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https://math.stackexchange.com/questions/3467845/triangle-area-problem-from-kangaroo-ksa-grade-9-10 | 1,580,008,190,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00337.warc.gz | 552,312,579 | 31,980 | # Triangle area problem from kangaroo KSA grade 9,10
Let ABC be a triangle with area $$S$$ such as in the image $$D$$ is the midpoint of $$BC$$ $$AP=2*AB,AQ=3*AD,AR=4*AC$$ what is the area of $$PQR$$ in terms of $$S$$
So what I tried is heron's formula, and a bit of angle chasing, but didn't get a very useful thing. I tried with geogebra and I think the awnser is simply just $$S$$
• Interesting ObTask: If $AD:DB=1:2$ show that $Q$ lies on the line $PR$. – Michael Hoppe Dec 8 '19 at 17:29
$$S_{ABD}=S_{ACD}=\frac{1}{2} S$$
Using the formula $$S_{XYZ} = \frac{1}{2} XY\cdot XZ \cdot sin(YXZ)$$, we get $$S_{QAP} = \frac{AP}{AB} \frac{AQ}{AD} S_{DAB} = 3 S$$ $$S_{QAR} = \frac{AQ}{AD} \frac{AR}{AC} S_{CAD} = 6 S$$ $$S_{APR} = \frac{AP}{AB} \frac{AR}{AC} S_{ABC} = 8 S$$ Now, $$S_{PQR} = S_{QAP}+S_{QAR}-S_{APR} = S$$
• That was my solution, too. – Michael Hoppe Dec 8 '19 at 11:51
Let X, Y, Z be the division points of RA. RQ is extended to meet YP at W. T, V are the division points of QA. ZV extended will meet the line through C parallel to BZ at U.
After adding the above, we find (1) UVDC, VZBD and UZBC are //gms; and (2) W(Q)RT(P) and A(U)P(T)Y(Z) are triangles meeting the midpoint theorem requirements. [The proofs of the above will be skipped.]
[PQR] = [red] = 0.5 [pink] = 0.5 [grey] = 0.5 [green] = [orange] = [ABC] = S.
A general solution:
Let$$AP=k_1AB$$, $$AQ=k_2AD$$ $$AR=k_3AC$$, and for $$0 let $$BD=tBC$$. (In our special case we have $$t=1/2$$.)
From the well known formula for the area $$F$$ of a triangle, $$F=\frac12ab\sin(\gamma)$$, we derive that \begin{align} F(AQP)&=k_1k_2F(ABD)\\ F(AQR)&=k_2k_3F(ADC)\\ F(APR)&=k_1k_3F(ABC). \end{align} Furthermore, $$F(ABD)=tF(ABC)$$ and $$F(ADC)=(1-t)F(ABC)$$. Putting that together we have \begin{align} F(PQR)&=F(AQP)+F(AQR)-F(APR)\\ &=k_1k_2tF(ABC)+k_2k_3(1-t)F(ABC)-k_1k_3F(ABC), \end{align} that is
$$\frac{F(PQR)}{F(ABC)}=tk_1k_2+(1-t)k_2k_3-k_1k_3.$$
(That ratio may be negative, in which case the orientation of $$PQR$$ changes, but that doesn't really matter.)
Now one might investigate some special cases , for example $$t=1/2$$ and $$k_1$$, $$k_2$$ and $$k_3$$ are in arithmetic progression, that is $$k_2=k_1+d$$ and $$k_3=k_1+2d$$. A short computation shows that the ratio is $$d^2$$, independent of $$k_1$$. That solves the original problem since $$k_1=2$$ (which is irrelevant anyway) and $$d=1$$.
Or, given $$t=1/2$$ again, the ratio vanishes if $$k_2$$ is the harmonic mean of $$k_1$$ and $$k_2$$.
A last one. Given $$k_2=k_1q$$ and $$k_2=k_1q^2$$ for positive $$q$$ the area of $$PQR$$ vanishes if $$t=\frac{q}{1+q}$$. | 1,028 | 2,621 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 46, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2020-05 | latest | en | 0.832692 |
https://termpapermaster.com/explain-why-you-made-your-decision-including-the-results-for-your-p-value-and-the-critical-value/ | 1,643,283,851,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305260.61/warc/CC-MAIN-20220127103059-20220127133059-00210.warc.gz | 593,399,786 | 13,411 | APA Format/ 3 Pages (I will provide the assignments from phase 1,2, and 3)
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Test the claim using α = 0.05 and assume your data is normally distributed and σ is unknown. | 712 | 3,563 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-05 | latest | en | 0.904455 |
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# Expressing Fractions as Whole & Whole Numbers as Fractions pgs. 47-50 (CCSS)
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Fractions: Expressing Fractions as Whole Numbers & Whole Numbers as Fractions
Are you ready for more fractions? We thought you might be so take a gander, won't you?
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express whole numbers as fractions. They'll get tons of practice and with your help, they'll master these skills in no time.
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\$1.50 | 1,370 | 5,280 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2018-05 | latest | en | 0.928071 |
https://biggboss3.net/qa/question-what-does-a-letter-after-a-number-mean.html | 1,606,780,400,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141515751.74/warc/CC-MAIN-20201130222609-20201201012609-00356.warc.gz | 213,606,170 | 8,129 | # Question: What Does A Letter After A Number Mean?
## What do you do when a variable is next to a number?
Any time you see a number and a variable, or two or more variables, placed right next to each other like this, it means that the number and the variable, or the many variables, are to be multiplied together.
When multiplying, we use exponents..
## What does letters mean in math?
In algebraic expressions, letters represent variables. These letters are actually numbers in disguise. In this expression, the variables are x and y. We call these letters “variables” because the numbers they represent can vary—that is, we can substitute one or more numbers for the letters in the expression.
## What does the letter T symbolize?
It is derived from the Semitic letters taw (ת, ܬ, ت) via the Greek letter τ (tau). In English, it is most commonly used to represent the voiceless alveolar plosive, a sound it also denotes in the International Phonetic Alphabet.
## What does R mean in math?
The set of real numbers is denoted using the symbol R or . Real numbers can be thought of as points on an infinitely long line called the number line or real line, where the points corresponding to integers are equally spaced.
## What is the U called in math?
more … The set made by combining the elements of two sets. So the union of sets A and B is the set of elements in A, or B, or both. The symbol is a special “U” like this: ∪
## What number does each letter represent?
Letters in the alphabet:Letter NumberLetter1A2B3C4D22 more rows
## What does letter I symbolize?
The Letter I is also the 9th Letter of The Alphabet and Is Symbolic that upcoming changes and blessings will be in regard to your purpose and Intentions. Do take care to Continue your positive thoughts and affirmations and actions. … You hold the power to change and renounce negative outcomes or thought patterns.
## What does a subscript of 0 mean?
Commonly, variables with a zero in the subscript are referred to as the variable name followed by “naught” (e.g. v0 would be read, “v-naught”). Subscripts are often used to refer to members of a mathematical sequence or set or elements of a vector.
## What letter can 6 Replace?
GHowever, in some countries, the number ‘6’ can be used to replace the letter ‘G’. This is simply because it appears similar. Then, there is also the letter ‘C’ which shares the shape of the number ‘6’.
## What number looks most like?
4The number ‘4’ is a favourite number for those who want an ‘A’ look alike on their number plates.
## What does equation mean?
An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an ‘equals’ sign. For example: 12.
## What numbers can you use as letters?
Here’s a list of the most common occasions when a number may be used to represent a letter (or the other way around).A >> 4.B >> 8 6 3 13.C >> 6 0 O.D >> 0 O.E >> 3.F >>G >> 8 9.H >> 4 11.More items…
## What does subscript mean in maths?
A small letter or number placed slightly lower than the normal text. Examples: • the number 1 here: A1 (pronounced “A sub 1” or just “A 1”) • the letter m here: xm (pronounced “x sub m” or just “x m”)
## What comes first variable or number?
The mathematical convention (the usual way of doing things) is to write the number before the variable when multiplying numbers by variables. In other words, we write 3x, not x3. If you do write x3 people will probably know what you mean, but you probably won’t be invited back to the convention.
## How do you work out variables?
Calculate total variable cost by multiplying the cost to make one unit of your product by the number of products you’ve developed. For example, if it costs \$60 to make one unit of your product, and you’ve made 20 units, your total variable cost is \$60 x 20, or \$1,200.
## What does a letter next to a number mean?
A Variable is a symbol for a number we don’t know yet. It is usually a letter like x or y. A number on its own is called a Constant. A Coefficient is a number used to multiply a variable (4x means 4 times x, so 4 is a coefficient)
## What does 831 mean?
I Love You831 means “I Love You.” The number 831 is a cyber term, used to mean “I Love You.” Each individual number in 831 has a specific definition: 8 = The total number of letters in the phrase “I Love You.” 3 = The total number of words in the phrase “I Love You.” 1 = The one meaning of the cyber term 831.
## What letter can 7 Replace?
The character “!” replaces the letter L, “3” poses as a backwards letter E, and “7” is the letter T, etc. Other examples of character/letter replacement include using “8” for the letter B, “9” for G, and the number 0 for the letter O. | 1,155 | 4,744 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2020-50 | latest | en | 0.903247 |
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p. 2-11 Example (The Matching Problem). Applications: (a) Taste Testing. (b) Gift Exchange. Let be all permutations = ( i 1 , …, i n ) of 1, 2, …, n . Thus, # = n !. Let A j = { : i j = j } and , Q : P ( A )=? (What would you expect when n is large?) By symmetry, Here, σ k = n k P ( A 1 ··· A k ) , A = n i =1 A i for k = 1, …, n . p. 2-12 Proportion: If A 1 , A 2 , …, is a partition of , i.e., 1. 2. A 1 , A 2 , …, are mutually exclusive, then, for any event A , P ( A )= i =1 P ( A A i ) . i =1 A i = , So, Note : approximation accurate to 3 decimal places if n 6. P ( A σ 1 σ 2 + +( 1) n +1 σ n = n k =1 ( 1) k +1 1 k ! , P ( A )=1 n k =0 ( 1) k 1 k ! 1 1 e P ( A c ) e 1
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p. 2-13 • Monotone Sequences Q : How to define probability in a continuous sample space? Definition: A sequence of events A 1 , A 2 , …, is called increasing if and decreasing if The limit of an increasing sequence is defined as and the limit of an decreasing sequence is A 1 A 2 ··· A n A n +1 A
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Ask a homework question - tutors are online | 496 | 1,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2018-05 | latest | en | 0.813624 |
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HOME > Mathematics > Anybody know how to solve this? 5x+2y>-10
# Anybody know how to solve this? 5x+2y>-10
[From: ] [author: ] [Date: 11-10-07] [Hit: ]
-5), with a slope of -2.5, and then shade in everything above it.......
The answer is supposed to be in y=mx+b form.
-
2y > -10 - 5x
y > -5 - 5/2 x
y > -(5/2) x - 5
The solution is the area that lies above the line y = -(5/2)x - 5, which is a line with a y-intercept of -5, and a slope of -5/2
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Yes ∃ someone such that ∀ function inequality problems such as 5x+2y>-10 they can find and graph the solution. I can prove this to you, I am that someone.
Lol, I'm just messing with you--I'm a math major and the way you worded this question, gramatically and logically, you're just asking if there is someone out there that can solve the problem, and of course there is. Btw, in advanced math speak, ∃=there exists, ∀=for all.
Anyway, here's my solution:
5x + 2y > -10
2y > -5x - 10
y > -2.5x - 5
To graph that, just draw a dashed line starting at (0,-5), with a slope of -2.5, and then shade in everything above it.
-
you cant get an inequality to turn into an equation and even if you meant to put it on a GRID then you need 2 equations
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5x + 2y > -10
2y > -5x - 10
y > (-5/2)x - 5
I don't really get what you're asking but that is how you solve for y
-
5x+2y>-10
2y>-5x-10
y>-5x/2-5
1
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https://brighterly.com/math/properties-of-addition/ | 1,726,314,910,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651579.22/warc/CC-MAIN-20240914093425-20240914123425-00188.warc.gz | 127,439,335 | 16,591 | # Properties of Addition – Definition, FAQs, Examples
Welcome to Brighterly, where we strive to illuminate the world of mathematics for our bright young learners. Today, we’re embarking on a colorful journey into the heart of arithmetic: the properties of addition. We’ll be your guides, unwrapping the mysteries of numbers and equations, transforming them into fun and digestible pieces of knowledge. Let’s start by asking ourselves, “What is addition?”
Addition is a fundamental pillar in the majestic architecture of mathematics. It’s the process of bringing numbers together, uniting them to form a larger number. Imagine you’re on a treasure hunt, collecting diamonds. Each diamond you find, you add to your bag. The total diamonds in your bag at any point is the sum of your addition. For instance, if you find 3 diamonds in one spot and 2 in another, you have a total of 5 diamonds (3 + 2 = 5). Addition is everywhere, from counting candies to calculating scores in a video game. It’s the first step we take into the vast universe of numbers. But there’s a fascinating framework that supports and guides the process of addition – the properties of addition.
## What Is Addition?
Addition is one of the fundamental operations in mathematics, and it’s something we use every day! Whether we are counting apples, adding up scores in a game, or calculating total expenses, we are using addition. In the simplest terms, addition is the process of combining two or more numbers to get a sum. For instance, if we add 2 and 3, we get 5 (2 + 3 = 5). It’s like putting together groups of objects and counting the total. But there’s more to addition than just putting numbers together. There are unique properties that make addition work the way it does. Let’s find out more about these.
## What are the Properties of Addition?
The properties of addition are rules that all numbers naturally follow. These are fundamental truths in mathematics, and they are consistent no matter what numbers you’re working with. These properties are not only crucial in understanding how addition works, but they also help us perform computations more efficiently and solve problems more easily.
## What are the Five Properties of Addition?
There are five basic properties of addition that govern the way numbers behave when added together: Closure, Commutative, Associative, Identity, and Inverse. These properties make addition predictable and reliable, allowing us to solve even complex mathematical problems with confidence.
## Closure Property of Addition
The Closure Property of Addition states that when you add two real numbers, the result is always a real number. For instance, if you add 4 and 3 (4 + 3), the answer is 7, which is still a real number. This means that the sum of any two numbers in a set is still part of the same set.
## Commutative Property of Addition
The Commutative Property of Addition tells us that the order in which we add numbers doesn’t change the sum. So, if you have two numbers, say 5 and 2, adding 5 + 2 gives the same result as adding 2 + 5. This property makes addition flexible and easy to work with, especially when dealing with multiple numbers.
## Associative Property of Addition
According to the Associative Property of Addition, when adding three or more numbers, the way the numbers are grouped doesn’t change the sum. So, if you’re adding 3 + (4 + 5), you’ll get the same result as (3 + 4) + 5. This property allows us to rearrange numbers when adding, which can often make calculations easier.
The Additive Identity Property of Addition states that if you add zero (0) to any number, the number stays the same. For example, if you add zero to 7 (7 + 0), the result is still 7. The number zero is called the “additive identity” because adding it doesn’t change the identity of the other number.
The Additive Inverse Property of Addition states that every number has an inverse that, when added to the original number, results in zero. For instance, the additive inverse of 6 is -6 because when you add them (6 + -6), the sum is zero. This property is vital in understanding concepts like subtraction and negative numbers.
## Solved Examples On Properties of Addition
Unfolding the magic of the properties of addition, let’s apply them in real-life scenarios with a dash of creativity and fun.
1. Closure Property: Let’s say our Brighterly adventurer, Max, has 7 marbles and finds 4 more during his exploration. How many marbles does he have now? Using addition, we add 7 and 4 to get 11 (7 + 4 = 11), which is also a real number. So, we see the closure property in action!
2. Commutative Property: Now, Max and his friend, Mia, decide to count their marble collections. Max has 5 marbles and Mia has 3. Whether they count Max’s marbles first and then Mia’s (5 + 3) or vice versa (3 + 5), they still end up with 8 marbles. This demonstrates the commutative property!
3. Associative Property: On another adventure, Max, Mia, and their friend, Leo, find some shiny pebbles. Max finds 2, Mia finds 3, and Leo finds 4. Regardless of how they group their pebbles when counting—(Max’s + Mia’s) + Leo’s or Max’s + (Mia’s + Leo’s)—the total is always 9. This is the associative property in action!
4. Additive Identity Property: Consider a day when Max doesn’t find any new marbles. He still has his previous collection of 7 marbles. Adding zero (his new finds) to 7 (his collection) still gives us 7. This demonstrates the additive identity property, with zero maintaining the identity of the initial number.
5. Additive Inverse Property: Imagine Max and Mia are playing a game. For every point Max scores, Mia loses a point and vice versa. If Max scores 6 points (+6) and Mia loses 6 points (-6), the total change in points is zero (+6 + -6 = 0), illustrating the additive inverse property.
## Practice Problems On Properties of Addition
Time to put your skills to the test with these practice problems! Remember, understanding the problem and identifying the right property are the first steps to finding the solution.
1. Show the closure property with your favorite numbers.
2. Using the numbers 4, 2, and 7, demonstrate the commutative property.
3. Illustrate the associative property with the numbers 5, 10, and 15.
4. Show the additive identity property using the number 9.
5. Display the additive inverse property with the number -5 and its inverse.
## Conclusion
In the grand adventure of mathematics, understanding the properties of addition is like acquiring a magical map. It guides you through the labyrinth of numbers, illuminating the path to the right solutions. These properties are your secret weapons, simplifying complex problems and enhancing your mathematical fluency.
At Brighterly, we believe in making this journey a joyful and enlightening one. We aim to transform the often-dreaded monster of math into a friendly guide, helping you navigate the fascinating landscapes of numbers, equations, and functions. Remember, every mathematical concept you master is a stepping stone to a brighter future. The properties of addition are just the beginning. Stay curious, keep practicing, and unlock the countless wonders of mathematics.
In the world of Brighterly, math isn’t just fun – it’s an exciting adventure, a magical journey, and a treasure hunt all rolled into one! So grab your explorer’s hat, put on your brightest smile, and step into the world of addition with confidence and joy!
## Frequently Asked Questions On Properties of Addition
In this section, we’ll answer some of the most commonly asked questions about the properties of addition. These responses will help clarify concepts and solidify your understanding.
### What is the Closure Property of Addition?
The Closure Property of Addition states that the sum of any two real numbers is always a real number. For example, if you add 3 and 5, the result is 8, which is also a real number. This property is called “closure” because the operation (in this case, addition) “closes” the numbers within the set of real numbers – the result doesn’t stray outside this set. This property is crucial in mathematics because it assures us that we’ll get a valid number that we can continue to work with whenever we perform addition.
### Why is the Commutative Property of Addition important?
The Commutative Property of Addition is essential because it gives us flexibility when adding numbers. This property says that changing the order of numbers doesn’t affect the sum. For example, 7 + 4 equals 4 + 7. This property can be particularly useful when performing mental math or when dealing with large numbers as it allows us to rearrange numbers to make addition easier.
### How does the Associative Property of Addition help in calculations?
The Associative Property of Addition can make calculations easier by allowing us to group numbers differently without changing the sum. For instance, if we are adding 1 + 2 + 3, we can group 1 and 2 first and then add 3, or we can group 2 and 3 first and then add 1. Both will yield the same result (6). This property is handy when dealing with large numbers or complex arithmetic expressions.
### What does the Additive Identity Property of Addition mean?
The Additive Identity Property of Addition refers to the fact that when you add zero to any number, the result is the number itself. In other words, zero is the “identity” for the addition operation because adding zero to a number doesn’t change its “identity.” For example, 5 + 0 equals 5. Understanding this property is crucial because it forms the basis for many mathematical principles and operations.
Sources: | 2,106 | 9,657 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2024-38 | latest | en | 0.906354 |
http://psychology.wikia.com/wiki/Statistical_dispersion?oldid=89916 | 1,462,161,155,000,000,000 | text/html | crawl-data/CC-MAIN-2016-18/segments/1461860122420.60/warc/CC-MAIN-20160428161522-00145-ip-10-239-7-51.ec2.internal.warc.gz | 233,697,441 | 20,573 | # Statistical dispersion
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In statistics, statistical dispersion (also called statistical variability or variation) is variability or spread in a variable or a probability distribution. Common examples of measures of statistical dispersion are the variance, standard deviation and interquartile range.
Dispersion is contrasted with location or central tendency, and together they are the most used properties of distributions.
## Measures of statistical dispersionEdit
A measure of statistical dispersion is a real number that is zero if all the data are identical, and increases as the data becomes more diverse. It cannot be less than zero.
Most measures of dispersion have the same scale as the quantity being measured. In other words, if the measurements have units, such as metres or seconds, the measure of dispersion has the same units. Such measures of dispersion include:
These are frequently used (together with scale factors) as estimators of scale parameters, in which capacity they are called estimates of scale.
All the above measures of statistical dispersion have the useful property that they are location-invariant, as well as linear in scale. So if a random variable X has a dispersion of SX then a linear transformation Y = aX + b for real a and b should have dispersion SY = |a|SX.
Other measures of dispersion are dimensionless (scale-free). In other words, they have no units even if the variable itself has units. These include:
There are other measures of dispersion:
Some measures of dispersion have specialized purposes, among them the Allan variance and the Hadamard variance.
For categorical variables, it is less common to measure dispersion by a single number. See qualitative variation. One measure which does so is the discrete entropy.
## Sources of statistical dispersionEdit
In the physical sciences, such variability may result only from random measurement errors: instrument measurements are often not perfectly precise, i.e., reproducible. One may assume that the quantity being measured is unchanging and stable, and that the variation between measurements is due to observational error.
In the biological sciences, this assumption is false: the variation observed might be intrinsic to the phenomenon: distinct members of a population differ greatly. This is also seen in the arena of manufactured products; even there, the meticulous scientist finds variation.
The simple model of a stable quantity is preferred when it is tenable. Each phenomenon must be examined to see if it warrants such a simplification. | 508 | 2,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-18 | latest | en | 0.939417 |
https://dnmtechs.com/are-0-and-0-equivalent/ | 1,701,306,489,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100164.15/warc/CC-MAIN-20231130000127-20231130030127-00674.warc.gz | 253,082,731 | 21,840 | # Are +0 and -0 Equivalent?
## Are +0 and -0 Equivalent?
When it comes to numbers, zero is often considered a neutral element. It sits between the positive and negative numbers, acting as a reference point for numerical values. However, in some cases, zero can exhibit peculiar behavior. One such example is the existence of both +0 and -0. At first glance, they may seem identical, but are they truly equivalent? Let’s delve deeper into this intriguing topic and explore the nuances of +0 and -0.
### The Concept of Zero
Zero, as a mathematical concept, represents the absence or lack of quantity. It is neither positive nor negative, making it a unique number in the numerical system. However, in the realm of computer science and floating-point arithmetic, zero can be further divided into positive zero (+0) and negative zero (-0). This division may seem counterintuitive, but it has its reasons.
### Understanding Floating-Point Arithmetic
In computer science, floating-point arithmetic is a method used to represent and perform calculations on real numbers. It involves expressing numbers in a scientific notation-like format, consisting of a sign, a significand (also known as a mantissa), and an exponent. This format allows for a wide range of values to be represented, including very large and very small numbers.
When it comes to floating-point arithmetic, zero can be represented in two ways: +0 and -0. These representations are used to differentiate between positive and negative zero, which may have implications in certain calculations and algorithms.
### Significance in Division and Limits
One area where the distinction between +0 and -0 becomes apparent is in division. When dividing a positive number by zero, the result is typically considered as positive infinity (+∞). However, when dividing a negative number by zero, the result is negative infinity (-∞). This distinction can be attributed to the presence of +0 and -0.
Additionally, the concept of limits in calculus also sheds light on the difference between +0 and -0. When evaluating the limit of a function as it approaches zero from the positive side, it is denoted as approaching +0. Conversely, when approaching zero from the negative side, it is denoted as approaching -0. This distinction helps in analyzing the behavior of functions and their limits.
### Practical Applications and Considerations
While the distinction between +0 and -0 may seem abstract, it has practical implications in certain fields. For example, in computer graphics, the direction of a vector can be determined by the sign of zero. This allows for more precise calculations and rendering of graphical elements.
However, it is important to note that most programming languages and applications treat +0 and -0 as equivalent. They are considered interchangeable in most scenarios, and their distinction is not crucial for everyday programming tasks. It is only in specific cases, such as numerical analysis or low-level programming, where the difference between +0 and -0 becomes significant.
```// Example JavaScript code illustrating the behavior of +0 and -0
console.log(1 / +0); // Output: Infinity
console.log(1 / -0); // Output: -Infinity
```
### Conclusion
While +0 and -0 may seem identical at first glance, they exhibit distinct behavior in certain mathematical operations and calculations. The division of positive and negative zero allows for more precise analysis and differentiation in specific scenarios. However, for most practical purposes, +0 and -0 can be considered equivalent. It is only in specialized fields and situations where their distinction becomes significant. Understanding the nuances of +0 and -0 can provide a deeper insight into the intricacies of numbers and their representation in computer science. | 750 | 3,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-50 | longest | en | 0.938368 |
https://artofproblemsolving.com/wiki/index.php/2008_iTest_Problems/Problem_3 | 1,701,692,001,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100529.8/warc/CC-MAIN-20231204115419-20231204145419-00824.warc.gz | 136,334,876 | 11,548 | # 2008 iTest Problems/Problem 3
## Problem
Michael plays catcher for his school's baseball team. He has always been a great player behind the plate, but this year as a junior, Michael's offense is really improving. His batting average is $.417$ after six games, and the team is $6-0$ (six wins and no losses). They are off to their best start in years.
On the way home from their sixth game, Michael notes to his father that the attendance seems to be increasing due to the team's great start, "There were $181$ people at the first game, then $197$ at the second, $203$ the third, $204$ the fourth, $212$ at the fifth, and there were $227$ at today's game." Just then, Michael's genius younger brother Tony, just seven-years-old, computes the average attendance of the six games. What is their average?
## Solution
The average of $n$ numbers is the sum of $n$ numbers divided by $n.$ Doing the following gives the answer as $\frac{181+197+203+204+212+227}{6}=\boxed{204}$. | 249 | 977 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 12, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2023-50 | latest | en | 0.981194 |
http://tasks.illustrativemathematics.org/content-standards/HSF/LE/A/2/tasks/638 | 1,701,425,558,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100286.10/warc/CC-MAIN-20231201084429-20231201114429-00094.warc.gz | 42,274,529 | 10,626 | # Snail Invasion
Alignments to Content Standards: F-LE.A.2 F-LE.A.4
In 1966, a Miami boy smuggled three Giant African Land Snails into the country. His grandmother eventually released them into the garden, and in seven years there were approximately 18,000 of them. The snails are very destructive and had to be eradicated. According to the USDA, it took 10 years and cost \$1 million to eradicate them. 1. Assuming the snail population grows exponentially, write an expression for the population,$P$, in terms of the number,$t$, of years since their release. 2. How long does it take for the population to double? 3. Assuming the cost of eradicating the snails is proportional to the population, how much would it have cost to eradicate them if 1. they had started the eradication program a year earlier? 2. they had let the population grow unchecked for another year? ## IM Commentary The purpose of this task is to give students experience modeling a real-world example of exponential growth, in a context that provides a vivid illustration of the power of exponential growth, for example the cost of inaction for a year. There is an opportunity for further discussion based on part (c), since the ratio of costs from one year to the next is the same in each part. Although the task doesn't point this out explicitly, the teacher can bring it out in discussion, and it illustrates a key property of exponential growth. Students might choose to model exponential growth using either$P(t)=ae^{rt}$or$P(t)=ab^t$. Each has its advantages and disadvantages. The base$b$is the factor by which the population increases each year, and so is useful in part (c). And this interpretation of$b$yields directly the fact that$3b^7 = 18,\!000$, so that it can be calculated without logarithms. On the other hand, students might be more confortable working with base$e$and fitting$P(t) = a e^{rt}$to two data points, without engaging initially in the reasoning above. Depending on the goal that a teacher has in mind when using this task, he or she might suggest using one form or the other, or might use the opportunity to compare results when different students use different methods. ## Solutions Solution: Solution using base$e$1. Since it is assumed that the growth is exponential, we write$P(t)=ae^{rt}$for some constants$a$and$r$. We are given two data points, namely that$P(0)=3$and$P(7) =18,000$. (This last value is only an approximation.) From the first data point we get$3 = P(0) =ae^{0t}=a$, so we must have$a=3$. Then$P(7)=18,000$gives $$18,000=3e^{r(7)}.$$ Dividing both sides by 3 and rewriting the equation based on the definition of natural logarithm gives$7r=\ln(6000)$, so$r=\frac{\ln(6000)}{7}\approx 1.24$. So$P(t)=3e^{1.24t}$for$0 \le t \le 7$. We use this function to model the population, bearing in mind that it is based on approximate data. 2. The doubling time is the same no matter what the starting value. It takes the same time for the population to double from 3 to 6, then from 6 to 12, and so on. We calculate the time for the population to double from 3 to 6. So we want to find$t$such that $$6=3e^{1.24t}.$$ Dividing both sides by 3 and using the definition of natural logarithm to rewrite the equation gives$\ln(2)=1.24 t$, so$t\approx 0.56$. It takes just over half a year for the population to double. 3. 1. After 6 years the population is $$P(6) = 3e^{1.24(6)}\approx 5100.$$ If it costs \$1 million to clean up 18,000 snails, then it would cost $\frac{5100}{18000}\times \$1\text{ million}\approx \$280,000$ to clean up 5100 snails. This makes sense since the number of snails doubles almost two times per year, so the population (and the cost of dealing with it) should be a little more than $\frac14$ what it is in the 7th year.
2. After 8 years the population is $$P(8) = 3e^{1.24(8)}\approx 61000.$$ So, since it costs \$1 million to clean up 18,000 snails, it would cost$\frac{61000}{18000}\times \$1\text{ million}\approx \$ 3,400,000$to clean up 61,000 snails. This makes sense since the number of snails doubles almost two times per year, so the number of snails (and the cost of dealing with them) should be a little less than 4 times what it is in the 7th year. Solution: Solution that does not use logarithms 1. Since it is assumed that the growth is exponential, we write$P(t)=ab^t$for some constants$a$and$b$. We are given two data points, namely that$P(0)=3$and$P(7) =18,000$. (This last value is only an approximation.) From the first data point we get$3 = P(0) =ab^0=a$, so we must have$a=3$. Then$P(7)=18,000$gives $$18,000=3b^7.$$ Dividing both sides by 3 and raising both sides to the power of$\frac{1}{7}$gives$b=6000^{\frac{1}{7}}$, so$b\approx 3.47$. So$P(t)=3(3.47)^t$for$0 \le t \le 7$. We use this function to model the population, bearing in mind that it is based on approximate data. 2. The doubling time is the same no matter what the starting value. It takes the same time for the population to double from 3 to 6, then from 6 to 12, and so on. We calculate the time for the population to double from 3 to 6. So we want to find$t$such that $$6=3(3.47)^t.$$ Dividing both sides by 3 gives$2=3.47^t$. Without logarithms, we can solve by graphing$y = 2$and$y = 3.47^t$on the same$xy$-plane and finding$t$where the graphs intersect. We find$t\approx 0.56$. It takes just over half a year for the population to double. 3. 1. After 6 years the population is $$P(6) = 3(3.47)^6\approx 5100.$$ If it costs \$1 million to clean up 18,000 snails, then it would cost $\frac{5100}{18000}\times \$1\text{ million}\approx \$280,000$ to clean up 5100 snails. This makes sense since the number of snails doubles almost two times per year, so the population (and the cost of dealing with it) should be a little more than $\frac14$ what it is in the 7th year.
2. After 8 years the population is $$P(8) = 3(3.47)^8\approx 61000.$$ So, since it costs \$1 million to clean up 18,000 snails, it would cost$\frac{61000}{18000}\times \$1\text{ million}\approx \$ 3,400,000\$ to clean up 61,000 snails. This makes sense since the number of snails doubles almost two times per year, so the number of snails (and the cost of dealing with them) should be a little less than 4 times what it is in the 7th year. | 1,767 | 6,240 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2023-50 | latest | en | 0.956396 |
http://www.schoollamp.com/class-10-tenth/cbse-10th-maths-previous-year-board-papers-solutions-2008-delhi-set-1-84 | 1,600,927,245,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00610.warc.gz | 206,320,055 | 15,360 | # 10th Maths Paper Solutions Set 1 : CBSE Delhi Previous Year 2008
General Instructions:
1. All questions are compulsory.
2. The question paper consists of 30 questions divided into four sections - A, B, C and D. Section A comprises of ten questions of 1 mark each, Section B comprises of five questions of 2marks each, Section C comprises of ten questions of 3 marks each and Section D comprises of five questions of 6marks each.
3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question.
There is no overall choice. However, an internal choice has been provided in one question of 2 marks each, three questions of 3 marks each and two questions of 6 marks each. You have to attempt only one of the alternatives in all such questions.
4. In question on construction, the drawing should be neat and as per the given measurements.
5. Use of calculators is not permitted.
Q1 :
Complete the missing entries in the following factor tree:
Let the missing numbers be x and y.
From the above factor tree, it is clear that:
y = 3 × 7 = 21
And, x = 2 × y = 2 × 21 = 42
Q2 :
If (x + a) is a factor of 2x2 + 2ax + 5x + 10, find a.
If x +a is a factor of 2x2 + 2ax + 5x + 10, then the remainder i.e., 10 − 5a = 0 ⇒ a = 2.
Q3 :
Show that x = −3 is a solution of x2 + 6x + 9 = 0.
L.H.S = x2 + 6x + 9 = (− 3)2 + 6 (− 3) + 9 ( x = 3)
= 9 − 18 + 9
= 0 = R.H.S
This proves that x = − 3 is a solution of x2 + 6x + 9 = 0.
Q4 :
The first term of an A.P. is p and its common difference is q. Find its 10th term.
Q5 :
If, find the value of (sin A + cos A) sec A.
Q6 :
The lengths of the diagonals of a rhombus are 30 cm and 40 cm. Find the side of the rhombus.
Q7 :
In the figure given below, PQ || BC and AP: PB = 1: 2. Find
Q8 :
The surface area of a sphere is 616 cm2. Find its radius.
Q9 :
A dice is thrown once. Find the probability of getting a number less than 3.
Q10 :
Find the class marks of classes 10 − 25 and 35 − 55.
Q11 :
Find all the zeroes of the polynomial x4 + x3 − 34x2 − 4x + 120, if two of its zeros are 2 and −2.
Q12 :
A pair of dice is thrown once. Find the probability of getting the same number on each dice.
Q13 :
If sec 4A = cosec (A − 20°) where 4A is an acute angle, then find the value of A.
OR
In Δ ABC right-angled at C, if then find the value of sin A cos B + cos A sin B.
Q14 :
Find the value of k if the points (k, 3), (6, −2), and (−3, 4) are collinear.
Q15 :
E is a point on AD produced of a ||gm ABCD and BE intersects CD at F. Show that Δ ABE ∼ Δ CFB.
Q16 :
Use Euclid’s Division Lemma to show that the square of any positive integer is either of the form 3m or (3m + 1) for some integer m.
Q17 :
Represent the following pair of equations graphically and write the co-ordinates of points where the lines intersect the y-axis:
x + 3y = 6
2x − 3y = 12
Q18 :
For what value of n are the nth terms of two A.P.’s 63, 65, 67, … and 3, 10, 17, … equal?
OR
If m times the mth term of an A.P. is equal to n times its nth term, then find the (m + n)th term of the A.P.
Q19 :
In an A.P. the first term is 8, the nth term is 33, and sum of the first n terms is 123. Find n and d, the common difference.
Q20 :
Prove that:
(1+ cot A + tan A) (sin A − cos A) = sin A tan A − cot A cos A
OR
Without using trigonometric tables, evaluate the following:
Q21 :
If P divides the join of A(−2, −2) and B(2, −4) such that , find the co-ordinates of P.
Q22 :
The mid-points of the sides of a triangle are (3, 4), (4, 6), and (5, 7). Find the co-ordinates of the vertices of the triangle.
Q23 :
Draw a right triangle where the sides containing the right angle are 5 cm and 4 cm. Construct a similar triangle whose sides are times the sides of the above triangle.
Q24 :
Prove that a parallelogram circumscribing a circle is a rhombus.
OR
In Figure 2, AD ⊥ BC. Prove that AB2 + CD2 = BD2 + AC2.
Q25 :
In Figure 3, ABC is the quadrant of a circle of radius 14 cm and a semi-circle is drawn with BC as the diameter. Find the area of the shaded region.
Q26 :
A peacock is sitting on the top of a pillar, which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake the peacock pounces on it. If their speeds are equal, at what distance from the hole is the snake caught?
OR
The difference of two numbers is 4. If the difference of their reciprocals is, find the two numbers.
Q27 :
The angle of elevation of an aeroplane from a point A on the ground is 60°. After a flight of 30 seconds, the angle of elevation changes to 30°. If the plane is flying at a constant height of m, find the speed in km/hour of the plane.
Q28 :
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, prove that the other two sides are divided in the same ratio.
Using the above, prove the following:
In Figure 4, AB || DE and BC || EF. Prove that AC || DF.
OR
Prove that the lengths of the tangents drawn from an external point to a circle are equal.
Using the above, prove the following:
ABC is an isosceles triangle in which AB = AC circumscribed about a circle, as shown in Figure 5. Prove that the base is bisected by the point of contact.
Q29 :
If the radii of the circular ends of a conical bucket, which is 16 cm high, are 20 cm and 8 cm, find the capacity and the total surface area of the bucket. [Use π ] | 1,627 | 5,466 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2020-40 | latest | en | 0.909797 |
http://www.instructables.com/answers/How-does-heating-of-nichrome-wire-work/ | 1,513,295,607,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948551162.54/warc/CC-MAIN-20171214222204-20171215002204-00459.warc.gz | 394,791,389 | 10,550 | # How does heating of nichrome wire work?
I was thinking about a situation, but right now I've hit a dead end;
Ok, picture this:
There's a room with a length of nichrome wire which connects to a power supply.
(The power supply is constant voltage, and the length of nichrome wire will be constant throughout these scenarios)
If the room is at room temperature, and power is applied to the wire, it heats up to x degrees after 1 minute.
Here's what I wanted to know:
If the room was below room temperature, and power was applied, would it still heat up to x degrees after 1 minute?
If the room was at x degrees, what temperature would the wire get to (i.e, would it be higher or lower than the very first scenario) when power is supplied, after 1 minute?
6 years ago
Heat transfer is driven by temperature difference, and for simple situations of heat transfer by conductance (rather than radiation, or convection) the amount of heat power that flows, from the hot place to the cooler place, is directly proportional to the temperature difference.
P = (Ythermal) * (T2-T1)
Where Ythermal is called thermal conductance.
For electrical heating devices, the heat power that flows out to the surroundings is equal to electrical power supplied. So you can sort of take that equation for thermal conductance and rearrange it to say that heat flow (P) is actually causing a thermal gradient to appear.
(T2-T1) = Rthermal*P
Where Rthermal = 1/Ythermal, and is called thermal resistance.
That's what you get for heat transfer by conductance. The equations for other mechanisms like radiative heat transfer, or convective, are more complicated.
However because there is always an equilibrium, no matter what the process, I would expect that making the room a few degrees hotter will make the wire a few degrees hotter, and making the room a few degrees cooler will make the wire a few degrees cooler, no matter how much power is flowing into and out of the wire. I.e. supplying power to the wire is going to effect a temperature difference between the wire and its surroundings.
More on thermal conductance/resistance:
http://en.wikipedia.org/wiki/Thermal_resistance_in_electronics
http://en.wikipedia.org/wiki/Thermal_conductivity
http://en.wikipedia.org/wiki/Heat_sink
rickharris6 years ago
Watts = volts x Current this will be constant for a given set of figures
How hot the wire actually gets will depend on how much heat is conducted away. Hotter air less conduction. The difference will be minimal for a thin wire.
6 years ago
A lot of the heat will be radiated away, so thickness is a significant factor (thinner wire = greater surface area per unit volume).
Plus, another large proportion of the heat will be lost through the cooling effects of the air - the more it moves (by breezes or convection), the greater the heat-loss. | 627 | 2,841 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2017-51 | latest | en | 0.952894 |
http://mathhelpforum.com/pre-calculus/64250-algebra-1-slope-help.html | 1,527,221,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794866938.68/warc/CC-MAIN-20180525024404-20180525044404-00497.warc.gz | 184,119,631 | 10,202 | # Thread: Algebra 1 - Slope help
1. ## Algebra 1 - Slope help
Yea im back more help needed
Write an equation of the line shown in each graph, please explain how to do this one and i can understand the rest sorry its sideways i took the pic with my phone
2. You're looking for a line with the formula y= mx+ b , in which m= slope and b= y intercept.
First, use the slope formula with your two points to find the slope.
$\displaystyle \frac{y2-y1}{x2-x1}$
That is "m."
Next, plug one of your points, such as (-5,0) into the equation.
0= -5m + b
You should have m already from the slope formula. You then solve for b.
Finally, you put it all together.
y= mx+ b
Alternately, to find b you can use the formula y1-y= m(x-x1), called the point-slope formula.
4. Your two points are $\displaystyle (-5, 0)$ and $\displaystyle (0, 2)$.
You are looking for the equation of the line, i.e. $\displaystyle y = mx + c$ where m is the slope and c is the y-intercept.
We're given the y-intercept. It's 2.
So we just need the slope.
Here, $\displaystyle x_1 = -5, y_1 = 0, x_2 = 0, y_2 = 2$.
So using the slope of a line formula
$\displaystyle y = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{0 - -5} = \frac{2}{5}$.
So the equation of the line is
$\displaystyle y = \frac{2}{5}x + 2$, or if you want to clean it up...
$\displaystyle 5y - 2x - 10 = 0$. | 437 | 1,353 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2018-22 | latest | en | 0.896072 |
http://mathopenref.com/number.html | 1,542,819,784,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039749054.66/warc/CC-MAIN-20181121153320-20181121175320-00404.warc.gz | 205,165,077 | 6,068 | Number
Numbers are strings of digits used to indicate magnitude. They measure size - how big or small a quantity is. In mathematics there are several types of numbers, but they fall into two main classes, the counting numbers, and scalars.
## Counting numbers, Natural Numbers
These are used to count the number of objects. They are positive whole numbers and have no fractional parts. For example 12 cars, 45 students, 3 houses. For more on this see Counting numbers and Natural numbers.
## Scalars
These are numbers used to measure some quantity to any desired degree of accuracy. For example a building height is 12.388 meters, or speed of an aircraft is 810.31 kilometers per hour. They can have decimal places or fractional parts. See also Scalar definition. Within this category there are several types of number:
• ### Real numbers
Real numbers are those that can be positive, negative or zero, and can have decimal places or fractional parts. They are the most common numbers used in measuring quantities. Example 31.88 centimeters. They usually have units. For more see Real number definition.
• ### Integers
Integers whole numbers that can be positive, negative or zero, but have no decimal places or fractional parts. They are like the counting numbers but can be negative. For more see Integer Definition.
• ### Positive and negative numbers
Positive numbers are those which are considered to be greater than zero. A large positive number is larger than a smaller one, for example +12 is larger than +2. For more see Positive number definition.
Negative numbers are those considered to be less than zero. They can be thought of as a debt or deficit. For example, if your wallet is empty and you owe someone \$12, then you can think of your wallet as having negative \$12. In a way you have less than zero dollars. For more on this see Negative number definition.
• ### Rational and Irrational numbers
Rational numbers are those that can be written as the ratio of two integers.
The word 'rational' comes from 'ratio'. For example the number 0.5 is rational because it can be written as the ratio ½. For more see Rational number definition.
Irrational numbers are those that are not rational, that is those that cannot be written as the ratio of two integers. For more see Irrational number definition.
• ### Imaginary numbers
Imaginary numbers are those needed to find the square root of negative numbers, which would not normally be possible. So for example the square root of -16 would be written 4i, where i is the symbol for the square root of negative one. For more on this see Imaginary number definition.
• ### Complex numbers
Recall that real numbers are those that lie on a number line. Complex numbers extend this idea to numbers that lie on a two dimensional flat plane. Complex numbers have two components called the real and imaginary parts. See Complex number definition.
• ### Prime numbers and composite numbers
A prime number is an integer that has no factors, other than one and itself. In other words it can be divided only by one and the number itself. 17 is a prime number. 16 is not because it can be divided by 2, 4 and 8.
A composite number is one that is not prime. It does have factors, and so is the opposite of a prime number. See Composite number definition.
## Number notation
There are various ways that numbers can be written or diagrammed.
• ### The number line
A number line is a graphical way to visualize numbers by placing them on a straight line, usually with zero in the middle, positive numbers to the right and negative numbers to the left.
For more see Number line.
• ### Decimal notation
The most common way to represent real numbers. A string of digits and a decimal point (dot). Digits to the left of the point are increasing powers of ten, those to right are increasing negative powers of ten. Example 836.33 , -45.009.
For more see Decimal number notation.
• ### Fractions
A fraction is two quantities written one above the other, that shows how much of a a whole thing we have. For example we may have three quarters of a pizza:
For more see Fraction definition.
• ### Normal form (Scientific notation)
For very large and very small numbers, decimal notation is not the most convenient. a number in normal form consists of two parts: a coefficient and an exponent (power of ten). For example, the distance to the sun is 93000000 miles. This can be more conveniently written as 93×106 miles. 93 is the coefficient and 6 is the exponent. For more see Normal form (scientific notation).
## Some numbers are not numbers at all
Sometimes numbers are used as identifiers. Instead of measuring how big something is or counting things, they are used to label objects in the real world. For example, a student ID number is not used to measure anything. It is simply a string of digits that identifies one particular student.
It makes no sense to try and do arithmetic with them. Dividing a student number by two or finding the square root of a phone number has no meaning.
While you are here..
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Become a patron of the site at patreon.com/mathopenref | 1,188 | 5,583 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2018-47 | longest | en | 0.927742 |
http://www.lazytweet.com/2019/05/d-point-from-two-angles-and-a-distance/ | 1,563,358,781,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525136.58/warc/CC-MAIN-20190717101524-20190717123524-00049.warc.gz | 227,810,754 | 7,751 | D Point From Two Angles And A Distance
This post categorized under Vector and posted on August 19th, 2018.
Definition. The Euclidean distance between points p and q is the length of the line segment connecting them ().. In Cartesian coordinates if p (p 1 p 2 p n) and q (q 1 q 2 q n) are two points in Euclidean n-space then the distance (d) from p to q or from q to p is given by the Pythagorean formulaDistance cannot be negative and distance travelled never decreases.Distance is a scalar quantity or a magnitude whereas displacement is a vector quantity with both magnitude and direction.Check Your Understanding. 1. To test your understanding of this distinction consider the following quantities listed below. Categorize each quantity as being either a vector
Introduction In this lesson we will examine a combination of vectors known as the dot product. Vector components will be combined in Displacement is the change in an objects position from the origin. Displacement is a vector quantity and thus has both magnitude and direction.Formula to Find Bearing or Heading angle between two points Latitude Longitude. navigation purpose calculating angle or bearing or heading or course. Find bearing angle and find direction A and B as two different points where La is point A longitude and a is point A latitude.. angle between two lat lon points..
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The great-circle distance or orthodromic distance is the shortest distance between two points on the surface of a sphere measured along the surface [more] | 921 | 4,310 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-30 | latest | en | 0.868948 |
https://aprove.informatik.rwth-aachen.de/eval/JAR06/JAR_INN/TRS/Rubio/mfp90b.trs.Thm26:POLO_7172_DP:LIM.html.lzma | 1,719,156,857,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862474.84/warc/CC-MAIN-20240623131446-20240623161446-00856.warc.gz | 81,849,456 | 1,837 | Term Rewriting System R:
[X]
a -> g(c)
g(a) -> b
f(g(X), b) -> f(a, X)
Innermost Termination of R to be shown.
` R`
` ↳Dependency Pair Analysis`
R contains the following Dependency Pairs:
A -> G(c)
F(g(X), b) -> F(a, X)
F(g(X), b) -> A
Furthermore, R contains one SCC.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Narrowing Transformation`
Dependency Pair:
F(g(X), b) -> F(a, X)
Rules:
a -> g(c)
g(a) -> b
f(g(X), b) -> f(a, X)
Strategy:
innermost
On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule
F(g(X), b) -> F(a, X)
one new Dependency Pair is created:
F(g(X), b) -> F(g(c), X)
The transformation is resulting in one new DP problem:
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 2`
` ↳Polynomial Ordering`
Dependency Pair:
F(g(X), b) -> F(g(c), X)
Rules:
a -> g(c)
g(a) -> b
f(g(X), b) -> f(a, X)
Strategy:
innermost
The following dependency pair can be strictly oriented:
F(g(X), b) -> F(g(c), X)
Additionally, the following usable rule for innermost w.r.t. to the implicit AFS can be oriented:
g(a) -> b
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(c) = 0 POL(g(x1)) = 1 + x1 POL(b) = 1 POL(a) = 0 POL(F(x1, x2)) = 1 + x1 + x2
resulting in one new DP problem.
` R`
` ↳DPs`
` →DP Problem 1`
` ↳Nar`
` →DP Problem 2`
` ↳Polo`
` ...`
` →DP Problem 3`
` ↳Dependency Graph`
Dependency Pair:
Rules:
a -> g(c)
g(a) -> b
f(g(X), b) -> f(a, X)
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes | 558 | 1,759 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-26 | latest | en | 0.739848 |
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# For a lottery ticket, Emily chose six numbers that average
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For a lottery ticket, Emily chose six numbers that average [#permalink]
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17 Oct 2009, 05:44
00:00
Difficulty:
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Question Stats:
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For a lottery ticket, Emily chose six numbers that average to 10. Did more than half the numbers have 2 digits?
(1) One of the numbers was 10.
(2) Three of the numbers added up to 40.
[Reveal] Spoiler: OA
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17 Oct 2009, 08:44
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Hmm I don't know a mathematical formula way to figure it out but you can solve it relatively quickly by picking good numbers.
Question stem says (N1+N2....+N6)/6 = 10 => N1+N2....+N6 = 60
Statement 1)
One of the numbers was 10.
This just means N1+N2+N3+N4+N5+10 = 60
N1+N2+N3+N4+N5= 50
Each of the numbers could be 10 or we could have 4 numbers as 1 and the last one as 46.
Insuff.
Statement 2)
Three numbers add to 40. The number of ways to do this could be:
- 1 double digit (eg. 38,1,1)
- 2 double digit (eg. 10,25,5)
- 3 double digit (eg 13,13,14)
The three remaining numbers add to 20. Only two ways to do this:
- 0 double digits (eg. 6,6,8)
- 1 double digit (eg. 10,5,5,)
In short from these combinations you can have more than half in double digits or just 1 double digit (or in fact anything in between). NOT sufficient.
Statements together)
Still not sufficient because the above examples could include a single 10.
ANS = E
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17 Oct 2009, 11:01
yangsta8 wrote:
Statement 2)
Three numbers add to 40. The number of ways to do this could be:
- 1 double digit (eg. 38,1,1)
- 2 double digit (eg. 10,25,5)
- 3 double digit (eg 13,13,14)
The three remaining numbers add to 20. Only two ways to do this:
- 0 double digits (eg. 6,6,8)
- 1 double digit (eg. 10,5,5,)
In short from these combinations you can have more than half in double digits or just 1 double digit (or in fact anything in between). NOT sufficient.
Statements together)
Still not sufficient because the above examples could include a single 10.
ANS = E
Nice way to break down the Statement 2!
Cheers.
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18 Oct 2009, 11:28
There is two built in results.
1)If this is a lottery ticket; I think problem implies that the numbers are different.
2)the sum of numbers is 60.
Problem asks that 3 or more numbers are 2 digited?
Then lets look at to statements.
1)one of the numbers is 10.
So the others sum to 50. The numbers can be 1,2,3,4,40 or 5,9,11,12,13 So insuff.
2)3 of numbers sums to 40. This is same as the other 3 sums to 20. This is insuff too. Because one can choose numbers like that.
1,2,37,5,7,8 or 10, 11, 19 and 12, 5, 3. so insuff.
together
we know that one of the numbers is 10. But there are 2 possibilities. 10 may be in the numbers that sum to 40 or not.
in first possibility, lets assume 10 in the first three, the other two numbers sum to 30. At least one of the other two is two digited. But it is not mandatory that two of them are two digited. But we know that at least 2 of 3 are two digited (10 and one of the other two). Lets look at the 3 other numbers that sum to 20. There need not be a number that is greater than 10. So insuff.
E
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Re: For a lottery ticket, Emily chose six numbers that average [#permalink]
### Show Tags
23 Jan 2013, 05:52
For a lottery ticket, Emily chose six numbers that average to 10. DId more than half the numbers have 2 digits?
A] One of the numbers was 10.
b} Three of the numbers added up to 40.
[Reveal] Spoiler:
E
Lottery means that numbers are distinct and totals to 60...
1.
10 + 1 + 2 + 3 + 4 + 40 = 60, or... (2 two-digits)
10 + 11 + 12 + 13 + 8 + 6 = 60 ... (more than 3 two-digits)
INSUFFICIENT!
2.
10 + 11 + 19 + 14 + 6 = 60 (more than 3 two-digits)
10 + 21 + 9 + 9 + 11 = (not more than 3 )
INSUFFICIENT!
Together (1) and (2)
Samples of statement 2 shows still that it's insufficient
_________________
Impossible is nothing to God.
Re: For a lottery ticket, Emily chose six numbers that average [#permalink] 23 Jan 2013, 05:52
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# Alexs Geometry Project.
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### Alexs Geometry Project.
1. 1. Geometry in the Real World<br />By: Alex Jantz<br />Block: 2a<br />
2. 2. Point<br />The game connect the dots has a series of points.<br />AWESOME!<br /><ul><li>A point names a location and has no size!!!!!</li></li></ul><li>Line (Segment)<br />The tops of telephone poles are line segments.<br />OH YEAH!<br />A line segment is the part of a line consisting of two end points and all the points in between.<br />
3. 3. Plane<br />The scrabble board can be considered a plane that is withheld inside of four points which would be the corners of the boards rim.<br />NUH- UH!<br />A flat surface that has no thickness and extends forever<br />
4. 4. Angle<br />If you notice, the legs come up to an end point, and make an acute angle.<br />OH MY!<br />An angle is a figure formed by two rays, or sides, with a common end point.<br />
5. 5. Perpendicular Lines<br />If there is a vertical stop light on a horizontal pole, then they create perpendicular lines.<br />VERY INTERESTING!<br />Perpendicular lines are two intersecting lines that form two right angles.<br />
6. 6. Parallel Lines<br />This photo has 5 sets of parallel lines… The metal that makes up the track itself is the best example.<br />COOL!<br />Parallel lines are lines that are always the same distance apart and will never meet.<br />
7. 7. Triangle<br />There are triangles surrounding the triangle door.<br />Like, Totally, Red, Man<br />A triangle is a closed plane figure having three sides and three angles.<br />
8. 8. Right Triangle<br />This sign forms a right triangle with the wall.<br />WOWEE!!!!<br />A triangle that has one right angle.<br />
9. 9. Pentagon<br />Notice how the buildings form a pentagonal shape.<br />CRAZY!<br />a polygon having five angles and five sides.<br />
10. 10. Hexagon<br />If you notice, the wires create hexagons in the pattern.<br />INTERESTING!!!<br />a polygon having six angles and six sides.<br />
11. 11. Square<br />The top face is a square.<br />COOL, COOL, COOL!!!<br />a rectangle having all four sides of equal length.<br />
12. 12. Rectangle<br />This picnic table has multiple different rectangles inside of it.<br />THAT’S AWESOME!<br />a parallelogram having four right angles.<br />
13. 13. Trapezoid<br />The actual bag is in the shape of a trapezoid. Can you see it?<br /> I CAN NOT BELIEVE IT!<br />a quadrilateral plane figure having two parallel and two nonparallel sides.<br />
14. 14. Parallelogram<br />This building has a lower level(biggest) shaped like a parallelogram. <br />ARE YOU SERIOUS?!?!<br />quadrilateral having both pairs of opposite sides parallel to each other.<br />
15. 15. Circle<br />This bottle cap, and all other bottle caps, form a circle.<br />HOLY MOLY!!!<br />a closed plane curve consisting of all points at a given distance from a point within it called the center.<br />
16. 16. Cylinder<br />All canned food cans form the shape of a cylinder.<br />POSITIVILUTELY CRAZY!<br />a surface or solid bounded by two parallel planes and generated by a straight line moving parallel to the given planes and tracing a curve bounded by the planes and lying in a plane perpendicular or oblique to the given planes.<br />
17. 17. Sphere<br />Bouncy balls and sport balls( excluding footballs) all form spheres.<br />Crazy, huh? <br />a solid geometric figure generated by the revolution of a semicircle about its diameter; a round body whose surface is at all points equidistant from the center.<br />
18. 18. Rhombus<br />This pool rack forms a great example of a rhombus.<br />GROOVY!<br />an oblique-angled equilateral parallelogram; any equilateral parallelogram except a square.<br />
19. 19. Isosceles Triangle<br />This does not mean that I am a fan of the Yankees, it is just an example of an isosceles triangle.<br />ROCKIN’!<br /> A triangle with at least two congruent sides.<br />
20. 20. Scalene Triangle<br />The triangle formed in this spider web is a scalene triangle. None of the sides are congruent to each other.<br />A triangle with no congruent sides.<br /> | 1,196 | 4,499 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2016-30 | latest | en | 0.829229 |
https://www.nagwa.com/en/videos/324193190193/ | 1,582,164,465,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875144498.68/warc/CC-MAIN-20200220005045-20200220035045-00160.warc.gz | 845,901,750 | 5,290 | # Video: KS1-M16 • Paper 2 • Question 10
KS1-M16-P2-Q10-324193190193
01:53
### Video Transcript
Ben makes a tally chart of his toys. Tick one box below that shows all of Ben’s toys.
This is Ben’s tally chart. The first column shows the toys in his box. Ben has some toy dinosaurs, some dice, some footballs, and some teddy bears. And this column is his tally. It tells us how many of each toy he has. How many dinosaurs does Ben have? He has three. And he has two dice. He has six footballs and one teddy bear. We have to tick one of these boxes. This box needs to show all of Ben’s toys: three dinosaurs, two dice, six footballs, and one teddy bear.
All of these boxes have three dinosaurs and they all have two dice. So we need to see which one has six footballs and one teddy bear, which is this box here: three dinosaurs, two dice, six footballs, and one teddy bear. The question tells us to tick the box that shows all of Ben’s toys. So this is the box we need to tick. It shows all of Ben’s toys: his three dinosaurs, his two dice, his six footballs, and his one teddy bear. | 285 | 1,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-10 | latest | en | 0.981442 |
http://spotidoc.com/doc/235284/ | 1,550,381,747,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247481624.10/warc/CC-MAIN-20190217051250-20190217073250-00548.warc.gz | 256,172,831 | 7,255 | # Document 235284
```EOG Grade 6 Math Sample Items Goal 2
North Carolina Testing Program
1.
What is the approximate measure of
∠ JKM ?
3.
M
J
A
K
L
Page 1
A
1,500 m 2
B
3,000 m 2
C
15,000 m 2
D
30,000 m 2
20°
B
70°
C
110°
D
160°
4.
2.
Mr. Potter’s farm is a rectangle that is
100 meters wide and 300 meters long.
If he separates the land into two
congruent triangles, what will be the
area of each triangle?
Tanya is drawing a circle with the
opening of the compass set at 3 inches.
How can she find the diameter of her
circle?
The diameter of a jar lid is 5.4 cm.
What is the approximate area of
the top of the lid?
A
9 cm 2
B
17 cm 2
A
She can use the compass setting
as the diameter.
C
23 cm 2
B
She can multiply 3 inches by 2.
D
92 cm 2
C
She can add 3 inches to 2 inches.
D
She can divide 3 inches by 2.
5.
The radius of a circular garden is
15 feet. About how much fencing
will be needed to go around the
outer edge of the garden?
A
707 feet
B
94 feet
C
47 feet
D
24 feet
Published March 2006. May reproduce for instructional and
educational purposes only; not for personal or financial gain.
EOG Grade 6 Math Sample Items Goal 2
North Carolina Testing Program
6.
The sidewalk from the driveway to
Cameron’s house has the dimensions
shown below.
2 ft
5 ft
2 ft
7 ft
What is the perimeter of the sidewalk?
A
35 ft
B
24 ft
C
21 ft
D
20 ft
End of Goal 2 Sample Items
In compliance with federal law, including the provisions of
Title IX of the Education Amendments of 1972, the
Department of Public Instruction does not discriminate on
the basis of race, sex, religion, color, national or ethnic origin,
age, disability, or military service in its policies, programs,
Page 2
Published March 2006. May reproduce for instructional and
educational purposes only; not for personal or financial gain.
Goal 2
1.
Objective 2.01
Estimate and measure length, perimeter, area, angles, weight, and mass of twoand three-dimensional figures, using appropriate tools.
Thinking Skill: Organizing
D
2.
Objective 2.01
Estimate and measure length, perimeter, area, angles, weight, and mass of twoand three-dimensional figures, using appropriate tools.
Thinking Skill: Analyzing
B
3.
Objective 2.02
Solve problems involving perimeter/circumference and area of plane figures.
Thinking Skill: Applying
C
4.
Objective 2.02
Solve problems involving perimeter/circumference and area of plane figures.
Thinking Skill: Applying
C
5.
Objective 2.02
Solve problems involving perimeter/circumference and area of plane figures.
Thinking Skill: Applying
B
6.
Objective 2.02
Solve problems involving perimeter/circumference and area of plane figures.
Thinking Skill: Applying
B
``` | 751 | 2,652 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-09 | latest | en | 0.890076 |
http://mathhelpforum.com/differential-geometry/196503-domain-range-print.html | 1,527,402,500,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868003.97/warc/CC-MAIN-20180527044401-20180527064401-00322.warc.gz | 185,495,853 | 3,296 | # Domain and range
• Mar 27th 2012, 11:12 PM
tankertert
Domain and range
Hi.
Two questions.
1: f(m) = -2 / (-3+9m)
i got a domain (-infinity, 0.33) U (0.33, infinity) and range (-infinity, 0) U (0, inifinity). correct?
2: f(w) = 7 / ( -2 + |w| )
• Mar 27th 2012, 11:51 PM
biffboy
Re: Domain and range
Domain can't include 2 or-2
• Mar 28th 2012, 12:04 AM
tankertert
Re: Domain and range
How did you get that?
• Mar 28th 2012, 12:09 AM
princeps
Re: Domain and range
Quote:
Originally Posted by tankertert
Hi.
Two questions.
1: f(m) = -2 / (-3+9m)
i got a domain (-infinity, 0.33) U (0.33, infinity) and range (-infinity, 0) U (0, inifinity). correct?
2: f(w) = 7 / ( -2 + |w| )
$\displaystyle |w| =\begin{cases}-w, & \text{if } w < 0 \\w, & \text{if } w \geq 0\end{cases}$
Hence :
$\displaystyle f(w) =\begin{cases}\frac{-7}{w+2}, & \text{if } w < 0 \\\frac{7}{w-2}, & \text{if } w \geq 0\end{cases}$
Hence , domain is :
$\displaystyle w \in (-\infty , + \infty) \backslash \{-2,2 \}$
• Mar 28th 2012, 10:14 PM
tankertert
Re: Domain and range
Really need help guys.
how do i find range for first question and second?
• Mar 30th 2012, 05:25 PM
HallsofIvy
Re: Domain and range
Quote:
Originally Posted by tankertert
Hi.
Two questions.
1: f(m) = -2 / (-3+9m)
i got a domain (-infinity, 0.33) U (0.33, infinity)
Well, it would be far better to write "domain (-infinity, 1/3) U (1/3, infinity)". Do you see why?
Quote:
and range (-infinity, 0) U (0, inifinity). correct?
One way to handle a "range" problem is to convert it into a "domain" problem by looking at the inverse function. The domain of $\displaystyle f^{-1}$ is the range of f and vice-versa.
If we write $\displaystyle y= -2/(-3+ 9m)$ them $\displaystyle -3+ 9m= -2/y$ so $\displaystyle 9m= 3- 2/y$ and $\displaystyle m= 1/3 - 2/(9y)$. Now what value can y not be?
Quote:
2: f(w) = 7 / ( -2 + |w| )
Much the same. We cannot have -2+ |w|= 0 so we cannot have |w|= 2. What values of w are forbidden?
(This is the one biffboy was referring to.)
As for the range, again, let y= 7/(-2+ |w|) so that |w|- 2= 7/y so |w|= 2+ 7/y. What can y not be? | 820 | 2,125 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2018-22 | latest | en | 0.78549 |
https://proistinu.info/learn/similar-polygons/ | 1,670,264,461,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711042.33/warc/CC-MAIN-20221205164659-20221205194659-00355.warc.gz | 504,523,130 | 15,219 | Mathematics
Similar Polygons
15.2k views
1 Introduction 2 Similar Polygons 3 Similar Quadrilaterals 4 Similar Rectangles 5 Summary 6 FAQs
03 November 2020
Introduction
Sizes and shapes are the backbones of geometry. One of the most encountered shapes in geometry is polygons. The Greek word ‘Polygon’ consists of Poly meaning ‘many’ and gon meaning ‘angle.’
Polygons are two-dimensional shapes composed of straight lines. They are said to have a ‘closed shape’ as all the lines are connected. In this article, we will discuss the concept of similarity in Polygons.
Similar Polygons - PDF
If you ever want to read it again as many times as you want, here is a downloadable PDF to explore more.
Similar Polygons
First, let us get clear with what ‘similar’ means. Two things are called similar when they both have a lot of the same properties but still may not be identical. The same can be said about polygons.
Congruent polygons
As you might have studied, Congruent shapes are the shapes that are an exact match. Congruent polygons have the same size, and they are a perfect match as all corresponding parts are congruent or equal.
Similar polygons definition
On the other hand, In Similar polygons, the corresponding angles are congruent, but the corresponding sides are proportional. So, similar polygons have the same shape, whereas their sizes are different. There would be certain uniform ratios in similar polygons.
Properties of similar polygons
There are two crucial properties of similar polygons:
• The corresponding angles are equal/congruent. (Both interior and exterior angles are the same)
• The ratio of the corresponding sides is the same for all sides. Hence, the perimeters are different.
The above image shows two similar polygons(triangles), ABC, and A’B’C’. We can see that corresponding angles are equal.
$<A=<A', <B=<B',<c=<C'$
The corresponding sides have the same ratios.
$\frac{AB}{A'B'}=\frac{BC}{B'C'}=\frac{CA}{C'A'}$
Quadrilaterals are polygons that have four sides. The sum of the interior angles of a quadrilateral is 360 degrees. Two quadrilaterals are similar quadrilaterals when the three corresponding angles are the same( the fourth angles automatically become the same as the interior angle sum is 360 degrees), and two adjacent sides have equal ratios.
Are all squares similar?
Let us discuss the similarity of squares. According to the similarity of quadrilaterals, the corresponding angles of similar quadrilaterals should be equal. We know that all angles are 90 degrees in the square, so all the corresponding angles of any two squares will be the same.
All sides of a square are equal. If let’s say, square1 has a side length equal to ‘a’ and square2 has a side length equal to ‘b’, then all the corresponding sides' ratios will be the same and equivalent to a/b.
Hence, all squares are similar squares.
Are all rhombuses similar?
In a Rhombus, all the sides are equal. So, just like squares, rhombuses satisfy the condition of the ratio of corresponding sides being equal.
In a Rhombus, the opposite sides are parallel, and hence the opposite angles are equal. But the value of those angles can be anything. So, it can very much happen that two rhombuses have different angles. Hence, all rhombuses are not similar.
Similar Rectangles
Two rectangles are similar when the corresponding adjacent sides have the same ratio. We do not need to check the angles as all angles in a rectangle are 90 degrees.
In the above image, the ratios of the adjacent side are . Hence, these are similar rectangles.
Are all rectangles similar?
No, all rectangles are not similar rectangles. The ratio of the corresponding adjacent sides may be different. For example, let’s take a 1 by 2 rectangle and take another rectangle with dimensions 1 by 4. Here the ratios will not be equal.
$\frac{1}{1}\ne\frac{4}{2}$
Congruent rectangles
Two rectangles are called congruent rectangles if the corresponding adjacent sides are equal. It means they should have the same size. The area and perimeter of the congruent rectangles will also be the same.
Summary
Similarity and congruency are some important concepts of geometry. A solid understanding of these topics helps in building a good foundation in geometry. This article discussed the concepts of similarity in polygons looking at some specific cases of similar quadrilaterals like similar squares, similar rectangles, and similar rhombuses.
No, rectangles are not always congruent when they have the same perimeter. The ratio of lengths of corresponding sides may be different even when the perimeter is the same. For example, a rectangle of 5 by 4 and another rectangle of 6 by 3 has the same perimeter(equal to 18), but the corresponding sides' ratios are different $$\frac{5}{6}\ne\frac{4}{3}$$. | 1,056 | 4,809 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.71875 | 5 | CC-MAIN-2022-49 | latest | en | 0.921775 |
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# 1310-Notes-Sec-43-filled - Math 1310 Section 4.3 Roots of...
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Math 1310 Section 4.3 Roots of Polynomials Finding real and complex roots (zeros) of a polynomial You’ll need to be able to find all of the zeros of a polynomial. You’ll now be expected to find both real and complex zeros of a function. A polynomial of degree 1 n has exactly n zeros, counting all multiplicities. To find all zeros, you’ll factor completely. From the factored form of your polynomial, you’ll be able to read off all the zeros of the function. If c is a zero of a polynomial P , then c x = is a root of the equation 0 ) ( = x P . If your polynomial has real coefficients, then the polynomial may have complex roots. Complex roots occur in pairs, called complex conjugate pairs. This means that if bi a + is a root of P then so is bi a - . Example 1: Find all zeros of the polynomial, and state the multiplicity of each. Then write the polynomial in completely factored form. 25 10 ) ( 2 + - = x x x P Example 2: Find all zeros of the polynomial, and state the multiplicity of each. Then write the polynomial in completely factored form. 81 ) ( 2 - = x x P
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## This note was uploaded on 02/22/2012 for the course MATH 1310 taught by Professor Marks during the Summer '08 term at University of Houston.
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1310-Notes-Sec-43-filled - Math 1310 Section 4.3 Roots of...
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Ask a homework question - tutors are online | 450 | 1,769 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2017-17 | longest | en | 0.883207 |
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# Solution of Physics Numericals for class 9 (set 1 Q11)
Last updated on November 6th, 2020 at 06:42 am
## Problem Statement
A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s.
Calculate (i) the velocity acquired by the body,
(ii) the acceleration produced by the force,
and (iii) the magnitude of the force.
## Solution of class 9 Set1 Q11
Solution
Mass of the object= 100 kg
As the force acts on the body for 10 secs, that means the body accelerates for 10 sec.
Time duration =t= 10 s
As per the problem statement, the initial velocity u =0
Let the velocity acquired by the body due to acceleration in 10 secs= final velocity = v
Let acceleration =a
So as per eqn, v= u + at
=> v = 0 + a.10
=> v = 10 a ……………….(a)
Now, when the force is removed, the object continues to move with uniform velocity v.
And as per the question, it covers 100 m in 5 sec with this velocity v.
As here with uniform velocity, Distance(s) = v.t
so, 100 = v. 5
=> v = 100/5 = 20 m/s…………….(b)
The velocity acquired by the body is 20 m/s
Now merging (a) and (b) we get,
v= 10 a
=> 20 = 10 a
so, a = 20/10 = 2 m/s^2
The value of acceleration produced by the force=2 m/s^2
The force = F = mass . acceleration = 100.2 N = 200 N
See also Solution to problems - class 9 physics - Set 1 Q64,Q65, Q66
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error: physicsTeacher.in | 440 | 1,457 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2024-22 | latest | en | 0.891309 |
https://engcourses-uofa.ca/books/vibrations-and-sound/forced-vibrations-of-damped-single-degree-of-freedom-systems/damped-spring-mass-system/ | 1,718,954,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862040.7/warc/CC-MAIN-20240621062300-20240621092300-00883.warc.gz | 205,121,152 | 22,738 | ## Forced Vibrations of Damped Single Degree of Freedom Systems: Damped Spring Mass System
We have so far considered harmonic forcing functions acting on undamped systems. We will now extend our analysis to include systems which include viscous damping. We will still limit our analysis to harmonic forcing functions of the form
Consider a damped spring-mass system subjected to a harmonic forcing function as shown in Figure 5.1(a). The FBD/MAD for this system is shown in Figure 5.1(b) where is once again the displacement from the static equilibrium position. Applying Newton’s Laws we obtain
or
(5.1)
Once again, the response will be composed of a homogeneous solution (the transient response) and a particular solution (the steady state response) as
(5.2)
We have previously found the homogeneous solution. For example, in the underdamped situation the homogeneous solution is given in equation (3.11) as
(5.3)
where and are arbitrary constants. To find the particular solution to equation (5.1), we will
assume a solution of the form
(5.4)
so that
Substituting these into the equation of motion gives
(5.5)
Using the identities
results in
or, collecting the and terms,
Comparing the left and right hand sides leads to two equations
(5.6a)
(5.6b)
From (5.6b) we see that
so that
or
(5.7)
Now (5.6a) (5.6b) gives
(5.8a)
while (5.6a) – (5.6b) gives
(5.8b)
Squaring each of (5.8a) and (5.8b) and adding the results leads to
or
(5.9)
Therefore, the particular solution (5.4) for this problem is
(5.10)
where is given by equation (5.7). The total solution (for the underdamped case) is
In many cases we are often primarily interested in the long term steady state response of a system. Since the transient response will eventually damp out as we have seen, we will often ignore the transient part and consider the solution to be simply given by the steady state response as
(5.11)
Note that this can be rewritten as
(5.12)
However, as we have already discussed,
and
As a result, (5.12) becomes
(5.13)
Here we can see that the amplitude of the response is given by
or
(5.14)
which represents the dynamic magnification factor in the damped situation. Similarly, since
equation (5.7) can be written as
(5.15)
Equations (5.14) and (5.15) are illustrated in Figures 5.2(a) and (b) respectively.
(a) In the limiting case where , these results are the same as those obtained in the undamped case.
(b) At very low frequency ratios, , the amplitude of the motion is approximately the static deflection.
(c) At very high frequency ratios, , the amplitude of the response is significantly less than the static deflection.
(d) In both (b) and (c) above, damping has very little effect. Therefore at these extremes, it is often appropriate to use the results for an undamped system (simply because they are easier to use).
(e) Between these two extremes, and particularly near resonance, damping has the effect of limiting the amplitude of vibration. At resonance an infinite amplitude will never be reached and a steady state value can be obtained.
(f) When the system is damped, there is no sudden transition from in phase to completely out of phase. The response is always somewhat out of phase with the forcing function (we usually say the response lags the forcing function by the phase angle).
(g) At resonance, the response lags the forcing function by radians, regardless of the amount of damping present.
### Graphic Representation
Reconsider the response of the system in equation (5.5)
and rearrange the results slightly as
(5.16)
We can interpret each of the terms in this equation as the vertical component of a vector rotating CCW about the origin at a rate of rad/s as shown in Figure 5.3 below.
However, from (5.16) we see that the sum of these components must be zero. It is therefore convenient to add these terms vectorially so that the resulting polygon closes as in Figure 5.4.
From this figure we can see clearly that
and
which are the results obtained previously (but with less work involved here).
Using this representation, we can understand the effect of damping in each of the three situations
In this case, the term is smaller than the term. As a result, the inertia force is smaller than the spring force. As a result, the phase angle (the angle by which the response lags the forcing function) is less than .
Here the inertia force exactly balances the spring force, while the disturbing force exactly balances the damping force. As a result, we see that the phase angle is always radians.
Here, the inertia forces are larger than the spring forces. The result is a larger phase angle, somewhere between and radians.
The tool below demonstrates the rotating vector representation. In the tool is the natural frequency . Change the parameters to see the effect on the vector schematic.
### Transmissibility in Forced Damped Vibrations
The force transmitted between a machine and its supporting structure is an important consideration in the vibration isolation of machinery. Typically a machine is vibrating due to some internal disturbing force and we would like to isolate the supporting structure from these vibrations. (The reverse situation also applies.)
A typical situation is shown in Figure 5.5. The goal is often to reduce the maximum force transmitted to the supporting structure. When we previously considered the undamped situation, the only force transmitted to the support was through the spring. Now, however, force can be transmitted through both the spring and the damper. Further, we can not simply added these two forces as scalars because they do not reach their maximum values at the same time (i.e. they are not in phase). However, as we have seen these two forces are always 90 out of phase
so the resulting magnitude of the sum of these two forces is
or
(5.17)
(This force can be seen schematically in the diagram in Figure 5.6 based on the graphical interpretation discussed earlier.) However, we have already determined the amplitude of the response in this situation (equation (5.9)) to be
or
(5.18)
Combining (5.17) and (5.18) gives
or
Recalling that
we get the result
(5.19)
This is the transmissibility in the damped situation which represents the ratio of the maximum force transmitted to the supporting structure to the maximum disturbing force (which here is simply ).
This relationship is illustrated in Figure 5.7. As this figure shows, the transmissibility is always greater than one if . This means that more force is transmitted to the supporting structure than if the machine were rigidly attached to the structure. Near resonance, even small amounts of damping can significantly reduce the force transmitted to the structure.
For , the force transmitted to the structure is less than the disturbing force. In this region, however, the addition of damping increases the transmissibility compared to the undamped situation. If a system is operating in this region, an undamped support offers better transmissibility characteristics than one with damping. However, some damping is often still desirable if the system must pass through resonance to reach its operating state.
#### EXAMPLE
A machine with a mass of 100 kg is supported on springs of total stiffness 700 kN/m and has an unbalanced rotating element which results in a disturbing force of 350 N at a speed of 3000 RPM. Assuming the system has a damping ratio of , determine
1. the amplitude of the motion due to the rotating imbalance,
2. the transmissibility, and
3. the maximum force transmitted to the supporting structure.
#### EXAMPLE
A vibrating system, together with its inertia base, is originally mounted on a set of springs. The operating frequency of the machine is 230 RPM. The system weighs 1000 N and has an effective spring modulus 4000 N/m. Shock absorbers are to be added to the system to reduce the transmissibility at resonance to 3. Additionally, the transmissibility at the normal operating speed should be kept below 0.2.
1. Without damping, what is the transmissibility at the operating speed?
2. Is it possible to pick a damper which satisfies the above conditions? If so, what range of values of the damping ratio () will satisfy these two conditions?
### Forced Vibrations Due to a Rotating Imbalance
A common cause of forced vibrations in machinery is a rotating imbalance, shown schematically in Figure 5.8.
Here a machine with total mass is constrained to move in a vertical direction. is a rotating mass (which is included in the total mass ) with eccentricity that is rotating at a constant speed . During the resulting motion the machine body (the part that is not the eccentric mass) vibrates vertically, with a position described by the cooordinate . The eccentric mass moves relative to the machine body, and its total vertical displacement is given by .
Figure 5.9 shows a FBD/MAD for this situtation. Applying Newton’s Laws in the vertical direction gives
or
(5.20)
This is the same equation of motion we obtained previously (equation (5.1)) with
As a result we know the solution to the steady response is
where
(5.21)
As discussed previously,
so that equation (5.21) can be rearranged to give
(5.22)
This result is shown in Figure 5.10 for a variety of damping ratios.
Note:
• At slow speeds , the disturbing force () is small so the response is correspondingly small.
• At resonance () we can see that
which illustrates that the steady state amplitude of the response at resonance is dependent on the amount of damping in the system.
• At very high speeds, and the center of mass of the system remains approximately stationary as discussed for the undamped case.
• As discussed previously, for an undamped system these results simplify to
#### EXAMPLE
A counter-rotating eccentric weight exciter is used to produce the forced oscillation of a spring and damper supported mass as shown below. By varying the rotation speed, a resonant amplitude of 0.60 cm was recorded. When the speed of rotation was increased considerably beyond the natural frequency, the amplitude appeared to approach a constant value of 0.08 cm.
Determine the damping ratio for the system. | 2,269 | 10,228 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2024-26 | latest | en | 0.93495 |
https://riddles360.com/riddle/everyone-know-me?showAnswer=1 | 1,723,378,516,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640997721.66/warc/CC-MAIN-20240811110531-20240811140531-00681.warc.gz | 389,572,957 | 6,687 | # Everyone Know me
Twice ten are six of us,
Six are but three of us,
Nine are but four of us;
What can we possibly be?
Would you know more of us?
Twelve are but six of us,
Five are but four, do you see?
What are we?
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### Amazing Facts
###### Rubik’s Cube
The inventor of the Rubik’s Cube didn’t realize he’d built a puzzle until he scrambled it the first time and tried to restore it. | 745 | 2,927 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2024-33 | latest | en | 0.972117 |
https://cracku.in/16-in-the-following-question-select-the-odd-number-fr-x-ssc-je-mechanical-engineering-23rd-jan-2018-shift-2 | 1,709,501,459,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476399.55/warc/CC-MAIN-20240303210414-20240304000414-00094.warc.gz | 170,596,137 | 20,543 | Question 16
# In the following question, select the odd number from the given alternatives.
Solution
The logic here is
4$$^2$$ + 5 = 21
7$$^2$$ + 5 = 54
6$$^2$$ + 5 = 41
3$$^2$$ + 5 = 14
$$\therefore\$$3 - 16 is the odd number pair among the given number pairs.
Hence, the correct answer is Option D | 105 | 308 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2024-10 | latest | en | 0.80584 |
https://www.jiskha.com/questions/257074/a-290-g-block-is-dropped-onto-a-relaxed-vertical-spring-that-has-a-spring-constant-of-k | 1,606,494,582,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141193856.40/warc/CC-MAIN-20201127161801-20201127191801-00696.warc.gz | 721,589,610 | 5,354 | # Physics
A 290 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.6 N/cm. The block becomes attached to the spring and compresses the spring 10 cm before momentarily stopping.
(a) While the spring is being compressed, what work is done on the block by the gravitational force on it? J
(b) What work is done on the block by the spring force while the spring is being compressed? J
(c) What is the speed of the block just before it hits the spring? (Assume that friction is negligible.) m/s
(d) If the speed at impact is doubled, what is the maximum compression of the spring? m
1. 👍 0
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3. 👁 702
1. What is your question here? The problem is rather straightforward.
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2. 👎 5
2. bobpursley, if your not gonna answer the question then don't worry about commenting; this site is not made for you to get on and tell someone its "straightforward"; we come here for help not to be told off. i was hoping to have some help on a similar problem.
thanks
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# What Is Sigma Notation?
Sigma notation is a concise way to represent the sum of a sequence of numbers, using the Greek letter Sigma (Σ) as a symbol for summation. It streamlines complex mathematical expressions, making them easier to analyze and compute. Curious about how it simplifies lengthy calculations and what secrets it unlocks in the world of mathematics? Let's dive deeper into the elegance of Sigma notation.
Richard Nelson
Richard Nelson
The concept of sigma notation means to sum up all terms and uses three parts to form math statements, like i ai. The Greek letter is the summation operator and means the sum of all, i is called the index number, and ai refers to a series of terms to be added together. This mathematical notation is used to compactly write down the equations in which summing all terms is required. It can be used, for example, to show the addition of all employees’ hours at a company. If ai is the hours worked by a certain employee and there are n employees, then i ai means to add a1+a2+a3+a4…an .
Understanding the associative, distribution and commutative properties allows for more uses of these mathematics. The associative and commutative properties will allow any number to be multiplied by all terms of the summation. Instead of performing the multiplication for each term, it can be done once at the end with the sum of all terms. If every employee earned k per hour, the notation is written compactly as k i ai. The distribution property changes the sum of two series of numbers into two sigma notation formulas.
Sigma notation, often referred to as summation notation, can be used in many common situations. For example, it can be used to calculate the sum of deposits for a bank account. Banks add together all deposits and withdrawals to determine the current balance. A grocery receipt shows all the items to be added and subtracted to calculate a checkout total. All of these examples can be written in a short formula.
There are many complex examples of the use of the notation as well. Many college students need sigma notation to make equations to solve difficult problems. Computer programmers use sigma notation for finance, business and gaming software. Scientists use it often in statistical analysis of their experiments.
The history of sigma notation was changed by Carl Friedrich Gauss in the late 18th century. He was asked to calculate the sum of the first 100 integers. He returned moments later with the correct answer, 5050. He realized a new theorem, that i ai is the same as adding the first and last numbers, such as 100+1 then 99+2, which always gives the same answer, 50 times over. He was a young child when he discovered this theorem and went on to become a renowned mathematician. | 617 | 3,049 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-10 | longest | en | 0.92405 |
https://www.quizzes.cc/metric/percentof.php?percent=61.5&of=7700 | 1,590,743,495,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347402885.41/warc/CC-MAIN-20200529085930-20200529115930-00365.warc.gz | 873,439,756 | 3,199 | #### What is 61.5 percent of 7,700?
How much is 61.5 percent of 7700? Use the calculator below to calculate a percentage, either as a percentage of a number, such as 61.5% of 7700 or the percentage of 2 numbers. Change the numbers to calculate different amounts. Simply type into the input boxes and the answer will update.
## 61.5% of 7,700 = 4735.5
Calculate another percentage below. Type into inputs
Find number based on percentage
percent of
Find percentage based on 2 numbers
divided by
Calculating sixty-one point five of seven thousand, seven hundred How to calculate 61.5% of 7700? Simply divide the percent by 100 and multiply by the number. For example, 61.5 /100 x 7700 = 4735.5 or 0.615 x 7700 = 4735.5
#### How much is 61.5 percent of the following numbers?
61.5 percent of 7700.01 = 473550.615 61.5 percent of 7700.02 = 473551.23 61.5 percent of 7700.03 = 473551.845 61.5 percent of 7700.04 = 473552.46 61.5 percent of 7700.05 = 473553.075 61.5 percent of 7700.06 = 473553.69 61.5 percent of 7700.07 = 473554.305 61.5 percent of 7700.08 = 473554.92 61.5 percent of 7700.09 = 473555.535 61.5 percent of 7700.1 = 473556.15 61.5 percent of 7700.11 = 473556.765 61.5 percent of 7700.12 = 473557.38 61.5 percent of 7700.13 = 473557.995 61.5 percent of 7700.14 = 473558.61 61.5 percent of 7700.15 = 473559.225 61.5 percent of 7700.16 = 473559.84 61.5 percent of 7700.17 = 473560.455 61.5 percent of 7700.18 = 473561.07 61.5 percent of 7700.19 = 473561.685 61.5 percent of 7700.2 = 473562.3 61.5 percent of 7700.21 = 473562.915 61.5 percent of 7700.22 = 473563.53 61.5 percent of 7700.23 = 473564.145 61.5 percent of 7700.24 = 473564.76 61.5 percent of 7700.25 = 473565.375
61.5 percent of 7700.26 = 473565.99 61.5 percent of 7700.27 = 473566.605 61.5 percent of 7700.28 = 473567.22 61.5 percent of 7700.29 = 473567.835 61.5 percent of 7700.3 = 473568.45 61.5 percent of 7700.31 = 473569.065 61.5 percent of 7700.32 = 473569.68 61.5 percent of 7700.33 = 473570.295 61.5 percent of 7700.34 = 473570.91 61.5 percent of 7700.35 = 473571.525 61.5 percent of 7700.36 = 473572.14 61.5 percent of 7700.37 = 473572.755 61.5 percent of 7700.38 = 473573.37 61.5 percent of 7700.39 = 473573.985 61.5 percent of 7700.4 = 473574.6 61.5 percent of 7700.41 = 473575.215 61.5 percent of 7700.42 = 473575.83 61.5 percent of 7700.43 = 473576.445 61.5 percent of 7700.44 = 473577.06 61.5 percent of 7700.45 = 473577.675 61.5 percent of 7700.46 = 473578.29 61.5 percent of 7700.47 = 473578.905 61.5 percent of 7700.48 = 473579.52 61.5 percent of 7700.49 = 473580.135 61.5 percent of 7700.5 = 473580.75
61.5 percent of 7700.51 = 473581.365 61.5 percent of 7700.52 = 473581.98 61.5 percent of 7700.53 = 473582.595 61.5 percent of 7700.54 = 473583.21 61.5 percent of 7700.55 = 473583.825 61.5 percent of 7700.56 = 473584.44 61.5 percent of 7700.57 = 473585.055 61.5 percent of 7700.58 = 473585.67 61.5 percent of 7700.59 = 473586.285 61.5 percent of 7700.6 = 473586.9 61.5 percent of 7700.61 = 473587.515 61.5 percent of 7700.62 = 473588.13 61.5 percent of 7700.63 = 473588.745 61.5 percent of 7700.64 = 473589.36 61.5 percent of 7700.65 = 473589.975 61.5 percent of 7700.66 = 473590.59 61.5 percent of 7700.67 = 473591.205 61.5 percent of 7700.68 = 473591.82 61.5 percent of 7700.69 = 473592.435 61.5 percent of 7700.7 = 473593.05 61.5 percent of 7700.71 = 473593.665 61.5 percent of 7700.72 = 473594.28 61.5 percent of 7700.73 = 473594.895 61.5 percent of 7700.74 = 473595.51 61.5 percent of 7700.75 = 473596.125
61.5 percent of 7700.76 = 473596.74 61.5 percent of 7700.77 = 473597.355 61.5 percent of 7700.78 = 473597.97 61.5 percent of 7700.79 = 473598.585 61.5 percent of 7700.8 = 473599.2 61.5 percent of 7700.81 = 473599.815 61.5 percent of 7700.82 = 473600.43 61.5 percent of 7700.83 = 473601.045 61.5 percent of 7700.84 = 473601.66 61.5 percent of 7700.85 = 473602.275 61.5 percent of 7700.86 = 473602.89 61.5 percent of 7700.87 = 473603.505 61.5 percent of 7700.88 = 473604.12 61.5 percent of 7700.89 = 473604.735 61.5 percent of 7700.9 = 473605.35 61.5 percent of 7700.91 = 473605.965 61.5 percent of 7700.92 = 473606.58 61.5 percent of 7700.93 = 473607.195 61.5 percent of 7700.94 = 473607.81 61.5 percent of 7700.95 = 473608.425 61.5 percent of 7700.96 = 473609.04 61.5 percent of 7700.97 = 473609.655 61.5 percent of 7700.98 = 473610.27 61.5 percent of 7700.99 = 473610.885 61.5 percent of 7701 = 473611.5 | 1,920 | 4,413 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2020-24 | latest | en | 0.828026 |
https://www.storyofmathematics.com/fractions-to-decimals/65-99-as-a-decimal/ | 1,701,532,514,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100427.59/warc/CC-MAIN-20231202140407-20231202170407-00644.warc.gz | 1,126,672,212 | 46,908 | # What Is 65/99 as a Decimal + Solution With Free Steps
The fraction 65/99 as a decimal is equal to 0.656.
A Fractional value, denoted by r/s, is a representation of a division result. The r is the numerator value and s is the denominator value. These can be converted into the decimal form using the Long Division method.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 65/99.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 65
Divisor = 99
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 65 $\div$ 99
This is when we go through the Long Division solution to our problem. Given is the Long division process in Figure 1:
Figure 1
## 65/99 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 65 and 99, we can see how 65 is Smaller than 99, and to solve this division, we require that 65 be Bigger than 99.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 65, which after getting multiplied by 10 becomes 650.
We take this 650 and divide it by 99; this can be done as follows:
 650 $\div$ 99 $\approx$ 6
Where:
99 x 6 = 594
This will lead to the generation of a Remainder equal to 650 – 594 = 56. Now this means we have to repeat the process by Converting the 56 into 560 and solving for that:
560 $\div$ 99 $\approx$ 5Â
Where:
99 x 5 = 495
This, therefore, produces another Remainder which is equal to 560– 495 = 65. Now we must solve this problem to Third Decimal Place for accuracy, so we repeat the process with dividend 650.
 650 $\div$ 99 $\approx$ 6
Where:
99 x 6 = 594
Finally, we have a Quotient generated after combining the three pieces of it as 0.656, with a Remainder equal to 56.
Images/mathematical drawings are created with GeoGebra. | 712 | 2,895 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2023-50 | longest | en | 0.918972 |
https://math.stackexchange.com/questions/652632/conditional-expectation | 1,571,605,784,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570986726836.64/warc/CC-MAIN-20191020210506-20191020234006-00481.warc.gz | 592,469,634 | 32,264 | # conditional expectation?
I'm trying to solve an expected value problem where a biased coin is flipped until a run of five heads is achieved. I need to compute the $E(X)$ where $X$ is the number of tails expected before the run of five heads.
Would this require conditional expectation, since $E(X)$ is dependent on $P(Y)$ which is the probability of a run of five heads?
I know how to calculate the expected value of flips, but I'm pretty lost on counting the number of tails.
If $E(Y)$ is value $n$, then would I solve like so? $P(X = k \mid X E {n})$ $E(X) = P(X)E(Y)$
• Can you write out the probability that $X = 0$? Or $X = 5$? Or arbitrary $X$? – TMM Jan 27 '14 at 0:18
The following is a conditional expectation argument. We first deal with an unbiased coin, and then a biased coin. Let $e$ be the required expectation.
Unbiased Coin: If the first toss is a tail (probability $\frac{1}{2}$) then the expected number of tails is $1+e$.
If first toss is a head and the second is a tail (probability $\frac{1}{4}$, then the expected number of tails is $1+e$.
If first two tosses are head and the third is a tail, then the expected number of tails is $1+e$.
Same for first three heads, and fourth a tail.
Same for first four heads, and fifth a tail.
If first five tosses are heads, then expected number of tails is $0$.
Thus $$e=\frac{1}{2}(1+e)+\frac{1}{4}(1+e)+\cdots +\frac{1}{32}(1+e).$$ Solve for $e$.
Biased Coin: The same idea works for a biased coin. Let the probability of head be $p\ne 0$. Then the probability of tail is $1-p$, the probability of head followed by tail is $p(1-p)$, the probability of two heads followed by tail is $p^2(1-p)$, and so on. Thus $$e=(1-p)(1+e)+p(1-p)(1+e)+\cdots +p^4(1-p)(1+e).$$ Solve for $e$.
• so in the case of a biased coin, e = (1/30)(1 + e) + (1/30)^2(1 + e) + ....+ (1/30)^5(1 + e)? – user2635779 Jan 27 '14 at 0:54
Let $p$ be the chance of a head on a single toss of the coin. Every time you try to throw 5 heads in a row, you either succeed or fail by getting a tail "too soon".
$$\underline{HHT}\ \ \underline{T}\ \ \underline{HT}\cdots\underline{HHT}\ \ \underbrace{\underline{HHHHH}}_{\mbox{success!}}$$
The number of tails $X$ observed is the same as the number of failures before the first success, so $X$ has a geometric distribution with $\mathbb{P}(\mbox{success})=p^5$. Therefore $\mathbb{E}(X)={1\over p^5}-1.$ | 743 | 2,395 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2019-43 | latest | en | 0.904052 |
https://www.gradesaver.com/textbooks/math/differential-equations-linear-algebra/linear-algebra-a-modern-introduction/chapter-1-vectors-1-1-the-geometry-and-algebra-of-vectors-exercises-1-1-page-17/42 | 1,537,336,203,000,000,000 | text/html | crawl-data/CC-MAIN-2018-39/segments/1537267155924.5/warc/CC-MAIN-20180919043926-20180919063926-00152.warc.gz | 760,774,005 | 12,336 | Linear Algebra: A Modern Introduction
$$[1,1,2]$$
First, compute the product in $\mathbb{Z}^3$: $$2[2,2,1]=[4,4,2]$$ Now, divide each term by $3$ and find the remainder, to convert the above to $\mathbb{Z}^3_3$: $$[1,1,2]$$ | 88 | 224 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-39 | longest | en | 0.699083 |
http://studylib.net/doc/6607186/solving-one-variable-equations | 1,531,834,934,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676589726.60/warc/CC-MAIN-20180717125344-20180717145344-00287.warc.gz | 349,353,008 | 12,654 | ```Algebra 1 Notes SOL A.4 Equations
Mr. Hannam
Name: __________________________________________ Date: _______________ Block: _______
Equations
An equation is an open sentence where two expressions are set _____________.
Equations may have one or more unknowns (_______________) which we may try to solve.
Solving an equation means finding the value(s) of variable(s) that make the equation
____________.
Equations that have the same solution(s) are called ____________________.
We use ______________________ operations to solve equations.
We justify the steps in solving equations by using field properties.
Solving Equations
“Isolate” the variable, justifying steps using field properties (properties of equality):
1) Put variables on one side of the = and numbers on the other by isolating x:
perform inverse operations (add, subtract, multiply, or divide)
2) Whatever you do to one side of the equation, you do to the other
3) Verify solutions
S
substitute solution in original equation (DO NOT SKIP THIS STEP!!!!!!)
Field Properties of Equality:
Property of Equality
Reflexive Property of Equality
Algebra (for real numbers a, b, c)
a=a
Symmetric Property of Equality
If a = b, then b = a
Transitive Property of Equality
If a = b and b = c, then a = c
Substitution Property of Equality
If a = b, then a can be substituted for b
Addition Property of Equality
If a = b, then a + c = b + c
Subtraction Property of Equality
If a = b, then a – c = b - c
Multiplication Property of
Equality
If a = b, then ac = bc
Division Property of Equality
If a = b and c 0, then
a b
c c
Example
Algebra 1 Notes SOL A.4 Equations
Solve the equations:
a) x – 4 = 6
x–4+4=6+4
x = 10
VERIFY: Is 10 a solution?
(10) – 4 = 6
6=6
Mr. Hannam Page 2
b) x + 3 = -5
x + 3 – 3 = -5 – 3
x = -8
VERIFY: Is -8 a solution?
(-8) + 3 = -5
-5 = -5
c) 4x = 16
4x ÷ 4 = 16 ÷ 4
x=4
VERIFY: Is 4 a solution?
4(4) = 16
16 = 16
Solve the equations, justifying steps:
a) Solve x + 4 = 10
1) x + 4 = 10
2) x + 4 – 4 = 10 – 4
3) x = 6
b) Solve
Given
Subtraction Property
of Equality
Simplify
2
x=4
3
2
x=4
Given
3
3 2
3
x 4 Mult. Property of Equality
2)
2 3
2
3) x = 6
Simplify
1)
Two-Step Equations
SAME! Put variables on one side of the = and numbers on the other:
How to Solve Two-Step Equations
1) Clear parentheses (distribute) and combine like terms if necessary
2) Do add/subtract steps first
3) Do multiply/divide steps
Examples:
x
5 11
2
x
5 5 11 5 ________________
2
x
16
________________
2
x
2 2 16
________________
2
x = 32
_________________
VERIFY:
a) 5x + 9 = 24
5x + 9 – 9 = 24 – 9 __________________
5x = 15
__________________
5x 15
__________________
5
5
x=3
__________________
VERIFY:
b)
c) 3x + 2x = 15
d) 4(x - 6) = 32
VERIFY:
VERIFY:
Algebra 1 Notes SOL A.4 Equations
Mr. Hannam Page 3
Practice - solve the equations, justifying steps with field properties:
a) x + 9 = 25
b) -4x = -20
a
46
3
e)
f) -1 =
z
3
37
3
x 3
5
d) 2x + 3 = -9
g) 7x – 4x = 21
h) 3(x + 2) = 9
c)
Multistep Equations and Equations with Variables on Both Sides
How to Solve Multi-Step Equations and Equations with Variables on Both Sides
1) Clear parentheses (distribute) and combine like terms if necessary
2) Add/subtract variable terms so that variable is on one side (NEW STEP!)
3) Do add/subtract steps first
4) Do multiply/divide steps
Example: Solve 7 – 3x = 4x – 7
1) 7 – 3x = 4x – 7
Given
2) 7 – 3x - 4x = 4x – 7 - 4x
________________________________________________
3) 7 – 7x = - 7
________________________________________________
4) 7 -7x - 7 = -7 - 7
________________________________________________
5) -7x = -14
________________________________________________
6)
________________________________________________
- 7x - 14
7
-7
7) x = 2
________________________________________________
Examples: Solve the equation, justifying steps…
1
a) 9x – 5 = (16x + 60)
b) 3(x + 12) – x = 4(2 – x) – 3x + 2x
4
9x – 5 = 4x + 15 ______________________
5x – 5 = 15
______________________
5x = 20
______________________
x=4
_____________________
Algebra 1 Notes SOL A.4 Equations
Mr. Hannam Page 4
Special cases:
Identity case (solutions are all real numbers):
No solution case:
2x + 6 = 2(x + 3)
3x = 3(x + 4)
2x + 6 = 2x + 6
Distribute
3x = 3x + 12 Distribute
0=0
Subtraction Prop. of =
0 = 12
Subtraction Prop. of =
Where did the variable go??
Where did the variable go??
When we get a TRUE statement at the end
when the variable “disappears,” EVERY x is
a solution (the two sides of the equation are
identical!).
When we get a FALSE statement at the end
when the variable “disappears,” there are NO
SOLUTIONS. There is no value of x that
makes the equation true.
ALL REAL NUMBERS are solutions.
Other Equations:
Solve: a)
x 3
4 2
Cross Product Property:
b)
If
x
2
x 1 3
a c
then ad = bc
b d
REMEMBER TO GROUP NUMERATORS AND DENOMINATORS (use parentheses)
You Try: Solve the equation if possible; if not, write “all real numbers” or “no solution”…
a) 8x + 5 = 6x + 1
b) x + 1 = 3x - 1
c) 9x = 6(x + 4)
d) 7(x + 7) = 5x + 59
e) 2 – 15x = 5(-3x + 2)
f) 12y + 6 = 6(2y + 1)
h) 40 + 14x = 2(-4x – 13)
i) 2(3x + 2) =
g) 5(x + 2) =
j)
x 9
2 3
3
(5 + 10x)
5
k)
4
8
x -8 2
l)
2
x
5 21 - x
1
(12x + 8)
2
``` | 1,937 | 5,231 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 5.03125 | 5 | CC-MAIN-2018-30 | latest | en | 0.801312 |
https://www.physicsforums.com/threads/is-this-acceptable.283521/ | 1,508,424,961,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823309.55/warc/CC-MAIN-20171019141046-20171019161046-00864.warc.gz | 954,527,570 | 15,607 | # Is this acceptable
1. Jan 7, 2009
### franky2727
prove R3=Udirect sum V
u={(a,2a,3a)} aER
v={(x,y,z)} x,y,z ER x+y+z=0
i solved the first bit UnV=0 but i'm having problems with the R3=U+W bit, my notes say i need to be able to "solve" a+x=b,,,2a+y=c,,,3a+z=d but what does this mean? is this simply putting the b,c,d all equal to one of the a,x,y or z?
i have done 6a=b+c+d is this adequate?
2. Jan 7, 2009
### tiny-tim
Hi franky2727!
I think they mean that, to prove that a general point (b,c,d) in R3 is in U + V, you have to find an (a,2a,3a) in U and an (x,y,z) in V which add to (b,c,d)
3. Jan 7, 2009
### Staff: Mentor
It might be helpful to think about the geometry here. The set {a, 2a, 3a} is a line in R3, while the equation x + y + z = 0 is a plane that passes through the origin, and whose normal is the vector (1, 1, 1).
What you're doing is showing that any arbitrary vector in R3 can be written as the sum of a vector along the line plus a vector that extends from the origin out to the plane somewhere.
4. Jan 8, 2009
### franky2727
you have to find an (a,2a,3a) in U and an (x,y,z) in V which add to (b,c,d)
well isnt that just a+x=b 2a+y=c 3a+z=d
5. Jan 8, 2009
### tiny-tim
Yup!
… but now you know why!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 458 | 1,345 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-43 | longest | en | 0.918005 |
https://www.jiskha.com/display.cgi?id=1216254023 | 1,503,272,457,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886106996.2/warc/CC-MAIN-20170820223702-20170821003702-00109.warc.gz | 938,319,883 | 3,691 | # Math 116 algebra please check
posted by .
Solve
0.3x + 8 < 1.1x-9
The solution is x > 85/7
The problem with the greater than sign has a line under it but I can get it to work correctly from my word document
• Math 116 algebra please check -
-0.8x < -17
8x > 170
4x > 85
Are you sure about that one?
## Respond to this Question
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10. ### Math
How do I graph inequality on a number line?
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# Calculate surface normal?
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Is there a function in DirectX8 to calculate the surface normal for a polygon? The only one I can find works for models only.
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There''s no function to do it, but the formula to calculate the normal of a polygon is pretty simple: Normal = Crossproduct of V and W, where V is a vector representing one side of the polygon and W is a vector representing another side.
Calculating surface normals at each vertex is a little more complex and involves averaging normalised polygon normals for each polygon the vertex is a part of.
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Never eat anything bigger than your own head.
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Just to clarify, if you have a triangle composed of points represented by vectors a,b,c, then you need to get 2 edges of the triangle (p,q) by:
p=b-a;
q=c-a;
Then take the Cross Product:
n = p cross q;
Then normalize the vector (if you need a unit vector):
normalize ( n );
This gives you the unit vector normal to the surface of the triangle.
If you want the vertex normal of a given vertex just sum up the triangle face normals of every triangle surrounding the given vertex, average them, and normalize the result.
for each vertex{ vertex_normal = 0,0,0; number_of_surrounding_faces = 0; for each triangle { if vertex is part of triangle { vertex_normal += triangle_normal; number_of_surrounding_faces++; } } vertex_normal /= number_of_surrounding_faces; normalize ( vertex_normal );}
Forgive the pseudo code, since I am at home for Thanksgiving without my computer. Of course if you are using the IDirect3dMesh class, it can do the vertex normal calculations for you.
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Sign me up! | 656 | 2,801 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-47 | latest | en | 0.867941 |
http://mrnussbaum.com/lunch-lady-lesson-plan-2/ | 1,524,421,905,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945637.51/warc/CC-MAIN-20180422174026-20180422194026-00356.warc.gz | 216,287,608 | 10,146 | Print Out a PDF of this Lesson
Mathematics
3-4
### Objective:
The Lunch Lady game will give students practice with calculating prices by adding decimals. Students will be able to correctly solve multi-digit addition problems involving decimals.
### Common Core Standards:
3.NBT.2 Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/or the relationship between addition and subtraction. 4.NBT.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm 4.OA.3 Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
### Warm-Up Activity:
Watch a short adding decimals math video as a way to engage your students and review the mathematical process of adding decimals prior to playing the “Lunch Lady” game. The adding decimals video can be accessed at the followed link.
http://mrnussbaum.com/decimals-video-5/
### Warm-Up Activity #2:
Another good warm-up activity prior to playing this game is to use the an interactive math workshop available on this site called – “Decimal Workshop”. In order to locate the interactive decimal workshop, go to the following link and select adding decimal. – http://mrnussbaum.com/decimals-workshop/
This activity could be done as a whole activity by allowing the students to work the problems out on dry erase boards. Then, the teacher can quickly identify which students are ready to play the game and which students might need extra practice adding decimals. The decimal workshop could also be used as an independent warm-up on individual student computers prior to playing the “Lunch Lady” game.
### Main Activity:
Now that your students have reviewed adding decimals they are ready to play the “Lunch Lady” game. Have then watch the instructional video at http://mrnussbaum.com/lunchlady/ prior to playing. This game could be played in a variety of ways. If you have lower level students that might need help, simply pair them with a higher level student in a computer lab situation. If you want to play the game as a whole class activity, ask all students to work each problem. Then, let them know that you will be randomly calling on students to answer. This will help to hold each student accountable, during a whole class version of this activity. | 500 | 2,465 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.28125 | 4 | CC-MAIN-2018-17 | latest | en | 0.900067 |
http://mathoverflow.net/feeds/question/75308 | 1,371,605,669,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368707439012/warc/CC-MAIN-20130516123039-00047-ip-10-60-113-184.ec2.internal.warc.gz | 160,922,034 | 2,854 | Not quite regular polyhedra - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:34:29Z http://mathoverflow.net/feeds/question/75308 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75308/not-quite-regular-polyhedra Not quite regular polyhedra Edmund Harriss 2011-09-13T13:03:15Z 2012-05-18T21:24:42Z <p>Take a naive interpretation of regular polyhedra:</p> <p>All vertices (including epsilon ball) congruent</p> <p>All edges congruent</p> <p>All faces congruent</p> <p>We can now find interesting families by removing one requirement. For example the uniform polyhedra have all vertices and edges congruent, but not all faces, and their duals have faces and edges congruent, but not vertices. </p> <p>Are there examples, or interesting families, of polyhedra where every pair of faces is congruent and every pair of vertices, but not every pair of edges?</p> http://mathoverflow.net/questions/75308/not-quite-regular-polyhedra/75311#75311 Answer by Douglas Zare for Not quite regular polyhedra Douglas Zare 2011-09-13T13:44:08Z 2011-09-13T13:44:08Z <p>Rhombic <a href="http://en.wikipedia.org/wiki/Disphenoid" rel="nofollow">disphenoids</a> are examples of polyhedra with identical vertices and faces, but distinguishable edges. These are irregular tetrahedra whose faces are scalene triangles. Their symmetry groups are isomorphic to \$C_2 \oplus C_2\$, and act transitively on the faces and vertices. You can make a disphenoid from an acute triangle by folding along the line segments connecting the midpoints of the sides. </p> <p>All tetrahedra whose sides have equal area are disphenoids. Also, ideal hyperbolic tetrahedra have the same symmetries as a disphenoid.</p> http://mathoverflow.net/questions/75308/not-quite-regular-polyhedra/76355#76355 Answer by Edmund Harriss for Not quite regular polyhedra Edmund Harriss 2011-09-25T18:42:20Z 2012-05-18T20:57:00Z <p>It turns out that these polyhedra that have congruent vertices and faces have a name. They are the <em>Noble Polyhedra</em>. If one insists that they also be convex the Noble polyhedra are the regular polyhedra plus the <a href="http://en.wikipedia.org/wiki/Disphenoid" rel="nofollow">disphenoids</a> mentioned in Douglas Zare's answer. </p> <p>When one allows intersecting faces, however, new collections turn up, such as the stephanoids, originally studied by Max Brüker:</p> <p>Max Brückner <em>Uber die gleichecking-gleichflachigen, diskontinuierlichen und nichtkonvexen Polyheder</em><br> Nova Acta Leop. 86(1906), No. 1, pp. 1 – 348 + 29 plates.<br> <a href="http://bulatov.org/polyhedra/bruckner1906/index.html" rel="nofollow">Images of the plates with pictures of the models.</a></p> <p>These shapes are also discussed and further developed by Branko Grünbaum:</p> <p>Polyhedra with hollow faces<br> <em>Proc. NATO-ASI Conf. on polytopes: abstract, convex and computational</em>, Toronto 1983, Ed. Bisztriczky, T. Et Al., Kluwer Academic (1994), p 43-70.</p> <p>Grünbaum's constructions do use generalisations of the definition of polyhedra. For a thorough discussion of these (including having polygons return to the same vertex, and coplanar faces) see the following paper, which also has a discussion of Noble Polyhedra. </p> <p>Grünbaum, B. Are your polyhedra the same as my polyhedra?<br> <em>Discrete and Computational Geometry: The Goodman-Pollack Festschrift</em> B. Aronov, S. Basu, J. Pach, and Sharir, M., eds. Springer, New York 2003, pp. 461 – 488.<br> <a href="http://www.math.washington.edu/~grunbaum/Your%20polyhedra-my%20polyhedra.pdf" rel="nofollow">http://www.math.washington.edu/~grunbaum/Your%20polyhedra-my%20polyhedra.pdf</a></p> <p>Interestingly the classification of all Noble Polyhedra is still an open problem.</p> http://mathoverflow.net/questions/75308/not-quite-regular-polyhedra/97354#97354 Answer by Patricia Hersh for Not quite regular polyhedra Patricia Hersh 2012-05-18T21:24:42Z 2012-05-18T21:24:42Z <p>Allan Edmonds has a few papers studying "equifacetal simplices", i.e. simplices in which any two facets must be congruent. Here are references:</p> <p>Edmonds, Allan, The center conjecture for equifacetal simplices. Adv. Geom. 9 (2009), no. 4, 563--576.</p> <p>Edmonds, Allan L., The partition problem for equifacetal simplices. Beitrage Algebra Geom. 50 (2009), no. 1, 195-213.</p> <p>Edmonds, Allan, The geometry of an equifacetal simplex. Mathematika 52 (2005), no. 1-2, 31-45.</p> | 1,341 | 4,462 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2013-20 | latest | en | 0.775455 |
https://admissions.jinnah.edu/faculty-of-engineering/complex-variables-and-transforms-mtme2053/ | 1,725,768,287,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650958.30/warc/CC-MAIN-20240908020844-20240908050844-00431.warc.gz | 73,381,163 | 7,147 | • To discuss the complex number system, different types of complex functions, analytic properties of complex numbers, theorems in complex analysis to carryout various mathematical operations in complex plane, roots of a complex equation.
• To discuss limits, continuity, differentiability, contour integrals, analytic functions and harmonic functions.
• Cauchy–Riemann equations in the Cartesian and polar coordinates, Cauchy’s integral formula, Cauchy–Goursat theorem, convergence of sequence and series, Taylor series, Laurents series.
• Integral transforms with a special focus on Laplace integral transform. Fourier transform.
[/vc_column_text][vc_custom_heading text=”COURSE LEARNING OUTCOMES (CLO)” use_theme_fonts=”yes”][vc_column_text]CLO-1: Define the complex number system, complex functions and integrals of complex functions (C1)
CLO-2: Explain the concept of limit, continuity, differentiability of complex valued functions (C2)
CLO-3: Apply the results/theorems in complex analysis to complex valued functions (C3)[/vc_column_text][vc_custom_heading text=”COURSE CONTENTS” use_theme_fonts=”yes”][vc_column_text css=”.vc_custom_1667208947336{margin-bottom: 0px !important;}”]
1. Introductory Concepts – Three Lectures
• Introduction to Complex Number System
• Argand diagram
• De Moivre’s theorem and its Application Problem Solving Techniques
2. Analyticity of Functions – Four Lectures
• Complex and Analytical Functions,
• Harmonic Function, Cauchy-Riemann Equations.
• Cauchy’s theorem and Cauchy’s Line Integral.
3. Singularities – Five Lectures
• Singularities, Poles, Residues.
• Contour Integration.
4. Laplace transform – Six Lectures
• Laplace transform definition,
• Laplace transforms of elementary functions
• Properties of Laplace transform, Periodic functions and their Laplace transforms,
• Inverse Laplace transform and its properties,
• Convolution theorem,
• Inverse Laplace transform by integral and partial fraction methods,
• Heaviside expansion formula,
• Solutions of ordinary differential equations by Laplace transform,
• Applications of Laplace transforms
5. Fourier series and Transform – Seven Lectures
• Fourier theorem and coefficients in Fourier series,
• Even and odd functions,
• Complex form of Fourier series,
• Fourier transform definition,
• Fourier transforms of simple functions,
• Magnitude and phase spectra,
• Fourier transform theorems,
• Inverse Fourier transform,
6. Solution of Differential Equations– Seven Lectures
• Series solution of differential equations,
• Validity of series solution, Ordinary point,
• Singular point, Forbenius method,
• Indicial equation,
• Bessel’s differential equation, its solution of first kind and recurrence formulae,
• Legendre differential equation and its solution,
• Rodrigues formula
[/vc_column_text][/vc_column][/vc_row] | 632 | 2,843 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-38 | latest | en | 0.675197 |
http://crypto.stackexchange.com/questions/1474/rsa-if-n-35-show-that-e-will-equal-d/1475 | 1,469,674,776,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827782.42/warc/CC-MAIN-20160723071027-00104-ip-10-185-27-174.ec2.internal.warc.gz | 47,087,804 | 17,373 | # RSA: If n=35, show that e will equal d
Show that if $n = 35$ is used as an RSA modulus then the encryption exponent $e$ always equals the decryption exponent $d$?
What I have so far:
$n = 35$
Therefore $p = 5$ and $q = 7$ or vice versa, which means Euler's phi function is
$\varphi = (5-1) \cdot (7-1) = 24$
Then to find $e$ we need the fact that $\gcd(e,\varphi) = 1$
$\gcd(e, 24) = 1$
Therefore the possible values of $e$ are only prime. $e = 5, 7, 11, 13, 17, 19, 23$
Then $d$ must follow the equation $d \equiv e^{-1} \mod \varphi$ or $(d\cdot e) \equiv 1 \mod 24$
Then I've tried for all those values which solve that equation. Why is that true? How can I explain this?
-
For extra credit, show that $n=91$ has the same property; that is, if $d=e$ is relatively prime to $\phi(91)$, then $(M^e)^d = M \mod 91$ for all $M$ (and hence, the encryption exponent $e$ always equals the decryption exponent $d$) – poncho Dec 15 '11 at 16:04
The number 24 has the curious property that, for any prime $p > 3$, $p^2 - 1$ is divisible by 24. (In fact, this holds for any odd number $p$ not divisible by 3.)
This follows simply from the fact that $p^2 - 1 = (p-1)(p+1)$; since $p$ is not divisible by either 2 or 3,
• the factors $p-1$ and $p+1$ must both be even,
• one of them must be divisible by 4, and
• one of them must be divisible by 3.
Thus, their product must be divisible by $2\cdot4\cdot3 = 24$.
- | 480 | 1,421 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.871926 |
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# help
0
33
3
+52
How many integers between 1 and 2009 inclusive can be expressed as a sum of five consecutive positive multiples of 7?
Aug 12, 2023
#1
0
Let the five consecutive multiples of 7 be 7n, 7(n+1), 7(n+2), 7(n+3), and 7(n+4). Then their sum is 35n+105=7(5n+15), so the integers which can be expressed as the sum of five consecutive positive multiples of 7 are just the multiples of 7 which are of the form 5n+15 for some integer n. The smallest such multiple is 7⋅2=14 and the largest is 7⋅287=2009, so there are 281 such integers.
Aug 12, 2023
#2
0
7 (n + n+1 + n+2 + n+3 + n +4)=35n + 70
35n + 70 <= 2009, solve for n
35n =2009 - 70=1939
n <= 1939 / 35 =55.40
n ==55 - number of 5 consecutive multiples of 7 between 1 and 2009.
Check : 7 + 14 + 21 + 28 + 35 = 105 - This is the smallest sum.
385 + 392 + 399 + 406 + 413 =1995 - this is the largest sum
The difference between any two consecutive sums of 5 multiples of 7 is =5 x 7 = 35.
[1995 - 105] / 35 + 1 =55
Aug 12, 2023
#3
+2
0
When you have a sequence of consecutive integers, the sum of the sequence can be found by multiplying the average of the first and last term by the number of terms. In this case, you have a sequence of five consecutive positive multiples of 7. The average of the first and last term would be (7 + 7*4) / 2 = 21.
So, the sum of the five consecutive positive multiples of 7 is:
Sum = Average * Number of Terms = 21 * 5 = 105.
Now, let's consider the possible values of the sum that can be obtained using this sequence. The minimum sum you can get is 105, which is when you use the sequence {7, 14, 21, 28, 35}. The maximum sum you can get is 105 + 4 * 7 = 133, which is when you use the sequence {28, 35, 42, 49, 56}.
Any integer between 105 and 133 (inclusive) can be expressed as the sum of five consecutive positive multiples of 7 using different sequences. Since both 105 and 133 are inclusive, there are a total of 133 - 105 + 1 = 29 integers between 1 and 2009 (inclusive) that can be expressed as the sum of five consecutive positive multiples of 7.
Aug 16, 2023 | 679 | 2,086 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-26 | latest | en | 0.934832 |
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### cdkrot's blog
By cdkrot, history, 6 years ago, translation,
Credits:
Div2A (Make a Trianle): Idea by Zlobober, development by vintage_Vlad_Makeev
Div2B (Equations of Mathematical Magic): Idea and development: Glebodin
Div1A (Oh Those Palindromes): Idea by Endagorion, development by Andreikkaa
Div1B (Labyrinth): Idea and development by qoo2p5
Div1C (Dwarves, Hats and Extrasensory Abilities): Idea and development by mingaleg
Div1D (Candies for Children): Idea by Endagorion, development by Flyrise
Div1E (Lasers and Mirrors): Idea and development by mingaleg
Div1F (String Journey): Original idea by GreenGrape, solution by vintage_Vlad_Makeev, development vintage_Vlad_Makeev and GreenGrape
• +62
| Write comment?
» 6 years ago, # | +37 D: I think the first part should be O(n). It is quite easy.
• » » 6 years ago, # ^ | 0 yes
» 6 years ago, # | ← Rev. 2 → +15 .
» 6 years ago, # | 0 Can anyone explain me intuition behind DIV2 B? I did not understand the tutorial that clearly? Thanks in advance !!:)
• » » 6 years ago, # ^ | +3 and a - x.Try looking yourself how they behave on some examples, like a = 3, a = 2 and some reasonable x's.A claim is that (always) and equality holds if and only if bits of x form a subset of bits of a.
• » » 6 years ago, # ^ | ← Rev. 2 → +9 a^x = a + x — 2 * (a & x) [ XOR property]So, a^x = a-x => a + x — 2 * (a & x) = a — x => a & x = xSo, this is true when we take any subset of ones (set bit) of 'a' => ans = (1 << (cnt of set bit in a))
• » » » 5 years ago, # ^ | 0 But wouldn't that give you some duplicate values? Like 100, hm can have abusers 0 and 00 which are the same.
» 6 years ago, # | ← Rev. 2 → +18 Regarding Div2B, another method to solve: Rewriting the equation we have:a = (a^x) + xNow use the property x+y = (x^y) + 2*(x&y) , where ^,& are bitwise xor,and respectively.Now the equation becomes:a = (a^x^x) + 2(x&(a^x))=> a = a + 2(x&(a^x))=> x&(a^x) = 0=> x&(!a) = 0 (Using a^x = ((a&!x) | (!a&x)))Hence we have bits of x should be a subset of a, hence number of solutions are 2^t where t is the number of set bits in a
» 6 years ago, # | 0 In Div2 D , Can someone explain why it can't use general bfs? Thank you!!
• » » 6 years ago, # ^ | ← Rev. 2 → +4 Consider we are now at (i,j).If the problem doesn't have the limit about move left and right,It's just a simple bfs problem.With limit,if one point can be reached by move (left,right) with many different ways, it is obviously that using less (left,right) move is optimal,because it maybe can reach more points.So we can just use bfs to record the less (left,right) move.See My Code 44337946
• » » » 3 years ago, # ^ | 0 in your code i saw that you have put the condition that the sum of left and right moves should be low to maintain the order of the priority queue , but the editorial says its sufficient to minimize either of the left or right moves and i actually did this but got a WA , can you explain why its wrong ?
• » » 6 years ago, # ^ | ← Rev. 2 → +1 Just using simple bfs , move down or up with affect the order about move left and right See this sample from ez_zh input20 73 65 2......*.****.*.****.*....*.*..*..*..*..*..*..*..*..*..*..*..*..*..*..***....******** output47
• » » » 6 years ago, # ^ | -18 Sample, not simple.
• » » 6 years ago, # ^ | +1 Because the fastest way to reach some block from the starting block may exceed the left and right constraints.
• » » 5 years ago, # ^ | 0 Because there are two kinds of edges.(i.e. Which costs zero and which costs one).So when you are doing bfs, you need to insert a node in the front of the queue if its dist equals to the previous dist,and insert in the back if its dist equals to the previous dist plus one. You can also use dijkstra by heap optimize or Bellmen-ford by queue optimize. Sorry for My poor English.:)
» 6 years ago, # | ← Rev. 2 → 0 Why this code 44315121 is getting runtime error? Question Div2.C.
• » » 6 years ago, # ^ | 0 That's div 2 C. You are a div 2 participant.
• » » » 6 years ago, # ^ | +3 That is the same question,however I changed my comment.Thank you.
» 6 years ago, # | ← Rev. 3 → +67 I've come up with a different approach for Div.1 D which runs in time, using the fact that has at most possible values, if both a and b are integers.Proof: If , , which has at most different values. Otherwise, b has at most different values. Lets call the number of taken candies of a single round num, which is n plus the number of people with a sweet tooth. Our purpose is to maximize num. To do this, we can check a bunch of num with same , which we'll call it r later, in the same time. Also, we'll call the number of children who took candies one more time than the others x, and k - r·num rem.Here are the constraints for num: num·r ≤ k num·(r + 1) > k n ≤ num ≤ 2n x ≤ rem ≤ 2x num - n - min(rem - x + 1, x) ≤ n - x Constraint 1 and 2 ensures the validity of r, constraint 3 ensures the validity of num, and constraint 4 and 5 ensures that there are enough space for children with a sweet tooth.We can use some simple math to deform these constraints so that num will appear at only one side of the inequation.Here's the original version: 44345319Here's the trimmed version so that it was the shortest on 2018-10-15: 44345626
• » » 6 years ago, # ^ | +8 Some questions: In definition of rem, did you mean k - r·num instead of k - r·sum? Given that there may be someone with a sweet tooth in x, why the 4th inequation isn't 0 ≤ rem ≤ 2x? What does the 5th inequation mean? I can't understand it. Thanks in advance.
• » » » 6 years ago, # ^ | 0 Yes, it was a typo. Thanks for pointing out. Because at least x candies were taken out by x people. num - n - min(rem - x + 1, x) is the number of people with a sweet tooth other than the x people.
• » » » » 6 years ago, # ^ | 0 Thanks for explanation. But I got confused when I looked into your code. The following inequation is used in your code:And the result conflicts with the 4th inequation you mentioned above. Do I misunderstand something in your code?Another question: in num - n - min(rem - x + 1, x), why we need +1 to rem - x? Just because the case where the last people with a sweet tooth?
• » » » » » 6 years ago, # ^ | 0 Sorry, I think the fourth in-equation should be x ≤ rem ≤ 2x.Reply to the second question: Because the last person may have a sweet tooth even if he took only 1 candy.
» 6 years ago, # | 0 Div 2 CNote that each palindrome have equal character at their ends. Suppose this character is c with x number of occurences. Then there are at most x(x+1)/2 palindromes with this character.Can someone elaborate this. How did we get 1+2+...+x if c has x number of occurences?
• » » 6 years ago, # ^ | +3 Since 'c' has x number of occurrences, thus palindrome with 'c' are x , palindrome which are 'cc' are x-1 , palindrome which are 'ccc' are x-2 and so on.so we get x+x-1+....+2+1 = x(x+1)/2 for 'c' which has frequency x.
• » » » 6 years ago, # ^ | 0 Your explanation is not correct. You are counting number of palindrome substrings of string 'cccc..' lengh x. If we have x characters 'c', then we can choose x * (x — 1) / 2 pairs of characters 'c' to be head and tail of a substring, or we can choose x substring "c", so in total we have x * (x + 1) / 2 ways to choose a substring which may be a palindrome substring. That why x * (x + 1) / 2 is the upper bound.
• » » » » 6 years ago, # ^ | +3 I think you misunderstood my explanation, I am saying that we have x ways to choose a substring "c", x-1 ways to choose a substring "cc", x-2 ways to choose "ccc" and so on till we have 1 way to choose "ccc.. (x times) ".so we get x+x-1+x-2+...+1 = x(x+1)/2
• » » » » » 6 years ago, # ^ | 0 I mean your explanation didn't show how to get the upper bound. You was counting number of substrings of string "cccc..." length x, and be cause there is only one charater 'c', so they were also palindrome substrings. The result you got was of the case when the upper bound was satisfied.
• » » » » » 6 years ago, # ^ | 0 Yeah, you wasn't wrong. Maybe i'm understanding the question in a different way
» 6 years ago, # | 0 Why this code is not working, please point out the error (Div-2 D) CODE
• » » 6 years ago, # ^ | 0 4 5 4 2 1 2 ...** .*... .***. .....Correct ans: 10. Your bfs will reach 2,3 but not with minimum left or right moves. That's why simple bfs with all edge weights same won't work as moving more up and down is feasible rather than moving left or right.
» 6 years ago, # | +3 Tutorial for Div2-A(1064A) has an error, answer should be max(0,c-(a+b)+1) instead of max(0,c-(a+b+1)) where c = largest side of given triangle.
• » » 6 years ago, # ^ | +8 Fixed it, thanks.
» 6 years ago, # | ← Rev. 2 → +73 For question F The way I see the fastest code is this: Use dp[i] to end the longest trip with i We insert all the suffixes of the last string (just dp[i]) into the hash table from long to short. If a string is inserted, stop inserting (the shorter suffix has already been inserted) So when we query whether the journey of the length r-l+1 ending in r (string [l, r]) is legal Just need to query whether [l, r-1] and [l+1, r] appear in the hash table. This does not require a suffix data structure, but I want to know what the time complexity is? O(n^1.5) or O(n)?
• » » 6 years ago, # ^ | +42 I think I have a proof for O(n). Assume that the dp is from left to right and dp[i] = length of the longest trip ending at i.Let S(i) be the set of the smallest dp[j] substrings ending at j for each j from 1 to i. Let c1(i) be the number of distinct strings of the set S(i). Call an index i type A if the second last string of the trip ending at [i-dp[i]+1, i] is equal to [i-dp[i]+1, i-1] and type B if it is [i-dp[i]+2, i]. It seems that c1(i) <= 2*i-dp[i+1]+O(1) (O(1) is some constant that I'm too lazy to find). We'll proceed by induction.Case 1: i is type A and i+1 is type ASince i+1 is type A, there is an occurrence of the string [i+1-dp[i+1]+1, i] before i+1-dp[i+1]+1. Thus, the substrings [i+1-dp[i+1]+1..i, i] do not contribute to c1(i), so c1(i) <= c1(i-1)+(i+1-dp[i+1]+1)-(i-dp[i]+1) <= 2*(i-1)-dp[i+1]+1+O(1).Case 2: i is type A and i+1 is type BSimilar argument to Case 1, c1(i) <= 2*(i-1)-dp[i+1]+2+O(1).Case 3: i is type BOnly [i-dp[i]+1, i] can be a new distinct string and dp[i+1]-1 <= dp[i], so c1(i) <= c2(i)+1+dp[i]-dp[i+1]+1 <= 2*(i-1)+2-dp[i+1]+O(1).
• » » » 6 years ago, # ^ | +25 I understand. Cool proof!
» 6 years ago, # | 0 In Div2 D / Div1 BR - L = j1 - j0 = const But it can be written this way L — R = j1 — j0 = constIf we optimise any one of L and R, How to count the number of moves for another one at any (i, j) which equation of the above two should we use ? I am getting correct answer for equation R — L = j1 — j0 if i am consdering cost 1 for left move and 0 for other moves. why is that ?
» 6 years ago, # | 0 Can someone explain how a+b is linear with respect to t in 1063D?
• » » 6 years ago, # ^ | 0 Actually we treat t as a constant, and a+b is parametric of z.Now we just have to say the parametrica = a0 + (t+1)zb = b0 + tz, z is element of integersis a line.Solve for z = (a-a0)/(t+1) = (b-b0)/tSince t, a0, b0 are constants this is actually just point-slope form of a line. (slopes are 1/(t+1) and 1/t)
» 6 years ago, # | 0 Why can't we use a simple DFS in Div2D? It gives wrong answer on test 13. Here is my submission
• » » 6 years ago, # ^ | ← Rev. 2 → 0 let, describe each cell as a struct { row, col, leftCount, rightCount }assume, you have a visited cell C and your current cell is CC; where (CC.row == C.row) and (C.col == CC.col+1)if(C.rightCount>(CC.rightCount + 1) || C.leftCount>CC.leftCount){ you have to call dfs with cell C again; why? because, now C has lesser leftCount or rightCount than before....so it may able to visit more cells than before} here is my code with bfs https://mirror.codeforces.com/contest/1064/submission/44414320
• » » 5 years ago, # ^ | 0 I did the same to you.
» 6 years ago, # | 0 Div2 E / Div1 CIs following approach right??initially b=1 , w = n+2if (white) print (b++, 1) ;;else print (w++, 1) ;;line -> (n+1,0) — (n+1,3)and why i'm getting "idleness limit exceeded on test 1" ??https://mirror.codeforces.com/problemset/submission/1064/44428642Thanks in advance.
• » » 6 years ago, # ^ | 0 You are supposed to give the interactor the position of point first, and then the interactor would tell you the color.
• » » » 6 years ago, # ^ | 0 OH!...Thanks. I never solved this type of problem!
» 6 years ago, # | ← Rev. 2 → 0 In Div2 C editorial, it is written that, It is easy to see, that the sorted string fulfills that bound Would anybody explain me it, please? I do not understand this.
• » » 6 years ago, # ^ | +3 The way I like to think of it is trying all size "windows" on a single letter.So imagine we have the stringa .... a .... a .... a where a's are evenly spaced outIn the optimal solution, the whole string is a palindrome and this satisfies the window of size 4. For example: abababa. Maximum there is 1 such window.Now lets look at window of size 3. We need [a ... a ... a] to be a palindrome. There are maximum of 2 such windows.Eventually we get 1+2+...cntA windows where cntA = amount of a's in the string.Each letter should contribute 1+2+3...cnt = (cnt)(cnt+1)/2 palindromes in an optimal solution.It is easy to see that a sorted string will always be optimal, because this property holds true. Actually any solution that "bunches" same letters together is optimal, so you can have ab O(n) solution (counting sort idea)
» 6 years ago, # | 0 In Div2-D why This Code fails but This Code passes. According to editorial minimizing either of L or R should work but if this is the case then the former code should not give WA verdict on test 20. What am I missing here?
• » » 6 years ago, # ^ | 0 Handle the case where l or r hits negative (they are invalid)
• » » » 6 years ago, # ^ | 0 l and r will always be positive because i am checking l>0 r>0 before pushing them in deque. Check out this. It gives WA not Runtime error. So l or r are always >= 0
• » » » » 6 years ago, # ^ | 0 It only confirms the node you are on meets this condition, not the neighbors you mark
• » » » » » 6 years ago, # ^ | 0 Actually the same code passes Test case 20 if i use queue instead of deque. Linksince the current node has l>0 and r>0 and in one move they can be only decreased by one so l>=0 and r>=0.It would be great if you could provide any test case on which my code with deque fails?
• » » » » » » 6 years ago, # ^ | 0 It's still incorrect if l > 0 and r > 0 you will only go to l=1,r=0 and l=0,r=1 cases, and miss the cases where l = r = 0The fact that queue passes 20 just means it happens to work in the test case, not that it is any more "correct"
• » » » » » » 6 years ago, # ^ | 0 Ok I just edited yours a resubmitted, http://mirror.codeforces.com/contest/1064/submission/44520194The l > 0 and r > 0 thing you said isnt even true, so you still have to fix the original issue of confirming the neighbors work
• » » » 6 years ago, # ^ | 0 Thanks brdy. I solved the issue and got it accepted.The issue was with my implementation of 0-1 bfs. 0-1 bfs generally involves relaxing edges but here when I was processing a node I was immediately marking its neighbours visited without considering the fact that since there may be more elements with the same L value in the queue there might exist a better path. 10 14 7 7 7 7 ******..****** **...*.******* **.*.......... ...**********. .************. .*...********. .*.*.*........ .*.*.*.******* .*.*.*.******* ...*...******* Correct output:53My output was 52 on this case as both (3,8) and (10,1) nodes will have L value 1 but since (3,8) entered the queue first it will mark (3,7) while there exists a better path from (10,1).
» 6 years ago, # | 0 Why does it surprise me that solution for interactive problem is binary search?
» 6 years ago, # | 0 For Div 1 D, since first part can be done in O(n), binary search on number of sweet tooth in second part also pass the TL. ( O( min( n, (k / n) * log(n) ) ) ), about 2e6 operations.
» 5 years ago, # | 0 For b, in div2, if a is 100, then the 2 subsets 0 and 00 are the same number right? So wouldnt the 2^k subsets have duplicate vakues?
» 5 years ago, # | 0 In div2 D, I didn't understand how 1-2 bfs is used to minimize the number of left and right moves used to reach a position. Can someone please explain
• » » 4 years ago, # ^ | 0 You are doing a 0-1 BFS where each (i,j) is connected to up to 4 neighbors, with cost of vertical edges being 1 and cost of horizontal edges being 0. This means that at the end of the search, costTo[i][j] = numLeftsUsed+numRightsUsed.Now since you are starting in column c and ending in column j, you must have numRightsUsed-numLeftsUsed = j-c. For instance if you start in column 1 and end in column 3, you might have used 0 lefts and 2 rights, or 1 lefts and 3 rights, or 2 lefts and 4 rights, etc.This lets you solve for costTo[i][j] in terms of just numLeftsUsed or numRightsUsed:numRightsUsed=j-c+numLeftsUsed so costTo[i][j] = numLeftsUsed + j — c + numLeftsUsed and similarly for expressing in terms of just numRightUsed.Minimizing costTo[i][j] = numLeftsUsed+numRightsUsed = j-c+2*numLeftsUsed is equivalent to minimizing numLeftsUsed.
• » » » 4 years ago, # ^ | 0 with 0-1 bfs, shouldn't you take horizontal edges with weight 1 instead?
» 4 years ago, # | 0 In,div2D,why this code is getting TLE?? https://mirror.codeforces.com/contest/1064/submission/85460883
» 3 years ago, # | 0 DIV2 D -->just look at this solution :: you will understand https://mirror.codeforces.com/contest/1064/submission/107079705
» 9 months ago, # | ← Rev. 3 → 0 Can somebody please tell me why am i getting TLE at test case 8 in my submission for Div.1 B question https://mirror.codeforces.com/contest/1063/submission/216168661 this is the link for my submission, i am doing a bfs traversal and with up down wights as 0 and left right weights as 1, thanks for your help | 5,638 | 18,255 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-18 | longest | en | 0.849857 |
https://ask.sagemath.org/question/35031/leading-coefficient-polynomial/?sort=oldest | 1,560,675,646,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627998084.36/warc/CC-MAIN-20190616082703-20190616104703-00271.warc.gz | 359,216,199 | 15,495 | This post is a wiki. Anyone with karma >750 is welcome to improve it.
Hello everybody,
I'm new to sagemath and python in general, and one of my course in Uni uses it... I have a vague and unclear tutorial the prof gave us and for now I know only the most basic commands.
I have to write a function that takes a polynomial of any degree and tells me the coefficient of the highest degree member (for example , 2x^4+3x^3 would be 2, 7x^3+2x^4+2 would be 7...).
I think the function would have to use "expand", "degree", and of course "coefficient". But i barely have any idea as how to write it.
If anyone could help me it would be great, I am kinda lost here...
Sorry for sloppy english and thanks in advance.
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Hint: if P is your polynomial, you should ask for its coefficients, you will get a list, the last element of that list is the leading coefficient.
more
For example,
P=7*x^3+2*x^4+2
you can extract its coefficients as follows:
P.coefficients()
to get
[[2, 0], [7, 3], [2, 4]]
So the last sublist contains the answer, in this case 2 is the coefficient of x^4.
To extract this automatically you can define a function that takes a polynomial as argument:
def get_highest_coeff(P):
m = P.coefficients()
ncoeffs = len(m)
return m[ncoeffs - 1][0]
And use it like this:
get_highest_coeff(P)
to get 2 as the answer. Hope this helps you get going!
more
If you have
P=7*x^3+2*x^4+2
then type P.[TAB] to see a list of possible "attributes" or "methods" attached to P. Other people have mentioned P.coefficients(). You might also hope that there is something like a leading_coefficient method, so try P.l[TAB] to find everything starting with the letter l. Then you should see P.leading_coefficient, so type
P.leading_coefficient?
to find out how to use it.
more
Sage trac ticket 21608 provides methods for "leading term", "leading coefficient", "leading monomial", and is available in Sage 7.5.beta6.
more | 524 | 2,005 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2019-26 | longest | en | 0.925696 |
https://mastr.clontz.org/diff-eq/C5/ | 1,571,754,845,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987822098.86/warc/CC-MAIN-20191022132135-20191022155635-00296.warc.gz | 616,167,924 | 4,121 | ## C5 - Homogeneous second-order linear IVP (ver. 1)
Find the solution to the given IVP.
$y''-2y'-15y = 0 \hspace{1em} y(0) = 8 , y'(0) = 16$
$y= 5 \, e^{\left(5 \, t\right)} + 3 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 2)
Find the solution to the given IVP.
$y''+6y'+5y = 0 \hspace{1em} y(0) = 0 , y'(0) = 8$
$y= 2 \, e^{\left(-t\right)} - 2 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 3)
Find the solution to the given IVP.
$y''+2y'-15y = 0 \hspace{1em} y(0) = 5 , y'(0) = -1$
$y= 3 \, e^{\left(3 \, t\right)} + 2 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 4)
Find the solution to the given IVP.
$y''+3y'-4y = 0 \hspace{1em} y(0) = -1 , y'(0) = -21$
$y= 4 \, e^{\left(-4 \, t\right)} - 5 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 5)
Find the solution to the given IVP.
$y''+y'-2y = 0 \hspace{1em} y(0) = -2 , y'(0) = 4$
$y= -2 \, e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 6)
Find the solution to the given IVP.
$y''-3y'-10y = 0 \hspace{1em} y(0) = 1 , y'(0) = 12$
$y= 2 \, e^{\left(5 \, t\right)} - e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 7)
Find the solution to the given IVP.
$y''-8y'+15y = 0 \hspace{1em} y(0) = 4 , y'(0) = 18$
$y= 3 \, e^{\left(5 \, t\right)} + e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 8)
Find the solution to the given IVP.
$y''-2y'-15y = 0 \hspace{1em} y(0) = 6 , y'(0) = 22$
$y= 5 \, e^{\left(5 \, t\right)} + e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 9)
Find the solution to the given IVP.
$y''+7y'+12y = 0 \hspace{1em} y(0) = -3 , y'(0) = 12$
$y= -3 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 10)
Find the solution to the given IVP.
$y''+8y'+15y = 0 \hspace{1em} y(0) = 1 , y'(0) = -13$
$y= -4 \, e^{\left(-3 \, t\right)} + 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 11)
Find the solution to the given IVP.
$y''+4y'+3y = 0 \hspace{1em} y(0) = -6 , y'(0) = 10$
$y= -4 \, e^{\left(-t\right)} - 2 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 12)
Find the solution to the given IVP.
$y''-5y'+6y = 0 \hspace{1em} y(0) = 7 , y'(0) = 17$
$y= 3 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 13)
Find the solution to the given IVP.
$y''-3y'+2y = 0 \hspace{1em} y(0) = -6 , y'(0) = -10$
$y= -4 \, e^{\left(2 \, t\right)} - 2 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 14)
Find the solution to the given IVP.
$y''+6y'+5y = 0 \hspace{1em} y(0) = 1 , y'(0) = -5$
$y= e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 15)
Find the solution to the given IVP.
$y''-5y'+4y = 0 \hspace{1em} y(0) = -1 , y'(0) = 5$
$y= 2 \, e^{\left(4 \, t\right)} - 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 16)
Find the solution to the given IVP.
$y''-4y'+3y = 0 \hspace{1em} y(0) = -1 , y'(0) = -11$
$y= -5 \, e^{\left(3 \, t\right)} + 4 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 17)
Find the solution to the given IVP.
$y''+0y'-9y = 0 \hspace{1em} y(0) = 1 , y'(0) = 9$
$y= 2 \, e^{\left(3 \, t\right)} - e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 18)
Find the solution to the given IVP.
$y''+8y'+15y = 0 \hspace{1em} y(0) = 5 , y'(0) = -25$
$y= 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 19)
Find the solution to the given IVP.
$y''-7y'+10y = 0 \hspace{1em} y(0) = -2 , y'(0) = -1$
$y= e^{\left(5 \, t\right)} - 3 \, e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 20)
Find the solution to the given IVP.
$y''+2y'-15y = 0 \hspace{1em} y(0) = 1 , y'(0) = -29$
$y= -3 \, e^{\left(3 \, t\right)} + 4 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 21)
Find the solution to the given IVP.
$y''+0y'-4y = 0 \hspace{1em} y(0) = 3 , y'(0) = -14$
$y= -2 \, e^{\left(2 \, t\right)} + 5 \, e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 22)
Find the solution to the given IVP.
$y''+3y'+2y = 0 \hspace{1em} y(0) = 5 , y'(0) = -10$
$y= 5 \, e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 23)
Find the solution to the given IVP.
$y''+6y'+5y = 0 \hspace{1em} y(0) = -4 , y'(0) = 24$
$y= e^{\left(-t\right)} - 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 24)
Find the solution to the given IVP.
$y''-5y'+6y = 0 \hspace{1em} y(0) = 3 , y'(0) = 10$
$y= 4 \, e^{\left(3 \, t\right)} - e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 25)
Find the solution to the given IVP.
$y''-8y'+15y = 0 \hspace{1em} y(0) = 4 , y'(0) = 22$
$y= 5 \, e^{\left(5 \, t\right)} - e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 26)
Find the solution to the given IVP.
$y''-6y'+5y = 0 \hspace{1em} y(0) = 6 , y'(0) = 10$
$y= e^{\left(5 \, t\right)} + 5 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 27)
Find the solution to the given IVP.
$y''+4y'-5y = 0 \hspace{1em} y(0) = -4 , y'(0) = -10$
$y= e^{\left(-5 \, t\right)} - 5 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 28)
Find the solution to the given IVP.
$y''+2y'-15y = 0 \hspace{1em} y(0) = 0 , y'(0) = -24$
$y= -3 \, e^{\left(3 \, t\right)} + 3 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 29)
Find the solution to the given IVP.
$y''-3y'-10y = 0 \hspace{1em} y(0) = -2 , y'(0) = 11$
$y= e^{\left(5 \, t\right)} - 3 \, e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 30)
Find the solution to the given IVP.
$y''-3y'+2y = 0 \hspace{1em} y(0) = -3 , y'(0) = -1$
$y= 2 \, e^{\left(2 \, t\right)} - 5 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 31)
Find the solution to the given IVP.
$y''+3y'-4y = 0 \hspace{1em} y(0) = 3 , y'(0) = 8$
$y= -e^{\left(-4 \, t\right)} + 4 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 32)
Find the solution to the given IVP.
$y''-7y'+12y = 0 \hspace{1em} y(0) = 2 , y'(0) = 11$
$y= 5 \, e^{\left(4 \, t\right)} - 3 \, e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 33)
Find the solution to the given IVP.
$y''-4y'-5y = 0 \hspace{1em} y(0) = -2 , y'(0) = 20$
$y= 3 \, e^{\left(5 \, t\right)} - 5 \, e^{\left(-t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 34)
Find the solution to the given IVP.
$y''-2y'-8y = 0 \hspace{1em} y(0) = 3 , y'(0) = 18$
$y= 4 \, e^{\left(4 \, t\right)} - e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 35)
Find the solution to the given IVP.
$y''-5y'+6y = 0 \hspace{1em} y(0) = 7 , y'(0) = 19$
$y= 5 \, e^{\left(3 \, t\right)} + 2 \, e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 36)
Find the solution to the given IVP.
$y''+8y'+15y = 0 \hspace{1em} y(0) = -7 , y'(0) = 25$
$y= -5 \, e^{\left(-3 \, t\right)} - 2 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 37)
Find the solution to the given IVP.
$y''+y'-20y = 0 \hspace{1em} y(0) = -1 , y'(0) = -4$
$y= -e^{\left(4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 38)
Find the solution to the given IVP.
$y''+y'-2y = 0 \hspace{1em} y(0) = 0 , y'(0) = 15$
$y= -5 \, e^{\left(-2 \, t\right)} + 5 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 39)
Find the solution to the given IVP.
$y''+9y'+20y = 0 \hspace{1em} y(0) = -8 , y'(0) = 37$
$y= -3 \, e^{\left(-4 \, t\right)} - 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 40)
Find the solution to the given IVP.
$y''+3y'+2y = 0 \hspace{1em} y(0) = 1 , y'(0) = -1$
$y= e^{\left(-t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 41)
Find the solution to the given IVP.
$y''+3y'-10y = 0 \hspace{1em} y(0) = 5 , y'(0) = -25$
$y= 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 42)
Find the solution to the given IVP.
$y''-6y'+8y = 0 \hspace{1em} y(0) = -6 , y'(0) = -22$
$y= -5 \, e^{\left(4 \, t\right)} - e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 43)
Find the solution to the given IVP.
$y''-5y'+6y = 0 \hspace{1em} y(0) = 5 , y'(0) = 14$
$y= 4 \, e^{\left(3 \, t\right)} + e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 44)
Find the solution to the given IVP.
$y''+4y'+3y = 0 \hspace{1em} y(0) = -2 , y'(0) = 12$
$y= 3 \, e^{\left(-t\right)} - 5 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 45)
Find the solution to the given IVP.
$y''-6y'+5y = 0 \hspace{1em} y(0) = -6 , y'(0) = -22$
$y= -4 \, e^{\left(5 \, t\right)} - 2 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 46)
Find the solution to the given IVP.
$y''+y'-6y = 0 \hspace{1em} y(0) = 1 , y'(0) = 22$
$y= 5 \, e^{\left(2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 47)
Find the solution to the given IVP.
$y''+6y'+5y = 0 \hspace{1em} y(0) = -8 , y'(0) = 28$
$y= -3 \, e^{\left(-t\right)} - 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 48)
Find the solution to the given IVP.
$y''-5y'+4y = 0 \hspace{1em} y(0) = 4 , y'(0) = 7$
$y= e^{\left(4 \, t\right)} + 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 49)
Find the solution to the given IVP.
$y''+5y'+6y = 0 \hspace{1em} y(0) = 1 , y'(0) = 2$
$y= 5 \, e^{\left(-2 \, t\right)} - 4 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 50)
Find the solution to the given IVP.
$y''-2y'-15y = 0 \hspace{1em} y(0) = -3 , y'(0) = -7$
$y= -2 \, e^{\left(5 \, t\right)} - e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 51)
Find the solution to the given IVP.
$y''-2y'-8y = 0 \hspace{1em} y(0) = 8 , y'(0) = 14$
$y= 5 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 52)
Find the solution to the given IVP.
$y''-5y'+6y = 0 \hspace{1em} y(0) = 1 , y'(0) = 2$
$y= e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 53)
Find the solution to the given IVP.
$y''+y'-20y = 0 \hspace{1em} y(0) = -4 , y'(0) = 29$
$y= e^{\left(4 \, t\right)} - 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 54)
Find the solution to the given IVP.
$y''-7y'+12y = 0 \hspace{1em} y(0) = -1 , y'(0) = 0$
$y= 3 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 55)
Find the solution to the given IVP.
$y''-4y'-5y = 0 \hspace{1em} y(0) = 2 , y'(0) = 10$
$y= 2 \, e^{\left(5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 56)
Find the solution to the given IVP.
$y''-1y'-12y = 0 \hspace{1em} y(0) = 1 , y'(0) = 18$
$y= 3 \, e^{\left(4 \, t\right)} - 2 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 57)
Find the solution to the given IVP.
$y''+0y'-16y = 0 \hspace{1em} y(0) = 3 , y'(0) = 4$
$y= 2 \, e^{\left(4 \, t\right)} + e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 58)
Find the solution to the given IVP.
$y''+6y'+8y = 0 \hspace{1em} y(0) = -7 , y'(0) = 18$
$y= -5 \, e^{\left(-2 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 59)
Find the solution to the given IVP.
$y''+6y'+8y = 0 \hspace{1em} y(0) = 1 , y'(0) = 2$
$y= 3 \, e^{\left(-2 \, t\right)} - 2 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 60)
Find the solution to the given IVP.
$y''+3y'+2y = 0 \hspace{1em} y(0) = 6 , y'(0) = -7$
$y= 5 \, e^{\left(-t\right)} + e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 61)
Find the solution to the given IVP.
$y''-3y'+2y = 0 \hspace{1em} y(0) = 3 , y'(0) = 3$
$y= 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 62)
Find the solution to the given IVP.
$y''-1y'-2y = 0 \hspace{1em} y(0) = -5 , y'(0) = -4$
$y= -3 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 63)
Find the solution to the given IVP.
$y''+3y'-10y = 0 \hspace{1em} y(0) = -5 , y'(0) = 4$
$y= -3 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 64)
Find the solution to the given IVP.
$y''-6y'+5y = 0 \hspace{1em} y(0) = 2 , y'(0) = -2$
$y= -e^{\left(5 \, t\right)} + 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 65)
Find the solution to the given IVP.
$y''+7y'+12y = 0 \hspace{1em} y(0) = 8 , y'(0) = -29$
$y= 3 \, e^{\left(-3 \, t\right)} + 5 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 66)
Find the solution to the given IVP.
$y''+8y'+15y = 0 \hspace{1em} y(0) = -5 , y'(0) = 17$
$y= -4 \, e^{\left(-3 \, t\right)} - e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 67)
Find the solution to the given IVP.
$y''+0y'-25y = 0 \hspace{1em} y(0) = -2 , y'(0) = 40$
$y= 3 \, e^{\left(5 \, t\right)} - 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 68)
Find the solution to the given IVP.
$y''-6y'+5y = 0 \hspace{1em} y(0) = 0 , y'(0) = 12$
$y= 3 \, e^{\left(5 \, t\right)} - 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 69)
Find the solution to the given IVP.
$y''+0y'-9y = 0 \hspace{1em} y(0) = -4 , y'(0) = 12$
$y= -4 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 70)
Find the solution to the given IVP.
$y''+4y'-5y = 0 \hspace{1em} y(0) = -2 , y'(0) = 28$
$y= -5 \, e^{\left(-5 \, t\right)} + 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 71)
Find the solution to the given IVP.
$y''+7y'+10y = 0 \hspace{1em} y(0) = 3 , y'(0) = -9$
$y= 2 \, e^{\left(-2 \, t\right)} + e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 72)
Find the solution to the given IVP.
$y''+0y'-16y = 0 \hspace{1em} y(0) = -1 , y'(0) = 4$
$y= -e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 73)
Find the solution to the given IVP.
$y''-7y'+12y = 0 \hspace{1em} y(0) = 1 , y'(0) = 1$
$y= -2 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 74)
Find the solution to the given IVP.
$y''+y'-6y = 0 \hspace{1em} y(0) = 4 , y'(0) = 8$
$y= 4 \, e^{\left(2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 75)
Find the solution to the given IVP.
$y''+7y'+10y = 0 \hspace{1em} y(0) = 6 , y'(0) = -15$
$y= 5 \, e^{\left(-2 \, t\right)} + e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 76)
Find the solution to the given IVP.
$y''-9y'+20y = 0 \hspace{1em} y(0) = 5 , y'(0) = 20$
$y= 5 \, e^{\left(4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 77)
Find the solution to the given IVP.
$y''-1y'-20y = 0 \hspace{1em} y(0) = -6 , y'(0) = 6$
$y= -2 \, e^{\left(5 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 78)
Find the solution to the given IVP.
$y''+2y'-8y = 0 \hspace{1em} y(0) = 2 , y'(0) = -26$
$y= -3 \, e^{\left(2 \, t\right)} + 5 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 79)
Find the solution to the given IVP.
$y''+6y'+8y = 0 \hspace{1em} y(0) = -8 , y'(0) = 26$
$y= -3 \, e^{\left(-2 \, t\right)} - 5 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 80)
Find the solution to the given IVP.
$y''-3y'-4y = 0 \hspace{1em} y(0) = 5 , y'(0) = -5$
$y= 5 \, e^{\left(-t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 81)
Find the solution to the given IVP.
$y''+9y'+20y = 0 \hspace{1em} y(0) = -9 , y'(0) = 41$
$y= -4 \, e^{\left(-4 \, t\right)} - 5 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 82)
Find the solution to the given IVP.
$y''+7y'+10y = 0 \hspace{1em} y(0) = 4 , y'(0) = -11$
$y= 3 \, e^{\left(-2 \, t\right)} + e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 83)
Find the solution to the given IVP.
$y''+4y'+3y = 0 \hspace{1em} y(0) = -7 , y'(0) = 15$
$y= -3 \, e^{\left(-t\right)} - 4 \, e^{\left(-3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 84)
Find the solution to the given IVP.
$y''+3y'-4y = 0 \hspace{1em} y(0) = -2 , y'(0) = 8$
$y= -2 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 85)
Find the solution to the given IVP.
$y''+6y'+8y = 0 \hspace{1em} y(0) = 5 , y'(0) = -18$
$y= e^{\left(-2 \, t\right)} + 4 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 86)
Find the solution to the given IVP.
$y''+0y'-4y = 0 \hspace{1em} y(0) = 5 , y'(0) = -2$
$y= 2 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-2 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 87)
Find the solution to the given IVP.
$y''-7y'+12y = 0 \hspace{1em} y(0) = 1 , y'(0) = -1$
$y= -4 \, e^{\left(4 \, t\right)} + 5 \, e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 88)
Find the solution to the given IVP.
$y''+2y'-8y = 0 \hspace{1em} y(0) = 7 , y'(0) = -4$
$y= 4 \, e^{\left(2 \, t\right)} + 3 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 89)
Find the solution to the given IVP.
$y''+y'-20y = 0 \hspace{1em} y(0) = -2 , y'(0) = 28$
$y= 2 \, e^{\left(4 \, t\right)} - 4 \, e^{\left(-5 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 90)
Find the solution to the given IVP.
$y''-4y'+3y = 0 \hspace{1em} y(0) = -5 , y'(0) = -9$
$y= -2 \, e^{\left(3 \, t\right)} - 3 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 91)
Find the solution to the given IVP.
$y''-1y'-20y = 0 \hspace{1em} y(0) = 5 , y'(0) = -20$
$y= 5 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 92)
Find the solution to the given IVP.
$y''+0y'-1y = 0 \hspace{1em} y(0) = 6 , y'(0) = 2$
$y= 2 \, e^{\left(-t\right)} + 4 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 93)
Find the solution to the given IVP.
$y''+3y'-4y = 0 \hspace{1em} y(0) = 1 , y'(0) = -24$
$y= 5 \, e^{\left(-4 \, t\right)} - 4 \, e^{t}$
## C5 - Homogeneous second-order linear IVP (ver. 94)
Find the solution to the given IVP.
$y''+y'-12y = 0 \hspace{1em} y(0) = -3 , y'(0) = 12$
$y= -3 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 95)
Find the solution to the given IVP.
$y''+2y'-8y = 0 \hspace{1em} y(0) = 0 , y'(0) = 6$
$y= e^{\left(2 \, t\right)} - e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 96)
Find the solution to the given IVP.
$y''-7y'+12y = 0 \hspace{1em} y(0) = -1 , y'(0) = -7$
$y= -4 \, e^{\left(4 \, t\right)} + 3 \, e^{\left(3 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 97)
Find the solution to the given IVP.
$y''-4y'-5y = 0 \hspace{1em} y(0) = 4 , y'(0) = 14$
$y= 3 \, e^{\left(5 \, t\right)} + e^{\left(-t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 98)
Find the solution to the given IVP.
$y''+6y'+8y = 0 \hspace{1em} y(0) = -3 , y'(0) = 14$
$y= e^{\left(-2 \, t\right)} - 4 \, e^{\left(-4 \, t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 99)
Find the solution to the given IVP.
$y''-2y'-3y = 0 \hspace{1em} y(0) = 4 , y'(0) = -8$
$y= -e^{\left(3 \, t\right)} + 5 \, e^{\left(-t\right)}$
## C5 - Homogeneous second-order linear IVP (ver. 100)
Find the solution to the given IVP.
$y''-1y'-2y = 0 \hspace{1em} y(0) = -7 , y'(0) = -8$
$y= -5 \, e^{\left(2 \, t\right)} - 2 \, e^{\left(-t\right)}$ | 9,142 | 20,128 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2019-43 | latest | en | 0.415996 |
https://testbook.com/question-answer/if-a4-1a4-50-a-0-then-find-the-value-o--5ea97584f60d5d62fd1fc23f | 1,725,814,570,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00604.warc.gz | 538,944,873 | 45,874 | # If a4 + 1/a4 = 50, a > 0 ,then find the value of a3 + 1/a3.
This question was previously asked in
SSC CGL Previous Paper 79 (Held On: 7 March 2020 Shift 3)
View all SSC CGL Papers >
1. $$\sqrt {2(1 + \sqrt {13} } ) + \left( { - 1 + 2\sqrt {13} } \right)$$
2. $$\sqrt {2(1 + \sqrt {13} } ) - \left( { - 1 - 2\sqrt {13} } \right)$$
3. $$\sqrt {2(1 + \sqrt {13} } )\left( { - 1 + 2\sqrt {13} } \right)$$
4. $$\sqrt {2(1 - \sqrt {13} } )\left( { - 1 + 2\sqrt {13} } \right)$$
Option 3 : $$\sqrt {2(1 + \sqrt {13} } )\left( { - 1 + 2\sqrt {13} } \right)$$
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## Detailed Solution
As we know,
a4 + 1/a4 = 50
⇒ (a2 + 1/a2)2 = a4 + 1/a4 + 2 = 50 + 2 = 52
⇒ a2 + 1/a2 = √52 = 2 √13
⇒ (a + 1/a)2 = a2 + 1/a2 + 2
⇒ (a + 1/a)2 = 2√13 + 2
$$\Rightarrow {\rm{}}\left( {a + \frac{1}{a}} \right) = \sqrt {2\left( {1\; + \;\sqrt {13} } \right)}$$
$$\Rightarrow {\rm{}}{a^3} + \frac{1}{{{a^3}}} = {\left( {a\; + \;\frac{1}{a}} \right)^3} - 3\left( {a + \frac{1}{a}} \right)$$
$$\Rightarrow {\rm{\;}}{a^3} + \frac{1}{{{a^3}}} = \left( {a + \frac{1}{a}} \right)[{\left( {a\; + \;\frac{1}{a}} \right)^2} - 3]$$
$$\Rightarrow {\rm{\;}}{a^3} + \frac{1}{{{a^3}}} = \sqrt {2\left( {1\; + \;\sqrt {13} } \right)} \left[ {2 + 2\sqrt {13} - 3} \right]$$
$$\Rightarrow {\rm{}}{a^3} + \frac{1}{{{a^3}}} = \sqrt {2\left( {1 + \sqrt {13} } \right)} \left[ { - 1 + 2\sqrt {13} } \right]$$
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-> Candidates should also use the SSC CGL previous year's papers for a good revision. | 866 | 1,967 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.590301 |
https://www.jiskha.com/display.cgi?id=1235707689 | 1,501,269,520,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500550977093.96/warc/CC-MAIN-20170728183650-20170728203650-00291.warc.gz | 815,871,993 | 4,375 | # math
posted by .
what number is 24% of 95.
Sorry but i don't understand how to get numbers using percents.
another problem im stuck with is
what percent of 90 is 60
• math -
24% is the same as 0.24
To get a fraction OF a number, multiply the fraction by the number. For example, 50% of 10 is 0.5 x 10.
60 is 2/3 of 90.
Convert 2/3 to percent
• math -
what number is 24% of 95
Change 24% to a decimal.
24% = 0.24
Multiply: 0.24 * 95 = ??
___________________________________
what percent of 90 is 60
This is a division problem.
60 / 90 = 0.66666 = 67%
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More Similar Questions
Post a New Question | 581 | 2,283 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2017-30 | longest | en | 0.890819 |
efarem.github.io | 1,555,779,374,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578529898.48/warc/CC-MAIN-20190420160858-20190420182000-00043.warc.gz | 57,024,890 | 2,835 | # Finally an O(1) Solution
I wrote a few bad solutions for Project Euler in PHP a while back and I figured it is time to go back and do it more efficiently, and in C++!
Problem 1 in O(1)!
So my first solution incremented variable `i` from `1 > n-1` each time testing if `i` was a multiple of 3 or 5 and adding to a total if it was. Fine… this works but `O(n)` doesn’t do too well if we’re testing cases up to say `10^9` plus PHP < 7.0 doesn’t like large integers without using bcmath.
So with a small amount of digging I found a simple arithmetic progression formula for calculating a sum of multiples `(n/2)*(a+b)` where a is the number in question b is the upper limit (highest number divisible by the number in question) and n is b divided by a.
So after running the sum of multiples below 1000 for 3:
`(333/2) * (3+999) = 166.5 * 1002 = 166,833`
And 5:
`(199/2) * (5+995) = 99.5 * 1000 = 99,500`
All you need to do is remove the sum of multiples that have been accounted for by 3 and 5. You can do this be subtracting the sum of multiples of 15:
`(66/2) * (15+990) = 33 * 1005 = 33,165`
From the sum of the previous two totals:
`166,833 + 99,500 - 33,165 = 233,168`
There you have it, with this in place my new solution achieves `O(1)`, tested for any case up to and including 10^9.
It still doesn’t pass the Hacker Rank tests, wtf!?
The division by 2 causes a rounding error when using `float` or even `long double` so I had to find another formula, luckily there is another `n*(x*(x+1)/2)` where n is the number we’re searching for multiples of and x is the upper limit divided by n.
Here’s a link to my solution on Github. | 479 | 1,645 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-18 | latest | en | 0.90192 |
https://atlanticgmat.com/if-the-positive-integer-n-is-added-to-each-of-the-integers-69-94-and-121-what-is-the-value-of-n/ | 1,721,844,614,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763518427.68/warc/CC-MAIN-20240724171328-20240724201328-00091.warc.gz | 88,676,194 | 52,898 | # If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
If the positive integer n is added to each of the integers 69, 94, and 121, what is the value of n?
1. 69 + n and 94 + n are the squares of two consecutive integers.
2. 94 + n and 121 + n are the squares of two consecutive integers.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are not sufficient.
Full explanation coming soon. Send us a note if you’d like this added to the express queue!
You’ll find tons of practice questions, explanations for GMAT Official Guide questions, and strategies on our GMAT Question of the Day page.
## Here are a few other extra challenging GMAT questions with in depth explanations:
Here’s a tough function question from the GMAT Prep tests 1 and 2:
For which of the following functions is f(a+b) = f(b) + f(a) for all positive numbers a and b?
And a very challenging word problem from the Official Guide. Almost no-one gets this one on the first try but there is a somewhat simple way through it:
Last Sunday a certain store sold copies of Newspaper A for \$1.00 each and copies of Newspaper B for \$1.25 each, and the store sold no other newspapers that day. If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?
Tanya’s letters from the GMAT Prep tests. This one often gets GMAT tutoring students caught up in a tangled net. With combinatorics it’s important to stay practical. We’ll take a look at how to do that in the explanation:
Tanya prepared 4 different letters to be sent to 4 different addresses. For each letter, she prepared an envelope with its correct address. If the 4 letters are to be put into the 4 envelopes at random, what is the probability that only 1 letter will be put into the envelope with its correct address?
Here’s an exponents puzzle that comes up a lot in GMAT tutoring sessions:
If n is a positive integer and n^2 is divisible by 72, then the largest positive integer that must divide n is
This is one of the most difficult questions in the GMAT universe. That said, there is a simple way to solve it that relies on a fundamental divisibility rule every GMAT studier should know:
For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) +1, then p is? | 664 | 2,793 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2024-30 | latest | en | 0.904451 |
https://math.stackexchange.com/questions/2827730/order-and-degree-of-expd3y-dx3-x-fracd2ydx2-y-0 | 1,563,531,913,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526210.32/warc/CC-MAIN-20190719095313-20190719121313-00458.warc.gz | 470,215,644 | 35,914 | # Order and degree of $\exp({{d^3y}/{dx^3}} )- x \frac{d^2y}{dx^2} + y=0$?
Order and degree of $\exp({{d^3y}/{dx^3}} )- x \frac{d^2y}{dx^2} + y=0$?
The order will be 3 since it involves $\frac{d^3y}{dx^3}$
For degree to be defined, the differential coefficients must have natural powers, not occur as an argument to some function.
Hence, the degree will not be defined since the differential coefficients occur as an argument of exponential function (or the log function in case we try to take the log)
Is this correct?
$$e^{y'''}=xy''-y\qquad \rightarrow \qquad y'''=\ln(xy''-y)$$
The ODE order is defined as $\ y^{(n)}\$ where $\ n\$ is the derivative and $\ n\ge0\$ it must be a natural number.
And the degree is defined as $\ (y^{(n)})^{k}\$ where $\ k\$ is the exponent and also $\ k>0\$ it must be a natural number
$\ y'''\$ is the highest derivative, $\ n=3\$ and $\ k=1\$
because the order $\ \ln(y')\$ isn't defined, as well $\ \ln(xy''-y)\$ too.
• But I've been taught that all the differential coefficients must be free from radical signs and should not appear as an argument of some other function. Can you please confirm whether the above should be true for all differential coefficients or only the coefficient of highest order ? – So Lo Jun 24 '18 at 11:21
• I think that degree should be not defined since the ODE cannot be written as a polynomial in differential coefficients – So Lo Jun 24 '18 at 11:32
• I investigated cause i wasn't sure of my answer and found these links, and you're right, the order it's not defined quora.com/… math.stackexchange.com/q/1117694/546265 – MR ASSASSINS117 Jun 24 '18 at 11:36
• thank you for helping – So Lo Jun 24 '18 at 12:02 | 496 | 1,691 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2019-30 | latest | en | 0.901138 |
https://convertoctopus.com/1716-grams-to-ounces | 1,628,064,896,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046154798.45/warc/CC-MAIN-20210804080449-20210804110449-00536.warc.gz | 188,850,763 | 7,925 | ## Conversion formula
The conversion factor from grams to ounces is 0.03527396194958, which means that 1 gram is equal to 0.03527396194958 ounces:
1 g = 0.03527396194958 oz
To convert 1716 grams into ounces we have to multiply 1716 by the conversion factor in order to get the mass amount from grams to ounces. We can also form a simple proportion to calculate the result:
1 g → 0.03527396194958 oz
1716 g → M(oz)
Solve the above proportion to obtain the mass M in ounces:
M(oz) = 1716 g × 0.03527396194958 oz
M(oz) = 60.53011870548 oz
The final result is:
1716 g → 60.53011870548 oz
We conclude that 1716 grams is equivalent to 60.53011870548 ounces:
1716 grams = 60.53011870548 ounces
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 ounce is equal to 0.016520701121795 × 1716 grams.
Another way is saying that 1716 grams is equal to 1 ÷ 0.016520701121795 ounces.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that one thousand seven hundred sixteen grams is approximately sixty point five three ounces:
1716 g ≅ 60.53 oz
An alternative is also that one ounce is approximately zero point zero one seven times one thousand seven hundred sixteen grams.
## Conversion table
### grams to ounces chart
For quick reference purposes, below is the conversion table you can use to convert from grams to ounces
grams (g) ounces (oz)
1717 grams 60.565 ounces
1718 grams 60.601 ounces
1719 grams 60.636 ounces
1720 grams 60.671 ounces
1721 grams 60.706 ounces
1722 grams 60.742 ounces
1723 grams 60.777 ounces
1724 grams 60.812 ounces
1725 grams 60.848 ounces
1726 grams 60.883 ounces | 480 | 1,741 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-31 | latest | en | 0.785317 |
https://gmatclub.com/forum/each-person-on-a-committee-with-40-members-voted-for-exactly-129999.html?oldest=1 | 1,506,210,545,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818689806.55/warc/CC-MAIN-20170923231842-20170924011842-00163.warc.gz | 675,948,710 | 54,881 | It is currently 23 Sep 2017, 16:49
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# Each person on a committee with 40 members voted for exactly
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Each person on a committee with 40 members voted for exactly [#permalink]
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01 Apr 2012, 05:29
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Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
[Reveal] Spoiler: OA
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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01 Apr 2012, 08:01
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Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
(2) Candidate H received 14 of the votes --> if F=15 and G=11 then F received the most of the votes but if F=13 and G=13 then F did not receiv the most votes. Not sufficient.
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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19 Sep 2012, 09:45
Bunuel wrote:
Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
(2) Candidate H received 14 of the votes --> if F=15 and G=11 then F received the most of the votes but if F=13 and G=13 then F did not receiv the most votes. Not sufficient.
Hi Bunuel,
Statement 1: Candidate F received 11 of the votes --> together G and H received 40-11=29 votes, Cant one of them get 10 and the other 19 ??? or 9 and 20!! That means we cannot say if F received most or least no of votes./???
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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19 Sep 2012, 09:53
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swathimn wrote:
Bunuel wrote:
Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
(2) Candidate H received 14 of the votes --> if F=15 and G=11 then F received the most of the votes but if F=13 and G=13 then F did not receiv the most votes. Not sufficient.
Hi Bunuel,
Statement 1: Candidate F received 11 of the votes --> together G and H received 40-11=29 votes, Cant one of them get 10 and the other 19 ??? or 9 and 20!! That means we cannot say if F received most or least no of votes./???
Hope it's clear.
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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19 Sep 2012, 23:16
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boomtangboy wrote:
Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
I got the answer incorrect but think the OA is wrong.
This is testing averages and distribution.
So, we have 40 votes to distribute to 3 candidates. The extreme way to get most votes is to have the other two candidates get just one vote each.
1,1,38
The most conservative way to get most votes is to distribute the votes equally and offset one vote to one candidate to make it the winner. 40/3=13 with a remainder of 1.
13,13,14
Therefore the candidate with the most votes v is like this 14 < v < 39
(1) if Candidate F has 11 then he surely did not get the v we are looking for. 11 could never be a value for the most vote. SUFFICIENT
(2) if Candidate H received 14.
If distribution is 14,13,13 then H got the most vote and not F.
If distribution is 14,1,25 then Either F or the other got the most vote.
INSUFFICIENT
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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21 Sep 2012, 11:20
Bunuel wrote:
swathimn wrote:
Bunuel wrote:
Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
(2) Candidate H received 14 of the votes --> if F=15 and G=11 then F received the most of the votes but if F=13 and G=13 then F did not receiv the most votes. Not sufficient.
Hi Bunuel,
Statement 1: Candidate F received 11 of the votes --> together G and H received 40-11=29 votes, Cant one of them get 10 and the other 19 ??? or 9 and 20!! That means we cannot say if F received most or least no of votes./???
Hope it's clear.
Thanks Bunuel!! You have been very helpful, All the posts from you are extremely useful!! I just wished there was someone who could show as much interest in verbal as well!! I don really see expert replies in verbal as a result of which you still have lots of doubts especially in reasoning abilities!! Thanks a lot, quant team is doing a great job
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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21 Sep 2012, 11:52
A...
ANS should be yes or no in DS questions.. A ans NO.. it should be the ans.
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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16 May 2013, 11:26
boomtangboy wrote:
Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
I got the answer incorrect but think the OA is wrong.
given 40 votes are cast, remaining 40-11 =29 even if divided equally among others 14, 15 or if one gets 0 even then f is not the max votes geter so A. is suff
stmt2: it saygs H got 14 remaining is 26, now f has cnances which we can't predict. so D is rules out..
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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13 Aug 2015, 23:23
boomtangboy wrote:
Each person on a committee with 40 members voted for exactly one of 3 candidates, F, G, or H. Did Candidate F receive the most votes from the 40 votes cast?
From state(2), we cannot answer because F and G can share the remaining 26 votes in any ratio.
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Re: Each person on a committee with 40 members voted for exactly [#permalink]
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Re: Each person on a committee with 40 members voted for exactly [#permalink] 31 Aug 2016, 07:26
Display posts from previous: Sort by | 2,878 | 10,064 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2017-39 | latest | en | 0.907352 |
https://wimcouwenberg.wordpress.com/2010/08/24/a-short-alternative-to-gordan/ | 1,529,444,783,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863206.9/warc/CC-MAIN-20180619212507-20180619232507-00020.warc.gz | 751,474,168 | 15,377 | ## A short alternative to Gordan
The nicely named “theorem of the alternative” of Paul Gordan states the following
If $V \subset \mathbb{R}^n$ is a linear subspace then either $V$ contains a vector with positive coordinates or the orthogonal complement $V^{\perp}$ contains a non-zero vector with non-negative coordinates.”
These options are mutually exclusive, hence the name of the theorem. Gordan gave a proof of this in 1873 which involved a very clever inductive argument. I just thought of a self contained very short proof of this theorem. In fact I will prove an equivalent statement
If $v_1, \dotsc, v_n \in \mathbb{R}^k$ then either there is a vector $v \in \mathbb{R}^k$ such that the inner product $\langle v, v_j \rangle$ is positive for all $j \in \{1, \dotsc, n\}$ or the convex hull $C$ of $v_1, \dotsc, v_n$ contains the origin.”
The equivalence can be seen as follows. Let $M$ be the $n \times k$ matrix with $v_j$ as its rows. Then the columns of $M$ span a subspace $V \subset \mathbb{R}^n$ and this transforms the second statement into the first and vice versa.
Now let $v \in C$ have minimal norm. If $|v| = 0$ then $C$ contains the origin. If $|v| > 0$ and $w$ is any point in the convex hull then the line segment from $v$ to $w$ lies in $C$ and contains no point with smaller norm than $v$. So for all $t \in (0,1]$ we have
$\dfrac{|(1-t) v + t w|^2 - |v|^2}{2t} = \langle v, w \rangle - |v|^2+ \dfrac{t}{2}|w-v|^2 \geq 0$.
This implies that $\langle v, w \rangle \geq |v|^2 > 0$ and in particular this holds for all $w \in \{v_1, \dotsc, v_n \}$. Done! | 496 | 1,586 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 27, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2018-26 | latest | en | 0.82099 |
https://stromcv.com/initial-momentum-calculator-61 | 1,675,876,725,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500837.65/warc/CC-MAIN-20230208155417-20230208185417-00276.warc.gz | 544,372,229 | 16,917 | # Initial momentum calculator
Total Initial Momentum Formula: Total Initial Momentum = M 1 × V 1 + M 2 × V 2 Where, V 1 = Velocity of Block1 V 2 = Velocity of Block2 M 1 = Mass of Block1 (kg) M 2 = Mass
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Initial momentum calculator uses Momentum = Mass*Initial velocity of Mass to calculate the Momentum, The Initial momentum formula is defined as the product of the mass and initial
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math is the study of numbers, shapes, and patterns. It is used in everyday life, from counting to measuring to more complex calculations. | 574 | 2,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-06 | latest | en | 0.879336 |
https://www.bayesianwitch.com/math-worksheet/common-core-first-grade-math-worksheets/ | 1,591,425,688,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590348511950.89/warc/CC-MAIN-20200606062649-20200606092649-00520.warc.gz | 620,231,071 | 8,726 | Staggering Common Core First Grade Math Worksheets
010 Math Worksheet Common Core First Grade
If you have read my article "Helping Your Child With Basic Arithmetic? Stay Away From Worksheets" then you know that I am not a fan of traditional worksheets. After writing that article, I found another credible teacher who has written many ezine articles expounding on the benefits of worksheets. I decided some clarification of position is in order. The primary problem with most math worksheets is that the problems are already written out and the child need only write the answers. For learning and practicing the basic skills of addition, subtraction, multiplication, and division, it is much more beneficial for the child to write out the entire fact and say the entire fact out loud. A child will learn a multiplication fact much faster if they are writing out 6 x 8 = 48 at the same time they are saying "six times eight is forty_eight" than if they just see 6 x 8 = ___ and only have to supply the 48. Rather than using worksheets, a better method is to use individual size white boards and have the child writing entire facts many times. Having a child writing 9 x 7 = 7 x 9 = 63 while saying "nine times seven is the same as seven times nine and is equal to sixty_three" is many times more successful than a worksheet with 9 x 7 = ___ and the student just thinks the answer once and then writes that answer on the duplicate problems.
This article offers some ideas and practical tips on how you can get your child working on printable worksheets, whether you are homeschooling or simply making sure that your child does, and understands, any homework they may have been given. Home Learning _ It is a lot easier to teach your child or to help them with school homework if they have had experience of learning in the home. Very often moms teach their children a lot of skills without actually realising that the fun they are having is a great way for a child to learn. Silly games like spotting the number of red cars while out on a shopping trip or playing about with words by making up silly rhymes all contribute to your child's education. The point is that you can still carry on with this type of learning activity and it will be a lot easier to incorporate printable worksheets into the fun and get your child working on them. Children love to draw and color and cut and paste so you can use this pleasure in a number of ways to make working on printable worksheets more enjoyable.
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https://software.intel.com/content/www/us/en/develop/documentation/onemkl-developer-reference-c/top/appendix-a-linear-solvers-basics/sparse-linear-systems/matrix-fundamentals.html | 1,627,563,649,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153857.70/warc/CC-MAIN-20210729105515-20210729135515-00337.warc.gz | 525,255,874 | 45,877 | Contents
Matrix Fundamentals
A matrix is a rectangular array of either real or complex numbers. A matrix is denoted by a capital letter; its elements are denoted by the same lower case letter with row/column subscripts. Thus, the value of the element in row
i
and column
j
in matrix
A
is denoted by
a
(
i
,
j
)
. For example, a 3 by 4 matrix
A
, is written as follows:
Note that with the above notation, we assume the standard Fortran programming language convention of starting array indices at 1 rather than the C programming language convention of starting them at 0.
A matrix in which all of the elements are real numbers is called a real matrix. A matrix that contains at least one complex number is called a complex matrix. A real or complex matrix
A
with the property that
a
(
i
,
j
) =
a
(
j
,
i
), is called a symmetric matrix. A complex matrix
A
with the property that
a
(
i
,
j
) =
conj
(
a
(
j
,
i
)), is called a Hermitian matrix. Note that programs that manipulate symmetric and Hermitian matrices need only store half of the matrix values, since the values of the non-stored elements can be quickly reconstructed from the stored values.
A matrix that has the same number of rows as it has columns is referred to as a square matrix. The elements in a square matrix that have same row index and column index are called the diagonal elements of the matrix, or simply the diagonal of the matrix.
The transpose of a matrix
A
is the matrix obtained by “flipping” the elements of the array about its diagonal. That is, we exchange the elements
a
(
i
,
j
)
and
a
(
j
,
i
)
. For a complex matrix, if we both flip the elements about the diagonal and then take the complex conjugate of the element, the resulting matrix is called the Hermitian transpose or conjugate transpose of the original matrix. The transpose and Hermitian transpose of a matrix
A
are denoted by
A
T
and
A
H
respectively.
A column vector, or simply a vector, is a
n
× 1
matrix, and a row vector is a
1 ×
n
matrix. A real or complex matrix
A
is said to be positive definite if the vector-matrix product
x
T
Ax
is greater than zero for all non-zero vectors
x
. A matrix that is not positive definite is referred to as indefinite.
An upper (or lower) triangular matrix, is a square matrix in which all elements below (or above) the diagonal are zero. A unit triangular matrix is an upper or lower triangular matrix with all 1's along the diagonal.
A matrix
P
is called a permutation matrix if, for any matrix
A
, the result of the matrix product
PA
is identical to
A
except for interchanging the rows of
A
. For a square matrix, it can be shown that if
PA
is a permutation of the rows of
A
, then
AP
T
is the same permutation of the columns of
A
. Additionally, it can be shown that the inverse of
P
is
P
T
.
In order to save space, a permutation matrix is usually stored as a linear array, called a permutation vector, rather than as an array. Specifically, if the permutation matrix maps the
i
-th row of a matrix to the
j
-th row, then the
i
-th element of the permutation vector is
j
.
A matrix with non-zero elements only on the diagonal is called a diagonal matrix. As is the case with a permutation matrix, it is usually stored as a vector of values, rather than as a matrix.
Product and Performance Information
1
Performance varies by use, configuration and other factors. Learn more at www.Intel.com/PerformanceIndex. | 851 | 3,405 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2021-31 | latest | en | 0.934518 |
https://study.com/academy/lesson/descartess-rule-of-signs-definition-example.html | 1,566,429,835,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027316549.78/warc/CC-MAIN-20190821220456-20190822002456-00500.warc.gz | 641,929,943 | 36,300 | # Descartes's Rule of Signs: Definition & Example
Instructor: Gerald Lemay
Gerald has taught engineering, math and science and has a doctorate in electrical engineering.
Finding the roots of a polynomial is a common task in math and science applications. In this lesson, we use a method called Descartes's Rule of Signs to predict the number of roots we expect to find.
## The Rule of Signs
The arena is packed! We hear Rene Day Cart (no relation to the famous mathematician-philosopher Rene Descartes) announcing the play-by-play: '… takes the puck, skates to the left, changes to the right, changes to the left, shoots and …'
In this lesson, we describe polynomials like Day Cart's play-by-play. Instead of reporting lefts and rights, we keep track of sign changes. The number of sign changes is related to the number of roots. We will start with a simple example and gradually build up our skills. By the end of the lesson, we will see how Descartes's rule of signs predicts the number of real roots to be at most equal to the number of sign changes.
## Real Roots
Just for fun, let's do a sportscast!
From the left, the curve climbs and crosses the horizontal axis (the x-axis). The curve continues to climb until it reaches a positive peak and then descends, crossing the vertical axis (the y-axis is the dashed vertical line). As the curve continues to move to the right, we see it crossing the x-axis, dipping below the x-axis and then climbing again where it once again crosses the x-axis.
The places on the x-axis where the curve crosses are the roots of the equation. On the right of the vertical axis, the roots are positive. See the blue circles?
For this curve, there is also a root to the left of the y-axis. This root has a negative value. See the green circle?
Finding the roots usually involves factoring or plotting a polynomial. What if we could learn something about the roots without factoring or plotting? Enter the Descartes's Rule of Signs. Counting the number of sign changes predicts the number of roots. We won't know where the curve crosses the x-axis but we will know how many roots to look for.
The equation for our example curve is
y = x^3 + x^2 - 10x + 8
On the right-hand side, we have a polynomial in x. The x terms have been ordered from highest exponent to lowest. Looking at the signs of each term, we see plus, plus, minus and then plus. We started with a plus, next term was still plus, then CHANGED to minus and then CHANGED to plus.
Two sign changes. How many positive roots did we see in the plot? Two. But what about the one negative real root?
To find the number of negative roots, replace the x in the equation with -x.
x^3 + x^2 - 10x + 8 becomes (-x)^3 + (-x)^2 - 10(-x) + 8 which equals
-x^3 + x^2 + 10x + 8.
How many sign changes now? We started with minus, CHANGED to plus, next term was still plus and the last term a plus. Total number of sign changes: 1. Total number of negative roots: 1. Aha!
The Descartes's rule of signs tells us:
• The number of positive real roots is at most equal to the number of sign changes. The number of roots could be less by multiples of two.
• After replacing x with -x, the number of negative real roots is at most equal to the number of sign changes. Again, the number of roots could be less by multiples of two.
What is this 'multiples of two' idea?
## The Multiples of Two
Let's do another example:
x^7 + x^6 - 9x^5 + x^4 + 24x^3 - 26x^2 - 16x + 24
A huge polynomial! Don't worry, no factoring; just counting sign changes.
The signs of the terms: plus, plus, minus, plus, plus, minus, plus, plus. We started with plus, then plus, CHANGED to minus, CHANGED back to plus, then plus, CHANGED to minus, CHANGED to plus, stayed at plus and ended with plus. A total of 4 signs changes. Thus, the number of positive real roots is 4 or 2 or 0. Do you see how the 'less by multiples of two' is used? How about the negative real roots?
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Did you know… We have over 200 college courses that prepare you to earn credit by exam that is accepted by over 1,500 colleges and universities. You can test out of the first two years of college and save thousands off your degree. Anyone can earn credit-by-exam regardless of age or education level. | 1,111 | 4,556 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.9375 | 5 | CC-MAIN-2019-35 | longest | en | 0.906138 |
https://alevelnotes.com/notes/physics/materials/hooke's-law | 1,701,331,868,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100172.28/warc/CC-MAIN-20231130062948-20231130092948-00176.warc.gz | 126,072,571 | 4,237 | # Hooke’s Law
• Hooke’s Law states that, for certain elastic materials, force is proportional to extension, when a sample is stretched. This means that the extension of the sample increases linearly with the amount of force applied. Materials that obey Hooke’s law are called Hookean Materials. Springs behave like Hookean Materials.
• Hooke’s law can be written as $$F = kx$$, where $$F$$ is Force, $$x$$ is extension, and $$k$$ is the Stiffness Constant of the sample. The stiffness constant describes the stiffness of a material, and is measured in $$Nm^{-1}$$ (or $$Kgs^{-2}$$).
$F = kx$
• Hooke’s law can be demonstrated with the use of Force-Extension graphs.
• However, no sample follows Hooke’s law indefinitely, and there comes a point, called the Limit of Proportionality, where there is no longer a linear relationship between force and extension. After yet more force is applied, the Elastic Limit will be reached. This means that the sample will no longer return to its original shape when the force ceases to be present. Eventually, the force will become so great that the material snaps. This is called the Yeild Point.
• Before the elastic limit is reached, the sample is experiencing Elastic Deformation, where it will return to its original shape when the load (force) has been removed. However, once the material passes that point, it experienced Plastic Deformation, where its shape is permanently changed.
• If two springs are used in series, the effective stiffness constant of both of them is less than either of them. In fact, it can be worked out by the formula: $$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$$. If two springs are in parallel, their effective stiffness constant is greater: $$k_{\text{eq}} = k_1 + k_2$$. | 422 | 1,764 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2023-50 | latest | en | 0.94483 |
https://techminded.wordpress.com/2016/01/17/castles-in-the-desert-games-of-chance-and-math/ | 1,555,929,472,000,000,000 | text/html | crawl-data/CC-MAIN-2019-18/segments/1555578551739.43/warc/CC-MAIN-20190422095521-20190422121521-00415.warc.gz | 566,553,717 | 22,377 | Home > Math > Castles in the Desert: Games of Chance and Math
## Castles in the Desert: Games of Chance and Math
Introduction
Casinos represent a mind boggling concentration of resources. It is not uncommon for casinos to be vast structures with hundreds of employees, restaurants, shopping , hotels, lavish finishes, advanced technology and extensive areas for gambling activities. All of these things are concentrated in the most unlikely of places, in a desert landscape where the blazing sun is plentiful and water is sparse. Under normal circumstances populations grow around areas high in resources like places where food can grow easily or that have access to large bodies of water. The casino hubs though, especially those in arid places like Nevada, grew because the business of gambling is extremely lucrative. The business is so lucrative that these companies can even build modern day castles in the desert.
The House Always Wins
The first thing to consider when looking at the math of gambling is the idea that the house always wins. The house, or the establishment that hosts the gambling activities, does not win every bet. There are times where individuals win both small and large sums of money. What we mean by the house always wins is that in the long term, over many, many betting transactions, the house is mathematically destined to win. For each winner there are many more losers. If the house always wins in the long run, you might ask: How often does the house win and by how much?
We’ll use a few concepts from an area of math called probability to gain an idea on how to answer that question. We’ll just scratch the surface, but it will give you an idea on how to learn more.
Basic Concepts
• Randomness
• Things happen without an order or predictable pattern
• Experiment
• An experiment is when we do a random act to get an outcome. So we might roll a dice to see which number appears.
• Event
• The result of an experiment. In the dice rolling example we might roll a three.
• Sample Space
• All of the possible outcomes or events. A cube shaped dice has six possible outcomes.
• Probability
• The chance something will happen. Usually expressed as a fraction. The probability of all events in a sample space adds up to 1. I have a one in six chance that a three will be rolled on a dice
• Independent Events
• Events are independent if they do not influence one another. Each time I roll a dice it does not matter what has been rolled before.
• Dependent Events
• Events are dependent if they influence one another. If I draw a card and do not replace it in the deck, then I can’t draw that card again. Specifically, if I had 10 cards and removed one, the next time I drew a card there would only be nine cards to choose from.
One Dice
Let’s think about one dice. How many ways are there to roll the dice? A cube has 6 sides so there are six ways to roll one dice.
Two Dice
What about if we roll two dice? Well there are 6 ways to roll the first dice and six ways to roll the second. The first conceptual challenge here is how do we count the events in the sample space? It is always easier to use real numbers so let’s consider that we roll 1 on the first dice. How many ways can the second dice be rolled? (6) We can repeat the thought experiment by rolling a 2 first, then a three first etc. We get something that looks like 6+6+6+6+6+6 or 6*6. We can also visualize this by organizing the outcomes in a table.
We can go further and look at how many times each sum appears in the table. We can use this information to determine the probability for each sum. For example a sum of two can be formed exactly one way (by rolling two ones) in all 36 combinations. Therefore the probability of the dice adding up to 2 is 1/36. The full distribution looks like this:
Craps
Craps is a gambling game where participants place bets on the results of rolling two dice. The rules of the game are as follows:
• The player (known as the shooter) rolls a pair of fair dice
• If the sum is 7 or 11 on the first throw, the shooter wins; this event is called a natural.
• If the sum is 2, 3, or 12 on the first throw, the shooter loses; this event is called craps.
• If the sum is 4, 5, 6, 8, 9, or 10 on the first throw, this number becomes the shooter’s point. The shooter continues rolling the dice until either she rolls the point again (in which case she wins) or rolls a 7 (in which case she loses).
Consider the case of winning on the first throw. There are 6 ways to roll a seven and two ways to roll an eleven. This means there are eight ways out of 36 to have a winning outcome. We can pair this information with two related bets on this outcome
The seven bet pays out 4:1. So if a dollar is bet, the player nets \$4 otherwise he loses his dollar. The gambler will win one in six tries in the long run or 16.7% of the time. So if the gambler places 1,000 one dollar bets he will win back \$668 (167 * \$4). However he will lose \$833 (833 * \$1) This means over the long run he walks out \$165 poorer.
The eleven bet pays out 15:1. So if a dollar is bet, the player nets \$15 otherwise he loses his dollar. Similar to the 7 bet calculation , the player has a 5.6% chance of an 11 bet succeeding. If he places 1,000 one dollar bets, he will win back \$840 (56 * \$15) and lose \$944 (944 * \$1). The means over the long run he walks out \$104 poorer.
In both of the previous cases the house was the long term winner. Even though the player may feel like they are winning at some points in time, over many betting cycles the house amasses a large influx of cash. The house’s winnings are always net positive in the long run and the player’s are always net negative.
We can use the simulator in the references below to gather the long term results of all craps bet types for our 1,000 one dollar bet scenario. Any way you go you will walk out poorer over the long run.
Further analysis can be done on other games of chance. The outcomes are similar with the house winning in all scenarios over the long run. Some games like Black Jack provide the house a smaller edge allowing for skilled players to reap rewards. One can even observe the flow of the game and use probability to make large bets when the hands left in the deck are in the player’s favor. This is a method called card counting. This can be multiplied by having multiple players gathering information on the deck. However this practice, known as team play, is prohibited and casinos will quickly escort out or take action against players found doing it.
Just such a scenario went down when a group of MIT students devised a plan to do team play against single deck Black Jack and reap huge winnings. Their exploits are detailed in the book “Bringing Down the House”. Casinos quickly caught on to their tactics and black listed them from playing. Outside of the US the young students faced threats of physical violence when they tried to employ their tactics.
While one could argue that individuals who counted solo without the help of a team were just being good players and mastering the game, the casinos were not in the business of losing money. They quickly adjusted their games of chance to include multiple decks. This nullified the counting advantages that single decks offered. Building castles in the desert is an expensive venture and the house is going to do everything in their power to ensure that they keep winning in the long run.
So the next time you see a grand structure where there would normally not be one, think a little about why resources are concentrating there. Math is a tool that can help you solve this riddle.
Handouts
• All Possible Outcomes of Rolling Two Dice
• Probability Distribution of Rolling Two Dice
References
Videos
Categories: Math Tags: , ,
1. January 17, 2016 at 3:20 pm
Your dedication to teaching and using your gift is amazing Charles, you poured a lot of work into this and one can tell…
2. January 28, 2016 at 3:08 pm
Another excellent article, Charles. It reminds me of a series of lectures on arithmatic by a U of Colorado physics professor. You have an amazing mind. If only you had a friend who could … 🙂
• January 28, 2016 at 6:17 pm
Yes giving my professional writer friends enough notice to give feedback and listening to my editor are both things I must work on 🙂 | 1,880 | 8,348 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2019-18 | latest | en | 0.931566 |
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# Independence number of edge-chromatic critical graphs
Yan Cao Guantao Chen Guangming Jing Songling Shan
Department of Mathemetics and Statistics, 25 Park Place, 14th Floor
Georgia State Univeristy, Atlanta, GA 30303
Department of Mathematics, 1326 Stevenson Center
Vanderbilt University, Nashville, TN 37240
October 3, 2020
###### Abstract
Let be a simple graph with maximum degree and chromatic index . A classic result of Vizing indicates that either or . The graph is called -critical if is connected, and for any , . Let be an -vertex -critical graph. Vizing conjectured that , the independence number of , is at most . The current best result on this conjecture, shown by Woodall, is that . We show that for any given , there exist positive constants and such that if is an -vertex -critical graph with minimum degree at least and maximum degree at least , then . In particular, we show that if is an -vertex -critical graph with minimum degree at least and , then
α(G)<⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩7n12,if d=3; 4n7,if d=4; d+2+3√(d−1)d2d+4+3√(d−1)dn<4n7,if d≥19.
Keywords: Chromatic index, Edge-chromatic critical, Vizing’s Independence Number Conjecture.
\usetikzlibrary
arrows \usetkzobjall
## 1 Introduction
All graphs considered are simple and finite. Let be a graph. We denote by and the vertex set and edge set of , respectively. For a vertex , is the set of neighbors of in , and is the degree of vertex in . We simply write and if is clear. For , denotes the graph obtained from by deleting the edge . We reserve the symbol for , the maximum degree of throughout this paper. The independence number of , denoted , is the largest size of an independent set in .
An edge -coloring of is a mapping from to the set of integers , called colors, such that no adjacent edges receive the same color under . The chromatic index of , denoted , is defined to be the smallest integer so that has an edge -coloring. We denote by the set of all edge -colorings of . In 1965, Vizing [7] showed that a graph of maximum degree has chromatic index either or . If , then is said to be of class 1; otherwise, it is said to be of class 2. Holyer [4] showed that it is NP-complete to determine whether an arbitrary graph is of class 1. Similar to vertex coloring, it is essential to color the “core” part of a graph and then extend the coloring to the whole graph without increasing the total number of colorings. This leads to the concept of edge-chromatic criticality. A graph is called edge-chromatic critical if for any proper subgraph , . We say is edge--critical or simply -critical if is edge-chromatic critical and . It is clear that is -critical if and only if is connected with and , for any edge . By this definition, every class 2 graph with maximum degree can be reduced to a -critical graph by removing edges or vertices. Vizing conjectured that -critical graphs have some special structural properties. In particular, he proposed the following conjectures.
###### Conjecture 1.1 (Vizing’s Independence Number Conjecture [8]).
Let be a -critical graph of order . Then .
###### Conjecture 1.2 (Vizing’s 2-factor Conjecture [6]).
Let be a -critical graph. Then contains a 2-factor; that is, a 2-regular subgraph of with .
###### Conjecture 1.3 (Vizing’s Average Degree Conjecture [6]).
Let be an -vertex -critical graph. Then the average degree of is at least .
The above conjectures do not hold for edge-chromatic critical class 1 graphs such as a length 2 path. Also, they do not hold for class 2 graphs which are not -critical. Partial results have been obtained for each of these conjectures. In this paper, we investigate Vizing’s Independence Number Conjecture. This conjecture was confirmed for special graph classes including graphs with many edges such as overfull graphs by Grünewald and Steffen [3], and -vertex -critical graphs with by Luo and Zhao [5]. Let be an -vertex -critical graph. Brinkmann et al. [1], in 2000, proved that ; and the upper bound is further improved when the maximum degree is between 3 to 10. Luo and Zhao [5], in 2008, by improving the result of Brinkmann et al., showed that when . In 2009, Woodall [9] further improved the upper bound of to . In this paper, by using new adjacency lemmas, we obtain the following results.
###### Theorem 1.4.
For any given , there exist positive constants and such that if is an -vertex -critical graph with minimum degree at least and maximum degree at least , then .
Theorem 1.4 is implied by the following result.
###### Theorem 1.5.
Let be an -vertex -critical graph with minimum degree at least and . Then
α(G)<⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩7n12,if d=3; 4n7,if d=4; d+2+3√(d−1)d2d+4+3√(d−1)dn,if d≥19.
When , . In fact, we suspect the following might be true.
###### Conjecture 1.6.
Let be a positive integer. Then there exists a constant depending only on such that if is an -vertex -critical graph with and , then
α(G)
The case for was confirmed by Woodall’s result [9], and the cases for are covered in Theorem 1.5.
The remainder of the paper is organized as follows. We introduce some edge-coloring notation and technique lemmas in Section 2, and we prove Theorem 1.5 in Section 3.
## 2 Technique Lemmas
In this section, we list the classic Vizing’s Adjacency Lemma and some new developed adjacency lemmas which will be used for proving Theorem 1.5.
###### Lemma 2.1 (Vizing’s Adjacency Lemma).
If is a -critical graph, then for any edge , is adjacent to at least vertices of degree with .
Let be a -critical graph, , , and . For any , the set of colors present at is , and the set of colors missing at is , where . Note that by the -criticality of , . Let be a positive integer. Define
σq(x,z) = |{u∈N(z)−{x}:d(u)≥q}|, Aφ(x,y,q) = {α∈φ(x):∃ u∈N(y) such that φ(yu)=α and d(u)
We simply write and if is specified and clear. By the definitions, we have that
Aφ(x,y,q)∪Bφ(x,y,q) = φ(x)∩φ(y), N(x,Mφ)∪N(x,Bφ)∪{y} = N(x).
For , let
φbad(v)={α∈[1,Δ]:either α∈¯¯¯¯φ(v) or ∃ v′∈N(v) so that φ(vv′)=α and d(v′)
For any , define
c0 = ⌈1−εε⌉, (1) f(ε) = 3c50+19c40+34c30+27c20+11c0+2ε, (2) N = (c0+2)(1λ+1)3c0+2, (3) D0 = max{f(ε),3c0+2λ2,N+1ε3}. (4)
###### Lemma 2.2 (Corollary 10, [2]).
Let , be a real number with , and let . If is a -critical graph, with , and , then except at most vertices in , for each other remaining vertex ,
###### Lemma 2.3 (Corollary 15, [2]).
Let be a -critical graph, and , and let with and . Then for any ,
Moreover, for any distinct ,
###### Lemma 2.4 (Lemma 13, [2]).
Let be a -critical graph, and , and let with and . For and , if , then for any , there exist and such that , and .
As a special case of Lemma 2.4, we get the following result.
###### Corollary 2.5.
Let be a -critical graph, and , and let with and . If and there exists such that , then for any and any , it holds that .
###### Lemma 2.6.
Let be a -critical graph, and , and let with and . Suppose , , and such that . Define . Then for any , there exists and such that , , and .
###### Proof.
A coloring is called valid if , and . Let and let . Assume that . We may assume that . Otherwise, . Let such that and . If , we recolor the edge using the color and let be the new coloring of . It is clear that for any edge with , . Furthermore, is a valid coloring. However, under , . So we assume that . Since and , there exits a color . Note that and by the -criticality of . Let , and be the paths induced by the two colors and which starts at and , respectively. We claim that . For otherwise, let be the new coloring of obtained by switching the colors and on the path . Then is a -coloring of such that . Now coloring the edge using the color gives a -coloring of , showing a contradiction to the assumption that . Thus, . This implies that is vertex-disjoint from . We let be the new coloring of obtained by switching the colors and on the path . We now have that . Since the switching of colors on does not affect the colors on the edges incident to , we still have that . Let be the new coloring of obtained from by recoloring the edge using the color , we see that . Because and to get , we only switched the two colors and on the path , and then changed the color on the edge from to , is a valid coloring. Furthermore, , for any edge which is incident to or the vertex such that , . Thus, we can use as a coloring for which satisfies that .
We now take , such that and . We take the color on the edge out and color the edge using the color , and we get a coloring of . We see that and . Since for any , for the specified vertex such that , we still have that , and . Since , by Lemma 2.3, and . By Lemma 2.4, we know that for any , there exists and such that , , and . Since , , , and , we see that with implies that . Furthermore, since and , we see that the assertion in Lemma 2.6 holds. ∎
Combining Lemma 2.4 and Lemma 2.6, we obtain the following result.
###### Corollary 2.7.
Let be a -critical graph, and , and let with and . If and , then for at least one and for any , .
## 3 Proof of Theorem 1.5
###### Proof of Theorem 1.5.
Let be a -critical graph of order with minimum degree at least and maximum degree . In fact, the proof only needs with defined in Equation (4), where by calculations with defined in (6), since by Equation (1) and Equation (6) , . By the definition of in (4),
D0 = max{f(ε),3c0+2λ2,N+1ε3} = N+1ε3≤(d2+2)((d+2)34+1)1.5d+2(d+22)3 < (d+2)4(d+2)4.5d+6<(d+2)5d+10.
Let be a largest independent set and let . Note that is not an independent set in . For otherwise is a bipartite graph which is not -critical. Define
ω=⎧⎪⎨⎪⎩2,if d=3,4; 3√(d−1)d,if d≥19, (5)
and define
q=d+2−ωd+2Δ, λ=ω32(d+2)3, andε=ωd+2. (6)
Denote by
X++={x∈X:d(x)=Δ},X+={x∈X:q≤d(x)<Δ},X−1={x∈X:εΔ≤d(x)
Since has minimum degree at least , by the definitions above, we see that , for , , and for , .
For each positive integer , define
g1(k)=(d+2)(Δ−k)k, andg2(k)=ωΔk−1. (7)
Clearly, and are both decreasing functions of , and
g1(q)=(d+2)(Δ−q)q=ω(d+2)d+2−ω. (8)
###### Claim 3.1.
Let and . Then .
Proof. Let . By calculation, the first order derivative of is . Since , . Thus, since if ,
g(k)≥g(q) = g2(q)−g1(q), = ωΔq−1−ωΔq>0.
Hence, .
Define three charge functions on as follows.
M0(x)=0,M1(x)=(d+2)d(x),M2(x)=(d+2)Δ,if x∈XM0(y)=(d+2+ω)Δ,M1(y)=ωΔ,M2(y)=0,if y∈Y.
Starting with the distribution , we redistribute charge according to the following Discharging Rule:
1. Step 0: Each vertex gives charge to each vertex . Denote the resulting charge by .
2. Step 1: Stating with , each vertex gives charge to each . Denote the resulting charge by .
3. Step 2: Stating with , for each vertex , if , distributes its remaining charge equally among all vertices (if any) in . Denote the resulting charge by .
###### Claim 3.2.
For each , if , then . Consequently, Theorem 1.5 holds.
Proof. By Step 0 of Discharging Rule,
M∗0(x)= ∑y∈N(x)(d+2)=(d+2)d(x)=M1(x), for each x∈X; M∗0(y)= M0(y)−∑x∈N(y)∩X(d+2) ≥ (d+2+ω)Δ−(d+2)Δ=ωΔ=M1(y), for each y∈Y.
Since is -critical and so it is not bipartite, there exists so that and thus . Hence,
∑v∈V(G)M1(v)<∑v∈V(G)M∗0(v)=∑v∈V(G)M0(v)=(d+2+ω)Δ|Y|.
For each , if , since is obtained based on by Steps 1 and 2 of Discharging Rule, then we have that
(d+2)Δ|X|=∑v∈V(G)M2(v)≤∑v∈V(G)M∗2(v)=∑v∈V(G)M1(v)<(d+2+ω)Δ|Y|.
The above inequality implies that . Plugging in the values of in gives the desired bounds on in Theorem 1.5.
By Claim 3.2, we only need to show that for each , holds. We show this by considering different cases according to which set belongs to.
###### Claim 3.3.
For each , .
Proof. Let , and let . By Lemma 2.1, is adjacent to at least neighbors of degree . Thus is adjacent to at most neighbors in .
By Step 2 in Discharging Rule, to show , it suffices to show that . By Step 1 in Discharging Rule, we have that
M∗1(y)=M1(y)−∑x∈N(y)∩X+g1(d(x)).
By Claim 3.1, for , . Since is decreasing and is the minimum value among the degrees of in , . Combining the arguments above, we get that
M∗1(y) = M1(y)−∑x∈N(y)∩X+g1(d(x))≥M1(y)−∑x∈N(y)∩X+g2(d(x)) ≥ M1(y)−|N(y)∩X+|ωΔk0−1≥M1(y)−(k0−1)ωΔk0−1 = ωΔ−ωΔ=0.
###### Claim 3.4.
For each , .
Proof. For each , by Step 0, we have that
M∗0(x)=∑y∈N(x)∩Y(d+2)=(d+2)Δ,
where we get since is an independent set in . The charge of keeps unchanged in Steps 1 and 2, thus .
For each , by Discharging Rule,
M∗2(x)=M∗1(x)=M1(x)+∑y∈N(x)g1(d(x))=(d+2)d(x)+∑y∈N(x)(d+2)(Δ−d(x))d(x)=(d+2)Δ.
The next claim will be used for showing that for each , .
###### Claim 3.5.
Let be a nonnegative integer and be a neighbor of , and . If , then gives at least
h(k,ℓ)=1k−ℓ−1(ωΔ−ℓg1(q))=1k−ℓ−1(ωΔ−ℓ(d+2)ωd+2−ω)
in Step 2.
Proof. Let be a set of neighbors of with degree , and let be a set, disjoint from , of neighbors of with degree at least , which exists since . Then in Steps 1 and 2, gives nothing to vertices in , and in Step 1, for each vertex , gives to . In Step 2, ’s remaining charge of at least is divided among ’s remaining neighbors. Denote the set of these remaining neighbors of by . For , being in , receives charge of at least . This is because of the following arguments. (a) For each , holds, and thus
ωΔ−ℓg1(q)k−ℓ−1 | 4,175 | 13,321 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2020-50 | latest | en | 0.875798 |
https://www.fishtanklearning.org/curriculum/math/algebra-1/exponents-and-exponential-functions/lesson-8/ | 1,627,350,324,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046152168.38/warc/CC-MAIN-20210727010203-20210727040203-00489.warc.gz | 793,140,194 | 23,797 | # Exponents and Exponential Functions
## Common Core Standards
### Core Standards
?
• N.RN.A.2 — Rewrite expressions involving radicals and rational exponents using the properties of exponents.
?
• 8.EE.A.1
• 8.EE.A.2
• 8.NS.A.1
## Criteria for Success
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1. Rewrite radical expressions by evaluating any possible roots from inside the radical. For example, simplify a square root so that no perfect squares remain inside the square root.
2. Understand that ${\sqrt[n]{a^n}=a}$.
## Tips for Teachers
?
Simplifying radicals will be a useful skill for when students compute with radicals in the upcoming lessons. It is helpful for students to be familiar with and to have readily in mind the perfect squares and cubes within 100.
## Anchor Problems
?
### Problem 1
Determine which value is greater without calculating the value of the radical expression.
a. ${11\sqrt3}$ _____ ${13\sqrt3}$
b. ${\sqrt{72}}$ _____ ${5\sqrt2}$
c. ${\sqrt{75}}$ _____ ${2\sqrt{27}}$
### Problem 2
Rewrite each expression as a radical with no perfect squares or cubes remaining inside the radical.
a. ${\sqrt{32x^5y^2}}$
b. ${2\sqrt[3]{540m^7n^5}}$
c. ${(24x^2)^{1\over2}}$
d. ${(24x^2)^{1\over3}}$
e. ${(24x^2)^{2\over3}}$
## Problem Set
?
The following resources include problems and activities aligned to the objective of the lesson that can be used to create your own problem set.
a. ${\sqrt{54x^8y^5}}$
b. ${\sqrt[3]{54x^8y^5}}$ | 442 | 1,461 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2021-31 | latest | en | 0.60752 |
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# 数据科学代写|经济统计代写Economic Statistics代考|ECON205 Frequency Distribution
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## 数据科学代写|经济统计代写Economic Statistics代考|Frequency Distribution
We begin the discussion of how tabular and graphical displays can be used to summarize categorical data with the definition of a frequency distribution.
FREQUENCY DISTRIBUTION
A frequency distribution is a tabular summary of data showing the number (frequency) of observations in each of several nonoverlapping categories or classes.
Let us use the following example to demonstrate the construction and interpretation of a frequency distribution for categorical data. Coca-Cola, Diet Coke, Dr. Pepper, Pepsi, and Sprite are five popular soft drinks. Assume that the data in Table $2.1$ show the soft drink selected in a sample of 50 soft drink purchases.
To develop a frequency distribution for these data, we count the number of times each soft drink appears in Table 2.1. Coca-Cola appears 19 times, Diet Coke appears 8 times, Dr. Pepper appears 5 times, Pepsi appears 13 times, and Sprite appears 5 times. These counts are summarized in the frequency distribution in Table 2.2.
This frequency distribution provides a summary of how the 50 soft drink purchases are distributed across the five soft drinks. This summary offers more insight than the original data shown in Table $2.1$. Viewing the frequency distribution, we see that CocaCola is the leader, Pepsi is second, Diet Coke is third, and Sprite and Dr. Pepper are tied for fourth. The frequency distribution summarizes information about the popularity of the five soft drinks.
## 数据科学代写|经济统计代写Economic Statistics代考|Relative Frequency and Percent Frequency Distributions
A frequency distribution shows the number (frequency) of observations in each of several nonoverlapping classes. However, we are often interested in the proportion, or percentage, of observations in each class. The relative frequency of a class equals the fraction or proportion of observations belonging to a class. For a data set with $n$ observations, the relative frequency of each class can be determined as follows:
RELATIVE FREQUENCY
$$\text { Relative frequency of a class }=\frac{\text { Frequency of the class }}{n}$$
The percent frequency of a class is the relative frequency multiplied by 100 .
A relative frequency distribution gives a tabular summary of data showing the relative frequency for each class. A percent frequency distribution summarizes the percent frequency of the data for each class. Table $2.3$ shows a relative frequency distribution and a percent frequency distribution for the soft drink data. In Table $2.3$ we see that the relative frequency for Coca-Cola is $19 / 50=.38$, the relative frequency for Diet Coke is $8 / 50=.16$, and so on. From the percent frequency distribution, we see that $38 \%$ of the purchases were Coca-Cola, $16 \%$ of the purchases were Diet Coke, and so on. We can also note that $38 \%+26 \%+16 \%=80 \%$ of the purchases were for the top three soft drinks.
## 数据科学代写|经济统计代写ECONOMIC STATISTICS代 考|FREQUENCY DISTRIBUTION
50 份软饮料购买样本中选择的软饮料。
## 数据科学代写|经济统计代写ECONOMIC STATISTICS代 考|RELATIVE FREQUENCY AND PERCENT FREQUENCY DISTRIBUTIONS
Relative frequency of a class $=\frac{\text { Frequency of the class }}{n}$
## Matlab代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 1,269 | 4,144 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2022-49 | latest | en | 0.778612 |
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# 201. Chris mixed 3 pounds of raisins with 4 pounds of nuts.
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201. Chris mixed 3 pounds of raisins with 4 pounds of nuts. [#permalink] 18 Feb 2011, 10:31
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201. Chris mixed 3 pounds of raisins with 4 pounds of nuts. If a pound of nuts costs 3 times as much as a pound of raisins, then the total cost of the raisins was what fraction of the total cost of the mixture?
(A) 1/7
(B) 1/5
(C) 1/4
(D) 1/3
(E) 3/7
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Re: 201. Chris mixed 3 pounds of raisins with 4 pounds of nuts. [#permalink] 18 Feb 2011, 10:35
Chris mixed 3 pounds of raisins with 4 pounds of nuts. If a pound of nuts costs 3 times as much as a pound of raisins, then the total cost of the raisins was what fraction of the total cost of the mixture?
1 lbs of raisin = $1 3 lbs of raisin =$3
1 lbs of nuts = $3 4 lbs of nuts =$12
Total value of mixture = 12+3 = 15
Fraction of the value of raisin = 3/15 = 1/5
Ans: "B"
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Re: 201. Chris mixed 3 pounds of raisins with 4 pounds of nuts. [#permalink] 18 Feb 2011, 10:57
4 nuts x 3 = 12
3 raisins
3/15 --> 1/5
Re: 201. Chris mixed 3 pounds of raisins with 4 pounds of nuts. [#permalink] 18 Feb 2011, 10:57
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http://www.comap.com/highschool/contests/himcm/2013problems.html | 1,508,251,986,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187822116.0/warc/CC-MAIN-20171017144041-20171017164041-00785.warc.gz | 411,106,029 | 7,924 | 2013 Problems
Problem A
Problem: Emergency Medical Response
The Emergency Service Coordinator (ESC) for a county is interested in locating the county's three ambulances to best maximize the number of residents that can be reached within 8 minutes of an emergency call. The county is divided into 6 zones and the average time required to travel from one zone to the next under semi-perfect conditions is summarized in the following Table 1.
Average Travel Times (min.) Zones 1 2 3 4 5 6 1 1 8 12 14 10 16 2 8 1 6 18 16 16 3 12 18 1.5 12 6 4 4 16 14 4 1 16 12 5 18 16 10 4 2 2 6 16 18 4 12 2 2
Table 1: Average travel times from Zone i to Zone j in semi-perfect conditions.
The population in zones 1, 2, 3, 4, 5 and 6 are given in Table 2 below:
Zones Population 1 50,000 2 80,000 3 30,000 4 55,000 5 35,000 6 20,000 Total 270,000
Table 2: Population in each Zone
1. Determine the locations for the three ambulances which would maximize the number of people who can be reached within 8 minutes of a 911 call. Can we cover everyone? If not, then how many people are left without coverage?
2. We now have only two ambulances since one has been set aside for an emergency call; where should we put them to maximize the number of people who can be reached within the 8 minute window? Can we cover everyone? If not, then how many people are left without coverage?
3. Two ambulances are now no longer available; where should the remaining ambulance be posted? Can we cover everyone? If not, then how many people are left without coverage?
4. If a catastrophic event occurs in one location with many people from all zones involved, could the ESC cover the situation? How do counties or cities design for those rare but catastrophic events?
5. In addition to the contest's format, prepare a short 1-2 page non-technical memo outlining your recommendations from your model and analysis finding for the ESC.
Problem B
Problem: Bank Service Problem
The bank manager is trying to improve customer satisfaction by offering better service. Management wants the average customer to wait less than 2 minutes for service and the average length of the queue (length of the waiting line) to be 2 persons or fewer. The bank estimates it serves about 150 customers per day. The existing arrival and service times are given in the tables below.
Time between arrival (min.) Probability 0 0.10 1 0.15 2 0.10 3 0.35 4 0.25 5 0.05
Table 1: Arrival times
Service Time (min.) Probability 1 0.25 2 0.20 3 0.40 4 0.15
Table 2: Service times
(1) Build a mathematical model of the system.
(2) Determine if the current customer service is satisfactory according to the manager guidelines. If not, determine, through modeling, the minimal changes for servers required to accomplish the manager's goal.
(3) In addition to the contest's format, prepare a short 1-2 page non-technical letter to the bank's management with your final recommendations. | 771 | 2,937 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2017-43 | latest | en | 0.918579 |
https://metanumbers.com/16455 | 1,632,144,920,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057039.7/warc/CC-MAIN-20210920131052-20210920161052-00648.warc.gz | 436,790,786 | 7,357 | # 16455 (number)
16,455 (sixteen thousand four hundred fifty-five) is an odd five-digits composite number following 16454 and preceding 16456. In scientific notation, it is written as 1.6455 × 104. The sum of its digits is 21. It has a total of 3 prime factors and 8 positive divisors. There are 8,768 positive integers (up to 16455) that are relatively prime to 16455.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 5
• Sum of Digits 21
• Digital Root 3
## Name
Short name 16 thousand 455 sixteen thousand four hundred fifty-five
## Notation
Scientific notation 1.6455 × 104 16.455 × 103
## Prime Factorization of 16455
Prime Factorization 3 × 5 × 1097
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 3 Total number of prime factors rad(n) 16455 Product of the distinct prime numbers λ(n) -1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) -1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 16,455 is 3 × 5 × 1097. Since it has a total of 3 prime factors, 16,455 is a composite number.
## Divisors of 16455
1, 3, 5, 15, 1097, 3291, 5485, 16455
8 divisors
Even divisors 0 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 8 Total number of the positive divisors of n σ(n) 26352 Sum of all the positive divisors of n s(n) 9897 Sum of the proper positive divisors of n A(n) 3294 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 128.277 Returns the nth root of the product of n divisors H(n) 4.99545 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 16,455 can be divided by 8 positive divisors (out of which 0 are even, and 8 are odd). The sum of these divisors (counting 16,455) is 26,352, the average is 3,294.
## Other Arithmetic Functions (n = 16455)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 8768 Total number of positive integers not greater than n that are coprime to n λ(n) 1096 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 1907 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 8,768 positive integers (less than 16,455) that are coprime with 16,455. And there are approximately 1,907 prime numbers less than or equal to 16,455.
## Divisibility of 16455
m n mod m 2 3 4 5 6 7 8 9 1 0 3 0 3 5 7 3
The number 16,455 is divisible by 3 and 5.
## Classification of 16455
• Arithmetic
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
• Sphenic
## Base conversion (16455)
Base System Value
2 Binary 100000001000111
3 Ternary 211120110
4 Quaternary 10001013
5 Quinary 1011310
6 Senary 204103
8 Octal 40107
10 Decimal 16455
12 Duodecimal 9633
20 Vigesimal 212f
36 Base36 cp3
## Basic calculations (n = 16455)
### Multiplication
n×y
n×2 32910 49365 65820 82275
### Division
n÷y
n÷2 8227.5 5485 4113.75 3291
### Exponentiation
ny
n2 270767025 4455471396375 73314781827350625 1206394734969054534375
### Nth Root
y√n
2√n 128.277 25.4351 11.3259 6.97043
## 16455 as geometric shapes
### Circle
Diameter 32910 103390 8.5064e+08
### Sphere
Volume 1.8663e+13 3.40256e+09 103390
### Square
Length = n
Perimeter 65820 2.70767e+08 23270.9
### Cube
Length = n
Surface area 1.6246e+09 4.45547e+12 28500.9
### Equilateral Triangle
Length = n
Perimeter 49365 1.17246e+08 14250.4
### Triangular Pyramid
Length = n
Surface area 4.68982e+08 5.25082e+11 13435.5
## Cryptographic Hash Functions
md5 ef9e8593306a8ce376d32b7792d94f05 816a9eea343995e11299ec3f9e4a142efa4b32a4 93e31df2f72724b3e4b46575c8eb074c5cbcfab4285cd5ffbf2607f8910b72ca 61972e145e74342a68f8a7fb5396ed9346f5dc816e61e8b25c347dbb2ad741bfdd614c0699a2c05381758ab0b7244ea49411464568f21fcb69f317b214e20b9f bd9caa1389eafc587de112a905925fda2a884587 | 1,463 | 4,140 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2021-39 | latest | en | 0.813918 |
https://www.physicsforums.com/threads/mean-theorim-question.273858/ | 1,547,765,667,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583659417.14/warc/CC-MAIN-20190117224929-20190118010929-00231.warc.gz | 925,160,336 | 13,689 | # Mean theorim question
1. Nov 21, 2008
### transgalactic
i know that h(x)=f(x)-g(x)
but on what basis they conclude that
h'(c)=f'(c)-g'(c)
http://img356.imageshack.us/img356/5427/69700925ii7.gif [Broken]
??
Last edited by a moderator: May 3, 2017
2. Nov 21, 2008
### Unco
If h(x) = f(x) for all x then h'(x) = f'(x).
Last edited: Nov 21, 2008
3. Nov 21, 2008
### transgalactic
for a fact
??
4. Nov 21, 2008
### Unco
Yes; consider the definition $$h'(x) = \lim_{t\to 0}\frac{h(x+t)-h(x)}{t}$$.
5. Nov 21, 2008
### Pere Callahan
If f(x)=g(x) for all x, then the two functions are the same. And for this very reason they have the same derivative.
6. Nov 21, 2008
### transgalactic
but f(x) is a curve and g(x) is a line
f(x) is not g(x)
and so is h(x) differs them all
http://img356.imageshack.us/img356/5427/69700925ii7.gif [Broken]
all the lines are different
h(c)=f(x)-g(x)
how they came to
h'(c)=f'(x)-g'(x)
Last edited by a moderator: May 3, 2017
7. Nov 21, 2008
### Unco
Now you're just confused by notation. Let's use some neutral letters.
If A and B are differentiable functions on [a,b] such that A(x) = B(x) for all x in [a,b], then A'(x) = B'(x).
Why? $$A'(x) = \lim_{t\to 0}\frac{A(x+t)-A(x)}{t} = \lim_{t\to 0}\frac{B(x+t)-B(x)}{t} = B'(x)$$.
Also, they said "h'(c)=f'(c)-g'(c)" NOT "h'(c)=f'(x)-g'(x)".
8. Nov 21, 2008
### transgalactic
when you say "A(x) = B(x) for all x in [a,b]"
that mean that they share each one of their points
which means that A(x) and B(x) are the same function then of course A'(x) = B'(x) (its the same function)
which is not true in the graph
Last edited: Nov 21, 2008
9. Nov 21, 2008
### Pere Callahan
Take A(x)=h(x) and B(x)=f(x)-g(x)
10. Nov 21, 2008
ahhhh thanks | 667 | 1,747 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2019-04 | latest | en | 0.846974 |
https://bookriff.com/what-is-the-basic-formula-for-algebra/ | 1,725,930,182,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651164.37/warc/CC-MAIN-20240909233606-20240910023606-00219.warc.gz | 122,063,224 | 8,128 | # BookRiff
If you don’t like to read, you haven’t found the right book
## What is the basic formula for algebra?
Basic Algebra Formula (a+b)2 = a2 + 2ab + b. a2 + b2 = (a – b)2 + 2ab. (a – b)2 = a2 – 2ab + b. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc.
What are the four rules of algebra?
They are:
• Commutative Rule of Multiplication.
• Associative Rule of Multiplication.
• Distributive Rule of Multiplication.
### What are the basic calculations?
Basic calculation is the addition or subtraction of numbers with sums less than 20 (e.g. 8 + 7, 15 – 7).
How do I start learning algebra?
How to Study Math: Algebra
1. Know Your Arithmetic. To learn algebra, you have to, have to, HAVE TO know your basic arithmetic.
2. Remember PEMDAS.
3. Get Positively Comfortable with Negative Numbers.
5. Don’t Let the Letters Scare You.
7. Be Sure to Answer the Right Question.
8. Work Practice Problems.
#### What are the basic math rules?
The order of operations is as follows: 1) simplify terms inside parentheses or brackets, 2) simplify exponents and roots, 3) perform multiplication and division, 4) perform addition and subtraction. Multiplication and division are given equal priority, as are addition and subtraction.
What are the formulas used in Algebra?
a 2 – b 2 = (a – b) (a + b). where a = 5 and b = 3. (a – b) (a + b). = (5 – 3) (5 + 3). = 2 ×× 8. = 16. Question 2: 4 3 ×× 4 2 =?. Solution: Using the exponential formula (a m ) (a n) = a m+n. where a = 4. 4 3 ×× 4 2. = 4 3+2. = 4 5. = 1024. Why Algebra Formula Needs?. Wondered how Algebra formulas are needed even outside your…
## What are all the math formulas?
Here is a list of Algebraic formulas – a 2 – b 2 = (a – b)(a + b) (a+b) 2 = a 2 + 2ab + b 2 a 2 + b 2 = (a – b) 2 + 2ab (a – b) 2 = a 2 – 2ab + b 2 (a + b + c) 2 = a 2 + b 2 + c 2 + 2ab + 2ac + 2bc (a – b – c) 2 = a 2 + b 2 + c 2 – 2ab – 2ac + 2bc (a + b) 3 = a 3 + 3a 2b + 3ab 2 + b 3 ; (a + b) 3 = a 3 + b 3 + 3ab(a + b)
What does basic algebra consist of?
Algebra consists of calculating items based upon fixed numbers; variables such as quantity, time and distance; and numbers that multiply variables. These components are used in equations, which show the relationship between particular numbers.
### What is the formula for Algebra 1?
MATHEMATICAL FORMULAE Algebra 1. (a+ b)2 = a2 +2ab+ b2; a2 + b2 =(a+b)2−2ab. 2. (a−b)2 = a 2−2ab+ b; a2 + b2 =(a−b)2+2ab. | 836 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.843581 |
https://gamedev.stackexchange.com/questions/27250/need-the-co-ordinates-of-innerpolygon | 1,624,230,608,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488257796.77/warc/CC-MAIN-20210620205203-20210620235203-00497.warc.gz | 257,182,831 | 39,283 | # Need the co-ordinates of innerPolygon
Let say I have this diagram,
Given that i have all the co-ordinates of outer polygon and the distance between inner and outer polygon is d is also given. How to calculate the inner polygon co-ordinates? Edit: I was able to solved the issue by getting the mid-points of all lines. From these mid-points I can move d distance, So I can get three points. No I have 3 points and 3 slopes. From this, I can get three new equations. Simultaneously, solving the equation get the 3 points.
• My first shot is to compute inner lines and their intersections. Inner lines can be computed from outer lines bz shifting them in correct direction (you can go for example counter clockwise and always shift line to the right). – zacharmarz Apr 12 '12 at 8:09
• Can you provide an example. – user960567 Apr 12 '12 at 8:24
• I think @zacharmarz was trying to say that you have to start from a P0 point on your poly, then march counterclockwise. For each P_i P_i+1 pair of points you need to find the inwards pointing vector (for counter clockwise it's in the positive semiplane if Pi.x<Pi+1.x, else in the negative semiplane determined by the Pi Pi+1 line) If you need more help, I could provide an explanation with a figure. – teodron Apr 12 '12 at 10:59
March counterclockwise on the edge list. For each Pi-Pi+1 edge, project Pi and Pi+1 inwards via the perpendicular on the Pi_Pi+1 line. Use a simple positive/negative half space check to get the correct inwards direction. ( perpendicular vectors in 2D, positive half plane/negative half plane).
Once you compute the normally displace Pi_Pi+1 line, you can check it for intersection with its homologous Pi+1_Pi+2 normally displaced line. The intersection is called P'i+1 and is a vertex of your required polygon.
Do note that the set of points that are d units away from the initial polygon and lie inside your polygon do not form a polygon, they form a "rounded corner" polyline. Note the little circle I drew. The arc between the two green and blue radii is part of this set. But the polygon you get is a "decent" approximation.
• See my edited post. – user960567 Apr 12 '12 at 11:43
I know this has already been answered, but I believe the following code may be helpful to readers looking for an implementation of the solution.
// This function is a bit tricky. Given a path ABC, it returns the
// coordinates of the outset point facing B on the left at a distance
// of 64.0.
// M
// - - - - - - X
// ^ / '
// | 64.0 / '
// X---->-----X ==> X--v-------X '
// A B \ A B \ .>'
// \ \<' 64.0
// \ \ .
// \ \ .
// C X C X
//
FTPoint FTContour::ComputeOutsetPoint(FTPoint A, FTPoint B, FTPoint C)
{
/* Build the rotation matrix from 'ba' vector */
FTPoint ba = (A - B).Normalise();
FTPoint bc = C - B;
/* Rotate bc to the left */
FTPoint tmp(bc.X() * -ba.X() + bc.Y() * -ba.Y(),
bc.X() * ba.Y() + bc.Y() * -ba.X());
/* Compute the vector bisecting 'abc' */
FTGL_DOUBLE norm = sqrt(tmp.X() * tmp.X() + tmp.Y() * tmp.Y());
FTGL_DOUBLE dist = 64.0 * sqrt((norm - tmp.X()) / (norm + tmp.X()));
tmp.X(tmp.Y() < 0.0 ? dist : -dist);
tmp.Y(64.0);
/* Rotate the new bc to the right */
return FTPoint(tmp.X() * -ba.X() + tmp.Y() * ba.Y(),
tmp.X() * -ba.Y() + tmp.Y() * -ba.X());
} | 932 | 3,614 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2021-25 | latest | en | 0.926482 |
https://socratic.org/questions/how-do-you-find-a-one-decimal-place-approximation-for-sqrt-37 | 1,624,135,441,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487649731.59/warc/CC-MAIN-20210619203250-20210619233250-00558.warc.gz | 476,568,188 | 6,112 | # How do you find a one-decimal place approximation for sqrt 37?
Oct 18, 2015
Use one step of Newton Raphson method to find:
$\sqrt{37} \approx \frac{73}{12} = 6.08 \dot{3} \approx 6.1$
#### Explanation:
To find the square root of a number $n$, choose a reasonable first approximation ${a}_{0}$ and use the formula:
${a}_{i + 1} = \frac{{a}_{i}^{2} + n}{2 {a}_{i}}$
Repeat to get more accuracy.
For our purposes, $n = 37$ and let ${a}_{0} = 6$ since ${6}^{2} = 36$.
Then:
${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}} = \frac{{6}^{2} + 37}{2 \cdot 6} = \frac{36 + 37}{12} = \frac{73}{12} = 6.08 \dot{3}$
We don't need any more steps to get the first decimal place, since we were pretty close to start.
Actually $\sqrt{37}$ is expressible as something called a continued fraction:
sqrt(37) = [6;bar(12)] = 6+1/(12+1/(12+1/(12+...)))
So you can also get approximations for $\sqrt{37}$ by just truncating this continued fraction and working out the value.
For example:
sqrt(37) ~~ [6;12,12] = 6+1/(12+1/12) = 6 + 12/145 = 882/145 ~~ 6.08276 | 381 | 1,052 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 12, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2021-25 | latest | en | 0.830121 |
https://www.coursehero.com/file/5885118/cp1-lect-3/ | 1,521,398,389,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645830.10/warc/CC-MAIN-20180318165408-20180318185408-00012.warc.gz | 811,661,853 | 102,571 | {[ promptMessage ]}
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# cp1_lect_3 - Taxonomy of Integers Integer Representat ion...
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5/10/10 Taxonomy of Integers Integer Representat ion Unsigne d Signed Signed Magnitud e One’s Complem ent Two’s Complem ent Mostl y used
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5/10/10 Unsigned Binary Integers 3-bits 5-bits 8-bits 0 000 00000 00000000 1 001 00001 00000001 2 010 00010 00000010 3 011 00011 00000011 4 100 00100 00000100 (011)2 = 0.22 + 1.21 + 1.20 = 2+1 = 3 k = number of bits Range is: 0 to 2k-1 Problem: How do we represent negative numbers? Applications: Counting, and Memory addressing in a computer Ex: k=5, Range is 0 to +31
5/10/10 Signed Magnitude Leading bit is the sign bit -4 10100 -3 10011 -2 10010 -1 10001 -0 10000 +0 00000 +1 00001 +2 00010 +3 00011 +4 00100 k=no. of bits Range is: -(2k-1 – 1) (2k- 1 – 1) Where a, b, and c can each take on 0/1 Ex: k=5 Range is : -15……..+15 Y = “abc” = (-1)a (b.21 + c.20)
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5/10/10 Signed Magnitude 41-4 Problems: Addition/Subtraction is not straightforward Try 2+(-3) 2 = 00010 -3 = 10011 -1(expected) = 10101 =-5 There are two zeroes! Application: No numbers, but for changing analog signals to digital signals
5/10/10 One’s Complement To convert -2 to +2 -4 11011 -3 11100 -2 11101 -1 11110 -0 11111 +0 00000 +1 00001 +2 00010 +3 00011 +4 00100 Range is: -(2k-1 – 1) … (2k-1 – 1) If MSB (most significant bit) is 1 then the number is negative (same as signed magnitude) Ex: k=8, Range is : -127…… +127
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5/10/10
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Ask a homework question - tutors are online | 717 | 2,151 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-13 | latest | en | 0.68086 |
https://zh.m.wikipedia.org/wiki/%E9%A6%AC%E7%88%BE%E5%8F%AF%E5%A4%AB%E9%8F%88 | 1,679,735,154,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945317.85/warc/CC-MAIN-20230325064253-20230325094253-00052.warc.gz | 1,210,440,845 | 37,842 | # 马尔可夫链
（重定向自馬爾可夫鏈
## 形式化定义
Xi的可能值构成的可数集S叫做该链的“状态空间”。
### 变种
• 具有连续索引。
• 时齐马尔可夫链（或静态马尔可夫链）是对于所有n
${\displaystyle \Pr(X_{n+1}=x\mid X_{n}=y)=\Pr(X_{n}=x\mid X_{n-1}=y)\,}$
• m阶马尔可夫链（或记忆为m的马尔可夫链），其中m有限，为满足
{\displaystyle {\begin{aligned}{}&\Pr(X_{n}=x_{n}\mid X_{n-1}=x_{n-1},X_{n-2}=x_{n-2},\dots ,X_{1}=x_{1})\\=&\Pr(X_{n}=x_{n}\mid X_{n-1}=x_{n-1},X_{n-2}=x_{n-2},\dots ,X_{n-m}=x_{n-m}){\text{ for }}n>m\end{aligned}}}
## 瞬态演变
n步从状态i到状态j的概率为
${\displaystyle p_{ij}^{(n)}=\Pr(X_{n}=j\mid X_{0}=i)\,}$
${\displaystyle p_{ij}=\Pr(X_{1}=j\mid X_{0}=i).\,}$
${\displaystyle p_{ij}^{(n)}=\Pr(X_{k+n}=j\mid X_{k}=i)\,}$
${\displaystyle p_{ij}=\Pr(X_{k+1}=j\mid X_{k}=i).\,}$
n步转移概率满足查普曼-科尔莫戈罗夫等式，对任意k使得0 < k < n
${\displaystyle p_{ij}^{(n)}=\sum _{r\in S}p_{ir}^{(k)}p_{rj}^{(n-k)}}$
${\displaystyle \Pr(X_{n}=j)=\sum _{r\in S}p_{rj}\Pr(X_{n-1}=r)=\sum _{r\in S}p_{rj}^{(n)}\Pr(X_{0}=r)}$
## 性质
### 可还原性
${\displaystyle P(X_{n+1}|X_{n})\,}$
${\displaystyle P(X_{n+2}|X_{n})=\int P(X_{n+2},X_{n+1}|X_{n})\,dX_{n+1}=\int P(X_{n+2}|X_{n+1})\,P(X_{n+1}|X_{n})\,dX_{n+1}}$
${\displaystyle P(X_{n+3}|X_{n})=\int P(X_{n+3}|X_{n+2})\int P(X_{n+2}|X_{n+1})\,P(X_{n+1}|X_{n})\,dX_{n+1}\,dX_{n+2}}$
### 周期性
${\displaystyle P(X_{n+1})=\int P(X_{n+1}|X_{n})\,P(X_{n})\,dX_{n}}$
${\displaystyle \pi (X)=\int P(X|Y)\,\pi (Y)\,dY}$
## 有限状态空间
${\displaystyle p_{ij}=\Pr(X_{n+1}=j\mid X_{n}=i).\,}$
### 稳定分布与特征向量和单纯形的关系
${\displaystyle \pi \mathbf {P} =\pi .\,}$
${\displaystyle \pi ={\frac {e}{\sum _{i}{e_{i}}}}}$
### 有限状态空间内的时齐马尔可夫链
${\displaystyle \lim _{k\rightarrow \infty }\mathbf {P} ^{k}=\pi ^{*}}$ ,
## 可反转马尔可夫链
{\displaystyle {\begin{aligned}\Pr(X_{n}=i\mid X_{n+1}=j)&={\frac {\Pr(X_{n}=i,X_{n+1}=j)}{\Pr(X_{n+1}=j)}}\\&={\frac {\Pr(X_{n}=i)\Pr(X_{n+1}=j\mid X_{n}=i)}{\Pr(X_{n+1}=j)}}.\end{aligned}}}
${\displaystyle \pi _{i}p_{ij}=\pi _{j}p_{ji}.\,}$
${\displaystyle \sum _{i}\pi _{i}p_{ij}=\pi _{j}\,}$
## 参考文献
### 引用
1. ^ Norris, James R. Markov chains. Cambridge University Press. 1998 [2015-03-19]. （原始内容存档于2021-05-06）.
2. ^ Prasad, NR; RC Ender; ST Reilly; G Nesgos. Allocation of resources on a minimized cost basis. 1974 IEEE Conference on Decision and Control including the 13th Symposium on Adaptive Processes. 1974, 13: 402–3. doi:10.1109/CDC.1974.270470. （原始内容存档于2015-02-12）.
3. Hamilton, James . A new approach to the economic analysis of nonstationary time series and the business cycle. Econometrica (Econometrica, Vol. 57, No. 2). 1989, 57 (2): 357–84. JSTOR 1912559. doi:10.2307/1912559.
### 来源
• A.A. Markov. "Rasprostranenie zakona bol'shih chisel na velichiny, zavisyaschie drug ot druga". Izvestiya Fiziko-matematicheskogo obschestva pri Kazanskom universitete, 2-ya seriya, tom 15, pp 135-156, 1906.
• A.A. Markov. "Extension of the limit theorems of probability theory to a sum of variables connected in a chain". reprinted in Appendix B of: R. Howard. Dynamic Probabilistic Systems, volume 1: Markov Chains. John Wiley and Sons, 1971.
• Leo Breiman. Probability. Original edition published by Addison-Wesley, 1968; reprinted by Society for Industrial and Applied Mathematics, 1992. ISBN 978-0-89871-296-4. （See Chapter 7.）
• J.L. Doob. Stochastic Processes. New York: John Wiley and Sons, 1953. ISBN 978-0-471-52369-7. | 1,546 | 3,272 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 60, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2023-14 | latest | en | 0.252839 |
http://mathhelpforum.com/algebra/83347-how-many-minutes.html | 1,481,062,637,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542002.53/warc/CC-MAIN-20161202170902-00221-ip-10-31-129-80.ec2.internal.warc.gz | 158,632,437 | 10,744 | 1. ## How Many Minutes?
Skipper can mow his lawn in 20 minutes less times with the power mower than with his hand mower. One day the power mower broke down 15 minutes after he started mowing, and he took 25 minutes more time to complete the job with the hand mower. How many minutes does Skipper take to mow the lawn with the power mower?
I wrestled with this question for more than 60 minutes and could not come up with the correct equation. Can someone help me? I dislike the way the question is worded. Do you agree?
2. Originally Posted by magentarita
Skipper can mow his lawn in 20 minutes less times with the power mower than with his hand mower. One day the power mower broke down 15 minutes after he started mowing, and he took 25 minutes more time to complete the job with the hand mower. How many minutes does Skipper take to mow the lawn with the power mower?
I wrestled with this question for more than 60 minutes and could not come up with the correct equation. Can someone help me? I dislike the way the question is worded. Do you agree?
Let x be the time (in minutes) it takes using the power mower, let y be the time (in minutes) it takes using the hand mower.
y = x + 20 .... (1)
After 15 minutes the fraction of time remaining to do the line and hence the fraction of lawn still to be mowed is $\frac{x - 15}{x}$. So the time it will take to finish the job using the hand mower is $\frac{x - 15}{x} \cdot y$.
But taking 25 minutes more to complete the job means that the lawn got mowed in 15 + 25 = 40 minutes. Hence:
[tex]15 + \frac{x - 15}{x} \cdot y = 40 \Rightarrow 15x + (x - 15)y = 40x
$\Rightarrow (x - 15)y = 25x$ .... (2)
Substitute equation (1) into equation (2):
$(x - 15) (x + 20) = 25x$.
You want the positive solution to this equation. I get x = 30.
3. ## ok..
Originally Posted by mr fantastic
Let x be the time (in minutes) it takes using the power mower, let y be the time (in minutes) it takes using the hand mower.
y = x + 20 .... (1)
After 15 minutes the fraction of time remaining to do the line and hence the fraction of lawn still to be mowed is $\frac{x - 15}{x}$. So the time it will take to finish the job using the hand mower is $\frac{x - 15}{x} \cdot y$.
But taking 25 minutes more to complete the job means that the lawn got mowed in 15 + 25 = 40 minutes. Hence:
[tex]15 + \frac{x - 15}{x} \cdot y = 40 \Rightarrow 15x + (x - 15)y = 40x
$\Rightarrow (x - 15)y = 25x$ .... (2)
Substitute equation (1) into equation (2):
$(x - 15) (x + 20) = 25x$.
You want the positive solution to this equation. I get x = 30.
I would have never come up with those two equations. First of all, I had no clue that this question is one involving a system of equations in two unknowns. Secondly, the wording is absolutely horrible, to say the least.
4. Originally Posted by magentarita
I would have never come up with those two equations. First of all, I had no clue that this question is one involving a system of equations in two unknowns. Secondly, the wording is absolutely horrible, to say the least.
I agree it could have been worded better. In particular:
One day the power mower broke down 15 minutes after he started mowing, and he took 25 minutes more time to complete the job with the hand mower
is ambiguous and could be interpretted several different ways. But the way I interpretted it is the only way that gives a sensible answer. | 905 | 3,397 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2016-50 | longest | en | 0.936965 |
https://www.gradesaver.com/textbooks/math/prealgebra/prealgebra-7th-edition/chapter-4-section-4-7-operations-on-mixed-numbers-exercise-set-page-298/25 | 1,532,112,762,000,000,000 | text/html | crawl-data/CC-MAIN-2018-30/segments/1531676591719.4/warc/CC-MAIN-20180720174340-20180720194340-00320.warc.gz | 894,176,867 | 14,839 | ## Prealgebra (7th Edition)
Published by Pearson
# Chapter 4 - Section 4.7 - Operations on Mixed Numbers - Exercise Set: 25
#### Answer
b. $10$
#### Work Step by Step
We can round $8\frac{1}{3}$ to $8$ and $1\frac{1}{2}$ to $2$ to get the expression: $8+2$, which equals $10$.
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. | 129 | 443 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2018-30 | latest | en | 0.853444 |
https://www.ilovephilosophy.com/t/riddles/40626/196 | 1,721,497,759,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517515.18/warc/CC-MAIN-20240720174732-20240720204732-00047.warc.gz | 700,439,496 | 3,907 | # Riddles
[tab]In the beginning A knows that (1) a = 12 or a = 15, and (2) B knows that b = 12 or b = 15.
Okay. But A does not know that B (2) knows, and B does not know that A (1) knows. So the statement above is not suited for the recursive conclusion.
But both A and B know all of the following statements and that each of them knows that the other one knows them:
(3) a = 24 - b or a = 27 - b and (4) b = 24 - a or b = 27 - a.
Now, from the first “no” of A and from (4) follows (5) b < 24, because if b >= 24, then A would be able to conclude a. This is the motor for the recursive conclusion.
Now, from the first “no” of B and from (3) and (5) follows (6) a > 3.
And so on.[/tab] | 224 | 691 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-30 | longest | en | 0.946225 |
https://pressbooks.library.upei.ca/montelpare/chapter/measures-of-association-part-2-the-kappa-statistic/ | 1,721,491,379,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515300.51/warc/CC-MAIN-20240720144323-20240720174323-00506.warc.gz | 412,006,288 | 25,554 | Measuring Correlation, Association, Reliability and Validity
# Part II: The Kappa Statistic to Measure Agreement
Given that the results of the McNemar Chi-Square statistic, calculated in the previous chapter, were not significant, then the question becomes, “if the outcome variables representing the results of a participant’s performance on each test are not statistically significant in their difference, does that necessarily mean that the outcome scores are in agreement?”
Since the Kappa statistic is a measure of agreement we can test this notion using the Kappa statistic applied to the fourfold or 2 x 2 table. Converse to the McNemar Chi-square which processes the data in the off-diagonal elements (cell “b” and cell “c”), the Kappa computations focus on the data in the major diagonal from upper left to lower right (cell “a” and cell “d”), examining whether counts along this diagonal differ significantly from what is expected to occur by chance. If no agreement exists between the counts on the major diagonal then we would expect the proportion of individuals scoring high or low on the lab and field tests to be similar.
Similar to the computation of the McNemar Chi-square, the Kappa statistic uses the data from the row and column probabilities of the 2 x 2 table. The exact computations for Kappa are specifically shown as follows:
1. COMPUTE ROW PROPORTIONS
Row 1 Proportion: p1. = (a+b) ÷ N; p1. = (23+12) ÷ 86 = 0.41
Row 2 Proportion: p2. = (c+d) ÷ N; p2. = (19+32) ÷ 86 = 0.59
Column 1 Proportion: p.1 = (a+c) ÷ N; p.1 = (23+19) ÷ 86 = 0.49
Column 1 Proportion: p.2 = (b+d) ÷ N; p.2 = (12+32) ÷ 86 = 0.51
1. COMPUTE THE $P_{i}$ TERMS
OBSERVED: $({\pi}) \textit{obs}$
$({\pi}) \textit{obs}$: the observed term of the main diagonal elements
$({\pi}) \textit{obs}$ = ((cell a) ÷ N) + ((cell d) ÷ N);
$({\pi}) \textit{obs}$ = ((23÷86) + (32 ÷ 86));
$({\pi}) \textit{obs}$ = (.27+.37);
$({\pi}) \textit{obs}$ = 0.64
EXPECTED: $({\pi}) \textit{exp}$
$({\pi}) \textit{exp}$: the expected term of the main diagonal elements
$({\pi}) \textit{exp}$ = ((p1. * p.1) + (p2. * p.2);
$({\pi}) \textit{exp}$ = ((0.41 * 0.49) + (0.59 * 0.51));
$({\pi}) \textit{exp}$ = (0.20 + 0.30);
$({\pi}) \textit{exp}$ = (0.50);
1. COMPUTE KAPPA $({\kappa})$
Kappa = (( $({\pi}) \textit{obs} - ({\pi}) \textit{exp}) ÷ (1- ({\pi}) \textit{exp}$ ))
Kappa = ((0.64 – 0.50) ÷ (1- 0.50))
Kappa = (0.14 ÷ 0.50)
Kappa = 0.28
The computed Kappa value is κ = 0.28. Our next task is then to determine if this is a true measure of agreement or an agreement that can happen by chance. Therefore, in order to evaluate this Kappa statistic we need to determine if the computed value is significantly different than 0.
We can do this by first computing the standard error of the Kappa statistic and then using this value to determine the z statistic for Kappa and comparing the value to the normal curve. Recall that 95% of scores on the normal curve are < ±1.96. Therefore, if our Zκ score is between -1.96 and +1.96 then we would accept the null hypothesis that κ=0.
To compute the standard error for our computed KAPPA SCORE we use the following procedure under the null hypothesis that Ho: k=0
1. COMPUTE THE SUM OF PROPORTIONS
p1. = 0.41; p.1 = 0.49; p2. = 0.59; p.2 = 0.51
sumP = (p1. * p.1 * (p1. + p.1)) + (p2. * p.2 * (p2. + p.2));
sumP = (0.41 * 0.49 * (0.41 + 0.49)) + (0.59 * 0.51 * (0.59 + 0.51));
sumP = (0.20 * (0.90)) + (0.30 * (1.10));
sumP = (0.18) + (0.33);
sumP = (0.51);
1. COMPUTE THE STANDARD ERROR
std error = 1/((1- ) * )*
std error = 1/((1- 0.5) * ) *
std error = 1/((0.5) * ) * 0.49
std error = 0.22 * 0.49
std error = 0.106
Use the following formula to compute Zκ which is the z score for Kappa, under the null hypothesis of Ho: k=0:
zKappa = (kappa/ stderr1) zKappa = (0.28/ 0.106) zKappa = 2.65
Considering that 2.65 is greater that 1.96 we can say that the zKappa is within the region of rejection in regard to the null hypothesis stated as Ho:k=0 and therefore we can say that there is agreement between the lab and field test.
Finally, we can also determine the significant difference of our Kappa estimate from 0 by using the standard error to compute the 95% confidence intervals for the Kappa statistic as follows:
1. COMPUTE THE STANDARD ERROR AND 95% CONFIDENCE INTERVAL
Use the following measurement terms taken from the McNemar Chi-square table:
p1. = 0.41 P11 = 23/86=0.27 p.1 = 0.49 p12 = 12/86=0.14 p2. = 0.59 p21 = 19/86 = 0.22 p.2 = 0.51 p22 = 32/86 =0.37
Aterm = (p11*(1-(p1. + p.1)*(1-kappa))**2 + p22*(1-(p2. + p.2)*(1-kappa))**2);
Aterm = (0.27*(1-(0.41 + 0.49)*(1-0.28))**2 + 0.37*(1-(0.59 + 0.51)*(1-0.28))**2);
Aterm = (0.27*(1-(0.9)*(1-0.28))**2 + 0.37*(1-(1.10)*(1-0.28))**2);
Aterm = (0.27*(1-(0.9)*(0.72))**2 + 0.37*(1-(1.10)*(0.72))**2);
Aterm = (0.27*(1-(0.648))**2 + 0.37*(1-(0.792))**2);
Aterm = (0.27*(0.352)**2 + 0.37*(0.208)**2);
Aterm = (0.27*(0.124) + 0.37*(0.043));
Aterm = (0.033 + 0.02);
Aterm = (0.049);
Bterm=((p12*(p.1 + p2.)2 + p21*(p.2 + p1.)2)*(1-kappa)2);
Bterm=((0.14*(0.49 + 0.59)2 + 0.22*(0.51 + 0.41)2)*(1-0.28)2);
Bterm=((0.14*(1.08)2 + 0.22*(0.92)2)*(0.72)2);
Bterm=((0.14*(1.17) + 0.22*(0.84))*(0.52));
Bterm=((0.16 + 0.185)*(0.52));
Bterm=(0.179);
Cterm=((kappa – *(1-kappa))**2);
Cterm=((0.28 – 0.5*(0.72))2);
Cterm=(0.0064)
A + B + C= (Aterm + Bterm + Cterm);
A + B + C= (0.049 + 0.179 + 0.0064);
A + B + C= 0.23
Compute the standard error used in the computation of the confidence interval:
stderr = = = = 0.01
ci95LL = (kappa – 1.96*(stderr));
ci95LL = (0.28 – 1.96 * 0.01);
ci95LL = (0.28 – 0.022);
ci95LL = (0.258)
ci95UL = (kappa + 1.96*(stderr2));
ci95UL = (0.28 + 1.96 * 0.01);
ci95UL = (0.28 + 0.022);
ci95UL = (0.302);
If the upper and lower limits of the 95% confidence interval do not include 0 then we can say that the Kappa value is significantly different from 0.
The SAS program to produce KAPPA in the 2 x 2 matrix was handled by the McNemar Chi-Square, where a=23, b=12, c= 19, d=32. Since the data were entered as cell summary data and not strings of raw data, the weight <dependent variable> format is used to read each cell value. The essential option is /AGREE which produces the Kappa measure of agreement.
PROC FREQ;
TABLES ROW*COL /AGREE;
WEIGHT OUTCOME;
RUN;
Statistics for Table of ROW BY COL
Simple Kappa Coefficient Kappa 0.2759 ASE 0.1024 95% Lower Conf Limit 0.0752 95% Upper Conf Limit 0.4767 | 2,392 | 6,541 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2024-30 | latest | en | 0.824375 |
https://themindofanaprilfool.com/2017/02/13/sharing-hearts-a-kinder-3-act/ | 1,563,728,546,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195527089.77/warc/CC-MAIN-20190721164644-20190721190644-00300.warc.gz | 573,551,506 | 22,773 | # Sharing Hearts – a Kinder 3 act
It’s that time of the year when our students are running to ___________(insert convenience store here—I prefer Target) to buy gobs of chocolates, candies, cards and all of the Valentine’s paraphernalia that is attributed to this holiday. And in the spirit of such heartsy, lovey, dovey emotions, Mrs Z. and I put together another 3 act lesson. What better way to embody Valentine’s Day then to share with the ones you love. In our case, it was sharing with your classmates.
## Act One –the fight.
Along with every Three Act lesson, we encourage the students to inquire about what they are seeing with the questions, “What do you notice?” and “What do you wonder?”
Notice – (Compiled between 2 classes)
• hands are fighting
• I heard the math wizard’s voice
• I see a box
• They were ready to share.
• Inside the box was candy
• Candies are hearts
• tried grabbing with hands
• Candies are different colors
• Why didn’t they get their own box of candies?
• Did they have enough to share?
• Are there other people there?
• How many candies are in the box?
• Why were they fighting?
• Why weren’t there 2 boxes?
A bonus concept for the kinder teachers was that there were a few talking points to review. Some students talked about what it means to share. One class focussed on sharing so that the results are fair. In one class, I had a
## Act 2 – How many hearts are in the box?
Act 2 consisted of us giving the students the information that there were 12 hearts in the box. Using this candy-hearts template, we had the students figure out how many candy hearts Jared and his mom should get. You can see the multiple ways that the students went about solving it.
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Mrs. Z called me over to listen in on one conversation. Take a look below at what we observed.
Terrific piece of evidence. This student observed that she knew that they were the same because they had the same number of cubes and had the same shape.
## Act 3 – The big reveal
Act 3 is usually my favorite because it reveals the answer. It’s really the best part of the lesson because you can hear the “YES!” or see the high fives spreading throughout the entire class. Here’s how we revealed the answer.
Overall the lesson went better than planned. There was another teacher that observed Mrs. Z doing this lesson and whispered, “this is dividing.” I nodded and told her that was true, however I knew that Mrs. Z had begun working on some addition. I also told her that you never know what our students are capable of until you give them that opportunity.
Keep spreading the LOVE of MATH!
Until next time…
Kristen | 628 | 2,680 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2019-30 | latest | en | 0.97388 |
https://im.kendallhunt.com/HS/teachers/2/1/16/practice.html | 1,679,899,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948609.41/warc/CC-MAIN-20230327060940-20230327090940-00635.warc.gz | 354,870,822 | 29,052 | # Lesson 16
More Symmetry
### Problem 1
For each figure, identify any angles of rotation that create symmetry.
### Problem 2
A triangle has rotation symmetry that can take any of its vertices to any of its other vertices. Select all conclusions that we can reach from this.
A:
All sides of the triangle have the same length.
B:
All angles of the triangle have the same measure.
C:
All rotations take one half of the triangle to the other half of the triangle.
### Problem 3
Select all the angles of rotation that produce symmetry for this flower.
A:
45 degrees
B:
90 degrees
C:
135 degrees
D:
180 degrees
E:
225 degrees
F:
270 degrees
### Problem 4
Identify any lines of symmetry the figure has.
### Solution
(From Unit 1, Lesson 15.)
### Problem 5
A triangle has a line of symmetry. Select all conclusions that must be true.
A:
All sides of the triangle have the same length.
B:
All angles of the triangle have the same measure.
C:
No sides of the triangle have the same length.
D:
No angles of the triangle have the same measure.
E:
Two sides of the triangle have the same length.
F:
Two angles of the triangle have the same measure.
### Solution
(From Unit 1, Lesson 15.)
### Problem 6
Here are 4 triangles that have each been transformed by a different transformation. Which transformation is not a rigid transformation?
A:
B:
C:
D:
### Solution
Match each directed line segment with the translation from Polygon $$P$$ to Polygon $$Q$$ by that directed line segment. | 369 | 1,520 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2023-14 | longest | en | 0.847606 |
https://sumidiot.wordpress.com/category/play/page/2/ | 1,529,797,380,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267865438.16/warc/CC-MAIN-20180623225824-20180624005824-00538.warc.gz | 703,760,993 | 42,646 | ## Archive for the ‘Play’ Category
### The Log of Zeta
November 21, 2009
Last time (too long ago, sorry), I finally got around to talking about the function $\zeta(s)$. Today, I think I’d like to talk about $\log \zeta(s)$.
First, though, I should mention Euler’s product formula. The goal is to re-write the infinite series $\zeta(s)=\sum_n n^{-s}$ as a product. For each prime $p$, we can tease out of the series above those terms where $n$ is $p$ to some power. That is, we can think about the sub-series $\sum_{k\geq 0} (p^k)^{-s}$. This is a geometric series that converges to $(1-p^{-s})^{-1}$. If we take two such series and multiply them together, which terms from our starting series for $\zeta(s)$ will we recover?
Suppose $p$ and $q$ are two primes, and consider the product
$\displaystyle\left(1+\frac{1}{p^s}+\frac{1}{p^{2s}}+\frac{1}{p^{3s}}+\cdots\right)\cdot \left(1+ \frac{1}{q^s}+\frac{1}{q^{2s}}+\frac{1}{q^{3s}}+\cdots\right)$.
The terms in $\sum_n n^{-s}$ that we obtain from this product are those where $n$ is a product of some power (possibly the 0 power) of $p$ and some power (again, possibly 0) of $q$. We could then think about another prime, say $r$, and it’s geometric series, and multiply it by the above, and obtain all the terms in $\sum_n n^{-s}$ where $n=p^aq^br^c$ for some $0\leq a,b,c$.
Continuing on, and recalling that every positive integer has a unique factorization as a product of primes, we obtain the magical formula (isn’t most of Euler’s work magical?):
$\displaystyle \sum_n \frac{1}{n^s}=\prod_p (1-p^{-s})^{-1}$,
where the product is indexed by the primes. What’s nice about this, for us, at the moment, is that logarithms work well with products and powers, but not soo well for sums. Recalling the Taylor polynomial
$\ln(1-x)=-1-x-x^2-x^3-\cdots=-\sum\limits_{n=0}^{\infty} x^n$
we find
$\displaystyle \ln \zeta(s)=\sum_p\sum_n \frac{1}{n}p^{-ns}$.
That was fun. Incidentally, it’s not much of a jump from here to show that the series of prime reciprocals diverges. I mentioned it in a post a while ago.
Let’s switch gears for a little bit. I’m going to define a function $J(x)$ on the positive reals. If you’re following Edwards’ book with me (as I jump all over the place), this will be almost exactly his $J(x)$, differing only at the integers (for today anyway :)). Let me start by setting $J(0)=0$. The function $J(x)$ will be a step function, which I’ll define to mean piecewise linear with slope 0 on each piece. So to define $J(x)$ I only need to tell you when to jump, and by how much. The rule is: when you get to an integer $n$, if $n=p^k$ for some prime $p$, then you jump up by $1/k$. So at $2,3,5,7,11,\ldots$ you jump by 1, at $4,9,25,49,\ldots$ you jump by 1/2, at $8, 27, \ldots$ you jump by 1/3, etc. Here’s a little picture of $J(x)$, with a few values highlighted:
Slightly more precisely, we can write
$\displaystyle J(x)=\sum_{p^n\leq x}\frac{1}{n}$.
Now, let me see if I can convince you that there is some justification in writing
$\displaystyle \ln \zeta(s)=s\int_0^{\infty} J(x)x^{-s-1}\ dx$.
Let’s work with the right-hand side. Since $J(x)=0$ near $x=0$, I’ll actually start my integral at $x=1$. I think the way to go about it is as follows:
$\begin{array}{rcl} \displaystyle s\int_1^{\infty}J(x)x^{-s-1}\ dx &=& \displaystyle s\sum_{m=1}^{\infty}\int_m^{m+1}J(x)x^{-s-1}\ dx \\ {} & = & \displaystyle \sum_{m=1}^{\infty}J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right).\end{array}$
Now suppose that $J(m)=J(m+1)$ for some $m$. Then the terms corresponding to $m$ and $m+1$ telescope as follows:
$\begin{array}{c} \displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+1)^s}\right)+J(m+1)\left(\frac{1}{(m+1)^s}-\frac{1}{(m+2)^s}\right)\\ =\displaystyle J(m)\left(\frac{1}{m^s}-\frac{1}{(m+2)^s}\right)\end{array}$.
If, also, $J(m)=J(m+2)$, this we can telescope another term into this one, and so on. So, really, the important $m$ in this sum are those where $J(x)$ jumps, which are the prime powers. Let $m_i=p_i^{n_i}$ be the $i$-th prime power (i.e., the point where $J(x)$ makes its $i$-th jump), starting with $m_1=2$. Then we can write
$\begin{array}{rcl} \displaystyle s\int_0^{\infty}J(x)x^{-s-1}\ dx & = & \displaystyle \sum_i J(m_i)\left(\frac{1}{m_i^s}-\frac{1}{m_{i+1}^s}\right) \\ & = & \displaystyle \frac{1}{2}J(2)+\sum_{i=2}^{\infty} -J(m_{i-1})\frac{1}{m_i^s}+J(m_i)\frac{1}{m_i^s} \\ & = & \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\left(J(m_i)-J(m_{i-1})\right) \\ & = & \displaystyle \frac{1}{2}+\sum_{i=2}^{\infty} \frac{1}{m_i^s}\cdot \frac{1}{n_i} \\ & = & \displaystyle \sum_{p}\sum_n \frac{1}{n}p^{-ns} \\ & = & \ln \zeta(s).\end{array}$
What good is writing $\ln \zeta(s)$ this way? I guess if you know something about Fourier inversion (which I don’t) then you get to say that
$\displaystyle J(x)=\frac{1}{2\pi i}\int_{a-i\infty}^{a+i\infty}\log \zeta(s)x^s\frac{ds}{s},$
for $a=\text{Re}(s)>1$. What good is that? I think I’ll have to read some more and tell you about it tomorrow, but it’ll turn out to be useful once we have yet another description of $\ln \zeta(s)$, in terms of the 0s of $\zeta(s)$ (finally getting to those 0s everybody cares so much about).
### Riemann’s Zeta Function
November 13, 2009
I guess it is about time to get to the zeta function side of this story, if we’re ever going to use Farey sequences to show how you could prove the Riemann hypothesis. I’ve been reading a bit of Edwards’ book, with the same title as this post, and thought I’d try to summarize the first chapter over the next few posts. I’m not sure that the content of the first chapter is hugely vital for the final goal of relating to the Farey sequences, but I wanted to try to learn some of it anyway.
I should mention, before talking about any of this, that I will not claim to know any complex analysis. The last complex analysis course I took was 5 years ago, I can’t be sure how much I paid attention at the time, and I haven’t used it since. I will be jumping over gaps of various sizes for quite a while in the upcoming posts. Perhaps sometimes I’ll mention that a gap is there. Mostly, though, what I’m after in this story is the outline, and how all of the parts fit together, as a big picture. Perhaps I’ll go back and fill in some gaps, as I understand more.
For today, let’s see if I can build up to an analytic expression of $\zeta(s)$. Our starting point is the function
$\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}$
which is defined for real values of $s$ larger than 1. The goal is to find a nice expression that is defined on more of the complex plane, but agrees with this definition on the reals larger than 1.
To get to that point, we’ll use the $\Gamma$ function:
$\Gamma(s)=\displaystyle\int_0^{\infty} e^{-x}x^{s-1}\ dx$
This is an analytic function defined everywhere except at the negative integers and 0. If you are only interested in real values of $s$, this function is a continuous extension of the factorial function, which is only defined on the positive integers. Actually, there is a shift involved, so $\Gamma(n)=(n-1)!$ (Edwards blames this shift on Legendre, whose reasons, he states, “are obscure”. Edwards uses $\Pi(s)=\Gamma(s+1)$ throughout, so I might make some $\pm 1$ errors). In particular, $s\Gamma(s)=\Gamma(s+1)$ when both sides make sense.
Another relation that we’ll use is that
$\pi=\Gamma(s)\Gamma(1-s)\sin(\pi s)$
If memory serves (from reading Artin’s lovely little book on the $\Gamma$ function), you can obtain this by determining that the derivative of the expression on the right is 0. This tells you that $\sin(\pi s)\Gamma(s)\Gamma(1-s)$ is a constant, and so you can pick your favorite value of $s$ to obtain the value of that constant. Anyway, the identity I’ll use is a rearrangement of the one above, namely
$\Gamma(s)\sin(\pi s)=\dfrac{\pi}{\Gamma(1-s)}$.
Now, in the expression
$\Gamma(s)=\displaystyle\int_0^{\infty}e^{-x}x^{s-1}\ dx$,
substitute $x=nu$ for some positive $n$. Messing about with that substitution for a few minutes (and then re-writing $u=x$ by abuse of notation), you can arrive at
$\dfrac{\Gamma(s)}{n^s}=\displaystyle\int_0^{\infty} e^{-nx}x^{s-1}\ dx$.
That $n^s$ in the denominator is useful for us, as that’s how it shows up in $\zeta(s)$. In particular, we can obtain
$\begin{array}{rcl}\Gamma(s)\zeta(s) &=& \displaystyle \Gamma(s)\sum_{n=1}^{\infty}\dfrac{1}{n^s}=\sum_{n=1}^{\infty} \frac{\Gamma(s)}{n^s} \\ &=& \displaystyle \sum_{n=1}^{\infty}\int_0^{\infty}e^{-nx}x^{s-1}\ dx \\ &=& \displaystyle \int_0^{\infty}x^{s-1}\sum_{n=1}^{\infty} e^{-nx}\ dx\\ &=& \displaystyle \int_0^{\infty} \dfrac{x^{s-1}}{e^x-1}\ dx\end{array}$
That last transition coming about by summing the geometric series. There are probably some things analysts like to check in here, moving infinite sums and improper integrals past each other… I’ll let them.
Ok, we’re nearly there. Next up, you do some complex line integral and show that
$\begin{array}{rcl} \displaystyle \int_{+\infty}^{+\infty} \dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x} &=& \displaystyle (e^{i\pi s}-e^{-i\pi s})\int_0^{\infty}\dfrac{x^{s-1}}{e^x-1}\ dx \\ &=& 2i\sin(\pi s)\Gamma(s)\zeta(s)\end{array}$.
That integral is a little weird, going “from $+\infty$ to $+\infty$“. Really, we take a path that “starts at $+\infty$“, swings around $0$, then goes back out to infinity. This is almost certainly one of the complex analysis things I should go back and learn more about. Since we are integrating $(-x)^s=e^{s\log(- x)}$ along positive real values of $x$, we’re working with logs along the negative real axis. Perhaps I’ll return to this integral in a future post.
Using the identity about $\sin(\pi s)\Gamma(s)$ from above, we obtain
$\zeta(s)=\dfrac{\Gamma(1-s)}{2\pi i}\displaystyle \int_{+\infty}^{+\infty}\dfrac{(-x)^s}{e^x-1}\ \dfrac{dx}{x}$.
We’re now at a point where we’ve (apparently) got an expression for $\zeta(s)$ that is analytic in the complex plane except for a simple pole at $s=1$.
### ?'(x)
November 11, 2009
Yesterday I talked about the function $?(x)$. If you recall (or rather, if you don’t… if you do recall, you can probably skip to the next paragraph), I defined this function by lining up two infinite binary trees. One tree was obtained by taking mediants, giving us the Stern-Brocot tree, while the other was obtained by taking midpoints, giving us a tree of “dyadic” fractions (denominators are powers of two). To evaluate $?(x)$, you find $x$ in the Stern-Brocot tree, and then $?(x)$ is the value in the corresponding position of the dyadic tree.
Instead of thinking about infinite binary trees, let us think about cutting each tree off at some level, $n$. The pictures from yesterday’s post are, if you ignore the vertical dots, the trees when $n=2$. In the dyadic tree, we will obtain all fractions $m/2^p$ where $0 is odd and $1\leq p\leq n+1$. Said another way, we will get all of the values $m/2^{n+1}$ where $0 (not necessarily odd anymore). If you think about it for a few moments, you’ll notice that these points are equally spaced in the interval, each being $1/2^{n+1}$ away from its neighbors. This is not the case for the Stern-Brocot tree.
In the truncated Stern-Brocot tree, points are scattered through the interval (0,1). We can see that they are scattered in a way that is symmetric around 1/2, but what else can we say? Think about choosing some number, $b$, of buckets, call them $B_0,\ldots,B_{b-1}$. Let $B_i$ collect all of the points from the truncated Stern-Brocot tree whose value lies between $i/b$ and $(i+1)/b$. Of course, I’m being a little loose with my endpoints, but hopefully the idea comes across. If you choose an appropriate number of buckets, depending on the depth of your truncated tree, you can arrange that no point is an endpoint of a bucket’s interval.
Anyway, when you’re done with that, count how many points are in each bucket. This defines a function from the interval $[0,1]$ to non-negative integers, where the value at all points in $B_i$ is the count of the number of Stern-Brocot points in the bucket $B_i$. Here’s an example graph, obtained by taking the depth 10 Stern-Brocot tree, and putting points into 59 buckets (unless I coded something up wrong :)):
You could next scale this graph vertically by a constant so that the total area of all of the rectangles is 1. That is, you could make the graph that of a probability distribution. Of course, the distribution you get will depend on not ony the depth of the tree, but also how many buckets you choose. Since I’m only vaguely sure what I’m talking about, let me not say much more about that.
A less graphical way to think about all of this is that, for each $n$, I define two random variables $X_n$ and $Y_n$. The first, $X_n$ picks a value from all of the nodes in the Stern-Brocot tree of depth $n$, assuming each is equally likely. The second is basically the same, using the dyadic tree instead. I’m not particularly well-versed in probability, but I guess you are allowed to take the limit of these probability distributions, let’s say $X=\lim X_n$, and similarly $Y=\lim Y_n$. This defines two probability functions on the interval $[0,1]$ (well, I suppose technically it probably defines a probability function on the rationals there…).
How are the two distributions related? Well, basically the same way as the trees are related by $?(x)$. In particular, we have
$P(a\leq X\leq b)=P(?(a)\leq Y\leq ?(b))$
Since $Y$ is just the uniform distribution (the dyadic points are equally distributed, recall), the probability on the right is just $?(b)-?(a)$. Let $\mu$ be the graph of the probability function for $X$, so that the probability on the left is the integral $\int_a^b \mu\ dx$. We have, then
$\displaystyle \int_a^b \mu(x)\ dx=?(b)-?(a)$
This looks so much like the fundamental theorem of calculus that we want to say that $?'(x)=\mu(x)$. You could also read it as saying that $?(x)$ is the cumulative distribution function of $\mu(x)$. Thinking back to my buckets from earlier, this means that if I define the value over the interval covered by $B_i$ to be the sum of the counts of all the values in all the buckets up to $B_i$, then I should get a graph that looks like $?(x)$. You be the judge:
Of course, we probably aren’t exactly allowed to say $?'(x)=\mu(x)$. Consider the rational $s/t$, and suppose that $?'(s/t)$ exists, i.e., that
$\displaystyle\lim_{x\to s/t}\dfrac{?(x)-?(s/t)}{x-s/t}$
exists. If this two sided limit exists, then we can calculate it by coming at it from either side. Let’s think about $x\to s/t^+$, that is, from the right. But let’s simplify some more, still. If the one-sided limit exists, then we can calculate it by calculating the limit of the difference quotient for any sequence of values $x$ converging to $s/t$. The sequence of values, on the right of $s/t$, that I want to consider are those neighbors, on the right, of $s/t$, in the Stern-Brocot sequences.
Let’s say that the first neighbor is the right child of $s/t$ in the Stern-Brocot tree, and call it $p/q$. Then all of the successive neighbors are the successive left children of $p/q$. These points are all of the points $(ms+p)/(mt+q)$, for positive values $m$. If we let $m\to \infty$, these weighted mediants converge to $s/t$.
So let’s consider the following:
$\displaystyle\lim_{m\to\infty}\dfrac{?\left(\frac{ms+p}{mt+q}\right)-?\left(\frac{s}{t}\right)}{\frac{ms+p}{mt+q}-\frac{s}{t}}$.
Let’s start simplifying in the denominator. That fraction is $(pt-sq)/(t(mt+q))$, but since $p/q$ and $s/t$ are neighbors in some Farey sequence, this difference is 1. So the denominator is $1/(t(mt+q))$.
In the numerator, we are evaluating $?(x)$ at weighted mediants of $s/t$ and $p/q$. The corresponding $?(x)$ values are then weighted averages. Playing with the formulas a little bit, you can determine that
$?\left(\frac{ms+p}{mt+q}\right)=\dfrac{(2^m-1)?(s/t)+?(p/q)}{2^m}$.
Combining all of the formulas so far, and simplifying, it turns out that the limit we are after is just
$\displaystyle\lim_{m\to\infty}\left(?(p/q)-?(s/t)\right)\cdot t\cdot (mt+q)\cdot 2^{-m}$
Of course, that power of two wins the race to infinity, and so this limit is 0.
We have shown then, that if $?'(p/q)$ exists, then it is 0. Since $?(x)$ isn’t a constant graph, we better have $?'(x)$ non-zero on the irrationals. Perhaps I’ll leave that to you.
I should, of course, mention that I learned about this reading the paper “On the Minkowski measure” by Linas Vepstas, available online here (along with a handful of other papers that look fun). Hopefully I haven’t mangled anything too badly.
### ?(x)
November 10, 2009
Sorry about the lack of a MaBloWriMo post yesterday. I got a little behind, left resources I wanted at the office, and did some other things.
If you remember, my original goal in this process was to relate the Farey sequences to the Riemann hypothesis. I have some vague understanding that the relationship is based on the distribution of the terms in the Farey sequence, as points in the unit interval. It’s getting on toward time to talk about the Riemann hypothesis side of this picture, so I can start relating the two. But before I do, there’s one more diversion (hopefully it isn’t too far off base) I’d like to make related to Farey sequences. I’ve still got some learning to do about it, but let’s see what I can get down today, laying some foundation for the next day or two, perhaps.
Recall that the way to construct one Farey sequence, $F_{n+1}$, from the previous one, $F_n$, is to throw in all of the mediants of successive terms whose denominators aren’t too big. What if we didn’t bother with that last part, and just left in all the mediants? All the numbers we’ll write down are still reduced, and we’ll obtain all of the fractions in $[0,1]$ this way, because they all show up in some Farey sequence, and we’re getting all the terms in all the Farey sequences, just sometimes a little earlier than Farey would.
Just to be clear, our first few sequences now look like this:
$0/1\quad 1/1$
$0/1\quad 1/2\quad 1/1$
$0/1\quad 1/3\quad 1/2\quad 2/3\quad 1/1$
$0/1\quad 1/4\quad 1/3\quad 2/5\quad 1/2\quad 3/5\quad 2/3\quad 3/4\quad 1/1$
and so on. We see that the $n$-th sequence in this list contains all of $F_n$, as well as some other numbers with bigger denominators. A nice property of these sequences, which isn’t shared with the Farey sequences, is that every other term in the $n$-th sequence is a new term that wasn’t in the $n-1$-st sequence. So when I go to make the next sequence in this process, I’m always taking the mediant of a “new” number and an “old” number. Moreover, each “new” number will be involved in two mediants… one taken “to the left” (with the “old” number on the left, making a smaller value) and one taken “to the right”.
This language let’s me write down a binary tree (every node has two branches down) as follows: Begin with 1/2 as the root (at the top, I’m throwing out 0/1 and 1/1). This is a new fraction in the second sequence above, so to make the third sequence, I added in the “left mediant” (1/3) and “right mediant” (2/3) from 1/2. To make the sequence after that, 1/3 gives a “left mediant” (1/4) and a “right mediant” (2/5), and similarly 2/3 spawns 3/5 and 3/4. So I build up a tree that looks like this:
$\begin{picture}(200,80) \put(85,65){1/2} \put(45,45){1/3}\put(125,45){2/3} \put(25,25){1/4} \put(65,25){2/5} \put(105,25){3/5} \put(145,25){3/4} \put(5,5){\vdots} \put(45,5){\vdots} \put(85,5){\vdots} \put(125,5){\vdots} \put(165,5){\vdots} \put(55,55){\line(2,1){20}} \put(35,35){\line(1,1){10}} \put(125,55){\line(-2,1){20}} \put(65,35){\line(-1,1){10}} \put(115,35){\line(1,1){10}} \put(145,35){\line(-1,1){10}} \end{picture}$
I’ll call this the Stern-Brocot tree, though the actual definition my be slightly different.
Another way I could build up a sequence of sequences, and generate another binary tree, is to use midpoints, instead of mediants. If my first sequence is still 0/1 and 1/1, then taking midpoints will always give me denominators that are powers of 2. My tree will now look like:
$\begin{picture}(200,80) \put(85,65){1/2} \put(45,45){1/4}\put(125,45){3/4} \put(25,25){1/8} \put(65,25){3/8} \put(105,25){5/8} \put(145,25){7/8} \put(5,5){\vdots} \put(45,5){\vdots} \put(85,5){\vdots} \put(125,5){\vdots} \put(165,5){\vdots} \put(55,55){\line(2,1){20}} \put(35,35){\line(1,1){10}} \put(125,55){\line(-2,1){20}} \put(65,35){\line(-1,1){10}} \put(115,35){\line(1,1){10}} \put(145,35){\line(-1,1){10}} \end{picture}$
I could now line up these trees, and determine a function that takes one rational number, finds it’s position in the first tree, and spits out the value in the same position from the second tree. For example, using our pictures above, we see that this function will take 3/5 to 5/8. This function is known as Minkowski’s question mark function, $?(x)$, and Wikipedia has a graph, so you should click that last link. I’ve only defined it on rationals, but you wave the “continuity” wand to extend my definition to a function on $[0,1]$.
So, to evaluate $?(x)$ for some rational, you have to find that rational in the Stern-Brocot tree. How do you do that? You know that you don’t have to go further down in the tree than your denominator, but that’s still a lot of fractions to look at, potentially. Perhaps there’s a better way?
I guess one way is to play the over-under game. Start at 1/2. If that’s your fraction, you are done. Otherwise, if your fraction is bigger, move right, and if your fraction is smaller, move left. Repeat until you get to your fraction.
But there’s more math we can say if we think about this process slightly differently.
The point of the tree being a tree is that there is only one way to get from the root, 1/2, to any fraction. Working your way down, you make a sequence of moves, either left ($L$) or right ($R$). So you generate some string – $LLRLRRRL$, for example. Of course, that notation gets out of hand quickly, so we might also use exponents to write this example as $L^2RLR^3L$. To such a string we can associate a sequence of fractions by writing down whatever fraction we are at when we change direction. Let us write this sequence as $p'_1/q'_1,p'_2/q'_2,\ldots$, and call these “turning points”. I’ve used primes here, because the numbers I’m actually interested in are one before I make a turn. Let me call that sequence $p_1/q_1,p_2/q_2,\ldots$, calling these values parents (so $p_k/q_k$ has a line directly down to $p'_k/q'_k$ in the Stern-Brocot tree, connecting the two). So, for example, the sequence $L^3R^2L^2$ has turning points $p'_1/q'_1=1/5$, $p'_2/q'_2=3/13$, and finally $p'_3/q'_3=7/31$ and associated parents $p_1/q_1=1/4$, $p_2/q_2=2/9$, and $5/22$.
What is the relationship among successive turning points and parents? Suppose you are at the turning point $p'_k/q'_k$, with parent $p_k/q_k$, looking to get to the next parent or turning point, and are going to do this with a sequence of $a$ moves (left or right, it doesn’t actually matter). The next parent will be the $(1,a-1)$ weighted mediant of the turning point $p'_k/q'_k$ with its parent $p_k/q_k$. That is, $p_{k+1}=p'_k+(a-1)p_k$, and $q_{k+1}=q'_k+(a-1)q_k$. Similarly, the next turning point will be the $(1,a)$ weighted mediant of $p'_k/q'_k$ with $p_k/q_k$. This also means that it is the $(1,1)$ weighted mediant of $p_k/q_k$ with $p_{k+1}/q_{k+1}$.
Combining all of these relationships, messing with the algebra just slightly, you can determine that $p_{k+1}=ap_k+p_{k-1}$ and $q_{k+1}=aq_k+q_{k-1}$. You may recall that these are exactly the relations between the convergents of a continued fraction (if you start putting subscripts on your $a$, writing it as $a_{k+1}$).
So, given a continued fraction $[0;a_1,\ldots,a_n]$ for a rational $x$ in $(0,1)$, begin at 1/1 (which isn’t in the tree, I know) and move left $a_1$ times (the first move getting you to 1/2), then move $a_2$ right, and so on. In your final step, you only move $a_n-1$ (if you wonder about the “-1”, you might think that the $n$-th convergent is the fraction you are after, and that is the $n$-th parent, not the $n$-th turning point), and you’ll arrive at $x$. On your way, all of the fractions that you hit, one before a turning point, will be the convergents to $x$.
That only gets us half-way into evaluating $?(x)$, for this rational $x$ you’ve picked. You first find its continued fraction and then, by the process above, you’ll see how to get down to $x$ from the top of your tree. That means $?(x)$ will be the value in our second tree (taking averages, the tree with powers of two in the denominator) that follows the same path. The initial $a_1$ lefts, from 1/1, will get us to a value of $1/2^{a_1}$. Then we are supposed to take $a_2$ rights. Taking 1 right will get us to $1/2^{a_1}+1/2^{a_1+1}$. Another right will then be $1/2^{a_1}+1/2^{a_1+1}+1/2^{a_1+2}$, which I’ll write $1/2^{a_1}+3/2^{a_1+2}$. Yet another right gets us to $1/2^{a_1}+7/(2^{a_1+3})$, and so on, until $1/2^{a_1}+(2^{a_2}-1)/2^{a_1+a_2}$. After that, we move left $a_3$ times. The analysis of where we end up is basically the same, you just have to remember that you are subtracting $1/2^{a_1+a_2+i}$s, instead of adding them.
Putting this all together, we obtain (remembering that in the last stage we only went $a_n-1$ moves.
$?([0;a_1,\ldots,a_n])=\dfrac{1}{2^{a_1}}+\dfrac{2^{a_2}-1}{2^{a_1+a_2}}-\dfrac{2^{a_3}-1}{2^{a_1+a_2+a_3}}+\cdots\pm \dfrac{2^{a_n-1}-1}{2^{a_1+\cdots+a_n-1}}$
Just as a warning, I may be off slightly in some of my formulas. I haven’t written down anything knowing that it was false, but I’m not convinced I obtain the correct formula at the end of the day (it looks slightly different from the formula at Wikipedia, for instance). If anybody sees an error, please let me know.
That’s probably plenty for today. Up next (I think), I want to take the derivative (even though I can’t) of $?(x)$, because Wikipedia thinks I’ll get something related to the density of the Farey sequences.
By the way, a paper I was looking at to figure out about these things was “The Minkowski Question Mark, $GL(2,\mathbb{Z})$ and the Modular Group (Expository)” by Linas Vepstas. The printed copy I have, from some time summer 2009, is incomplete, but it looks to be available online.
### The Only LFTs You’ll Ever Need
November 8, 2009
Let us consider linear fractional transformations $\frac{az+b}{cz+d}$ where the coefficients are all integers and $ad-bc=1$. Since multiplying all of the coefficients by a constant doesn’t change the transformation, we may multiply everything by -1 whenever we want (for instance, to make a chosen coefficient positive). Recall, also, that writing the transformation as a matrix, $\left(\begin{smallmatrix}a&b\\ c&d\end{smallmatrix}\right)$ is not an entirely unreasonable thing to do, as composites of transformations are correctly determined by matrix multiplication. I’ll be talking about sequence of these matrices, so let my original matrix be $\left(\begin{smallmatrix}a_0&b_0\\ c_0&d_0\end{smallmatrix}\right)$
If $c_0=0$, then the requirement $a_0d_0-b_0c_0=1$ means that $a_0=d_0=\pm 1$, and we may choose $a_0=d_0=1$. Then our transformation is simply $\left(\begin{smallmatrix}1 & b_0 \\ 0 & 1\end{smallmatrix}\right)$. Since $b_0$ is an integer, we could write this as $\left(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix}\right)^{b_0}$.
If $c_0\neq 0$, then there is a $q_0\in \mathbb{Z}$ so that $0\leq a_0-q_0c_0. I’ll leave you to check the following:
$L_0 = \begin{pmatrix}1 & q_0\\ 0 & 1\end{pmatrix}\begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix}c_0 & d_0 \\ a_0-q_0c_0 & b_0-q_0d_0\end{pmatrix}$
Writing this right-most matrix as $L_1=\left(\begin{smallmatrix}a_1 & b_1\\ c_1 & d_1\end{smallmatrix}\right)$, we are looking at another matrix with integer coefficients whose determinant is 1. So we may repeat the process, finding integers $q_k$ and LFTs $L_k$ with $L_k=\left(\begin{smallmatrix}1&q_k\\ 0&1\end{smallmatrix}\right)\left(\begin{smallmatrix}0&1\\ 1&0\end{smallmatrix}\right)L_{k+1}$. The process must stop eventually, because $c_0>c_1>c_2>\cdots$ are all non-negative integers. Eventually, one of them will be 0, and we already dealt with transformations like that in the previous paragraph.
Let $T=\left(\begin{smallmatrix}1&1\\ 0&1\end{smallmatrix}\right)$, and note that $T^b=\left(\begin{smallmatrix}1&b\\ 0&1\end{smallmatrix}\right)$, which we’ve used before. Also, let $\overline{S}=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ (you might note that $\overline{S}^2$ is the identity matrix). Then our relation above is $L_k=T^{q_k}\overline{S}L_{k+1}$. That is,
$\begin{array}{rcl}\begin{pmatrix}a&b\\ c&d\end{pmatrix}=L_0 &=& T^{q_0}\overline{S}L_1 \\ &=&T^{q_0}\overline{S}T^{q_1}\overline{S}L_2\\ &=&\cdots=T^{q_0}\overline{S}T^{q_1}\overline{S}\cdots T^{q_k}.\end{array}$
The collection of matrices that we have been working with, 2×2 matrices with integer coefficients whose determinant is 1, tend to get called the modular group and we have just shown that $T$ and $\overline{S}$ can be multiplied together to give you any element of the group. More officially, they “generate” the group. You might fuss and say that technically these matrices are really the special linear group $SL_2(\mathbb{Z})$, and note that the modular group is really the quotient of this group where you say two matrices are the same if one is -1 times the other.
One interesting thing about this process is that it is really just the process you go through to find the greatest common divisor of $a_0$ and $c_0$. The standard algorithm (Euclid’s) is to note that, if $c_0\neq 0$, there is an integer $q_0$ such that $a_0=q_0c_0+c_1$ where $0\leq c_1. The next step is to repeat the process, writing $c_0=q_1c_1+c_2$, with $0\leq c_2. Iterating, you find $q_k$ so that $c_{k-1}=q_kc_k+c_{k+1}$, and eventually a $c_n$ is 0, and the process stops.
You might be a little worried (justifiably) that this process only seems to look at $a_0$ and $c_0$, leaving $b_0$ and $d_0$ out of the picture. If you look at what we did with the matrices, our process took $\left(\begin{smallmatrix}a_0&b_0\\ c_0&d_0\end{smallmatrix}\right)$ and wrote it as $T^{q_0}\overline{S}\left(\begin{smallmatrix}a_1 & b_1\\ c_1 & d_1\end{smallmatrix}\right)$, and the $b_1$ and $d_1$ depended on $b_0$ and $d_0$. In the final step of the process, where the $c$-coefficient is 0, we ended up with a $d$-coefficient of 1, but the $b$-coefficient will be a function of the values $b_0$ and $d_0$, so those values do show up and get used, eventually. Of course, since $a_0d_0-b_0c_0=1$, knowing one of $b_0$ or $d_0$ is enough to determine the other (assuming you know $a_0$ and $c_0$). You should be thinking that the $b_i$ and $d_i$ are the extra variables that run around in Euclid’s alrgorithm when you do the “extended” version.
We’ve been talking about all of this in terms of matrices, but remember that the matrices represent linear fractional transformations. The matrix $T$ is the transformation $z+1$, and the power $T^b$ is then $z+b$. These are just horizontal translations. The matrix $\overline{S}$ is the transformation $1/z$. With our relations among the $L_i$ above, we see that we could write
$\begin{array}{rcl}\dfrac{a_0z+b_0}{c_0z+d_0} &=& T^{q_0}\overline{S}L_1 = q_0 + \frac{1}{L_1} \\ &=& q_0+\frac{1}{T^{q_1}\overline{S}L_2}=q_0+\cfrac{1}{q_1+\cfrac{1}{L_2}} \\ &=& q_0+\cfrac{1}{q_1+\cfrac{1}{\ddots+\cfrac{1}{q_k+z}}}\end{array}$
Hurray for continued fractions! Of course, I feel like I set something up backwards. Plugging in $z=0$ will give me a continued fraction for $b_0/d_0$, but I was saying that the $q_k$ come from thinking about the greatest common divisor of $a_0$ and $c_0$. Ah well.
[Update 20091203: The error I’m concerned about above stems from some loose dealings with the end of the iterative procedure. Letting $L_i=\left(\begin{smallmatrix}a_i & b_i\\ c_i & d_i\end{smallmatrix}\right)$, we will get to something like $L_n=T^{q_n}\overline{S}L_{n+1}$ where $L_{n+1}=\left(\begin{smallmatrix}1 & b_{n+1} \\ 0 & 1\end{smallmatrix}\right)$, i.e. $L_{n+1}=T^{b_{n+1}}$. So we end up saying $L_0=T^{q_0}\overline{S}T^{q_1}\overline{S}\cdots T^{q_n}\overline{S}T^{b_{n+1}}$. And that $b_{n+1}$ does depend on the initial $b_0$, presumably in a manner such that plugging in $z=0$ makes the fraction make sense.]
What I’ve said is that every 2×2 matrix with coefficients in $\mathbb{Z}$ and determinant 1 can be written in terms of $T$ and $\overline{S}$. However, you might want to use $S=\left(\begin{smallmatrix}0&-1\\ 1&0\end{smallmatrix}\right)=-1/z$ instead of $\overline{S}$. Going through essentially the same process as before, with some minus signs sprinkled in appropriately, one can determine that $T$ and $S$ can be used, instead of $T$ and $\overline{S}$, to generate any of the matrices we are considering. What is the advantage of $S$ over $\overline{S}$? One way to state the advantage, for our story, is that applying $S$ to points in the upper half plane leaves them in the upper half plane (and similaly for the lower half), whereas $\overline{S}$ flips points in upper half plane to the lower half plane. We should think of this an advantage, because all of our Ford circles lie in the upper half plane. If you go back to yesterday’s discussion, you’ll note that I used $S$ in the factorization.
So, anyway, $T$ and $S$ are the only LFTs you need. You can quickly check that $S^2=1$ and also (not quite as quickly) that $(TS)^3=1$. If you write $U=TS$, then I guess you get to say that the modular group is generated by $S$ and $U$, and can establish an isomorphism between the modular group and the free product of the cyclic group of order 2 with the cyclic group of order 3. That’s something.
### LFTs and Ford Circles
November 7, 2009
Given 4 complex numbers, $a,b,c,d$, we may consider the linear fractional transformation (LFT)
$\dfrac{az+b}{cz+d}$.
Well, 4 numbers are enough to make a matrix, $\left(\begin{smallmatrix} a & b\\c & d\end{smallmatrix}\right)$. Is there any better reason to relate the linear fractional transformation with this matrix?
Suppose you have two matrices, $\left(\begin{smallmatrix}a&b \\ c&d\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}a'&b'\\c'&d'\end{smallmatrix}\right)$. Then the product is as follows:
$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}a'&b'\\c'&d'\end{pmatrix}=\begin{pmatrix}aa'+bc' & ab'+bd' \\ ca'+dc' & cb'+dd'\end{pmatrix}$.
If you take the composite of the two linear fractional transformations, i.e.,
$\dfrac{a\cdot \frac{a'z+b'}{c'z+d'}+b}{c\cdot \frac{a'z+b'}{c'z+d'}+d}$
and then play around simplifying that expression for a few minutes, you obtain the LFT
$\dfrac{(aa'+bc')z+(ab'+bd')}{(ca'+dc')z+(cb'+dd')}$,
which is precisely the LFT corresponding to the product matrix above. So, if nothing else, writing LFTs as matrices this way won’t lead us astray when thinking about composites.
This idea is not without its confusion, for me anyway. Generally when you think about a 2×2 matrix of complex values, you are thinking about that matrix as a linear map $\mathbb{C}^2\to\mathbb{C}^2$, which is not what we are doing above. Instead, I guess we are saying that the “monoid” (group without inverses) of 2×2 matrices, $M_2(\mathbb{C})$, acts (in the technical sense) on $\mathbb{C}$ as linear fractional transformations. My guess is that there are even better ways to say what is going on.
I think it is also important to keep in mind that two different matrices may correspond to the same LFT. For example, $\left(\begin{smallmatrix}1&2\\ 0&2\end{smallmatrix}\right)$ represents the same LFT as $\left(\begin{smallmatrix} 1/2 & 1\\ 0 & 1\end{smallmatrix}\right)$. More generally, if $\lambda$ is any complex value (nonzero), then $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$ represents the same LFT as $\left(\begin{smallmatrix}\lambda a&\lambda b\\ \lambda c & \lambda d\end{smallmatrix}\right)$. I guess one can think of $M_2(\mathbb{C})$ as a $\mathbb{C}$-vector space (isomorphic to $\mathbb{C}^4$), and then think of its projective space (the quotient where two “vectors” (matrices here) are the same when they differ by a scalar (complex) multiple), which I’ll denote $P(M_2(\mathbb{C}))$. Then I think I’m saying that the action of $M_2(\mathbb{C})$ on $\mathbb{C}$ actually is an action of the quotient, $P(M_2(\mathbb{C}))$. I’m not sure if this is a useful viewpoint (or, indeed, correct).
Yesterday, when I was talking about how to picture what an LFT does to $\mathbb{C}$, I wrote down a factorization of the LFT as a composite. Our new notation gives us another way to write that factorization (recall $\alpha=-(ad-bc)/c^2$, and that we had assumed $c\neq 0$):
$\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}\alpha & a/c\\ 0&1\end{pmatrix}\begin{pmatrix}0&1\\ 1&0\end{pmatrix}\begin{pmatrix}1&d/c\\ 0&1\end{pmatrix}$.
As is frequently useful, we will assume that $ad-bc\neq 0$ (indeed, this factorization seems to require it – I think I’m missing something somewhere, anybody see it?). Notice that $ad-bc$ is the determinant of the matrix representing our LFT. We may then multiply all entries in our matrix (without changing the LFT, as discussed above) by $1/(ad-bc)$, and obtain a matrix with determinant 1. Let’s do that, making $\alpha=-1/c^2$.
Yesterday, when I was working on the factorization above, I only had something like $\epsilon$ idea where I was going. I think today I’ve got about twice that, so I want to re-write the factorization. Let me write it as
$\begin{pmatrix}1 & a/c\\ 0 & 1\end{pmatrix}\begin{pmatrix}1&0 \\ 0& c^2\end{pmatrix}\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}\begin{pmatrix}1 & d/c\\ 0 & 1\end{pmatrix}$.
So what’s the connection with Ford circles? Recall that for a reduced fraction $h/k$, the associated Ford circle is the circle centered at $(h/k,1/(2k^2))$ with radius $1/(2k^2)$. Following Rademacher (and, presumably, others), let us say that the “fraction” $1/0$ also gets a Ford “circle”, the line $y=1$ in the plane. This isn’t such a nasty thing to do, as it has the tangency properties I talked about when talking about Ford circles. Anyway, let us think about applying our transformation $\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)$, as the composite given above, and see what happens to this line $y=1$. We’ll assume that $a,b,c$, and $d$ are all integers.
The first step, $\left(\begin{smallmatrix}1&d/c\\ 0 &1\end{smallmatrix}\right)$ is the linear translation $z+d/c$. Since $d/c$ is real (since $c,d$ are integers), this translation is a horizontal shift, which leaves $y=1$ unchanged.
Next up, $\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$, which is $-1/z$. Thinking of a point on the line $y=1$, you can quickly determine that its polar coordinates are $(\csc \theta,\theta)$. The transformation $-1/z$ is the composite of: (1) inversion with respect to the unit circle (the point becomes $(\sin \theta,\theta)$), (2) reflection across the horizontal axis (giving $(\sin \theta,-\theta)$), and finally (3) multiplication by -1 (giving $(-\sin \theta,-\theta)$, since these are polar coordinates). This final point is $(\sin \theta,\pi-\theta)=(\sin \pi-\theta,\pi-\theta)$. As $\theta$ varies through $(0,\pi)$, $\pi-\theta$ also varies through this interval, and so we get the graph of the polar curve $r=\sin \theta$. If you know your polar curves, you know what this looks like…
So, the first two transformations take the line $y=1$ to the circle with center (0,1/2) and radius 1/2. The next in our composite is multiplication by $1/c^2$, which is just a scaling (since $c\in \mathbb{R}$). This scaling takes our circle to the circle with center $(0,1/(2c^2))$ and radius $1/(2c^2)$. Finally, the last transformation is another horizontal translation, leaving our circle centered at $(a/c,1/(2c^2))$. We recognize this as the Ford circle for the fraction $a/c$ (as long as that fraction is reduced).
Wasn’t that fun? If you want to think about it some more, you might convince yourself that any point above the line $y=1$ will get moved to a point inside the Farey circle resulting from this process.
Anyway, enough out of me. Hopefully tomorrow I’ll have slightly more of an idea what I’m talking about. Don’t count on it though.
### Linear Fractional Transformations
November 6, 2009
a.k.a. Möbius Transformations, are a type of function. I’ll talk about them as functions from the complex plane to itself. Such functions are given by a formula
$\dfrac{az+b}{cz+d}$
where $a,b,c,d$ are complex values. If $c$ and $d$ are both 0, this isn’t much of a function, so we’ll assume at least one isn’t 0.
I’d like to talk about what these functions do, how to have some hope of picturing them as transformations $\mathbb{C}\to\mathbb{C}$. To do this, let’s consider some easy cases first.
If $c=0$ (and so by assumption $d\neq 0$), then we may write the function $\frac{a}{d}z+\frac{b}{d}$, or simply $a'z+b'$ for some complex values $a',b'$. This is now a linear (some might say affine) transformation of the complex plane. Think about it as the composite $z\mapsto a'z\mapsto a'z+b$, where the first map multiplies by $a'$, and the second adds $b'$. Multiplying by a complex value $a'$ is the same as scaling by the real value $|a'|$ (the “norm” of $a'$, distance from $a'$ to the origin) and then rotating by the “argument” of $a'$. If you think about $a'$ as a point $(r,\theta)$ in polar coordinates, then the argument of $a'$ is $\theta$ (or so), and so multiplication by $a'$ is multiplication by the real value $r$ (which is just a stretching (or shrinking) of the complex plane away from (toward) the origin if $r>1$ (resp. $0\leq r<1$)) and then rotation by the angle $\theta$. The second transformation in the composite, “add $b'$“, just shifts the entire plane (as a “rigid transformation”) in the direction of $b'$.
So the case when $c=0$ is just a linear transformation, which aren’t too difficult to picture. Another important case is $1/z$, so the coefficients are $a=0,b=1,c=1,d=0$. To talk about what this does, let’s first talk about “inversion” with respect to a fixed circle.
Let $C$ be a circle with radius $r$, in the plane, and $z$ any point in the plane. Let $O$ denote the center of the circle and $d$ the distance from $O$ to $z$. The inversion of $z$, with respect to $C$, is the point on the line through $O$ and $z$ (in the direction of $z$ from $O$) whose distance from $O$ is $d'=r^2/d$. This means that points near the center of $C$ are sent far away, and vice versa. Points on $C$ are unchanged. Technically I guess we should say that this function isn’t defined at $O$, but people like to say it is taken to “the point at infinity” and, conversely, that inversion takes $\infty$ to $O$. These things can be made precise.
You might notice that doing inversion twice in a row gets you right back where you started. It also turns out that If $C'$ is another circle in the plane, not passing through $O$, then the inversion of all of its points is another circle. If $C'$ passes through $O$, then the inversion works out to be a line. Since doing inversion twice is the identity, inversion takes lines to circles through $O$. If you’re thinking about the comments about $\infty$ above, this makes sense because every line “goes to $\infty$“, and so the inversion of a line will go through the inversion of $\infty$, which I said should be $O$.
All of this talk about inversion was to describe the function $1/z$. This function is the composite of inversion with respect to the unit circle centered at the origin followed by a reflection across the horizontal axis (real line). Don’t believe me? The equation $d'=r^2/d$ defining the relationship between distances when doing the inversion can be re-written as $dd'=r^2$. If we’re doing inversion with respect to a unit circle, then $dd'=1$. This means that when we multiply $z$ with its inversion with respect to the unit circle, call it $z'$, the result will be a point with norm 1 (i.e., a point on the unit circle). Next up, multiplying $z$ by $z'$ produces a point whose angle from the positive real axis (which I called the argument before, the $\theta$ from polar coordinates) is the sum of the angles for $z$ and $z'$. Since we did the reflection across the horizontal axis, the argument for $z'$ is precisely the negative of the argument for $z$, meaning their sum (the argument of their product) is 0. So $zz'$ is a point on the unit circle making an angle of 0 with the positive real line, i.e., $zz'=1$. That makes $z'=1/z$, as promised.
Let’s get back to the general setup, with the function
$\dfrac{az+b}{cz+d}$
and let’s assume $c\neq 0$ (since we already handled the case $c=0$, it’s just a linear transformation). For some notational convenience, let me let $\alpha=-(ad-bc)/c^2$. Consider the following composite:
$\begin{array}{rcl} z & \xrightarrow{w\mapsto w+\frac{d}{c}} & z+\dfrac{d}{c} \\ {} & \xrightarrow{w\mapsto \frac{1}{w}} & \dfrac{c}{cz+d} \\ {} & \xrightarrow{w\mapsto \alpha w+\frac{a}{c}} & \dfrac{\alpha c}{cz+d}+\dfrac{a}{c} \end{array}$
If you check all of these steps, and then play around simplifying the final expression, then you obtain the original formula above. So we can think of any linear fractional transformation as a composite of some linear functions and an inversion, and we know how to picture all of those steps.
That’s maybe enough for today. It’s certainly enough for me for today. Before I go, I’ll leave you with a video that might be helpful, and is pretty either way.
### Ford Circles
November 5, 2009
Now that we’ve got some foundations laid for Farey sequences and rational approximations, let’s put pictures to some of those claims, via Ford circles.
For every rational $h/k$, in reduced terms, draw a circle in the plane with center $(h/k,1/(2k^2))$ and radius $1/(2k^2)$. I’m not going to try to come up with a better picture for you than the one used on the Wikipedia page for Ford circles. (I tried embedding the picture below, but that didn’t work. So I’ll leave it to you to click the link.)
What’s so great about this construction? Well, you may recognize those radii as the error term in our discussion of rational approximations. The statement that for every irrational $\omega$ there are infinitely many reduced fractions $h/k$ such that $|\omega-h/k|<1/(2k^2)$ is the same as saying that the vertical line $x=\omega$ intersects infinitely many of the Ford circles. The improved bound $1/(\sqrt{5}k^2)$ can also be found by (very carefully) analyzing vertical lines passing through Ford circles (more specifically, the gaps between them), but I think for now I’m going to let that statement be for now. If you want, the proof via Ford circles is in [Rademacher], and a proof via continued fractions is in [Hardy and Wright]. The result is known as Hurwitz’s theorem.
What else can be said about Ford circles? Well, yesterday I was talking about neighbors in Farey sequences. It turns out that $h/k$ and $H/K$ are neighbors in a Farey sequence if and only if the Ford circles at $h/k$ and at $H/K$ are tangent to each other.
To see why this is true, consider the circles at $h/k$ and $H/K$, with $h/k so $Hk-hK>0$). Since we know the location of their centers, we can find an expression for the distance between those centers. The two circles will intersect iff the distance between the centers is no bigger than the sum of the radii of the two circles. Re-arranging terms (and working with squares to avoid square roots), you might look at $d^2-(r_k+r_K)^2$, where $d$ is the distance between the centers, and $r_k$ and $r_K$ denote the (hopefully obvious) radii. The two circles will be tangent if this is 0, and won’t intersect if the difference is positive. More care needs to be used when talking about what happens if the difference is negative, but we’ll see momentarily that this doesn’t happen, so let’s forget it. So, it turns out that $d^2-(r_k+r_K)^2$ is $((Hk-hK)^2-1)/(k^2K^2)$. Since all the letters there are positive integers, we see that $d^2-(r_k+r_K)^2\geq 0$, with equality iff $Hk-hK=1$. This says, based on our work yesterday, precisely that two Ford circles intersect iff the fractions they correspond to are neighbors in a Farey sequence, and when this happens, they intersect at a single point (they are tangent to eachother).
This justifies the picture from above, then. The circles never overlap, and some of them are tangent.
It might be worth it to go back and justify why any vertical line at an irrational value slices through infinitely many of the Ford circles. To that end, let $\omega\in (0,1)$ be irrational. The two circles at $0/1$ and $1/1$ both have radius 1/2, and so certainly $x=\omega$ hits at least one Ford circle, kicking off an induction argument.
Suppose that $x=\omega$ hits the Ford circle $h/k$. For the sake of argument, let us suppose that $\omega>h/k$, so the vertical line lies to the right of the center of the circle. Not much changes if the opposite inequality is true, so let’s just stick with this one. I aim to show that $x=\omega$ hits a circle with a smaller radius, i.e. goes through a circle corresponding to a fraction with a bigger denominator than $k$. I claim that it hits one of the circles tangent to the circle at $h/k$. By the above, we know that these are the circles corresponding to neighbors of $h/k$ in some Farey sequence. Since I’m thinking about circles to the right of $h/k$, these will be successors in the Farey sequence. Yesterday we said that all of these successors are $(m,1)$-weighted mediants of $h/k$ with $H/K$ where $H/K$ is the successor of $h/k$ in $F_k$ (the first Farey sequence that $h/k$ shows up in).
Consider, then, the circles corresponding to the (1,1), (2,1), (3,1),… weighted mediants of $h/k$ with $H/K$. Let me call the circles $C_1,C_2,\ldots$, using $C$ for the circle at $h/k$. It isn’t too much work to check the necessary inequalities to show that the center of $C_1$ is further right than the right-most point on $C$. I’ve already said, even if not directly, that for each $n$, $C_n$ is tangent to $C_{n+1}$. So, if you want, you may start at the top of $C_1$, and move left along the tops of all the circles in turn, jumping from one to the next at points of tangency. How far can you go doing this? Well, if you take the limit as $m\to\infty$ of the $(m,1)$-weighted mediant of $h/k$ with $H/K$, you get $h/k$ (l’Hospital!). So, the tops of the circles form an unbroken path from the top of $C_1$ to as close to $h/k$, on the right, as we care to get. The vertical line $x=\omega$, which passes through the right side of $C$, must surely pass through this path, and therefore at least one of the $C_n$.
That’s probably just about enough with Ford circles for now. I’m not sure what I’ll be writing about tomorrow. Hopefully something else fun. I would like to note that, as I was writing this post, I was excited to find a connection to continued fractions. I knew a connection should be there (and I’ll probably find plenty more later), and found it while asking “how can we tell which of the $C_n$ does $x=\omega$ pass through?”. I remembered from reading about continued fractions, $[a_0;a_1,a_2,\ldots]$, that the “convergents”, $[a_0;a_1,\ldots,a_n]$ are rationals $p_n/q_n$ that give “best rational approximations”. Those should be the circles that $x=\omega$ passes through, or so. And so if you pass through $p_n/q_n$, the next circle you should get to is $p_{n+1}/q_{n+1}$. Well, it turns out, there is a nice relationship between successive convergents. Namely, $p_{n+1}=a_{n+1}p_n+p_{n-1}$ and $q_{n+1}=a_{n+1}q_n+q_{n-1}$. Look at that! It’s the $(a_{n+1},1)$-weighted mediant of $p_n/q_n$ with $p_{n-1}/q_{n-1}$! Fantastic. Could this story get any better? I’m glad I’ve decided to find out this month.
### Neighbors in Farey Sequences
November 5, 2009
I was reading about Ford circles today, and was planning on writing about them. But then I noticed that one of the things I wanted to say was based on some things I didn’t say about Farey sequences when I talked about them a few days ago. So I thought I’d go back and fill in that gap, and save Ford circles for tomorrow.
So what did I skip? Well, I mentioned that to get from one Farey sequence to the next, you throw in mediants whose denominators aren’t too big. Is there a way, given a fraction $h/n$ in $F_n$ to figure out what terms in $F_{n-1}$ it was the mediant of? Well, we could, of course, list the terms in $F_{n-1}$, in order, and then pick the two that surround the value $h/n$. I’m after another way.
Since $h/n$ is supposed to be reduced, $h$ and $n$ are relatively prime, and so there are integers $s$ and $t$ such that $hs-nt=1$ (this usually (in my mind) gets written with a “+”, not a “-“, but just change the sign of your $t$). Once you’ve found such a pair $(s,t)$, all of the other pairs that satisfy the same equation are of the form $(s+mn,t+mh)$, where $m$ is any integer. Among the $s+mn$, there is only one in the interval $[0,n)$. Abusing notation a bit, or assuming you picked well to start, let’s call this value $s$, and its partner $t$. I claim that $t/s$ is the term directly before $h/n$ in $F_n$.
Before talking about $t/s$, I should have checked that $s\neq 0$. If this is not the case, then $hs-nt=1$ means $n=1$, in which case we are in $F_1$ which is just 0/1 and 1/1. Not much interesting going on there. So let’s go ahead and assume $s>0$.
Now, the linear equation $hs-nt=1$ gets re-written $h/n-t/s=1/(ns)$ to show that $h/n>t/s$ (so the fractions are at least in the correct order). Messing about with inequalities some more, you can check that $0\leq t, and since $s$ and $t$ are relatively prime (they satisfy the linear equation), the fraction $t/s$ is actually in $F_n$. Let $a/b$ be the predecessor of $h/n$ in $F_n$ (we want to show $a=t,b=s$). Since it is a neighbor, we showed last time that then $hb-na=1$. But now we’re saying that all solutions of the equation $hx-ny=1$ have a fixed form. That is, $b=s+mn$ and $a=t+mh$ for some integer $m$. But if $m\neq 0$ then $b$ is not in $[0,n)$, and so $a/b$ is not in $F_n$. So $m=0$ and $a=t,b=s$, and we’re done.
Man, ok, so, after all that, we now know how to find the term before $h/n$ in $F_n$. I’ll call it $t/s$, sticking with our notation above. What about the term after $h/n$? You could play mostly the same games as above, I suppose. Or, if the successor is $T/S$, then we know that $h=t+T$ and $n=s+S$, since we made $h/n$ as a mediant. So $T=h-t$ and $S=n-s$, easy enough. [Update 20091109: You have to be careful with this. The above method for finding the successor to $h/n$ in $F_n$ only works because you are in $F_n$. If you look for successors in $h/k$ in $F_n$, where $k, this method, using the predecessor, is wrong. For example, 1/4 is the predecessor of 1/3 in $F_5$, but the fraction (1-1)/(3-4) suggested by the previous method is clearly NOT the sucessor of 1/3 in $F_5$]
So we know the neighbors of $h/n$ in $F_n$. What is the next Farey sequence when $h/n$ gets a new neighbor? Let’s stick to the next time it gets a new neighbor “on the left” (less than itself). We know that when that happens, it had to be because we made the mediant with the old neighbor, our friend $t/s$. So the next neighbor is $(t+h)/(s+n)$, which doesn’t show up until $F_{s+n}$. The newest neighbor after that will be the mediant with this new neighbor, i.e., $(t+2h)/(s+2n)$, in $F_{s+2n}$. Continuing on, we see that all of the neighbors are of the form $(t+mh)/(s+mn)$, for some positive integer $m$. I called these $(1,m)$-weighted mediants (of $t/s$ and $h/n$) yesterday.
The story goes pretty much the same on the right side. I called the neighbor on the right, in $F_n$, $T/S$. The next neighbors will be $(h+T)/(n+S)$, $(2h+T)/(2n+S)$, etc. These are the $(m,1)$-weighted mediants of $h/n$ and $T/S$.
We can put both of these into the same statement by noticing that the $(1,-m)$ mediant of $t/s$ with $h/n$ is the same as the $(m-1,1)$ mediant of $h/n$ with $T/S$. In other words, all of the neighbors of $h/n$, in all of the Farey sequences, can be written as $(1,m)$ weighted mediants of $t/s$ with $h/n$. They will have numerators of the form $a=t+mh$ and denominators of the form $b=s+mn$. Guess what? These are exactly all of the solutions to $hx-ny=1$. Go back a few paragraphs, it’s the same thing I said there.
In summary, the neighbors to $h/n$, in any Farey sequence, are in one-to-one correspondence with solutions to $hx-ny=1$.
If I knew what I were doing, I could probably have gotten to this statement a bit more quickly. But I don’t remember the last time I knew what I was doing, so there you have it.
### Farey Sequences
November 3, 2009
Ok, so if I’m going to talk about Farey sequences and the Riemann hypothesis for MaBloWriMo, then Farey sequences are probably a good place to start.
Let us say that the $n$-th Farey sequence is the sequence of fractions $h/k$ where $0\leq h\leq k\leq n$ and $h$ and $k$ are relatively prime (so that $h/k$ is in “lowest terms”). Let’s denote this sequence by $F_n$. The first few are:
$F_1: 0/1\quad 1/1$
$F_2: 0/1 \quad 1/2\quad 1/1$
$F_3: 0/1\quad 1/3\quad 1/2\quad 2/3\quad 1/1$
$F_4: 0/1\quad 1/4\quad 1/3\quad 1/2\quad 2/3\quad 3/4\quad 1/1$.
There’s sort of a nice way to get from one Farey sequence to the next: add fractions incorrectly.
The mediant of the fractions $h_1/k_1$ and $h_2/k_2$ is defined to be the fraction $(h_1+h_2)/(k_1+k_2)$. It turns out that the mediant of two fractions always lies between them, which you can check by taking differences and noticing they have the correct sign. Now, to make the sequence $F_{n+1}$, knowing the sequence $F_n$, you simply include all the existing terms, and any mediants of successive terms that have a denominator no bigger than $n+1$. So, for example, consider $F_4$ above. We consider the fractions obtained by adding to this sequence the mediants:
$0/1\quad 1/5\quad 1/4\quad 2/7\quad 1/3\quad 2/5\quad 1/2\quad 3/5\quad 2/3\quad 5/7\quad 3/4 \quad 4/5\quad 1/1$
and then we throw out any fraction whose denominator is bigger than 5, giving us
$F_5: 0/1\quad 1/5\quad 1/4 \quad 1/3\quad 2/5\quad 1/2\quad 3/5\quad 2/3\quad 3/4 \quad 4/5\quad 1/1$.
There’s a little something to worry about in this process. Well, probably a few little somethings. The one I’ve got in mind is: are the mediants always reduced? However, you might also ask: are we sure we get all of the fractions this way?
To begin to answer these, let me change the subject. Look at any of the Farey sequences listed above, and pick any two successive terms in that sequence. Call them $h_1/k_1$ and $h_2/k_2$, and let’s say that we’ve picked our indices so that $h_1/k_1 < h_2/k_2$. Now compute $h_2/k_2 - h_1/k_1$. Done? You got $1/(k_1k_2)$, didn’t you? This happens no matter what Farey sequence you are in, and no matter which successive terms you take. You always find that, with the notation above, $h_2k_1-h_1k_2=1$. This is fairly interesting.
We already knew that, for example, $h_1$ and $k_1$ were relatively prime. And for a pair $h_1,k_1$ of relatively prime integers, there always exist integers $s,t$ such that $h_1s+k_1t=1$. Conversely, if you can find such a pair of integers, then $h_1$ and $k_1$ are relatively prime. What’s perhaps a little surprising is that, in this case, the $s$ and $t$ happen to show up in the next term of the Farey sequence: $s=-k_2$ and $t=h_2$.
Armed with the observation that the difference between successive terms has a 1 in the numerator, in even just the first Farey sequence, $F_1$, we can proceed by induction on $n$ to show that if $h_1/k_1 < h_2/k_2$ are successive terms in $F_n$ then: (1) $h_2k_1-h_1k_2=1$, and also (2) the mediant of these fractions is reduced. Therefore, if the mediant has a denominator no bigger than $n+1$, it will get included in $F_{n+1}$.
Does this process of expanding one Farey sequence to the next by throwing in mediants actually give us all of the terms of the next Farey sequence? That is, if $m/n$ is a fraction in reduced form, is it the mediant of two successive fractions from $F_{n-1}$?
Let me answer that by considering what I’ll call weighted mediants (a name I just made up, it may or may not show up in your favorite reference). Let me say that the $(a,b)$-weighted mediant of fractions $h_1/k_1 < h_2/k_2$ is the fraction $(ah_1+bh_2)/(ak_1+bk_2)$. I’ll assume $a,b,h_1,h_2,k_1,k_2>0$. So, for example, the normal mediant, defined above, is just the $(1,1)$-weighted mediant. I claim that: (1) every weighted mediant lies between the two original fractions, (2) every fraction between two successive terms in a Farey sequence can be obtained as a weighted mediant.
Claim (1) is easy, just take appropriate differences and note that they have the appropriate sign, as before. For claim (2), suppose $h_1/k_1, where $h_1/k_1,h_2/k_2$ are successive terms in a Farey sequence, so that $h_2k_1-h_1k_2=1$. Our goal is to find $a,b$ so that $p=ah_1+bh_2$ and $q=ak_1+bk_2$. The assumption $h_2k_1-h_1k_2=1$ means that we can, indeed, find integers satisfying this condition, namely: $a=-k_2p+h_2q$ and $b=k_1p-h_1q$.
So, back to the question about obtaining every fraction in $F_n$ by taking mediants from $F_{n-1}$. We need to show that if $m/n<1$ and $m$ is relatively prime to $n$, then $m/n$ is the mediant of two successive fractions in $F_{n-1}$. Well, $m/n$ certainly lies between two of them, say $h_1/k_1$ and $h_2/k_2$ in keeping with our notation, and we have just shown that it may therefore be obtained as some weighted mediant (we want it to be the $(1,1)$-weighted mediant). Since the denominator of the $(a,b)$-weighted mediant is $ak_1+bk_2$, the smallest denominator is obtained only with the $(1,1)$-weighted (i.e., normal) mediant. This denominator must be at least $n$, because otherwise the fraction would show up in some Farey sequence before $F_n$. As $m/n$ is obtainable as a weighted mediant, it must be the $(1,1)$-weighted mediant, which is what we wanted to show. Let’s call it a day. | 18,917 | 62,398 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 759, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2018-26 | latest | en | 0.899055 |
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# Binomial distribution, probability density function, cumulative distribution function, mean and variance
This calculator calculates probability density function, cumulative distribution function, mean and variance of a binomial distribution for given n and p
## This page exists due to the efforts of the following people:
TimurArticle : Binomial distribution, probability density function, cumulative distribution function, mean and variance - Author, Translator ru - enCalculator : Binomial distribution - Author, Translator ru - en
In probability theory and statistics, the binomial distribution with parameters n and p is the discrete probability distribution of the number of successes in a sequence of n independent experiments, each asking a yes–no question, and each with its own boolean-valued outcome: a random variable containing single bit of information: success/yes/true/one (with probability p) or failure/no/false/zero (with probability q = 1 − p). Wikipedia
Probability density function: $f(x;p,n) = C^{n}_x (p)^x (1-p)^{n-x}$,
Cumulative distribution function: $F(x;p,n) = \sum_{i=0}^{x}{C^{n}_i (p)^{i}(1 - p)^{(n-i)}}$
where
$C^{n}_x=\frac{n!}{(n-x)!x!}$ - binomial coefficient
Mean, or expected value of a binomial distribution is equal to $\mu_x=np$, and the variance is equal to $\sigma^{2}_{x}=np(1-p)$
If the number n is rather big, then binominal distribution practically equal to the normal distribution with the expected value np and dispersion npq.
This calculator calculates probability density function, cumulative distribution function, mean and variance for given p and n.
### Binomial distribution
Digits after the decimal point: 4
Expected value
Variance
Probability density function
Cumulative distribution function | 405 | 1,820 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 5, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2019-09 | longest | en | 0.744533 |
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### "10 people and 10 hats (an old problem)"
When drawing inferences using probability theory it is crucial to understand whether events are dependent or independent. In gambling games this is the difference between a slot machine where the last spin does not influence the outcome of the next spin and blackjack where each card that is dealt changes the composition of the deck of remaining cards.
In investing, misunderstanding this concept can lead to disaster as when the beautiful, well-trimmed hedge you have constructed from seemingly uncorrelated assets turns into a garden of weeds as the correlation you either didn't see or didn't understand makes itself apparent.
From Standard Wisdom:
Problem: 10 people walk into a party and give their hats to the coat and hat check guy. When the party finishes, their hats are returned in no specific order and with no specific intent. What is the probability that no one gets their own hat back? (This is an old problem, and a standard one in combinatorics and probability.)
First, an approximate solution: probability that a person gets his hat back is 1/10, so the probability that the person does not get his hat back is 9/10. So the probability that no one gets their hat back is 0.9^10 = 0.34868. Why is this solution imperfect? Well, because the events aren’t really independent. If one person gets their hat back, it increases the chance that the remaining people will get their hats back (easy to see that with two people 😉 ). We can’t really multiple the event probabilities, if the events are not independent.
The solution: The correct solution comes from combinatorics. Total number of ways in which hats can be returned is 10! (10 factorial = 10 x 9 x 8 x … 1). If we can count the number of ways in which no guest receives their hat back, then we can deduce the probability by dividing that number by 10!. To count that, we can use the principle of inclusion and exclusion. Say A1 is the set of cases in which the first guest receives his own hat back. Then, we can enumerate:
The cases in which some guest(s) receive their own hats back are: A1 U A2 U … U A10...
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posted by carlos
A rectangular prism has dimensions 5 by 6 by x. If the total surface area of the prism is 157.9 square units, what is the value of x?
What is the volume of the above prism?
1. MsPi_3.14159265
Surface Area = 2wL +2LH + 2Hw
and you have a length of 5, a width of 6, so your height is X
157.9 = (2)(6)(5) + 2(5)(x) + 2(x)(6)
Simplify, gather like terms and solve for x
then to find the volume sub your values into Volume = LWH
2. carlos
thanks
3. MsPi_3.14159265
You are very welcome!
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( 5x - 4 ) 6 9 ) ( q + 5 ) ( b - 4 ) +! | 5,997 | 25,902 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-39 | latest | en | 0.811199 |
https://elmath.org/index.php?id=display_topic&topic=2 | 1,701,510,643,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100381.14/warc/CC-MAIN-20231202073445-20231202103445-00447.warc.gz | 272,032,412 | 4,348 | # Wieferich primes and Mersenne primes
For elMath.org: Miroslav Kures Institute of Mathematics, Brno University of Technology Contact: Creation date: 2007-12-28
## 1 Introduction
Let n ∈ N; Mersenne numbers are defined by Mn=2n−1. A binary expression of Mersenne numbers consists of units only.
Proposition 1 If Mn is a prime number, then n is a prime number. (The reverse implication need not hold.)
Let n=kl be a composite number. Then 2n−1=2kl−1=(2l−1)(2l(k−1)+2l(k−2)+...+2l+1), i.e. 2n−1 is a composite number, too. On the other hand, 211−1=2047=23·89.
Thus, we monitor especially Mersenne numbers Mp, where p is a prime number. If such Mp is also a prime, we call it the Mersenne prime. It is not known, how many Mersenne primes exist not even if it is a finite number. Up to now, 44 Mersenne primes are known, the latest one is M32582657, it was discovered in 2006 and it represents a greatest known prime at all.
## 2 Perfect numbers
A number n ∈ N is called a perfect number if it is equal to the sum of all its divisors, excluding n itself. (For example, 6 and 28 are perfect numbers.)
Proposition 2 n is an even perfect number if and only if it has form n=2p−1(2p−1) and 2p−1=Mp is a (Mersenne) prime. (It is not known whether or not there exists an odd perfect number.)
## 3 Are Mersenne numbers square free?
We call a Mersenne number Mn divisible by square, if there exists a prime q such that q2 divides Mp. If there is no such a prime, we call Mn square free. If n is a composite number, then Mn can be divisible by square, e.g. for n=6 we have M6=63=32·7. It is remarkable that M1092 is divisible by 10932 and M3510 is divisible by 35112, see [1]. (1093 and 3511 are Wieferich primes satisfying 2q−1 ≡ 1 ( mod q2).)
If p is a prime, then the question if Mp is square free is open. However, the following result is known.
Proposition 3 Let p,q be primes. If q2 divides Mp, then q is a Wieferich prime.
Evidently, p and q must be odd. Let q2 divides Mp, i.e.
2p ≡ 1 ( mod q2),
it follows q also divides Mp, i.e.
2p ≡ 1 ( mod q).
It follows from this equation, that p must be the order of 2 (in Fq), because p is a prime. From the Little Fermat Theorem follows that
2q−1 ≡ 1 ( mod q)
and it means that p divides q−1. However, q−1 is even and that is why q−1=2kp for some k ∈ N. Hence
2p = 2[(q−1)/2k] ≡ 1 ( mod q2);
after the exponentiation to the 2k-th power
2q−1 ≡ 1 ( mod q2);
it means q is a Wieferich prime. Nevertheless, it was proved that squares of Wieferich primes 1093 and 3511 never divide Mp. Therefore aspirants for it are only squares of possible new Wieferich primes.
## 4 Wieferich primes and ECC
The interesting application in elliptic curve cryptography is indicated by the following theorem from [2].
Proposition 4 For a Mersenne prime q=Mp=2p−1, binomials xp+2s and xp−2s, where s ≠ 0 ( mod p), is irreducible in Fq[x] if p is not a Wieferich prime.
## References
[1]
Guy, R. K., The primes 1093 and 3511, The Mathematics Student 35 (1967), 204-206
[2]
Baktir, S., Sunar, B., Frequency domain finite field arithmetic for elliptic curve cryptography, preprint
Author: Miroslav Kures | 990 | 3,165 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2023-50 | latest | en | 0.866982 |
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VIEWS: 168 PAGES: 7
• pg 1
``` Chapter 4 Hewitt (p. 43)
Newton’s First Law of Motion – Inertia
Mrs. Corey Conceptual Physics
What causes motion?
4.1 Aristotle on Motion
Describe Aristotle’s Idea (______ Century BC)
Two kinds of motion:
*Natural motion –
*Violent Motion –
What was commonly thought (for almost 2000 years) to cause an object to “move against nature”?
What was thought to be the natural state of objects?
Because the Earth is so large, most “thinkers” before the 16th century believed what about the Earth?
4.2 Copernicus and the Moving Earth
Describe Nicolaus Copernicus’ theory of the moving Earth.
4.3 Galileo on Motion
Did Galileo support Copernicus’ theory?
Define the following terms:
Force –
Friction –
Rough surfaces - more? less? friction than smooth surfaces.
If no friction, a moving object would move for how long?
If there is friction, what keeps an object moving?
Explain Galileo’s experiments with inclined planes.
Define Inertia:
Galileo’s ideas disagreed? agreed? with Aristotle’s ideas.
Aristotle’s idea – objects seek a state of __________.
Define gravity:
Question 4.3 A ball is rolled across the top of a pool table and slowly rolls to a stop. How would Aristotle interpret this
behavior? How would Galileo interpret it?
4.4 Newton’s Law of Inertia
Isaac Newton (b. 1642) developed famous laws of motion, which replaced Aristotle’s 2000 year-old ideas.
***Newton’s First Law (Law of Inertia)*** p. 46
Every object continues in its state of ____________, or of ___________ in a straight line at constant speed, unless
it is compelled to change that state by ____________exerted on it.
4.4 Question
1. If suddenly the force of gravity of the sun stopped acting on the planets, in what kind of path would the planets move?
4.5 Mass – A Measure of Inertia
Why do some things have more inertia than others?
Define mass –
Question 4.5 What is a measure of the inertia of an object?
Mass is Not Volume
Volume – Units -
Mass – Units –
Weight - Units-
Question 4.5a Does a 2-kilogram iron block have twice as much inertia as a 1-kilogram block of iron?
Twice as much mass?
Twice as much volume?
Twice as much weight (when weighed at the same location)?
Question 4.5b Does a 2-kilogram bunch of bananas have twice as much inertia as a 1-kilogram loaf of bread?
Twice as much mass?
Twice as much volume?
Twice as much weight (when weighed in the same location)?
1 kilogram weighs __________ Newtons Units:
SI (metric system)
Unit of mass – kilogram (Kg)
Unit of Force – Newton (N) On earth, 1 Kg = 2.2 lb
Question 4.5 On earth, a 1-kg bag of nails weighs 9.8 N. On the moon, does a 1-Kg bag of nails weigh 9.8N?
Question 4.5a Find the mass of a 5 pound bag of sugar on earth.
Question 4.5b How many Newtons is a 5 pound bag of sugar?
4.6 Net Force
In the absence of a net force, objects in motion do what?
Define net force:
4.7 Equilibrium – When Net Force Equals Zero
What forces act on a book when it is on a table?
Define support force:
Normal force:
Equilibrium:
Question 4.7 When you step on a bathroom scale, the downward force supplied by your feet and the upward force
supplied by the floor compress a calibrated spring. The compression of the spring gives your weight. In effect, the scale
measures the floor’s support force. What will each scale read if you stand on two scales with your weight divided equally
between them? What happens if you stand with more of your weight on one foot than the other?
We will not cover in this course.
4.9 The Moving Earth Again
If earth moves, why can you drop straight down from a tree limb and land directly below? Earth’s speed around the sun is
~ 30 km/s
Key Terms – Know these. You defined them in your note outlines.
Review Questions (p. 56 & 57) #1-15
1. What was the distinction that Aristotle made between natural motion and violent motion? (4.1)
2. Why was Copernicus reluctant to publish his ideas? (4.2)
3. What is the effect of friction on a moving object? How is an object able to maintain a constant speed when friction
acts upon it? (4.3)
4. The speed of a ball increases as it rolls down an incline, and the speed decreases as the ball rolls up an incline. What
happens to the speed on a smooth horizontal surface? (4.3)
5. Galileo found that a ball rolling down one incline will pick up enough speed to roll up another. How high will it roll
compared to its initial height? (4.3)
6. Does the law of inertia pertain to moving objects, objects at rest, or both? List 2 examples of each. (4.4)
7. The law of inertia states that no force is required to maintain motion. Why, then, do you have to keep pedaling your
bicycle to maintain motion? (4.4)
8. If you were in a spaceship and fired a cannonball into frictionless space, how much force would have to be exerted on
the ball to keep it going? (4.4)
9. Does a 2-Kg rock have twice the mass of a 1-Kg rock? Twice the inertia? Twice the weight (when weighed in the
same location)? (4.5)
10. Does a liter of molten lead have the same volume as a liter of apple juice? Does it have the same mass? (4.5)
11. Why do physics types say that mass is more fundamental than weight? (4.5)
12. An elephant and a mouse would both have the same weight – zero-in gravitation-free space. If they were moving
toward you with the same speed, would they bump into you with the same effect? Explain. (4.5)
13. What is the weight of 2 Kg of yogurt? (4.5)
14. What is the net force or, equivalently, the resultant force acting on an object in equilibrium? (4.6)
15. Forces of 10N and 15N in the same direction act on an object. What is the net force on the object? (4.6)
16. If forces of 10 N and 15N act in opposite directions on an object, what is the net force? (4.6)
17. How does the tension in your arms compare when you let yourself dangle motionless by both arms and by one arm?
(4.7)
19. If you hold a coin above your head while in a bus that is not moving, the coin will land at your feet when you drop it.
Where will it land if the bus is moving in a straight line at constant speed? Explain. (4.9)
20. In the cabin of a jetliner that cruises at 600 km/h, a pillow drops from an overhead rack into your lap below. Since
the jet is moving so fast, why doesn’t the pillow slam into the rear of the compartment when it drops? What is the
horizontal speed of the pillow relative to the ground? Relative to you inside the jet? (4.9)
Plug and Chug (p.57)
21. If a woman has a mass of 50 kg, calculate her weight in Newtons.
22. Calculate in Newtons the weight of a 2000-kg elephant.
23. Calculate in Newtons the weight of a 2.5-kg melon. What is its weight in pounds?
24. An apple weighs about 1 N. What is its mass in kilograms? What is its weight in pounds?
25. Susie Small finds she weighs 300 N. Calculate her mass.
Think and Explain (p.57-58)
26. Many automobile passengers have suffered neck injuries when struck by cars from behind. How does Newton’s law
of inertia apply here? How do headrests help to guard against this type of injury?
27. Suppose you place a ball in the middle of a wagon, and then accelerate the wagon forward. Describe the motion of
the ball relative to a) the ground and b) the wagon.
28. When a junked car is crushed into a compact cube, does its mass change? Its volume? Its weight?
29. If an elephant were chasing you, its enormous mass would be most threatening. But if you zigzagged, its mass would
30. When you compress a sponge, which quantity changes: mass, inertia, volume, or weight?
31. a. A massive ball is suspended by a string from above, and slowly pulled by a string from below. Is the string
tension greater in the upper or the lower string? Which string is more likely to break? Which property – mass or weight –
is important here?
b. If the string is instead snapped downward, which string is more likely to break? Which property – mass or weight – is
important this time?
32. If the head of a hammer is loose, and you wish to tighten it by banging it against the top of a work bench, why is it
best to hold it with the handle down rather than with the head down? Explain in terms of inertia.
33. The little girl in the figure (p.58) hangs at rest from the ends of a rope. How does the reading on the scale compare
with her weight?
35. As the earth rotates about its axis, it takes three hours for the United States to pass beneath a point above the earth that
is stationary relative to the sun. What is wrong with this scheme: To travel from Washington, D.C. to San Francisco
using very little fuel, simply ascend in a helicopter high over Washington, D.C. and wait three hours until San Francisco
passes below?
Think and Solve (p.58)
38. A medium-size American automobile has a weight of about 3000 pounds. What is its mass in kilograms?
39. If a woman weighs 500 N on Earth, what would she weigh on Jupiter, where the acceleration of gravity is 26 m/s2?
```
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posted by .
m<AOB=4x-2,m<BOC=5x+10,m<COD=2x+14
How do you solve the equation? And how do you find the measurement of each of the 3 given angles??
Similar Questions
1. geometry
If m<AOC=90 and m<AOB=20. Find m<BOC.
2. Geometry
how do you find x for: <AOB=x+3, <AOC=2x+11, <BOC=4x-7
3. Geometry
m<AOB=4x-2,m<BOC=5x+10,m<COD=2x+14 How do you solve the equation?
4. geometry
find the value value of X m<AOB=4x-2, m<BOC=5x+10 m<COD=3x-8
5. Geometry
in each of the following relationships among marked angles are given below the figure. find the measures of the marked angles a. m(<DOC)=3?
6. Geometry
Solve for x. Find the angle measures to check your work. m<AOB = 4x - 2 m<BOC = 5x + 10 m<COD = 2x + 14 How do I solve for x?
7. Geometry
If angle AOB = 28, Angle BOC = 3X-2, and Angle AOD = 6x... and if Angle AOD = a total of 110 degrees. Then what is the measurement of Angle BOC and Angle AOD
8. math
In each of the following, relationships among marked angles are given below the figure. Find the measures of the marked angles. b. m(∠AOB ) is 30° less than 2m (∠BOC ) c. m(∠AOB ) – m(∠BOC ) = 50
9. st pats
if the diagram below, AOB=p, If BOC is two times AOB, COD is four times, AOB and DOA is five times AOB, Find the values of All four angles
10. geometry
If m AOB = m COD = 90 ,find m COB + m AOD.
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Anonymous asked in Social ScienceEconomics · 3 years ago
# Need help finding predicted exchange rate?
Price of big mac in Indonesia = 14,600 rupiah
Actual exchange rate = 9,541 rupiah / US\$
Price of big mac in USA = \$3.06
Predicted exchange rate = ? rupiah / US\$
The formula that i got from the textbook is:
Real exchange rate = Nominal exchange rate * Domestic price / Foreign Price
I am kind of confused. Is Nominal Exchange rate = Predicted exchange rate?
Relevance
• Anonymous
3 years ago
In the terminolgy given in the question, predicted exchange rate is for the real exchange rate. In the question the actual exchange rate has been given This is also called the nominal exchange rate.
So, your predicted exchange rate is the Real exchange rate
= Nominal exchange rate * Domestic price / Foreign Price
=9,541 * (14,600 / 3) =
you can calculate now.
But I suspect thereis some problem with your textbook formula. The purchasing power parity ratio is \$3= rupiah 14600, or \$1 is equivalent to rupia 4866.67.
Thus , Domestic price / Foreign Price = 4866.67 Multiplying this with the nominal exchange rate of 9,541 is a huge sum which cannot be said to be the real exchange rate!!!
I think you check the text book again. The formula should have been (though strictly still not accurate):
Real exchange rate = Nominal exchange rate *Foreign price / Domestic Price.
With my formula above, the real exchange rate =
9,541 * (3 /14,600) = close to 1.9
This means that the real exchange rate of rupaih is 1.9 times the actual/ noimal exchange rate. In other words, the real exchange value of rupiah would be 9,541/ 2 or about 5026 rupiah per dollar.
It is easy to see why it is so. You can buy a big mac abroad at \$3 which is equivalent to 3* 9541= 28,623 rupaih as against big mac oprice of 14,600 rupiah at home. So, the real value of rupiah is 28,623/ 14,600 = 1.9 times higher than the nominal exchange rate. So, you should be giving 1.9 times fewer rupaih per dollar than the nominal/ ctual/ official/ market exchange rate. The real worth of rupiah is 9,541/ 2 or about 5026 rupiah per dollar.
See notes below:
1. Nominal and real exchange rates
The nominal exchange rate e is the price in domestic currency of one unit of a foreign currency.
The real exchange rate (RER) is defined as
RER = e * (P*/P), where P is the domestic price level and P * the foreign price level. P and P * must have the same arbitrary value in some chosen base year. Hence in the base year, RER = e.
The RER is only a theoretical ideal. In practice, there are many foreign currencies and price level values to take into consideration. Correspondingly, the model calculations become increasingly more complex. Furthermore, the model is based on purchasing power parity (PPP), which implies a constant RER. The empirical determination of a constant RER value could never be realised, due to limitations on data collection. PPP would imply that the RER is the rate at which an organization can trade goods and services of one economy (e.g. country) for those of another. For example, if the price of a good increases 10% in the UK, and the Japanese currency simultaneously appreciates 10% against the UK currency, then the price of the good remains constant for someone in Japan. The people in the UK, however, would still have to deal with the 10% increase in domestic prices. It is also worth mentioning that government-enacted tariffs can affect the actual rate of exchange, helping to reduce price pressures. PPP appears to hold only in the long term (3–5 years) when prices eventually correct towards parity.
More recent approaches in modelling the RER employ a set of macroeconomic variables, such as relative productivity and the real interest rate differential.
Nominal exchange rate = (Real exchange rate 1)* (expected inflation 1) -1.
Next, Start with the domestic (U.S.) and foreign (German) prices of a good, say a gallon of gas:
P = price in \$ of domestic good (say, one gallon of gas cost \$1.20)
Pf = price in units of foreign currency of the same good in a foreign country (say, a gallon of gas in Germany cost DM 2.0)
Is gas more expensive in the U.S. or in Germany? Clearly, the answer depends on the exchange rate between \$ and DM.
The price in dollars (P\$f) of a unit of the foreign good (a gallon of gas in Germany) is equal to its price in the foreign currency (Pf = DM 2) divided by the exchange rate of the dollar relative to the foreign currency e (DM/\$) = 1.5
P\$f = Pf / e = 2 / 1.5 = 1.33
The relative price of the domestic good to the foreign good (price of gas in the U.S. to the price of gas in Germany expressed in \$) is:
P / (Pf / e) = 1.20 / 1.33 = 0.90
where e is the (spot) exchange rate. Gasoline in the U.S. is about 10% cheaper than it is in Germany. We can call this the real exchange rate for gasoline. If all goods were the same in the U.S. and Germany and there were no barries or costs of trade, then we would expect the price of goods to be the same when expressed in the same currency. That is, if the price of gasoline in the U.S. rose to \$1.33 while the nominal exchange rate and the price of gasoline remained unchanged, then the real exchange rate for gasoline would be 1.0. In fact, if the real exchange rate were less than one and trade could take place costlessly, buyers in Germany would only buy in the cheap country (US), driving up prices there until foreign and domestic prices were equal. If Germans could fill up their tanks in the U.S., the prices of gas, expressed in a common currency, should be the same, leading the real exchange rate of gasoline to stay around one. However, we do know that it is hard to drive across the ocean to fill one's tank of gasoline. Remember though, that if there were no tunnel tolls, we might be tempted to drive to New Jersey where gasoline is cheaper than in New York.
Often we use price indexes, like CPI or GDP deflators, representing baskets of goods to compare exchange rate adjusted prices in different countries.. In this case the ratio of domestic prices to foreign prices in domestic currency is referred to as the real exchange rate. A real exchange rate calculated with price indexes does not tell us anything about difference in the absolute price levels in the two countries. If both price indexes have a vaule of 100 in some base year, the real exchange rate is equal to the nominal exchange rate in that year. However, movements in the real exchange rate from year to year tell us about changes in the purchasing power of one currency compared to the other.
Real Exchange rate index = P / (Pf / e) = eP / Pf
The theory of price equalization around the world can be applied to the baskets of goods underlying aggregate price indexes. We call this purchasing power parity, (or PPP) since the purchasing power of a dollar is predicted to be the same in both countries. The PPP condition is:
P = Pf /e (= P\$f )
In the gasoline example we did not satisfy the PPP condition:
P =1.20 < Pf /e =1.33 (divergence from PPP)
So how, can we reach a PPP equilibrium when the relative price differs from unity ? There are three alternative ways the equilibrium can be restored:
1. German prices could fall from DM 2 to DM 1.8 so that:
P = 1.20 = Pf /e = 1.8 / 1.5
2. US prices may go up from \$1.20 to \$ 1.33 so that:
P = 1.33 = Pf /e = 2.0 / 1.5
3. The DM/\$ exchange rate could appreciate from e = 1.5 to e = 1.67 so that:
P = 1.20 = Pf /e = 2.00 / 1.67
In practice, all of the three effects will be at work in reality.
What are the forces that might lead to convergence towards PPP?
What is the evidence on the PPP ? If the PPP holds, the real exchange rate should be equal to one and constant over time.
If the PPP holds:
e P / Pf = P / P = 1
However, we find, when we compute real exchange rate using consumer price indexes, that it varies a lot: prices of gasoline in particular, and goods in general, are often much different in Germany and the US, and between any two other countries, as well.
At least in the short-run, PPP is a poor description of reality. | 1,999 | 8,119 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2020-05 | longest | en | 0.90813 |
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MAX octagonal area (Posted on 2012-11-28)
The octagon ABCDEFGH is inscribed in a circle, with the vertices around the circumference in the given order. Given that the polygon ACEG is a square of area 5 and the polygon BDFH is a rectangle of area 4, find the maximum possible area of the octagon.
No Solution Yet Submitted by Danish Ahmed Khan No Rating
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Different approach, same answer Comment 2 of 2 |
Since ACEG is a square with area 5, diameter AE=sqrt(10).
Since BDFH is a rectangle and BF=sqrt(10) set DF=4/BD, then use Pyth to get an easily soluble double quadratic and find BD=2sqrt(2) and DF=sqrt(2).
On the assumption of symmetry (BD parallel to AE) we're set.
The area of the octagon is twice the area of the pentagon ABCDE, which is comprised of trapezoid ABDE and triangle BCD.
ABDE has parallel sides AE=sqrt(10) and BD=2sqrt(2) and height DF/2 = sqrt(2)/2.
BCD has base BD and altitude = sqrt(10)/2 - sqrt(2)/2.
Do the arithmetic and find area octagon = 3*sqrt(5)
Posted by xdog on 2012-11-28 22:01:43
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A person is on the roof of a building, 46.0 m above the ground. His teacher, who is 1.80 meters tall, is walking toward the building at a constant speed of 1.20 m/s. If the person wishes to drop the egg on his teacher's head, where should the teacher be when the person releases the egg?
So I know that I need to set two equations equal to each other to get the position the teacher needs to be for the egg to drop on him. So should I use $x = v_{y}_{0}t + \frac{1}{2} a_{x}t^{2}$ for the egg, and $x = x_{0} + v_{x}_{0}t + \frac{1}{2} a_{x}t^{2}$?
Thanks
Related Introductory Physics Homework Help News on Phys.org
Fermat
Homework Helper
How far does the egg have to fall before impacting on the head of the 1.8 m tall teacher ?
How long will it take for the egg to fall this distance ?
How far can teacher walk in this time ?
1. $46 - 1.8 = 44.2 m$
2. $44.2 = \frac{1}{2} (9.8) t^{2}, t = 3$
3. $x = 0 + 1.20(3) = 3.6 m$ | 308 | 948 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2020-05 | longest | en | 0.910121 |
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posted by .
If a pastry chef has 13 lbs of apple slices to make 8 apple pies if slices are distributed equally what will be the wright of the apple slices used to make eavh pie
• math -
13 / 8 = ?
• math -
2ounces
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Density and specific gravity metrics in biomass research 3 bark are harder to find, but miles and smith 2009 have compiled sg values for 156 north american tree species. This physics video tutorial provides a basic introduction into specific gravity and density. When released, does the ball sink to the bottom or float to the surface. Specific gravity is an expression of density in relation to the density of a standard or reference usually water. Jul 27, 2015 step by step density practice problems to help you pass chemistry duration. Urine specific gravity usg measures the concentration of particles in urine and the density of urine compared with the density of water. The buoyant force is also equal to density of the fluid volume displaced gravity. Calculate the bulk density of a 400 cm3 soil sample that weighs 575 g oven dry weight. Most important the specific gravity is called relative. Porosity and bulk density bulk density is the weight per unit volume of a soil sample.
Once you know density, you know mass per unit volume, and from that you can determine the weight. Practice questions you make a cake that has a mass of 300 grams and fits. Specific gravity, ratio of the density of a substance to that of a standard substance. Density is defined as the ratio of mass grams to a unit volume cm3 of matter. Density and specific gravity confusion often results over the meaning of the two terms, density and specific gravity. Two objects m1 and m2 each with a mass of 6 kg and 9 kg separated by a distance of 5. The specific gravity of an object is the ratio between the density of an object to a reference liquid. Chem 2990 density and specific gravity calculations dr. Specific gravity practice problem student doctor network. The usual standard of comparison for solids and liquids is water at 4 c 39. Specific gravity ratio of the density mass of a unit volume of a substance to the density mass of the same unit volume of a reference substance. Calculating specific gravity of rocks and minerals problem 8. Density and specific gravity metrics in biomass research. Converting from volume to weight is accomplished simply by taking the known volume ft3 of the ingredient and multiplying by the specific gravity of the ingredient and again multiplying by the density of water 62.
Specific gravity for liquids is nearly always measured with respect to water at its densest at 4 c or 39. If the specific gravity ratio for a given substance is less than 1, that means the material will float in the reference substance. The specific gravity of an aqueous salt solution is determined by filling a pycnometer with distilled water. These problems are meant to be easy in the beginning and then gradually become more challenging. Once it is in ethyl alcohol the object has an apparent weight of 8g. If one side of the cube measures 4 cm, what is the density of the aluminum. It contains practice problems on calculating the mass of air a rectangular room and how to calculate the weight of a balloon filled with air. Experiment 4 specific gravity determination purpose. So we measure the density of different polymers such as pvc,hdpe and ps by using displacement method astm d 792.
Parabolic motion, work and kinetic energy, linear momentum, linear and angular motion problems and solutions. Specific gravity weight of substance weight of an equal volume of water example. Specific gravity and density of mixtures fluids physics problems. This, however, is just the reciprocal of density, and actually represents the same thing. Specific gravity is the ratio of the mass of unit volume of soil at a stated temperature to the mass of the same volume of gasfree distilled water at a stated temperature. There is three times as much mass of fluid 2 as there is of fluid 1. You have a rock with a volume of 15cm3 and a mass of 45 g. A pressure vessel of volume 100 liters is filled with 1 kg of liquid water and 80 liters of steam with a specific volume of 0. Edward walker density, the ratio of massvoume, has many applications in the chemical industry. Ap physics 2 help fluids fluid statics density and specific gravity. What is the specific gravity of a substance that weights 75 g and occupies a volume of 150 ml. The specific gravity of a substance is the ratio of its density to a specified reference substance.
Density, specific weight,specific gravity, viscosity. This lets you solve for volume displaced, and then calculate the density of the object. You can also calculate the specific gravity of a material, such as gasoline, if you know its density. Step by step density practice problems to help you pass chemistry duration. Specific gravity weight of substance weight of an equal volume of water. It explains how to calculate the specific gravity of a substance and how to calculate the density of a. Determining density of solids equations mettler toledo balance manual. Calculating specific gravity of rocks and minerals. Specific gravity is a measure of density relative to the density of a reference substance.
This articles is all about properties of fluid mass density, specific weight, specific volume, specific gravity, pdf. Using physics, you can show how mass and volume are related to density. Density and specific gravity ap physics 2 varsity tutors. Both density and specific gravity describe mass and may be used to compare different substances. When it was placed in a graduated cylinder containing 20.
Free practice questions for ap physics 2 density and specific gravity. Density practice problem worksheet 1 a block of aluminum occupies a volume of 15. Specific gravity relative density, specific gravity for gases, specific weight, calculation examples. Calculation of specific gravity specific gravity ratio of the density mass of a unit volume of a substance to the density mass of the same unit volume of a reference substance. Then, it is refilled with an identical volume of salt water. D 792 07 standard test methods for density and speci. Calculate the bulk density of a 400 cm3 soil sample that weighs 600 g and that is 10% moisture. Specific weight or weight density of a fluid is the ratio between the weight of a. In the imperial measurement system, you have to convert the mass in slugs to weight in pounds. Why do we need a term that already defined a property for a substance in your case,density and now were comparing it to the density of another substance, and created a whole new term for it. Properties of fluidmass density, specific weight, volume.
Density specific gravity an overview sciencedirect. Simpson, research forest products technologist forest products laboratory, madison, wisconsin introduction knowledge about the density of wood is useful for estimating shipping weights. Jan 11, 2018 specific gravity is a concept that shows up in a variety of industrial applications, particularly as it relates to fluid dynamics. If a material has a specific gravity less than 1, it will float on water. This lab is performed to determine the specific gravity of soil by using a pycnometer. Usually, our reference liquid is water, which has a density of 1 gml or 1 gcm3. Nov 11, 2017 this physics video tutorial provides a basic introduction into specific gravity and density. The density plays an important role to commercialize the polymer in the market because plastic sold w.
Gases are commonly compared with dry air, which has a. The density of wood depends on specific gravity and moisture content. This standard is issued under the fixed designation c 127. Since the density of water at 4oc is practically 1. Density and specific gravity calculations weber state university. Again, the specific gravity is the density of the substance divided by the density of water, so this shows that the specific gravity does not change when measurements are made in different units, so long as the density of the object and the density of water are in the same units. Density and specific gravity in physics problems dummies. Chem 2990 density and specific gravity calculations. For example, if youve ever taken your car in for service and the mechanic showed you how small plastic balls floated in your transmission fluid, youve seen specific gravity in action. Calculation of specific gravity ptcb practice test questions. An 12g object is dropped into ethyl alcohol specific gravity. Volume of water is 35 cm 3 and mass of water is 60 gram, what is the density of the water. Density and specific gravity practice problems serc. Specific gravity formula definition, equations, examples.
Sometimes specific volume will be used to related an objects mass to its volume. Specific volume and density problem physics forums. Force of gravity and gravitational field problems and solutions. Rock and mineral density rock and mineral specific gravity you can download the questions acrobat pdf 25kb jul24 09 if you. Sample problem density l a volume of 1m3 is occupied by two fluids. Density and specific gravity midstate technical college. Buoyancy problem set 1 a stone weighs 105 lb in air.
By the end of this practice, you should be able to. Also, the density of the object determines this factor. The forces on a particle in a liquid are buoyant force and its weight mg. It contains many examples involving calculations with volume, mass, and density. The specific gravity of an aqueous salt solution is determined by filling a pycnometer with distilled.
However, if the specific gravity is expressed at different temperatures, it will no longer be equal to the density. Calculate density and specific gravity using the english and metric systems. The subscript x refers to the material in question. Density specific gravity specific gravity is defined by astm d4439 from volume 04. Specific gravity and density of mixtures fluids physics. Walker consider the following data and perform the indicated calculations. Density is a physical characteristic, or property, of matter. Density versus specific gravity the density has various units of measure, while specific gravity has no dimension and is therefore a constant value for each substance when measured under controlled conditions.
You have a different rock with a volume of 30cm3 and a mass of 60g. The specific gravity is a ratio between the mass of a given volume and the mass of an equal volume of water at 4oc. Specific gravity is the density of a substance divided by the density of water. We can start with newtons 2nd law for this problem. Specific gravity, moisture content, and density relationship for wood william t. Specific gravity is the density of a material divided by the density of water at. Density specific gravity an overview sciencedirect topics. The last characteristic of a fluid that is dependent on density is the specific weight of a fluid. The density is directly related to the mass of the object unit.
As the density of liquid hydrocarbons changes about 0. Thus, whereas the density of water may be variously expressed as 1 gml, gl, or 62. Mass of an metal is 120 gram and volume of an metal is 60 cm 3. On this page i put together a collection of density problems to help you better understand calculations involving density. The reference material could be anything, but the most common reference is pure water. Density problems and solutions solved problems in basic.
Density can be used to quantify the dissolved solids in liquids. Calculate the density, specific volume, and specific gravity of a solid that occupies 3 ft 3 and weighs 9 lb. Answer to calculate the density, specific volume, and specific gravity of a solid that occupies 3 ft 3 and weighs 9 lb. Relative density, or specific gravity, is the ratio of the density mass of a unit volume of a substance to the density of a given reference material.
Density means concentration of matter, measured as mass per unit volume hodgman, 1950, or, where p is used to represent density, m represents mass, and v represents volume. The relationship between mass and volume is an important aspect of the characterization and specification of both solids. Calculate the density and specific gravity of the object. From this information, calculate the density of mercury. Hey guys im still a little shaky on fluid problems, so heres one that i hope some one can clarify. Density is mass divided by volume, whereas specific gravity is the density of a material being studied, divided by the density of water under the experimental conditions close to but not exactly 1. Determine its specific gravity and indicate whether it would float or sink in water. Unless otherwise stated, answers should be in gml or the equivalent gcm 3.
138 876 927 713 343 100 965 860 491 603 747 424 468 1473 244 470 324 1026 320 59 32 738 874 121 1076 1358 57 58 | 2,618 | 12,801 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2024-22 | latest | en | 0.910574 |
https://www.printablemultiplication.com/multiplication-flash-cards-at-dollar-tree/ | 1,638,563,687,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362918.89/warc/CC-MAIN-20211203182358-20211203212358-00546.warc.gz | 1,068,977,143 | 10,750 | # Multiplication Flash Cards At Dollar Tree
Studying multiplication after counting, addition, and subtraction is ideal. Young children find out arithmetic by way of a all-natural progression. This advancement of discovering arithmetic is truly the following: counting, addition, subtraction, multiplication, lastly section. This document contributes to the issue why discover arithmetic with this series? Moreover, why find out multiplication following counting, addition, and subtraction before division?
## The subsequent details answer these concerns:
1. Young children understand counting initial by associating visible physical objects making use of their fingertips. A tangible illustration: How many apples are available in the basket? More abstract instance is when aged are you?
2. From counting numbers, the subsequent logical step is addition followed by subtraction. Addition and subtraction tables can be quite useful instructing assists for youngsters because they are visual resources making the move from counting much easier.
3. Which will be learned up coming, multiplication or section? Multiplication is shorthand for addition. At this time, young children have got a organization grasp of addition. As a result, multiplication is definitely the following reasonable kind of arithmetic to learn.
## Review fundamentals of multiplication. Also, review the basic principles how to use a multiplication table.
We will assessment a multiplication instance. By using a Multiplication Table, increase several times three and get a response a dozen: 4 x 3 = 12. The intersection of row about three and line four of the Multiplication Table is 12; a dozen will be the response. For kids beginning to find out multiplication, this is easy. They could use addition to resolve the problem thus affirming that multiplication is shorthand for addition. Case in point: 4 by 3 = 4 4 4 = 12. It is really an outstanding guide to the Multiplication Table. An added benefit, the Multiplication Table is graphic and reflects back to discovering addition.
## Where by will we commence studying multiplication using the Multiplication Table?
1. First, get acquainted with the table.
2. Start with multiplying by 1. Start off at row # 1. Go on to line number one. The intersection of row one particular and line the initial one is the best solution: 1.
3. Perform repeatedly these techniques for multiplying by one particular. Grow row 1 by posts 1 by way of twelve. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Perform repeatedly these methods for multiplying by two. Grow row two by columns 1 via 5. The solutions are 2, 4, 6, 8, and 10 respectively.
5. We will bounce forward. Repeat these steps for multiplying by 5 various. Grow row 5 by posts one by way of twelve. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now allow us to improve the quantity of problems. Repeat these actions for multiplying by 3. Grow row three by posts a single through 12. The answers are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 respectively.
7. Should you be more comfortable with multiplication to date, try a test. Fix the next multiplication difficulties in your thoughts then assess your answers towards the Multiplication Table: grow six as well as 2, flourish nine and a few, multiply a single and 11, increase several and several, and multiply several and two. The issue responses are 12, 27, 11, 16, and 14 correspondingly.
When you received several away from five issues correct, create your own multiplication exams. Determine the solutions in your mind, and appearance them while using Multiplication Table. | 826 | 3,687 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.65625 | 5 | CC-MAIN-2021-49 | latest | en | 0.92696 |
https://subwiki.org/wiki/Relation_theory | 1,653,161,631,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662540268.46/warc/CC-MAIN-20220521174536-20220521204536-00514.warc.gz | 630,763,162 | 16,731 | # Relation theory
This article treats relations from the perspective of combinatorics, in other words, as a subject matter in discrete mathematics, with special attention to finite structures and concrete set-theoretic constructions, many of which arise quite naturally in applications. This approach to relation theory, or the theory of relations, is distinguished from, though closely related to, its study from the perspectives of abstract algebra on the one hand and formal logic on the other.
## Preliminaries
Two definitions of the relation concept are common in the literature. Although it is usually clear in context which definition is being used at a given time, it tends to become less clear as contexts collide, or as discussion moves from one context to another.
The same sort of ambiguity arose in the development of the function concept and it may save some effort to follow the pattern of resolution that worked itself out there.
When we speak of a function we are thinking of a mathematical object whose articulation requires three pieces of data, specifying the set the set and a particular subset of their cartesian product So far so good.
Let us write to express what has been said so far.
When it comes to parsing the notation everyone takes the part to specify the type of the function, that is, the pair but is used equivocally to denote both the triple and the subset that forms one part of it. One way to resolve the ambiguity is to formalize a distinction between a function and its graph, letting
Another tactic treats the whole notation as sufficient denotation for the triple, letting denote
In categorical and computational contexts, at least initially, the type is regarded as an essential attribute or an integral part of the function itself. In other contexts it may be desirable to use a more abstract concept of function, treating a function as a mathematical object that appears in connection with many different types.
Following the pattern of the functional case, let the notation bring to mind a mathematical object that is specified by three pieces of data, the set the set and a particular subset of their cartesian product As before we have two choices, either let or let denote and choose another name for the triple.
## Definition
It is convenient to begin with the definition of a -place relation, where is a positive integer.
Definition. A -place relation over the nonempty sets is a -tuple where is a subset of the cartesian product
## Remarks
Though usage varies as usage will, there are several bits of optional language that are frequently useful in discussing relations. The sets are called the domains of the relation with being the domain. If all of the are the same set then is more simply described as a -place relation over The set is called the graph of the relation on analogy with the graph of a function. If the sequence of sets is constant throughout a given discussion or is otherwise determinate in context, then the relation is determined by its graph making it acceptable to denote the relation by referring to its graph. Other synonyms for the adjective -place are -adic and -ary, all of which leads to the integer being called the dimension, adicity, or arity of the relation
## Local incidence properties
A local incidence property (LIP) of a relation is a property that depends in turn on the properties of special subsets of that are known as its local flags. The local flags of a relation are defined in the following way:
Let be a -place relation
Select a relational domain and one of its elements Then is a subset of that is referred to as the flag of with at or the -flag of an object that has the following definition:
Any property of the local flag is said to be a local incidence property of with respect to the locus
A -adic relation is said to be -regular at if and only if every flag of with at has the property where is taken to vary over the theme of the fixed domain
Expressed in symbols, is -regular at if and only if is true for all in
## Regional incidence properties
The definition of a local flag can be broadened from a point in to a subset of arriving at the definition of a regional flag in the following way:
Suppose that and choose a subset Then is a subset of that is said to be the flag of with at or the -flag of an object which has the following definition:
## Numerical incidence properties
A numerical incidence property (NIP) of a relation is a local incidence property that depends on the cardinalities of its local flags.
For example, is said to be -regular at if and only if the cardinality of the local flag is for all in or, to write it in symbols, if and only if for all
In a similar fashion, one can define the NIPs, -regular at -regular at and so on. For ease of reference, a few of these definitions are recorded here:
Returning to 2-adic relations, it is useful to describe some familiar classes of objects in terms of their local and numerical incidence properties. Let be an arbitrary 2-adic relation. The following properties of can be defined:
If is tubular at then is called a partial function or a prefunction from to This is sometimes indicated by giving an alternate name, say, and writing
Just by way of formalizing the definition:
If is a prefunction that happens to be total at then is called a function from to indicated by writing To say that a relation is totally tubular at is to say that it is -regular at Thus, we may formalize the following definition:
In the case of a function one has the following additional definitions:
## Variations
Because the concept of a relation has been developed quite literally from the beginnings of logic and mathematics, and because it has incorporated contributions from a diversity of thinkers from many different times and intellectual climes, there is a wide variety of terminology that the reader may run across in connection with the subject.
One dimension of variation is reflected in the names that are given to -place relations, for with some writers using the Greek forms, medadic, monadic, dyadic, triadic, -adic, and other writers using the Latin forms, nullary, unary, binary, ternary, -ary.
The number of relational domains may be referred to as the adicity, arity, or dimension of the relation. Accordingly, one finds a relation on a finite number of domains described as a polyadic relation or a finitary relation, but others count infinitary relations among the polyadic. If the number of domains is finite, say equal to then the relation may be described as a -adic relation, a -ary relation, or a -dimensional relation, respectively.
A more conceptual than nominal variation depends on whether one uses terms like predicate, relation, and even term to refer to the formal object proper or else to the allied syntactic items that are used to denote them. Compounded with this variation is still another, frequently associated with philosophical differences over the status in reality accorded formal objects. Among those who speak of numbers, functions, properties, relations, and sets as being real, that is to say, as having objective properties, there are divergences as to whether some things are more real than others, especially whether particulars or properties are equally real or else one is derivative in relationship to the other. Historically speaking, just about every combination of modalities has been used by one school of thought or another, but it suffices here merely to indicate how the options are generated.
## Examples
See the articles on relations, relation composition, relation reduction, sign relations, and triadic relations for concrete examples of relations.
Many relations of the greatest interest in mathematics are triadic relations, but this fact is somewhat disguised by the circumstance that many of them are referred to as binary operations, and because the most familiar of these have very specific properties that are dictated by their axioms. This makes it practical to study these operations for quite some time by focusing on their dyadic aspects before being forced to consider their proper characters as triadic relations.
## References
• Peirce, C.S., “Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole's Calculus of Logic”, Memoirs of the American Academy of Arts and Sciences, 9, 317–378, 1870. Reprinted, Collected Papers CP 3.45–149, Chronological Edition CE 2, 359–429.
• Ulam, S.M. and Bednarek, A.R., “On the Theory of Relational Structures and Schemata for Parallel Computation”, pp. 477–508 in A.R. Bednarek and Françoise Ulam (eds.), Analogies Between Analogies : The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, University of California Press, Berkeley, CA, 1990.
## Bibliography
• Barr, Michael, and Wells, Charles (1990), Category Theory for Computing Science, Prentice Hall, Hemel Hempstead, UK.
• Bourbaki, Nicolas (1994), Elements of the History of Mathematics, John Meldrum (trans.), Springer-Verlag, Berlin, Germany.
• Carnap, Rudolf (1958), Introduction to Symbolic Logic with Applications, Dover Publications, New York, NY.
• Chang, C.C., and Keisler, H.J. (1973), Model Theory, North-Holland, Amsterdam, Netherlands.
• van Dalen, Dirk (1980), Logic and Structure, 2nd edition, Springer-Verlag, Berlin, Germany.
• Devlin, Keith J. (1993), The Joy of Sets : Fundamentals of Contemporary Set Theory, 2nd edition, Springer-Verlag, New York, NY.
• Halmos, Paul Richard (1960), Naive Set Theory, D. Van Nostrand Company, Princeton, NJ.
• van Heijenoort, Jean (1967/1977), From Frege to Gödel : A Source Book in Mathematical Logic, 1879–1931, Harvard University Press, Cambridge, MA, 1967. Reprinted with corrections, 1977.
• Kelley, John L. (1955), General Topology, Van Nostrand Reinhold, New York, NY.
• Kneale, William; and Kneale, Martha (1962/1975), The Development of Logic, Oxford University Press, Oxford, UK, 1962. Reprinted with corrections, 1975.
• Lawvere, Francis William; and Rosebrugh, Robert (2003), Sets for Mathematics, Cambridge University Press, Cambridge, UK.
• Lawvere, Francis William; and Schanuel, Stephen H. (1997/2000), Conceptual Mathematics : A First Introduction to Categories, Cambridge University Press, Cambridge, UK, 1997. Reprinted with corrections, 2000.
• Manin, Yu. I. (1977), A Course in Mathematical Logic, Neal Koblitz (trans.), Springer-Verlag, New York, NY.
• Mathematical Society of Japan (1993), Encyclopedic Dictionary of Mathematics, 2nd edition, 2 vols., Kiyosi Itô (ed.), MIT Press, Cambridge, MA.
• Mili, A., Desharnais, J., Mili, F., with Frappier, M. (1994), Computer Program Construction, Oxford University Press, New York, NY. (Introduction to Tarskian relation theory and relational programming.)
• Mitchell, John C. (1996), Foundations for Programming Languages, MIT Press, Cambridge, MA.
• Peirce, Charles Sanders (1870), ``Description of a Notation for the Logic of Relatives, Resulting from an Amplification of the Conceptions of Boole's Calculus of Logic", Memoirs of the American Academy of Arts and Sciences 9 (1870), 317–378. Reprinted (CP 3.45–149), (CE 2, 359–429).
• Peirce, Charles Sanders (1931–1935, 1958), Collected Papers of Charles Sanders Peirce, vols. 1–6, Charles Hartshorne and Paul Weiss (eds.), vols. 7–8, Arthur W. Burks (ed.), Harvard University Press, Cambridge, MA. Cited as (CP volume.paragraph).
• Peirce, Charles Sanders (1981–), Writings of Charles S. Peirce : A Chronological Edition, Peirce Edition Project (eds.), Indiana University Press, Bloomington and Indianapolis, IN. Cited as (CE volume, page).
• Poizat, Bruno (2000), A Course in Model Theory : An Introduction to Contemporary Mathematical Logic, Moses Klein (trans.), Springer-Verlag, New York, NY.
• Quine, Willard Van Orman (1940/1981), Mathematical Logic, 1940. Revised edition, Harvard University Press, Cambridge, MA, 1951. New preface, 1981.
• Royce, Josiah (1961), The Principles of Logic, Philosophical Library, New York, NY.
• Runes, Dagobert D. (ed., 1962), Dictionary of Philosophy, Littlefield, Adams, and Company, Totowa, NJ.
• Styazhkin, N.I. (1969), History of Mathematical Logic from Leibniz to Peano, MIT Press, Cambridge, MA.
• Suppes, Patrick (1957/1999), Introduction to Logic, 1st published 1957. Reprinted, Dover Publications, New York, NY, 1999.
• Suppes, Patrick (1960/1972), Axiomatic Set Theory, 1st published 1960. Reprinted, Dover Publications, New York, NY, 1972.
• Tarski, Alfred (1956/1983), Logic, Semantics, Metamathematics : Papers from 1923 to 1938, J.H. Woodger (trans.), Oxford University Press, 1956. 2nd edition, J. Corcoran (ed.), Hackett Publishing, Indianapolis, IN, 1983.
• Ulam, Stanislaw Marcin; and Bednarek, A.R. (1977), “On the Theory of Relational Structures and Schemata for Parallel Computation”, pp. 477–508 in A.R. Bednarek and Françoise Ulam (eds.), Analogies Between Analogies : The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, University of California Press, Berkeley, CA, 1990.
• Ulam, Stanislaw Marcin (1990), Analogies Between Analogies : The Mathematical Reports of S.M. Ulam and His Los Alamos Collaborators, A.R. Bednarek and Françoise Ulam (eds.), University of California Press, Berkeley, CA.
• Ullman, Jeffrey D. (1980), Principles of Database Systems, Computer Science Press, Rockville, MD.
• Venetus, Paulus (1472/1984), Logica Parva : Translation of the 1472 Edition with Introduction and Notes, Alan R. Perreiah (trans.), Philosophia Verlag, Munich, Germany.
## Syllabus
### Relational concepts
Relation theory Relative term Sign relation Triadic relation
## Document history
Portions of the above article were adapted from the following sources under the GNU Free Documentation License, under other applicable licenses, or by permission of the copyright holders. | 3,350 | 14,018 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 146, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2022-21 | latest | en | 0.943288 |
https://practicaldev-herokuapp-com.global.ssl.fastly.net/colinfay/advent-of-code-2020-02-with-r-javascript-1ai2 | 1,723,106,499,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640723918.41/warc/CC-MAIN-20240808062406-20240808092406-00154.warc.gz | 362,130,542 | 21,666 | ## DEV Community
Colin Fay
Posted on • Originally published at colinfay.me on
# Advent of Code 2020-02 with R & JavaScript
Solving Advent of Code 2020-01 with R and JavaScript.
[Disclaimer] Obviously, this post contains a big spoiler about Advent of Code.
[Disclaimer bis] I’m no JavaScript expert so this might not be the perfect solution. TBH, that’s also the case for the R solution.
The JavaScript code has been written in the same RMarkdown as the R code. It runs thanks to the `{bubble}` package:https://github.com/ColinFay/bubble
## Instructions
### Step 1
• Inputs are of the form `1-3 a: abcde`, for `min`-`max``letter`:`code`, where `letter` must be found at least `min` and no more than `max` in `code`
How to: split the input in four columns, for `min`,`max`,`letter`, and`code`, count the number of `letter` in `code`, and then make sure this count is >= min and <= max.
### Step 2
• Inputs are of the form `1-3 a: abcde`, for `position1` `position2` `letter`:`code`, where `code[position1] == letter | code[position2] == letter` but not `code[position1] == letter & code[position2] == letter`, nor `!code[position1] == letter & !code[position2] == letter`
## R solution
### Part one
``````# Read
library(dplyr, warn.conflicts = FALSE)
library(tidyr)
library(purrr)
ipt %>%
# Create the four columns
separate(V1, c("min", "max", "letter", "code")) %>%
pmap_dbl(~{
# If the letter is not in the input, return 0
if (!stringr::str_detect(..4, ..3)) return(0)
# Count the n of letter in code
count <- stringr::str_count(..4, ..3)
# Is this count between the boundaries?
count >= as.numeric(..1) & count <= as.numeric(..2)
}) %>% sum()
## [1] 607
``````
### Part two
``````ipt %>%
# Create the four columns
separate(V1, c("min", "max", "letter", "code")) %>%
pmap_dbl(~{
#browser()
# Split the code
code <- strsplit(..4, "")[[1]]
# code[position1] == letter
p1_match <- code[as.numeric(..1)] == ..3
# code[position1] == letter
p2_match <- code[as.numeric(..2)] == ..3
# No match
if (!p1_match & !p2_match) return(0)
# Two matches
if (p1_match && p2_match) return(0)
return(1)
}) %>% sum()
## [1] 321
``````
## JS solution
### Part one & Two
// Reading the file and converting it to int
``````const fs = require('fs')
var ipt = fs.readFileSync("2020-02-aoc.txt", 'utf8').split("\n").filter(x => x.length != 0);
ipt = ipt.map(x => x.split(/[^A-Za-z0-9]/));
``````
// Doing the combn
``````var res1 = ipt.map(x => {
if (!x[4].includes(x[2])) return 0
var match = x[4].match(new RegExp(x[2], "g")).length
return match >= parseInt(x[0]) & match <= parseInt(x[1])
})
``````
// Solution
``````// https://developer.mozilla.org/fr/docs/Web/JavaScript/Reference/Objets_globaux/Array/reduce
const reducer = (accumulator, currentValue) => accumulator + currentValue;
res1.reduce(reducer)
## undefined
## undefined
## 607
``````
// Part 2
``````var res2 = ipt.map(x => {
// Split the code
var code = x[4].split("")
// code[position1] == letter
var p1_match = code[parseInt(x[0]) - 1] === x[2]
// code[position1] == letter
var p2_match = code[parseInt(x[1]) - 1] === x[2]
// No match
if (!p1_match && !p2_match) return 0
// Two matches
if (p1_match && p2_match) return 0
return 1
})
``````
// Solution
``````res2.reduce(reducer)
## 321
`````` | 1,027 | 3,272 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2024-33 | latest | en | 0.661002 |
https://testbook.com/question-answer/all-members-of-the-frame-shown-below-has-equal-fle--59608cef0ed3fe08d02369f5 | 1,631,852,964,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780054023.35/warc/CC-MAIN-20210917024943-20210917054943-00264.warc.gz | 587,039,858 | 29,951 | All members of the frame shown below has equal flexural rigidity EI. Calculate the rotation of joint O if moment M is applied?
This question was previously asked in
Gujarat Engineering Service 2019 Official Paper (Civil Part 2)
View all GPSC Engineering Services Papers >
1. $$\frac{{ML}}{{13EI}}$$
2. $$\frac{{2ML}}{{13EI}}$$
3. $$\frac{{3ML}}{{13EI}}$$
4. $$\frac{{ML}}{{12EI}}$$
Option 2 : $$\frac{{2ML}}{{13EI}}$$
Free
Gujarat Engineering Service 2019 Official Paper (Civil Part 1)
622
150 Questions 150 Marks 90 Mins
Detailed Solution
Concept :-
Stiffness value, K when far end is fixed = 4EI/L
Stiffness value, K when far end is roller = 3EI/L
Stiffness value, K when far end is guided roller = EI/L
Calculation:-
$$\begin{array}{l} {K_{OA}} = \frac{{4EI}}{L}\\ {K_{OC}} = \frac{{3EI}}{{2L}}\\ {K_{OB}} = \frac{{EI}}{L}\\ \therefore \sum K = \frac{{4EI}}{L} + \frac{{3EI}}{{2L}} + \frac{{EI}}{L} = \frac{{13EI}}{{2L}} \end{array}$$
Now, Rotation, $$\theta = \frac{M}{{\sum K}} = \frac{{M \times 2L}}{{13EI}} = \frac{{2ML}}{{13EI}}$$
For stiffness coefficient when far end is guided roller support refer :-
Advanced Structural Analysis, Prof. Devdas MenonDepartment of Civil Engineering, Indian Institute of Technology, Madras
Module - 5.3 ,Lecture - 29Matrix Analysis of Beams and GridsPage No. - 9 | 433 | 1,317 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2021-39 | latest | en | 0.660118 |
https://electronicsarea.com/dielectric-constant-relative-permittivity-capacitor/ | 1,718,417,734,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861578.89/warc/CC-MAIN-20240614235857-20240615025857-00831.warc.gz | 207,684,455 | 12,469 | ## Dielectric Constant / Relative Permittivity
The dielectric constant or Relative Permittivity is a dimensionless physical constant (Dielectric constant has no units) that describes how an electric field affects a material. The dielectric constant is the ratio of the permittivity of a substance to the permittivity of free space.
Capacity of a capacitor depends on the dielectric constant. It is known that the value of the capacity of a capacitor is given by the following formula: C = Q / V. Where:
• C: Capacitor capacity
• Q: capacitor charge
• V: Potential difference (voltage) between the capacitor plates
The capacitance of a capacitor can also be obtained differently by including the value of the dielectric constant.
## How to obtain the Capacity, knowing the dielectric constant and its physical dimensions?
A capacitor is formed by two parallel plates. If there is a vacuum between these plates, the value of the capacity is: C = εo a / d. where:
• a = area of each plate in m2
• d = distance between plates in meters
• εo = dielectric constant in vacuum, whose value is: 8.85 x 10-12 farad / meter
If a dielectric is introduced between the plates, the capacitance will increase by a factor εr. So the capacity is:
C = εo εr a / d or C = ε a / d, where:
• ε = εo εr
• εr is the relative dielectric constant and depends on the physical properties of the medium used.
• ε is the absolute dielectric constant.
There is a great difference between values of the dielectric constants of different materials. Some important examples of dielectric constants are shown in the following table.
## Table of dielectric constants
Table of dielectric constants (20 °C)
Example of capacity and charge calculation of parallel plate’s capacitor.
We have a parallel plates capacitor separated by vacuum. The plates are 1 mm apart and have an area of 2 x 10-6 meters. If the capacitor is connected to a voltage of 250 volts:
1. What is the capacitance of the capacitor?
2. What is the charge on each plate?
3. What is the capacitance to be if it has a paper dielectric?
1. Capacitance:
Using the formula: C = εo εr a / d.
C = (8.85 x 10-12) (1) (2 x 10-6) / (1 x 10-3) = 1.77 x 10-14 Farads.
2. Charge on each plate:
Using the formula: C = Q / V, we obtain that Q = C x V.
Q = (1.77 x 10-14) x 250 = 4.425 x 10-12 Coulombs.
3. Capacitance with paper dielectric (εr = 2.5)
Using the formula: C = εo εr a / d
C = (8.85 x 10-12) (2.5) (2 x 10-6) / (1 x 10-3) = 4.425 x 10-14 Farads.
You can see the increase in capacitor capacity with a higher value dielectric. | 697 | 2,578 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2024-26 | latest | en | 0.874743 |
https://testbook.com/question-answer/the-answer-of-the-operation-101112--605dd1b8539b87cd2e14587c | 1,632,518,934,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057580.39/warc/CC-MAIN-20210924201616-20210924231616-00440.warc.gz | 584,570,596 | 31,087 | # The answer of the operation (10111)2 × (1110)2 in hex equivalent is
This question was previously asked in
NIMCET 2013 Official Paper
View all NIMCET Papers >
1. 150
2. 14C
3. 142
4. 13E
Option 3 : 142
Free
NIMCET 2020 Official Paper
1342
120 Questions 480 Marks 120 Mins
## Detailed Solution
Concept:
Convert Binary to Decimal:
For binary number with n digits;
Dn-1 …. D3D2D1D0
The decimal number is equal to the sum of binary digits (Dn) times their power of 2 (2n)
Decimal = D0 × 20 + D1 × 21 + D2 × 22 + ….+Dn-1 × 2n-1
Calculation:
To Find: Hex equivalent of (10111)2 × (1110)2
To do this, at first translate it to decimal here so :
(10111)2 = 1 × 20 + 1 × 21+ 1 × 22 + 0 × 23 + 1 × 24 = 23
(1110)2 = 0 × 20 + 1 × 21+ 1 × 22 + 1 × 23 = 14
Now, (10111)2 × (1110)2 = 23 × 14 = 322
Converting 32210 in hex equivalent here so:
The whole part of a number is obtained by dividing on the basis of new
Division by 16 Quotient Remainder 322/16 20 2 20/16 1 4
∴ 32210 = 14216 | 417 | 990 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2021-39 | latest | en | 0.538645 |
http://www.adaic.org/resources/add_content/standards/05rm/html/RM-4-5-5.html | 1,544,867,811,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376826842.56/warc/CC-MAIN-20181215083318-20181215105318-00402.warc.gz | 318,466,335 | 5,014 | # 4.5.5 Multiplying Operators
#### Static Semantics
1
The multiplying operators * (multiplication), / (division), mod (modulus), and rem (remainder) are predefined for every specific integer type T
2
function "*" (Left, Right : Treturn T
function "/" (Left, Right : Treturn T
function "mod"(Left, Right : Treturn T
function "rem"(Left, Right : Treturn T
3
Signed integer multiplication has its conventional meaning.
4
Signed integer division and remainder are defined by the relation:
5
A = (A/B)*B + (A rem B)
6
where (A rem B) has the sign of A and an absolute value less than the absolute value of B. Signed integer division satisfies the identity:
7
(-A)/B = -(A/B) = A/(-B)
8
The signed integer modulus operator is defined such that the result of A mod B has the sign of B and an absolute value less than the absolute value of B; in addition, for some signed integer value N, this result satisfies the relation:
9
A = B*N + (A mod B)
10
The multiplying operators on modular types are defined in terms of the corresponding signed integer operators, followed by a reduction modulo the modulus if the result is outside the base range of the type (which is only possible for the "*" operator).
11
Multiplication and division operators are predefined for every specific floating point type T
12
function "*"(Left, Right : Treturn T
function "/"(Left, Right : Treturn T
13
The following multiplication and division operators, with an operand of the predefined type Integer, are predefined for every specific fixed point type T
14
function "*"(Left : T; Right : Integer) return T
function "*"(Left : Integer; Right : Treturn T
function "/"(Left : T; Right : Integer) return T
15
All of the above multiplying operators are usable with an operand of an appropriate universal numeric type. The following additional multiplying operators for root_real are predefined, and are usable when both operands are of an appropriate universal or root numeric type, and the result is allowed to be of type root_real, as in a number_declaration:
16
function "*"(Left, Right : root_realreturn root_real
function "/"(Left, Right : root_realreturn root_real
17
function "*"(Left : root_real; Right : root_integerreturn root_real
function "*"(Left : root_integer; Right : root_realreturn root_real
function "/"(Left : root_real; Right : root_integerreturn root_real
18
Multiplication and division between any two fixed point types are provided by the following two predefined operators:
19
function "*"(Left, Right : universal_fixedreturn universal_fixed
function "/"(Left, Right : universal_fixedreturn universal_fixed
#### Name Resolution Rules
19.1/2
The above two fixed-fixed multiplying operators shall not be used in a context where the expected type for the result is itself universal_fixed — the context has to identify some other numeric type to which the result is to be converted, either explicitly or implicitly. Unless the predefined universal operator is identified using an expanded name with prefix denoting the package Standard, an explicit conversion is required on the result when using the above fixed-fixed multiplication operator if either operand is of a type having a user-defined primitive multiplication operator such that:
19.2/2
• it is declared immediately within the same declaration list as the type; and
19.3/2
• both of its formal parameters are of a fixed-point type.
19.4/2
A corresponding requirement applies to the universal fixed-fixed division operator.
#### Legality Rules
20/2
This paragraph was deleted.
#### Dynamic Semantics
21
The multiplication and division operators for real types have their conventional meaning. For floating point types, the accuracy of the result is determined by the precision of the result type. For decimal fixed point types, the result is truncated toward zero if the mathematical result is between two multiples of the small of the specific result type (possibly determined by context); for ordinary fixed point types, if the mathematical result is between two multiples of the small, it is unspecified which of the two is the result.
22
The exception Constraint_Error is raised by integer division, rem, and mod if the right operand is zero. Similarly, for a real type T with T'Machine_Overflows True, division by zero raises Constraint_Error.
NOTES
23
16 For positive A and B, A/B is the quotient and A rem B is the remainder when A is divided by B. The following relations are satisfied by the rem operator:
24
A rem (-B) = A rem B
(-A) rem B = -(A rem B)
25
17 For any signed integer K, the following identity holds:
26
A mod B = (A + K*B) mod B
27
The relations between signed integer division, remainder, and modulus are illustrated by the following table:
28
A B A/B A rem B A mod B A B A/B A rem B A mod B
29
10 5 2 0 0 -10 5 -2 0 0
11 5 2 1 1 -11 5 -2 -1 4
12 5 2 2 2 -12 5 -2 -2 3
13 5 2 3 3 -13 5 -2 -3 2
14 5 2 4 4 -14 5 -2 -4 1
30
A B A/B A rem B A mod B A B A/B A rem B A mod B
10 -5 -2 0 0 -10 -5 2 0 0
11 -5 -2 1 -4 -11 -5 2 -1 -1
12 -5 -2 2 -3 -12 -5 2 -2 -2
13 -5 -2 3 -2 -13 -5 2 -3 -3
14 -5 -2 4 -1 -14 -5 2 -4 -4
#### Examples
31
Examples of expressions involving multiplying operators:
32
I : Integer := 1;
J : Integer := 2;
K : Integer := 3;
33
X : Real := 1.0; -- see 3.5.7
Y : Real := 2.0;
34
F : Fraction := 0.25; -- see 3.5.9
G : Fraction := 0.5;
35
Expression Value Result Type
I*J 2 same as I and J, that is, Integer
K/J 1 same as K and J, that is, Integer
mod J 1 same as K and J, that is, Integer
X/Y 0.5 same as X and Y, that is, Real
F/2 0.125 same as F, that is, Fraction
3*F 0.75 same as F, that is, Fraction
0.75*G 0.375 universal_fixed, implicitly convertible
to any fixed point type
Fraction(F*G) 0.125 Fraction, as stated by the conversion
Real(J)*Y 4.0 Real, the type of both operands after
conversion of J | 1,942 | 6,572 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2018-51 | latest | en | 0.634866 |
https://internetbeep.com/7th-grade-math-games/ | 1,601,320,146,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401604940.65/warc/CC-MAIN-20200928171446-20200928201446-00574.warc.gz | 407,997,098 | 11,786 | Are you looking for free 7th grade math games? The activities provided on this website are designed to reinforce math concepts in a fun and interactive way.
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Integers Math Game (New)
Destroy lots of virtual monsters when playing this fun and interactive game about integers.
7th Grade Halloween Math Game (New)
In this exciting 7th Grade Halloween Math Game, students calculate percents, the whole, and the part as they develop percent proportions. Play this fun, online Halloween math game and practice your math skills to destroy a lot of monsters. For each correct answer, you will enter a bonus round where you can earn points by smashing monsters.
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Another exciting math racing game about subtracting integers.
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Play this exciting Halloween Math Game to determine unit rates.
Match addition and subtraction problems with the correct integers in as few attempts as possible.
Multiplying and Dividing Integers
Operations with Integers
Review the rules for adding, subtracting, multiplying, and dividing integers when playing this cool jeopardy game.
This is an interesting Math Pirate game about adding and subtracting integers.
Absolute Value Millionaire Game
Play this fun millionaire game about the absolute value.
Writing Algebraic Expressions
Play this attention-grabbing millionaire game about writing algebraic expressions.
Solving One-Step Equations
Play this fun basketball math game alone, with a friend, or in two teams.
One-Step Equations with Addition and Subtraction
This is an exciting soccer math game about solving linear equations.
Changing Fractions to Percents
In this concentration game, students will match different fractions with the equivalent percents.
Measuring and Classifying Angles
Fun jeopardy game about angles. It can be played alone or in teams.
Coordinate Plane Game
Interesting game about locating points in the coordinate plane and identifying the coordinates.
Coordinate Plane Game
Coordinate Plane Jeopardy
Fun jeopardy game about the coordinate plane. It makes an excellent classroom activity.
3D Shapes Game (Concentration)
In this concentration game, students will match pictures of three-dimensional shapes with the correct words. If there is a match, the problems remain on the page; if not, the cards are turned over.
Another interactive jeopardy game that 7th graders can play at the end of the year as a review activity.
7th Grade Numbers and Operations Review
Students can play this cool jeopardy game to review operations with rational numbers and important facts about absolute value.
Equation Word Search
This is an interactive word search game that students can solve directly on the computer screen. It helps them memorize important vocabulary words related to solving equations.
How to get better at math,
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