0b>\",\"WARC-IP-Address\":\"192.0.78.25\",\"WARC-Target-URI\":\"https://wpblog.wyzant.com/resources/lessons/math/geometry/triangles/angles_of_triangles/\",\"WARC-Payload-Digest\":\"sha1:TWN4YYJ3CIPNZZ6MYJ2W62IHIZOA3CLZ\",\"WARC-Block-Digest\":\"sha1:TRTXGEMHL3YS4GZCMHG63HXEO2OLGNLC\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-49/CC-MAIN-2021-49_segments_1637964362992.98_warc_CC-MAIN-20211204124328-20211204154328-00015.warc.gz\"}"}
https://www.askiitians.com/rd-sharma-class-8-solutions/	[ "#### Thank you for registering.\n\nOne of our academic counsellors will contact you within 1 working day.\n\nClick to Chat\n\n1800-5470-145\n\n+91 7353221155\n\nCART 0\n\n• 0\nMY CART (5)\n\nUse Coupon: CART20 and get 20% off on all online Study Material\n\nITEM\nDETAILS\nMRP\nDISCOUNT\nFINAL PRICE\nTotal Price: Rs.\n\nThere are no items in this cart.\nContinue Shopping\n\n# RD Sharma Class 8 Solutions\n\nRD Sharma Solutions for Class 8 Maths include answers to all the exercises of RD Sharma Class 8 Maths book in a stepwise manner. RD Sharma is one of the best reference books when it comes to Maths. With this book, Class 8 students can revise Maths concepts easily and prepare themselves for their exams. Class 8 RD Sharma book includes 27 Chapters with various topics overlapping with the CBSE Class 8 Maths syllabus. This is why many teachers and students recommend this book. Students can now access RD Sharma free solutions for Class 8 Maths from the askIITians website. Our team has prepared these solutions so that the students can solve their doubts instantly.\n\nRD Sharma Class 8 Maths book includes a variety of questions that can help students strengthen their conceptual understanding. But there can be times when students are unable to get the correct answer. This halts their learning right away and they have to wait to get their doubts solved. But with online RD Sharma Class 8 Solutions by askIITians, students can continue their learning without any hindrance. They can refer to the solutions whenever they get stuck on a question and identify their mistakes. This not only saves their time but also makes learning much easier for them. askIITians provides chapter-wise solutions for RD Sharma Class 8 Maths book. Download our free RD Sharma solutions right now and start practising.\n\n## RD Sharma Solutions for Class 8 Maths All Chapters\n\nRD Sharma Class 8 Maths book includes 27 Chapters on topics like cubes, cube roots, time and work, percentage, mensuration, visualising shapes, factorisation, division of algebraic expressions and profit and loss. It also includes topics like graphs, probability, data handling, powers, rational numbers, inverse and direct variations, linear equations in one variable and compound interest. Concepts of shapes such as polygons, quadrilaterals, and special types of quadrilaterals like parallelogram, rhombus, square and trapezium are split into three different chapters for an in-depth understanding of the students.\n\nHere, you can access the free RD Sharma Solutions for Class 8 Maths book chapter-wise so that there is no confusion. We have organised the solutions for every chapter on the basis of exercises to simplify the process for you.\n\n### R.D. Sharma Class 8 Solutions\n\nChapter 1: Rational Numbers\n\nChapter 15: Understanding Shapes I Polygons\n\nChapter 2: Powers\n\nChapter 16: Understanding Shapes II Quadrilaterals\n\nChapter 3: Squares and Square Roots\n\nChapter 17: Understanding Shapes Special III Types Quadrilaterals\n\nChapter 4: Cubes and Cube Roots\n\nChapter 18: Practical Geometry\n\nChapter 5: Playing With Numbers\n\nChapter 19: Visualising Shapes\n\nChapter 6: Algebraic Expressions and Identities\n\nChapter 20: Area of Trapezium and Polygon\n\nChapter 7: Factorization\n\nChapter 21: Volumes Surface Area Cuboid Cube\n\nChapter 8: Division of Algebraic Expressions\n\nChapter 22: Surface Area and Volume of Right Circular Cylinder\n\nChapter 9: Linear Equation In One Variable\n\nChapter 23: Classification and Tabulation of Data\n\nChapter 10: Direct and Inverse Variations\n\nChapter 24: Graphical Representation of Data As Histograms\n\nChapter 11: Time and Work\n\nChapter 25: Pictorial Representation of Data As Pie Charts Or Circle Graphs\n\nChapter 12: Percentage\n\nChapter 26: Data Handling Probability\n\nChapter 13: Profit Loss Discount and Value Added Tax\n\nChapter 27: Introduction to Graphs\n\nChapter 14: Compound Interest\n\n## Why RD Sharma Solutions for Class 8 Maths are Important for Students?\n\nMaths is an important subject, especially in Class 8 and Class 9 as the topics that you learn in these classes lay the foundation for higher-level topics. Solving Maths problems enhances your analytical and critical thinking which is important for students these days as they have to appear in different entrance tests, olympiads and other competitive examinations. RD Sharma Class 8 Maths book includes a variety of questions that can help the students solidify their Maths concepts from an early stage and make this subject their strength. Here are a few reasons why RD Sharma solutions by askIITians can benefit the students:\n\n• The online RD Sharma Class 8 Solutions are created by askIITians experts for self-learning of the students. These solutions are self-explanatory so students can understand them easily without any guidance.\n• With the help of RD Sharma solutions, students can enhance their conceptual understanding of Class 8 Maths. Our experts have solved every question in a detailed stepwise manner so that students have no problems in understanding the concepts.\n• All the solutions of RD Sharma Class 8 Maths chapters are available to the students free of cost. Students can download the RD Sharma free solutions from our website and refer to them any time they want.\n• These solutions are created considering the learning curve of the students. All the solutions are verified by our experts which means there is no chance of mistakes.\n• Students can study RD Sharma solutions at their own pace and as per their learning schedule. They do not need to solve all the exercise questions all at once.\n\nRD Sharma Class 8 Solutions are a must for the students who want to score full marks in their school examination like CBSE. Along with this, students who want to participate in various competitive examinations like olympiads, quizzes, aptitude tests etc can prepare themselves with the help of RD Sharma problems. In case of any confusions, our RD Shamra solutions will be of great help!\n\nDownload RD Sharma Free Solutions for Class 8 Maths by askIITians today and start practising with full confidence!\n\n## RD Sharma Class 8 Maths Solutions FAQs\n\n#1 Is RD Sharma good for Class 8?\n\nOf course! The variety of problems in RD Sharma Class 8 Maths book allow them to strengthen their conceptual knowledge. With practice, Class 8 students can master every Math concept in their syllabus and score full marks in Maths which seems to be the toughest of all to many.\n\n#2 How many chapters are there in Class 8 RD Sharma?\n\nThere are 27 Chapters in the Class 8 RD Sharma textbook for Mathematics. These include topics like Compound Interest, Understanding Shapes – I (Polygons), Understanding Shapes – II (Quadrilaterals), Understanding Shapes – III (Special Types of Quadrilaterals), Practical Geometry (Constructions), Visualising Shapes, Mensuration – I (Area of a Trapezium and a Polygon), Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube), Mensuration – III (Surface Area and Volume of a Right Circular Cylinder), Data Handling – I (Classification and Tabulation of Data), Data Handling – II (Graphical Representation of Data as Histogram), Data Handling – III (Pictorial Representation of Data as Pie Charts or Circle Graphs) and more!\n\n#3 Can online RD Sharma Class 8 solutions help in CBSE Class 8 Maths exam?\n\nYes, RD Sharma Class 8 Maths book includes concepts from the CBSE syllabus. There are different types of problems for every topic. Practising such questions can help students prepare well for their CBSE Class 8 Maths exam.\n\n#4 From where can I find RD Sharma free solutions for Class 8 Maths?\n\nStudents can find free solutions for RD Sharma Class 8 Maths from the askIITians website. We provide chapter-wise solutions for RD Sharma. These solutions are prepared and verified by our experts.\n\n#5 What if I need additional help in understanding Class 8 Maths RD Sharma concepts?\n\nStudents who require guidance on Class 8 Maths concepts can enrol in our online coaching classes where our Maths experts teach every concept step by step in live classes. We also provide the latest study resources for Class 8 Maths including chapter notes, important formulae, worksheets, mindmaps, flashcards and more!" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.90065265,"math_prob":0.5928591,"size":8680,"snap":"2021-31-2021-39","text_gpt3_token_len":1858,"char_repetition_ratio":0.17646381,"word_repetition_ratio":0.04869186,"special_character_ratio":0.2077189,"punctuation_ratio":0.13240857,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.95306146,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-07-29T14:50:44Z\",\"WARC-Record-ID\":\"<urn:uuid:791214c4-f7e1-458b-9d12-01bf9878ba84>\",\"Content-Length\":\"254794\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:1c011555-f6ca-4a72-8b0d-dba2b6ce21a5>\",\"WARC-Concurrent-To\":\"<urn:uuid:ee379631-56d9-411b-9fc8-fa2c226b6c14>\",\"WARC-IP-Address\":\"3.7.36.8\",\"WARC-Target-URI\":\"https://www.askiitians.com/rd-sharma-class-8-solutions/\",\"WARC-Payload-Digest\":\"sha1:YR2RVB4IC7K7VTOHNVLOKPXFBVPOZZXM\",\"WARC-Block-Digest\":\"sha1:BGBT63BC2S67X7K47A7CNFFNGWPURESK\",\"WARC-Identified-Payload-Type\":\"application/xhtml+xml\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-31/CC-MAIN-2021-31_segments_1627046153860.57_warc_CC-MAIN-20210729140649-20210729170649-00420.warc.gz\"}"}
https://zbmath.org/?q=an%3A0752.11045	[ "# zbMATH — the first resource for mathematics\n\nStark units and Kolyvagin’s “Euler systems”. (English) Zbl 0752.11045\nIn his important work on the conjecture of Birch and Swinnerton-Dyer, V. Kolyvagin [The Grothendieck Festschrift, Vol. II, Prog. Math. 87, 435-483 (1990; Zbl 0742.14017)] introduced the concept of an Euler system. Loosely speaking, an Euler system is a set of points on an algebraic group satisfying two conditions: a norm property and a congruence property. In the present paper, the author shows that the norm property implies a weak form of the congruence property that is sufficient for most applications. The proof is fairly formal, relying on the introduction of a “universal Euler system”.\nThe main importance of the result is that the units predicted to exist by Stark’s conjecture satisfy the norm condition, hence the congruence condition, so they conjecturally yield a new family of Euler systems, which can be used to study the structure of ideal class groups as in the work of F. Thaine [Ann. Math., II. Ser. 128, 1-18 (1988; Zbl 0665.12003)], the author [Invent. Math. 89, 511-526 (1987; Zbl 0628.12007)], and V. Kolyvagin.\n\n##### MSC:\n 11R27 Units and factorization 11G16 Elliptic and modular units 11R29 Class numbers, class groups, discriminants 14G05 Rational points\nFull Text:" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8130566,"math_prob":0.7296139,"size":1617,"snap":"2021-31-2021-39","text_gpt3_token_len":466,"char_repetition_ratio":0.11221327,"word_repetition_ratio":0.008163265,"special_character_ratio":0.2943723,"punctuation_ratio":0.2005988,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9742465,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-26T22:40:44Z\",\"WARC-Record-ID\":\"<urn:uuid:a6cb1b56-fa17-46c5-9998-48887ddec103>\",\"Content-Length\":\"49231\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c3f70e1d-c139-4e31-a672-4995422e4ce7>\",\"WARC-Concurrent-To\":\"<urn:uuid:6d2962b2-8090-42d8-a36b-a02ae06ae7f6>\",\"WARC-IP-Address\":\"141.66.194.2\",\"WARC-Target-URI\":\"https://zbmath.org/?q=an%3A0752.11045\",\"WARC-Payload-Digest\":\"sha1:4MB2AYIMRNXBZG3T7L3SIJNTUHZ6YDY6\",\"WARC-Block-Digest\":\"sha1:4PBQI2Q5EG6MTXFXXQ5JEVRRATFFRVLA\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-39/CC-MAIN-2021-39_segments_1631780057973.90_warc_CC-MAIN-20210926205414-20210926235414-00139.warc.gz\"}"}
https://www.colorhexa.com/06e312	[ "# #06e312 Color Information\n\nIn a RGB color space, hex #06e312 is composed of 2.4% red, 89% green and 7.1% blue. Whereas in a CMYK color space, it is composed of 97.4% cyan, 0% magenta, 92.1% yellow and 11% black. It has a hue angle of 123.3 degrees, a saturation of 94.8% and a lightness of 45.7%. #06e312 color hex could be obtained by blending #0cff24 with #00c700. Closest websafe color is: #00cc00.\n\n• R 2\n• G 89\n• B 7\nRGB color chart\n• C 97\n• M 0\n• Y 92\n• K 11\nCMYK color chart\n\n#06e312 color description : Vivid lime green.\n\n# #06e312 Color Conversion\n\nThe hexadecimal color #06e312 has RGB values of R:6, G:227, B:18 and CMYK values of C:0.97, M:0, Y:0.92, K:0.11. Its decimal value is 451346.\n\nHex triplet RGB Decimal 06e312 `#06e312` 6, 227, 18 `rgb(6,227,18)` 2.4, 89, 7.1 `rgb(2.4%,89%,7.1%)` 97, 0, 92, 11 123.3°, 94.8, 45.7 `hsl(123.3,94.8%,45.7%)` 123.3°, 97.4, 89 00cc00 `#00cc00`\nCIE-LAB 79.051, -78.396, 74.453 27.652, 55.017, 9.734 0.299, 0.595, 55.017 79.051, 108.116, 136.478 79.051, -74.456, 95.565 74.174, -63.26, 44.141 00000110, 11100011, 00010010\n\n# Color Schemes with #06e312\n\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #e306d7\n``#e306d7` `rgb(227,6,215)``\nComplementary Color\n• #69e306\n``#69e306` `rgb(105,227,6)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #06e381\n``#06e381` `rgb(6,227,129)``\nAnalogous Color\n• #e30669\n``#e30669` `rgb(227,6,105)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #8106e3\n``#8106e3` `rgb(129,6,227)``\nSplit Complementary Color\n• #e31206\n``#e31206` `rgb(227,18,6)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #1206e3\n``#1206e3` `rgb(18,6,227)``\nTriadic Color\n• #d7e306\n``#d7e306` `rgb(215,227,6)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #1206e3\n``#1206e3` `rgb(18,6,227)``\n• #e306d7\n``#e306d7` `rgb(227,6,215)``\nTetradic Color\n• #04980c\n``#04980c` `rgb(4,152,12)``\n• #05b10e\n``#05b10e` `rgb(5,177,14)``\n• #05ca10\n``#05ca10` `rgb(5,202,16)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #0af917\n``#0af917` `rgb(10,249,23)``\n• #23f92e\n``#23f92e` `rgb(35,249,46)``\n• #3cfa46\n``#3cfa46` `rgb(60,250,70)``\nMonochromatic Color\n\n# Alternatives to #06e312\n\nBelow, you can see some colors close to #06e312. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #31e306\n``#31e306` `rgb(49,227,6)``\n• #1fe306\n``#1fe306` `rgb(31,227,6)``\n• #0ce306\n``#0ce306` `rgb(12,227,6)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #06e324\n``#06e324` `rgb(6,227,36)``\n• #06e337\n``#06e337` `rgb(6,227,55)``\n• #06e349\n``#06e349` `rgb(6,227,73)``\nSimilar Colors\n\n# #06e312 Preview\n\nText with hexadecimal color #06e312\n\nThis text has a font color of #06e312.\n\n``<span style=\"color:#06e312;\">Text here</span>``\n#06e312 background color\n\nThis paragraph has a background color of #06e312.\n\n``<p style=\"background-color:#06e312;\">Content here</p>``\n#06e312 border color\n\nThis element has a border color of #06e312.\n\n``<div style=\"border:1px solid #06e312;\">Content here</div>``\nCSS codes\n``.text {color:#06e312;}``\n``.background {background-color:#06e312;}``\n``.border {border:1px solid #06e312;}``\n\n# Shades and Tints of #06e312\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, #001101 is the darkest color, while #fdfffd is the lightest one.\n\n• #001101\n``#001101` `rgb(0,17,1)``\n• #012403\n``#012403` `rgb(1,36,3)``\n• #013704\n``#013704` `rgb(1,55,4)``\n• #024a06\n``#024a06` `rgb(2,74,6)``\n• #025d07\n``#025d07` `rgb(2,93,7)``\n• #037009\n``#037009` `rgb(3,112,9)``\n• #03830a\n``#03830a` `rgb(3,131,10)``\n• #04970c\n``#04970c` `rgb(4,151,12)``\n• #04aa0d\n``#04aa0d` `rgb(4,170,13)``\n• #05bd0f\n``#05bd0f` `rgb(5,189,15)``\n• #05d010\n``#05d010` `rgb(5,208,16)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\n• #07f614\n``#07f614` `rgb(7,246,20)``\nShade Color Variation\n• #17f924\n``#17f924` `rgb(23,249,36)``\n• #2af936\n``#2af936` `rgb(42,249,54)``\n• #3efa48\n``#3efa48` `rgb(62,250,72)``\n• #51fa5a\n``#51fa5a` `rgb(81,250,90)``\n• #64fb6c\n``#64fb6c` `rgb(100,251,108)``\n• #77fb7e\n``#77fb7e` `rgb(119,251,126)``\n• #8afc90\n``#8afc90` `rgb(138,252,144)``\n• #9dfca2\n``#9dfca2` `rgb(157,252,162)``\n• #b0fdb4\n``#b0fdb4` `rgb(176,253,180)``\n• #c3fdc6\n``#c3fdc6` `rgb(195,253,198)``\n• #d6fed9\n``#d6fed9` `rgb(214,254,217)``\n• #eafeeb\n``#eafeeb` `rgb(234,254,235)``\n• #fdfffd\n``#fdfffd` `rgb(253,255,253)``\nTint Color Variation\n\n# Tones of #06e312\n\nA tone is produced by adding gray to any pure hue. In this case, #727772 is the less saturated color, while #06e312 is the most saturated one.\n\n• #727772\n``#727772` `rgb(114,119,114)``\n• #69806a\n``#69806a` `rgb(105,128,106)``\n• #608962\n``#608962` `rgb(96,137,98)``\n• #57925a\n``#57925a` `rgb(87,146,90)``\n• #4e9b52\n``#4e9b52` `rgb(78,155,82)``\n• #45a44a\n``#45a44a` `rgb(69,164,74)``\n• #3cad42\n``#3cad42` `rgb(60,173,66)``\n• #33b63a\n``#33b63a` `rgb(51,182,58)``\n• #2abf32\n``#2abf32` `rgb(42,191,50)``\n• #21c82a\n``#21c82a` `rgb(33,200,42)``\n• #18d122\n``#18d122` `rgb(24,209,34)``\n• #0fda1a\n``#0fda1a` `rgb(15,218,26)``\n• #06e312\n``#06e312` `rgb(6,227,18)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #06e312 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" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.502236,"math_prob":0.7240383,"size":3660,"snap":"2019-35-2019-39","text_gpt3_token_len":1605,"char_repetition_ratio":0.12773523,"word_repetition_ratio":0.011090573,"special_character_ratio":0.56092894,"punctuation_ratio":0.23489933,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9908747,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-08-26T02:25:44Z\",\"WARC-Record-ID\":\"<urn:uuid:abfedb27-daab-4bf5-a8e2-a18203d42c18>\",\"Content-Length\":\"36216\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:ceb0684f-f661-4c58-a013-442171d5ef5e>\",\"WARC-Concurrent-To\":\"<urn:uuid:ad0fc5d3-f249-4ef7-abe1-da27a004481f>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/06e312\",\"WARC-Payload-Digest\":\"sha1:6MAW44KIXZWNBGEU6IIQ57XRY5ARI5NY\",\"WARC-Block-Digest\":\"sha1:X2RBLDBUKIKW6TXBCOB2OMUYGTUVHX37\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-35/CC-MAIN-2019-35_segments_1566027330962.67_warc_CC-MAIN-20190826022215-20190826044215-00207.warc.gz\"}"}
https://www.physicsforums.com/threads/expansion-of-van-der-waals-for-small-pressure.718202/	[ "# Expansion of Van der waals for small pressure\n\nGold Member\n\n## Homework Statement\n\nIn the VDW eqn of state, ##(P + a/v^2)(v-b)=RT## write Pv as a function of {P,T} and by expanding the result in powers of P or otherwise show that the first terms of a Virial expansion in powers of P for a VDW gas are given by $$Pv = RT + \\left(b - \\frac{a}{RT}\\right)P + O(P^2)$$\n\nTaylor expansion\n\n## The Attempt at a Solution\n\nI have already solved this (I think) via a method belonging in the 'otherwise' category. However, I wish to also solve it via the method outlined. The hint is that we can express y =Pv, x=P and write ##y = y(0,T) + y'(0,T)x + O(x^2)##\nThis can be rewritten as $$Pv = Pv(0,T) + \\frac{d}{dP} (Pv)\|_{P=0}P + O(P^2)$$, so essentially we are expanding the VDW eqn about small pressures. The first term on the RHS I think should be RT, which makes sense, yet I don't see how it comes about from subbing P=0 into VDW. The differentiation of the second term gives $$b - a\\frac{d}{dP} \\frac{1}{v} + ab\\frac{d}{dP} \\frac{1}{v^2} = b -a \\frac{d}{dv}\\frac{1}{v}\\frac{dv}{dP} + ab\\frac{d}{dv} \\frac{1}{v^2} \\frac{dv}{dP}$$ but I am not sure how to continue from here.\nMany thanks.\n\nChestermiller\nMentor\nTry rewriting the equation as:\n\n$$Pv+\\frac{a}{v}-Pb-\\frac{ab}{v^2}=RT$$\n\nNext, reexpress this as:\n$$Pv+\\frac{aP}{Pv}-Pb-\\frac{abP^2}{(Pv)^2}=RT$$\n\nNow, for convenience, substitute Pv = x:\n\n$$x+\\frac{aP}{x}-Pb-\\frac{abP^2}{x^2}=RT$$\n\nYou know that, at P = 0, x = RT\n\nSo do your Taylor series expansion about this point. Take the derivative of the equation with respect to P, and then solve for the derivative of x with respect to P. Evaluate this derivative at P=0, x = RT.\n\nChet\n\n•", null, "1 person" ]	[ null, "data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8642055,"math_prob":0.9992152,"size":1168,"snap":"2021-21-2021-25","text_gpt3_token_len":401,"char_repetition_ratio":0.11512028,"word_repetition_ratio":0.0,"special_character_ratio":0.33904108,"punctuation_ratio":0.064150944,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99996305,"pos_list":[0,1,2],"im_url_duplicate_count":[null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-05-09T19:45:41Z\",\"WARC-Record-ID\":\"<urn:uuid:fe75fbec-7fed-4d67-a5ba-2001cadfd5eb>\",\"Content-Length\":\"65868\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:aadca489-71ea-4d4c-a87c-de407baee27d>\",\"WARC-Concurrent-To\":\"<urn:uuid:ffcefe48-75e4-42a3-89c3-5ae741c0a421>\",\"WARC-IP-Address\":\"104.26.14.132\",\"WARC-Target-URI\":\"https://www.physicsforums.com/threads/expansion-of-van-der-waals-for-small-pressure.718202/\",\"WARC-Payload-Digest\":\"sha1:N4PT2I5RJ2MDFMOVRZ773FAY6MEJYV5H\",\"WARC-Block-Digest\":\"sha1:2DQLBM353XVYUXYO5YR7TLOR6MM6NCRG\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-21/CC-MAIN-2021-21_segments_1620243989012.26_warc_CC-MAIN-20210509183309-20210509213309-00391.warc.gz\"}"}
http://ntiertraining.com/training-course/machine-learning-with-apache-spark/	[ "866-526-3921 [email protected]\n\nThis course teaches machine learning at scale with the popular Apache Spark framework.\n\nNo machine learning knowledge is assumed. For each machine learning concept, we first discuss the foundations, its applicability and limitations. Then we explain the implementation and use, covering specific use cases. This is achieved through a combination of 50 percent lecture, 50 percent lab work.\n\nPlease note that this course does not include in-depth coverage of the maths/stats behind machine learning.\n\nThis course is taught using Spark and Python.\n\nAudience: This class is intended for data scientists and software engineers.\nCourse Duration: 3 Days\nPrerequisites:\n\nWe assume no previous knowledge of Machine Learning. We teach popular Machine Learning algorithms from scratch.\n\nPrior to taking this course, students should have:\n\n• A working knowledge of Apache Spark\n• A programming background\n• Familiarity with Python would be a plus, but not required\nHardware and Software Requirements:\n\nA working Spark environment will be provided for students (no software needs to be installed on student machines). Participants will need an SSH client and a browser.\n\nCourse Objectives:\n• Learn popular machine learning algorithms, their applicability and limitations\n• Practice the application of these methods in the Spark machine learning environment\n• Learn practical use cases and limitations of algorithms\nCourse Outline:\n• Section One – Machine Learning (ML) Overview\n• Machine Learning landscape\n• Machine Learning applications\n• Understanding ML algorithms and models (supervised and unsupervised)\n• Section Two – ML in Python and Spark\n• Spark ML Overview\n• Introduction to Jupyter notebooks\n• Lab – Working with Jupyter + Python + Spark\n• Lab – Spark ML utilities\n• Section Three – Machine Learning Concepts\n• Statistics Primer\n• Covariance, Correlation, Covariance Matrix\n• Errors, Residuals\n• Overfitting / Underfitting\n• Cross validation, bootstrapping\n• Confusion Matrix\n• ROC curve, Area Under Curve (AUC)\n• Lab – Basic stats\n• Section Four – Feature Engineering (FE)\n• Preparing data for ML\n• Extracting features, enhancing data\n• Data cleanup\n• Visualizing Data\n• Lab – Data cleanup\n• Lab – Visualizing data\n• Section Five – Linear regression\n• Simple Linear Regression\n• Multiple Linear Regression\n• Running LR\n• Evaluating LR model performance\n• Lab\n• Use case – House price estimates\n• Section Six – Logistic Regression\n• Understanding Logistic Regression\n• Calculating Logistic Regression\n• Evaluating model performance\n• Lab\n• Use case –Credit card application, college admissions\n• Section Seven – Classification: SVM (Supervised Vector Machines)\n• SVM concepts and theory\n• SVM with kernel\n• Lab\n• Use case – Customer churn data\n• Section Eight – Classification: Decision Trees and Random Forests\n• Theory behind trees\n• Classification and Regression Trees (CART)\n• Random Forest concepts\n• Labs\n• Use case – Predicting loan defaults, estimating election contributions\n• Section Nine – Classification: Naive Bayes\n• Theory\n• Lab\n• Use case – Spam filtering\n• Section 10 – Clustering (K-Means)\n• Theory behind K-Means\n• Running K-Means algorithm\n• Estimating the performance\n• Lab\n• Use case – Grouping cars data, grouping shopping data\n• Section 11 – Principal Component Analysis (PCA)\n• Understanding PCA concepts\n• PCA applications\n• Running a PCA algorithm\n• Evaluating results\n• Lab\n• Use case – Analyzing retail shopping data\n• Section 12 – Recommendation (Collaborative Filtering)\n• Recommender systems overview\n• Collaborative Filtering concepts\n• Lab\n• Use case – Movie recommendations, music recommendations\n• Section 13 – Final Workshop (time permitting)\n• Students will analyze a couple of datasets and run ML algorithms\n• This is done as a group exercise. Each group will present findings to the class" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.74638206,"math_prob":0.6773937,"size":3794,"snap":"2019-51-2020-05","text_gpt3_token_len":811,"char_repetition_ratio":0.13562006,"word_repetition_ratio":0.055829227,"special_character_ratio":0.20980495,"punctuation_ratio":0.062730625,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9852204,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-15T21:37:39Z\",\"WARC-Record-ID\":\"<urn:uuid:65027cbc-9cc5-4222-8cb6-35f7e83d158f>\",\"Content-Length\":\"68170\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:c1de28ad-714d-4b34-b77e-17dd6fc2f6f0>\",\"WARC-Concurrent-To\":\"<urn:uuid:199f8387-a09c-470b-8308-d6fb64a10a27>\",\"WARC-IP-Address\":\"104.24.111.225\",\"WARC-Target-URI\":\"http://ntiertraining.com/training-course/machine-learning-with-apache-spark/\",\"WARC-Payload-Digest\":\"sha1:SSHAGBXGOTLUXE37XIEAP5SYUU7MBHQC\",\"WARC-Block-Digest\":\"sha1:XZVFNX36JJZGYBGRQTLSQXINJCY72IAB\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575541310866.82_warc_CC-MAIN-20191215201305-20191215225305-00399.warc.gz\"}"}
http://taorith.com/unsorted/queuing-theory-examples-problems-of-power-63757621.html	[ "# Queuing theory examples problems of power", null, "Pinedo and Introduction to stochastic networks, by R. Service systems are usually classified in terms of their number of channels, or numbers of servers. Gaurav Gupta The server has an exponential service time distribution with a mean service rate of 4 customers per minute, i. Expected cost of system per hour. See our User Agreement and Privacy Policy. If you do not see its contents the file may be temporarily unavailable at the journal website or you do not have a PDF plug-in installed and enabled in your browser.\n\n• Queueing theory\n• Queueing Theory — Worked Examples and Problems (pdf) Paperity\n\n• EXAMPLES OF THE POWER OF QUEUEING THEORY.\n\n7 example, the. For an example involving power management in data centers, consider the problem. Queueing Theory — Worked Examples and Problems. Journal of the Operational Research Society, May Andrew D.\n\nYoung. we analyze the basic features of queuing theory and its applications. Keywords theatres etc., all have Queuing problems.\n\nQueues Bank is an example of unlimited queue length . Queuing is Model VIII – Power supply Model. Model IX.\nAdditional queueing theory information can be found here and here.\n\nFor two servers working at the original rate the output is as below. For this very simple queueing system there are exact formulae that give the statistics above under the assumption that the system has reached a steady state - that is that the system has been running long enough so as to settle down into some kind of equilibrium position.\n\nVideo: Queuing theory examples problems of power #40 Queuing Theory Example of (M/M/1):(Infinite/FCFS) in Hindi//O.R.\n\nWhat would be the proportion time the petrol pump is idle? Are you sure you want to Yes No.\n\n## Queueing theory", null, "DATA GRAMMAR PLURAL The next four chapters consider geometric, dynamic, integer and stochastic programming respectively. If you continue browsing the site, you agree to the use of cookies on this website. Chapter 2 considers classical optimization techniques; Chapters 3 and 4 are devoted to linear programming problems. Pankaj Gangwar Suriender Singh Prajapati Other important arrival processes are scheduled arrivals; batch arrivals; and time dependent arrival rates i.\n1.\n\nIntroduction.-The best way of opening this discussion on the Theory of Queues would use the terminology appropriate to these homely examples. Let me begin with a straightforward problem of power-series expansion.", null, "It is interesting. Queuing Theory. Ingredients of Queuing Problem: 1: Queue input Example: Imagine customers arriving at a fa- cility at times. is a jth power the infinite sum. Department of Electrical Engineering Basic concepts; Source models; Service models (demo); Single-queue systems; Priority/shared.\n\nPoisson traffic; Batch arrivals; Example applications – voice, video, file transfer.\n\n## Queueing Theory — Worked Examples and Problems (pdf) Paperity\n\npriorities; Again, to deal with the segmentation problem, we approximate as follows: For each packet: .\nIncrease in number of buses and reservation counters requires additional resources. To analyse this sub-system we need information relating to:.", null, "On an average one customer arrives 12 minutes and each customer takes 6 minutes for getting served. The first line of the output says that the results are from a formula.\n\nWhat is average time for the hob to be in the regrinding section?", null, "MASTER PASSWORD MOZILLA THUNDERBIRD How many jobs are ahead of the average set just brought in? How many more booths should be established to reduce the waiting time less than or equal to half of the present waiting time. Certainly, the student of OR or management science using this book would need a supplementary textbook on the theory the author recommends Page. Simulating for hours to reduce the overall elapsed time required and looking at just the total system cost per hour item 22 in the above outputs we have the following:. Queues form because resources are limited.Video: Queuing theory examples problems of power Queuing Theory - Single Server Infinite QueueAs before we can investigate how the system might behave with more servers. Start on.\n\n#### 5 thoughts on “Queuing theory examples problems of power”\n\n1.", null, "Zologar:\n\nServices do not occur when the queue is empty i.\n\n2.", null, "Shakara:\n\nQueuing theory deals with problems which involve queuing or waiting.\n\n3.", null, "Kajim:\n\nThe length of service by the clerk has an exponential distribution. If we choose to adopt the approximation approach we get:.\n\n4.", null, "Akinogis:\n\nIf service facilities are increased, then the question arise how much to increase?\n\n5.", null, "Tygolrajas:" ]	[ null, "http://taorith.com/media/queuing-theory-examples-problems-of-power-5.jpg", null, "http://taorith.com/media/queuing-theory-examples-problems-of-power.jpg", null, "http://taorith.com/media/queuing-theory-examples-problems-of-power-4.jpg", null, "http://taorith.com/media/queuing-theory-examples-problems-of-power-3.jpg", null, "http://taorith.com/media/queuing-theory-examples-problems-of-power-2.jpg", null, "http://1.gravatar.com/avatar/1cb1c39857f5eef49897f849251861a9", null, "http://1.gravatar.com/avatar/1cb1c39857f5eef49897f849251861a9", null, "http://1.gravatar.com/avatar/1cb1c39857f5eef49897f849251861a9", null, "http://1.gravatar.com/avatar/1cb1c39857f5eef49897f849251861a9", null, "http://1.gravatar.com/avatar/1cb1c39857f5eef49897f849251861a9", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8829008,"math_prob":0.73143995,"size":4436,"snap":"2019-51-2020-05","text_gpt3_token_len":899,"char_repetition_ratio":0.113267146,"word_repetition_ratio":0.014164306,"special_character_ratio":0.19229035,"punctuation_ratio":0.11567164,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9772309,"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,2,null,2,null,2,null,2,null,2,null,null,null,null,null,null,null,null,null,null,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-12-10T23:15:56Z\",\"WARC-Record-ID\":\"<urn:uuid:d095be7b-139b-428c-a3dd-9247e2451cbd>\",\"Content-Length\":\"21898\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:caab2277-7273-430c-84ba-b76857c00509>\",\"WARC-Concurrent-To\":\"<urn:uuid:8a94ea25-5026-494b-ac80-854fc20a3833>\",\"WARC-IP-Address\":\"212.32.230.212\",\"WARC-Target-URI\":\"http://taorith.com/unsorted/queuing-theory-examples-problems-of-power-63757621.html\",\"WARC-Payload-Digest\":\"sha1:PIF75UYKPWQJBXVFLRQDUOVSMEOXC2ZZ\",\"WARC-Block-Digest\":\"sha1:ZBXWOUXT34LCWNWG5R7IME7AHUCMJ6LD\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-51/CC-MAIN-2019-51_segments_1575540529006.88_warc_CC-MAIN-20191210205200-20191210233200-00375.warc.gz\"}"}
https://cpes.vt.edu/library/viewnugget/633	[ "LIBRARY\n\n# Equivalent Electric Circuit for a wave energy converter", null, "Fig. 1 Mechanical Diagram of WEC system including: Buoy, MMR, Mechanical Coupling, Generator, and Power Converter.\nAn equivalent circuit provides a faster and reliable way to analyze and design the wave energy converters (WECs). A WEC as a mechanical - electrical composite system which converts ocean wave energy to electrical power is shown in Fig. 1. The WEC system starts from buoy-wave interaction coupling with a rotational mechanical system, and then a mechanical motion rectifier (MMR) drives the permanent magnet synchronous generator (PMSG) which delivers energy to a power electronics converter.\n\nThrough approximation and circuit synthesis methods, wave-buoy behavior can be linearized as a RLC circuit network in Fig 2. The nonlinearity of MMR sets can be modeled as transformers and diodes. Along with a generator circuit model and power converter, a general methodology is shown to convert a multidisciplinary system into electrical components, which can be an aid to the electrical designers in the wave energy area.\n\nTwo applications will be introduced in this work. One is to apply a reliability analysis of each component. Under irregular wave conditions, a wave force is modeled as a current source that defines the input of the system in Fig. 2. Each rotational speed and torque values can be monitored real-time as voltage and current waveforms, so the maximum stress of each component can be recorded. Another is applied to optimize the systems output power. Based on an equivalent circuit, the system's output power capability can, thus be analyzed under different wave periods and gear ratio in Fig. 3.", null, "Fig. 2 Equivalent circuit of the WEC and its output waveforms under irregular wave condition.", null, "Fig. 3 Application of Equivalent Circuit for maximum output power under different wave frequency and gear ratio." ]	[ null, "https://cpes.vt.edu/modules/library/img/nuggets/2017/2017_D2.1_Fig1.JPG", null, "https://cpes.vt.edu/modules/library/img/nuggets/2017/2017_D2.1_Fig2.JPG", null, "https://cpes.vt.edu/modules/library/img/nuggets/2017/2017_D2.1_Fig3.JPG", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.87821937,"math_prob":0.981422,"size":1082,"snap":"2021-43-2021-49","text_gpt3_token_len":213,"char_repetition_ratio":0.10946196,"word_repetition_ratio":0.0,"special_character_ratio":0.18299446,"punctuation_ratio":0.092307694,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.98538935,"pos_list":[0,1,2,3,4,5,6],"im_url_duplicate_count":[null,4,null,4,null,4,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-10-18T13:42:53Z\",\"WARC-Record-ID\":\"<urn:uuid:744022dd-c337-4e16-a31d-55a7e0a7ab22>\",\"Content-Length\":\"80204\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d3a34aa9-72b1-42e0-834e-e98f26608b92>\",\"WARC-Concurrent-To\":\"<urn:uuid:718d3492-151f-4cc2-acaf-95e0d60bcf16>\",\"WARC-IP-Address\":\"128.173.90.245\",\"WARC-Target-URI\":\"https://cpes.vt.edu/library/viewnugget/633\",\"WARC-Payload-Digest\":\"sha1:U4K3S2SHEQTXBAY2VVCPEYMLDZDDAJ2G\",\"WARC-Block-Digest\":\"sha1:7HTIVEISZF3CGKF2TBSEHDOIDTH27FLE\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2021/CC-MAIN-2021-43/CC-MAIN-2021-43_segments_1634323585203.61_warc_CC-MAIN-20211018124412-20211018154412-00364.warc.gz\"}"}
https://www.colorhexa.com/3b4227	[ "# #3b4227 Color Information\n\nIn a RGB color space, hex #3b4227 is composed of 23.1% red, 25.9% green and 15.3% blue. Whereas in a CMYK color space, it is composed of 10.6% cyan, 0% magenta, 40.9% yellow and 74.1% black. It has a hue angle of 75.6 degrees, a saturation of 25.7% and a lightness of 20.6%. #3b4227 color hex could be obtained by blending #76844e with #000000. Closest websafe color is: #333333.\n\n• R 23\n• G 26\n• B 15\nRGB color chart\n• C 11\n• M 0\n• Y 41\n• K 74\nCMYK color chart\n\n#3b4227 color description : Very dark desaturated green.\n\n# #3b4227 Color Conversion\n\nThe hexadecimal color #3b4227 has RGB values of R:59, G:66, B:39 and CMYK values of C:0.11, M:0, Y:0.41, K:0.74. Its decimal value is 3883559.\n\nHex triplet RGB Decimal 3b4227 `#3b4227` 59, 66, 39 `rgb(59,66,39)` 23.1, 25.9, 15.3 `rgb(23.1%,25.9%,15.3%)` 11, 0, 41, 74 75.6°, 25.7, 20.6 `hsl(75.6,25.7%,20.6%)` 75.6°, 40.9, 25.9 333333 `#333333`\nCIE-LAB 26.657, -8.255, 15.498 4.118, 4.973, 2.662 0.35, 0.423, 4.973 26.657, 17.559, 118.042 26.657, -2.718, 16.596 22.3, -6.061, 8.531 00111011, 01000010, 00100111\n\n# Color Schemes with #3b4227\n\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #2e2742\n``#2e2742` `rgb(46,39,66)``\nComplementary Color\n• #423c27\n``#423c27` `rgb(66,60,39)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #2e4227\n``#2e4227` `rgb(46,66,39)``\nAnalogous Color\n• #3c2742\n``#3c2742` `rgb(60,39,66)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #272e42\n``#272e42` `rgb(39,46,66)``\nSplit Complementary Color\n• #42273b\n``#42273b` `rgb(66,39,59)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #273b42\n``#273b42` `rgb(39,59,66)``\nTriadic Color\n• #422e27\n``#422e27` `rgb(66,46,39)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #273b42\n``#273b42` `rgb(39,59,66)``\n• #2e2742\n``#2e2742` `rgb(46,39,66)``\nTetradic Color\n• #10120b\n``#10120b` `rgb(16,18,11)``\n• #1e2214\n``#1e2214` `rgb(30,34,20)``\n• #2d321e\n``#2d321e` `rgb(45,50,30)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #495230\n``#495230` `rgb(73,82,48)``\n• #58623a\n``#58623a` `rgb(88,98,58)``\n• #667243\n``#667243` `rgb(102,114,67)``\nMonochromatic Color\n\n# Alternatives to #3b4227\n\nBelow, you can see some colors close to #3b4227. Having a set of related colors can be useful if you need an inspirational alternative to your original color choice.\n\n• #424227\n``#424227` `rgb(66,66,39)``\n• #404227\n``#404227` `rgb(64,66,39)``\n• #3d4227\n``#3d4227` `rgb(61,66,39)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #394227\n``#394227` `rgb(57,66,39)``\n• #374227\n``#374227` `rgb(55,66,39)``\n• #344227\n``#344227` `rgb(52,66,39)``\nSimilar Colors\n\n# #3b4227 Preview\n\nText with hexadecimal color #3b4227\n\nThis text has a font color of #3b4227.\n\n``<span style=\"color:#3b4227;\">Text here</span>``\n#3b4227 background color\n\nThis paragraph has a background color of #3b4227.\n\n``<p style=\"background-color:#3b4227;\">Content here</p>``\n#3b4227 border color\n\nThis element has a border color of #3b4227.\n\n``<div style=\"border:1px solid #3b4227;\">Content here</div>``\nCSS codes\n``.text {color:#3b4227;}``\n``.background {background-color:#3b4227;}``\n``.border {border:1px solid #3b4227;}``\n\n# Shades and Tints of #3b4227\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, #040403 is the darkest color, while #f9faf7 is the lightest one.\n\n• #040403\n``#040403` `rgb(4,4,3)``\n• #0f110a\n``#0f110a` `rgb(15,17,10)``\n• #1a1d11\n``#1a1d11` `rgb(26,29,17)``\n• #252918\n``#252918` `rgb(37,41,24)``\n• #303620\n``#303620` `rgb(48,54,32)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #464e2e\n``#464e2e` `rgb(70,78,46)``\n• #515b36\n``#515b36` `rgb(81,91,54)``\n• #5c673d\n``#5c673d` `rgb(92,103,61)``\n• #677344\n``#677344` `rgb(103,115,68)``\n• #72804b\n``#72804b` `rgb(114,128,75)``\n• #7d8c53\n``#7d8c53` `rgb(125,140,83)``\n• #88985a\n``#88985a` `rgb(136,152,90)``\nShade Color Variation\n• #92a363\n``#92a363` `rgb(146,163,99)``\n• #9baa6f\n``#9baa6f` `rgb(155,170,111)``\n• #a4b17c\n``#a4b17c` `rgb(164,177,124)``\n• #acb988\n``#acb988` `rgb(172,185,136)``\n• #b5c094\n``#b5c094` `rgb(181,192,148)``\n• #bdc7a1\n``#bdc7a1` `rgb(189,199,161)``\n• #c6cfad\n``#c6cfad` `rgb(198,207,173)``\n• #ced6b9\n``#ced6b9` `rgb(206,214,185)``\n• #d7ddc6\n``#d7ddc6` `rgb(215,221,198)``\n• #e0e4d2\n``#e0e4d2` `rgb(224,228,210)``\n• #e8ecde\n``#e8ecde` `rgb(232,236,222)``\n• #f1f3eb\n``#f1f3eb` `rgb(241,243,235)``\n• #f9faf7\n``#f9faf7` `rgb(249,250,247)``\nTint Color Variation\n\n# Tones of #3b4227\n\nA tone is produced by adding gray to any pure hue. In this case, #353633 is the less saturated color, while #4d6603 is the most saturated one.\n\n• #353633\n``#353633` `rgb(53,54,51)``\n• #373a2f\n``#373a2f` `rgb(55,58,47)``\n• #393e2b\n``#393e2b` `rgb(57,62,43)``\n• #3b4227\n``#3b4227` `rgb(59,66,39)``\n• #3d4623\n``#3d4623` `rgb(61,70,35)``\n• #3f4a1f\n``#3f4a1f` `rgb(63,74,31)``\n• #414e1b\n``#414e1b` `rgb(65,78,27)``\n• #435217\n``#435217` `rgb(67,82,23)``\n• #455613\n``#455613` `rgb(69,86,19)``\n• #475a0f\n``#475a0f` `rgb(71,90,15)``\n• #495e0b\n``#495e0b` `rgb(73,94,11)``\n• #4b6207\n``#4b6207` `rgb(75,98,7)``\n• #4d6603\n``#4d6603` `rgb(77,102,3)``\nTone Color Variation\n\n# Color Blindness Simulator\n\nBelow, you can see how #3b4227 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" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.54135025,"math_prob":0.651491,"size":3683,"snap":"2019-13-2019-22","text_gpt3_token_len":1675,"char_repetition_ratio":0.12530579,"word_repetition_ratio":0.011070111,"special_character_ratio":0.5661146,"punctuation_ratio":0.23756906,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.99083585,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2019-05-24T15:38:13Z\",\"WARC-Record-ID\":\"<urn:uuid:9d3a2e37-3b77-4d50-9bf7-9f7a749ae013>\",\"Content-Length\":\"36368\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:9de1dd86-28da-46bf-95a0-91a8dcd2c505>\",\"WARC-Concurrent-To\":\"<urn:uuid:bab0ed47-7f45-4336-b9ed-52886b279758>\",\"WARC-IP-Address\":\"178.32.117.56\",\"WARC-Target-URI\":\"https://www.colorhexa.com/3b4227\",\"WARC-Payload-Digest\":\"sha1:JNQJ6TNLCCVXPAD3JM24YHOQ4QSIXLZW\",\"WARC-Block-Digest\":\"sha1:6LQKVTUE2SE5JNPLJ7KYF42UK4XU37FM\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2019/CC-MAIN-2019-22/CC-MAIN-2019-22_segments_1558232257660.45_warc_CC-MAIN-20190524144504-20190524170504-00460.warc.gz\"}"}
https://openstax.org/books/college-algebra/pages/1-3-radicals-and-rational-exponents	[ "College Algebra\n\nCollege Algebra1.3 Radicals and Rational Exponents\n\n### Learning Objectives\n\nIn this section students will:\n• Evaluate square roots.\n• Use the product rule to simplify square roots.\n• Use the quotient rule to simplify square roots.\n• Add and subtract square roots.\n• Rationalize denominators.\n• Use rational roots.\n\nA hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1, and use the Pythagorean Theorem.\n\nFigure 1\n$a 2 + b 2 = c 2 5 2 + 12 2 = c 2 169 = c 2 a 2 + b 2 = c 2 5 2 + 12 2 = c 2 169 = c 2$\n\nNow, we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one.\n\n### Evaluating Square Roots\n\nWhen the square root of a number is squared, the result is the original number. Since$4 2 =16, 4 2 =16,$the square root of$16 16$is$4. 4.$The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root.\n\nIn general terms, if$a a$is a positive real number, then the square root of$a a$is a number that, when multiplied by itself, gives$a. a.$The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals$a. a.$The square root obtained using a calculator is the principal square root.\n\nThe principal square root of$a a$is written as$a . a .$The symbol is called a radical, the term under the symbol is called the radicand, and the entire expression is called a radical expression.", null, "### Principal Square Root\n\nThe principal square root of$a a$is the nonnegative number that, when multiplied by itself, equals$a. a.$It is written as a radical expression, with a symbol called a radical over the term called the radicand:$a . a .$\n\n### Q&A\n\nDoes$25 =±5? 25 =±5?$\n\nNo. Although both$5 2 5 2$and$(−5) 2 (−5) 2$are$25, 25,$the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is$25 =5. 25 =5.$\n\n### Example 1\n\n#### Evaluating Square Roots\n\nEvaluate each expression.\n\n1. $100 100$\n2. $16 16$\n3. $25+144 25+144$\n4. $49 − 81 49 − 81$\n\n### Q&A\n\nFor$25+144 , 25+144 ,$can we find the square roots before adding?\n\nNo.$25 + 144 =5+12=17. 25 + 144 =5+12=17.$This is not equivalent to$25+144 =13. 25+144 =13.$The order of operations requires us to add the terms in the radicand before finding the square root.\n\nTry It #1\n\nEvaluate each expression.\n\n1. $225 225$\n2. $81 81$\n3. $25−9 25−9$\n4. $36 + 121 36 + 121$\n\n### Using the Product Rule to Simplify Square Roots\n\nTo simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers into the product of two separate rational expressions. For instance, we can rewrite$15 15$as$3 ⋅ 5 . 3 ⋅ 5 .$We can also use the product rule to express the product of multiple radical expressions as a single radical expression.\n\n### The Product Rule for Simplifying Square Roots\n\nIf$a a$and$b b$are nonnegative, the square root of the product$ab ab$is equal to the product of the square roots of$a a$and$b. b.$\n\n$ab = a ⋅ b ab = a ⋅ b$\n\n### How To\n\nGiven a square root radical expression, use the product rule to simplify it.\n\n1. Factor any perfect squares from the radicand.\n3. Simplify.\n\n### Example 2\n\n#### Using the Product Rule to Simplify Square Roots\n\n1. $300 300$\n2. $162 a 5 b 4 162 a 5 b 4$\nTry It #2\n\nSimplify$50 x 2 y 3 z . 50 x 2 y 3 z .$\n\n### How To\n\nGiven the product of multiple radical expressions, use the product rule to combine them into one radical expression.\n\n1. Express the product of multiple radical expressions as a single radical expression.\n2. Simplify.\n\n### Example 3\n\n#### Using the Product Rule to Simplify the Product of Multiple Square Roots\n\n$12 ⋅ 3 12 ⋅ 3$\n\nTry It #3\n\nSimplify$50x ⋅ 2x 50x ⋅ 2x$assuming$x>0. x>0.$\n\n### Using the Quotient Rule to Simplify Square Roots\n\nJust as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. We can rewrite$5 2 5 2$as$5 2 . 5 2 .$\n\n### The Quotient Rule for Simplifying Square Roots\n\nThe square root of the quotient$a b a b$is equal to the quotient of the square roots of$a a$and$b, b,$where$b≠0. b≠0.$\n\n$a b = a b a b = a b$\n\n### How To\n\nGiven a radical expression, use the quotient rule to simplify it.\n\n1. Write the radical expression as the quotient of two radical expressions.\n2. Simplify the numerator and denominator.\n\n### Example 4\n\n#### Using the Quotient Rule to Simplify Square Roots\n\n$5 36 5 36$\n\nTry It #4\n\nSimplify$2 x 2 9 y 4 . 2 x 2 9 y 4 .$\n\n### Example 5\n\n#### Using the Quotient Rule to Simplify an Expression with Two Square Roots\n\n$234 x 11 y 26 x 7 y 234 x 11 y 26 x 7 y$\n\nTry It #5\n\nSimplify$9 a 5 b 14 3 a 4 b 5 . 9 a 5 b 14 3 a 4 b 5 .$\n\n### Adding and Subtracting Square Roots\n\nWe can add or subtract radical expressions only when they have the same radicand and when they have the same radical type such as square roots. For example, the sum of$2 2$and$3 2 3 2$is$4 2 . 4 2 .$However, it is often possible to simplify radical expressions, and that may change the radicand. The radical expression$18 18$can be written with a$2 2$in the radicand, as$3 2 , 3 2 ,$so$2 + 18 = 2 +3 2 =4 2 . 2 + 18 = 2 +3 2 =4 2 .$\n\n### How To\n\nGiven a radical expression requiring addition or subtraction of square roots, solve.\n\n### Example 6\n\nAdd$5 12 +2 3 . 5 12 +2 3 .$\n\nTry It #6\n\nAdd$5 +6 20 . 5 +6 20 .$\n\n### Example 7\n\n#### Subtracting Square Roots\n\nSubtract$20 72 a 3 b 4 c −14 8 a 3 b 4 c . 20 72 a 3 b 4 c −14 8 a 3 b 4 c .$\n\nTry It #7\n\nSubtract$3 80x −4 45x . 3 80x −4 45x .$\n\n### Rationalizing Denominators\n\nWhen an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator.\n\nWe know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical.\n\nFor a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is$b c , b c ,$multiply by$c c . c c .$\n\nFor a denominator containing the sum or difference of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is$a+b c , a+b c ,$then the conjugate is$a−b c . a−b c .$\n\n### How To\n\nGiven an expression with a single square root radical term in the denominator, rationalize the denominator.\n\n1. Multiply the numerator and denominator by the radical in the denominator.\n2. Simplify.\n\n### Example 8\n\n#### Rationalizing a Denominator Containing a Single Term\n\nWrite$2 3 3 10 2 3 3 10$in simplest form.\n\nTry It #8\n\nWrite$12 3 2 12 3 2$in simplest form.\n\n### How To\n\nGiven an expression with a radical term and a constant in the denominator, rationalize the denominator.\n\n1. Find the conjugate of the denominator.\n2. Multiply the numerator and denominator by the conjugate.\n3. Use the distributive property.\n4. Simplify.\n\n### Example 9\n\n#### Rationalizing a Denominator Containing Two Terms\n\nWrite$4 1+ 5 4 1+ 5$in simplest form.\n\nTry It #9\n\nWrite$7 2+ 3 7 2+ 3$in simplest form.\n\n### Using Rational Roots\n\nAlthough square roots are the most common rational roots, we can also find cube roots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number.\n\n#### Understanding nth Roots\n\nSuppose we know that$a 3 =8. a 3 =8.$We want to find what number raised to the 3rd power is equal to 8. Since$2 3 =8, 2 3 =8,$we say that 2 is the cube root of 8.\n\nThe nth root of$a a$is a number that, when raised to the nth power, gives$a. a.$For example,$−3 −3$is the 5th root of$−243 −243$because$(−3) 5 =−243. (−3) 5 =−243.$If$a a$is a real number with at least one nth root, then the principal nth root of$a a$is the number with the same sign as$a a$that, when raised to the nth power, equals$a. a.$\n\nThe principal nth root of$a a$is written as$a n , a n ,$where$n n$is a positive integer greater than or equal to 2. In the radical expression,$n n$is called the index of the radical.\n\n### Principal $nn$th Root\n\nIf$a a$is a real number with at least one nth root, then the principal nth root of$a, a,$written as$a n , a n ,$is the number with the same sign as$a a$that, when raised to the nth power, equals$a. a.$The index of the radical is$n. n.$\n\n### Example 10\n\n#### Simplifying nth Roots\n\nSimplify each of the following:\n\n1. $−32 5 −32 5$\n2. $4 4 ⋅ 1,024 4 4 4 ⋅ 1,024 4$\n3. $− 8 x 6 125 3 − 8 x 6 125 3$\n4. $8 3 4 − 48 4 8 3 4 − 48 4$\n\n### Try It #10\n\nSimplify.\n\n1. $−216 3 −216 3$\n2. $3 80 4 5 4 3 80 4 5 4$\n3. $6 9,000 3 +7 576 3 6 9,000 3 +7 576 3$\n\n#### Using Rational Exponents\n\nRadical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index$n n$is even, then$a a$cannot be negative.\n\n$a 1 n = a n a 1 n = a n$\n\nWe can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an nth root. The numerator tells us the power and the denominator tells us the root.\n\n$a m n = ( a n ) m = a m n a m n = ( a n ) m = a m n$\n\nAll of the properties of exponents that we learned for integer exponents also hold for rational exponents.\n\n### Rational Exponents\n\nRational exponents are another way to express principal nth roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is\n\n$a m n = ( a n ) m = a m n a m n = ( a n ) m = a m n$\n\n### How To\n\nGiven an expression with a rational exponent, write the expression as a radical.\n\n1. Determine the power by looking at the numerator of the exponent.\n2. Determine the root by looking at the denominator of the exponent.\n3. Using the base as the radicand, raise the radicand to the power and use the root as the index.\n\n### Example 11\n\n#### Writing Rational Exponents as Radicals\n\nWrite$343 2 3 343 2 3$as a radical. Simplify.\n\n### Try It #11\n\nWrite$9 5 2 9 5 2$as a radical. Simplify.\n\n### Example 12\n\n#### Writing Radicals as Rational Exponents\n\nWrite$4 a 2 7 4 a 2 7$using a rational exponent.\n\n### Try It #12\n\nWrite$x (5y) 9 x (5y) 9$using a rational exponent.\n\n### Example 13\n\n#### Simplifying Rational Exponents\n\nSimplify:\n\n1. $5( 2 x 3 4 )( 3 x 1 5 ) 5( 2 x 3 4 )( 3 x 1 5 )$\n2. $( 16 9 ) − 1 2 ( 16 9 ) − 1 2$\n\n### Try It #13\n\nSimplify$( 8x ) 1 3 ( 14 x 6 5 ). ( 8x ) 1 3 ( 14 x 6 5 ).$\n\n### Media\n\nAccess these online resources for additional instruction and practice with radicals and rational exponents.\n\n### 1.3 Section Exercises\n\n#### Verbal\n\n1.\n\nWhat does it mean when a radical does not have an index? Is the expression equal to the radicand? Explain.\n\n2.\n\nWhere would radicals come in the order of operations? Explain why.\n\n3.\n\nEvery number will have two square roots. What is the principal square root?\n\n4.\n\nCan a radical with a negative radicand have a real square root? Why or why not?\n\n#### Numeric\n\nFor the following exercises, simplify each expression.\n\n5.\n\n$256 256$\n\n6.\n\n$256 256$\n\n7.\n\n$4( 9+16 ) 4( 9+16 )$\n\n8.\n\n$289 − 121 289 − 121$\n\n9.\n\n$196 196$\n\n10.\n\n$1 1$\n\n11.\n\n$98 98$\n\n12.\n\n$27 64 27 64$\n\n13.\n\n$81 5 81 5$\n\n14.\n\n$800 800$\n\n15.\n\n$169 + 144 169 + 144$\n\n16.\n\n$8 50 8 50$\n\n17.\n\n$18 162 18 162$\n\n18.\n\n$192 192$\n\n19.\n\n$14 6 −6 24 14 6 −6 24$\n\n20.\n\n$15 5 +7 45 15 5 +7 45$\n\n21.\n\n$150 150$\n\n22.\n\n$96 100 96 100$\n\n23.\n\n$( 42 )( 30 ) ( 42 )( 30 )$\n\n24.\n\n$12 3 −4 75 12 3 −4 75$\n\n25.\n\n$4 225 4 225$\n\n26.\n\n$405 324 405 324$\n\n27.\n\n$360 361 360 361$\n\n28.\n\n$5 1+ 3 5 1+ 3$\n\n29.\n\n$8 1− 17 8 1− 17$\n\n30.\n\n$16 4 16 4$\n\n31.\n\n$128 3 +3 2 3 128 3 +3 2 3$\n\n32.\n\n$−32 243 5 −32 243 5$\n\n33.\n\n$15 125 4 5 4 15 125 4 5 4$\n\n34.\n\n$3 −432 3 + 16 3 3 −432 3 + 16 3$\n\n#### Algebraic\n\nFor the following exercises, simplify each expression.\n\n35.\n\n$400 x 4 400 x 4$\n\n36.\n\n$4 y 2 4 y 2$\n\n37.\n\n$49p 49p$\n\n38.\n\n$( 144 p 2 q 6 ) 1 2 ( 144 p 2 q 6 ) 1 2$\n\n39.\n\n$m 5 2 289 m 5 2 289$\n\n40.\n\n$9 3 m 2 + 27 9 3 m 2 + 27$\n\n41.\n\n$3 a b 2 −b a 3 a b 2 −b a$\n\n42.\n\n$4 2n 16 n 4 4 2n 16 n 4$\n\n43.\n\n$225 x 3 49x 225 x 3 49x$\n\n44.\n\n$3 44z + 99z 3 44z + 99z$\n\n45.\n\n$50 y 8 50 y 8$\n\n46.\n\n$490b c 2 490b c 2$\n\n47.\n\n$32 14d 32 14d$\n\n48.\n\n$q 3 2 63p q 3 2 63p$\n\n49.\n\n$8 1− 3x 8 1− 3x$\n\n50.\n\n$20 121 d 4 20 121 d 4$\n\n51.\n\n$w 3 2 32 − w 3 2 50 w 3 2 32 − w 3 2 50$\n\n52.\n\n$108 x 4 + 27 x 4 108 x 4 + 27 x 4$\n\n53.\n\n$12x 2+2 3 12x 2+2 3$\n\n54.\n\n$147 k 3 147 k 3$\n\n55.\n\n$125 n 10 125 n 10$\n\n56.\n\n$42q 36 q 3 42q 36 q 3$\n\n57.\n\n$81m 361 m 2 81m 361 m 2$\n\n58.\n\n$72c −2 2c 72c −2 2c$\n\n59.\n\n$144 324 d 2 144 324 d 2$\n\n60.\n\n$24 x 6 3 + 81 x 6 3 24 x 6 3 + 81 x 6 3$\n\n61.\n\n$162 x 6 16 x 4 4 162 x 6 16 x 4 4$\n\n62.\n\n$64y 3 64y 3$\n\n63.\n\n$128 z 3 3 − −16 z 3 3 128 z 3 3 − −16 z 3 3$\n\n64.\n\n$1,024 c 10 5 1,024 c 10 5$\n\n#### Real-World Applications\n\n65.\n\nA guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by evaluating$90,000+160,000 . 90,000+160,000 .$What is the length of the guy wire?\n\n66.\n\nA car accelerates at a rate ofwhere t is the time in seconds after the car moves from rest. Simplify the expression.\n\n#### Extensions\n\nFor the following exercises, simplify each expression.\n\n67.\n\n$8 − 16 4− 2 − 2 1 2 8 − 16 4− 2 − 2 1 2$\n\n68.\n\n$4 3 2 − 16 3 2 8 1 3 4 3 2 − 16 3 2 8 1 3$\n\n69.\n\n$m n 3 a 2 c −3 ⋅ a −7 n −2 m 2 c 4 m n 3 a 2 c −3 ⋅ a −7 n −2 m 2 c 4$\n\n70.\n\n$a a− c a a− c$\n\n71.\n\n$x 64y +4 y 128y x 64y +4 y 128y$\n\n72.\n\n$( 250 x 2 100 b 3 )( 7 b 125x ) ( 250 x 2 100 b 3 )( 7 b 125x )$\n\n73.\n\n$64 3 + 256 4 64 + 256 64 3 + 256 4 64 + 256$" ]	[ null, "https://openstax.org/resources/461131596b00446489b1c74e97b32bf1dfee7a7b", null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.8644621,"math_prob":0.99962723,"size":12067,"snap":"2020-10-2020-16","text_gpt3_token_len":2617,"char_repetition_ratio":0.21371135,"word_repetition_ratio":0.10313901,"special_character_ratio":0.1927571,"punctuation_ratio":0.09376422,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.999734,"pos_list":[0,1,2],"im_url_duplicate_count":[null,1,null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2020-04-10T05:54:50Z\",\"WARC-Record-ID\":\"<urn:uuid:c480c1da-8e12-4e97-89ca-10c9dc5ae00b>\",\"Content-Length\":\"383256\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:729aa3e9-a4db-4dc8-a6ba-ae3989e7bea0>\",\"WARC-Concurrent-To\":\"<urn:uuid:7946c37f-f525-450f-9d1f-327c2a7fee2e>\",\"WARC-IP-Address\":\"13.249.39.18\",\"WARC-Target-URI\":\"https://openstax.org/books/college-algebra/pages/1-3-radicals-and-rational-exponents\",\"WARC-Payload-Digest\":\"sha1:RUMX34PBRTYE2D3GTZZFPGLP2ELRLOE3\",\"WARC-Block-Digest\":\"sha1:WBEPS2S5CFZALDBOGS4LGV3MPA6IQFLK\",\"WARC-Identified-Payload-Type\":\"text/html\",\"warc_filename\":\"/cc_download/warc_2020/CC-MAIN-2020-16/CC-MAIN-2020-16_segments_1585371886991.92_warc_CC-MAIN-20200410043735-20200410074235-00245.warc.gz\"}"}
https://www.dasjs.com/rxjs-crash-course-do-and-filter-operator/	[ "# RxJS Crash Course – do and filter Operator\n\n0\n\n### .do Operator\n\nThe next operator we’re going to look at is do operator. This will allow us to execute code without affecting the underlying observable. So, we create observable of a couple of strings, and then we’ll print out their names as they first appear.\n\nThen we’ll transform them to uppercase and then we’ll print them again. When we actually run this code you can see that it gives us a glimpse into the observable data at any given point in time.\n\nExample of .do() function\n\nSo here we see the value and the upper case value.\n\n### .filter() operator\n\nNow let’s take a look at filter this does pretty much what you’d expect. You give it a condition and only values meeting that condition make it through. So, here we have an observable of a bunch of positive and negative numbers and we’ll filter it to only show the positive numbers.\n\nSo, the only value is greater than or equal to zero will be displayed and we reload the page and we only see the positive numbers.\n\nExample of .filter function\n\nIf we flip greater than to less than then we get only the negative numbers displayed using this same example.\n\n### .first() operator\n\nI’m going to show you the first and last operator the first operator will just take the very first element from the observable.\nIn this case, the first element is negative three so that’s what we see here on the screen.\n\n### .last() operator\n\nWe can also do the inverse of this with the last operator and negative two is the last element so we’ll see it updated to negative two.\n\nSee you at next session of the lecture.\n\n0" ]	[ null ]	{"ft_lang_label":"__label__en","ft_lang_prob":0.88117814,"math_prob":0.9297561,"size":1531,"snap":"2021-31-2021-39","text_gpt3_token_len":323,"char_repetition_ratio":0.13228552,"word_repetition_ratio":0.0,"special_character_ratio":0.21162638,"punctuation_ratio":0.08201893,"nsfw_num_words":0,"has_unicode_error":false,"math_prob_llama3":0.9681633,"pos_list":[0],"im_url_duplicate_count":[null],"WARC_HEADER":"{\"WARC-Type\":\"response\",\"WARC-Date\":\"2021-09-24T21:55:44Z\",\"WARC-Record-ID\":\"<urn:uuid:b77df818-5b5e-4cc4-9796-4f46e2091237>\",\"Content-Length\":\"87287\",\"Content-Type\":\"application/http; msgtype=response\",\"WARC-Warcinfo-ID\":\"<urn:uuid:d7bba82c-b45f-432c-9ccf-af7364677511>\",\"WARC-Concurrent-To\":\"<urn:uuid:07becac7-87b9-43bb-9c42-2381fb457913>\",\"WARC-IP-Address\":\"104.21.72.251\",\"WARC-Target-URI\":\"https://www.dasjs.com/rxjs-crash-course-do-and-filter-operator/\",\"WARC-Payload-Digest\":\"sha1:7X6TP7D4YYGIMKZZQUGPDRFAOG7BDVKD\",\"WARC-Block-Digest\":\"sha1:FEEVZYWSQ5AIJPC5LGHN7FMPL2YJTN3Q\",\"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-20210924201616-20210924231616-00044.warc.gz\"}"}