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https://www.gradesaver.com/textbooks/math/calculus/thomas-calculus-13th-edition/chapter-8-techniques-of-integration-section-8-3-trigonometric-integrals-exercises-8-3-page-462/14 | 1,679,891,340,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296946637.95/warc/CC-MAIN-20230327025922-20230327055922-00504.warc.gz | 852,970,992 | 13,824 | ## Thomas' Calculus 13th Edition
$\frac{\pi}{4}$
Recall the identity $\cos 2x=1-2\sin^{2} x$ which gives $\sin^{2}x=\frac{1-\cos 2x}{2}$. Therefore, $\int \sin^{2}x\,dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos 2x\,dx$ $= \frac{x}{2}-\frac{1}{4}\sin 2x+C$ Then $\int_{0}^{\pi/2} \sin^{2}x\,dx=[\frac{x}{2}-\frac{1}{4}\sin 2x]^{\pi/2}_{0}=(\frac{\pi}{4}-0)-0=\frac{\pi}{4}$ | 185 | 370 | {"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} | 3.890625 | 4 | CC-MAIN-2023-14 | latest | en | 0.32245 |
https://slideplayer.com/slide/5677869/ | 1,709,383,610,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475825.14/warc/CC-MAIN-20240302120344-20240302150344-00461.warc.gz | 539,631,567 | 22,106 | # Linear Inequalities in Two Variables
## Presentation on theme: "Linear Inequalities in Two Variables"— Presentation transcript:
Linear Inequalities in Two Variables
The graph of the solutions of a linear inequality in two variables is a half-plane. Procedure for Graphing Linear Inequalities 1. Rewrite the linear inequality as a linear equation. Change the inequality symbol ( < , > , , or ) to an equal sign. 2. The next step is to graph that line. If the inequality symbol has an equal sign ( or ), draw a solid line. If the inequality symbol does not have an equal sign (< , >), draw a dashed line. Pick any ordered pair that is not on the graphed line. This will be your test point. Substitute the coordinates of that ordered pair into the original inequality. If the inequality is true, shade the side of the line where the test point is. If the inequality is false, shade the other side of the line. Next Slide
x y 0 -4 Example 1. Graph: 3x +2y < 6 Solution:
x axis y axis Example 1. Graph: 3x +2y < 6 Solution: Write the inequality as a linear equation, then graph. 3x + 2y = 6 Pick a test point not on the line, say (0,0). Since this is true, we can shade the side where the test point is. x y 0 -4 3 0 y axis x axis
Write the inequality as a linear equation, and graph.
Example 2. Graph y > 2x Solution: x axis y axis Write the inequality as a linear equation, and graph. y = 2x Pick a test point, say (0,4). Since this is false, we must shade the side opposite the test point. Note: There is another method to determining which side of the line to shade. If the inequality is y> or y≥, shade above the line. If the inequality is y< or y≤, shade below the line. This example is y>2x, therefore we shade above the line. Using this method takes a little work to get y by itself but then you don’t have to worry about the test point. y axis x axis
Since this is false, shade on the opposite side of the test point.
x axis y axis Since this is false, shade on the opposite side of the test point. Note: If the inequality is x> or x≥, shade to the right of the line. If the inequality is x< or x≤, shade to the left of the line. This example is x≤−4, therefore we shade to the left of the line. x axis y axis
Because our inequality is y >, shade above the dashed line.
x axis y axis 1st, write the inequality as linear equation to graph the line. Either use the y intercept and the slope or choose values for x. If you choose values for x, always choose x=0 and also choose values that are multiples of the denominator in the fraction. Because our inequality is y >, shade above the dashed line. x axis y axis
The shaded area will be to the left.
x axis y axis Recall that the word “and” indicates the intersection of the solutions sets of each inequality. The shaded area will be to the left. Answer The shaded area will be below the line. The intersection is the area that satisfied both inequalities. (shaded twice) x axis y axis
In the previous sections, we solved systems of equations such as:
The solution set of the system is the intersection of the solution sets of the individual lines. (i.e., where the two lines meet.) The previous example contained the word “and”, which means the intersection. Another way of asking for the intersection is using a system such as: The solution set of the system is the intersection of the solution sets of the individual inequalities. (i.e., where the graph is shaded twice.) Next Slide
Example 6. Solve the following system by graphing.
Solution: x axis y axis Next Slide Since y >, shade above. The solution is the area which was shaded twice. We will darken that area with another color to show the intersection more clearly. Since y ≥, shade above.
Solve the following system by graphing.
Your Turn Problem #6. Solve the following system by graphing. x axis y axis Answer:
Example 7. Solve the following system by graphing.
Solution: x axis y axis Therefore, our graph will only be shaded in Quadrant I, the upper right-hand quadrant. The solution is the area in Quadrant I which satisfies all four inequalities, shown here in the light blue color. Easier to graph last two inequalities using the intercept method. Then use test points to determine shaded area. Next Slide
Your Turn Problem #7. Graph: Answer: The End. B.R. 1-25-07 y axis
x axis y axis (4,0) (0,4) (7,0) (0,6) Answer: The End. B.R.
Download ppt "Linear Inequalities in Two Variables"
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https://anhngq.wordpress.com/2009/02/11/solving-initial-value-problem-for-wave-equation-via-fourier-transform/ | 1,643,006,001,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304515.74/warc/CC-MAIN-20220124054039-20220124084039-00666.warc.gz | 154,939,951 | 34,080 | # Ngô Quốc Anh
## February 11, 2009
### Solving initial value problem for wave equation via Fourier transform
Filed under: Các Bài Tập Nhỏ, Linh Tinh, Nghiên Cứu Khoa Học, PDEs — Tags: — Ngô Quốc Anh @ 12:43
In this topic, we will show you how can we use Fourier transform to solve initial value problem for wave equation in $\mathbb R$. Following is the problem
$\displaystyle \left\{ \begin{gathered} {u_{tt}} = {u_{xx}}, \qquad x \in \mathbb{R}, \hfill \\ u\left( {x,0} \right) = \varphi \left( x \right), \hfill \\ {u_t}\left( {x,0} \right) = \psi \left( x \right). \hfill \\ \end{gathered} \right.$
From the equation by taking Fourier transform to the both sides, we obtain
$\displaystyle \widehat{{u_{tt}}\left( {\eta ,t} \right)} - {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0$
This is an ODE, the solution is given by
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = A\sin \left( {\eta t} \right) + B\cos \left( {\eta t} \right)$
From the initial date we get
$\widehat{u\left( {\eta ,0} \right)} = \widehat{\varphi \left( \eta \right)}$
which implies that
$\widehat{\varphi \left( \eta \right)} = B$.
From the initial date we see that
$\widehat{{u_t}\left( {\eta ,0} \right)} = \widehat{\psi \left( \eta \right)} = A\eta$.
Thus, we obtain
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = \frac{{\widehat{\psi \left( \eta \right)}}} {\eta}\sin \left( {\eta t} \right) + \widehat{\varphi \left( \eta \right)}\cos \left( {\eta t} \right)$.
Note that
$\displaystyle \cos \left( {\eta t} \right) = \frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}, \quad \frac{{\sin \left( {\eta t} \right)}} {\eta } = \frac{{{e^{i\eta t}} - {e^{ - i\eta t}}}} {{2i\eta }} = \frac{1} {2}\int\limits_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta }$.
Moreover,
$\displaystyle \widehat{\delta ( {x - \alpha t} )} = \int\limits_{ - \infty }^\infty {{e^{ - i\eta x}}\delta ( {x - \alpha t} )dx} = {e^{ - i\alpha \eta t}}\int\limits_{ - \infty }^\infty {{e^{ - i\eta ( {x - \alpha t} )}}\delta ( {x - \alpha t} )dx} = {e^{ - i\alpha \eta t}}$.
Then
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = \left( {\frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}} \right)\widehat{\varphi \left( \eta \right)} + \left( {\frac{1} {2}\int\limits_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta } } \right)\widehat{\psi \left( \eta \right)}$.
Since
$\displaystyle \left( {\frac{{{e^{i\eta t}} + {e^{ - i\eta t}}}} {2}} \right)\widehat{\varphi \left( \eta \right)} = \frac{{\widehat{\delta \left( {x + t} \right)*\varphi \left( x \right)} + \widehat{\delta \left( {x - t} \right)*\varphi \left( x \right)}}} {2}$
and
$\displaystyle \left( {\frac{1} {2}\int_{ - t}^t {\frac{d} {{d\theta }}{e^{i\eta \theta }}d\theta } } \right)\widehat{\psi \left( \eta \right)} = \frac{1} {2}\int\limits_{ - t}^t {\widehat{\delta \left( {x + \theta } \right)*\psi \left( x \right)}d\theta }$
then
$\displaystyle \widehat{u\left( {\eta ,t} \right)} = \frac{{\widehat{\delta \left( {x + t} \right)*\varphi \left( x \right)} + \widehat{\delta \left( {x - t} \right)*\varphi \left( x \right)}}} {2} + \frac{1} {2}\int\limits_{ - t}^t {\widehat{\delta \left( {x + \theta } \right)*\psi \left( x \right)}d\theta }$.
Thus,
$\displaystyle u\left( {x,t} \right) = \frac{{\delta \left( {x + t} \right)*\varphi \left( x \right) + \delta \left( {x - t} \right)*\varphi \left( x \right)}} {2} + \frac{1} {2}\int\limits_{ - t}^t {\delta \left( {x + \theta } \right)*\psi \left( x \right)d\theta }$
or equivalently,
$\displaystyle u\left( {x,t} \right) = \frac{{\varphi \left( {x + t} \right) + \varphi \left( {x - t} \right)}} {2} + \frac{1} {2}\int\limits_{x - t}^{x + t} {\psi \left( y \right)dy}$.
This is the so-call D’ Alembert formula.
1. Bài này nếu giải bằng hàm Green thì thế nào NQA nhỉ ?
Comment by viettran — March 24, 2009 @ 20:27
• Nếu thế cần phải tìm hàm Green đã, việc này chắc ko dễ 😦
Comment by Ngô Quốc Anh — December 14, 2009 @ 17:25
2. hi , my name nguyen , i come from USA , nice to meet you ,
Comment by vo khanh nguyen — December 31, 2009 @ 16:54
• Thanks for coming to my blog 🙂
Comment by Ngô Quốc Anh — January 2, 2010 @ 15:20
3. In the odd-dimensional space ($n \geq 3$)
http://arxiv.org/abs/0904.3252
and in the even case, implies by using Hadamard’s method of descent.
Comment by Tuan Minh — January 14, 2010 @ 23:40
• Aha, that’s a good and new work, thanks Minh 🙂
Comment by Ngô Quốc Anh — January 14, 2010 @ 23:44
4. In fact, this idea is mentioned in the well-known book of Stein in the case $n=3$ (by using FT/Bessel function), Stein also gave an exercise for the general case. The Torchinsky’s inverse formula for $\dfrac{\sin(R\xi)}{|\xi|}$ is nice!
Comment by Tuan Minh — January 15, 2010 @ 1:53
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 1,826 | 4,852 | {"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": 19, "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-2022-05 | latest | en | 0.541982 |
https://libraryguides.centennialcollege.ca/c.php?g=645085&p=5124707 | 1,726,203,183,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651507.67/warc/CC-MAIN-20240913034233-20240913064233-00606.warc.gz | 331,578,783 | 12,542 | # Math help from the Learning Centre
This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.
## Introduction
Question: Would you rather receive $10,000 a day for a month or one penny ($0.01) that doubles every day for a month?
You might choose the $10,000 at first, but you'll see that with the magic of compounding, the doubling penny option would earn you a whopping$10,737,418 while the $10,000 every day would only make you$310,000 richer. This means that the penny would earn almost 35X the amount that the $10,000 did! ## Terminology & Formulas Interest that is earned on both the initial principal amount and the interest accrued in previous periods during the investment/loan term is called compound interest. In other words, it can be seen as "interest on interest" - hence the name compound interest. Unlike simple interest, compound interest grows exponentially (while simple interest grows linearly), making it a common method of earning interest when investing or borrowing money. To calculate compound interest, we have to consider the original amount of money, called principal (present value), the time (period over which the money is being used), the length and number of compounding periods (details below), and the (compound) interest rate. Taking these factors into account, we can use the following formula to calculate the maturity/future value of an amount earning compound interest: $$S=P(1+i)^n$$, where: • $$S$$ is the maturity/future value, • $$P$$ is the principal/present value, • $$i$$ is the periodic rate of interest, and • $$n$$ is the total number of compounding periods. Tip: We can also rearrange the above formula to solve for $$P$$, the principal (present value): $$P=S(1+i)^{-n}$$ Compounding Periods The following table shows common compounding frequencies/periods, their durations, how many times each period occurs per year, and how to calculate the interest rates per compounding period given the annual interest rate: ## Application Examples The following examples will demonstrate how we can use the formulas introduced in the previous section. Example 1 Example: Marina is deciding between two investment opportunities for the next six years. Investment A pays 8.4% p.a. compounded bi-weekly and Investment B pays 7.5% p.a. compounded monthly. If she has$25,000 to invest, which investment option should she choose and how much more interest will she earn for that option?
Solution
First, we can identify the information that is given in the problem. For both options, we have:
$$P=25,000, t=6$$ years
Investment A Investment B Given information: $$i=\frac{0.084}{26}, n=26 \times 6 = 156$$ Calculation: $$S=25000(1+\frac{0.084}{26})^{156}=41349.63$$ Maturity Value: $$41,349.63$$ Interest Earned: $$41,349.63-25,000=$$ $$16,349.63$$ Given information: $$i=\frac{0.075}{12}=0.00625,n=12 \times 6=72$$ Calculation: $$S=25000(1+0.00625)^{72}=39152.94$$ Maturity Value: $$39,152.94$$ Interest Earned: $$39,152.94-25,000=$$ $$14,152.94$$
Comparing the two maturity values we get for both options, we find that Marina should choose Investment A, and she will earn $$16,349.63-14,152.94=$$ $$2196.69$$ more in interest with this option.
Example 2
Example: What amount must be invested at 4.75% compounded quarterly for 6 years to reach a maturity value of $15,000? Solution Since we're looking for the amount invested, we are determining the principal amount or $$P$$. First, we can identify the information that is given in the problem: $$S=15,000, i=\frac{0.0475}{4}=0.011875, n=6\times 4=24$$ Substituting these values into the compound interest formula gives: $$15000=P(1+0.011875)^{24}$$ Then, we can isolate for $$P$$: $$P=\frac{15000}{(1+0.011875)^{24}}$$ $$=11299.17$$ So, $$11, 299.17$$ must be invested to reach the required maturity value. Example 3 Example: Debts of$1250 due 6 months from today and \$4725 due 4 years from today are to be settled by a single payment 1.5 years from today. If money is worth 8.75% compounded bi-annually, what is the size of the single equivalent payment?
Solution
See the video below for the solution!
Tip: This solution will use the ideas of equivalent values and setting focal dates (To review these concepts, see Equivalent Values). | 1,118 | 4,377 | {"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} | 4.84375 | 5 | CC-MAIN-2024-38 | latest | en | 0.936535 |
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Activity Discussion Math Math Reply To: Math
• ### Aashutosh
Member
June 2, 2021 at 8:22 pm
0
Pythagoras theorem states that “The square of the hypotenuse side is equal to the sum of squares of the other two sides in a right-angle triangle“. The longest side of the right-angle triangle is called the hypotenuse. Because it is opposite to the angle of 90 degrees. The other two sides are perpendicular and base.
Let us consider the right-angle triangle ABC with 90 degrees at B and draw a perpendicular BD meeting AC at D as shown in the photo.
According to the question: △ADB ~ △ABC
Therefore, AD/AB=AB/AC (corresponding sides of similar triangles)
Or, AB^2 = AD × AC ……………………………..……..(1)
Also, △BDC ~△ABC
Therefore, CD/BC=BC/AC (corresponding sides of similar triangles)
Or, BC^2= CD × AC ……………………………………..(2)
Adding the equations (1) and (2) we get,
AB^2 + BC^2 = AD × AC + CD × AC
AB^2 + BC^2 = AC (AD + CD)
Since, AD + CD = AC
Therefore, AC^2 = AB^2 + BC^2
Hence, the Pythagoras theorem is proved.
+ | 306 | 1,049 | {"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} | 4.21875 | 4 | CC-MAIN-2023-40 | latest | en | 0.855592 |
https://themathmompuzzles.blogspot.com/2014/01/a-puzzle-that-99-will-fail-to-answer.html?showComment=1403624209783 | 1,618,493,083,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038085599.55/warc/CC-MAIN-20210415125840-20210415155840-00121.warc.gz | 618,953,374 | 22,772 | ## Friday, January 10, 2014
### A puzzle that 99% will fail to answer.
This came from my son who got it from someone else at school.
I solved it quickly and texted him the answer just as I was getting into my car for work.
He wrote: "Wrong" and sent me a ridiculous solution.
I typed: "Ohh boy, how could you miss it! I will beat you up at home:)" and started driving.
Stuck in traffic I thought back to it and the 99% that are getting it wrong and suddenly realized that his answer is in fact correct!
I texted: "Apology! Now that my head has cooled off I see that you were right."
He wrote: "Hahahahahaha".
No one at my work got it right either. Could you?
There is at least two solutions. Let's call them: my and my son's. Both are kind of correct. But his is more original.
Your thoughts and suggestions are accepted any time until midnight Eastern Time on Sunday, on our Family Puzzle Marathon.
Jody said...
26?
Mike said...
5=26! 2+5+3=10; 3+10+4=17; 4+17+5=26....
Leah said...
My first thought was to solve it as an algorithm and I got 5=26. x*x+1 (2*2+1=5, 3*3+1=10, 4*4+1=17)
But then, duh! 2=5, so 5=2! Admittedly, my son noticed that answer before me!
Anonymous said...
5=2; isn't an equal sign reflexive!?
Laura W said...
SteveGoodman18 said...
Most will answer 26, as the pattern appears to be 1 more than the square of the number. However, if the reflexive property holds, if 2 = 5, then 5 = 2.
tom said...
The two answers I like are both 26.
Solution #1 is "the square, plus one."
Solution #2 is just the series, +5, +7, and then +9.
renee said...
I'm going with 26 (x=y2 + 1), which I'm also guessing was your answer.
Well, the trap that you presumably fell into is noticing the pattern in each pair of numbers (call them A and B) - each 'B' number is 1 higher than the square of the 'A' number (so b = a^2 +1).
However, since the first thing we are told is '2 = 5' we therefore know that '5 = 2' ^_^
Jerome said...
2 or 26
25 because the pattern is n^2 + 1
2 because an equal sign means that if 2 is equivalent to 5 then 5 must be equivalent to 2.
I don't consider either one of them correct. Either the number system is getting tortured (2 and 5 are not equivalent) or the equal sign is. The whole premise of 2 being the same as 5 or 5 the same as 26 is a disobedience of the standard meaning of =.
But then I belong to the Bah Humbug school of thought. I might even be a charter member.
SN said...
Solution 1:
5 = 26 (square the number and add 1 to it)
Solution 2:
5 = 2 (from line 1)
Anonymous said...
Maria.
I imagine that my answer is like yours. I got 19. Each number increases by 5 plus 2 above the preceding one.
Gurubandhu
Katrina said...
I solved it as 26. If you look at the sequence, the answer seems to be adding the previous equations numbers, plus the current equations number i.e. 2 = 5, 3 = 10 (2+3+5), 4=17 (3+4+10), 5 = 26 (4+5+17).
Showing this to my 11 yo, he said that 5 = 2 because it said it in the first equality.
Anonymous said...
Nicolas said...
I would be tempted to say that since the first line states 2=5, we should also have 5=2.
Anonymous said...
5 could equal 2 since you already said that once. But the next number could also be 26 since each number on the right is obtained by adding the two numbers on the left to it. 10 = 5 +(2+3), 17 = 10 + (3+4), so 26 = 17 + (4+5).
So 2 or 26, take your pick.
Anonymous said...
My answer is 26. And I got it by squaring the first number plus one.
Jessica K. via email
Anonymous said...
Kim via email:
Hi Maria,
Been busy, but saw today's email, and gave it a look. Plus, I see Ilya (who I know) is only one answer behind me!
I posted online, but I seem to recall that when I was doing these every week, my posts didn't always go through, so here's my answer:
26. In each case, the number to the right of the equals sign = the sum of the numbers in the equation above it and the number to the left of the equals sign (e.g., 2+5+3 = 10; 3+10+4=17, so 4+17+5=26)
BTW, my answer is also works for x^2+1.
Which is not a coincidence.
What we have is:
x --> x^2 + 1
so the next line is x+1 --> (x+1)^2 + 1 = x^2 + 2x + 2
and it would also equal (summing the three components), x + x^2 + 1 + (x+1) -> x^2 + 2x + 2!
So, I'm assuming that's what you got. What did your son get?
a bit later in another email:
The number on the left side is the number of prime numbers to add.
The first five prime numbers are 2, 3, 5, 7, 11
N sum of first n prime numbers
2 5
3 10
4 17
5 28
and a few hours later:
Still thinking about you. I have one other possible answer, and I bet THIS is your son's:
We know from the first equation that 2=5. Therefore, 5=2.
Susan said...
I keep coming up with 26.
A. If the column on the right is the addition of progression of odd numbers, starting with +5, then +7, the next would be 17+9, or 26.
B. If the pattern is Y=X^2 +1, then 5^2 + 1 is 26.
Jerome's wife said...
I tried both directions in attempting to come to the sums of each number; the horizontal and the vertical summations. The horizontal attempt fell through as I could not establish a pattern, but vertically, travelling downward from each number to the next I did see a pattern. The answer is 5 = 26. I noticed that from sum 5 to sum 10 were 5 steps or increments and from sum 10 to sum 17 were 7 steps or increments. Following through to arrive at the answer of 26 I added 9 steps or increments to 17. I knew this was correct because then testing the theory by noticing that there are 2 increments between 5 and 7 increments, it is logical to assume that there is also 2 more increments, to make the steps look like 5,7 and 9. Then 9 is correct by adding 17 to 9 equals 26.
Ilya said...
The notation with equality sign is somewhat strange, I interpreted it as "function of", i.e. F(2)=5, etc. One solution that fits is F(x)=x*x+1, leading to F(5)=26.
Anonymous said...
My first answer is 26. All the numbers are squared and increased by 1.
While trying to think of another answer, the only one that makes sense at the moment is: if 2 = 5 then 5 should be equal to 2.
Lulu
Annie said...
I would say there are two answers:
If it is looked at as a progression then the answer would be 26. (Add 5, then 7, then 9).
If each statement is looked at as a separate fact, then the answer would be 2, 2=5, so 5= 2!
Maria said...
Ok. As you see there are two very different answers. 26 and 2.
The easier one is 2 and it is harder to find it. But it is written up there on the first line that 2=5, so it should be that 5=2.
2 = 5
3 = 10
4 = 17
5 = ?
We see that mathematically it is wrong because 2 is not equal to 5 etc. So someone is lying: either numbers or the "=" sign. We are either in a kingdom where = is not equal but more like correspondence sign but numbers are real numbers. Or we are in a kingdom where numbers are not what we expect them to be but "=" is equal.
In the first kingdom, we all rushed to find the pattern that led us to the answer 26. And there are a few ways to find it!
In the second kingdom , "=" is "=" and if 2=5 then 5=2!
Pretty amazing that quite a few of you saw both answers. It is very difficult to change state of mind and move between those kingdoms. Those who got both answers will receive 2 puzzle points, those who got one will receive 1 point.
Gurubandhu - I am sorry but I do not understand the 19th. Your solution still comes to 26 if I understood it correctly.
Have a great week!
Jerome said...
I have to confess that I never would have found the reflexive property had you not said there were 2 answers. These problems you come up with are just wonderful even if I do crab about them sometimes.
Anonymous said...
32
mohamed el fouhil said...
If 2=5
then 5=2
mohamed el fouhil said...
If 2=5
Then 5=2
Anonymous said...
Interesting! I came up with 26, but, it's an odd use of equal signs, as has been pointed out. Missed the reflexivity solution completely.
As far as the "other solution" goes, I kept looking to the odd language on the original poster image. In fact if we're being quite logical about it, "99% will fail to answer" means that only 1% (presumably of people who see the puzzle) will actually hazard an answer, whether correct or not. It doesn't say that the 99% failed to answer _correctly_, after all.
I also notice that the puzzle starts with the word "if" but there's no "then," or perhaps the "then" is implied. Otherwise a wise guy could say "there's no answer because there's no question!"
All-in-all an interesting mishmash of conventions and assumption-baiting.
Margaret and Fiona
Unknown said...
In fact other than 2 and 26 even 24 is a possible answer!! Let me explain it..
2=5 -----> 2+3=5 and 3 is a prime number
3=10----->3+7=10 and 7 is a prime number and observe the pattern of prime numbers added here .. after 3, 5 is not added.
4=17----> 4+13=17 13 is a prime number and in between 7 and 13 , prime number 11 is left.
now, 5=?
5=5+19=24.. leaving prime number 17 in between 13 and 19..
But, answer 2 is more convincing. | 2,583 | 9,062 | {"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} | 3.859375 | 4 | CC-MAIN-2021-17 | latest | en | 0.977077 |
http://www.creatingrealmathematicians.com/year-planner-years-3-4 | 1,656,510,257,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103639050.36/warc/CC-MAIN-20220629115352-20220629145352-00283.warc.gz | 76,605,995 | 13,942 | ### Year Planner. Years 3/4
Year Planner Years 3 & 4 Terms 1-4
Each term revisits concepts and teaching and learning takes individual students to higher level of understanding.
Individual student growth documented against VELS Progression Points for each topic each term.
Addition Strategies include Counting forwards on 100’s Chart, Open Number Lines, Partial Sums addition, Counting on Addition using Place Value (Jump strategy). Subtraction Strategies include Counting Backwards on 100’s Chart, Open Number Lines, Counting Up or Counting Back to find difference, Counting Up using Place Value. Multiplication Strategies include Making Arrays, Multiplication as Repeated Addition. Partial Products Multiplication and the Area Model for Multiplication (box method). Division Strategies include division as sharing, what to do with left overs, sharing one place value at a time (left to right). Inverse of multiplication (2x?=6). Fractions Making explicit links to division and multiplication. (½ of 6 is 3, 6÷2=3, 2x3=6, 3+3=6, 6-3-3=0). Denominator is number of groups shared between. Numerator number of groups selected. (Group is the whole & One is the whole). Shape Location Chance Data Measurement. Decimals & Percent Recognise and describe polygons students sort lines, shapes and solids according to key features. Recognise and describe the directions of lines as vertical, horizontal or diagonal. They use grid references (for example, B5 on a street directory) to specify location and compass bearings to describe directions. Locate and identify places on maps and diagrams. They give travel directions and describe positions using simple compass directions (for example, N for North) and grid references on a street directory. Investigate natural variability in chance events and order them from least likely to most likely. Compare the likelihood of everyday events (for example, the chances of rain and snow). They describe the fairness of events in qualitative terms (likely, unlikely) Conduct experiments and collect data to construct simple frequency They use simple two-way tables (karnaugh maps) to sort non-numerical data. Measure the attributes of everyday objects and events using formal (for example, metres and centimetres) and informal units(for example, pencil lengths). Recognise and name common three-dimensional shapes such as spheres, prisms and pyramid. Identify edges, vertices and faces. Use nets to create three-dimensional shapes and explore them by counting edges, faces and vertices They use local and larger-scale maps to locate places and describe suitable routes between them. Plan and conduct chance experiments (for example, using colours on a spinner) and display the results of these experiments. Interpret timetables and calendars in relation to familiar events. Tell the time using analogue and digital clocks and relate familiar activities to the calendar Visualise and draw simple solids as they appear from different positions. Use two-dimensional nets, cross-sections and simple projections to represent simple three-dimensional shapes. Recognise different types of data: non-numerical (categories), separate numbers (discrete) or points on an unbroken number line (continuous). Explore the concept of angle as turn (for example, using clock hands) and as parts of shapes and objects (for example, at the vertices of polygons). Recognise angles are the result of rotation of lines with a common end-point Investigate simple transformations (reflections, slides and turns) to create tessellations and designs. Follow instructions to produce simple tessellations (for example, with triangles, rectangles, hexagons) and puzzles such as tangrams. Use a column or bar graph to display the results of an experiment (for example, the frequencies of possible categories). · Contexts include Whole Number, Length, Time, Mass, Volume, Money. · Money used to develop concepts of Decimals within Operations. · All strategies emphasise use of Place Value components. · Teachers need to make explicit the links between the four operations and Fractions. · All activities emphasise the use of Number in student relevant authentic open contexts. · Differentiation built into every lesson extending understanding for all students.
Warm Up Activities emphasise the following throughout the year.
Patterns (Counting and Shape), Probability games, Automatic recall of simple number facts (Doubling, adding two one digit numbers, compliments of ten, odd and even, etc), Matching Analogue, Digital and Written time.
Problem Solving activities aligned to topics under investigation presented to students weekly to further develop Proficiency Strands of Australian Curriculum (Understanding, Fluency, Problem Solving & Reasoning). Resources include Exemplars, Pictures/Numbers/Words.com)
Literacy emphasised for each topic. Students to become fluent in vocabulary of each topic.
Written Share/Reflections of developed understandings are standard practice and completed at the end of every lesson.
All lessons to explicitly identify to students the Learning Intentions for each lesson.
## Mathematics - Level 3
### Learning focus
As students work towards the achievement of Level 3 standards in Mathematics, they recognise and explore patterns in numbers and shape. They increasingly use mathematical terms and symbols to describe computations, measurements and characteristics of objects.
In Number, students use structured materials to explore place value and order of numbers to tens of thousands. They skip count to create number patterns. They use materials to develop concepts of decimals to hundredths. They use suitable fraction material to develop concepts of equivalent fraction and to compare fraction sizes. They apply number skills to everyday contexts such as shopping. They extend addition and subtraction computations to three digit numbers. They learn to multiply and divide by single digit numbers.
In Space, students sort lines, shapes and solids according to key features. They use nets to create three-dimensional shapes and explore them by counting edges, faces and vertices. They visualise and draw simple solids as they appear from different positions. They investigate simple transformations (reflections, slides and turns) to create tessellations and designs. They explore the concept of angle as turn (for example, using clock hands) and as parts of shapes and objects (for example, at the vertices of polygons). They use grid references (for example, A5 on a street directory) to specify location and compass bearings to describe directions. They use local and larger-scale maps to locate places and describe suitable routes between them.
In Measurement, chance and data, students measure the attributes of everyday objects and events using formal (for example, metres and centimetres) and informal units(for example, pencil lengths). Students tell the time using analogue and digital clocks and relate familiar activities to the calendar. Students investigate natural variability in chance events and order them from least likely to most likely. Students conduct experiments and collect data to construct simple frequency graphs. They use simple two-way tables (karnaugh maps) to sort non-numerical data.
In Structure, students use structured material (in tens, hundreds and thousands) to develop ideas about multiplication by replication and division by sharing. They recognise the possibility of remainders when dividing. They learn to use number properties to support computations (for example, they use the commutative and associative properties for adding or multiplying three numbers in any order or combination). They investigate the distributive property to develop methods of multiplication and division by single digit whole numbers. They learn to use and describe simple algorithms for computations. They use simple rules to generate number patterns (for example, ‘the next term in the sequence is two more than the previous term’). They create and complete number sentences using whole numbers, decimals and fractions.
When Working mathematically, students use mathematical symbols (for example, brackets, division and inequality, the words and, or and not). Students develop and test ideas (conjectures) across the content of mathematical experience. For example:
• in Number, the size and type of numbers resulting from computations
• in Space, the effects of transformations of shapes
• in Measurement, chance and data, the outcomes of random experiments and inferences from collected samples.
Students learn to recognise practical applications of mathematics in daily life, including shopping, travel and time of day. They identify the mathematical nature of problems for investigation. They choose and use learned facts, procedures and strategies to find solutions. They use a range of tools for mathematical work, including calculators, computer drawing packages and measuring tools.
### National Statements of Learning
This learning focus statement, with the following elaboration, incorporates the Year 3 National Statement of Learning for Mathematics.
Elaboration:
They recognise angles … as parts of shapes and objects …
### Standards
#### Number
At Level 3, students use place value (as the idea that ‘ten of these is one of those’) to determine the size and order of whole numbers to tens of thousands, and decimals to hundredths. They round numbers up and down to the nearest unit, ten, hundred, or thousand. They develop fraction notation and compare simple common fractions such as 3/4 > 2/3 using physical models. They skip count forwards and backwards, from various starting points using multiples of 2, 3, 4, 5, 10 and 100.
They estimate the results of computations and recognise whether these are likely to be over-estimates or under-estimates. They compute with numbers up to 30 using all four operations. They provide automatic recall of multiplication facts up to 10 × 10.
They devise and use written methods for:
• whole number problems of addition and subtraction involving numbers up to 999
• multiplication by single digits (using recall of multiplication tables) and multiples and powers of ten (for example, 5 × 100, 5 × 70 )
• division by a single-digit divisor (based on inverse relations in multiplication tables).
They devise and use algorithms for the addition and subtraction of numbers to two decimal places, including situations involving money. They add and subtract simple common fractions with the assistance of physical models.
#### Space
At Level 3, students recognise and describe the directions of lines as vertical, horizontal or diagonal. They recognise angles are the result of rotation of lines with a common end-point. They recognise and describe polygons. They recognise and name common three-dimensional shapes such as spheres, prisms and pyramids. They identify edges, vertices and faces. They use two-dimensional nets, cross-sections and simple projections to represent simple three-dimensional shapes. They follow instructions to produce simple tessellations (for example, with triangles, rectangles, hexagons) and puzzles such as tangrams. They locate and identify places on maps and diagrams. They give travel directions and describe positions using simple compass directions (for example, N for North) and grid references on a street directory.
#### Measurement, chance and data
At Level 3, students estimate and measure length, area, volume, capacity, mass and time using appropriate instruments. They recognise and use different units of measurement including informal (for example, paces), formal (for example, centimetres) and standard metric measures (for example, metre) in appropriate contexts. They read linear scales (for example, tape measures) and circular scales (for example, bathroom scales) in measurement contexts. They read digital time displays and analogue clock times at five-minute intervals. They interpret timetables and calendars in relation to familiar events. They compare the likelihood of everyday events (for example, the chances of rain and snow). They describe the fairness of events in qualitative terms. They plan and conduct chance experiments (for example, using colours on a spinner) and display the results of these experiments. They recognise different types of data: non-numerical (categories), separate numbers (discrete), or points on an unbroken number line (continuous).They use a column or bar graph to display the results of an experiment (for example, the frequencies of possible categories).
#### Structure
At Level 3, students recognise that the sharing of a collection into equal-sized parts (division) frequently leaves a remainder. They investigate sequences of decimal numbers generated using multiplication or division by 10. They understand the meaning of the ‘=’ in mathematical statements and technology displays (for example, to indicate either the result of a computation or equivalence). They use number properties in combination to facilitate computations (for example, 7 + 10 + 13 = 10 + 7 + 13 = 10 + 20). They multiply using the distributive property of multiplication over addition (for example, 13 × 5 = (10 + 3) × 5 = 10 × 5 + 3 × 5). They list all possible outcomes of a simple chance event. They use lists, venn diagrams and grids to show the possible combinations of two attributes. They recognise samples as subsets of the population under consideration (for example, pets owned by class members as a subset of pets owned by all children). They construct number sentences with missing numbers and solve them.
#### Working mathematically
At Level 3, students apply number skills to everyday contexts such as shopping, with appropriate rounding to the nearest five cents. They recognise the mathematical structure of problems and use appropriate strategies (for example, recognition of sameness, difference and repetition) to find solutions.
Students test the truth of mathematical statements and generalisations. For example, in:
• number (which shapes can be easily used to show fractions)
• computations (whether products will be odd or even, the patterns of remainders from division)
• number patterns (the patterns of ones digits of multiples, terminating or repeating decimals resulting from division)
• shape properties (which shapes have symmetry, which solids can be stacked)
• transformations (the effects of slides, reflections and turns on a shape)
• measurement (the relationship between size and capacity of a container).
Students use calculators to explore number patterns and check the accuracy of estimations. They use a variety of computer software to create diagrams, shapes, tessellations and to organise and present data. | 2,940 | 14,866 | {"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-2022-27 | longest | en | 0.855621 |
https://schoollearningcommons.info/question/how-to-prove-a-2-b-2-a-2-b-2-ab-24945330-98/ | 1,632,715,984,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780058263.20/warc/CC-MAIN-20210927030035-20210927060035-00295.warc.gz | 519,760,616 | 13,157 | ## How to prove (a /2+ b)^2(a /2- b)^2= ab
Question
How to prove (a /2+ b)^2(a /2- b)^2=
ab
in progress 0
1 month 2021-08-21T22:25:28+00:00 1 Answer 0 views 0
We can show that the right side of the equation is equal to the left side, by expanding it out and then factorising:
( a+b)2+(a−b)2
= (a+b)(a+b)+(a−b)(a−b)
= (a2+ab+ab+b2)+(a2−ab−ab+b2)
= a2+2ab+b2+a2−2ab+b2
= 2a2+2b2
= 2(a2+b2)
Therefore, it is shown that
2(a2+b2)=(a+b)2+(a−b)2 | 210 | 453 | {"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-2021-39 | latest | en | 0.694171 |
https://mathoverflow.net/questions/240166/inequalities-for-marginals-of-distribution-on-hyperplane | 1,718,476,608,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00696.warc.gz | 344,710,598 | 24,854 | # Inequalities for marginals of distribution on hyperplane
Let $H = \{ (a,b,c) \in \mathbb{Z}_{\geq 0}^3 : a+b+c=n \}$. If we have a probability distribution on $H$, we can take its marginals onto the $a$, $b$ and $c$ variables and obtain three probability distributions $\alpha$, $\beta$ and $\gamma$ in $\mathbb{R}^{n+1}$. The set of possible $\alpha$, $\beta$ and $\gamma$ is a polytope in $\mathbb{R}^{3n+3}$. Does anyone know a list of defining inequalities for it?
For those who don't like the probability language: Let $p_{ijk} \in \mathbb{R}_{\geq 0}^H$ with $\sum p_{ijk}=1$. Set $\alpha_i = \sum_{j} p_{ij(n-i-j)}$, $\beta_j = \sum_k p_{(n-j-k)jk}$ and $\gamma_k = \sum_i p_{i(n-i-k)k}$. What are the inequalities defining the possible vectors $\alpha$, $\beta$ and $\gamma$?
• The polytope in question (call it $P_n$) is the convex hull of $e_a + f_b + g_c$, where $e_0,\dots,e_n,f_0,\dots,f_n,g_0,\dots,g_n$ is a basis for ${\mathbb R}^{3n+3}$ and $(a,b,c)$ ranges over $H$. There appear to be some recursive relations that allow one to describe $P_{n+1}$ or $P_{2n}$ in terms of $P_n$ which may be helpful. Commented Jun 7, 2016 at 16:59
• The defining inequalities correspond to weights $r_0,\dots,r_n,s_0,\dots,s_n,t_0,\dots,t_n$ such that $r_a + s_b + t_c \geq 0$ for all $(a,b,c) \in H$, with equality holding in some "spanning" set of $H$. This looks somewhat combinatorially unappetising, though. Commented Jun 7, 2016 at 17:02
• For n=2. If the variables are $\alpha_0, \alpha_1, \alpha_2, \beta_0, \beta_1, \beta_2, \gamma_0, \gamma_1, \gamma_2$, then the polytope is defined by 6 inequalities and 4 equalities:$$0\le \alpha_2,\beta_2,\gamma_2,d,e,f;$$ $$\alpha_1=e+f;\ \ \beta_1=d+f;\ \ \gamma_1=d+e;$$ $$\alpha_0+\beta_0+\gamma_0=\alpha_2+\beta_2+\gamma_2+1$$ where $d,e,f$ abbreviate $\alpha_0-\beta_2-\gamma_2$, $\beta_0-\alpha_2-\gamma_2$, $\gamma_0-\alpha_2-\beta_2$.
– user44143
Commented Jan 15, 2018 at 5:18
While the literal question I asked is still open, Luke Pebody has proven an important relevant result: If $$\alpha_0 \geq \alpha_1 \geq \cdots \geq \alpha_n$$, and the same for $$\beta$$ and $$\gamma$$, and $$\sum \alpha_j + \sum \beta_j + \sum \gamma_k = 3$$, then there is a probability distribution $$\pi$$ with marginals $$\alpha$$, $$\beta$$, $$\gamma$$. This constructs a large polytope inside the one which this question asks for a characterization of. | 845 | 2,401 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.59375 | 4 | CC-MAIN-2024-26 | latest | en | 0.777733 |
https://en.m.wikipedia.org/wiki/Implicit_surface | 1,657,213,760,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104495692.77/warc/CC-MAIN-20220707154329-20220707184329-00410.warc.gz | 288,786,052 | 23,959 | # Implicit surface
In mathematics, an implicit surface is a surface in Euclidean space defined by an equation
Implicit surface torus (R=40, a=15).
Implicit surface of genus 2.
Implicit non-algebraic surface (wineglass).
${\displaystyle F(x,y,z)=0.}$
An implicit surface is the set of zeros of a function of three variables. Implicit means that the equation is not solved for x or y or z.
The graph of a function is usually described by an equation ${\displaystyle z=f(x,y)}$ and is called an explicit representation. The third essential description of a surface is the parametric one: ${\displaystyle (x(s,t),y(s,t),z(s,t))}$, where the x-, y- and z-coordinates of surface points are represented by three functions ${\displaystyle x(s,t)\,,y(s,t)\,,z(s,t)}$ depending on common parameters ${\displaystyle s,t}$. Generally the change of representations is simple only when the explicit representation ${\displaystyle z=f(x,y)}$ is given: ${\displaystyle z-f(x,y)=0}$ (implicit), ${\displaystyle (s,t,f(s,t))}$ (parametric).
Examples:
1. plane ${\displaystyle x+2y-3z+1=0.}$
2. sphere ${\displaystyle x^{2}+y^{2}+z^{2}-4=0.}$
3. torus ${\displaystyle (x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(x^{2}+y^{2})=0.}$
4. Surface of genus 2: ${\displaystyle 2y(y^{2}-3x^{2})(1-z^{2})+(x^{2}+y^{2})^{2}-(9z^{2}-1)(1-z^{2})=0}$ (see diagram).
5. Surface of revolution ${\displaystyle x^{2}+y^{2}-(\ln(z+3.2))^{2}-0.02=0}$ (see diagram wineglass).
For a plane, a sphere, and a torus there exist simple parametric representations. This is not true for the fourth example.
The implicit function theorem describes conditions under which an equation ${\displaystyle F(x,y,z)=0}$ can be solved (at least implicitly) for x, y or z. But in general the solution may not be made explicit. This theorem is the key to the computation of essential geometric features of a surface: tangent planes, surface normals, curvatures (see below). But they have an essential drawback: their visualization is difficult.
If ${\displaystyle F(x,y,z)}$ is polynomial in x, y and z, the surface is called algebraic. Example 5 is non-algebraic.
Despite difficulty of visualization, implicit surfaces provide relatively simple techniques to generate theoretically (e.g. Steiner surface) and practically (see below) interesting surfaces.
## Formulas
Throughout the following considerations the implicit surface is represented by an equation ${\displaystyle F(x,y,z)=0}$ where function ${\displaystyle F}$ meets the necessary conditions of differentiability. The partial derivatives of ${\displaystyle F}$ are ${\displaystyle F_{x},F_{y},F_{z},F_{xx},\ldots }$ .
### Tangent plane and normal vector
A surface point ${\displaystyle (x_{0},y_{0},z_{0})}$ is called regular if and only if the gradient of ${\displaystyle F}$ at ${\displaystyle (x_{0},y_{0},z_{0})}$ is not the zero vector ${\displaystyle (0,0,0)}$ , meaning
${\displaystyle (F_{x}(x_{0},y_{0},z_{0}),F_{y}(x_{0},y_{0},z_{0}),F_{z}(x_{0},y_{0},z_{0}))\neq (0,0,0)}$ .
If the surface point ${\displaystyle (x_{0},y_{0},z_{0})}$ is not regular, it is called singular.
The equation of the tangent plane at a regular point ${\displaystyle (x_{0},y_{0},z_{0})}$ is
${\displaystyle F_{x}(x_{0},y_{0},z_{0})(x-x_{0})+F_{y}(x_{0},y_{0},z_{0})(y-y_{0})+F_{z}(x_{0},y_{0},z_{0})(z-z_{0})=0,}$
and a normal vector is
${\displaystyle \mathbf {n} (x_{0},y_{0},z_{0})=(F_{x}(x_{0},y_{0},z_{0}),F_{y}(x_{0},y_{0},z_{0}),F_{z}(x_{0},y_{0},z_{0}))^{T}.}$
### Normal curvature
In order to keep the formula simple the arguments ${\displaystyle (x_{0},y_{0},z_{0})}$ are omitted:
${\displaystyle \kappa _{n}={\frac {\mathbf {v} ^{\top }H_{F}\mathbf {v} }{\|\operatorname {grad} F\|}}}$
is the normal curvature of the surface at a regular point for the unit tangent direction ${\displaystyle \mathbf {v} }$ . ${\displaystyle H_{F}}$ is the Hessian matrix of ${\displaystyle F}$ (matrix of the second derivatives).
The proof of this formula relies (as in the case of an implicit curve) on the implicit function theorem and the formula for the normal curvature of a parametric surface.
## Applications of implicit surfaces
As in the case of implicit curves it is an easy task to generate implicit surfaces with desired shapes by applying algebraic operations (addition, multiplication) on simple primitives.
Equipotential surface of 4 point charges
### Equipotential surface of point charges
The electrical potential of a point charge ${\displaystyle q_{i}}$ at point ${\displaystyle \mathbf {p} _{i}=(x_{i},y_{i},z_{i})}$ generates at point ${\displaystyle \mathbf {p} =(x,y,z)}$ the potential (omitting physical constants)
${\displaystyle F_{i}(x,y,z)={\frac {q_{i}}{\|\mathbf {p} -\mathbf {p} _{i}\|}}.}$
The equipotential surface for the potential value ${\displaystyle c}$ is the implicit surface ${\displaystyle F_{i}(x,y,z)-c=0}$ which is a sphere with center at point ${\displaystyle \mathbf {p} _{i}}$ .
The potential of ${\displaystyle 4}$ point charges is represented by
${\displaystyle F(x,y,z)={\frac {q_{1}}{\|\mathbf {p} -\mathbf {p} _{1}\|}}+{\frac {q_{2}}{\|\mathbf {p} -\mathbf {p} _{2}\|}}+{\frac {q_{3}}{\|\mathbf {p} -\mathbf {p} _{3}\|}}+{\frac {q_{4}}{\|\mathbf {p} -\mathbf {p} _{4}\|}}.}$
For the picture the four charges equal 1 and are located at the points ${\displaystyle (\pm 1,\pm 1,0)}$ . The displayed surface is the equipotential surface (implicit surface) ${\displaystyle F(x,y,z)-2.8=0}$ .
### Constant distance product surface
A Cassini oval can be defined as the point set for which the product of the distances to two given points is constant (in contrast, for an ellipse the sum is constant). In a similar way implicit surfaces can be defined by a constant distance product to several fixed points.
In the diagram metamorphoses the upper left surface is generated by this rule: With
{\displaystyle {\begin{aligned}F(x,y,z)={}&{\Big (}{\sqrt {(x-1)^{2}+y^{2}+z^{2}}}\cdot {\sqrt {(x+1)^{2}+y^{2}+z^{2}}}\\&\qquad \cdot {\sqrt {x^{2}+(y-1)^{2}+z^{2}}}\cdot {\sqrt {x^{2}+(y+1)^{2}+z^{2}}}{\Big )}\end{aligned}}}
the constant distance product surface ${\displaystyle F(x,y,z)-1.1=0}$ is displayed.
Metamorphoses between two implicit surfaces: a torus and a constant distance product surface.
### Metamorphoses of implicit surfaces
A further simple method to generate new implicit surfaces is called metamorphosis of implicit surfaces:
For two implicit surfaces ${\displaystyle F_{1}(x,y,z)=0,F_{2}(x,y,z)=0}$ (in the diagram: a constant distance product surface and a torus) one defines new surfaces using the design parameter ${\displaystyle \mu \in [0,1]}$ :
${\displaystyle F(x,y,z)=\mu F_{1}(x,y,z)+(1-\mu )F_{2}(x,y,z)=0}$
In the diagram the design parameter is successively ${\displaystyle \mu =0,\,0.33,\,0.66,\,1}$ .
Approximation of three tori (parallel projection)
POV-Ray image (central projection) of an approximation of three tori.
### Smooth approximations of several implicit surfaces
${\displaystyle \Pi }$ -surfaces [1] can be used to approximate any given smooth and bounded object in ${\displaystyle R^{3}}$ whose surface is defined by a single polynomial as a product of subsidiary polynomials. In other words, we can design any smooth object with a single algebraic surface. Let us denote the defining polynomials as ${\displaystyle f_{i}\in \mathbb {R} [x_{1},\ldots ,x_{n}](i=1,\ldots ,k)}$ . Then, the approximating object is defined by the polynomial
${\displaystyle F(x,y,z)=\prod _{i}f_{i}(x,y,z)-r}$ [1]
where ${\displaystyle r\in \mathbb {R} }$ stands for the blending parameter that controls the approximating error.
Analogously to the smooth approximation with implicit curves, the equation
${\displaystyle F(x,y,z)=F_{1}(x,y,z)\cdot F_{2}(x,y,z)\cdot F_{3}(x,y,z)-r=0}$
represents for suitable parameters ${\displaystyle c}$ smooth approximations of three intersecting tori with equations
{\displaystyle {\begin{aligned}F_{1}=(x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(x^{2}+y^{2})=0,\\[3pt]F_{2}=(x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(x^{2}+z^{2})=0,\\[3pt]F_{3}=(x^{2}+y^{2}+z^{2}+R^{2}-a^{2})^{2}-4R^{2}(y^{2}+z^{2})=0.\end{aligned}}}
(In the diagram the parameters are ${\displaystyle R=1,\,a=0.2,\,r=0.01.}$ )
POV-Ray image: metamorphoses between a sphere and a constant distance product surface (6 points).
## Visualization of implicit surfaces
There are various algorithms for rendering implicit surfaces,[2] including the marching cubes algorithm.[3] Essentially there are two ideas for visualizing an implicit surface: One generates a net of polygons which is visualized (see surface triangulation) and the second relies on ray tracing which determines intersection points of rays with the surface.[4] The intersection points can be approximated by sphere tracing, using a signed distance function to find the distance to the surface.[5] | 2,740 | 8,899 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 59, "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-2022-27 | latest | en | 0.789309 |
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# How many joules of energy does a 100 watt light bulb have?
012
###### 2013-04-22 02:39:45
welll it really depends on the volasity of light energy but from my calculatios it is 900
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## Related Questions
there are 100 joules in an energy efficient light bulb 75 joules go towards the light and 25 joules go towards the heat
1) The power use of bulbs is specified in units of power, not energy. In other words, watts - which means joules per second. 2) Light-bulbs come in different capacities. 3) It is incorrect to say that the energy (or power) "is in" the light-bulb. The light-bulb simply transforms one type of energy to another, using electrical energy which is NOT stored in the light-bulb.
60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).60 watts means the same as 60 joules per second. This is the energy used; all of the energy will be given off by the light bulb (but not all of it will be radiated as visible light).
Watt means joules/second. It refers to the amount of energy a device uses, in this case. Multiply the power (in watts) by the time (in seconds) to get the energy (in joules).
It's 100 watts times 3600 seconds, that's 360,000 joules of energy. A joule is 1 watt for 1 second.
100 watts running for 60 seconds uses 100 x 60 Joules of energy. A Joule is the same as a watt-second.
Every second a 150 Watt bulb converts 150 Joules from electricity into heat and light. The number of Watts tells you how many Joules pass per second.
100 watts means 100 Joules/Second. So in 24 hours, the bulb would use 24*60*60*100 Joules. so that's 8,640,000 joules
Almost 90 % of electrical energy provided to an incandescent light bulb goes as heat and rest as light. A 100 Watt bulb puts out 100 Joules of heat per second. So - for one minute it would put out 6000 Joules (100 Watts X 60 seconds). 1 BTU (British Thermal Unit) of heat = 1055.056 Joules. So a 100 watt bulb, burning for one minute would put out 5.68 BTUs of heat. ( 6000 Joules / 1055.056 Joules) = 5.68 BTUs. Same bulb burning for one hour would generate 341 BTUs of heat.
A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.A joule is a joule, whether it be electrical energy or light energy - although commonly, lamps are not 100% efficient.On the other hand, you can't convert joules directly to watts. Watts means joules per second (joules / second), or equivalently, joules is watts times seconds.
You can't calculate how many volts with that information; you could calculate the energy - 60 watts for 15 minutes is equivalent to 54,000 joules.
1 watt= 1 joules/1 sec60 w = 60joules/minuteor=3600joules/hour
A light bulb produces light on demand when it is connected to a sufficient source of electrical energy. Many types of light bulbs also produce heat as a byproduct.
This is a very good question. A Watt is a unit of power, or energy with respect to (divided by) time, and is defined as 1 Joule per second. A 100 Watt light bulb consumes energy at a rate of 100 Joules (J) per second (s). There are 60 seconds in every minute, and 60 minutes in every hour. So also, there are 3600 seconds in every hour (60 X 60). 100 J/s X 3600s = 360,000 J (the s cancels out in the division) Your bulb consumes 360,000 Joules in one hour.
Your question is rather like asking "How many miles per hour do you do in a week?" You don't consume watts over time, it's a measure of how many joules of energy you consume over time.
Yes, there are many examples. Electrical energy to light energy (light bulb). Mechanical energy to heat energy (rub your hands together). Chemical energy to light energy (a glowstick), and on and on.
Joule is the unit of energy. Watt is the unit of power. One Joule provides one Watt for one second. A 100 Watt lamp uses 100 Joules per second.AnswerPower is the rate of doing work. Work is measured in joules, so power is measured in joules per second. However, in SI, a joule per second is given a special name: the watt. So a 100-W lamp consumes 100 J of energy per second.
Those numbers describe the power used by the two bulbs, in other words how many joules of electrical energy they use per second. The 100 watt bulb uses 40 watts more.
With current (as of 2013) technology, from best to worst efficiency, the light bulbs are basically:LED light bulbs (most efficient)Fluorescent lightsThe old-fashioned incandescent lights (worst)
In a traditional light bulb, the electrical energy is converted to heat. The filament gets hot and emits the thermal energy as light. The electrical energy itself is not directly converted to light but goes through the thermal energy stage.There are many kinds of lights and more complicated processes which are not described in this brief answer.
Energy (Joules) is equal to the mass multiplied by the speed of light squared (E=mc^2).
###### Home ElectricitySciencePhysicsUnits of MeasureMath and ArithmeticEnergyCalorie Count
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# Series Sum ISC 2014 Theory
A class SeriesSum is designed to calculate the sum of the following series:
Sum = x2 / 1! + x4 / 3! + x6 / 5! + … + xn / (n – 1)!
Some of the members of the class are given below:
Class name: SeriesSum
Data members/instance variables:
x: to store an integer number.
n: to store number of terms.
sum: double variable to store the sum of the series.
Member functions:
SeriesSum(int xx, int nn): constructor to assign x = xx and n = nn.
double findFact(int m): to return the factorial of m using recursive technique.
double findPower(int x, int y): to return x raised to the power of y using recursive technique.
void calculate(): to calculate the sum of the series by invoking the recursive functions respectively.
void display(): to display the sum of the series.
(a) Specify the class SeriesSum, giving details of the constructor, double findFact(int), double findPower(int, int), void calculate() and void display(). Define the main() function to create an object and call the functions accordingly to enable the task.
(b) State the two differences between iteration and recursion.
``````import java.io.*;
class SeriesSum{
private int x;
private int n;
private double sum;
public SeriesSum(int xx, int nn){
x = xx;
n = nn;
sum = 0.0;
}
public double findFact(int m){
if(m <= 1)
return 1.0;
else
return m * findFact(m - 1);
}
public double findPower(int x, int y){
if(y == 0)
return 1.0;
else
return x * findPower(x, y - 1);
}
public void calculate(){
for(int i = 1; i <= n; i++)
sum += findPower(x, i * 2) / findFact(i * 2 - 1);
}
public void display(){
System.out.println("Sum = " + sum);
}
public static void main(String[] args)throws IOException{
System.out.print("x = "); | 452 | 1,722 | {"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-2021-25 | latest | en | 0.58354 |
https://plainmath.net/91156/distance-between-1-6-and-4-7 | 1,670,026,151,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710916.70/warc/CC-MAIN-20221202215443-20221203005443-00411.warc.gz | 499,317,596 | 11,477 | # Distance between (1,-6) and (4,7)?
Distance between (1,-6) and (4,7)?
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The formula for calculating the distance between two points is:
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Substituting the values from the points in the problem gives:
$d=\sqrt{{\left(4-1\right)}^{2}+{\left(7--6\right)}^{2}}$
$d=\sqrt{{3}^{2}+{13}^{2}}$
$d=\sqrt{178}$
$d\cong 13.342$ | 207 | 616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 16, "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} | 3.953125 | 4 | CC-MAIN-2022-49 | latest | en | 0.843472 |
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Pairs of primes (Posted on 2019-07-27)
Imagine a bag containing cards representing all n-digit odd numbers. A random card is drawn and two new numbers are created by preceding the drawn number by each of its even neighbors.
What is the probability that each of those 2 numbers is prime?
Examples:
For n=1 there are 5 cards i.e. 1,3,5,7 and 9. Clearly only numbers 3 and 9 qualifiy since fboth 23 and 43 are primes and so are 89 and 109 & there are no other answers. So for n=1 p=0.4 is the probability we were looking for.
For n=2 I will not provide the answer but will show you one of the qualifying numbers e.g. 69, since both 6869 and 7069 are prime.
Now evaluate the correct probabilities for n=2,3, ...8,9 (or as far as your resources allow) - and you will get a sequence for which you may be credited @ OEIS.
So this time you get a task both challenging and rewarding!
GOOD LUCK...
No Solution Yet Submitted by Ady TZIDON No Rating
Comments: ( Back to comment list | You must be logged in to post comments.)
re: via computer - the start of the soln | Comment 5 of 10 |
(In reply to via computer - the start of the soln by Steven Lord)
Your first three values agree with mine. Thereafter your probabilities go up while mine go down. Going up seems counterintuitive, as primes get rarer as the order of magnitude increases.
I found only 56 occurrences among the 4500 4-digit numbers. Your probability of .0376 implies .0376*4500 ~= 169. Perhaps you could modify your program to list the 169 found, to see if I missed some or some of yours involved non-primes.
I'm thinking that if the problem is on your side (too many found), the problem might lie in the primality test:
m=(1.*i)/(1.*j)
While m is defined as an integer, the decimal points make the division take place in floating point, or real. As the conversion is implicit, there's no "kind" specification. Perhaps in the division of 8-digit or higher numbers, the integer part of the quotient is not actually the true quotient, if only single precision is used.
Edited on July 28, 2019, 11:10 am
Posted by Charlie on 2019-07-28 07:58:12
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## Measurement and Central Tendency Assignment Help
The measurement of central tendency is recognized as the summary of statistic which represents the unique value or the center point of a dataset. These measures aim to indicate locations where most of the values in a distribution drop and they are also known as the central position of a distribution. The Measurement and Central Tendency is an essential aspect of statistics and students who are engaged in studying statistics often come across the task of completing assignments on this topic. When students find themselves helpless, they contact the writers of BookMyEssay for getting expert Measurement and Central Tendency assignment help. The assignment writing from our side tends to be absolutely flawless, and they help the students in securing impressive grades in their assignments. Moreover, the Measurement and Central Tendency assignment help from our side remains unique, and this makes our students confident and increases their belief in our services.
## Understanding the Concept of Measures of Central Tendency
In statistics, the three important common measures of Central Tendency are the arithmetic mean, the median, and the mode. Though all of these three measures do calculate the position of the central point, yet they make use of unique methods. However, selecting the most exceptional measure of central tendency is dependent on the kind of data you have. You can calculate the fundamental trend for either theoretical distribution, like the normal distribution or for a fixed set of values. This type of tendency is commonly contrasted with its variability or dispersion, and central tendency and dispersion are habitually branded characteristics of distributions. An analyst can judge whether or not a data has got a strong or a meek central tendency grounded in its dispersion.
Mean – The mean is considered the arithmetic average and it is also viewed as the measure of central tendency with which most of you are familiar. You can quickly calculate the mean, and for calculating your mean, you have to include all the values plus divide it by the observations present in your dataset. Calculating the mean consists of all the values present in the data and if you change any value, then the mean changes too.
Median – The median is considered the middle value and it is that value which splits the dataset into two halves. For finding out the median, you must order the data from the smallest to the largest and then discover the data point which has got a similar amount of values. The process of locating the mean does differ depending on whether or not your dataset has gained an even or odd number of values.
Mode – The mode is acknowledged as the value which happens in your data set most frequently. On a bar diagram, the mode is considered the highest bar, and if the data does have many values which are tied to occur most frequently, then you have got multimodal distribution. Again, if there is no repetition of value, then it can be said that the data doesn’t have any mode.
### Discovering the Best
When you have got asymmetrical distribution for incessant data then the mean, the median, plus the mode are similar. In these cases, the analysts use the mean as it involves all the data in the calculations. Again, when you have a skewed distribution, then the median is considered the best. And mode is used when categorical data is involved.
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https://arcade-book.readthedocs.io/en/latest/chapters/25_recursion/recursion.html | 1,539,869,992,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583511872.19/warc/CC-MAIN-20181018130914-20181018152414-00098.warc.gz | 618,347,338 | 12,716 | # 28. Recursion¶
A child couldn't sleep, so her mother told her a story about a little frog,
who couldn't sleep, so the frog's mother told her a story about a little bear,
who couldn't sleep, so the bear's mother told her a story about a little weasel...
who fell asleep.
...and the little bear fell asleep;
...and the little frog fell asleep;
...and the child fell asleep.
Recursion is an object or process that is defined in terms of itself. Mathematical patterns such as factorials and the Fibonacci series are recursive. Documents that can contain other documents, which themselves can contain other documents, are recursive. Fractal images, and even certain biological processes are recursive in how they work.
## 28.1. Where is Recursion Used?¶
Documents, such as web pages, are naturally recursive. For example, Figure 20.1 shows a simple web document.
That web document can be contained in a “box,” which can help layout the page as shown in Figure 20.2.
This works recursively. Each box can contain a web page, that can have a box, which could contain another web page as shown in Figure 20.3.
Recursive functions are often used with advanced searching and sorting algorithms. We’ll show some of that here and if you take a “data structures” class you will see a lot more of it.
Even if a person does not become a programmer, understanding the concept of recursive systems is important. If there is a business need for recursive table structures, documents, or something else, it is important to know how to specify this to the programmer up front.
For example, a person might specify that a web program for recipes needs the ability to support ingredients and directions. A person familiar with recursion might state that each ingredient could itself be a recipes with other ingredients (that could be recipes.) The second system is considerably more powerful.
## 28.2. How is Recursion Coded?¶
In prior chapters, we have used functions that call other functions. For example:
Functions calling other functions
1 2 3 4 5 6 7 8 def f(): g() print("f") def g(): print("g") f()
It is also possible for a function to call itself. A function that calls itself is using a concept called recursion. For example:
Recursion
1 2 3 4 5 def f(): print("Hello") f() f()
The example above will print Hello and then call the f() function again. Which will cause another Hello to be printed out and another call to the f() function. This will continue until the computer runs out of something called stack space. When this happens, Python will output a long error that ends with:
RuntimeError: maximum recursion depth exceeded
The computer is telling you, the programmer, that you have gone too far down the rabbit hole.
## 28.3. Controlling Recursion Depth¶
To successfully use recursion, there needs to be a way to prevent the function from endlessly calling itself over and over again. The example below counts how many times it has been called, and uses an if statement to exit once the function has called itself ten times.
Controlling recursion levels
1 2 3 4 5 6 7 8 9 10 11 def f(level): # Print the level we are at print("Recursion call, level",level) # If we haven't reached level ten... if level < 10: # Call this function again # and add one to the level f(level+1) # Start the recursive calls at level 1 f(1)
Output
1 2 3 4 5 6 7 8 9 10 Recursion call, level 1 Recursion call, level 2 Recursion call, level 3 Recursion call, level 4 Recursion call, level 5 Recursion call, level 6 Recursion call, level 7 Recursion call, level 8 Recursion call, level 9 Recursion call, level 10
## 28.4. Recursion In Mathematics¶
### 28.4.1. Recursion Factorial Calculation¶
Any code that can be done recursively can be done without using recursion. Some programmers feel that the recursive code is easier to understand.
Calculating the factorial of a number is a classic example of using recursion. Factorials are useful in probability and statistics. For example:
Recursively, this can be described as:
Below are two example functions that calculate . The first one is non-recursive, the second is recursive.
Non-recursive factorial
1 2 3 4 5 6 7 # This program calculates a factorial # WITHOUT using recursion def factorial_nonrecursive(n): answer = 1 for i in range(2, n + 1): answer = answer * i return answer
Recursive factorial
1 2 3 4 5 6 7 # This program calculates a factorial # WITH recursion def factorial_recursive(n): if n == 1: return 1 elif n > 1: return n * factorial_recursive(n - 1)
The functions do nothing by themselves. Below is an example where we put it all together. This example also adds some print statements inside the function so we can see what is happening.
Trying out recursive functions
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 # This program calculates a factorial # WITHOUT using recursion def factorial_nonrecursive(n): answer = 1 for i in range(2, n + 1): print(i, "*", answer, "=", i * answer) answer = answer * i return answer print("I can calculate a factorial!") user_input = input("Enter a number:") n = int(user_input) answer = factorial_nonrecursive(n) print(answer) # This program calculates a factorial # WITH recursion def factorial_recursive(n): if n == 1: return 1 else: x = factorial_recursive(n - 1) print( n, "*", x, "=", n * x ) return n * x print("I can calculate a factorial!") user_input = input("Enter a number:") n = int(user_input) answer = factorial_recursive(n) print(answer)
Output
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 I can calculate a factorial! Enter a number:7 2 * 1 = 2 3 * 2 = 6 4 * 6 = 24 5 * 24 = 120 6 * 120 = 720 7 * 720 = 5040 5040 I can calculate a factorial! Enter a number:7 2 * 1 = 2 3 * 2 = 6 4 * 6 = 24 5 * 24 = 120 6 * 120 = 720 7 * 720 = 5040 5040
### 28.4.2. Recursive Expressions¶
Say you have a mathematical expression like this:
$$f_{n} = \begin{cases} 6 & \text{if } n = 1, \\ \frac{1}{2}f_{n-1}+4 & \text{if } n > 1. \end{cases}$$
Looks complicated, but it just means that if $$n=1$$ we are working with $$f_{1}$$. That function returns a 6.
For $$f_{2}$$ we return $$\frac{1}{2}f_{1}+4$$.
1 def f(n):
Then we need to add that first case:
1 2 3 def f(n): if n == 1: return 6
See how closely if follows the mathematical notation? Now for the rest:
1 2 3 4 5 def f(n): if n == 1: return 6 elif n > 1: return (1 / 2) * f(n - 1) + 4
Converting these types of mathematical expressions to code is straight forward. But we’d better try it out in a full example:
1 2 3 4 5 6 7 8 9 10 11 12 13 def f(n): if n == 1: return 6 elif n > 1: return (1 / 2) * f(n - 1) + 4 def main(): result = f(10) print(result) main()
## 28.5. Recursive Graphics¶
### 28.5.1. Recursive Rectangles¶
Recursion is great to work with structured documents that are themselves recursive. For example, a web document can have a table divided into rows and columns to help with layout. One row might be the header, another row the main body, and finally the footer. Inside a table cell, might be another table. And inside of that can exist yet another table.
Another example is e-mail. It is possible to attach another person’s e-mail to a your own e-mail. But that e-mail could have another e-mail attached to it, and so on.
Can we visually see recursion in action in one of our Pygame programs? Yes! Figure 19.4 shows an example program that draws a rectangle, and recursively keeps drawing rectangles inside of it. Each rectangle is 20% smaller than the parent rectangle. Look at the code. Pay close attention to the recursive call in the recursive_draw function.
recursive_rectangles.py
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 """ Recursive Rectangles """ import arcade SCREEN_WIDTH = 800 SCREEN_HEIGHT = 500 def draw_rectangle(x, y, width, height): """ Recursively draw a rectangle, each one a percentage smaller """ # Draw it arcade.draw_rectangle_outline(x, y, width, height, arcade.color.BLACK) # As long as we have a width bigger than 1, recursively call this function with a smaller rectangle if width > 1: # Draw the rectangle 90% of our current size draw_rectangle(x, y, width * .9, height * .9) class MyWindow(arcade.Window): """ Main application class. """ def __init__(self, width, height): super().__init__(width, height) arcade.set_background_color(arcade.color.WHITE) def on_draw(self): """ Render the screen. """ arcade.start_render() # Find the center of our screen center_x = SCREEN_WIDTH / 2 center_y = SCREEN_HEIGHT / 2 # Start our recursive calls draw_rectangle(center_x, center_y, SCREEN_WIDTH, SCREEN_HEIGHT) def main(): MyWindow(SCREEN_WIDTH, SCREEN_HEIGHT) arcade.run() if __name__ == "__main__": main()
### 28.5.2. Fractals¶
Fractals are defined recursively. Here is a very simple fractal, showing how it changes depending on how “deep” the recursion goes.
Here is the source code for the “H” fractal:
recursive_h.py
You can explore fractals on-line:
If you want to program your own fractals, you can get ideas of easy fractals by looking at Chapter 8 of The Nature of Code by Daniel Shiffman.
## 28.6. Recursive Mazes¶
There are maze generation algorithms. Wikipedia has a nice Maze generation algorithm article that details some. One way is the recursive division method.
The algorithm is described below. Images are from Wikipedia.
This method results in mazes with long straight walls crossing their space, making it easier to see which areas to avoid.
Here is sample Python code that creates a maze using this method:
Recursive Maze Example
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 import random # These constants are used to determine what should be stored in the grid if we have an empty # space or a filled space. EMPTY = " " WALL = "XXX" # Maze must have an ODD number of rows and columns. # Walls go on EVEN rows/columns. # Openings go on ODD rows/columns MAZE_HEIGHT = 51 MAZE_WIDTH = 51 def create_grid(width, height): """ Create an empty grid. """ grid = [] for row in range(height): grid.append([]) for column in range(width): grid[row].append(EMPTY) return grid def print_maze(maze): """ Print the maze. """ # Loop each row, but do it in reverse so 0 is at the bottom like we expect for row in range(len(maze) - 1, -1, -1): # Print the row/y number print(f"{row:3} - ", end="") # Loop the row and print the content for column in range(len(maze[row])): print(f"{maze[row][column]}", end="") # Go down a line print() # Print the column/x at the bottom print(" ", end="") for column in range(len(maze[0])): print(f"{column:3}", end="") print() def create_outside_walls(maze): """ Create outside border walls.""" # Create left and right walls for row in range(len(maze)): maze[row][0] = WALL maze[row][len(maze[row])-1] = WALL # Create top and bottom walls for column in range(1, len(maze[0]) - 1): maze[0][column] = WALL maze[len(maze[0]) - 1][column] = WALL def create_maze(maze, top, bottom, left, right): """ Recursive function to divide up the maze in four sections and create three gaps. Walls can only go on even numbered rows/columns. Gaps can only go on odd numbered rows/columns. Maze must have an ODD number of rows and columns. """ # Figure out where to divide horizontally start_range = bottom + 2 end_range = top - 1 y = random.randrange(start_range, end_range, 2) # Do the division for column in range(left + 1, right): maze[y][column] = WALL # Figure out where to divide vertically start_range = left + 2 end_range = right - 1 x = random.randrange(start_range, end_range, 2) # Do the division for row in range(bottom + 1, top): maze[row][x] = WALL # Now we'll make a gap on 3 of the 4 walls. # Figure out which wall does NOT get a gap. wall = random.randrange(4) if wall != 0: gap = random.randrange(left + 1, x, 2) maze[y][gap] = EMPTY if wall != 1: gap = random.randrange(x + 1, right, 2) maze[y][gap] = EMPTY if wall != 2: gap = random.randrange(bottom + 1, y, 2) maze[gap][x] = EMPTY if wall != 3: gap = random.randrange(y + 1, top, 2) maze[gap][x] = EMPTY # Print what's going on print(f"Top/Bottom: {top}, {bottom} Left/Right: {left}, {right} Divide: {x}, {y}") print_maze(maze) print() # If there's enough space, to a recursive call. if top > y + 3 and x > left + 3: create_maze(maze, top, y, left, x) if top > y + 3 and x + 3 < right: create_maze(maze, top, y, x, right) if bottom + 3 < y and x + 3 < right: create_maze(maze, y, bottom, x, right) if bottom + 3 < y and x > left + 3: create_maze(maze, y, bottom, left, x) def main(): # Create the blank grid maze = create_grid(MAZE_WIDTH, MAZE_HEIGHT) # Fill in the outside walls create_outside_walls(maze) # Start the recursive process create_maze(maze, MAZE_HEIGHT - 1, 0, 0, MAZE_WIDTH - 1) if __name__ == "__main__": main() | 3,794 | 13,212 | {"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} | 3.90625 | 4 | CC-MAIN-2018-43 | latest | en | 0.93039 |
https://socratic.org/questions/what-is-the-distance-between-4-3-and-2-4 | 1,579,754,743,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250608295.52/warc/CC-MAIN-20200123041345-20200123070345-00204.warc.gz | 654,390,713 | 6,211 | # What is the distance between (-4, 3) and (-2, 4) ?
Dec 5, 2015
$\sqrt{5}$
#### Explanation:
The distance between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$
is given by the Pythagorean Theorem as
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
In this case
$\textcolor{w h i t e}{\text{XXX}} d = \sqrt{{2}^{2} + {\left(- 1\right)}^{2}} = \sqrt{5}$
The relationship between the point can be seen in the image below: | 206 | 535 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "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} | 3.984375 | 4 | CC-MAIN-2020-05 | longest | en | 0.762427 |
http://www.physicsforums.com/showthread.php?t=165482 | 1,369,332,317,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368703662159/warc/CC-MAIN-20130516112742-00098-ip-10-60-113-184.ec2.internal.warc.gz | 649,327,166 | 11,733 | ## Change of integration order?
1. The problem statement, all variables and given/known data
On this site,
http://www.math.ohio-state.edu/~gerl..._with_function
they changed integration order with e^(-ikt) integrated wrt t to t is now a constant and it's now integrated wrt k. How can one do this?
As stated from this sentence "By interchanging integration order and letting one has"
PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
$$f(x) \propto \int e^{ikx} \int e^{-ikt} f(t) dt dk = \int \int e^{ikx} e^{-ikt} f(t) dt dk = \int \int e^{ikx} e^{-ikt} f(t) dk dt$$ There's nothing wrong with changing the order of integration here, think back of integrals as Riemann sums, changing the order in which you sum terms doesn't matter. (You were going to integrate both exp(ikx) and exp(-ikt) over k anyway, so why not do that first and then integrate over t on which only the latter term depends).
Right, but why is the f (hat) function in terms of t? f is in terms of x and k only.
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## Change of integration order?
k and t are "dummy" variables. They can call them whatever they wish!
So only x counts. This brings about the question, what is the point of the fourier transform.
The point? Um, well, from my relatively limited experience in partial differential equations, in Fourier space differentiation and antidifferentiation break down into multiplication and division. Which is always good I suppose.
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Quote by pivoxa15 So only x counts. This brings about the question, what is the point of the fourier transform.
I'm not sure what you mean by that. Certainly in the integral you show, f is a function of x only. However, you have not said what it is you want to do with the Fourier Transform (the function inside the integral). Surely "the point of the Fourier transform" depends on that!
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Take equation (234) from here and there, replace $$\hat{f}(k)$$ by equation (233).
Then you almost have equation (235), except that you have to change
the order of integration.
Quote by pivoxa15 Right, but why is the f(hat) function in terms of t? f is in terms of x and k only.
The function $$\hat{f}(k)$$ is indeed a function of k, but it is not the function f, whose variable is x. Maybe you are a little confused about the meaning of $$\hat{f}(k)$$.
Think of the Fourier expansion as sort of linear combination of basis-vectors.
Let me try to explain:
Take the vectors in $$R^3$$.
Every vector can be represented as a linear combination of the basis-vectors $$(1,0,0)$$, $$(0,1,0)$$ and $$(0,0,1)$$.
For example:
Take the vector $$v=(9,8,4)$$. This vector can be written as a linear combination of the basisvectors vectors
$$u_1=(1,0,0)$$
$$u_2=(0,1,0)$$
$$u_3=(0,0,1)$$
in the following way:
$$(9,8,4) = 9 \cdot (1,0,0) + 8 \cdot (0,1,0) + 4 \cdot (0,0,1)$$
$$(9,8,4) = 9 \cdot u_1 + 8 \cdot u_2 + 4 \cdot u_3$$
Let's call the numbers 9,8,4 in front of the basis-vectors coefficients.
$$(9,8,4) = c_1 \cdot u_1 + c_2 \cdot u_2 + c_3 \cdot u_3$$
where $$c_1 = 9$$, $$c_2=8$$ and $$c_3=4$$
What if I choose different basis vectors, for example:
$$b_1=(3,0,0)$$
$$b_2=(0,2,0)$$
$$b_3=(0,0,2)$$
Then our vector $$v=(9,8,4)$$ can be written as:
$$(9,8,4) = 3 \cdot b_1 + 4 \cdot b_2 + 2 \cdot b_3$$
Thus, the coefficients are 3,4,2 for our new basis-vectors
$$b_1$$,$$b_2$$ and $$b_3$$.
In general, you can write a vector $$v$$ as a linear combination of
basis-vectors $$b_k$$, where in front of the basis-vectors you have the coefficients $$c_k$$.
$$v = \sum_{k=1}^{3} c_k b_k$$
In our last example we had
$$b_1=(3,0,0)$$
$$b_2=(0,2,0)$$
$$b_3=(0,0,2)$$
together with the coefficients:
$$c_1=3$$
$$c_2=4$$
$$c_3=2$$
Just check the formula
$$v = \sum_{k=1}^{3} c_k b_k$$
by plugging in the above values:
$$v = c_1 \cdot b_1 + c_2 \cdot b_2 + c_3 \cdot b3$$
$$v = 3 \cdot b_1 + 4 \cdot b_2 + 2 \cdot b_3$$
$$=3 \cdot (3,0,0) + 4 \cdot (0,2,0) + 2 \cdot (0,0,2)$$
----------
So, every vector can be represented by some basis vectors $$b_k$$
as
$$v = \sum_{k=1}^{3} c_k b_k$$
More general, if we have a vector with n entries instead of 3, we write
$$v = \sum_{k=1}^{n} c_k b_k = c_1 b_1 + c_2 b_2 + ... + c_n b_n$$
Now, let's make a step from the discrete to the continuous case.
Say we have a function $$f(x)$$ and we also want to ask, whether it's possible to represent $$f(x)$$ as a linear combination of ''basis-vectors''.
Question: Is it possible to write
$$f(x) = \sum_{k=1}^n c_k b_k$$
Let us be more specific and ask: Can I write $$f(x)$$ as a sum
of $$b_k = e^{ikx}$$? So my new basis-vectors are $$b_k=e^{ikx}$$.
The question then becomes:
Question: Is it possible to write
$$f(x) = \sum_{k=1}^n c_k e^{ikx}$$
Indeed, it is possible, with a correction for the values of k.
Instead of going from k=1 to n, we use infinitely many basis-vectors, and
write $$k=- \infty$$ to $$k=+ \infty$$.
Corrected version:
$$f(x) = \sum_{k=-\infty}^{+\infty} c_k e^{ikx}$$
The only question is, how do the coefficients
$$c_k$$ look like?
Have a look at Wikipedia
or here on page 2 of the pdf. It shows how the coefficients can be calculated.
To finally get to your Fourier integral, we replace the sum by an integral
$$f(x) = \int_{-\infty}^{+\infty} c(k) e^{ikx} dk$$
Now, on your ohio-website, $$c(k)$$ is $$\hat{f}(k)$$, see equation (234) on the ohio-website.
Thus, $$\hat{f}(k)$$ plays the role of the coefficients.
----------
Note 1 (on how to get from the Fourier series to the Fourier integral):
A Fourier series can sometimes be used to represent a function over an interval. If a function is defined over the entire real line, it may still have a Fourier series representation if it is periodic. If it is not periodic, then it cannot be represented by a Fourier series for all x. In such case we may still be able to represent the function in terms of sines and cosines, except that now the Fourier series becomes a Fourier integral. The motivation comes from formally considering Fourier series for functions of period 2T and letting T tend to infinity.
The quote is taken from here.
Also see here
for the transition from the Fourier series to the Fourier integral.
Note 2: Summary:
In summary, consider the Fourier integral as a linear combination of basis vectors $$e^{ikx}$$ with the
coefficients $$c(k)$$ (or $$\hat{f}(k)$$).
Note 3 (on applications of the Fourier transform:
Have a look at What is a Fourier Transform and what is it used for?.
Applications of Fourier Transform
in Communications, Astronomy, Geology and Optics
Note 4 (functions as basis-vectors?)
You might ask "Why can I consider the functions $$e^{ikx}$$ as
basis-vectors?"
The functions $$e^{ikx}$$ fulfill some properties similar to those of
basis vectors (1,0,0),(0,1,0),(0,0,1)from $$R^3$$.
The functions are orthonormal, that is they are
orthogonal to each other and they are normalized to 1 (delta-function?).
See
Fourier Analysis on page 3.
here on page 8.
Orthonormal functions: Definition on Wolfram mathworld
http://mathworld.wolfram.com/OrthonormalBasis.html
Note 5: Some examples of coefficients
Examples of Fourier transforms can be found here
Note 6: Java Applets
Approximation of a function by a Fourier transform
This applet shows you how you can approximate a function by a Fourier transform. The more
coefficients you use, the better the approximation becomes.
Applet: Rectangular pulse approximation by Fourier Transform
This applet shows how you approximate a rectangular shaped pulse. If you don't use enough
"basis-vectors" (bandwidth is limited), then the pulse will not look rectangular anymore. This has applications in
electronics where you want to transmit a signal but the bandwidth is limited.
I see. From all the Fourier transforms I have seen they use x as the integration variable in f (hat) which is confusing. f(hat) is not related to x in f(x) the original function but merely computes the coefficients of the basis vectors e^(ikx) for each k from -infinity to +infinity.
One thing I find fascinating is that there is a dirac delta function lurking inside every fourier transform. | 2,454 | 8,324 | {"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} | 3.515625 | 4 | CC-MAIN-2013-20 | latest | en | 0.91773 |
https://math.stackexchange.com/questions/2373613/calculating-the-number-of-irreducible-polynomials-over-a-finite-field | 1,561,366,504,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999298.86/warc/CC-MAIN-20190624084256-20190624110256-00201.warc.gz | 516,993,475 | 35,932 | # Calculating the number of irreducible polynomials over a finite field
I am trying to find the number of irreducible polynomials of degree $n$ over $\mathbb{F}_p$. Here is what I have done:
(1). Let $K=\mathbb{F}_{p^n}$. Let $M(n,p)$ the number of monic irreducible polynomials of degree $n$ over $\mathbb{F}_p$.
(2). Any root of a monic irreducible polynomial of degree $n$ is a primitive element of $K|\mathbb{F}_p$. Conversely, any primitive element of $K|\mathbb{F}_p$ is a root of a monic irreducible polynomial of degree $n$. Therefore, $$n\cdot M(n,p)=\text{ Number of primitive elements of }K|\mathbb{F}_p$$
(3). But we know that $K^{\times}$ is a cyclic group of order $p^n-1$. The primitive elements of $K|\mathbb{F}_p$ are precisely the generators of the cyclic group $K^{\times}$. The number of such generators are $\varphi(p^n-1)$.
(4). Combining (2) and (3) we conclude that $$M(n,p)=\frac{\varphi(p^n-1)}{n}$$
(5). Any irreducible polynomial of degree $n$ is a nonzero constant multiple of a monic irreducible polynomial of degree $n$. The number of such constants is $p-1$. Therefore the number of such polynomials is $$(p-1)\cdot M(n,p)=\frac{(p-1)\varphi(p^n-1)}{n}$$
So the final answer is $$\frac{(p-1)\varphi(p^n-1)}{n}$$
But my answer does not match with the usual way of solving this question which involves the Mobius inversion formula. What am I doing wrong?
First of all, very well-asked question! which makes it much easier to find the most helpful response.
It seems that you are conflating two different properties into one:
• a "primitive element" of a field extension $L/K$, which is an element $\alpha\in L$ such that $L=K(\alpha)$;
• a "primitive element" of a cyclic group, which is a generator of that group.
These two concepts do not correspond exactly with each other.
For example, consider $\Bbb F_9$, which is a degree-2 extension of $\Bbb F_3$. There are 6 elements of $\Bbb F_9$ that are not in $\Bbb F_3$, and any of them serves as a primitive element for the extension $\Bbb F_9/\Bbb F_3$ (in the first sense above). On the other hand, the multiplicative group $\Bbb F_9^\times$ is cyclic of order 8, and thus has $\phi(8)=4$ generators; all of these are primitive elements in the field-extension sense, but there are two other primitive elements that are not generators of the multiplicative group.
We can be super concrete here. Write $\Bbb F_9$ as $\Bbb F_3[x]/\langle x^2+1\rangle$, valid since $x^2+1$ is a degree-2 irreducible polynomial over $\Bbb F_3$. Then the elements of $\Bbb F_9$ are $a+bx$ where $a,b\in\{0,1,2\}$, with addition and multiplication defined as usual for polynomials except that $x^2=-1=2$. In this field, $x$ (or, more pedantically, the image of $x$ under the quotient map from $\Bbb F_3[x]$ to $\Bbb F_3[x]/\langle x^2+1\rangle$) is certainly a primitive element for the field extension. However, $x^4 = (x^2)^2 = 2^2 = 1$, and so $x$ does not generate the multiplicative group $\Bbb F_9^\times$. | 889 | 2,985 | {"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} | 3.71875 | 4 | CC-MAIN-2019-26 | longest | en | 0.844805 |
http://www.multiplication.com/learn/more/7/x/9 | 1,527,073,417,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865595.47/warc/CC-MAIN-20180523102355-20180523122355-00329.warc.gz | 443,452,347 | 8,807 | # Learn a Fact: 7 x 9
Follow steps 1-6 to master this fact.
## More Tips to Remember 7 x 9 = 63
Step 1Put your hands on the table in front of you. Step 2Your fingers represent the numbers 1 through 10. Step 3Each finger to the left of the curled finger represents 10. Since one finger is to the left of the curled finger, there are 6 tens or 60. (The first number of the answer is 6)Each finger to the right of the curled finger represents one. Count 1, 2, 3. (Or 61, 62, 63) 7 x 9 = 63
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8
X 0
0 | 204 | 595 | {"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} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.807995 |
https://everything.explained.today/Superquadrics/ | 1,627,082,261,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046150067.51/warc/CC-MAIN-20210723210216-20210724000216-00631.warc.gz | 264,441,268 | 6,311 | In mathematics, the superquadrics or super-quadrics (also superquadratics) are a family of geometric shapes defined by formulas that resemble those of ellipsoids and other quadrics, except that the squaring operations are replaced by arbitrary powers. They can be seen as the three-dimensional relatives of the superellipses. The term may refer to the solid object or to its surface, depending on the context. The equations below specify the surface; the solid is specified by replacing the equality signs by less-than-or-equal signs.
The superquadrics include many shapes that resemble cubes, octahedra, cylinders, lozenges and spindles, with rounded or sharp corners. Because of their flexibility and relative simplicity, they are popular geometric modeling tools, especially in computer graphics.
Some authors, such as Alan Barr, define "superquadrics" as including both the superellipsoids and the supertoroids.[1] [2] However, the (proper) supertoroids are not superquadrics as defined above; and, while some superquadrics are superellipsoids, neither family is contained in the other.Comprehensive coverage of geometrical properties of superquadrics and a method of their recovery from range images is covered in a monograph.[3]
## Formulas
### Implicit equation
The surface of the basic superquadric is given by
\left|x\right|r+\left|y\right|s+\left|z\right|t=1
where r, s, and t are positive real numbers that determine the main features of the superquadric. Namely:
• less than 1: a pointy octahedron modified to have concave faces and sharp edges.
• exactly 1: a regular octahedron.
• between 1 and 2: an octahedron modified to have convex faces, blunt edges and blunt corners.
• exactly 2: a sphere
• greater than 2: a cube modified to have rounded edges and corners.
• infinite (in the limit): a cube
Each exponent can be varied independently to obtain combined shapes. For example, if r=s=2, and t=4, one obtains a solid of revolution which resembles an ellipsoid with round cross-section but flattened ends. This formula is a special case of the superellipsoid's formula if (and only if) r = s.
If any exponent is allowed to be negative, the shape extends to infinity. Such shapes are sometimes called super-hyperboloids.
The basic shape above spans from -1 to +1 along each coordinate axis. The general superquadric is the result of scaling this basic shape by different amounts A, B, C along each axis. Its general equation is
\left| x A
\right|r+\left|
y B
\right|s+\left|
z C
\right|t=1.
### Parametric description
Parametric equations in terms of surface parameters u and v (equivalent to longitude and latitude if m equals 2) are
\begin{align} x(u,v)&{}=Ag\left(v,
2 r
\right)g\left(u,
2 r
\right)\\ y(u,v)&{}=Bg\left(v,
2 s
\right)f\left(u,
2 s
\right)\\ z(u,v)&{}=Cf\left(v,
2 t
\right)\\ &-
\pi 2
\lev\le
\pi 2
,-\pi\leu<\pi, \end{align}
where the auxiliary functions are
\begin{align} f(\omega,m)&{}=sgn(\sin\omega)\left|\sin\omega\right|m\\ g(\omega,m)&{}=sgn(\cos\omega)\left|\cos\omega\right|m \end{align}
and the sign function sgn(x) is
sgn(x)=\begin{cases} -1,&x<0\\ 0,&x=0\\ +1,&x>0. \end{cases}
## Plotting code
The following GNU Octave code generates a mesh approximation of a superquadric:function superquadric(epsilon,a) n = 50; etamax = pi/2; etamin = -pi/2; wmax = pi; wmin = -pi; deta = (etamax-etamin)/n; dw = (wmax-wmin)/n; [i,j] = meshgrid(1:n+1,1:n+1) eta = etamin + (i-1) * deta; w = wmin + (j-1) * dw; x = a(1) .* sign(cos(eta)) .* abs(cos(eta)).^epsilon(1) .* sign(cos(w)) .* abs(cos(w)).^epsilon(1); y = a(2) .* sign(cos(eta)) .* abs(cos(eta)).^epsilon(2) .* sign(sin(w)) .* abs(sin(w)).^epsilon(2); z = a(3) .* sign(sin(eta)) .* abs(sin(eta)).^epsilon(3);
mesh(x,y,z);end | 1,087 | 3,767 | {"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} | 3.8125 | 4 | CC-MAIN-2021-31 | latest | en | 0.921658 |
https://www.physicsforums.com/threads/method-od-undetermined-coeff-help-needed.168892/ | 1,511,344,729,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934806543.24/warc/CC-MAIN-20171122084446-20171122104446-00073.warc.gz | 808,944,168 | 15,156 | # Method od undetermined coeff help needed.
1. May 5, 2007
### HappMatt
1. The problem statement, all variables and given/known data
Ive been working on this problem for way too long and i can get it almost right except im missing a factor of (1/12)e^4x and for the life of me i cant figure out why. Ive been mostly trying to use the methode of undeermined coefficients to no luck and so have also tried variation of parameters and still something is not going right. I have run the equation through my 89 and i have the answer i just cant seem to get it by hand.
2. Relevant equations
y''-2y'-8y=3*e^(4x)-5x^2
complimentary solution=yc=C1*e^(4x)+C28*e^(-2x)
as for finding the particular solution(yp) i think this is where my problem is.
the actual solution as given by my ti 89 is y=((x/2)
+C1-(1/12))*e^(4x)+C2*e^(-2x)+(5x^2)/8-(5x)/16+15/64
3. The attempt at a solution
using yp=Axe^(4x) +Bx^2+Cx+d i got it all right except that im missing the (1/12)*e^4x value and ive tried to many varriations to list them all here but i have litterally spent hours on this and i probally should have posted this back when i was still at only 3 hours worth of time into it but im well beyond that now and its time for a little rest so i can wake up and keep hammering at this one till i get it.
2. May 5, 2007
### Pseudo Statistic
Just a word of advice.
When dealing with undetermined co-efficients, you'd probably find it best to, rather than tackle both right hand terms (3e^(4x) - 5x^2) at the same time, deal with them as two seperate problems; that is, guess their particular solution for:
y'' - 2y' - 8y = 3e^(4x)
Then do the same for y'' - 2y' - 8y = -5x^2 and add up your solutions.
However, it seems like that's not the problem you're having; when you say you're missing this "1/12 e^4x", that doesn't quite add up, since 1/12 e^4x contributes to the homogeneous solution.
That is, unless you're given some initial conditions you're not telling us about? ;)
3. May 5, 2007
### HappMatt
thanks for the advice Psuedo as far as seperating the particular solution Ive tried that and recived the same results. As for the (1/12)*e^(4x) Im mostly sure that is part of the particular solution since the homogeneous solution is just C2e^4x +C1e^(-2x)
4. May 5, 2007
### HappMatt
tried this probleb again for the 10000th and i finally got the answer using variation of parameters.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook | 714 | 2,485 | {"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} | 3.6875 | 4 | CC-MAIN-2017-47 | longest | en | 0.930269 |
https://www.physicsforums.com/threads/why-did-they-use-only-meter-resistance-for-tau-of-inductor.848049/ | 1,529,703,092,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864795.68/warc/CC-MAIN-20180622201448-20180622221448-00008.warc.gz | 867,525,402 | 16,258 | # Homework Help: Why did they use only meter resistance for tau of inductor
1. Dec 13, 2015
### nickmanc86
1. The problem statement, all variables and given/known datahttp://imgur.com/uLZdBJC
27. The switch has been closed for
about 1 h. It is then opened at the time defined as t 0 s.
a. Determine the time required for the current iL to drop to
10 μA.
b. Find the voltage VL at t 10μs
c. Calculate VL at t 5tau.
2. Relevant equations
So I completely worked through this problem. Got Eth (20V) and Rth (1.66MΩ). Used 5H for the inductance and plugged into the formula for IL getting iL = 20V/1.66MΩ * e -t/tau where I assumed tau to be 5H/1.66MΩ= 3μs. However the answer comes out incorrect .55μs.
Looking at the solution for the book it calculates tau using only the resistance of the meter 10MΩ and gets a tau of 5μs. This of course gives the same answer as the back of the book. I do not understand why they use only the meters resistance for the tau calculation.
For part B i use the simplified release formula for VLof VL = -Vi * e-t/tau . I simply took my Vi to be Eth or 20V with my tau of 3μs and solved for VL.
However they proceed to use Vi = I*Rm where Vi = (20/1.66MΩ)*10MΩ =120V giving them VL = 120V * e-t/tau using their tau of 5μs.
I do not understand why they calculate tau using only Rmeter and then subsequently find Vi using Eth/Rth * Rmeter.
3. My Work
A)
Rth = 2MΩ||10MΩ = 1.66MΩ
Eth = 24V*10MΩ/12MΩ = 20V
Tau = L/R = 5H/1.66MΩ = 3μs
Plugged into
IL = E/R * e-t/tau -> IL = .55μs
B)
Vi = 20V
Tau = L/R = 5H/1.66MΩ = 3μs
Plugged into
VL = -Vi *e-t/tau = -0.713V
2. Dec 13, 2015
### Staff: Mentor
Are you sure the problem statement is that the switch has been closed and is then opened at t=0? I could be wrong, but that will result in a *very* large voltage being developed across the inductor, limited only by the parasitic capacitance of the inductor and the circuit (which are not specified).
The problem would make much more sense if the switch were closed at t=0...
3. Dec 13, 2015
### nickmanc86
I double checked the problem in the book and it definitely states the switch was closed for 1h and then opened at a time t = 0s. I suppose it is possible there is a typo in the book (it apparently has many). However, correct or not, they proceed to solve it as stated and end up with those answers. Their initial voltage for part B is 120V which is high relative to the voltage source. Unfortunately out of my realm of understanding (basic circuits FTW).
4. Dec 13, 2015
### Staff: Mentor
When the switch opens the only available path for the current is via the 10 MΩ meter...
5. Dec 13, 2015
### nickmanc86
Oh my, of course. Wow I feel dumb now. Thank you very much that definitely makes sense. | 828 | 2,745 | {"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} | 3.8125 | 4 | CC-MAIN-2018-26 | latest | en | 0.92068 |
https://convertoctopus.com/274-feet-to-yards | 1,619,167,956,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039568689.89/warc/CC-MAIN-20210423070953-20210423100953-00456.warc.gz | 294,645,168 | 7,562 | ## Conversion formula
The conversion factor from feet to yards is 0.33333333333333, which means that 1 foot is equal to 0.33333333333333 yards:
1 ft = 0.33333333333333 yd
To convert 274 feet into yards we have to multiply 274 by the conversion factor in order to get the length amount from feet to yards. We can also form a simple proportion to calculate the result:
1 ft → 0.33333333333333 yd
274 ft → L(yd)
Solve the above proportion to obtain the length L in yards:
L(yd) = 274 ft × 0.33333333333333 yd
L(yd) = 91.333333333333 yd
The final result is:
274 ft → 91.333333333333 yd
We conclude that 274 feet is equivalent to 91.333333333333 yards:
274 feet = 91.333333333333 yards
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 yard is equal to 0.010948905109489 × 274 feet.
Another way is saying that 274 feet is equal to 1 ÷ 0.010948905109489 yards.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that two hundred seventy-four feet is approximately ninety-one point three three three yards:
274 ft ≅ 91.333 yd
An alternative is also that one yard is approximately zero point zero one one times two hundred seventy-four feet.
## Conversion table
### feet to yards chart
For quick reference purposes, below is the conversion table you can use to convert from feet to yards
feet (ft) yards (yd)
275 feet 91.667 yards
276 feet 92 yards
277 feet 92.333 yards
278 feet 92.667 yards
279 feet 93 yards
280 feet 93.333 yards
281 feet 93.667 yards
282 feet 94 yards
283 feet 94.333 yards
284 feet 94.667 yards | 461 | 1,669 | {"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} | 4.1875 | 4 | CC-MAIN-2021-17 | latest | en | 0.767378 |
https://math.stackexchange.com/questions/639972/method-of-characteristics-and-initial-value-problem | 1,643,212,302,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320304954.18/warc/CC-MAIN-20220126131707-20220126161707-00375.warc.gz | 446,520,607 | 33,802 | # Method of Characteristics and Initial Value Problem
$u_t + 3u_x = 2t$, $u(x,0)=\sin(x/2)$.
I used the method of characteristics to get the answer, $u(x,t)=t^2 + 2\sin^{-1}(x-3t)$.
Does this satisfy the initial condition? I checked for the first equation and it does; however I do not think it satisfies the intial value when $t=0.$ Am I correct in saying so?
• Yes you are corect to saying that it does not satisfy the IC. It is clear when you put $t=0$ in the solution you have found. Jan 16 '14 at 3:48
The solution is $u=\frac{2}{3}\cos(\frac{x}{2})+\sin(\frac{x-3t}{2})-\frac{2}{3}\cos(\frac{x-3t}{2})$.
I think your u(x,t) is not correct: apart from some factors of 2 different, I have $\sin$ and not $\sin^{-1}$.
Also, worth recognizing that you can solve this with just change of variables, $x \rightarrow x + 3 t$.
answer for above question is $$u(x,t)=t^2+\sin(x−3t/2)$$ | 286 | 886 | {"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} | 3.59375 | 4 | CC-MAIN-2022-05 | latest | en | 0.899414 |
https://www.khanacademy.org/math/mr-class-10/x5cfe2ca097f0f62c:surface-area-and-volume/x5cfe2ca097f0f62c:areas-of-combination-of-plane-figures/v/area-of-shaded-region | 1,686,266,349,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224655143.72/warc/CC-MAIN-20230608204017-20230608234017-00718.warc.gz | 885,971,197 | 60,522 | If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
## Class 10 (Marathi)
### Course: Class 10 (Marathi)>Unit 11
Lesson 6: Areas of combination of plane figures
# Area of a shaded region
Here's a fun one: find the area of a shaded region where you first determine the area of a square and then the area of a circle. Created by Sal Khan.
## Want to join the conversation?
• This may not sound very smart but why did you multiple 3*3
• Sal multiplied 3 and 3 because the formula for getting area is A = r^2 pi. If our radius is 3, and if part of the formula is r^2, to get the radius to the second power you multiply 3 and 3 .
• at what was that green thing
• It is due to an incomplete answer. Once you finish typing your answer, assuming it is an acceptable form for the particular question, the green guy goes away:)
• Why he didn't multiply it by 4 like:
100 - 4(3^2)pi? isn't this going to give us all four sides? o.O #confused
PS: Oh I get it, I get it now :D. (3^2)pi will give us the entire area of full circle :D
• What if your book doesn't give the area of the shape?
• what is this website he's using called?
• it still doesn't make sence. | 350 | 1,324 | {"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} | 3.6875 | 4 | CC-MAIN-2023-23 | latest | en | 0.935451 |
https://www.physicsforums.com/threads/physics-problem-with-spring-constant-and-kinetic-friction.790440/ | 1,508,825,320,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187828178.96/warc/CC-MAIN-20171024052836-20171024072836-00327.warc.gz | 965,717,625 | 17,924 | # Physics Problem with Spring Constant and Kinetic Friction
1. Jan 3, 2015
### Q7heng
1. The problem statement, all variables and given/known data
A person attaches a spring to an cubic object that weighs 36kg and pulls this object along a table made of material X horizontally with a steady speed of 1.3m/s. The spring stretches a distance of 3.4cm. Find the ratio of the spring constant to the kinetic friction coefficient.
2. Relevant equations
I suppose FkkN
Fspring=-kx
F=ma
Not sure if any other ones are needed, but I couldn't find a way to solve this and get a reasonable answer.
2. Jan 3, 2015
### Bystander
What does this information tell you?
3. Jan 4, 2015
### Q7heng
I'm not sure, but I tried to apply it to the F=MA formula... Am I on track?
4. Jan 4, 2015
### Bystander
Okay, we'll work it through that way: "Steady speed" means what in terms of "A?"
5. Jan 4, 2015
### Q7heng
Acceleration=0 at steady speed, but the trouble I'm having is converting everything to Newtons, since spring constant is Newtons/meter, and kinetic friction coefficient is Force of Kinetic Friction/Normal Force. The problem didn't give any units related to those calculations, maybe there is a way to convert it but I have yet figured it out/learned it.
6. Jan 4, 2015
### Bystander
This looks perfectly useful. What's the normal force?
Nothing wrong with this.
You aren't required to generate a numerical answer for every problem on the planet. Sometimes it's just a matter of coming up with a symbolic expression.
7. Jan 4, 2015
### haruspex
All good so far. Now, what is the relationship between F, Fk and Fspring?
8. Jan 5, 2015
### Q7heng
F=Fk+Fspring right?
If that is so then:
F=MA, A=0, and F=0
So 0=Fk+Fspring
0=-kx+μkN
kx=μkN
k/μk=N/x, since we are trying to find the ratio between the spring constant and the kinetic friction coefficient
Since N, normal force, is 36*9.8N right now, and x, distance, is 3.4cm or 0.034m right now, then it is 36*9.8/0.034=10,376.47, which is the ratio of spring constant to kinetic friction, is that correct?
9. Jan 5, 2015
### haruspex
Yes, that all looks right. But you should include the units in the answer.
10. Jan 5, 2015
Thanks!!! | 647 | 2,202 | {"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} | 3.671875 | 4 | CC-MAIN-2017-43 | longest | en | 0.911221 |
https://www.queryhome.com/puzzle/12075/if-3-4-3-4-4-35-5-4-4-6-4-12-7-4-60-then-8-4 | 1,542,088,054,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039741219.9/warc/CC-MAIN-20181113041552-20181113063552-00308.warc.gz | 970,446,273 | 31,146 | # If 3+4=3; 4+4=35; 5+4=4; 6+4=12; 7+4=60 then 8+4=?
2,945 views
If
3+4=3
4+4=35
5+4=4
6+4=12
7+4=60
Then
8+4=?
posted Jan 10, 2016
Ans will be 5
Look
3+4=3 in this
Multiply these 3 and 4 and when u get the two digit ans, square the both, higher digit square will be substracted by lower digit square
Now 3*4=12
2^ - 1^ = 4-1=3
Here ^ is used by me as square
4+4=35
4*4=16
6^ - 1^=36-1=35
Solve all like this and finally
8+4
8*4=32
3^ - 2^=9-4=5
And 5 is ans
Ans given by 8601084829
Krishna P. Sharma
answer Aug 13, 2016 by anonymous
Wow wonderful, touch one to crack
Similar Puzzles
If
2+3=8,
3+7=27,
4+5=32,
5+8=60,
6+7=72
then
7+8=??
–1 vote
If 2+3=3, 4+4=5, 4+5=7, 5+7=10, 5+8=12, 6+5=8
then (3+7)+2 = ? | 374 | 716 | {"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-2018-47 | latest | en | 0.568094 |
https://www.exactlywhatistime.com/time-from/107-weeks-from-today | 1,718,895,503,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861957.99/warc/CC-MAIN-20240620141245-20240620171245-00886.warc.gz | 665,817,556 | 7,701 | # What's the date 107 weeks from today?
## Thursday July 09, 2026
0
107 weeks from today will be 09 Jul 2026, a Thursday. For half-year projections, double-check whether 107 weeks forward remains within 2024. It seems that 107 weeks from now will bring us back to pre 2023. Please include this into our planning as it impacts both the calendar and fiscal year transitions. For extensive calculations like this, I begin by segmenting the year, then multiple 107 by days to get 749 total days ahead. Then either count 749 from June (but that will take forever). You can also estimate 107 weeks to 3.566666666666667 months, count from July and get closer to Thursday July 09, 2026.
## How we calculated 107 weeks from today
All of our day calculators are measured and QA'd by our engineer. Read more about the Git process here. But here's how adding 107 weeks to today's date gets calculated on each visit:
• Started with date inputs: starting point: 20 Jun, Units to add: 107 weeks, and year: 2024
• Noted your current time of year: 10 days in middle of June
• Added 107 weeks from current day: 20 Jun, factoring in there are 10 days left in before July
• Did NOT factor in workdays: In this calculation, we kept weekend. See below for just workdays or the 2024 fiscal calendar.
### Tips to get your solution: July 09
Thursday Thursday July 09, 2026 is the 190 day of the year or 52.05% through 2026.
• Current date: 20 Jun
• Day of the week: Thursday
• New Date: Thursday July 09, 2026
• New Date Day of the week: Thursday
• June is the end of Q2 so if you're counting dates ahead, you might end up in H2 of the fiscal year.
• This calculation crosses at least one month. Remeber, this will change our day of the week.
• The solution crosses into a different year..
## Ways to calculate 107 weeks from today
1. Just calculate it: Start with a time from today calculator. 107 weeks is easiest solved on a calculator. For ours, we've already factored in the days in + all number of days in each month and the number of days in 2024. Simply add your weeks and choose the length of time, then click "calculate". This calculation does not factor in workdays or holidays (see below!).
2. Use June's calendar: Begin by identifying on a calendar, note that it’s Thursday, and the total days in July (trust me, you’ll need this for smaller calculations) and days until next year (double trust me, you'll need this for larger calculations). From there, count forward 107 times by weeks, adding weeks from 20 Jun.
3. Use Excel: Regardless of unit type, I use day calculations here. Type =TODAY()+107 into the cell. If you want to add weeks, multiply your day by 7 and months/years will take their own calculation due to the changing days of the week. To find 107 weeks workdays, convert to days but use =WORKDAY(TODAY(), [number of days], [holidays]) into the cell. [number of days] is how many working days you want to add, and [holidays] is an optional range of cells that contain dates of holidays to exclude.
## 107 working weeks from today
107 weeks is Thursday July 09, 2026 or could be Monday May 03, 2027 if you only want workdays. This calculation takes 107 weeks and only adds by the number of workdays in a week. Remember, removing the weekend from our calculation will drastically change our original Thursday July 09, 2026 date.
Work weeks Solution
Monday
Tuesday
Wednesday
Thursday
July 09
Friday
Saturday
Sunday
## In 107 weeks, the average person Spent...
• 160885.2 hours Sleeping
• 21391.44 hours Eating and drinking
• 35053.2 hours Household activities
• 10426.08 hours Housework
• 11504.64 hours Food preparation and cleanup
• 3595.2 hours Lawn and garden care
• 62916.0 hours Working and work-related activities
• 57882.72 hours Working
• 94733.52 hours Leisure and sports
• 51411.36 hours Watching television
## What happened on July 09 (107 weeks from now) over the years?
### On July 09:
• 1952 American educator, Presbyterian minister and TV host Fred Rogers (24) weds Sara Joanne Byrd
• 1877 First ever Wimbledon tennis championship begins - first official lawn tennis tournament - men's singles only | 1,041 | 4,141 | {"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-2024-26 | latest | en | 0.937656 |
https://mocktestpro.in/mcq/fluid-mechanics-mcq-surface-tension-capillarity-vapour-pressure-and-cavitation/ | 1,685,258,701,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224643585.23/warc/CC-MAIN-20230528051321-20230528081321-00447.warc.gz | 444,016,778 | 23,678 | Engineering Questions with Answers - Multiple Choice Questions
# Fluid Mechanics MCQ – Surface Tension, Capillarity, Vapour Pressure and Cavitation
1 - Question
Calculate the magnitude of capillary effect in millimeters in a glass tube of 7mm diameter, when immersed in mercury. The temperature of the liquid is 25℃ and the values of surface tension of mercury at 25℃ is 0.51 N/m. The angle of contact for mercury is 130°.
a) 140
b) 280
c) 170
d) 210
Explanation: Capillarity rise or fall
h=4*cosθ*σ/ρ*g*d
=4*cos130*0.51/13600*9.81*0.007
=140 mm.
2 - Question
Determine the minimum size of glass tube that can be used to measure water level if the capillary rise in the tube is restricted to 5mm. Consider surface tension of water in contact with air as 0.073 N/m
a) 5.95mm
b) 11.9mm
c) 2.97mm
d) 4.46mm
Explanation: d=4*cosθ*σ/ρ*g*h
=4*1*0.073/1000*9.81*0.005
=5.95mm.
3 - Question
A An oil of vicosity 7 poise is used for lubrication between shaft and sleeve. The diameter of shaft is 0.6 m and it rotates is 360 rpm. Calculate the power lost in oil for a sleeve length of 160mm. The thickness of oil film is 1.0mm
a) 25.31 kW
b) 50.62 kW
c) 37.97 kW
d) 12.65 kW
Explanation: Power lost= torque * angular velocity
= shear stress * area* radius* angular velocity
Shear Stress = viscosity* velocity gradient
Power lost= 7916.8*3.142*0.3*0.3*0.3*2*3.142*60
= 25.31 kW.
4 - Question
Find the capillarity rise or fall if a capillary tube of diameter .03m is immersed in hypothetical fluid with specific gravity 6.5, surface tension 0.25 N/m and angle of contact 147°.
a) 0.44mm fall
b) 0.88mm fall
c) 0.44mm rise
d) 0.88mm rise
Explanation: h=4*cosθ*σ/ρ*g*d
=4*cos147*0.25/6.5*1000*9.81*0.03
=-0.44 mm i.e 0.44 mm fall.
5 - Question
Will capillary rise occur and if it occurs what will be capillary rise if glass capillarity tube is immersed in water and experiment is carried out by astronauts in space.
a) Capillarity rise will not occur
b) Capillarity rise will occur infinitely and will come out in form of fountain
c) Capillarity rise will occur finitely and will be the whole length of tube
d) None of the mentioned
Explanation: Capillary rise is given by
h=4*cosθ*σ/ρ*g*d
hence rise is inversely proportional to g
In space g is 0 m/s2
Hence, capillarity rise will occur finitely and will be the whole length of tube.
6 - Question
The surface tension of fluid in contact with air at 25℃ is 0.51N/m. The pressure inside a droplet is to be 0.05 N/cm2 greater than outside pressure. Determine the diameter of the droplet of water.
a) 4.08mm
b) 8.16mm
c) 2.04mm
d) None of the mentioned
Explanation: P=4*σ/d
d= 4*.51/500
=4.08 mm.
7 - Question
If a fluid of certain surface tension and diameter is used to create a soap bubble and a liquid jet. Which of the two, bubble or liquid jet, will have greater pressure difference on the inside and outside.
a) Liquid jet
b) Soap bubble
c) Both will have same pressure differrence
d) None of the mentioned
Explanation: For soap bubble,
P=8*σ/d
For liquid jet,
P=2*σ/d
Hence, soap bubble will be having more pressure difference.
8 - Question
Capillarity fall is reduced if we take the appartus (capillary tube immersed in fluid having acute angle of contact) considerable distance inside the earth( i.e below the earth crust).
a) True
b) False
Explanation: Capillary rise is given by
h=4*cosθ*σ/ρ*g*d
Inside the earth, g (acceleration due to gravity) decreases. Hence, capillary rise will increase compared to that on the earth’s surface.
9 - Question
For liquid fluids will capillarity rise (or fall) increase or decrease with rise in temperature.
a) Increase
b) Decrease
c) Remain constant
d) First decrease then increase
Explanation: Capillary rise is given by
h=4*cosθ*σ/ρ*g*d
As temperature increases, σ(surface tension) decreases. Therefore, correspondingly rise(or fall) will decrease as their is direct proportional relation between the two.
10 - Question
Cavitation is more pronounced in rough pipes than smooth surfaced pipes.
a) True
b) False | 1,199 | 4,023 | {"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} | 3.8125 | 4 | CC-MAIN-2023-23 | latest | en | 0.793148 |
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# If x^2 - 3x + k = -30, what is the value of k?
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If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 06:34
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If x^2 - 3x + k = -30, what is the value of k?
(1) 2 is an element of the solution set of the given equation.
(2) 5 is an element of the solution set of the given equation.
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 09:10
Bunuel wrote:
If x^2 - 3x + k = -30, what is the value of k?
(1) 2 is an element of the solution set of the given equation.
(2) 5 is an element of the solution set of the given equation.
x^2 - 3x + k = -30
we need value of x to find the value of k
from 1:
x=2
we can find the value of k by replacing the value of x in given eq.
sufficient.
from 2:
x=5
we can find the value of k by replacing the value of x in given eq.
sufficient
hence D
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 09:47
x^2 - 3x + k = -30
x^2 - 3x + (k + 30) = 0
sum of roots = 3
product of roots = (k+30)
1) 2 + a = 3 => a = 1.
So, k+30 = 2 => k = 28
2) 5 + b = 3 => b = -2.
So, k + 30 = -10 => k = -40
EIther sufficient.
(D)
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Joined: 04 Apr 2018
Posts: 21
Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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10 Apr 2018, 23:21
(1). We know that 2 is the solution set, hence simply put that in the given equation & "k" can easily be deduced. Hence, SUFFICIENT.
(2). Similarly, this is also SUFFICIENT.
Hence, both are individually self sufficent to asnwer the question. Hence D
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Joined: 24 Apr 2016
Posts: 30
Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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11 Apr 2018, 06:04
1
This one stumped me for awhile and I'm still not 100% sure. I thought for DS questions 1) and 2) are hints for the same answer. But if X=2 then k=-28. But if X=5 then k= -40. Am I calculating this incorrectly?
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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11 Apr 2018, 06:07
Don't know lol....
Math Expert
Joined: 02 Sep 2009
Posts: 56276
Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink]
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11 Apr 2018, 06:08
dracobook wrote:
This one stumped me for awhile and I'm still not 100% sure. I thought for DS questions 1) and 2) are hints for the same answer. But if X=2 then k=-28. But if X=5 then k= -40. Am I calculating this incorrectly?
Sent from my Pixel 2 XL using GMAT Club Forum mobile app
You are right. The question is flawed. Archived.
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Re: If x^2 - 3x + k = -30, what is the value of k? [#permalink] 11 Apr 2018, 06:08
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https://byjusexamprep.com/syllogism-for-ssc-exams-i-03d9a2c3-b92c-11e5-a60c-80817d4bb946 | 1,631,876,941,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00592.warc.gz | 193,134,483 | 76,029 | Tricks to solve Syllogism easily in Reasoning Section!
By Parnab Mallick|Updated : January 6th, 2021
The Reasoning section of SSC Exams includes questions from the topic “Syllogisms”. Syllogism is very important not only for SSC CGL but for all the SSC Exams. Students generally find difficulty in understanding Syllogism. In this article, we have explained the basics of syllogism with the help of Venn diagram. Sit with pen and paper while going through this article.
The Reasoning section of SSC Exams includes questions from the topic “Syllogisms”. Syllogism is very important not only for SSC CGL but for all the SSC Exams. Students generally find difficulty in understanding Syllogism. In this article, we have explained the basics of syllogism with the help of Venn diagram. Sit with pen and paper while going through this article.
Syllogism Concepts for SSC Exams
Syllogism comes under verbal reasoning section and is frequently asked in many competitive exams. These types of questions contain two or more statements and these statements are followed by the number of conclusions. You have to find which conclusions logically follows from the given statements.
The best method of solving the Syllogism’s problem is through Venn Diagrams. There are four ways in which the relationship could be made.
Category 1
All A are B – Means the whole circle representing A lies within the circle representing B.
Here we can also make a conclusion: Some B are A. Some A are B.
For example: All boys are men.
Here we can also make a conclusion: Some men are boys. Some boys are men.
All apples are fruits.
Here we can also make a conclusion: Some fruits are apples. Some apples are fruits.
Category 2
No A is B – means that circles representing A and B does not intersect at all.
For example : No ball is bat.
No door is wall.
Category 3
Some A are B
Means that some part of the circle represented by A is within the circle represented by B.
This type of (category 3) statement gives the following conclusions:
(i) Some A are B also indicates that - Some A are not B
(ii) Some A are B also indicates that – All A are B.
(iii) Some A are B also indicates that – All B are A.
(iv) Some A are B also indicates that – All A are B and All B are A.
For e.g.: Some mobiles are phones.
(i)
Category 4.
Some A are not B
Means that some portion of circle A has no intersection with circle B while the remaining portion of circle A is uncertain whether this portion touches B or not.
(i) Some A are not B also indicates that – Some A are B.
(ii) Some A are not B also indicates that – No A is B.
Types of Sentences Conclusions All A are B Some B's are A Some A's are B Some A are B All A are B All B are A Some B are A Some A are not B No A is B No B is A Some A are not B Some A are B All B are A No A is B
Important Points –
1. At least statement – At least statement is same as some statement.
For ex:
Statement: All kids are innocent.
Here we can make conclusion: At least some innocent are kids (Some innocent are kids).
2. Some not statement: Some not statement is opposite to “All type” statement. If All being true then Some not being false
For e.g.
1. Statement: Some pens are pencils. No pencils are jug. Some jug is pens.
Here we can make a conclusion: Some pens are not pencils, which is true. In above figure, green shaded part shows; some pens are not pencils, because in statement it is already given No pencils is jug.
Complementary Pairs: (Either & or) – Either and or cases only takes place in complementary pairs.
Conclusions: (i) Some A are B. (ii) No A are B.
From the given above conclusions, it is easy to understand that one of the given conclusions must be true, which is represented by option either (i) or (ii). These types of pairs are called complementary pairs.
Note: ‘All A are B’ & ‘Some A are not B’ are also complementary pairs.
Note: It is important to note that, in complementary pairs, one of the two conclusions is true and other will be false simultaneously.
For example –
Statement: All A are B. Some B are C.
Conclusion: I. All C are A. II. Some C are not A.
Here we can make a conclusion, either I or either II follows.
Possibility cases in Syllogism – In possibilities cases, we have to create all possibilities to find whether the given conclusion is possible or not. If it is possible and satisfies the given statement than given conclusion will follow otherwise conclusion will not follow.
1. E.g.
Statement: All A are B. Some B are C.
Conclusion: All A being C is a possibility.
Conclusion is true.
Possibility figure –
2. E.g.
Statements: No stone is a white. Some white are papers.
Conclusions: I. All stones being paper is a possibility.
Possibility figure:
Conclusion is true.
3. E.g.
Statements: Some mouse is cat.
All mouse are pets. No pet is animal.
Conclusions: I. All mouse being animal is a possibility.
The conclusion is false because possibility figure is not possible.
If we say all mouse being an animal is a possibility is true than given statements No pet is animal will be wrong. Here in the statement, it is given No pet is animal and All mouse is pet. So we can make also conclusion here that no mouse are animal is true.
Important Rule:
The statement itself is not a conclusion – Conclusion has to be different from the statement.
E.g. Statement - All A are B
Conclusion - All are B. (invalid) Conclusion does not follow.
Conclusion - Some A are B (follow) Conclusion follows.
Note: If statement and conclusion are same then the conclusion does not follow. This rules also follow in possibilities case
You can now take up the below-given Reasoning Quiz on Syllogisms to clear your doubts related to the topic:
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Parnab Mallick is an educator and mentor with an expertise in SSC and Railway exams. He tries to make students’ life easy by guiding them the right path and knowledge to cater to their dream govt. job. He lives with the notion ‘We all have equal opportunity to be unequal’
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SSE dayaram meenaJul 15, 2021 | 1,569 | 6,495 | {"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-2021-39 | longest | en | 0.958912 |
http://mathhelpforum.com/pre-calculus/119405-quadratic-factors-cubic-polynomial.html | 1,526,903,983,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794864063.9/warc/CC-MAIN-20180521102758-20180521122758-00285.warc.gz | 185,826,809 | 12,127 | 1. Quadratic factors of a cubic polynomial
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
2. Hello Detanon
Originally Posted by Detanon
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
Suppose the other factor is $\displaystyle (rx+s)$. Then:
$\displaystyle (x^2+bx-2)(rx+s) = x^3+(2b-1)x^2 -p$
Comparing coefficients:
$\displaystyle x^3: r = 1$
So:
$\displaystyle (x^2+bx-2)(x+s) = x^3+(2b-1)x^2 -p$
Comparing coefficients:
$\displaystyle x^2: (s+b) = (2b-1) \Rightarrow s = b-1$ (1)
$\displaystyle x: -2+sb = 0 \Rightarrow sb = 2$ (2)
So, from (1) and (2):
$\displaystyle b(b-1)=2$
$\displaystyle \Rightarrow b^2-b-2=0$
$\displaystyle \Rightarrow (b-2)(b+1)=0$
$\displaystyle \Rightarrow b = 2, -1$
$\displaystyle \Rightarrow s = 1, -2$
Constant term:
$\displaystyle p = 2s$
$\displaystyle \Rightarrow p = 2, -4$
3. Sorry, but I really dont get that.
I dont get why you suppose the other facotr is (rx+s)
and what does $\displaystyle x^3: r$
mean?
Is there a different method to do it?
4. Hello Detanon
Originally Posted by Detanon
Sorry, but I really dont get that.
I dont get why you suppose the other facotr is (rx+s)
Because when one factor of a cubic is a quadratic, the other is linear; i.e. it is of the form $\displaystyle (rx + s)$ for some values of $\displaystyle r$ and $\displaystyle s$.
and what does $\displaystyle x^3: r$mean?
It doesn't mean anything. You have taken it out of context. Read the whole thing. I am comparing coefficients. (Do you know what thay means?) First I am looking at the coefficient of $\displaystyle x^3$. Later on, I look at the coefficients of $\displaystyle x^2, x$ and the constant. I have used a colon (':') each time to show which coefficient I am looking at.
Is there a different method to do it?
No, not a method that is as sensible and straightforward as the one I have shown you.
Read it through again - it's really very simple.
5. I still dont get it.. Is there another method where dividing is involved? I thnik thats the one I get.
6. Hello Detanon
I really don't know what your problem is here. Multiply out the brackets from
$\displaystyle (x^2+bx-2)(rx+s)$
and then note that the result has to be identically equal to
$\displaystyle x^3 +(2b-1)x^2 -p$
(Can you do that?)
In this way you get the equations for $\displaystyle r, s, b$ and $\displaystyle p$ that I have shown you. I repeat: if you understand the basic rules of algebra, then trust me, this is really very simple. (And certainly much easier than trying to do an algebraic long division.)
7. Originally Posted by Detanon
If x² + bx -2 is factor of xcubed + (2b-1)x² - p, find two possible values of p.
Im not sure what method to use to solve this. do i divide them?
If $\displaystyle x^2 + bx - 2$ is a factor then it should be obvious, after a little thought, that the other factor has to be $\displaystyle x + \frac{p}{2}$. This is the only way that you can get the $\displaystyle x^3$ term and the constant term of $\displaystyle -p$.
So $\displaystyle x^3 + (2b - 1) x^2 - p = (x^2 + bx - 2)\left(x + \frac{p}{2}\right)$. Expand this out. You require the coefficient of $\displaystyle x^2$ to be $\displaystyle 2b - 1$ and the coeffcient of $\displaystyle x$ to be zero:
$\displaystyle \frac{p}{2} + b = 2b - 1 \Rightarrow p = 2b - 2$ .... (1)
$\displaystyle \frac{bp}{2} - 2 = 0 \Rightarrow bp = 4$ .... (2)
Solve equations (1) and (2) simultaneously for b and p.
For checking purposes:
Spoiler:
b = 2, -1 and p = 2, -4
8. Where do you get
Constant term:
p = 2s
from??
9. Hello Detanon
Originally Posted by Detanon
Where do you get
Constant term:
p = 2s
from??
The constant term when we expand:
$\displaystyle (x^2+bx-2)(rx+s)$ is $\displaystyle -2s$, and this is equal to the constant term in $\displaystyle x^3 +(2b-1)x^2 -p$
Hence $\displaystyle p = 2s$ | 1,240 | 4,011 | {"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} | 4.3125 | 4 | CC-MAIN-2018-22 | latest | en | 0.836116 |
https://math.stackexchange.com/questions/3172943/number-of-matrices-with-each-row-and-column-having-exactly-one-1 | 1,561,542,965,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560628000266.39/warc/CC-MAIN-20190626094111-20190626120111-00226.warc.gz | 512,959,215 | 35,924 | Number of matrices with each row and column having exactly one 1.
Consider a square matrix of order $$n = 5$$ such that $$a_{ij} = 0 ~ \forall ~ i+j = n+1; a_{ij} \in \{0,1\}$$. In each row as well as in each column there is only one non zero element. Then number of such matrices is?
First we note that right diagonal has only $$0$$s.
Then I tried it this way: We choose a place for one among each row and mark the other places in that column and same row as forbidden (i.e. no more one's).
So for first column we have 4 choices then 3 choices then 2 then 1 and then 2s.
Thus, Number of ways = $$4\times 3 \times 2\times 1\times 2 = 48$$
But its erroneous because the number of choices change if we place 1 above 0 in each attempt.
What's the correct way to solve this question?
Flip the array uoside down. Then $$a_{ii}=0$$, so the positions of the nonzero $$a_{ij}$$ form a derangement of the numbers from $$1$$ to $$5$$.
• Suppose the nonzero cells are $a_{1a},a_{2b},a_{3c},a_{4d}$ and $a_{5e}$. Then $a,b,c,d,e$ are five different numbers, and $1\neq a,2\neq b,3\neq c,4\neq d,5\neq e$. – Empy2 Apr 3 at 8:47
• The only one in row 1 is $a_{1a}$. The only one in column $a$ is $a_{1a}$. – Empy2 Apr 3 at 9:49 | 406 | 1,221 | {"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": 8, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2019-26 | longest | en | 0.864078 |
http://mathoverflow.net/questions/61125/system-of-two-second-order-differential-equations | 1,469,681,712,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257827791.21/warc/CC-MAIN-20160723071027-00085-ip-10-185-27-174.ec2.internal.warc.gz | 160,288,893 | 16,449 | # system of two second order differential equations
Hi everyone, I know that this system dont have analytical solutions. I want to get numerical solutions, but in function of some constants $A_i$. Mathematica can help me, but if somebody have idea? This equations describe a physical model
$$A_6 x + A_4 (y')^2 - 2 A_2 x'' - A_3xy'' + A_4yy'' = 0$$
$$A_5 - A_3 (x')^2 - A_3xx'' + A_4yx'' - 2A_1y'' = 0$$
$'=(d/dt)$, $''=(d^2/dt^2)$, $A_i$-known constants. The initial conditions are:
$$x(0)=a, y(0)=0, x'(0)=0, y'(0)=0$$
Thank you in advance!!!
-
What is the "physical model" this system describes; i.e., where does it come from? Knowing this might help answer your question. – drbobmeister Apr 10 '11 at 18:03
## 2 Answers
You can reduce the number of parameters quite a bit, for starters. Set $$x = \alpha u , \; y = \beta v, \; t = \gamma \tau$$
and write $\dot w = \frac{d}{d \tau} w, \ddot w = \frac{d^2}{d\tau^2} w$. By choosing the constants $\alpha, \beta, \gamma$ properly, you should be able to nondimensionalize the system to something like $$c u + (\dot v)^2 - \ddot u - d u \ddot v + v \ddot v = 0$$ $$1 - (\dot u)^2 - u \ddot u + d^{-1} \ddot u v - \ddot v = 0$$ $$u(0) = \tilde a, \; v(0) = \dot u(0) = \dot v(0) = 0.$$ So there are only three independent constants in the system, not 7.
-
Thank you a lot professor Engler. In my new post - New system of two second order differential equations I got maybe simple system but I am not sure can I get a solutions like this way. If you can help me it will be great. In some way, if the number of constants can reduce, can I final get the solutions [x(...,t),y(,...t)]? Thank you again! – reptil Apr 13 '11 at 8:37
If you can take this problem into consideration to reduce number of constants to get semi-analytical solutions for my new post, I will be very grateful (New system of two second order differential equations). Of course, to lead system to may cause long with analytical aid. But I have 5 constants and Runge-Kutte can not help me. Just for special cause, where constants have numerical value. Thank you professor Engler. – reptil Apr 13 '11 at 14:17
It is unlikely that there is an analytic solution but you may be able to make some progress by rewriting as a first-order system. For example, with the equations as in Hans Engler's answer, you can define $w=\dot{u}$ and $z=\dot{v}$, and get a system of equations
$\dot{u} = w$
$\dot{v} = z$
$\dot{w} = \frac{cu + z^2 + (v-du)(1-w^2)}{1-(du-v)(u-v/d)}$
$\dot{z} = \frac{(v/d-u)(cu+z^2) + 1 - w^2}{1-(du-v)(u-v/d)}$
with
$u(0)=\tilde{a}$, $v(0)=w(0)=z(0)=0$.
This looks more complicated than the original set of equations, but you have the advantage that it's first-order and autonomous, and hence amenable to the techniques applicable to first-order autonomous nonlinear dynamical systems, such as linearization about fixed points, analysis of periodic orbits, energy theorems etc.
-
Thank you for good suggestion. If you can describe some of method on simple example or take me to some literature. In my new post - New system of two second order differential equations I got maybe simple system but I am not sure can I get a solutions like this way. If you can help me it will be great. In some way, if the number of constants can reduce, can I final get the solutions [x(...,t),y(,...t)]? Thank you a lot! – reptil Apr 13 '11 at 8:36
What do you think to find dw/dz and eliminate t? Can you help me to describe my new system of two second order differential equations to find w and z like here? Yes iti is trivial but I didn't understand how Hans Engler got just 1 in front of almost all (du/dt,d^2v/dt^2...) – reptil Apr 14 '11 at 11:19 | 1,122 | 3,683 | {"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} | 3.796875 | 4 | CC-MAIN-2016-30 | latest | en | 0.850732 |
https://math.stackexchange.com/questions/1416175/ternary-expansion-and-cantor-set | 1,563,608,528,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526489.6/warc/CC-MAIN-20190720070937-20190720092937-00339.warc.gz | 474,247,133 | 35,396 | # Ternary expansion and Cantor set
If $x$ has a ternary expansion $\sum \limits_{k=1}^{\infty}\dfrac{c_k}{3^k}$ where each $c_k\in \{0,2\}$ then $x$ belongs to Cantor set.
Proof: Suppose $x$ has a ternary expansion $\sum \limits_{k=1}^{\infty}\dfrac{c_k}{3^k}$ where each $c_k\in \{0,2\}$. We will show $x\in C$ by induction. Clearly $x\in C_0=[0,1]$ since $0\leqslant x \leqslant 1$. Next, if $c_1=0$, then $$x\leqslant (0.0222\dots)_3=(0.1)_3=\frac{1}{3}$$ On the other hand, if $c_1=2$, then $$x\geqslant (0.2000\dots)_3=(0.2)_3=\frac{2}{3},$$and hence, $x\in C_1$. Now we approach the inductive step.
Assume $x\in C_k$ for some $k\in \mathbb{N}$. Then $x$ is in any of the $2^k$ subintervals of $C_k$, each of the length $\frac{1}{3^k}$. WLOG, say $x\in [a,b]$. Since we remove the interior of the middle-thirds, we have two disjoint closed intervals, $$\left[a, a+\frac{1}{3^{k+1}}\right] \quad\text{and} \quad \left[b-\frac{1}{3^{k+1}}, b\right]$$If $c_{k+1}=0$, then $x=\sum \limits_{n=1}^{\infty}\dfrac{c_n}{3^n}\leqslant \sum \limits_{n=1}^{k}\dfrac{c_n}{3^n}+\dfrac{1}{3^{k+1}}$ and $x\in [a,b]$.
Why in this case the right side of inequality is $\leqslant a+\dfrac{1}{3^{k+1}}$?
• Because $\sum\limits_{n=k+2}^\infty\frac2{3^n}=\frac1{3^{k+1}}$. – Did Aug 31 '15 at 20:12
• @Did, I know this. I am asking why $\sum \limits_{n=1}^{k}\dfrac{c_n}{3^n}\leqslant a$ – ZFR Aug 31 '15 at 20:21
• Actually, by construction, $\sum\limits_{n=1}^k\frac{c_n}{3^n}=a$... – Did Aug 31 '15 at 20:25
• Is it also possible to argue that $x$ is removed in the $k$-th iteration only if $\alpha_k = 1$ for all ternary expansions of $x$? – David Nov 5 '15 at 19:37 | 711 | 1,659 | {"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} | 4.34375 | 4 | CC-MAIN-2019-30 | latest | en | 0.526084 |
https://blog.c0nrad.io/posts/scattering-angles/ | 1,611,528,601,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703557462.87/warc/CC-MAIN-20210124204052-20210124234052-00301.warc.gz | 236,031,974 | 5,498 | ## Learning and learning
Aug 30, 2020 - 3 minute read
# Scattering Angles
Calculating scattering angles from fully elastic spherical collision.
## Background
In one of my first blog posts, I talked about the difficulty I had with calculating the resulting angles from a collision of two spheres.
I could bounce objects off each other, but not at an angle.
Well, fast forward to today (5 months later), I’m now in school for physics (undergrad)! And a little bit better prepared to tackle the problem.
I reached out to my professor of classical mechanics about the above problem, and he suggested I read Ch. 14 of Taylor’s Mechanics on Collision Theory. The chapter was pretty interesting, and it was stuff I wanted to learn sooner or later. I’m hoping soon I can start modeling particle collisions (a long term goal of mine, I’m about halfway through “Elementary Particles” by Griffiths).
Anyways, in Taylor’s book, example 14.5 talks about hard sphere scattering, and it turns out that the angle of incidence and reflection are equal. Kind of reminds me of bouncing photons off a medium.
To be honest, I don’t quite understand why the angles should be equal. There’s probably some beautiful action/Lagrangian minimization that shows that it must be the path. I’ll solve that another day.
But, poking around some more, I found out that wikipedia lists the equations I needed:
$$\bm{v_1 \prime} = \bm{v_1} - \frac{2m_2}{m_1 + m_2} \frac{ < \bm{v_1} - \bm{v_2}, \bm{x_1} - \bm{x_2}>}{ || \bm{x_1} - \bm{x_2} ||^2} ( \bm{x_1} - \bm{x_2})$$
$$\bm{v_2 \prime} = \bm{v_1} - \frac{2m_1}{m_1 + m_2} \frac{ < \bm{v_2} - \bm{v_1}, \bm{x_2} - \bm{x_1}>}{ || \bm{x_2} - \bm{x_1} ||^2} ( \bm{x_2} - \bm{x_1})$$
With this, I could put a bunch of balls into a container and watch them bounce! I think I might use this in the future to do some Statistical Mechanics/Boltzman entropy modeling fun later:
## Scattering
Anyways, I figured it’d be fun to build something similar to Rutherford’s Scattering Experiment:
(Code)
It turned out pretty nice.
From this I was able to plot the scattering angle against the impact parameter (b, how much offset the two items are).
(Code)
Pretty neat stuff. It’s cool how often stuff bounces back. I imagine this is probably similar to what Rutherford saw all those years ago.
### What was difficult
• In terminal, character height is twice as much as character width, so the points appears to move much faster in one direction. So instead the renderer now treats one character height as 2 units of length
• I’m using a very simple time-stepping integrator, and some of the incident angles are pretty sensitive, so if the ball moves to fast I’ll get nonsense collision angles since the circles will be on top of each other
• The correct thing to do would be to reverse the simulation till the two spheres are at the point of collision, and then calculate the angles
• Being lazy, I just set the timestep to 1000 frames/second, in the future I’ll go back and do it correct
Otherwise not much. I wish I understood the math behind the collisions better.
### What went well
• I really prefer doing graphics in terminal compared to rendering in 3d with webgl, being able to quickly iterate is very nice.
### Future
Probably nothing for this project. Up next I want to calculate the Clebsh-Gordan coefficients. Eventually I’ll come back to this for elementary particle collisions. First will probably be relativistic collisions. | 877 | 3,469 | {"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} | 3.765625 | 4 | CC-MAIN-2021-04 | longest | en | 0.899301 |
https://ddxart.co.uk/the-goldbach-conjecture/ | 1,631,869,714,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055632.65/warc/CC-MAIN-20210917090202-20210917120202-00578.warc.gz | 235,816,442 | 8,881 | # The Goldbach Conjecture
Mathematicians like to prove things. Things that most of us would take for granted, statements like 1+1=2 are of no use to a mathematician unless it they can be ‘proved’ – demonstrated to be true in every circumstance. In Bertrand Russell and Alfred North Wallace’s great work, The Principles of Mathematics, several hundred pages are devoted to proving that 1+1=2. Why? Because in other places mathematicians will want to use the statement that 1+1=2 as part of the proof of something far more complex, and any proof is only as good as the parts that make it up. The history of mathematics is littered with proofs that relied on something that was ‘self-evident’, only to have it turn out years later that the proof is useless because the self-evident something turned out not always to be true.
It’s every mathematician’s dream to produce a proof of something that has never been proved before (or to demolish someone else’s proof). Many of the propositions that have been put forward but not yet proved are exceedingly complex but some are so simple that you’d think it would take no more than a few minutes to work out why they ‘must’ be true.
So it was that back in the 18th century, a German mathematician named Christian Goldbach suggested that every even number other than 2 could be produced by adding together two prime numbers – numbers that can only be divided by themselves and 1 without leaving a remainder. It doesn’t take long to demonstrate that it is true for even numbers of reasonable size and, in the centuries following, Goldbach’s ‘conjecture’ was found to be true of bigger and bigger even numbers. The advent of computers has led to the analysis of numbers of ridiculous size being considered and, yes, every one of them can be produced by adding together two prime numbers.
But no-one has ever proved that it will be true for every conceivable number, and there are plenty of cases of statements that turned out to be true over and over again, with bigger and bigger numbers, only for an exception to be found – often after many years.
So two centuries later Goldbach’s Conjecture remains just that, a conjecture. In practice, unless you’re going to be working with numbers that have 20 or more digits, you needn’t worry, because that’s how far the conjecture has been shown to be true. But don’t try telling a mathematician that the conjecture ‘must’ be true after all this time – or they may ask you to prove it!
Anyway this little illustration is hardly the most visually exciting of the collection but it does make the point – for the first few even numbers at least – and there are some interesting patterns if you look. Simply follow the lines from any pair of prime numbers, one on each side, and where they meet they are added together to produce an even number.
Before you ask, the number 1 isn’t included since mathematicians don’t consider it a prime number – you can work out for yourself why if you look back at the definition of a prime number given above.
And finally, a little problem to ponder. If the Goldbach Conjecture is ever proved to be true, it will also prove that any number is actually the sum of three prime numbers. Can you see how? | 695 | 3,221 | {"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-2021-39 | latest | en | 0.973207 |
https://physics.com.hk/2021/08/10/ex-1-21-the-dumbbell-3-2/ | 1,675,619,890,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500273.30/warc/CC-MAIN-20230205161658-20230205191658-00356.warc.gz | 477,335,245 | 16,816 | # Ex 1.21 The dumbbell, 3.2
Structure and Interpretation of Classical Mechanics
.
c. Make a change of coordinates to a coordinate system with center of mass coordinates $\displaystyle{x_{cm}}$, $\displaystyle{y_{cm}}$, angle $\displaystyle{\theta}$, distance between the particles $\displaystyle{c}$, and tension force $\displaystyle{F}$. Write the Lagrangian in these coordinates, and write the Lagrange equations.
~~~
[guess]
\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} y_{cm} &= \frac{m_0 y_0 + m_1 y_1}{m_0 + m_1} \\ x_{cm} &= \frac{m_0 x_0 + m_1 x_1}{m_0 + m_1} \\ \end{aligned}}
\displaystyle{ \begin{aligned} \ddot y_{cm} &= \frac{F \sin \theta - F \sin \theta}{m_0 + m_1} = 0 \\ \ddot x_{cm} &= \frac{F \cos \theta - F \cos \theta}{m_0 + m_1} = 0 \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} y_0 &= y_{cm} - \frac{m_1}{M} c(t) \sin \theta \\ x_0 &= x_{cm} - \frac{m_1}{M} c(t) \cos \theta \\ y_1 &= y_{cm} + \frac{m_0}{M} c(t) \sin \theta \\ x_1 &= x_{cm} + \frac{m_0}{M} c(t) \cos \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} x_1 - x_0 &= \frac{m_0}{M} c(t) \cos \theta + \frac{m_1}{M} c(t) \cos \theta \\ &= c(t) \cos \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} \dot x_1 - \dot x_0 &= \dot c(t) \cos \theta - c(t) \dot \theta \sin \theta \\ \end{aligned}}
\displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= \ddot c(t) \cos \theta - \dot c(t) \dot \theta \sin \theta - \dot c(t) \dot \theta \sin \theta - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
\displaystyle{ \begin{aligned} y_1 - y_0 &= c(t) \sin \theta \\ \dot y_1 - \dot y_0 &= \dot c(t) \sin \theta + c(t) \dot \theta \cos \theta \\ \ddot y_1 - \ddot y_0 &= \ddot c(t) \sin \theta + \dot c(t) \dot \theta \cos \theta + \dot c(t) \dot \theta \cos \theta + c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} m_0 \ddot y_0 &= F \sin \theta \\ m_0 \ddot x_0 &= F \cos \theta \\ m_1 \ddot y_1 &= -F \sin \theta \\ m_1 \ddot x_1 &= -F \cos \theta \\ \end{aligned}}
When $\displaystyle{\dot c(t) = 0}$ and $\displaystyle{\ddot c(t) = 0}$,
\displaystyle{ \begin{aligned} \ddot x_1 - \ddot x_0 &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
\displaystyle{ \begin{aligned} \ddot y_1 - \ddot y_0 &= ... \\ - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} \tan \theta &= \frac{ c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta }{- c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta} \\ &= \frac{ \ddot \theta - \dot \theta^2 \tan \theta }{- \ddot \theta \tan \theta - \dot \theta^2} \\ \end{aligned}}
\displaystyle{ \begin{aligned} \tan \theta \left( - \ddot \theta \tan \theta - \dot \theta^2 \right) &= \ddot \theta - \dot \theta^2 \tan \theta \\ &... \\ 0 &= \ddot \theta (1 + \tan^2 \theta) \\ \ddot \theta &= 0 \\ \end{aligned}}
.
\displaystyle{ \begin{aligned} - \left( \frac{1}{m_1} + \frac{1}{m_0} \right) F \sin \theta &= c(t) \ddot \theta \cos \theta - c(t) \dot \theta^2 \sin \theta \\ - \left( \frac{1}{m_1} + \frac{1}{m_1} \right) F \cos \theta &= - c(t) \ddot \theta \sin \theta - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
Let $\displaystyle{\frac{1}{\mu} = \left( \frac{1}{m_1} + \frac{1}{m_1} \right)}$ and since $\displaystyle{\ddot \theta = 0}$,
\displaystyle{ \begin{aligned} - \frac{1}{\mu} F \sin \theta &= - c(t) \dot \theta^2 \sin \theta \\ - \frac{1}{\mu} F \cos \theta &= - c(t) \dot \theta^2 \cos \theta\\ \end{aligned}}
Since $\displaystyle{\sin \theta}$ and $\displaystyle{\cos \theta}$ cannot be both zero at the same time,
\displaystyle{ \begin{aligned} - \frac{1}{\mu} F &= - c(t) \dot \theta^2 \\ \end{aligned}}
Put $\displaystyle{c(t) = l}$,
\displaystyle{ \begin{aligned} \frac{1}{\mu} F &= l \dot \theta^2 \\ \dot \theta^2 &= \frac{1}{l \mu} F \\ \end{aligned}}
[guess]
— Me@2021-08-08 05:41:21 PM
.
. | 1,750 | 4,361 | {"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": 29, "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-2023-06 | latest | en | 0.356996 |
http://math.stackexchange.com/questions/265897/whether-twin-primes-satisfy-this-one | 1,466,820,119,000,000,000 | text/html | crawl-data/CC-MAIN-2016-26/segments/1466783391766.5/warc/CC-MAIN-20160624154951-00135-ip-10-164-35-72.ec2.internal.warc.gz | 188,713,330 | 18,105 | # Whether twin primes satisfy this one?
It seems that difference of squares of any twin primes $+1$ will always lead to
number which might be
a) A square of a twin prime
b) Itself a twin prime
$C$ = ($A^2$-$B^2$ )+$1$ ------> $(1)$
Where
$C$ --- > might be a twin prime or square of a twin prime,
$A$ and $B$ are twin primes where $A$ is > $B$
My questions is whether eqn ($1$) is true?
-
When you say $A$ and $B$ are twin primes with $A > B$, do you mean $A = B + 2$, or that $A$ can be any twin prime greater than $B$? – Michael Albanese Dec 27 '12 at 12:57
If $A$, $B$ are twin primes, they differ by 2, so the conjecture seems to be that
if $C = 2(A + B) + 1$, with $A$, $B$ twin primes, then $C$ is either a twin prime or square of a twin prime.
That's quickly falsified by taking $A = 101, B = 103$. For then $C = 409$ which is neither a twin prime nor the square of one.
-
Similarly 137, 139, 553 and many (most?) larger examples – Henry Dec 27 '12 at 13:02
$\,(103^2-101^2)+1=409\,$ , which is neither of (a)-(b), though it is a prime.
$\,(4801^2-4799^2)+1=19201=7\cdot 13\cdot 211\,$ , which is neither of (a)-(b) and not even a prime
-
Again the number 19201 is a product of primes? – Shan Dec 27 '12 at 13:14
Well, any natural number is a prime or a product of primes except $\,1\,$....is this what you actually meant to ask? – DonAntonio Dec 27 '12 at 13:15
I agree..need to work more on this.. – Shan Dec 27 '12 at 13:19
Some counterexamples:
$$\begin{array}{c|c} p & (p+2)^2-p^2+1 \\\hline 137 & 7^1 \times 79^1 \\ 179 & 7^1 \times 103^1 \\ 197 & 13 \times 61^1 \\ 227 & 11^1\times 83^1 \\ 269 & 23^1\times 47^1 \\ 347 & 7^1\times 199^1 \\ 431 & 7^1\times 13^1\times 19^1 \end{array}$$
- | 635 | 1,716 | {"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} | 3.5625 | 4 | CC-MAIN-2016-26 | latest | en | 0.83293 |
http://www.ck12.org/book/Texas-Instruments-Calculus-Student-Edition/r1/section/5.1/anchor-content/ | 1,455,264,555,000,000,000 | text/html | crawl-data/CC-MAIN-2016-07/segments/1454701163512.72/warc/CC-MAIN-20160205193923-00217-ip-10-236-182-209.ec2.internal.warc.gz | 343,395,521 | 28,486 | <img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Calculus Student Edition Go to the latest version.
# 5.1: Exploring Area under the Curve
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Calculus, Chapter 4, Lesson 3.
## Problem 1 – Explore and discover
Graph the curve $y = x^2$.
Your challenge is to think of at least two ways to estimate the area bounded by the curve $y = x^2$ and the $x-$axis on the interval [0, 1] using rectangles. Use the following guidlines:
• all rectangles must have the same width
• you must build all your rectangles using the same methods
• the base of each rectangle must lie on the $x-$axis
Graph $y = x^2$ and set your window to [-0.1, 1] for $x$ and [-0.2, 1.3] for $y$. Draw your first and second method on the graphs below. For each method calculate the following:
• Number of rectangles
• Height and width of each one
• Area of each
• Sum of the area
• Which method did a better job?
• How could you improve on it?
In the following problem, you will examine three common techniques that use rectangles to find the approximate area under a curve. Perhaps you discovered some of these techniques during your exploration in the above problem. The first problem uses rectangles whose right-endpoints lie on the curve $y = x^2$.
## Problem 2 – Using five right-endpoint rectangles
Divide the interval [0, 1] into five equal pieces. Enter the information for each interval or rectangle in the table below. Remember that the right endpoint is the $x-$value and the height is the $y-$value of the right endpoint on the curve.
Interval Right Endpoint Height Area
• Calculate this sum.
• Now add up the numbers in the Area column.
The formula that can be used to express the total area is:
$R_5 &= 0 . 2 \cdot f1(0 . 2) + 0 . 2 \cdot f1(0 . 4) + 0 . 2 \cdot f1(0 . 6) + 0 . 2 \cdot f1(0 . 8) + 0 . 2 \cdot f1(1 . 0)\\& \qquad \qquad \qquad \qquad \qquad \qquad \text{or}\\R_5 &= 0 . 2 [f1(0 . 2) + f1(0 . 4) + f1(0 . 6) + f1(0 . 8) + f1(1 . 0)]$
• Are these two numbers the same or different?
Another way to find the area of the rectangles is using sigma notation.
• Write the notation in the $\sum_{x=1}^5x^2$ form. Adjust what is being summed.
To sum it on the calculator, use Home > F3:Calc > 4:Sigma for the command with the format:
$\sum$(expression, variable, lower limit, upper limit)
• Does this agree with the answer for the area you found previously?
## Problem 3 – Using five left-endpoint rectangles
This problem uses rectangles whose left-endpoints lie on the curve $y = x^2$.
Divide the interval [0, 1] into five equal pieces. Enter the information for each interval in the table below. Remember that the left endpoint is the $x-$value and the height is the $y-$value of the left endpoint on the curve.
Interval Left Endpoint Height Area
• Calculate this sum.
• Now add up the numbers in the Area column.
The formula that can be used to express this area is:
$L_5 &= 0 . 2 \cdot f1(0) + 0 . 2 \cdot f1(0 . 2) + 0 . 2 \cdot f1(0 . 4) + 0 . 2 \cdot f1(0 . 6) + 0 . 2 \cdot f1(0 . 8)\\& \qquad \qquad \qquad \qquad \qquad \text{or}\\L_5 &= 0 . 2 [f1(0) + f1(0 . 2) + f1(0 . 4) + f1(0 . 6) + f1(0 . 8)]$
• Are these two numbers the same or different?
• What is the sigma notation for the area of the rectangles?
• Use the calculator to find the sum. Does this result agree with the answer above?
## Problem 4 – Using five midpoint rectangles
We will now investigate a midpoint approximation. How would you draw five rectangles, with equal width, such that their midpoints lie on the curve $y = x^2$?
Divide the interval [0, 1] into five equal pieces. Enter the information for each interval or rectangle in the table below. Remember that the midpoint is the $x-$value and the height is the $y-$value of the midpoint on the curve.
Interval Midpoint Height Area
• Calculate this sum.
• Now add up the numbers in the Area column.
• Are these two numbers the same or different?
The formula that can be used to express this area is:
$M_5 &= 0 . 2 \cdot f1(0 . 1) + 0 . 2 \cdot f1(0 . 3) + 0 . 2 \cdot f1(0 . 5) + 0 . 2 \cdot f1(0 . 7) + 0 . 2 \cdot f1(0 . 9)\\ & \qquad \qquad \qquad \qquad \qquad \text{or}\\M_5 &= 0 . 2 [f1(0 . 1) + f1(0 . 3) + f1(0 . 5) + f1(0 . 7) + f1(0 . 9)]$
• What is the sigma notation for the area of the rectangles?
• Use the calculator to find the sum. Does this result agree with the answer above?
## Problem 5- Summarize your findings
In this activity, you explored three different methods for approximating the area under a curve. The exact area under the curve $y = x^2$ on the interval [0, 1] is $\frac{1}{3}$ or 0.333.
• Which approximation produced the best estimate for the actual area under the curve?
• Describe which factors contribute to left, right, and midpoint rectangles giving overestimates versus underestimates.
• What can you do to ensure that all three of these techniques produce an answer that is very close to $\frac{1}{3}$? Test your conjecture by using evaluating a sum that produces a much more accurate answer.
Feb 23, 2012
Nov 04, 2014 | 1,568 | 5,270 | {"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": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 24, "texerror": 0} | 4.6875 | 5 | CC-MAIN-2016-07 | latest | en | 0.821074 |
https://numberworld.info/root-of-32052 | 1,653,793,527,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652663035797.93/warc/CC-MAIN-20220529011010-20220529041010-00247.warc.gz | 487,678,751 | 2,962 | # Root of 32052
#### [Root of thirty-two thousand fifty-two]
square root
179.0307
cube root
31.7652
fourth root
13.3802
fifth root
7.9647
In mathematics extracting a root is known as the determination of the unknown "x" in the equation $y=x^n$ The outcome of the extraction of the root is seen as a mathematical root. In the case of "n is equivalent to 2", one talks about a square root or sometimes a second root also, another possibility could be that n = 3 then one would consider it a cube root or simply third root. Considering n beeing greater than 3, the root is declared as the fourth root, fifth root and so on.
In maths, the square root of 32052 is represented as this: $$\sqrt[]{32052}=179.03072362028$$
Also it is possible to write every root down as a power: $$\sqrt[n]{x}=x^\frac{1}{n}$$
The square root of 32052 is 179.03072362028. The cube root of 32052 is 31.765208577536. The fourth root of 32052 is 13.380236306593 and the fifth root is 7.9647294273138.
Look Up | 285 | 983 | {"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} | 3.6875 | 4 | CC-MAIN-2022-21 | latest | en | 0.908355 |
https://cracku.in/79-common-factor-of-12a4b6-18a6c2-36a2b2-is-x-ssc-chsl-11-jan-2017-afternoon-shift | 1,718,270,987,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861372.90/warc/CC-MAIN-20240613091959-20240613121959-00700.warc.gz | 169,516,648 | 21,721 | Question 79
# Common factor of $$12a^4b^6, 18a^6c^2, 36a^2b^2$$ is
Solution
Factors of :Â
$$12a^4b^6$$ = $$(2 \times 6) \times (a^2 \times a^2) \times b^6$$
$$18a^6c^2$$ = $$(3 \times 6) \times (a^2 \times a^4) \times c^2$$
$$36a^2b^2$$ = $$(6 \times 6) \times (a^2) \times b^2$$
The common factor in the 3 terms = $$6a^2$$
=> Ans - (D) | 175 | 345 | {"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} | 3.890625 | 4 | CC-MAIN-2024-26 | latest | en | 0.543417 |
https://www.jiskha.com/display.cgi?id=1164059050 | 1,516,385,496,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084888077.41/warc/CC-MAIN-20180119164824-20180119184824-00076.warc.gz | 928,424,281 | 4,097 | # physics
posted by .
A mass of 20 kg on a plane inclined at 40 degrees. A string attached to that mass goes up the plane, passed over a pullley and is attached to mass of 30 kg that hangs verticalyy. a) find the acceleration and it's dirction b) the tension in the string. Assume no friction.
I first drew the picture. How would I find the equation since F=ma doesn't include everything?
F=ma (combined with Newton's third law) give you all you need to solve this problem.
The force on the hanging mass in the downward direction is:
F2 = m2 * g - T
where m2 = 30 kg
The force on the other mass in the direction parallel to the plane in which the string is pulling is:
F1 = -m1 * g sin(40°) + T
where m1 = 20 kg
Begause the string is assumed to be of fixed length the acceleration of mass 1 in the direction the string is pulling must be the same as the acceleration of mass 2 in the downward direction.
This means that:
F1/m1 = F2/m2
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More Similar Questions | 878 | 3,547 | {"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 | 0.908985 |
https://www.statology.org/testing-the-significance-of-a-regression-slope/ | 1,585,845,378,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370506988.10/warc/CC-MAIN-20200402143006-20200402173006-00255.warc.gz | 1,155,693,769 | 9,894 | # Testing the Significance of a Regression Slope
## Conducting a Simple Linear Regression
Suppose we have the following dataset that shows the square feet and price of 12 different houses:
We want to know if there is a significant relationship between square feet and price. To get an idea of what the data looks like, we first create a scatterplot with square feet on the x-axis and price on the y-axis:
We can clearly see that there is a positive correlation between square feet and price. As square feet increases, the price of the house tends to increase as well.
However, to know if there is a statistically significant relationship between square feet and price, we need to run a simple linear regression.
So, we run a simple linear regression using square feet as the predictor and price as the response and get the following output:
Whether you run a simple linear regression in Excel, SPSS, R, or some other software, you will get a similar output to the one shown above.
Recall that a simple linear regression will produce the line of best fit, which is the equation for the line that best “fits” the data on our scatterplot. This line of best fit is defined as:
ŷ = b0 + b1
where ŷ is the predicted value of the response variable, b0 is the y-intercept, b1 is the regression coefficient, and x is the value of the predictor variable.
The value for b0 is given by the coefficient for the intercept, which is 47588.70.
The value for b1 is given by the coefficient for the predictor variable Square Feet, which is 93.57.
Thus, the line of best fit in this example is ŷ = 47588.70+ 93.57x
Here is how to interpret this line of best fit:
• b0When the value for square feet is zero, the average expected value for price is \$47,588.70. (In this case, it doesn’t really make sense to interpret the intercept, since a house can never have zero square feet)
• b1For each additional square foot, the average expected increase in price is \$93.57.
So, now we know that for each additional square foot, the average expected increase in price is \$93.57. To find out if this increase is statistically significant, we need to conduct a hypothesis test for B1 or construct a confidence interval for B1.
Note: A hypothesis test and a confidence interval will always give the same results.
## Constructing a Confidence Interval for a Regression Slope
To construct a confidence interval for a regression slope, we use the following formula:
b1 +/- (t1-∝/2, n-2) * (standard error of b1)
where:
• b1 is the slope coefficient given in the regression output
• (t1-∝/2, n-2) is the t critical value for confidence level 1-∝ with n-2 degrees of freedom where is the total number of observations in our dataset
• (standard error of b1) is the standard error of b1 given in the regression output
For our example, here is how to construct a 95% confidence interval for B1:
• b1 is 93.57 from the regression output.
• Since we are using a 95% confidence interval, ∝ = .05 and n-2 = 12-2 = 10, thus t.975, 10 is 2.228 according to the t-distribution table
• (standard error of b1) is 11.45 from the regression output
Thus, our 95% confidence interval for Bis:
93.57 +/- (2.228) * (11.45) = (68.06 , 119.08)
This means we are 95% confident that the true average increase in price for each additional square foot is between \$68.06 and \$119.08.
Notice that \$0 is not in this interval, so the relationship between square feet and price is statistically significant at the 95% confidence level.
## Conducting a Hypothesis Test for a Regression Slope
To conduct a hypothesis test for a regression slope, we follow the standard five steps for any hypothesis test:
Step 1. State the hypotheses.
The null hypothesis (H0): B1 = 0
The alternative hypothesis: (Ha): B1 ≠ 0
Step 2. Determine a significance level to use.
Since we constructed a 95% confidence interval in the previous example, we will use the equivalent approach here and choose to use a .05 level of significance.
Step 3. Find the test statistic and the corresponding p-value.
In this case, the test statistic is = coefficient of b1 / standard error of b1 with n-2 degrees of freedom. We can find these values from the regression output:
Thus, test statistic = 92.89 / 13.88 =6.69.
Using the T Score to P Value Calculator with a t score of 6.69 with 10 degrees of freedom and a two-tailed test, the p-value = 0.000.
Step 4. Reject or fail to reject the null hypothesis.
Since the p-value is less than our significance level of .05, we reject the null hypothesis.
Step 5. Interpret the results.
Since we rejected the null hypothesis, we have sufficient evidence to say that the true average increase in price for each additional square foot is not zero. | 1,177 | 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} | 4.0625 | 4 | CC-MAIN-2020-16 | latest | en | 0.878242 |
http://mathhelpforum.com/number-theory/174792-repeating-decimals-1-7-a.html | 1,481,222,286,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698542648.35/warc/CC-MAIN-20161202170902-00095-ip-10-31-129-80.ec2.internal.warc.gz | 174,070,234 | 10,046 | 1. ## Repeating Decimals 1/7
Hello,
How would I go about explaining why all the fractions of seven consist of the same 6 digits cyclically permuted, and how would I explain why the first three digits and last three digits of the fraction add to 999? I believe it has something to do with Fermat's Little Theorem, but I'm not sure exactly how to explain these phenomenon. Any help is appreciated!
2. Hi,
in a fast attemp of solving the problem I noticed the following fact (that might be a rough proof).
First of all, division (when decimals exist) is based in the last digit a [0...9] with 7.
if it less than 7 must be done: $a0$
ie. $[10, 20, 30, 40, 50, 60]$
Next note the following:
$10 mod 7=3$ ... $10 floor 7=1$
Next we append to 3 the zero so we can find the remainding term:
$30 mod 7=2$ ... $30 floor 7=4$
...
$20 mod 7=6$ ... $20 floor 7=2$
...
$60 mod 7=4$ ... $60 floor 7=8$
...
$40 mod 7=5$ ... $40 floor 7=5$
...
$50 mod 7=1$ ... $50 floor 7=7$
where $floor$ is the integer part of the division.
There is a period at modulos of 7 with 10, 20, 30, 40, 50, 60
if you intefere more with that fact you can make a more elegant proof
3. Note that this can also be done via Primitive Root tecniques. | 382 | 1,216 | {"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": 15, "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} | 4.03125 | 4 | CC-MAIN-2016-50 | longest | en | 0.874086 |
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To arrive at its destination on time the bus should have
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To arrive at its destination on time the bus should have maintained a speed of $$V$$ kmh throughout the journey. Instead, after going the first third of the distance at $$V$$ kmh, the bus increased its speed and went the rest of the distance at $$(1.2)*V$$ kmh. As a result, the bus arrived at its destination $$X$$ minutes earlier than planned. What was the actual duration of the trip?
1. $$V = 60$$
2. $$X = 20$$
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30 Oct 2008, 02:48
B.
If d is the total distance covered, then,
(d/v)*(1/3 + 2/3*1/1.2) = d/V - X
or, d/V = 9X/60
If we know X, we will get to know d/v and hence actual time taken and that will be d/V - X
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30 Oct 2008, 13:22
B
X = (d/3)/v + (2d/3)/1.2v
X = d/v(1/3+2/3/1.2)
All we need to find is d/v
stmt2 is Sufficient
Re: DS : BUS [#permalink] 30 Oct 2008, 13:22
Display posts from previous: Sort by | 705 | 2,364 | {"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} | 4.21875 | 4 | CC-MAIN-2016-50 | latest | en | 0.914267 |
https://mv-organizing.com/what-is-the-probability-of-rolling-a-2-or-a-3-on-a-six-sided-die/ | 1,660,600,058,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572212.96/warc/CC-MAIN-20220815205848-20220815235848-00595.warc.gz | 378,056,246 | 12,672 | # What is the probability of rolling a 2 or a 3 on a six-sided die?
## What is the probability of rolling a 2 or a 3 on a six-sided die?
Two (6-sided) dice roll probability table
Roll a… Probability
2 1/36 (2.778%)
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
What is the probability of rolling the same number exactly 3 times with 5 six-sided dice?
What is the probability of rolling the same number exactly three times with five six-sided dice? 1/5 5/1296. MOHAMMED KHAJA MUHIDDIN.
What is the probability of rolling a 5 with two dice?
Probabilities for the two dice
Total Number of combinations Probability
3 2 5.56%
4 3 8.33%
5 4 11.11%
6 5 13.89%
### What is the probability of rolling a 2 on a 6 sided die?
16
How many different outcomes are possible for 6 rolls of a die?
We can view the outcomes as two separate outcomes, that is, the outcome of rolling die number one and the outcome of rolling die number two. For each of 6 outcomes for the first die the second die may have any of 6 outcomes, so the total is 6+6+6+6+6+6=36, or more compactly, 6⋅6=36.
What is the probability of rolling a 1 and then a 2 on a six sided dice?
The probability of rolling a 1 when there are six possible options is 1/6. The probability of rolling a 2 when there are six possible options is 1/6. 1/6 * 1/6 = 1/36.
## What are the odds of rolling a 1 with two dice?
The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on.
What is the probability of rolling a sum of 4 with two dice?
3/36
When rolling two dice What is the probability that one die is twice the other?
6/36
### What are the outcomes of rolling two dice?
Rolling two six-sided dice: Each die has 6 equally likely outcomes, so the sample space is 6 • 6 or 36 equally likely outcomes….
First coin Second coin outcome
H T HT
T H TH
T T TT
What is the probability of rolling a 3 twice?
Since the occurrence of 3 on first roll is independent of the occurrence of an odd number on the second roll, Probability of complete event =(probability of showing a 3 on the first roll) x (probability of an odd number on the second roll)=1/6×1/2=1/12.
What is the probability of rolling a 4 twice with two rolls of a number cube?
What is the probability of showing a 4 on both rolls? The probability of the first throw being a 4 is 1/6. The second throw is independent, so it’s also 1/6. For both throws to be a 4 would be 1/6 times 1/6 or 1/36.
## Is there a 100 sided die?
Zocchihedron is the trademark of a 100-sided die invented by Lou Zocchi, which debuted in 1985. Rather than being a polyhedron, it is more like a ball with 100 flattened planes. It is sometimes called “Zocchi’s Golfball”.
What are the odds of rolling double sixes 3 times in a row?
The probability of doing that 3 times in a row is 1/36 to the third power, which would be 1/46656, or about 0.002% of the time. Since there are six sides to a die, you multiply the (1/46656) by 6 to find the probability of rolling the same doubles 3 times in a row. That would be 6/46656, or about 0.013% of the time.
What is the probability of rolling two six sided dice and obtaining an odd number on at least one die?
(a) At least one of the dice shows an even number? P(at least one is even) = 1 – P(both are odd). And the probability that the first die shows an odd number is 1/2, as is the probability that the second does….Probability.
Outcome Probability Product
5 1/6 5/6
6 1/6 6/6
Total: 21/6
### What is the probability of not rolling any 6’s in four rolls of a balanced die?
a) Consider the complement problem, there is a 5/6 probability of not rolling a six for any given die, and since the four dice are independent, the probability of not rolling a six is (5/6)4 = 54/64 = 625/1296. The probability of rolling at least one six is therefore 1 − 625/1296 = 671/1296 ≈ . 517.
What is the probability of rolling a 4 or an odd number?
There are two cases, one where you first get the four, or you get an odd then a four. The probability of just rolling a 4 first is 16. If you roll one odd number before the 4, that has a chance of 12∗16 Now, you can roll 2 or 6 infinitely many times and it won’t matter.
What is the probability of rolling an even number from a fair die?
The probability of rolling an even number is three out of six, or three-sixths.
## What is P A the probability that the six-sided die is an even number?
For one roll, the outcomes are 1,2,3,4,5,6 of which 2,4,6 are even, so the probability is 3/6=1/2.
What is the probability of rolling an odd number?
The probability when rolling a regular six-sided dice that the score is an odd number is three-sixths or three out of six. Both three and six are divisible by three. Therefore, this fraction could be simplified to one-half. Three divided by three is equal to one.
What is the probability of a dice landing on an even number?
The probability of landing on an even number is 2/6 = 1/3, but the probability of landing on an odd number is 4/6 = 2/3.
### Is 1 a odd number?
An odd number is an integer when divided by two, either leaves a remainder or the result is a fraction. One is the first odd positive number but it does not leave a remainder 1. Some examples of odd numbers are 1, 3, 5, 7, 9, and 11. Since odd numbers are integers, negative numbers can be odd.
What is the probability of rolling a 3 or an even number?
Explanation: The probability of rolling a 3 is 16 . The probability of rolling an even number is 36 . This is when you first roll a 3 and then an even number.
What is the probability of rolling an even number and then an odd number?
Answer: Size of sample spaces= 36. Number of desired outcomes=9. P(first is even and second is odd)=1/4.
## What is the probability of rolling a prime number?
We find this number by multiplying 6 x 6. The logic is there are six sides to each die, so for each number on one die you can pair with six different numbers on the other die. Therefore, the probability of rolling a prime number on two dice is 15/36, which reduces to 5/12 (E).
How do you convert odd numbers to even numbers?
An even number can only be formed by the sum of either 2 odd numbers (odd + odd = even), or 2 even numbers (even + even = even). An odd number can only be formed by the sum of an odd and even number (odd + even = odd, or even + odd = odd).
What is the odd numbers from 1 to 100?
The odd numbers from 1 to 100 are: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99.
### What is the largest odd number?
The largest two-digit odd number we could make is 97. We could not make 99 because we only have one 9. If you pulled out an odd number first this should always go in the ‘ones’ column.
Which is the smallest odd whole number?
The smallest even natural number is 2 and smallest odd natural number is 1.
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https://www.sudokulovers.com/solution/082439050040781020900625000060004002204098600100206040000512008000967230020843060/ | 1,656,924,593,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00219.warc.gz | 1,079,192,286 | 18,216 | 8
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This Sudoku Puzzle has 58 steps and it is solved using Locked Candidates Type 1 (Pointing), Locked Candidates Type 2 (Claiming), Skyscraper, Discontinuous Nice Loop, Naked Single, AIC, Hidden Single, Naked Triple, undefined, Full House, Naked Pair techniques.
Try To Solve This Puzzle
## Solution Steps:
1. Locked Candidates Type 1 (Pointing): 1 in b1 => r3c789<>1
2. Locked Candidates Type 2 (Claiming): 1 in c8 => r4c7,r5c9<>1
3. Skyscraper: 9 in r4c8,r6c2 (connected by r7c28) => r4c3,r6c79<>9
4. Discontinuous Nice Loop: 3/7/9 r7c3 =6= r7c1 -6- r1c1 -7- r9c1 -5- r2c1 =5= r2c3 =6= r7c3 => r7c3<>3, r7c3<>7, r7c3<>9
5. Row 7 / Column 3 → 6 (Naked Single)
6. AIC: 8 8- r3c8 =8= r3c7 =4= r3c9 -4- r8c9 =4= r8c1 =8= r8c3 -8- r6c3 =8= r6c7 -8 => r3c7,r4c8<>8
7. Row 3 / Column 8 → 8 (Hidden Single)
8. Discontinuous Nice Loop: 3 r2c1 -3- r7c1 =3= r7c2 =9= r9c3 -9- r9c9 =9= r2c9 =6= r2c1 => r2c1<>3
9. Naked Triple: 5,6,7 in r129c1 => r48c1<>5, r47c1<>7
10. Discontinuous Nice Loop: 3 r3c3 -3- r2c3 -5- r2c1 =5= r9c1 -5- r8c2 -1- r3c2 =1= r3c3 => r3c3<>3
11. Discontinuous Nice Loop: 3/5/7 r6c7 =8= r6c3 =9= r9c3 -9- r9c9 =9= r2c9 -9- r2c7 -3- r2c3 =3= r3c2 -3- r7c2 =3= r7c1 -3- r4c1 -8- r4c7 =8= r6c7 => r6c7<>3, r6c7<>5, r6c7<>7
12. Row 6 / Column 7 → 8 (Naked Single)
13. AIC: 7 7- r7c8 -9- r4c8 =9= r4c7 =5= r9c7 -5- r9c1 -7 => r7c2,r9c79<>7
14. W-Wing: 9/3 in r2c7,r7c2 connected by 3 in r2c3,r3c2 => r7c7<>9
15. XY-Chain: 3 3- r4c1 -8- r8c1 -4- r7c1 -3- r7c2 -9- r7c8 -7- r5c8 -1- r5c4 -3 => r4c4,r5c2<>3
16. Row 4 / Column 4 → 1 (Naked Single)
17. Row 5 / Column 4 → 3 (Full House)
18. Row 5 / Column 8 → 1 (Hidden Single)
19. Discontinuous Nice Loop: 3/5/7 r6c3 =9= r6c2 -9- r7c2 -3- r3c2 =3= r2c3 -3- r2c7 -9- r2c9 =9= r9c9 -9- r9c3 =9= r6c3 => r6c3<>3, r6c3<>5, r6c3<>7
20. Row 6 / Column 3 → 9 (Naked Single)
21. Row 7 / Column 2 → 9 (Hidden Single)
22. Row 7 / Column 8 → 7 (Naked Single)
23. Row 4 / Column 8 → 9 (Full House)
24. Row 7 / Column 7 → 4 (Naked Single)
25. Row 7 / Column 1 → 3 (Full House)
26. Row 4 / Column 1 → 8 (Naked Single)
27. Row 8 / Column 1 → 4 (Naked Single)
28. Row 3 / Column 9 → 4 (Hidden Single)
29. Row 8 / Column 3 → 8 (Hidden Single)
30. X-Wing: 5 r58 c29 => r6c29,r9c9<>5
31. Row 6 / Column 5 → 5 (Hidden Single)
32. Row 4 / Column 5 → 7 (Full House)
33. Locked Candidates Type 1 (Pointing): 7 in b4 => r3c2<>7
34. Locked Candidates Type 1 (Pointing): 7 in b6 => r1c9<>7
35. Naked Pair: 3,5 in r24c3 => r9c3<>5
36. Skyscraper: 3 in r3c2,r4c3 (connected by r34c7) => r2c3,r6c2<>3
37. Row 2 / Column 3 → 5 (Naked Single)
38. Row 6 / Column 2 → 7 (Naked Single)
39. Row 6 / Column 9 → 3 (Full House)
40. Row 2 / Column 1 → 6 (Naked Single)
41. Row 4 / Column 3 → 3 (Naked Single)
42. Row 5 / Column 2 → 5 (Full House)
43. Row 4 / Column 7 → 5 (Full House)
44. Row 5 / Column 9 → 7 (Full House)
45. Row 1 / Column 1 → 7 (Naked Single)
46. Row 9 / Column 1 → 5 (Full House)
47. Row 2 / Column 9 → 9 (Naked Single)
48. Row 2 / Column 7 → 3 (Full House)
49. Row 8 / Column 2 → 1 (Naked Single)
50. Row 3 / Column 2 → 3 (Full House)
51. Row 3 / Column 3 → 1 (Full House)
52. Row 3 / Column 7 → 7 (Full House)
53. Row 8 / Column 9 → 5 (Full House)
54. Row 9 / Column 3 → 7 (Full House)
55. Row 1 / Column 7 → 1 (Naked Single)
56. Row 1 / Column 9 → 6 (Full House)
57. Row 9 / Column 9 → 1 (Full House)
58. Row 9 / Column 7 → 9 (Full House) | 1,736 | 3,462 | {"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} | 4 | 4 | CC-MAIN-2022-27 | latest | en | 0.592266 |
https://brainly.in/question/59714 | 1,484,929,328,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280835.60/warc/CC-MAIN-20170116095120-00032-ip-10-171-10-70.ec2.internal.warc.gz | 808,236,735 | 9,838 | # A number has two digits whose sum is 8. If 18 is added to the number, its digits are reversed. What is that number.Help me please.
2
2014-12-03T20:36:29+05:30
Let the no be 10x + y
given, x + y = 8 ------------ (1)
10x +y + 18 = 10y + x
9x - 9y + 18 =0
x - y +2 = 0
x - y = -2 ----- (2)
from (1) & (2)
x + y = 8
x - y = -2
-----------------
2x = 6
x = 3
sub x = 3 in (1)
3 + y = 8
y = 5
number 10x + y = 10(3) + 5
= 30+5
= 35
required number is 35.
2014-12-03T20:37:03+05:30
The number is 62.
6+2=8 +18= 26
hence the number is reversed | 249 | 540 | {"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} | 4.03125 | 4 | CC-MAIN-2017-04 | latest | en | 0.720026 |
https://hamronotes.com/lessons/number-systems/decimal-number-system/ | 1,653,496,338,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662588661.65/warc/CC-MAIN-20220525151311-20220525181311-00695.warc.gz | 360,726,793 | 16,796 | # Computer Science XI
## Decimal Number System
In our daily life, we use a system based on digits to represent numbers. The system that uses the decimal numbers or digit symbols 0 to 9 is called as the decimal number system. This system is said to have a base, or radix, of ten. Sequence of digit symbols are used to represent numbers greater than 9. When a number is written as a sequence of decimal digits, its value can be interpreted using the positional value of each digit in the number. The positional number system is a system of writing numbers where the value of a digit depends not only on the digit, but also on its placement within a number. In the positional number system, each decimal digit is weighted relative to its position in the number. This means that each digit in the number is multiplied by ten raised to a power corresponding to that digit’s position. Thus the value of the decimal sequence 948 is: 94810 = 9 x 102 + 4 x 101 + 8 x 100 Fractional values are represented in the same manner, but the exponents are negative for digits on the right side of the decimal point. Thus the value of the fractional decimal sequence 948.23 is: 948.2310 = 9 X 102 + 4 X 101 + 8 X 100 + 2 X 10-1 + 3X10-2 In general, for the decimal representation of X = {…. x2x1x0 . x-1x-2x-3 ……} the value of X is X= Sixi10i where i = …… 2, 1, 0, -1, -2, …… | 356 | 1,358 | {"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-2022-21 | latest | en | 0.862173 |
http://math.stackexchange.com/questions/219586/triangulation-algorithm-for-mobile-geolocation-detection | 1,469,330,186,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257823935.18/warc/CC-MAIN-20160723071023-00128-ip-10-185-27-174.ec2.internal.warc.gz | 152,533,187 | 16,821 | # Triangulation algorithm for mobile geolocation detection
Please help with algorithm for getting radius (L) and center decimal coordinates (X) of blue circle.
Drawing here
Following points are given:
c1 = 56.963022,24.640274 r1 = 6km
c2 = 56.878324,24.490585 r2 = 5km
c3 = 56.870819,24.746017 r3 = 8km
Answer should be in following format:
X = XX.XXXXXX;
L = X.XX km
PS. I need this for creating software for my Master's work and I am a web programmer, not mathematician, so please try not to use very scientific language :)
Thanks!
-
For small distances like this, you can consider the earth to be flat. We will use $R$ for the radius of the earth, $\lambda$ for latitude and $\phi$ for longitude. If you pick one of the points to be the origin, the distance in $y$ (north/south) is $R (\lambda_2-\lambda_1)\frac \pi{180}$ and the distance in $x$ (east/west) is $R \cos \lambda (\phi_2-\phi_1)\frac \pi{180}$, where the $\frac \pi{180}$ factor comes because your angles are degrees. Then you can use the formlae in mathworld to find the points where the circles intersect. The center of your blue circle is the centroid of the triangle. | 325 | 1,147 | {"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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2016-30 | latest | en | 0.826003 |
https://alumniagri.in/task/a-rectangle-of-area-144-cm-square-has-its-length-equal-to-x-17962839 | 1,708,659,345,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474360.86/warc/CC-MAIN-20240223021632-20240223051632-00402.warc.gz | 95,219,173 | 4,762 | Math, asked by Darshananand3443, 10 days ago
# A rectangle of area 144 cm square has its length equal to x cm to write down its breath in terms of x given that perimeter is 50 to write down the equation in text to solve to determine the dimension of the rectangle how to determine this?
0
Length = 43.37cm
Width =3.32cm
Step-by-step explanation:
A = LB
L = x
A =LB
A = 144cm²
Bx = 144 ⇒⇒eq. i
Also,
Perimeter = 2L + 2B
= 2x + 2B
50cm = 2x + 2B ⇒⇒eq.ii
Solving eq.i and eq.ii,
144 = Bx
50 = 2B + 2x
From (i., x= 144÷B
Substituting into (ii.
50 = 2B +2(144÷B)
Multiply thru by B
50B = 2B² +144
2B² - 50B +144 =0
B = 3.32 or 21.68
x = 144÷3.32 or 144÷21.68
x = 43.37 or 6.64
∴∴ The length x = 43.37cm and the width is 3.32cm
Since x=43.37 and B=3.32 solve the question best. Thank you!!
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Math, 3 months ago | 400 | 996 | {"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} | 4.15625 | 4 | CC-MAIN-2024-10 | latest | en | 0.78271 |
https://www.physicsforums.com/threads/momentum-force-and-impulse-help-with-balance-pan-problem.842699/ | 1,722,863,355,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640447331.19/warc/CC-MAIN-20240805114033-20240805144033-00855.warc.gz | 733,087,923 | 20,070 | # Momentum, Force, and Impulse: help with balance pan problem
• toboldlygo
In summary: So you'll need to double your calculated value of Δv.That's it. So you'll need to double your calculated value of Δv.In summary, to solve this problem, the velocity of the beads before and after hitting the pan must be taken into account, as well as the difference between the two. By doubling the calculated value of Δv, the correct answer for the mass needed in the other pan of the balance can be obtained.
toboldlygo
## Homework Statement
[/B]
A stream of elastic glass beads, each with a mass of 0.53 g, comes out of a horizontal tube at a rate of 99 per second. The beads fall a distance of 0.49 m to a balance pan and bounce back to their original height. How much mass must be placed in the other pan of the balance to keep the pointer at zero?
## Homework Equations
impulse = force * Δp ==> F = mΔv/Δt
conservation of energy: potential energy = kinetic energy ==> mgh = 0.5mv2
## The Attempt at a Solution
I think I'm supposed to find the velocity of the initial and final moments (when the beads exit the tube, and as they hit the pan, respectively), then use that to find the force (using the impulse equation), and then find mass from that.
So, I set mgh = 0.5mv2, and using the velocity that I obtained, I solved for F (I used 1/99 as Δt), and then I divided that number by 9.81 m/s/s to get my answer. I don't know where I'm going wrong, though. Any help would be much appreciated!
toboldlygo said:
So, I set mgh = 0.5mv2, and using the velocity that I obtained, I solved for F (I used 1/99 as Δt), and then I divided that number by 9.81 m/s/s to get my answer.
(I suspect you used the wrong value for Δv.)
haruspex said:
(I suspect you used the wrong value for Δv.)
Here's my work. I've attached it to the reply.
#### Attachments
• Doc - 11-11-15, 8-35 PM.pdf
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toboldlygo said:
Here's my work. I've attached it to the reply.
My suspicion is confirmed.
What happens to the bead after it hits the pan?
After it hits the pan, the beads bounce back up to their original height. But how does that factor into the velocity? The velocity of the bead is 3.1006 m/s before it hits the pan, and then zero when it does. Isn't that a change of 3.1006 m/s?
toboldlygo said:
After it hits the pan, the beads bounce back up to their original height. But how does that factor into the velocity? The velocity of the bead is 3.1006 m/s before it hits the pan, and then zero when it does. Isn't that a change of 3.1006 m/s?
If its velocity is zero after hitting the pan, how does it regain its original height?
haruspex said:
If its velocity is zero after hitting the pan, how does it regain its original height?
It couldn't, if its velocity were zero. Does that mean it hits the pan with twice the velocity? There can't be zero change in velocity, could there? I know what you're saying now, but I don't know how I would factor that into a solution.
toboldlygo said:
It couldn't, if its velocity were zero. Does that mean it hits the pan with twice the velocity? There can't be zero change in velocity, could there? I know what you're saying now, but I don't know how I would factor that into a solution.
Velocity and momentum are vectors; signs matter. If the velocity before impact is -x (up positive) what will it be after bouncing? What is the difference of the two?
toboldlygo
haruspex said:
Velocity and momentum are vectors; signs matter. If the velocity before impact is -x (up positive) what will it be after bouncing? What is the difference of the two?
Oh, that makes sense! So if I were to subtract initial from final, I'd get x-(-x), which is just 2x, right? Or did I completely miss the mark?
toboldlygo said:
Oh, that makes sense! So if I were to subtract initial from final, I'd get x-(-x), which is just 2x, right? Or did I completely miss the mark?
That's it.
toboldlygo
## What is momentum?
Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. It is a vector quantity, meaning it has both magnitude and direction.
## What is force?
Force is a push or pull that causes an object to accelerate or change its motion. It is also a vector quantity and is measured in Newtons (N).
## What is impulse?
Impulse is a change in momentum and is equal to the force applied to an object multiplied by the time it is applied. It is also a vector quantity and has the same direction as the force.
## What is the relationship between momentum, force, and impulse?
The relationship between momentum, force, and impulse is described by Newton's Second Law of Motion, which states that the net force acting on an object is equal to the rate of change of its momentum. In other words, the greater the force or the longer it is applied, the greater the change in momentum.
## How can I use the balance pan problem to understand momentum, force, and impulse?
The balance pan problem is a common physics demonstration that involves balancing two unequal masses on a lever. By adjusting the position of the masses, you can observe how changes in force and momentum affect the balance of the system. This can help you understand the relationship between these concepts and how they apply in real-world situations.
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1K | 1,472 | 5,859 | {"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-2024-33 | latest | en | 0.951411 |
https://datagenetics.com/blog/june62020/index.html | 1,670,071,296,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710931.81/warc/CC-MAIN-20221203111902-20221203141902-00811.warc.gz | 225,377,683 | 3,570 | # Perfect Roots Puzzle
What is the lowest number (other than the trivial answer of one), which is a perfect square, perfect cube, and perfect fifth power?
Specifically what is the lowest N that satisfies the equation:
N = a2 = b3 = c5
Where a,b,c are whole numbers.
## Trivia Diversion
Interesting piece of trivia: The last digit of any number raised to the fifth power does not change. e.g.:
65 = 7,776
1195 = 23,863,536,599
145 = 537,824
1,0075 = 1,035,493,442,021,807
Advertisement:
## Solution
We could brute force this, but we don’t really know what the range is, plus it seems wasteful when we can apply some simple math to get a solution.
First of all, we want to get all the exponents of all the numbers the same.
N = a2 = b3 = c5 = xz
We can represent N as some value x raised to the power z. To find a suitable value for z, we can use the lowest common multiple of the powers 2,3,5. As these are all relatively prime, the answer is 30.
N = x30
Because of the way powers work, we can split this common multiple into the components we need:
N = (x2)15 = (x3)10 = (x5)6
Which of course is equivalent to:
N = (x15)2 = (x10)3 = (x6)5
We want the lowest value, so after the trivial x = 1, we can insert x = 2, which gives the answer: 1,073,741,824
The smallest number, greater than one, which is a perfect square, cube, and fifth power is 1,073,741,824
1,073,741,824 = (215)2 = (210)3 = (26)5
1,073,741,824 = 32,7682 = 1,0243 = 645
This answer is quiet a large number because 2,3,5 are relatively prime. If we changed the question to the smaller number with perfect roots for powers of 2,3,6 then the answer would be just 64. | 496 | 1,636 | {"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-2022-49 | latest | en | 0.921786 |
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Standard Gibbs Free Energy and Equilibrium Constant for JEE
Last updated date: 06th Aug 2024
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Gibbs Free Energy or Free Enthalpy
What is Gibbs free energy? The Gibbs free energy is also known as the Gibbs function. Gibbs energy, or free enthalpy, is a thermodynamic potential that is used to calculate the maximum amount of work done in any provided thermodynamic process when the temperature and pressure of the system are held constant. The letter $G$ represents Gibbs's free energy of any system. Due to the fact that it is also a kind of energy, its value is often expressed in either Joules or Kilojoules. The Gibbs free energy is equal to the maximum amount of work that a system is capable of doing in its environment, while simultaneously undergoing a spontaneous change or reaction (at constant temperature and pressure).
Gibbs energy is a thermodynamic property discovered by the American scientist Josiah Willard Gibbs in 1876. At that time, Gibbs was conducting experiments to determine whether or not a process could take place simultaneously and spontaneously at a given temperature. Gibbs came up with the concept for Gibbs energy while trying to predict the behaviour of systems when they were combined together. The free energy calculated by Gibbs was sometimes referred to as the available energy.
Gibbs Free Energy Equation
The Gibbs free energy is calculated by the difference in enthalpy between the system and the product of the entropy and temperature. The equation may be written as follows:
G=H-TS
where
G= Gibbs free energy
H= Enthalpy of the system
T= Temperature of the system
S= Entropy of the system
As Gibbs free energy is a state function, it does not depend on the path of the process (i.e., it is a path independent entity).
$\Delta G=\Delta H-\Delta \left( TS \right)$
If the system or reaction is isothermal in nature, assuming constant temperature throughout the reaction, $\Delta T=0$, then the following is true:
$\Delta G=\Delta H-\left( T\Delta S \right)$
The Gibbs Helmholtz equation is the name given to this particular equation.
$\Delta G\,\,>\,0$; the reaction is non-spontaneous and endergonic.
$\Delta G\,\,<\,0$; the reaction is spontaneous and exergonic.
$\Delta G\,\,=\,0$; the reaction is at equilibrium.
What is the Equilibrium Constant?
The value of reaction quotient of a chemical reaction at chemical equilibrium is known as the equilibrium constant. Chemical equilibrium is a state that is approached by a dynamic chemical system after sufficient time has elapsed. At this point, the system's composition has no measurable tendency towards further change.
The equilibrium constant does not depend on the initial analytical concentrations of the reactant or product species in the mixture, for any particular set of reactions. Therefore, the composition of a system at equilibrium may be determined by using the initial composition of the system in conjunction with the known values of the equilibrium constants, provided that the initial composition of the system is known.
On the other hand, the value of the equilibrium constant may be affected by various reaction factors such as temperature, solvent, and ionic strength.
The thermodynamically accurate expression for the constant of equilibrium connects the activities of all of the species that are involved in the reaction. Although the idea of activity is too advanced for the typical general chemistry course, to simplify experimental measurements, the equilibrium constant of concentration approximates the activities of solutes and gases in dilute solutions with their respective molarities. However, the molarities of solids, pure liquids, and solvents cannot be used as a substitute for estimating their activities. On the other hand, the value of these actions is defined to be equal to 1.
The expression for the equilibrium constant is written as K in terms of activity for the hypothetical reactions given below:
$bB+cC\rightleftharpoons dD+eE$
${K_{eq}}={K_C}=\dfrac {{\left [D\right]^d}~{\left [E\right]^e}}{{\left [B\right]^b}~\left [C\right]^c}$
The equilibrium constant formula generally represents the concentration and pressure of species at equilibrium.
Standard Gibbs Free Energy
It is the change in Gibbs free energy that occurs during the formation of one mole of a compound from its component elements, when those elements are in their standard states at the most stable form of the element, that is, at ${{\Delta }_{f}}G$.
As there is no change occurring, the standard Gibbs free energy change of formation for all elements in their standard states equals zero.
${\Delta_f}G={\Delta_f}{G^o}+R~T~ln{Q_f}$
Where ${Q_f}$ stands for the response quotient.
Conditions for Standard State of the System
• All gases involved in the process have a partial pressure of 100 kpa.
• 1 M is the concentration of all aqueous solutions.
• The measurements are often done at a temperature of $\,C\,\,\left( 298K \right)$.
• Pure liquids have a total (hydrostatic) pressure of 1 atm.
For the pure solid at 1 atm pressure.
In equilibrium, ${{\Delta }_{f}}G$ is equal to zero, and Qf is to K; therefore, the equation becomes:
${{\Delta }_{f}}G=-R~T ~ln K$
where $K$ is the equilibrium constant.
Consider a spontaneous chemical process in a closed system at constant temperature and pressure, without non-PV work. The Clausius inequality $\Delta S > \dfrac{Q}{T_{surr}}$ for such a system converts into a condition for the change in Gibbs free energy.
$\Delta G < 0$
Relation Between Gibbs Free Energy and Equilibrium Constant
The standard free energy change of the reaction, $\Delta {G^o}$(which is equal to the difference in the free energies of formation of the products and reactants both in their standard states), is connected to the reaction's free energy change in any state, (where equilibrium has not been achieved).
$\Delta G=\Delta G{}^\circ+RT~ln Q$
Here, Q is the reaction quotient.
At equilibrium,
$~\Delta G\,=\,0$, and Q becomes equal to the equilibrium constant, K at equilibrium. Hence, the equation will become:
$\Delta G{}^\circ =2.303RT\log {{K}_{eq}}$
Here, R is equal to 8.314 J mol-1 k-1 or $\,0.008314~kJ~mo{{l}^{-1}}~{{k}^{-1}}$.
T corresponds to the temperature on the Kelvin scale.
The relation between the equilibrium constant and Gibbs free energy can be derived using the following equation:
${K_{eq}}={K_C}=\dfrac {{\left [D\right]^d}~{\left [E\right]^e}}{{\left [B\right]^b}~\left [C\right]^c}$
Relationship between Gibbs Free Energy and EMF of a Cell
The Gibbs energy change $~\Delta G$ of galvanic cells is proportional to the amount of electrical work the cell does.
$\Delta G=-nF{{E}_{cell}}$
where $n\,\,=$ no. of moles of electron involved
$F\,\,=$ constant
$E\,\,=$ EMF of the cell
$F\,\,=\,\,1\,{\text{Faraday}}\,\,=\,\,96500\,\,\text{coulombs}$
If reactants and products are in their standard states,
$\Delta G{}^\circ =nFE{{{}^\circ }_{cell}}$
$E{{{}^\circ }_{cell}}$ and equilibrium constant, ${{K}_{eq}}$
$\Delta G=\Delta G{}^\circ +RT\ln Q$
$0=\Delta G{}^\circ +RT \ln\text{ Keq}$
$ln\text{ Keq}=-\frac{\Delta G{}^\circ }{RT}$
${{K}_{eq}}={{e}^{-\,\frac{\Delta G{}^\circ }{RT}}}$
$\Delta \mathbf{G}{}^\circ$ and Equilibrium Condition of Spontaneous, Non-Spontaneous and Equilibrium
$\Delta \mathbf{G}{}^\circ$ Reaction ${{K}_{eq}}$ + Non-spontaneous << 1 - Spontaneous >>1 0 At equilibrium 0
Table 1: Conditions for Equilibrium for Various Processes
Conclusion
Gibbs energy is a state function, which is related to entropy, temperature and enthalpy changes of the system, and measure of its capacity to do useful work. The decrease in Gibbs free energy during a process gives the maximum possible useful work that can be obtained from the system. The free energy of a system is a measure of its capacity to do useful work. When $\Delta G$ is negative, the given reaction is spontaneous. For a reversible reaction, $\Delta G$ is negative for the forward as well as backward reaction. $\Delta G$ is zero at equilibrium and there is no further free energy change.
Competitive Exams after 12th Science
FAQs on Standard Gibbs Free Energy and Equilibrium Constant for JEE
1. Is Gibbs's free energy an example of a state function? Explain.
Gibbs's free energy is an example of a state function because it is explained by other state functions. Gibbs free energy is defined by other state functions (G=H-TS), enthalpy, and entropy. Additionally, Gibbs free energy is a state property because it is defined by final-initial state and is not dependent on the path it undergoes during the change. Gibbs free energy is also an extensive property.
2. What is the relationship between KP and KC?
The equilibrium constants for a perfect gaseous mixture are KP and KC. KP is the appropriate equilibrium constant to use when expressing equilibrium concentrations in terms of atmospheric pressure, while KC is the appropriate equilibrium constant to use when expressing equilibrium concentrations in terms of molarity. For a general equilibrium reaction,
$\mathrm{aA}+\mathrm{bB} \rightleftharpoons \mathrm{cC}+\mathrm{dD}$
The connection between Kp and Kc may be expressed as follows:
$K_{p}=K_{c}(R T)^{\Delta n g}$
Here, $\Delta n g$= no. of moles of gaseous products - no. of moles of gaseous reactants for balanced chemical equations. | 2,296 | 9,478 | {"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": 2, "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-2024-33 | latest | en | 0.938203 |
http://mathhelpforum.com/number-theory/197893-find-number-primitive-roots-z_101-z_12-a.html | 1,566,246,028,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027314959.58/warc/CC-MAIN-20190819201207-20190819223207-00222.warc.gz | 133,219,061 | 9,936 | # Thread: Find the number of primitive roots in Z_101, Z_12
1. ## Find the number of primitive roots in Z_101, Z_12
Find the number of primitive roots in $\displaystyle \mathbb{Z}_{101}$, $\displaystyle \mathbb{Z}_{12}$.
I cannot find an example from my lecture notes about primitive roots in $\displaystyle \mathbb{Z}_n$. What is the definition of primitive roots in $\displaystyle \mathbb{Z}_n$?
The answers are 40 and none respectively.
2. ## Re: Find the number of primitive roots in Z_101, Z_12
A primitive root in $\displaystyle \mathbb Z_n^\times$ is an integer $\displaystyle k$ such that for each $\displaystyle i\in\{1,2,\ldots,n-1\}$ such that $\displaystyle \gcd(i,n)=1,$ there exists an integer $\displaystyle m$ such that $\displaystyle i\equiv k^m\mod n.$ (It follows that $\displaystyle \gcd(k,n)=1.)$
When $\displaystyle n=p$ is a prime, $\displaystyle \mathbb Z_p^\times$ always has primitive roots. Indeed $\displaystyle \mathbb Z_p^\times$ is a cyclic group of order $\displaystyle p-1$ generated by any primitive root; hence the number of primitive roots is $\displaystyle \varphi(p-1).$ This answers your question for $\displaystyle \mathbb Z_{101}^\times.$
If $\displaystyle n$ is not prime, things are a little complicated. For $\displaystyle n=12,$ though, you can easily see that $\displaystyle k^2\equiv1\mod{12}$ for $\displaystyle k=1,5,7,11,$ which are the integers coprime with $\displaystyle 12.$ Hence $\displaystyle \mathbb Z_{12}^\times$ has no primitive roots since there is are no integers $\displaystyle k,m$ such that $\displaystyle k^m\equiv5\mod{12}$ even though $\displaystyle \gcd(5,12)=1.$
3. ## Re: Find the number of primitive roots in Z_101, Z_12
Thank you. I see. This is really the same as primitive roots in $\displaystyle \mathbb{U}_{101}$ and $\displaystyle \mathbb{U}_{12}$, just in different notation. | 531 | 1,865 | {"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} | 3.828125 | 4 | CC-MAIN-2019-35 | latest | en | 0.732065 |
https://math.answers.com/other-math/A_bicycle_is_on_sale_at_12_more_than_half_of_the_regular_price._If_the_sale_price_is_75_find_the_regular_price | 1,716,165,490,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058009.3/warc/CC-MAIN-20240519224339-20240520014339-00609.warc.gz | 338,267,942 | 49,303 | 0
# A bicycle is on sale at 12 more than half of the regular price. If the sale price is 75 find the regular price?
Updated: 4/28/2022
Wiki User
14y ago
If 75 is 12 more than half of the regular price then simply deduct 12 from 75, which is 63 and then double it, which is 126.
Wiki User
14y ago
Earn +20 pts
Q: A bicycle is on sale at 12 more than half of the regular price. If the sale price is 75 find the regular price?
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It is 1.5 times the unit price. | 672 | 2,797 | {"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-2024-22 | latest | en | 0.952397 |
http://www.physicsforums.com/showthread.php?t=422543 | 1,409,668,381,000,000,000 | text/html | crawl-data/CC-MAIN-2014-35/segments/1409535922087.15/warc/CC-MAIN-20140909045602-00232-ip-10-180-136-8.ec2.internal.warc.gz | 1,260,723,877 | 11,311 | # I need to calculate δR: R is Ricci scalar
by sourena
Tags: δr, ricci, scalar
P: 14 I need to calculate δR: R is Ricci scalar
Sci Advisor HW Helper P: 1,275 This should be shown in most GR textbooks...
P: 14
Quote by nicksauce This should be shown in most GR textbooks...
This is not a homework.
P: 894 I need to calculate δR: R is Ricci scalar This should help: http://en.wikipedia.org/wiki/Einstei...e_Ricci_scalar
P: 14
Thank you for your time and care, but I need to obtain this:
δR=Rμσ δg+gμσ δg^μσ -∇μ ∇σ δg^μσ
in f(R) gravity:
http://en.wikipedia.org/wiki/F(R)_gravity
Quote by JustinLevy This should help: http://en.wikipedia.org/wiki/Einstei...e_Ricci_scalar
P: 1,411 Do you have any problem with these equations:
P: 14 No, I don't have problem with these equations, but I have problem to calculate this equation from them: δR=Rab δgab+gab δgab -∇a ∇b δgab
P: 1,411 First step: The term $$\nabla_\sigma \left( g^{\mu\nu}\delta\Gamma^{\sigma}_{\mu\nu}-g^{\mu\sigma}\delta\Gamma^{\rho}_{\rho_\mu}\right)$$ is of the form $$\nabla_\sigma A^\sigma$$. But $$\nabla_\sigma A^\sigma=|\det g]^{-\frac12}\partial_\sigma (A^\sigma |\det g|^{\frac12})$$ for any vector field $$A^\sigma$$. This is handy. P.S. For some reason my display is not displaying correctly one character in your first term on the right. So, I do not know what is exactly the formula you would like to have.
P: 894
Alright, I tried to work it out, but it looks like I got a sign error somewhere. I wrote everything out in more detail than was probably necessary, so where this occurs can stand out.
Quote by sourena No, I don't have problem with these equations, but I have problem to calculate this equation from them: δR=Rab δgab+gab δgab -∇a ∇b δgab
Okay, so we at least have a starting point we can agree on:
$$\delta R = R_{ab} \delta g^{ab} + g^{ab} \delta R_{ab}$$
$$\delta R_{ab} &= [\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}]$$
As in wikipedia noting that $$\delta \Gamma^\lambda_{\mu\nu}\,$$ is actually the difference of two connections, it should transform as a tensor. Therefore, it can be written as
$$\delta \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda d}\left(\nabla_\mu\delta g_{d\nu}+\nabla_\nu\delta g_{d\mu}-\nabla_d\delta g_{\mu\nu} \right)$$
and substituting in the equation, after playing with it in excruciating detail one finds:
\begin{align*} \delta R_{ab} &= [\nabla_c \delta \Gamma^c_{ab} - \nabla_b \delta \Gamma^c_{c a}] \\ &= [\nabla_c \frac{1}{2}g^{c d}\left(\nabla_a\delta g_{db}+\nabla_b\delta g_{da}-\nabla_d\delta g_{ab} \right) -\nabla_b \frac{1}{2}g^{c d}\left(\nabla_c\delta g_{da}+\nabla_a\delta g_{dc}-\nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}g^{c d}[\left(\nabla_c\nabla_a\delta g_{db}+\nabla_c\nabla_b\delta g_{da}-\nabla_c\nabla_d\delta g_{ab} \right) -\left(\nabla_b\nabla_c\delta g_{da}+\nabla_b\nabla_a\delta g_{dc}-\nabla_b\nabla_d\delta g_{ca} \right)] \end{align*}
swap the derivative order on the fourth term
http://en.wikipedia.org/wiki/Covaria...ative#Examples
$$\nabla_b\nabla_c\delta g_{da} = \nabla_c\nabla_b\delta g_{da} + R^{e}{}_{dbc} \delta g_{ea} + R^{e}{}_{abc} \delta g_{de}$$
\begin{align*} g^{ab}\delta R_{ab} &= g^{ab} \frac{1}{2}g^{c d}[\left(\nabla_c\nabla_a\delta g_{db}-\nabla_c\nabla_d\delta g_{ab} \right) -\left(R^{e}{}_{dbc} \delta g_{ea} + R^{e}{}_{abc} \delta g_{de} + \nabla_b\nabla_a\delta g_{dc}-\nabla_b\nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}g^{c d}[\left(\nabla_c\nabla^b \delta g_{db}-\nabla_c\nabla_d g^{ab} \delta g_{ab} \right) -\left(R^{e}{}_d{}^a{}_c \delta g_{ea} + R^{eb}{}_{bc} \delta g_{de} + \nabla_b\nabla^b \delta g_{dc}-\nabla^a \nabla_d\delta g_{ca} \right)] \\ &= \frac{1}{2}[\left(\nabla^d\nabla^b \delta g_{db}-\nabla^d\nabla_d g^{ab} \delta g_{ab} \right) -\left(R^{eca}{}_c \delta g_{ea} + R^{eb}{}_b{}^d \delta g_{de} + \nabla_b\nabla^b g^{c d} \delta g_{cd} -\nabla^a \nabla^c\delta g_{ca} \right)] \\ &= \frac{1}{2}[\left(\nabla^d\nabla^b \delta g_{db}-\nabla^d\nabla_d g^{ab} \delta g_{ab} \right) -\left(R^{ea}\delta g_{ea} - R^{de} \delta g_{de} + \nabla_b\nabla^b g^{c d} \delta g_{cd} -\nabla^a \nabla^c\delta g_{ca} \right)] \\ &= \nabla^a\nabla^b \delta g_{ab} - g^{ab} \nabla^c\nabla_c \delta g_{ab} \end{align*}
Hmm...
maybe there isn't an error. Does
$\delta g_{ab} = - \delta g^{ab}$ ?
I'm too tired to check right now.
P: 1,411
Quote by JustinLevy Hmm... maybe there isn't an error. Does $\delta g_{ab} = - \delta g^{ab}$ ? I'm too tired to check right now.
Not exactly.
Use:
$$0=\delta ( \delta^a_b) =\delta (g^{ac}g_{cb})= ....$$
Now use the Leibniz rule, calculate what you need.
P: 894 Thanks. Alright, so $$0=\delta ( \delta^a_b) =\delta (g^{ac}g_{cb})= g_{cb} \delta g^{ac} + g^{ac} \delta g_{cb}$$ $$g_{cb} \delta g^{ac} = - g^{ac} \delta g_{cb}$$ Thus manipulating the previous posts result gives \begin{align*} g^{ab}\delta R_{ab} &= \nabla^a\nabla^b \delta g_{ab} - g^{ab} \nabla^c\nabla_c \delta g_{ab} \\ &= \nabla^a\nabla_c g^{bc} \delta g_{ab} - \nabla^c\nabla_c g^{ab} \delta g_{ab} \\ &= - \nabla^a\nabla_c g_{ab} \delta g^{bc} + \nabla^c\nabla_c g_{ab} \delta g^{ab} \\ \end{align*} which is what sourena wanted to find: $$g^{ab}\delta R_{ab} = g_{ab} \nabla^c\nabla_c \delta g^{ab} - \nabla_a\nabla_b \delta g^{ab}$$ ... arkajad, I assume your post #8 hint leads to the result faster, but I'm not sure how to apply that. Once we've converted from covariant to ordinary coordinate derivative, how do we recognize the result as terms with two covariant derivatives?
P: 1,411 The trick with ordinary derivatives is useful when you calculate under the integral. There, in variational calculus, when you assume that variations vanish at the boundary (usually at infinity), you need a true divergence of a vector field, and not some "covariant one". You discard such terms converting volume integral into surface integrals using ordinary rules of the calculus.
P: 14 Dear JustinLevy Sorry for being late to answer your posts. I value your hard work to obtain this expression a great deal. Thank you very much for your time and care.
P: 14 Dear arkajad I wanted to calculate this: $$g^{ab}\delta R_{ab} = g_{ab} \nabla^c\nabla_c \delta g^{ab} - \nabla_a\nabla_b \delta g^{ab}$$ or $$\delta R = R_{ab} \delta g^{ab} + g_{ab} \nabla^c\nabla_c \delta g^{ab} - \nabla_a\nabla_b \delta g^{ab}$$
Sci Advisor P: 912 Do you know how to derive $$\nabla_{\mu}\delta\Gamma^{\lambda}_{\nu\rho} = \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu}\nabla_{\rho}\delta g_{\nu\alpha} - \nabla_{\mu}\nabla_{\alpha}\delta g_{\nu\rho}]$$ ? This can be shown directly by varying the connection, but you can also guess the form of it; varying the connection gives a tensor, and the partial derivatives in it can only become covariant ones (convince yourself if you're not!). You can plug this in the Palatini equation, which gives $$\delta R_{\mu\nu\rho}^{\ \ \ \ \lambda} = \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu\rho}\delta g_{\nu\alpha}-\nabla_{\mu}\nabla_{\alpha}\delta g_{\nu\rho}] - \frac{1}{2}g^{\lambda\alpha}[\nabla_{\nu}\nabla_{\mu}\delta g_{\rho\alpha}+\nabla_{\nu}\nabla_{\rho}\delta g_{\mu\alpha} - \nabla_{\nu}\nabla_{\alpha}\delta g_{\mu\rho}]$$ Does this help?
Sci Advisor P: 912 Do you know how to derive $$\nabla_{\mu}\delta\Gamma^{\lambda}_{\nu\rho} = \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\na_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu}\na_{\rho}\delta g_{\nu\alpha} - \nabla_{\mu}\na_{\alpha}\delta g_{\nu\rho}]$$ This can be shown directly, but you can also guess the form of it; varying the connection gives a tensor, and the partial derivatives in it can only become covariant ones (convince yourself if you're not!). You can plug this in the Palatini equation, which gives $$\delta R_{\mu\nu\rho}^{\ \ \ \ \lambda} &=& \frac{1}{2}g^{\lambda\alpha}[\nabla_{\mu}\nabla_{\nu}\delta g_{\rho\alpha} + \nabla_{\mu}\nabla_{\rho}\delta g_{\nu\alpha}-\nabla_{\mu}\nabla_{\alpha}\delta g_{\nu\rho}] - \frac{1}{2}g^{\lambda\alpha}[\nabla_{\nu}\nabla_{\mu}\delta g_{\rho\alpha}+\nabla_{\nu}\nabla_{\rho}\delta g_{\mu\alpha} - \nabla_{\nu}\nabla_{\alpha}\delta g_{\mu\rho}]$$
P: 3,967 The equation and mathematics posted in this thread are very impressive, but I always wonder how useful they are (or what the point of them is) if these sort of calculations seemingly can not be used to answer "simple" questions like the ones posed here http://www.physicsforums.com/showthread.php?t=431712 and here http://www.physicsforums.com/showthr...38#post2901038 ? This is one reason I have never made the effort to really try and learn the apparatus of tensors, because of the seeming limited applicability of these formalisms. Is it that tensors are "overkill" to solve the questions posed in those links (which have not yet been solved), like using a sledgehammer to crack a walnut, or is it just that the people that like to play with tensors never look at or try to solve the "simple" questions? My real question is this. If a person makes an effort to learn tensors would they able to answer those "simple" questions and is it worth the effort to learn these formidable looking formalisms, if all you want to do is solve the sort of simple questions posed in those threads? (which up to now are seemingly unsolvable).
P: 1,411 There is another simple question: "What is the purpose of life?" For an engineer his purpose can be, for instance, to learn about elasticity and to apply his knowledge. And when you start learning elasticity stuff - you will soon find that you can't go too far without tensors. So it all depends on your purpose.
Related Discussions Special & General Relativity 4 Special & General Relativity 3 Advanced Physics Homework 1 Special & General Relativity 8 Special & General Relativity 2 | 3,482 | 9,858 | {"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} | 3.53125 | 4 | CC-MAIN-2014-35 | latest | en | 0.879545 |
https://socratic.org/questions/water-is-leaking-out-of-an-inverted-conical-tank-at-a-rate-of-10-000-cm3-min-at- | 1,656,156,811,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103034930.3/warc/CC-MAIN-20220625095705-20220625125705-00297.warc.gz | 566,450,442 | 6,775 | # Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank?
Let $V$ be the volume of water in the tank, in $c {m}^{3}$; let $h$ be the depth/height of the water, in cm; and let $r$ be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that $\setminus \frac{h}{r} = \setminus \frac{6}{2} = 3$ so that $h = 3 r$.
The volume of the inverted cone of water is then $V = \setminus \frac{1}{3} \setminus \pi {r}^{2} h = \setminus \pi {r}^{3}$.
Now differentiate both sides with respect to time $t$ (in minutes) to get $\setminus \frac{\mathrm{dV}}{\mathrm{dt}} = 3 \setminus \pi {r}^{2} \setminus \cdot \setminus \frac{\mathrm{dr}}{\mathrm{dt}}$ (the Chain Rule is used in this step).
If ${V}_{i}$ is the volume of water that has been pumped in, then $\setminus \frac{\mathrm{dV}}{\mathrm{dt}} = \setminus \frac{{\mathrm{dV}}_{i}}{\mathrm{dt}} - 10000 = 3 \setminus \pi \setminus \cdot {\left(\setminus \frac{200}{3}\right)}^{2} \setminus \cdot 20$ (when the height/depth of water is 2 meters, the radius of the water is $\setminus \frac{200}{3}$ cm).
Therefore $\setminus \frac{{\mathrm{dV}}_{i}}{\mathrm{dt}} = \setminus \frac{800000 \setminus \pi}{3} + 10000 \setminus \approx 847758 \setminus \setminus \frac{\setminus m b \otimes {\left\{c m\right\}}^{3}}{\min}$. | 573 | 1,745 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 13, "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-2022-27 | latest | en | 0.761484 |
http://jwilson.coe.uga.edu/EMAT6680/Cooper/asn4tc/proof.html | 1,542,552,073,000,000,000 | text/html | crawl-data/CC-MAIN-2018-47/segments/1542039744381.73/warc/CC-MAIN-20181118135147-20181118161147-00122.warc.gz | 163,990,553 | 3,938 | Assignment 4: Similar Triangles
by
Tom Cooper
I set out to do problem #8, but I would like to report on a variation.
8. Take an acute triangle ABC. Construct H and the segments HA, HB, and HC. Construct the midpoints of HA, HB, and HC. Connect the midpoints to form a triangle. Prove that this triangle is similar to triangle ABC and congruent to the medial triangle. Construct G, H, C, and I for this triangle. Compare.
Instead of letting H be the Orthocenter of ABC, I started with the circumcenter. Since I use C in my triangle ABC, I refer to the circumcenter by H.
I then constructed the circumcenter H and the segments HA, HB, and HC. I constructed the midpoints of HA, HB, and HC, and I connected the midpoints to form a triangle.
It appears that triangle ABC is similar to triangle DEF. Is it?
Claim: Triangle ABC is similar to triangle DEF
Proof (Using Classical Euclidean Geometry):
1. Angle AHC = Angle DHF 1. They are the same angle.
2. (AH)/(DH) =(HC)/(HF)=2 2. D and F are midpoints of AH and HC, respectively.
3. Triangle AHC is similar to Triangle DHF 3. SAS
4. Angle HAC = Angle HDF and 4. Corresponding angles of similar triangles.
Angle HFD = Angle HCA
A similar argument shows that triangle ABH is similar to triangle DEH and triangle BHC is similar to triangle EHF.
Thus, by corresponding parts of similar triangles, angle BAH = angle EDH, angle ABH = angle DEH, angle HBC = angle HEF, and angle BCH = angle EFH.
Thus we have,
Therefore, triangle ABC is similar to triangle DEF by AAA.
Note that this proof will hold even when H is not inside of triangle ABC.
In fact, the proof will hold even when H is not the circumcenter. That gives the following generalization:
Claim: Given a triangle ABC and any point H, construct the segments HA, HB, and HC. Then construct the midpoints of HA, HB, and HC and construct a triangle with these three midpoints as vertices. Then the resulting triangle will be similar to triangle ABC with one forth of the area.
Explore this yourself with Geometer's Sketchpad.
I will use bold letters and endpoints to indicate vectors. For example BA is the vector from B to A.
Proof (Using Vector Algebra):
BA + AH = BH and ED + DH = EH
or equivalently,
BA = BH - AH and ED = EH - DH
Since E and D are the midpoints of BH and AH respectively, we have | 591 | 2,432 | {"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-2018-47 | latest | en | 0.905899 |
https://www.physicsforums.com/threads/basics-question-of-inductor-to-powering-high-voltage-lamp.667408/ | 1,627,535,480,000,000,000 | text/html | crawl-data/CC-MAIN-2021-31/segments/1627046153816.3/warc/CC-MAIN-20210729043158-20210729073158-00004.warc.gz | 968,901,382 | 26,946 | # Basics question of Inductor to powering high voltage lamp
I was trying to study the inductor role in Flyback converters and all, they seem to rely on the principle that the inductor is first energized while the switch is closed and then, it acts as a voltage source itself when the switch is opened.
All that is good, but I am confused with the POLARITY.
http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE5803
The simulations here seem to show all the current directions from negative to positive. Are they talking about the flow of electrons?
Either ways, I don't see when the inductor gets 65volts across it to power the neon lamp.
This is what I understood: First the inductor has a voltage of 10 when the switch is closed with A, then it goes reducing to zero. After the switch is closed with B, the voltage is formed across the inductor to power the neon-lamp.
But all the currents shown are negative to positive?
Can someone clear me on this? I am aware of the equations V = L(di/dt). So when the constant current becomes zero, voltage is there across the inductor, right? But the polarity?
sophiecentaur
Gold Member
2020 Award
Assume a positive supply voltage, a series L, a switch and a load connected to ground.
When you (try to) break the circuit, the voltage across the inductor will change from a small forward voltage (due to the finite resistance of the winding) to a high reverse voltage. This high, inverse voltage will be such as to maintain the current that is flowing out of the inductor into the circuit it feeds. It will provide a (positive) voltage at the switch terminal that is much higher than the original value. If a high voltage neon has been connected between L and ground, the voltage will then be enough to strike the neon. The more current the load takes, the higher the voltage when the switch breaks the circuit - which you said when referring to Ldi/dt
When exactly does the inductor in the link develop a voltage large enough to power a 65V neon lamp, from a 10 V source?
In the page of the time constant, it has 10 V just when the switch is closed. Then it goes reducing till zero as time passes.
After that the next page has these two situations (shown in the attachment). It seems to shift its polarity once the connection is made. Even after that it seems to be showing current going from negative to positive!
So which is the correct polarity of voltage at the inductor, and what about the current direction?
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sophiecentaur
Gold Member
2020 Award
Your circuit is not the one I was discussing. 'My' switch and neon are the other side of the inductor, so my wording was not appropriate.
In your circuit, in order to maintain the same current through the resistor, when the switch is opened, the volts at the 'top' of the inductor will, as you show, go negative with a large amplitude.
Could your confusion about the direction of the current be to do with the fact that the inductor is an energy source (like a battery) whilst the voltage spike is being formed. For current to flow out of it, through the resistor, the potential that end must be maintained and this is achieved by virtue of the emf, generated in the inductor. When the battery is connected, the negative end of the battery is connected to ground. Whilst the inductor is functioning, the ground connection is replaced by the neon , which will only strike when there is 65V across it. Point B will be 65V negative for this to happen. (total volts on inductor will be 75V - note that on my circuit, you only need 55V across the inductor as there is already 10V from the supply)
You also mention the timing of all this. There will be finite capacitance in the circuit (mostly the self capacitance of the coil) and this will introduce a time constant which will cause a finite time for the voltage reversal. If the neon were replaced with a capacitor, you would have a series of oscillations as the capacitor discharges back through the inductor - and so on - until the energy is dissipated in the load resistor.
NascentOxygen
Staff Emeritus
When exactly does the inductor in the link develop a voltage large enough to power a 65V neon lamp, from a 10 V source?
The high voltage develops the instant that the current begins to change significantly. Because when current changes rapidly, di/dt is large and there's the large voltage you are asking about.
Let's simplify things even more, using just a battery and the inductor. You connect the battery across the inductor, and current flows, small at first but increasing with time. You then decide to disconnect the battery. Instant removal of the battery would cause the current to instantly fall to zero. But that means di/dt would be infinite, so an infinite voltage would appear across the inductor terminals. An infinite (or, at least, very high) voltage here would jump across the gap of air between the inductor terminals and the battery you are valiantly trying to drag off the inductor, this voltage revealing itself as a blue spark. And if you do the experiment, that's exactly what you'll see. The faster you try to remove the battery from the inductor, the bigger the spark that results.
The spark represents current flow, meaning the current you tried to interrupt and cause to drop to zero is actually not instantly falling to zero — some current manages to continue flowing by jumping across the air gap. Therefore, di/dt is not infinite (and negative), but it's still a large value.
If you have a neon connected across the inductor terminals, current will preferentially flow through the ionised neon gas rather than jump through air.
sophiecentaur
Gold Member
2020 Award
Here's a question for 'the student' - What happens if there is no spark or 'neon'? (If the insulation is good enough and there clearly can't be infinite voltage.)
NascentOxygen
Staff Emeritus
Well, this student says that if that's the case then you haven't rapidly interrupted the current, but rather gradually/steadily wound it back to zero.
sophiecentaur
Gold Member
2020 Award
What was it that limited how fast you could 'wind it back'. ""
NascentOxygen
Staff Emeritus
What was it that limited how fast you could 'wind it back'. ""
Apparently the lab that day had issued me with a slow switch. The mechanical separation of its contacts was slow (and, indeed, smooth and gentle) enough that di/dt was limited during the separation process to a value that induced no voltage large enough to ionize any air across the separating points.
sophiecentaur
Gold Member
2020 Award
No one likes a smartypants.
NascentOxygen
Staff Emeritus
No one likes a smartypants.
Oh, come on now. Enough of that talk! I'm sure lots of people like you. :tongue:
P.S. I didn't say I believed the instructor's claim of the switch generating no spark.
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sophiecentaur
My confusion lies with the direction of current.
1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?
2. How is ground replaced? As I see it first the neon lamp was connected between point B and ground, but no circuit was being made (because switch was at A), after the switch closes with B the circuit is made, and ground stays where it was.
The diagram sort of explains what is going in my mind. The arrow in the loop being the current direction
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davenn
Gold Member
sophiecentaur
My confusion lies with the direction of current.
1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?
2. How is ground replaced? As I see it first the neon lamp was connected between point B and ground, but no circuit was being made (because switch was at A), after the switch closes with B the circuit is made, and ground stays where it was.
The diagram sort of explains what is going in my mind. The arrow in the loop being the current direction
The confusion is that you are using conventional current flow positive to negative
where I suspect sophicentaur is using electron current flow which is negative to positive
so when its negative at the top of the inductor, current flow is anticlockwise from top of inductor down through neon through resistor and back to inductor
hopefully I havent misread that ;)
Dave
NascentOxygen
Staff Emeritus
My confusion lies with the direction of current.
With the switch set to position A, current flows through the inductor from top to bottom. As expected, the top is + because it's connected to the battery + terminal. We agree on this.
1. You said when the switch closes with position B, the voltage at the 'top' of the inductor is negative as shown. So shouldn't the direction of current be clockwise (from the diagram) because it has to go from the more positive terminal to the more negative right?
When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.
Let's look at the elements apart from the inductor. How to get current to flow in a path that takes it DOWN through the resistor then UP through the neon? This can only happen if the top of the resistor is + with respect to the top of the neon. (Remember, current always flows from + to –.) BUT, we can see that the top of the resistor is the same point as the bottom of the coil, and the top of the neon is the same point as the top of the coil.
So, if the coil is to maintain current unchanged when the switch moves to position B, we have shown that the bottom of the inductor must go + and its top must be – . And indeed it does!
What we haven't examined is the magnitude of this flyback voltage. In this discussion all I've considered is its polarity.
Remember: The voltage across an inductor can change instantly. It's the current through an inductor that takes time to change.
davenn
Gold Member
Have read a number of times and trying to make heads and tails of it ??
................
When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.
is not this delayed current flow still part of that which flowed from the battery ?
if so then after switching to 'B' its magnitude and direction ( for that brief time) is going to roughly be the same ?
therefore you are not going to have current flowing through the neon as the voltage isnt high enough to strike/light the neon ?
Let's look at the elements apart from the inductor. How to get current to flow in a path that takes it DOWN through the resistor then UP through the neon? This can only happen if the top of the resistor is + with respect to the top of the neon. (Remember, current always flows from + to –.) BUT, we can see that the top of the resistor is the same point as the bottom of the coil, and the top of the neon is the same point as the top of the coil.
So, if the coil is to maintain current unchanged when the switch moves to position B, we have shown that the bottom of the inductor must go + and its top must be – . And indeed it does!
What we haven't examined is the magnitude of this flyback voltage. In this discussion all I've considered is its polarity.
Remember: The voltage across an inductor can change instantly. It's the current through an inductor that takes time to change.
isnt the voltage/current spike going to be caused by the collapse of the magnetic field
and its going to flow in the opposite direction ?
its that spike thats going to be high enough to strike/light the neon ?
just wanna clarify this for myself too :)
Dave
NascentOxygen
Staff Emeritus
The voltage across an inductor is no indication of the direction or magnitude of current through that inductor. To determine the current through an inductor you need to know its history.
davenn
Gold Member
The voltage across an inductor is no indication of the direction or magnitude of current through that inductor. To determine the current through an inductor you need to know its history.
not sure who or what particular comment thats aimed at ??
I agree, and we do know the history :) so not sure what point you are making ?
Dave
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NascentOxygen
Staff Emeritus
is not this delayed current flow still part of that which flowed from the battery ?
Um, delayed current? That is your word, and I don't know what you mean. I have never mentioned delayed current. Current doesn't get delayed, or bank up, or anything like that.
if so then after switching to 'B' its magnitude and direction ( for that brief time) is going to roughly be the same ?
Yes, inductor current never changes instantly, for to do so would require infinite voltage.
therefore you are not going to have current flowing through the neon as the voltage isnt high enough to strike/light the neon ?
I think this is a question you should answer. You agree that there is an interval of time after the battery has been disconnected when current continues flowing through the inductor. A fundamental principle is that current always flows in a closed loop. You say no current is flowing in the neon. We agree the battery is now out of the picture, so we can forget the battery. So can you describe the closed loop that you perceive the inductor current completing during this time.
isnt the voltage/current spike going to be caused by the collapse of the magnetic field
Yes.
and its going to flow in the opposite direction ?
Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.
its that spike thats going to be high enough to strike/light the neon ?
Yes.
NascentOxygen
Staff Emeritus
not sure who or what particular comment thats aimed at ??
I am lost in all this discussion of so many posts. Can someone just clarify if my logic in the diagram in my previous post is right?
Also, at the various time constants the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?
I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?
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sophiecentaur
Gold Member
2020 Award
I am lost in all this discussion of so many posts. Can someone just clarify if my logic in the diagram in my previous post is right?
Also, at the various time constants the values of voltage across the inductor are said to be ranging from 10V to 0V. (see diagram in this post). That is why I am confused as to WHEN exactly does the inductor get a value of 65V across it?
I understand V = L(di/dt) and that this happens when we switch open the switch suddenly. But what exactly allows us to give a specific value of voltage across the inductor when it switches from point A to B?
Firstly, what does the table in the diagram refer to? I don't think the heading "time constant" means what it says. Is it supposed to be 'time'? Could you explain what it all means? This looks more like a h/w question which supplies the information that you need to solve it. Are you expected to plot a graph of the volts across the L and extrapolate to the high volts you need (75V) after the graph has crossed the Y axis and intercepts -75. It's a strange circuit because it requires more volts from the L than necessary - if it were connected as I described a few days ago - you'd only have to get to 55V. But the question setter was after an answer from what he gave you, I suppose. Plot the graph and see what you get. When you are given data which makes no sense to you, a graph can often help.
To work out the "when" question (precisely), we need to know more than the information given. As I mentioned way back, the rate of change of current will be due to the Inductance value and also the parasitic Capacitance involved. You basically have a resonant circuit which has current flowing through it and, ignoring any other paths in the circuit, when you open the switch, the current in the Inductor will divert into charging the capacitance and the resonant circuit will ring at a frequency given by 1/2∏√(LC). There will be losses which will cause the oscillation to decay, of course and which will limit the peak voltage. You don't know those losses so you don't know when the volts will reach the 75V you need for your neon to strike. Neither do you know the value of the parasitic C but you could work out what you want if you knew (or could measure) the R and C values.
If you wanted to, you could put a capacitor across the L which would swamp the parasitic one and then you would have an idea of the (much lower resulting) resonant frequency and that would tell you how long before the peak was reached.
But all this may not be relevant to the answer you want. If you want a more useful answer to homework problems then I suggest you make it more clear what it's all about and the post in the right forum - where you won't find people batting your problem about in quite the same way. You can see a "no homework" message at the top of the list of topics in each forum. It tells you what to do.
davenn
Gold Member
Um, delayed current? That is your word, and I don't know what you mean. I have never mentioned delayed current. Current doesn't get delayed, or bank up, or anything like that.
poor choice of words ... was referring to the brief period of time the current still flows
Yes, inductor current never changes instantly, for to do so would require infinite voltage.
OK can accept that
I think this is a question you should answer. You agree that there is an interval of time after the battery has been disconnected when current continues flowing through the inductor. A fundamental principle is that current always flows in a closed loop. You say no current is flowing in the neon. We agree the battery is now out of the picture, so we can forget the battery. So can you describe the closed loop that you perceive the inductor current completing during this time.
there is NO closed loop, its an open cct at the neon, but you said the current is flowing up through the neon.....
When the switch is moved to position B, the current continues to flow unchanged (at least for a while) through the inductor in the same direction as it has been, because that's a property of inductance: it resists current changes. So we have current continuing to flow through the inductor from top to bottom, though now the battery has disappeared out of the picture. This current flows in a closed path (as it always must do), down through the inductor and up through the neon bulb.
So if that "residual" current is still flowing, and we know its not enough to strike/light the neon so as said I believe its an open cct.
I want you to tell me how you have current flowing in an open cct please :)
Are you suggesting inductor current instantly reverses direction? It doesn't. Inductor current continues as it was, before starting to decrease.
no Im not, we have already established that
Dave
The time constant discussion was for the "charging" of the inductor, as the current is building up from 0 to Max Amps.
When the switch is moved from A to B - this is now a V= L * di/dt discussion, not based on the time constant. It is also interesting to note - that you are discharging energy from the inductor - so a faster switch may see a bigger arc ( higher arc Voltage) but faster decrease in current ( higher V - Lower I).
Nevertheless - the 2 questions are not yet understood - Why does the V across the inductor "reverse" and how does the V get to 65 V.
Since the Inductor current can not change instantly - when the switch is opened the inductor will "push" current in the came direction, to do this the polarity at the bottom as shown will be +.
As for getting to 65V - it will be best to "do the math" ..... so throw a few more data points in there. Make L=50mH and R=50Ohms. ... I max (Switch in position A for more than 5 TC... then I = 0.2A. - Now - the key here is that there is no "perfect" switch, the better the switch ( how fast and cleanly it opens) the higher the voltage will be created by the inductor. Let's consider the point when the contacts in the switch begin to separate - at first they are separated by a very small gap - for example enough that 1V can jump the gap... also once you start the arc, the ionized air is more conductive. Assume we know or can control - how long it takes for the switch to extinguish this arc ( 1nS - 1 mS ? How "fast" will the switch need to be to get the 65+V to ignite the Neon Lamp.
I would not let the diagrams - current representation throw you off too much. The current convention is meaningless unless you agree on a convention or do real math ( since again the Physicists like to think of the literal electrons ( negative charges flowing in the negative direction) and EEs - prefer the + to - Current flow abstraction.
sophiecentaur
Gold Member
2020 Award
The time constant discussion was for the "charging" of the inductor, as the current is building up from 0 to Max Amps.
When the switch is moved from A to B - this is now a V= L * di/dt discussion, not based on the time constant. It is also interesting to note - that you are discharging energy from the inductor - so a faster switch may see a bigger arc ( higher arc Voltage) but faster decrease in current ( higher V - Lower I).
Nevertheless - the 2 questions are not yet understood - Why does the V across the inductor "reverse" and how does the V get to 65 V.
Since the Inductor current can not change instantly - when the switch is opened the inductor will "push" current in the came direction, to do this the polarity at the bottom as shown will be +.
As for getting to 65V - it will be best to "do the math" ..... so throw a few more data points in there. Make L=50mH and R=50Ohms. ... I max (Switch in position A for more than 5 TC... then I = 0.2A. - Now - the key here is that there is no "perfect" switch, the better the switch ( how fast and cleanly it opens) the higher the voltage will be created by the inductor. Let's consider the point when the contacts in the switch begin to separate - at first they are separated by a very small gap - for example enough that 1V can jump the gap... also once you start the arc, the ionized air is more conductive. Assume we know or can control - how long it takes for the switch to extinguish this arc ( 1nS - 1 mS ? How "fast" will the switch need to be to get the 65+V to ignite the Neon Lamp.
I would not let the diagrams - current representation throw you off too much. The current convention is meaningless unless you agree on a convention or do real math ( since again the Physicists like to think of the literal electrons ( negative charges flowing in the negative direction) and EEs - prefer the + to - Current flow abstraction.
The rate of change of current is limited, ultimately, by the stray Capacitance.
But the data (table) given in the question is what will yield the answer to it. All the rest is arm waving because we do not know enough of the details about the actual circuit. The slope of the graph gives the rate of change of the volts - i.e. it will give you the rise time of the pulse.
NascentOxygen
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https://merupulu.com/class-8-maths-chapter-10-visualising-solid-shapes/ | 1,723,392,339,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641002566.67/warc/CC-MAIN-20240811141716-20240811171716-00519.warc.gz | 317,260,633 | 60,642 | HomeNCERT SOLUTIONS8th CLASSClass 8 Maths Chapter 10 Visualising Solid Shapes
# Class 8 Maths Chapter 10 Visualising Solid Shapes
## NCERT Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.1 Solutions
NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Exercise 10.1
Ex 10.1 Class 8 Maths Question 1.
For each of the given solid, the two views are given. Match for each solid the corresponding top and front views. The first one is done for you.
Solution:
(a) A bottle → (iii) → (iv)
(b) A weight → (i) → (v)
(c) A flask → (iv) → (ii)
(d) Cup and saucer → (v) → (iii)
(e) Container → (ii) → (i)
Ex 10.1 Class 8 Maths Question 2.
For each of the given solid, the three views are given. Identify for each solid the corresponding top, front and side views.
Solution:
(a) An Almirah → (i) Front → (ii) Side → (iii) Top
(b) A Match box → (i) Side → (it) Front → (iii) Top
(c) A Television → (i) Front → (ii) Side → (iii) Top
(d) A Car → (i) Front → (ii) Side → (iii) Top
Ex 10.1 Class 8 Maths Question 3.
For each given solid, identify the top view, front view and side view.
Solution:
(a) (i) Top → (ii) Front → (iii) Side
(b) (i) Side → (ii) Front → (iii) Top
(c) (i) Top → (ii) Side → (iii) Front
(d) (i) Side → (ii) Front → (iii) Top
(e) (i) Front → (ii) Top → (iii) Side
Ex 10.1 Class 8 Maths Question 4.
Draw the front view, side view and top view of the given objects.
Solution:
# Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2
## NCERT Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.2 Solutions
NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Exercise 10.2
Ex 10.2 Class 8 Maths Question 1.
Look at the given map of a city.
(a) Colour the map as follows: Blue-water, red- fire station, orange-library, yellow-schools, Green-park, Pink-College, Purple-Hospital, Brown-Cemetery.
(b) Mark a green ‘X’ at the intersection of Road ‘C’ and Nehru Road, Green ‘Y’ at the intersection of Gandhi Road and Road A.
(c) In red, draw a short street route from Library to the bus depot.
(d) Which is further east, the city park or the market?
(e) Which is further south, the primary school or the Sr. Secondary School?
Ex 10.2 Class 8 Maths Question 2.
Draw a map of your classroom using proper scale and symbols for different objects.
Ex 10.2 Class 8 Maths Question 3.
Draw a map of your school compound using proper scale and symbols for various features like a playground, main building, garden etc.
Ex 10.2 Class 8 Maths Question 4.
Draw a map giving instructions to your friend so that she reaches your house without any difficulty.
Solution:
Question 1 to Question 4 each all activities. You can try yourself.
# Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3
## NCERT Class 8 Maths Chapter 10 Visualising Solid Shapes Ex 10.3 Solutions
NCERT Solutions for Class 8 Maths Chapter 10 Visualising Solid Shapes Exercise 10.3
Ex 10.3 Class 8 Maths Question 1.
Can a polyhedron have for its faces
(i) 3 triangles?
(ii) 4 triangles?
(iii) a square and four triangles?
Solution:
(i) No, because polyhedron must have edges meeting at vertices which are points.
(ii) Yes, because all the edges are meeting at the vertices.
(iii) Yes, because all the eight edges meet at the vertices.
Ex 10.3 Class 8 Maths Question 2.
Is it possible to have a polyhedron with any given number of faces?
(Hint: Think of a pyramid)
Solution:
Yes, it is possible if the number of faces is greater than or equal to 4.
Example: Pyramid which has 4 faces.
Ex 10.3 Class 8 Maths Question 3.
Which are prisms among the following?
Solution:
Only (ii) unsharpened pencil and (iv) a box are the prism.
Ex 10.3 Class 8 Maths Question 4.
(i) How are prisms and cylinders alike?
(ii) How are pyramids and cones alike?
Solution:
(i) If the number of sides in a prism is increased to certain extent, then the prism will take the shape of cylinder.
(ii) If the number of sides of the pyramid is increased to same extent, then the pyramid becomes a cone.
Ex 10.3 Class 8 Maths Question 5.
Is a square prism same as a cube? Explain.
Solution:
Every square prism cannot be cube. It may be cuboid also.
Ex 10.3 Class 8 Maths Question 6.
Verify Euler’s formula for these solids.
Solution:
(i) Faces = 7
Sides = 15
Vertices = 10
Euler’s formula: F + V – E = 2
⇒ 7 + 10 – 15 = 2
⇒ 2 = 2
Hence, Euler’s formula is verified.
(ii) Faces = 9
Sides = 16
Vertices = 9
Euler’s Formula: F + V – E = 2
⇒ 9 + 9 – 16 = 2
⇒ 2 = 2
Hence, Euler’s formula is verified.
Ex 10.3 Class 8 Maths Question 7.
Using Euler’s formula find the unknown.
Solution:
Using Eulers Formula: F + V – E = 2
Ex 10.3 Class 8 Maths Question 8.
Can a polyhedron have 10 faces, 20 edges and 15 vertices?
Solution:
Here faces = 10, Edges = 20, Vertices = 15
According to Euler’s Formula:
F + V – E = 2
⇒ 10 + 15 – 20 = 25 – 20
⇒ 5 ≠ 2
A polyhedron do not have 10 Faces, 20 Edges and 15 Vertices.
## NCERT Solutions for Class 8 Maths
### సీడీపీవో, ఈవో పోస్టులకు పరీక్ష షెడ్యూలు రిలీజ్
error: Content is protected !! | 1,570 | 5,036 | {"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-2024-33 | latest | en | 0.790172 |
http://mathhelpforum.com/advanced-applied-math/26243-reversing-equation-print.html | 1,527,109,845,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794865809.59/warc/CC-MAIN-20180523200115-20180523220115-00094.warc.gz | 186,835,754 | 3,027 | # Reversing an equation
• Jan 16th 2008, 05:39 PM
victurbo
Reversing an equation
Need to figure out how to reverse this equation for work...
C = 4.42 - .113 * A + 0.0406 * B
C = Dwell time in a vessel in minutes
A = Screw speed drawing material out of a vessel in RPM's
B = Level in the tank in inches
How do I solve for "B" if I know what cook time I want to target?
Thanks for any assistance in advance. My brain must be broken as this seemed simple until I actually sat down and tried to figure it out.
• Jan 16th 2008, 06:02 PM
mr fantastic
Quote:
Originally Posted by victurbo
Need to figure out how to reverse this equation for work...
C = 4.42 - .113 * A + 0.0406 * B
C = Dwell time in a vessel in minutes
A = Screw speed drawing material out of a vessel in RPM's
B = Level in the tank in inches
How do I solve for "B" if I know what cook time I want to target?
Thanks for any assistance in advance. My brain must be broken as this seemed simple until I actually sat down and tried to figure it out.
What do you mean by "cook time"?
• Jan 16th 2008, 07:16 PM
victurbo
Cook Time = Dwell Time
Sorry for the confusion. It's the order of transposition I'm struggling with. Algebra was a long time ago.
• Jan 16th 2008, 08:52 PM
mr fantastic
Quote:
Originally Posted by victurbo
Cook Time = Dwell Time
Sorry for the confusion. It's the order of transposition I'm struggling with. Algebra was a long time ago.
Quote:
Originally Posted by victurbo
Need to figure out how to reverse this equation for work...
C = 4.42 - .113 * A + 0.0406 * B
C = Dwell time in a vessel in minutes
A = Screw speed drawing material out of a vessel in RPM's
B = Level in the tank in inches
How do I solve for "B" if I know what cook time I want to target? [snip]
C = 4.42 - .113 * A + 0.0406 * B
therefore C - 4.42 = - .113 * A + 0.0406 * B
therefore C - 4.42 + .113 * A = 0.0406 * B
therefore (C - 4.42 + .113 * A)/0.0406 = B.
B = (C - 4.42 + .113 * A)/0.04
Of course, you won't be able to get a value for B unless you also know the value of A .....
• Jan 17th 2008, 06:11 AM
victurbo
That did it
Thanks for the help. | 663 | 2,131 | {"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} | 4.03125 | 4 | CC-MAIN-2018-22 | latest | en | 0.930224 |
https://fistofawesome.com/interesting/what-is-the-square-of-288/ | 1,680,333,385,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.56/warc/CC-MAIN-20230401063607-20230401093607-00596.warc.gz | 312,029,272 | 11,465 | What is the square of 288?
What is the square of 288?
The square root of 288 is 16.970.
What is the square of 504?
We know that the square root of 504 is 22.450.
Is 157 a square number?
The square root of 157 is expressed as √157 in the radical form and as (157)½ or (157)0.5 in the exponent form….Square Root of 157 in radical form: √157.
1. What is the Square Root of 157?
2. How to find the Square Root of 157?
3. Is the Square Root of 157 Irrational?
4. FAQs
IS 128 a perfect square?
128 is not a perfect square.
What is the square of 648?
The square root of 648 is 25.45584.
How do you find the square root of 5184?
So, 5184=2×2×2×2×2×2×3×3×3×3 . So, the square root of 5184 can be written as √5184=2×2×2×3×3=72 . Hence, the square root of 5184 is 72.
What does 2i mean?
2i is an imaginary number because it has the form ‘bi’ Remember, ‘i’ is the imaginary unit and is equal to the square root of -1. Even though ‘i’ is NOT a variable, we can multiply it as if it were. So i • i gives us i2. Squaring √ (-1) cancels out the square root, leaving us with just -1.
What is the meaning of 2 root 3?
What is 3 root 3 simplified?
Since √3√3 is equal to 1 , you simply rearranged the way it was written. The value of the simplified fraction stays the same.
Is 2i a real number?
What’s a pure number? Is 2i a real zero? Therefore it must have no real roots, and 2 imaginary roots. By setting the function equal to zero and solving for x, we can see that this is true. So the roots are 2i and -2i, both of which are imaginary. How do you tell if a root is real? The discriminant (EMBFQ)
What does 2i mean in math terms?
2i is an imaginary number because it has the form ‘bi’. Remember, ‘i’ is the imaginary unit and is equal to the square root of -1. Even though ‘i’ is NOT a variable, we can multiply it as if it were. So i • i gives us i 2. Squaring √ (-1) cancels out the square root, leaving us with just -1. So i 2 is equal to -1.
What is 2i equal to?
What is 2i equal to? The absolute value of the complex number, 2i, is 2. We can put the complex number, 2i, in the form a + bi by letting a = 0. What is 10 squared mean? -10 squared it 100. Any negative number that is squared is positive. Therefore, -10 squared is equal to 10 squared. -10 and 10 are the two square roots of 100.
How do you simplify (2i)?
Tap for more steps… Multiplyby . Raise to the powerof . Raise to the powerof . Use the power ruleto combineexponents. Add and . Rewrite as . Multiplyby . Cookies & Privacy | 747 | 2,506 | {"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-2023-14 | longest | en | 0.928516 |
https://en.m.wikibooks.org/wiki/Partial_Differential_Equations/Parallel_Plate_Flow:_Realistic_IC | 1,529,563,095,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864039.24/warc/CC-MAIN-20180621055646-20180621075646-00611.warc.gz | 604,913,521 | 22,555 | # Advanced Mathematics for Engineers and Scientists/Parallel Plate Flow: Realistic IC
(Redirected from Partial Differential Equations/Parallel Plate Flow: Realistic IC)
## Parallel Plate Flow: Realistic ICEdit
The initial velocity profile chosen in the last problem agreed with intuition but honestly came out of thin air. A more realistic development follows.
The problem stated that (to come up with an IC) the fluid was under a pressure difference for some time, so that the flow became steady aka flowing steadily. "Steady" is another way of saying "not changing with time", and "not changing with time" is another way of saying that:
${\displaystyle {\frac {\partial u}{\partial t}}=0\,}$
Putting this into the PDE from the previous section:
${\displaystyle {\frac {\partial u}{\partial t}}=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}-{\frac {P_{x}}{\rho }}\,}$
${\displaystyle {\Big \Downarrow }}$
${\displaystyle 0=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}-{\frac {P_{x}}{\rho }}\quad \Rightarrow \quad {\frac {P_{x}}{\nu \rho }}={\frac {d^{2}u}{dy^{2}}}\,}$
Independent of ${\displaystyle t}$ , the PDE became an ODE with variables separated and thus we can integrate.
${\displaystyle {\frac {d^{2}u}{dy^{2}}}={\frac {P_{x}}{\nu \rho }}\quad \Rightarrow \quad u={\frac {P_{x}}{2\nu \rho }}y^{2}+C_{1}y+C_{0}\,}$
The no slip condition results in the following BCs: ${\displaystyle u=0}$ at ${\displaystyle y=0}$ and ${\displaystyle y=1}$ . We can plug the BC values into the integrated ODE and resolve the Cs.
${\displaystyle 0={\frac {P_{x}}{2\nu \rho }}\cdot 0^{2}+C_{1}\cdot 0+C_{0}\quad \Rightarrow \quad C_{0}=0\qquad {\mbox{(Bottom plate BC: u = 0, y = 0)}}\,}$
${\displaystyle 0={\frac {P_{x}}{2\nu \rho }}\cdot 1^{2}+C_{1}\cdot 1\quad \Rightarrow \quad C_{1}=-{\frac {P_{x}}{2\nu \rho }}\qquad {\mbox{(Top plate BC: u = 0, y = 1)}}\,}$
Inserting the Cs and and simplifying yields:
${\displaystyle u={\frac {P_{x}}{2\nu \rho }}(y^{2}-y)\,}$
For the sake of example, take ${\displaystyle P_{x}/(2\nu \rho )=-4}$ (recall that a negative pressure gradient causes left to right flow). Also note that this is a constant gradient or slope. This gives a parabola which starts at ${\displaystyle 0}$ , increases to a maximum of ${\displaystyle 1}$ at ${\displaystyle y=1/2}$ , and returns to ${\displaystyle 0}$ at ${\displaystyle y=1}$ .
This parabola looks pretty much identical to the sinusoid previously used (you must zoom in to see a difference). However, even more so on the narrow domain of interest, the two are very different functions (look at their taylor expansions, for example). Using the parabola instead of the sine function results in a much more involved solution.
So this derives the steady state flow, which we will use as an improved, realistic IC. Recall that the problem is about a fluid that's initially in motion that is coming to a stop due to the absence of a driving force. The IBVP (Initial Boundary Value Problem) is now subtly different:
${\displaystyle {\frac {\partial u}{\partial t}}=\nu {\frac {\partial ^{2}u}{\partial y^{2}}}\qquad {\mbox{(PDE)}}\,}$
${\displaystyle u(y,0)=4y-4y^{2}\qquad {\mbox{(IC)}}\,}$
${\displaystyle u(0,t)=0\,}$
${\displaystyle u(1,t)=0\qquad {\mbox{(BCs)}}\,}$
### SeparationEdit
Since the only difference from the problem in the last section is the IC, the variables may be separated and the BCs applied with no difference, giving:
${\displaystyle u(y,t)=e^{-(n\pi )^{2}\nu t}B\sin(n\pi y)\,}$
But now we're stuck (after applying the BCs)! Applying the IC makes the ${\displaystyle e^{-(n\pi )^{2}\nu t}}$ term go away as t = 0, which is the IC. However, then the IC function can't be made to match:
${\displaystyle u(y,0)=4y-4y^{2}\,}$
${\displaystyle B\sin(n\pi y)=4y-4y^{2}\,}$
What went wrong? It was the assumption that ${\displaystyle u(y,t)=Y(y)T(t)}$ . The fact that the IC couldn't be fulfilled means that the assumption was wrong. It should be apparent now why the IC was chosen to be ${\displaystyle \sin(\pi y)}$ in the previous section.
We can proceed however, thanks to the linearity of the problem. Another detour is necessary, it gets long.
Linearity (the superposition principle specifically) says that if ${\displaystyle u_{1}}$ is a solution to the BVP (not the whole IBVP, only the BVP, Boundary Value Problem, the BCs applied) and so is another ${\displaystyle u_{2}}$ , then a linear combination, ${\displaystyle C_{1}u_{1}+C_{2}u_{2}}$ , is also a solution.
Let's take a step back and suppose that the IC was
${\displaystyle u(y,0)=\sin(\pi y)+1/5\sin(3\pi y).}$
This is no longer a realistic flow problem but it contains the first two terms of what is called a Fourier sine expansion, see these examples of Fourier sine expansions. We are going to generalize this below. Let's now use this expression and equate it to the half way solution (BCs applied) with ${\displaystyle e^{-(n\pi )^{2}\nu t}}$ being eliminated as t = 0:
${\displaystyle B\sin(n\pi y)=\sin(\pi y)+{\frac {1}{5}}\sin(3\pi y)\,}$
And it still can't match. However, observe that the individual terms in the IC can. We simply set the constants to values making both sides match:
${\displaystyle B_{1}\sin(n_{1}\pi y)=\sin(\pi y)\Rightarrow n_{1}=1\ {\mbox{and}}\ B_{1}=1\,}$
${\displaystyle B_{3}\sin(n_{3}\pi y)={\frac {1}{5}}\sin(3\pi y)\Rightarrow n_{3}=3\ {\mbox{and}}\ B_{3}={\frac {1}{5}}\,}$
Note the subscripts are used to identify each term: they reflect the integer ${\displaystyle n}$ from the separation constant. Solutions may be obtained for each individual term of the IC, identified with ${\displaystyle n}$ :
${\displaystyle u_{1}(y,t)=e^{-(1\pi )^{2}\nu t}1\sin(1\pi y)=e^{-\pi ^{2}\nu t}\sin(\pi y)}$ ${\displaystyle n=1\ {\mbox{and}}\ B=1\,}$
${\displaystyle u_{3}(y,t)=e^{-(3\pi )^{2}\nu t}{\frac {1}{5}}\sin(3\pi y)=e^{-9\pi ^{2}\nu t}{\frac {1}{5}}\sin(3\pi y)\,}$ ${\displaystyle n=3\ {\mbox{and}}\ B={\frac {1}{5}}\,}$
Linearity states that the sum of these two solutions is also a solution to the BVP (no need for new constants):
${\displaystyle u_{1+3}(y,t)=e^{-\pi ^{2}\nu t}\sin(\pi y)+{\frac {1}{5}}e^{-9\pi ^{2}\nu t}\sin(3\pi y)\,}$
So we added the solutions and got a new solution... what is this good for? Try setting ${\displaystyle t=0}$ :
${\displaystyle u_{1+3}(y,0)=e^{-\pi ^{2}\nu \cdot 0}\sin(\pi y)+{\frac {1}{5}}e^{-9\pi ^{2}\nu \cdot 0}\sin(3\pi y)=\sin(\pi y)+{\frac {1}{5}}\sin(3\pi y)\,}$
Each component solution satisfies the BVP, and the sum of these just happened to satisfy our surrogate IC. The IBVP with IC ${\displaystyle u(y,0)=\sin(\pi y)+1/5\sin(3\pi y)}$ is now solved. It would work the same way for any linear combination of sine functions whose half frequencies are ${\displaystyle n\pi }$ . "Linear combination" means a sum of terms, each multiplied by a constant. The sum is assumed to converge and be term by term differentiable.
Let's do what we just did in a more generalized fashion. First, we make our IC a linear combination of sines (with ${\displaystyle e^{-(n\pi )^{2}\nu t}}$ eliminated as t = 0), in fact, infinitely many of them. But each successive term has to 'converge', it can't stray wildly all over the place.
${\displaystyle u(y,0)=\sum _{n=1}^{\infty }B_{n}\sin(n\pi y)\qquad {\mbox{(IC: an arbitrary linear combination of sines)}}\,}$
Second, find the n and B for each term assuming t = 0 (the IC), then plug them back into each term making no assumptions about t, leaving t as is.
${\displaystyle u_{n}(y,t)=e^{-(n\pi )^{2}\nu t}B_{n}\sin(n\pi y)\qquad {\mbox{(solution of}}\ n^{th}\ {\mbox{term)}}\,}$
Third, sum up all the terms with their individual n and Bs.
${\displaystyle u(y,t)=\sum _{n=1}^{\infty }u_{n}(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}B_{n}\sin(n\pi y)\qquad {\mbox{(sum of solutions)}}\,}$
Fourth, plug t = 0 into the sum of terms and recover the IC from the first step.
${\displaystyle u(y,0)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu \cdot 0}B_{n}\sin(n\pi y)=\sum _{n=1}^{\infty }B_{n}\sin(n\pi y)\qquad {\mbox{(IC recovered)}}\,}$
So we went full circle on this example but found the n and Bs because we were able to equate/satisfy each term with the IC. Now we can solve the problem if the IC is a linear combination of sine functions. But the IC for this problem isn't such a sum, it's just a stupid parabola. Or is it?
### Series ConstructionEdit
In the 19th century, a man named Joseph Fourier took a break from helping Napoleon take over the world to ask an important question while studying this same BVP (concerning heat flow): can a function be expressed as a sum of sinusoids, similar to a taylor series? The short answer is yes, if a few reasonable conditions apply as we have already indicated. The long answer follows, and this section is a longer answer.
A function meeting certain criteria may indeed be expanded into a sum of sines, cosines, or both. In our case, all that is needed to accomplish this expansion is to find the coefficients ${\displaystyle B_{n}}$ . A little trick involving an integral makes this possible.
The sine function has a very important property called orthogonality. There are many flavors of this, which will be served in the next chapter. Relevant to this problem is the following:
${\displaystyle \int _{0}^{1}2\sin(m\pi y)\sin(n\pi y)\,dy={\begin{cases}1,&m=n\\0,&m\neq n\end{cases}}}$
A quick hint may help. Orthogonality literally means two lines at a right angle to each other. These lines could be vectors, each with its own tuple of coordinates. If those two vectors are at a right angle to each other, multiplying and summing their coordinate tuples always yields zero (in Euclidean space). The method of multiplying and summing is also used to determine whether two functions are orthogonal. Using this definition, our multiplied and integrated functions above are orthogonal most of the time, but not always.
Let's call the IC ${\displaystyle \phi (y)}$ to generalize it. We equate the IC with its expansion, meaning the linear combination of sines, and then apply some craftiness. And remember that our goal is to reproduce a parabolic function from linearly combined sines:
${\displaystyle \sum _{n=1}^{\infty }B_{n}\sin(n\pi y)=\phi (y)\,}$
${\displaystyle 2\sin(m\pi y)\cdot \sum _{n=1}^{\infty }B_{n}\sin(n\pi y)=2\sin(m\pi y)\cdot \phi (y)\,}$
${\displaystyle \sum _{n=1}^{\infty }B_{n}\cdot 2\sin(m\pi y)\sin(n\pi y)=2\sin(m\pi y)\phi (y)\,}$
${\displaystyle \int _{0}^{1}\sum _{n=1}^{\infty }B_{n}\cdot 2\sin(m\pi y)\sin(n\pi y)dy=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$
${\displaystyle \sum _{n=1}^{\infty }B_{n}\int _{0}^{1}2\sin(m\pi y)\sin(n\pi y)dy=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$
${\displaystyle B_{m}=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$
In the last step, all of the terms in the sum became ${\displaystyle 0}$ except for the ${\displaystyle m^{th}}$ term where ${\displaystyle m=n}$ , the only case where we get ${\displaystyle 1}$ for the otherwise orthogonal sine functions. This isolates and explicitly defines ${\displaystyle B_{m}}$ which is the same as ${\displaystyle B_{n}}$ as m = n. The expansion for ${\displaystyle \phi (y)}$ is then:
${\displaystyle \phi (y)=\sum _{m=1}^{\infty }(\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy)\cdot \sin(m\pi y)\,}$
Or equivalently:
${\displaystyle \phi (y)=\sum _{m=1}^{\infty }B_{m}\sin(m\pi y)\qquad ;\ B_{m}=\int _{0}^{1}2\sin(m\pi y)\phi (y)\ dy\,}$
Many important details have been left out for later in a devoted chapter; one noteworthy detail is that this expansion is only approximating the parabola (very superficially) on the interval ${\displaystyle 0\leq y\leq 1}$ , not say from ${\displaystyle -\infty }$ to ${\displaystyle +\infty }$ .
This expansion may finally be combined with the sum of sines solution to the BVP developed previously. Note that the last equation looks very similar to ${\displaystyle u(y,0)}$ . Following from this:
${\displaystyle u(y,0)=\sum _{n=1}^{\infty }B_{n}\sin(n\pi y)\,}$
${\displaystyle u(y,0)=\sum _{n=1}^{\infty }\int _{0}^{1}2\sin(n\pi y)\phi (y)\ dy\cdot \sin(n\pi y)\qquad \,}$
So the expansion will satisfy the IC given as ${\displaystyle \phi (y)}$ (surprised?). The full solution for the problem with arbitrary IC is then:
${\displaystyle u(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}B_{n}\sin(n\pi y)\,}$
${\displaystyle u(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}\int _{0}^{1}2\sin(n\pi y)\phi (y)\ dy\cdot \sin(n\pi y)\,}$
In this problem specifically, the IC is ${\displaystyle \phi (y)=4y-4y^{2}}$ , so:
${\displaystyle B_{n}=\int _{0}^{1}2\sin(n\pi y)(4y-4y^{2})dy={\frac {8}{n^{3}\pi ^{3}}}(2-2\cos(n\pi )-n\pi \sin(n\pi ))\,}$
Sines and cosines appear from the integration dependent only on ${\displaystyle n\pi }$ . Since ${\displaystyle n}$ is an integer, these can be made more aesthetic.
${\displaystyle \sin(n\pi )=0\qquad ;\ n{\mbox{ is an integer.}}\,}$
${\displaystyle \cos(n\pi )=(-1)^{n}\qquad ;\ n{\mbox{ is an integer.}}\,}$
${\displaystyle \qquad {\Big \Downarrow }}$
${\displaystyle B_{n}={\frac {16-16(-1)^{n}}{n^{3}\pi ^{3}}}\,}$
Note that for even ${\displaystyle n}$ , ${\displaystyle B_{n}=0}$ . Putting everything together finally completes the solution to the IBVP:
${\displaystyle u(y,t)=\sum _{n=1}^{\infty }e^{-(n\pi )^{2}\nu t}{\frac {16-16(-1)^{n}}{n^{3}\pi ^{3}}}\sin(n\pi y)\,}$
There are many interesting things to observe. To begin with, ${\displaystyle u(y,t)}$ is not a product of a function of ${\displaystyle y}$ and a function of ${\displaystyle t}$ . Such a solution was assumed in the beginning, proved to be wrong, but eventually happened to yield a solution anyway thanks to linearity and what is called a Fourier sine expansion.
A careful look at the procedure reveals something that may be disturbing: this lengthy solution is strictly valid for the given BCs. Thanks to the definition of ${\displaystyle \phi (y)}$ , the solution is generic as far as the IC is concerned (the IC doesn't even need to match the BCs), however a slight change in either BC would mandate starting over almost from the beginning.
The parabolic IC, which looks very similar to the sine function used in the previous section, is wholly to blame (or thank once you understand the beauty of a Fourier series!) for the infinite sum. It is interesting to approximate the first several numeric values of the sequence ${\displaystyle B_{n}}$ :
{\displaystyle {\begin{aligned}B_{1}&\approx 1.03205\\B_{3}&\approx 0.03822\\B_{5}&\approx 0.00826\\B_{7}&\approx 0.00301\,\end{aligned}}}
Recall that the even terms are all ${\displaystyle 0}$ . The first term by far dominates, this makes sense since the first term already looks very, very similar to the parabola. Recall that ${\displaystyle n^{2}}$ appears in an exponential, making the higher terms even smaller for time not too close to ${\displaystyle 0}$ . | 4,753 | 14,919 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 99, "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-2018-26 | latest | en | 0.754273 |
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## Posts tagged with mathmistakes
### Some Common Math Errors
November6
Errors #1: This image shows a student’s working. What went wrong? Well, obviously x should = 17, not the square root of 17. Perhaps a better way to explain the error is like this: When solving equations (expressions with an equal sign) like this, remember the algebraic rule: “Whatever you do to one side, do […]
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NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?] | 318 | 1,052 | {"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} | 3.53125 | 4 | CC-MAIN-2019-51 | latest | en | 0.924437 |
https://www.mathhomeworkanswers.org/269286/verticles-rectangle-coordinates-fourth-vertext-rectangle?show=269311 | 1,580,092,158,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251694176.67/warc/CC-MAIN-20200127020458-20200127050458-00189.warc.gz | 975,996,122 | 20,995 | A.(-7,0)
B. (-6,0)
C. (-6,-1)
D. (-5,1)
Two sides of the rectangle must be perpendicular—but which two? The product of the slopes of these lines must be -1 to be perpendicular. Label the points: A(-5,3), B(1,-1), C(-1,-4). D is the vertex whose coords we have to find.
Slope of AB=(3-(-1))/(-5-1)=-4/6=-⅔.
Slope of BC=(-1-(-4))/(1-(-1))=3/2.
Slope of AC=(3-(-4))/(-5-(-1))=-7/4.
Clearly AB and BC are perpendicular because (-2/3)(3/2)=-1.
So DC is parallel to AB and perpendicular to AD.
B is as far over to the right from A as D is to the left of C (x direction).
B is as far below A as D is above C (y direction).
B is 1-(-5)=6 units right of A and 3-(-1)=4 units below A.
D is 6 units left of C at x=-1-6=-7.
D is 4 units above D at y=-4+4=0.
by Top Rated User (716k points) | 288 | 792 | {"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} | 4.03125 | 4 | CC-MAIN-2020-05 | longest | en | 0.901421 |
http://www.xpmath.com/forums/showthread.php?s=2175b1df346fae53d46579a86f211755&mode=hybrid&t=1662 | 1,611,016,599,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703517159.7/warc/CC-MAIN-20210118220236-20210119010236-00115.warc.gz | 187,712,849 | 11,358 | I have a math question!!? - XP Math - Forums
XP Math - Forums I have a math question!!?
07-15-2007 #1 Blondie Guest Posts: n/a I have a math question!!? How many liters of bedding will it take to fill something of the following dimensions: 11''wide 22''long and 1'' high? I would be so greatful if you could tell me.
07-15-2007 #2 sugarhigh4life Guest Posts: n/a 242to fill sumthin its volume so u multiply them all
07-15-2007 #3 Al Bunn Guest Posts: n/a 22 * 11 * 1 = 242 cubic inches = 3.97 liters
07-15-2007 #4 yvikas70 Guest Posts: n/a I am assuming you are trying to find the volume. The formula for the volume is w x h x l. so the answer would be 11 x 1 x 22 = 242 cu. inches. now you have to convert this to liters. the conversion from cubic inches to liters is { 1 cubic inch = 0.016387 liter} so 242 cu inches = 3.965 liters of bedding
07-15-2007 #5 Diminati Guest Posts: n/a We have to calculate the volume in cubic inches first.22 * 11 = 242 cubic inches.Next convert cubic inches to liters:1 cubic inch = 0.016387064 liters242 * 0.016387 = 3.97 liters
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http://www.wisegeek.com/what-is-a-mobius-strip.htm | 1,529,522,955,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863834.46/warc/CC-MAIN-20180620182802-20180620202802-00407.warc.gz | 527,191,598 | 21,202 | Category:
# What is a Mobius Strip?
Article Details
• Written By: Michael Pollick
• Edited By: Niki Foster
2003-2018
Conjecture Corporation
India is building what will become the world’s biggest statue; it will be twice as tall as the Statue of Liberty. more...
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If you took a strip of paper and taped the ends together, you would most likely end up with a belt. It would be a loop with an inside surface and an outside surface. But what if you took that same strip of paper and gave it a half twist before taping the ends together? The result would be a fascinating geometric oddity called a mobius strip.
A mobius strip is an example of non-Euclidean geometry made real. Most of the time, non-Euclidean designs can only be imagined, or drawn like optical illusions. They could never exist outside of an M.C. Escher dream world. Yet the mobius strip is indeed a three dimensional object with only one side. The strangeness doesn't end there, however.
To construct a mobius strip, you will need a length of paper at least two inches wide for best results. A strip of newspaper cut lengthwise will suffice. Take the two ends of the strip in both hands and give one end a half-twist. Bring the two ends together and bind them with tape.
What you should have is a belt of paper with a half-twist. This is now an official mobius strip. Find a pair of scissors and a marking pen to perform the rest of the experiment.
The first principle to demonstrate with a mobius strip is the concept of a single surface. Using a marking pen, start drawing a line down the middle of the mobius strip without stopping. Your continuous line should eventually meet up with your original starting point. This proves that the mobius strip does indeed only have one side. Performing the same action on a normal paper loop would only mark the inside or outside surface.
Using the scissors, cut along the line created by the pen. Instead of becoming two separate loops, a mobius strip will form a single loop twice as large as the original. Cutting down the new mobius strip will result in two interlocking loops. If you use a wider strip of paper, the mobius strip will continue to form continuous or interlocking loops. You can also vary the experiment by cutting the loop into three equal sections or sections of varying lengths.
A mobius strip is an excellent way to introduce students to the worlds of science and geometry. The experiments are simple enough for young children to perform, but the science behind the illusion should fascinate older students as well.
## Recommended
anon348761 Post 6 This is mind blowing. TreeMan Post 5 Because mobius strips only have one side that repeats over and over again if you were to follow its path, I wonder if maybe there would be a way to utilize that somehow. I don't know what it would be exactly. I have seen videos, though, of people who make mobius strip music. There are small "music boxes" you can buy that use a strip of paper with holes punched into it to make sound. The concept is the same as a player piano or record player. If you twist the paper into a mobius strip, the song will play in a loop until you stop it. Like I said, though, I don't know how you would turn this into a practical use. JimmyT Post 4 @matthewc23 - I'd be interested in someone explaining more about non-Euclidean geometry, too. As far as uses of a mobius strip go, I would say its shape really limited the way it can be used in functional things like machinery or tools. Mobius strips have a very striking shape, though, and I have seen them used all the time in artwork and things like mobius strip bracelets. As a matter of fact, I was just at a museum not too long ago, and they had a sculpture garden outside. Someone had used a huge metal beam and twisted it into a mobius strip. It was an interesting piece. matthewc23 Post 3 @titans62 - I had the same thought you did until I actually got a piece of paper and tried it for myself. Because of the twist in the middle, the ends of the line do not meet up after the first pass around the mobius strip. You have to go around twice, and then the lines do match up, and you have a mark on both of the original sides of the paper. I tried out the rest of the experiment, and both of those parts work, too. What is the significance of non-Euclidean geometry? I am not a math person by any means, so I'm sure I don't even fully grasp why a mobius strip is important. It is just a neat experiment as far as I am concerned. Do mobius strips have any sort of practical use in daily life? Is there anything that uses the shape? titans62 Post 2 I have always seen mobius strips, but never understood the significance of them. The seem to be a popular display in math departments. I don't have any tape around me to try out the experiments that were described in the article, but I don't understand how you can prove a mobius strip has only one side by drawing a line on it. Even if you drew the line, you would still have an empty side. Wouldn't that prove that it is two sided? I don't know about cutting down the middle and what effect that would have, so I'll have to assume the article is right. That kind of mental thinking is a little too complex without having a mobius strip in my hand. | 1,208 | 5,369 | {"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-2018-26 | latest | en | 0.945607 |
http://novius.co/kalaj/order-of-operations-homework-2547.php | 1,550,664,869,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247494741.0/warc/CC-MAIN-20190220105613-20190220131613-00635.warc.gz | 196,868,211 | 4,181 | ## Order of operations homework
Relate this to the Order of Operations that we use in mathematics and how it has an order in which we do things so.Here we review what the proper order of operations when solving an equation with multiple mathematical expressions.
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Homework Unlocked offers homework help to parents teach their kids the order of operations when solving an equation with multiple mathematical expressions.Homework and Practice 1-4 Order of Operations LESSON Evaluate each expression. 1. 7 3 4 2. (45 10) 2 3. (20 16) 2 10 4. 22 3 4 5. 9 (7 1) 2 6. (9 2) 8 (1 3).
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This is a package of 7 homework sheets practicing the order of operations using GEMDAS.CMP offers mathematical help for each grade level associated with CMP. | 541 | 2,583 | {"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} | 3.8125 | 4 | CC-MAIN-2019-09 | latest | en | 0.921173 |
https://math.stackexchange.com/questions/162337/dimension-of-tangent-space-via-curves | 1,563,864,758,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195529007.88/warc/CC-MAIN-20190723064353-20190723090353-00501.warc.gz | 470,240,707 | 36,094 | dimension of tangent space - via curves
Lets have smooth $n-$ manifold $M\subset R_N$ and define its tangent space at $p\in M$ to be set of equivalence classes of smooth curves with $\gamma: I\to M$, $\gamma(0)=p$ with relation $\gamma_1\sim\gamma_2$ if $\frac{d\gamma_1(0)}{dt}=\frac{d\gamma_2(0)}{dt}$. How to prove that $T_p(M)$ is $n$ dimensional vector space?
• How are you defining $\frac{d \gamma_1(0)}{dt}$ without defining tangent spaces? – Qiaochu Yuan Jun 24 '12 at 10:10
• You first have to establish the operations of addition and scalar multiplication. Do you know how to define these? – Olivier Bégassat Jun 24 '12 at 10:11
• I think that $\frac{d\gamma}{d t}(0)=(\varphi\circ\gamma)'(0)$ for a chart $(U,\varphi)$. – JBC Jun 24 '12 at 10:12
Let $(U,\varphi)$ a chart in $p$. Let $\theta_\varphi:T_pM\rightarrow\mathbb R^n$ defined by $\theta_\varphi([\gamma])=(\varphi\circ\gamma)'(0)$.
Then $\theta_\varphi$ is clearly injective (by the definition of your equivalence relation). We check that it's surjective : let $v\in\mathbb R^n$, let $\gamma:t\mapsto\varphi^{-1}(tv+\varphi(p))$ then $v=\theta_\varphi([\gamma])$.
So, $\theta_\varphi$ is a bijection and we can use it to transfer the vector space structure from $\mathbb R^n$ to $T_pM$ :
If $\xi,\eta\in T_pM$, we define $\xi+\eta=\theta_\varphi^{-1}(\theta_\varphi(\xi)+\theta_\varphi(\eta))$ and if $\lambda\in\mathbb R$, we define $\lambda\xi=\theta_\varphi^{-1}(\lambda\theta_\varphi(\xi))$.
You can easily check that you get a space vector and that $\theta_\varphi$ is a linear isomorphism.
This construction is independant of the choice of $(U,\varphi)$ : if $(V,\psi)$ is another chart containing $p$, then $\theta_\varphi\circ\theta_\psi^{-1}(v)=d_{\psi(m)}(\varphi\circ\psi^{-1})(v)$. | 587 | 1,767 | {"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} | 3.65625 | 4 | CC-MAIN-2019-30 | latest | en | 0.658036 |
https://congresosistemasdesalud.com/qa/question-how-many-60-degrees-does-it-take-to-make-a-full-turn.html | 1,611,112,489,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703519883.54/warc/CC-MAIN-20210120023125-20210120053125-00282.warc.gz | 278,802,986 | 8,749 | # Question: How Many 60 Degrees Does It Take To Make A Full Turn?
## Is 90 degrees a half turn?
There are 360° in a full turn, 180° in a half turn and 90° in a quarter turn..
## How many angles does it take to make a full turn?
There are 360 degrees in one Full Rotation (one complete circle around).
## What are the 7 types of angles?
Types of Angles – Acute, Right, Obtuse, Straight and Reflex…Acute angle.Right angle.Obtuse angle.Straight angle.Reflex angle.
## Why is there 360 degrees in a full turn?
The Mesopotamians passed their base-60 numerical system to the ancient Egyptians, who used it to divide a circle into 360 degrees, Mary Blocksma writes in her book Reading the Numbers. The 360-degree circle worked out great: The Egyptians loved perfect triangles, and exactly six of them fit into a circle.
## What angle is 45?
A straight angle measures 180°. … An angle can be measured using a protractor, and the angle of measure 90 degrees is called a right angle. In a right angle, the two arms are perpendicular to each other. When the right angle is divided into two equal parts each angle measures 45°.
## What type of angle is 180 degrees?
straight anglesAngles that are 180 degrees (θ = 180°) are known as straight angles. Angles between 180 and 360 degrees (180°< θ < 360°) are called reflex angles. Angles that are 360 degrees (θ = 360°) are full turn.
## What type of angle is 350 degrees?
Reflex angles are the types of angles whose degree measurement is more than 180° but less than 360°. Common examples of reflex angles are; 200°, 220°, 250°, 300°, 350° etc. A complete angle is equal to 360°.
## What are the six types of angles?
Types of AnglesAcute Angle: An angle whose measure is less than 90° is called an acute angle.Right Angle: An angle whose measure is 90° is called right angle.Obtuse Angle: An angle whose measure is greater than 90° but less than 180° is called an obtuse angle. … Straight Angle: … Reflex Angle: … Zero Angle:
## How many spins is a 720?
However, consider what happens instead when the cup is rotated, not through just one 360° turn, but two 360° turns for a total rotation of 720°.
## What does 90 degrees look like?
In geometry and trigonometry, a right angle is an angle of exactly 90° (degrees), corresponding to a quarter turn. … The term is a calque of Latin angulus rectus; here rectus means “upright”, referring to the vertical perpendicular to a horizontal base line.
## How many 45 angles would it take to turn in a complete circle?
So to find how many 45 degree angles could fit in 360 lets divide. So this means that 8, 45 degree angles can fit in 360. And therefore create a full circle.
## Is a circle 360 degrees?
A full circle is 360 degrees because of the Babylonian’s usage of the sexagesimal system, number of days in a year and also because 360 is highly composite.
## Why is a triangle 180 degrees?
A triangle’s angles add up to 180 degrees because one exterior angle is equal to the sum of the other two angles in the triangle. In other words, the other two angles in the triangle (the ones that add up to form the exterior angle) must combine with the third angle to make a 180 angle.
## How many 180 angles does it take to make a full turn?
Step-by-step explanation: you do the equation 180 + 180 and get 360 or you can multiply 180 x 2 and get 360. Hope this helps you!
## How many 120 angles does it take to make a full turn?
A full turn represents 360 degrees. Therefore 360/120 = 3. So 3 turns of 120 degree angles would result in a full turn.
## How many 45 degrees make a full turn?
3. How many 45 angles does it take to make a full turn? A full turn is 360° there since 45×8=360 45 × 8 = 360 there are eight 45° in a full turn.
## Do a full 360?
360 is to turn all the way around so as to face in the same direction as you just were. 180 is to turn around so to face in the opposite direction as you were ie backwards. … Some people say 360 by accident, but that would mean you did a full circle (180+180) gets u going the exact way u were going before.
## How do you type an angle symbol?
Click the “Start” button on your desktop and type “Character Map” in the search field. Click “Character Map” in the search results to open the utility.Select “Symbol” in the Font list.Locate the angle symbol in the symbol list. … Paste the angle symbol in your document.
## How much of a circle does a 100 degree angle turn through?
The measure of a circle is 360 . If an angle makes a 100 degree turn through the circle then the angle has turned through of the circle: The angle turns through 5/18 of the circle.
## How many 30 angle does it take to make a full turn?
Answer: 12 turns. 360 degrees is a full turn, so if you multiply 30 degreed by 12, you go around in a circle.
## How many 360 angles does it take to make a full turn?
Comparing Revolutions, Degrees, and Radianswordsrevdeghalf turn1/2180°three-quarter turn3/4270°full turn1360°twelfth turn1/1230°11 more rows
## What is a 33 degree angle called?
In geometry, there are three types of angles: acute angle-an angle between 0 and 90 degrees. right angle-an 90 degree angle. obtuse angle-an angle between 90 and 180 degrees.
## How many 90 degrees angles does it take to make a full turn?
Your a…” Comment on Kaylen’s post “There are 360 in a full angle. 360/3 = 120.
## What does 180 degrees look like?
Straight Angles An angle with a measure of 180 degrees is called a straight angle. It looks just like a line. | 1,403 | 5,498 | {"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} | 4.0625 | 4 | CC-MAIN-2021-04 | latest | en | 0.894978 |
http://mathhelpforum.com/calculus/139012-chain-rules-partial-derivatives-print.html | 1,513,159,638,000,000,000 | text/html | crawl-data/CC-MAIN-2017-51/segments/1512948522343.41/warc/CC-MAIN-20171213084839-20171213104839-00222.warc.gz | 179,677,327 | 3,646 | # Chain Rules - Partial Derivatives
Printable View
• Apr 13th 2010, 03:30 PM
Belowzero78
Chain Rules - Partial Derivatives
Let $z =f(x,y) = (8+25x)y^{tan(x)})$ , where: $x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t)$, and $y = (2 - t)e^{t} - 12(t-1)$. Use the chain rule to find $\frac{dz}{dt}$ at $t = 1$.
$\frac{dz}{dt} = ?$
The Chain rule formula I used is the following: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
And, when $t = 1$, $x = \frac{\pi}{4}$
When $t = 1$, $y = e$
$\frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)}$
$\frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1}$. This is one correct?
$\frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t}$
$\frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}}$
• Apr 13th 2010, 06:36 PM
dwsmith
Quote:
Originally Posted by Belowzero78
Let $z =f(x,y) = (8+25x)y^{tan(x)})$ , where: $x = (\frac{12 - e}{2e})(t-1) + \tan^{-1}(t)$, and $y = (2 - t)e^{t} - 12(t-1)$. Use the chain rule to find $\frac{dz}{dt}$ at $t = 1$.
$\frac{dz}{dt} = ?$
The Chain rule formula I used is the following: $\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{\partial y}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$
In your first partial, you should an x wrt t not y wrt t but you did the correct derivatives in your work.
And, when $t = 1$, $x = \frac{\pi}{4}$
When $t = 1$, $y = e$
$\frac{\partial z}{\partial x} = (8 + 25x)y^{\sec^{2}(x)} + 25y^{\tan(x)}$
$\frac{\partial z}{\partial y} = (8 + 25x)\tan(x)y^{\tan(x) - 1}$. This is one correct?
$\frac{\partial y}{\partial t} = (2 - t)e^{t} + e^{t}$
$\frac{\partial x}{\partial t} = \frac{6}{e} - \frac{1}{2} + \frac{1}{1+ t^{2}}$
Looks fine besides the typo; however, I didn't check your derivatives. I am little confused on what you are asking. You want to know what $\frac{dz}{dt}$ is but you have done everything as long as your math is correct.
• Apr 13th 2010, 07:32 PM
Belowzero78
Could someone please show me what the partial is for: $\frac{\partial z}{\partial x}$?
I'm really confused about the y^tan(x) part when taking the derivative of x.
Would it be: $\frac{\partial z}{\partial x} = (8 + 25x)y^{\tan(x)}\ln(y)\sec^{2}(x) + 25y^{\tan(x)}$ ?
• Apr 13th 2010, 07:50 PM
dwsmith
For example, $y^{tan(x)}$ can be solved like this:
$y=u$ and $tanx=v$
$\big[\frac{u'*v}{u}+ln(u)*v'\big]*u^v$
Therefore, $25y^{tan(x)}=25\frac{y^{tan(x)}*ln(y)}{cos^2(x)}$
If this example doesn't help you with, then just ask.
• Apr 13th 2010, 07:52 PM
Belowzero78
Ok, after this how do i evaluate dz/dt? Like what stuff do i plug in?
• Apr 13th 2010, 07:54 PM
dwsmith
Add your solutions $\frac{\partial z}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial t}
$
of the partial derivatives you obtain. | 1,123 | 2,929 | {"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": 39, "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-2017-51 | longest | en | 0.743122 |
http://www.physicsforums.com/showthread.php?t=605432 | 1,369,017,156,000,000,000 | text/html | crawl-data/CC-MAIN-2013-20/segments/1368698207393/warc/CC-MAIN-20130516095647-00072-ip-10-60-113-184.ec2.internal.warc.gz | 651,120,862 | 7,638 | ## Collision of 2 balls in air
1. The problem statement, all variables and given/known data
Two elastic balls of mass m1 = 0.3kg and m2 = 0.2kg are hanging on a thin rope so that their centre of gravity is on the same height. ball 1 goes out of balance raising it up to h0 = 0.05m
Then letting it go hitting ball 2. Calculate the maximum height of both balls reaching after the collision.
m1 = 0.3kg
m2 = 0.2kg
h0 = 0.05m
g = 10m/s2
and since m2 is not moving it's V2 = 0m/s
h1 = ?
h2 = ?
2. Relevant equations
I'm not 100% sure,
but I do know I am supposed to use the elastic collision equation:
m1*V1+m2*V2 = m1*V1'+m2*V2' since V2 = 0m/s it's:
m1*V1 = m1*V1'+m2*V2'
And I'm guessing kinetic energy:
ΔEk = 4 * m1*m2/(m1+m2)2 * Ek1
Ek1 = 1/2*m1*V12
3. The attempt at a solution
First I made a drawing of it to understand it better.
I just don't see how am I supposed to collide those 2 equations and then the height.
And what am i supposed to do with g = 10m/s2 ??
PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
By raising the ball to a ht. of ho, you give it a potential energy change of mg(ho), which will be converted to Kinetic energy when the ball comes down mgho = 1/2m(v1^2), thus u get v1. Find final velocities. use potential energy again after that.
Quote by vanihba By raising the ball to a ht. of ho, you give it a potential energy change of mg(ho), which will be converted to Kinetic energy when the ball comes down mgho = 1/2m(v1^2), thus u get v1. Find final velocities. use potential energy again after that.
I've managed to finish the problem after your help. Thank you!
h1 = 0.0125m
h2 = 0.072m
Tags collision | 548 | 1,747 | {"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} | 3.90625 | 4 | CC-MAIN-2013-20 | latest | en | 0.909038 |
https://lotterychecker.co.uk/a-realistic-look-at-lottery-strategies-and-winning/ | 1,624,336,393,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623488507640.82/warc/CC-MAIN-20210622033023-20210622063023-00421.warc.gz | 320,987,364 | 8,468 | I was motivated to write this article when I read the article, the "expected" to win the lottery. The fact that I am a statistician, I was naturally interested in the issue, but apparently need not duly published. You can not use the expected value as a way to win the lottery. Should be used only to help you decide when to play the lottery. There are many ways you can show the expected value. For these purposes I will use a simplified version of the expected value. We intend to consider the number of people who play the lottery and the chance to win another, and because there is no way to tell how many people will not have special training. The simplest version of the expected value, you should take the prize wins a prize divided by the odds to reach the value of each prize value. 649 For example, the odds of lottery picking 3 6 beats quantity is 1 / 57 Then, by multiplying the value of the dollar rates for the prize (\$ 2. 00 in this case). That gives us value. 0351, you must repeat these steps until all jackpot prizes. When you reach the jackpot should sum all the values that invent, and to reduce their number will increase in dollar amount, use 1 ticket to play (most games cost \$ 1. 00). This will give you another value should be multiplied by the total number of combinations that the lottery (a lottery ticket 649 is 13,983,816). What will it take for the jackpot lottery just over \$ 12. 1 million reasons for me. This method is used only as a guide, the best time to play is because it helped to increase yields. However, you can use it as a means to increase yields. Only proven way to increase the odds is to buy more tickets. So how can you buy more tickets to go outside. It's really simple, do not buy tickets until the jackpot is high enough to justify buying one. If the odds are 1 / 14 million, it is usually 5 or 6 months to achieve a good jackpot. If you were to save \$ 2 a week for 6 months, you need to spend \$ 48 ticket. This will increase your returns 1 290000, and you use an extra dime.
Knowing when to play a lottery is the key to winning the lottery. Online Lottery in the strategy, we calculate the best time to play all the big ticket. | 504 | 2,186 | {"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-2021-25 | latest | en | 0.956222 |
https://www.jiskha.com/display.cgi?id=1353525402 | 1,502,897,053,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102307.32/warc/CC-MAIN-20170816144701-20170816164701-00677.warc.gz | 931,346,891 | 3,726 | # math
posted by .
rectangle room is 3.75 m by 2.84 m what is the length of the diagonal? how to calcuate?
• math -
d^2 = 3.75^2 + 2.84^2
d = 4.70
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More Similar Questions | 580 | 2,156 | {"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} | 3.890625 | 4 | CC-MAIN-2017-34 | latest | en | 0.8816 |
http://gambatria.blogspot.com/2011/09/total-opposites-slots-and-atms.html | 1,503,040,112,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886104612.83/warc/CC-MAIN-20170818063421-20170818083421-00450.warc.gz | 169,813,851 | 12,601 | ## Tuesday, September 20, 2011
### Total Opposites: Slots and ATMs
I’m writing this article from my new place in Las Vegas! I’ve got boxes everywhere and can’t find half of my stuff, but none of that will deter me from writing my weekly column.
I figure my new beginning here in Las Vegas is a perfect idea for a topic. What should you do when you’re just beginning to go to the casino?
I know one thing you shouldn’t do – play the slot machines. No game requires less knowledge than slots, so that’s why a lot of beginners wind up there. Few, if any, games give you less of a chance to win.
That’s why I think a good place to begin is with video poker. No, video poker is NOT a slot machine. It may look a little like a slot machine, but just because it has a computer screen and some buttons doesn’t make it a slot machine.
Perhaps we should consider ATM machines to be slots as well?
Video poker machines work on a totally different premise than slots. Video poker machines use random number generators to simulate dealing actual playing cards. Slot machines use random number generators to simulate nothing – they simply use it to pick which symbols will appear, but none of it is based on any actual anything.
You see 20 different symbols but that doesn’t mean they will appear with equal frequency. With video poker, you have 52 cards and each one should appear with equal probability.
It is this difference that makes all the difference. Because we know that a video poker machine simulates an actual deck of cards, we can create computer programs and math models to tell us absolutely everything about the machine.
Of course, we don’t know exactly which card will show up when, but we use probabilities to tell us the likelihood of any given card and, in turn, any given hand from showing up. It is from this we are able to develop actual strategies for which cards to hold and which cards to discard.
To start with, we know there are "only" 2,598,960 possible initial 5-card deals from a standard 52-card deck. Each one of these combinations has an equal likelihood of being dealt to the player. For each of these, we know there are 32 different ways to play the hand, ranging from discarding no cards to discarding them all.
Obviously, many of these ways would make little sense. If you are dealt three 6’s, a 10 and a 2, you’re not going to discard the 6’s. But, to be absolutely sure, our computer program takes a look at all 32 ways and determines which is the best way to play the hand.
The value it assigns to each is called the "expected value" or EV for short. The EV is calculated by looking at every possible draw that can occur given which cards were initially dealt and which ones were held.
The program sums up the payouts for each of these hands and divides by the total number of draws. We then look at the expected value for each of the 32 possible draws. Whichever has the highest EV is the proper way to play that hand.
By looking at the results of all 2.6 million hands, we are able to summarize the strategy into what is called a strategy table. This is what a player must learn in order to play each hand correctly.
As some of you are reading this, you may think this all sounds very complicated and not very beginnerish. But, it really is far less complex than it sounds. Most of the hard work is done by people like me.
Your part is to learn the strategy table and to use the information from it to play each hand. About 75% of the hands will be fairly obvious and the remaining 25% may take some memorization to get correct.
I suppose the alternative is to just keep playing slots. As I questioned earlier, if you think video poker is slots, you might as well consider a slot machine to be an ATM – even if they are complete opposites. The ATM gives you your money and the slot machines take it away! | 847 | 3,836 | {"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-2017-34 | longest | en | 0.92401 |
https://www.studysmarter.co.uk/explanations/math/pure-maths/evaluating-and-graphing-polynomials/ | 1,721,002,211,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514654.12/warc/CC-MAIN-20240714220017-20240715010017-00366.warc.gz | 890,585,307 | 79,086 | # Evaluating and Graphing Polynomials
When working with polynomial functions, you might be asked to evaluate them for a specific value of the variable. Do you know what to do in this case? Also, according to their degree, polynomial functions will have different types of graphs. Do you know how to solve polynomials by graphing and how to recognise the different types of polynomial graphs? In this article, we will answer all these questions using practical examples.
#### Create learning materials about Evaluating and Graphing Polynomials with our free learning app!
• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams
Firstly, let's recall what do we mean by polynomials.
Polynomials are expressions involving multiple terms that contain a variable raised to a series of positive whole-number exponents. Each term may also be multiplied by coefficients.
Polynomial functions are functions that follow the standard form:
$f\left(x\right)={\mathbit{a}}_{\mathbf{n}}{\mathbit{x}}^{\mathbf{n}}+{a}_{\left(n-1\right)}{x}^{\left(n-1\right)}+{a}_{\left(n-2\right)}{x}^{\left(n-2\right)}+...+{a}_{1}x+{\mathbit{a}}_{\mathbf{0}}$
As you can see, polynomial functions in standard form are written in decreasing order, from the biggest exponent to the smallest. Also from the equation above, you can notice the following characteristics of a polynomial:
• ${a}_{n}{x}^{n}$ is the leading term of the polynomial, since this is the term with the highest exponent.
• ${a}_{n}$ is the leading coefficient, because this is the coefficient of the leading term of the polynomial.
• The highest power or exponent present in a polynomial is called the degree of the polynomial. In this case, the degree is n.
• ${a}_{0}$ is a constant because this term is not accompanied by a variable.
$f\left(x\right)=3{x}^{2}+2x+5$ is a polynomial of degree 2
$f\left(x\right)=2{x}^{-1}+5$ is not a polynomial because it has a negative exponent
## Evaluating polynomials
There are a couple of methods that you can use to evaluate polynomials, which are direct substitution and synthetic substitution.
### Direct substitution
To evaluate a polynomial using direct substitution, all you need to do is substitute x with a number to find its solution.
Evaluate $f\left(x\right)=3{x}^{2}+2x+5$ when$x=4$ using direct substitution
$f\left(4\right)=3{\left(4\right)}^{2}+2\left(4\right)+5$
$f\left(4\right)=3×16+8+5$
$f\left(4\right)=48+13$
$\mathbit{f}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{61}$
### Synthetic substitution
Using synthetic substitution, the process is different. Let's evaluate the polynomial function from the previous example to explain the steps that you need to follow using this alternative method.
Evaluate $f\left(x\right)=3{x}^{2}+2x+5$ when $x=4$ using synthetic substitution
The steps to follow with synthetic substitution are:
Step 1. Write down the coefficients of each term in the polynomial function in descending order. To the left-hand side of the leading coefficient, write down the value of x that you are evaluating the polynomial function with, as shown below.
Remember to add any missing terms with a coefficient of zero (0).
Step 2. Bring down the leading coefficient below the horizontal line. Multiply the leading coefficient by the value of x. Write down the result of the multiplication just under the following coefficient. Then, add the values in the second column taking into account their signs. Write down the result of the addition directly below the horizontal line.
Bring down the leading coefficient, which is 3
Multiply the leading coefficient times the value of x: $3×4=12$, and put the result under the following coefficient
Add the values in the second column: $2+12=14$
Step 3. Multiply the result of the addition by the value of x, and put the result of the multiplication just under the following coefficient. Then, add the values taking into account their signs. Repeat this step for all the coefficients. The final value below the horizontal line will be the value of $f\left(x\right)$, in this case, $f\left(4\right)$.
Multiply 14 times the value of x: $14×4=56$, and put the result under the following coefficient
Add the values in the third column: $5+56=61$
We do not need to repeat this step, because there are no more coefficients left.
$\mathbit{f}\mathbf{\left(}\mathbf{4}\mathbf{\right)}\mathbf{=}\mathbf{61}$Notice that this is the same result that we obtained using the previous method.
## Graphing polynomials
Polynomial graphs are graphical representations of polynomial functions.
### Types of polynomial graphs
There are different types of polynomial graphs according to their degree:
Notice that the degree of a polynomial matches the number of direction changes in their graph and the number of zeros or x-intercepts.
Degree 1 - Linear Degree 2 - Quadratic Degree 3 - Cubic Degree 4 - Quartic Degree 5 - Quintic Degree 6 -Hexic
## How to solve polynomials by graphing?
As the behaviour of higher exponent functions is not as predictable as lines or parabolas, to get a more accurate representation of their curve, we need to use some key features.
### Key features of polynomial graphs
1. Find the zeros: The zeros of a function are the values of x that make the function equal to zero. They are also known as x-intercepts.
To find the zeros of a function, you need to set the function equal to zero and use whatever method required (factoring, division of polynomials, completing the square or quadratic formula) to find the solutions for x. Please refer to the Factoring Polynomials article if you need a reminder of this.
After doing polynomial division and factoring of the polynomial function ${x}^{3}+6{x}^{2}+5x-12=0$, we get the result $\left(x-1\right)\left(x+3\right)\left(x+4\right)=0$.
Based on this, the zeros or x-intercepts are:
$x=1$, $x=-3$ and $x=-4$
If a zero appears as part of the solution twice (it is repeated), then the curve of the function will touch the x-axis at that value of x, and then bounce off the x-axis changing its direction.
2. Find the y-intercept: Substitute x = 0 in the original polynomial function. The result will be the y-coordinate where the curve crosses the y-axis.
$f\left(x\right)={x}^{3}+6{x}^{2}+5x-12$
$f\left(0\right)={0}^{3}+6{\left(0\right)}^{2}+5\left(0\right)-12$
$f\left(0\right)=-12$
The point where the curve of the function crosses the y-axis is (0, -12)
3. End Behavior: The curves of polynomials that have a degree of 2 or more are continuous and smooth lines that can have maximum or minimum points where they change direction in the middle section of the curve, and on either end of the curve, they tend to go towards positive or negative infinity.
How do you determine the end behavior of a function?
Leading Coefficient Test: As mentioned before, the leading term of a polynomial is the term with the highest exponent. You will need to look at whether its exponent is even or odd and the sign of its coefficient to help you determine the end behaviour of the curve.
• Odd function (i.e. ${x}^{3},{x}^{5},{x}^{7}$)
a) Positive leading coefficient: In this case, the function will point downwards on the left and point up on the right end of the curve.
End Behavior - odd function and positive coefficient, Marilú García De Taylor - StudySmarter Originals
b) Negative leading coefficient: In this case, the function will point up on the left and point down on the right end of the curve.
End Behavior - odd function and negative coefficient, Marilú García De Taylor - StudySmarter Originals
• Even function (i.e. ${x}^{2},{x}^{4},{x}^{6}$)
a) Positive leading coefficient: In this case, the function will point upwards on both ends of the curve.
End Behavior - even function and positive coefficient, Marilú García De Taylor - StudySmarter Originals
b) Negative leading coefficient: In this case, the function will point downwards on both ends of the curve.
End Behaviour - even function and negative coefficient, Marilú García De Taylor - StudySmarter Originals
To sum up the end behavior of the polynomial graph we have the table below:
Odd function Even function Sign of the Leading Coefficient Positive Negative Positive Negative End Behavior Left Right Left Right Left Right Left Right ↓ ↑ ↑ ↓ ↑ ↑ ↓ ↓
$f\left(x\right)={x}^{3}+6{x}^{2}+5x-12$
The leading term of the polynomial function is ${x}^{3}$, which means that it is an odd function with a positive leading coefficient. Therefore the end behaviour of the curve will be like this:
Left Right ↓ ↑
4. Sketch the curve of the function.
Sketch of a polynomial graph example, Marilú García De Taylor - StudySmarter Originals
### Alternative method for graphing polynomials
Another way of graphing polynomials in which you do not need to have the polynomial in factored form is as follows:
1. Make a table of values using direct substitution of a few values of x.
2. Plot the points on the coordinate plane to determine the middle section of the graph.
3. Connect the points with a smooth and continuous curve.
4. Use the leading coefficient test to determine the end behaviour of the graph.
## Evaluating and Graphing Polynomials - Key takeaways
• The methods that you can use to evaluate polynomials are direct substitution and synthetic substitution.
• To evaluate a polynomial using direct substitution, substitute x with a number to find its solution.
• Polynomial graphs are graphical representations of polynomial functions.
• The degree of a polynomial matches the number of direction changes in their graph and the number of zeros or x-intercepts.
• Factoring a polynomial allows you to find the zeros or x-intercepts, which are the values of x where the graph intercepts the x-axis.
• You will need to look at whether the exponent of the leading term of the polynomial is even or odd and the sign of its coefficient, to help you determine the end behaviour of the curve.
#### Flashcards in Evaluating and Graphing Polynomials 10
###### Learn with 10 Evaluating and Graphing Polynomials flashcards in the free StudySmarter app
We have 14,000 flashcards about Dynamic Landscapes.
What are polynomials in Math?
Polynomials are expressions involving multiple terms that contain a variable raised to a series of positive whole-number exponents. Each term may also be multiplied by coefficients.
What is evaluating the value of a polynomial?
Evaluating the value of a polynomial consists of finding the solution of the polynomial for a specific value of x.
How do you evaluate a polynomial example?
You can evaluate a polynomial using two methods: direct substitution or synthetic substitution.
What does a polynomial graph look like?
According to their degree, polynomial functions will have different types of graphs. The curves of polynomials that have a degree of 2 or more are continuous and smooth lines that can have maximum or minimum points where they change direction in the middle section of the curve, and on either end of the curve, they tend to go towards positive or negative infinity. The degree of a polynomial matches the number of direction changes in their graph and the number of zeros or x-intercepts.
How do you solve polynomials by graphing?
To graph polynomial functions follow these steps:
1. Find the zeros using whatever method required (factoring, division of polynomials, completing the square or quadratic formula).
2. Find the y-intercept.
3. Carry out the leading coefficient test to find the end behaviour of the polynomial function.
4. Sketch the function.
Alternatively, you can make a table of values using direct substitution of a few values of x, and plot the points on the coordinate plane to determine the middle section of the graph. Then, you can connect the points with a smooth and continuous curve, and use the leading coefficient test to determine the end behaviour of the graph. In this case, you do not need to have the polynomial in factored form.
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• Checked by StudySmarter Editorial Team | 2,937 | 12,909 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 33, "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} | 4.6875 | 5 | CC-MAIN-2024-30 | latest | en | 0.909251 |
https://math.stackexchange.com/questions/3709302/calculating-matrix-exponential-for-n-times-n-jordan-block | 1,603,781,717,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107893402.83/warc/CC-MAIN-20201027052750-20201027082750-00535.warc.gz | 418,314,957 | 28,836 | # Calculating matrix exponential for $n \times n$ Jordan block [duplicate]
I want to calculate exponential of the matrix which on diagonal has some $$a \in \mathbb{R}$$ and ones above. The $$n\times n$$ matrix looks like following
$$A = \left( \begin{matrix} a & 1 & 0 & 0 & \cdots & 0 \\ 0 & a & 1 & 0 & \cdots & 0 \\ 0 & 0 & a & 1 & 0 & 0 \\ 0 & 0 & 0 & a & \ddots & 0 \\ 0 & 0 & 0 & 0 & \ddots & 1 \\ 0 & 0 & 0 & 0 & 0 & a \end{matrix} \right)$$
I tried to do it by counting determinant of matrix $$A-\lambda I$$ by the following algorithm :
1. Divide last row by $$a-\lambda$$, so the $$n$$-th row is just $$0$$ and $$1$$ in the $$n$$-th column.
2. Subtract $$(n-1)$$-th row by $$n$$-th row. Then the $$1$$ in the $$n$$-th column and $$n-1$$ row disappears.
3. Divide $$n-1$$ row by $$a-\lambda$$, so the $$(n-1)-$$ th row is just $$0$$ and $$1$$ in the $$(n-1)$$-th column.
And so on so on. By algorithm above we get matrix with only ones at diagonal, so the determinant of that matrix is just $$(a - \lambda)^n$$. So we have $$n$$-th fold eigenvalue equals to $$\lambda$$. I now I have a problem with derivation of eigenvectors of matrix $$A$$. Can you give me some advice? Is there any simplest way to calculate that?
Maybe a simpler way to calculate the matrix exponential is to write \begin{align} A=a\mathbb{I}+B \end{align} where $$\mathbb{I}$$ is the unit matrix and $$B$$ has only one's above the diagonal. Since $$\mathbb{I}$$ and $$B$$ commute, you get \begin{align} \exp(A)=\exp(a\mathbb{I})\cdot \exp(B) \end{align} Calculating the exponential of $$a\mathbb{I}$$ is straightforward and calculating the exponential of $$B$$ is also not too difficult, since $$B$$ is nilpotent, i.e $$B^k=0$$ for some appropriate $$k$$.
• Okey, so calculating $aI$ is really simple. But B is a nilpotent matrix, but using your notation the $k$ that $B^k=0$ equals to $n$. $B^2$ is a matrix $B$ "moved by one unit" to right side, $B^3$ is moved by two etc. But it's nightmare to write this in formal way $n-th$ times. There is no other way that strictly calculation ? – John Jun 7 at 13:37
• I would simply prove by induction how the $k$-th power of $B$ looks like. I don't think that this calculation is a 'nightmare'. – Jake28 Jun 7 at 14:11 | 742 | 2,248 | {"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": 35, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2020-45 | latest | en | 0.827186 |
https://www.tutordale.com/geometry-unit-5-study-guide-answers/ | 1,721,389,568,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514902.63/warc/CC-MAIN-20240719105029-20240719135029-00520.warc.gz | 904,598,254 | 52,503 | Saturday, July 13, 2024
# Geometry Unit 5 Study Guide Answers
## Geometrytrig Name Unit 5 Study Guide Date Section
Geometry Final Exam Review – Study Guide
Geometry/Trig Name: _____________ Unit 5 Study Guide page 2 Date: ______________ Section 5 -5 Definitions: Trapezoid, Base, Leg, Isosceles Trapezoid, Median of a Trapezoid Theorems: 5 -18: Base angles of an isosceles triangle are congruent. 5 -19: The median of a trapezoid 1. is parallel to the bases. 2. has the length equal to the average of the base lengths. Know How To: Determine whether a given quadrilateral is a trapezoid determine whether a trapezoid is isosceles calculate the length of the median of a trapezoid apply theorem 5 -19 by setting up and solving algebraic equations. Suggested Exercises: p. 192 C. E #1 -11 p. 192 -3 #1 -11, 14 -18 Other Suggested Exercises: p. 197 -8 Chapter Review #1 -8, 13 -16, 19 -22 p. 199 Chapter Test Additionally: There will be problems on the test that involve systems of equations and factoring.
## Study Guides & Test Prep
How to Study for Math:Always go into an assessment feeling confident that you did all you could to prepare, and that you were able to get the answers correctly in the study materials. The hardest part of an assessment is recognizing WHAT the problem is asking. While studying, take time to understand WHY you are doing the formula/equation on that particular problem .
• Review the notes before you start the study guide problems.
• Complete the study guides with the aide of your notes initially, checking your answers along the way.
• THEN, do the problems in a test environment = take away all distractions, notes and study aides. Just you, a calculator and some paper. See if you can do the work correctly in the testing environment you’ve created for yourself. Now, if you are not getting the answers correct, then you need to go BACK to step 1, rinse, and repeat.
• MIDTERM 2018
## Geometry Unit 5 Test Study Guide Review The Following
Geometry â Unit 5 Test Study Guide Review the following …
Geometry â Unit 5 Test Study Guide Review the following …
< strong> Geometry< /strong> < strong> Unit< /strong> 5 < strong> Test< /strong> < strong> Study< /strong> < strong> Guide< /strong> < strong> Review< /strong> < strong> the< /strong> < strong> following< /strong> items: ü Class Notes ü Homework Assignments ü Socrative Quiz Website m.socrative.com Room 50487 ü Quiz Problems & Solutions
Also Check: How Did Geography Influence The Greek Way Of Life | 623 | 2,501 | {"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-2024-30 | latest | en | 0.869366 |
https://support.minitab.com/en-us/minitab/20/help-and-how-to/statistics/basic-statistics/how-to/2-proportions/perform-the-analysis/select-the-analysis-options/ | 1,619,177,271,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00212.warc.gz | 642,197,421 | 4,690 | # Select the analysis options for 2 Proportions
Stat > Basic Statistics > 2 Proportions > Options
Specify the confidence level for the confidence interval, specify the hypothesized difference, define the alternative hypothesis, or specify whether to use the pooled estimate of the proportion.
## Confidence level
In Confidence level, enter the level of confidence for the confidence interval.
Usually, a confidence level of 95% works well. A 95% confidence level indicates that, if you take 100 random samples from the population, the confidence intervals for approximately 95 of the samples will contain the population difference.
For a given set of data, a lower confidence level produces a narrower confidence interval, and a higher confidence level produces a wider confidence interval. The width of the interval also tends to decrease with larger sample sizes. Therefore, you may want to use a confidence level other than 95%, depending on your sample size.
• If your sample size is small, a 95% confidence interval may be too wide to be useful. Using a lower confidence level, such as 90%, produces a narrower interval. However, the likelihood that the interval contains the population difference decreases.
• If your sample size is large, consider using a higher confidence level, such as 99%. With a large sample, a 99% confidence level may still produce a reasonably narrow interval, while also increasing the likelihood that the interval contains the population difference.
## Hypothesized difference
Enter a value in Hypothesized difference. The hypothesized difference defines your null hypothesis. Think of this value as a target value or a reference value. For example, a company tests whether the proportion of defective parts from a new supplier is different by 0.01 (1%) from the proportion from the current supplier (pnew – pcurrent = 0.01).
## Alternative hypothesis
From Alternative hypothesis, select the hypothesis that you want to test:
Difference < hypothesized difference
Use this one-sided test to determine whether the difference between the population proportions of sample 1 and sample 2 is less than the hypothesized proportion, and get an upper bound. This one-sided test has greater power than a two-sided test, but it cannot detect whether the difference is greater than the hypothesized difference.
For example, an engineer uses this one-sided test to determine whether the difference between the proportions of defective parts from two grades of material is less than 0. This one-sided test has greater power to detect whether the difference in proportions of defective parts is less than 0, but it cannot detect whether the difference is greater than 0.
Difference ≠ hypothesized difference
Use this two-sided test to determine whether the difference in population proportions differs from the hypothesized difference, and to get a two-sided confidence interval. This two-sided test can detect differences that are less than or greater than the hypothesized difference, but it has less power than a one-sided test.
For example, a bank manager tests whether the proportion of customers who have savings accounts differs at two locations. Because any difference in the proportions is important, the manager uses this two-sided test to determine whether the proportion at one location is greater than or less than the other location.
Difference > hypothesized difference
Use this one-sided test to determine whether the difference between the population proportions of sample 1 and sample 2 is greater than the hypothesized difference, and to get a lower bound. This one-sided test has greater power than a two-sided test, but it cannot detect whether the difference is less than the hypothesized difference.
For example, a logistics analyst uses this one-sided test to determine whether the difference in the proportions of on-time packages for two locations is greater than 0. This one-sided test has greater power to detect whether the difference in on-time deliveries is greater than 0, but it cannot detect whether the difference is less than 0.
For more information on selecting a one-sided or two-sided alternative hypothesis, go to About the null and alternative hypotheses.
## Test method
From Test method, select the method for estimating the proportion. When the samples are large and equal, the default method of estimating the proportions separately is preferred. If the samples are equal but small the default method is less accurate.
When you select Use the pooled estimate of the proportion, the Hypothesized difference must be equal to 0 and Minitab does not calculate a confidence interval based on the pooled estimate of the proportion. Minitab still displays a confidence interval, but it is calculated based on the default method of estimating the proportions separately.
By using this site you agree to the use of cookies for analytics and personalized content. Read our policy | 944 | 4,957 | {"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-2021-17 | latest | en | 0.845329 |
http://caml.inria.fr/pub/old_caml_site/oreilly-book/html/book-ora041.html | 1,529,860,011,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267866984.71/warc/CC-MAIN-20180624160817-20180624180817-00009.warc.gz | 51,803,260 | 7,059 | ## Exercises
### Binary Trees
We represent binary trees in the form of vectors. If a tree a has height h, then the length of the vector will be 2(h+1)-1. If a node has position i, then the left subtree of this node lies in the interval of indices [i+1 , i+1+2h], and its right subtree lies in the interval [i+1+2h+1 , 2(h+1)-1]. This representation is useful when the tree is almost completely filled. The type 'a of labels for nodes in the tree is assumed to contain a special value indicating that the node does not exist. Thus, we represent labeled trees by the by vectors of type 'a array.
1. Write a function , taking as input a binary tree of type 'a bin_tree
```# let` `fill_array` `tree` `tab` `empty` ``=`` `` `` `` `let` `rec` `aux` `i` `p` ``=`` `function` `` `` `` `` `` `` `Empty` `->` `tab`.`(i)` ``<-`` `empty` `` `` `` `` `` ``|`` `Node` `(l`,`e`,`r)` `->` `` `` `` `` `` `` `` `` `` `tab`.`(i)` ``<-`` `e` `;` `` `` `` `` `` `` `` `` `aux` `(i`+``1`)` `(p`/``2`)` `l` `;` `` `` `` `` `` `` `` `` `aux` `(i`+`p)` `(p`/``2`)` `r` `` `` `` `` `in` `aux` ``0`` `(((Array.length` `tab)`+``1`)`/``2`)` `tree` `;;`val fill_array : 'a bin_tree -> 'a array -> 'a -> unit = <fun>`
```
```# type` `'a` `bin_tree` ``=`` `` `` `` `` `` `Empty` `` `` ``|`` `Node` `of` `'a` `bin_tree` ``*`` `'a` ``*`` `'a` `bin_tree` `;;`type 'a bin_tree = | Empty | Node of 'a bin_tree * 'a * 'a bin_tree`
```
(defined on page ??) and an array (which one assumes to be large enough). The function stores the labels contained in the tree in the array, located according to the discipline described above.
2. Write a function to create a leaf (tree of height 0).
```# let` `leaf` `empty` ``=`` ``[|`` `empty` ``|]`` `` `;;`val leaf : 'a -> 'a array = <fun>`
```
3. Write a function to construct a new tree from a label and two other trees.
```# let` `node` `elt` `left` `right` ``=`` `` `` `` `let` `ll` ``=`` `Array.length` `left` `and` `lr` ``=`` `Array.length` `right` `in` `` `` `let` `l` ``=`` `max` `ll` `lr` `in` `` `` `let` `res` ``=`` `Array.create` `(`2``*`l`+``1`)` `elt` `in` `` `` `` `Array.blit` `left` ``0`` `res` ``1`` `ll` `;` `` `` `Array.blit` `right` ``0`` `res` `(ll`+``1`)` `lr` `;` `` `` `res` `` `;;`val node : 'a -> 'a array -> 'a array -> 'a array = <fun>`
```
4. Write a conversion function from the type 'a bin_tree to an array.
```# let` `rec` `make_array` `empty` ``=`` `function` `` `` `` `` `` `Empty` `->` `leaf` `empty` `` `` ``|`` `Node` `(l`,`e`,`r)` `->` `node` `e` `(make_array` `empty` `l)` `(make_array` `empty` `r)` `;;`val make_array : 'a -> 'a bin_tree -> 'a array = <fun>`
```
5. Define an infix traversal function for these trees.
```# let` `infix` `tab` `empty` `f` ``=`` `` `` `` `let` `rec` `aux` `i` `p` ``=`` `` `` `` `` `` `if` `tab`.`(i)`<>`empty` `then` `(` `aux` `(i`+``1`)` `(p`/``2`)` `;` `f` `tab`.`(i)` `;` `aux` `(i`+`p)` `(p`/``2`)` `)` `` `` `in` `aux` ``0`` `(((Array.length` `tab)`+``1`)`/``2`)` `;;`val infix : 'a array -> 'a -> ('a -> 'b) -> unit = <fun>`
```
6. Use it to display the tree.
```# let` `print_tab_int` `tab` `empty` ``=`` `` `` `` `infix` `tab` `empty` `(fun` `x` `->` `print_int` `x` `;` `print_string` ``" - "`)` `;;`val print_tab_int : int array -> int -> unit = <fun>`
```
7. What can you say about prefix traversal of these trees? Prefix traversal of the tree corresponds to left-to-right traversal of the array.
```# let` `prefix` `tab` `empty` `f` ``=`` `` `` `` `for` `i`=``0`` `to` `(Array.length` `tab)`-``1`` `do` `if` `tab`.`(i)`<>`empty` `then` `f` `tab`.`(i)` `done` `;;`val prefix : 'a array -> 'a -> ('a -> unit) -> unit = <fun>`
```
### Spelling Corrector
The exercise uses the lexical tree , from the exercise of chapter 2, page ??, to build a spelling corrector.
1. Construct a dictionary from a file in ASCII in which each line contains one word. For this, one will write a function which takes a file name as argument and returns the corresponding dictionary.
```# type` `noeud_lex` ``=`` `Lettre` `of` `char` ``*`` `bool` ``*`` `arbre_lex` `` `and` `arbre_lex` ``=`` `noeud_lex` `` `list;;`type noeud_lex = | Lettre of char * bool * arbre_lex``type arbre_lex = noeud_lex list`# type` `mot` ``=`` `string;;`type mot = string`# let` `rec` `` `existe` `m` `d` ``=`` `` `` `` `let` `` `aux` `sm` `i` `n` ``=`` `` `` `` `` `` `match` `d` `with` `` `` `` `` `` `` `[]` `->` `false` `` `` `` `` ``|`` `(Lettre` `(c`,`b`,`l))::q` `->` `` `` `` `` `` `` `` `` `` `` `if` `c` ``=`` `sm`.[`i`]`` `then` `` `` `` `` `` `` `` `` `` `` `` `` `if` `n` ``=`` ``1`` `then` `b` `` `` `` `` `` `` `` `` `` `` `` `` `else` `existe` `(String.sub` `sm` `(i`+``1`)` `(n`-``1`))` `l` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `` `else` `existe` `sm` `q` `` `` `in` `aux` `m` ``0`` `(String.length` `m);;`val existe : string -> arbre_lex -> bool = <fun>`# let` `rec` `ajoute` `m` `d` ``=`` `` `` `` `let` `aux` `sm` `i` `n` ``=`` `` `` `` `` `` `if` `n` ``=`` ``0`` `then` `d` `else` `` `` `` `` `` `match` `d` `with` `` `` `` `` `` `` `` `[]` `->` ``[`Lettre` `(sm`.[`i`],`` `n` ``=`` ``1``,`` `ajoute` `(String.sub` `sm` `(i`+``1`)` `(n`-``1`))` `[])`]`` `` `` `` `` ``|`` `(Lettre(c`,`b`,`l))::q` `->` `` `` `` `` `` `` `` `` `` `if` `c` ``=`` `sm`.[`i`]`` `then` `` `` `` `` `` `` `` `` `` `` `` `if` `n` ``=`` ``1`` `then` `(Lettre(c`,`true`,`l))::q` `` `` `` `` `` `` `` `` `` `` `` `else` `Lettre(c`,`b`,`ajoute` `(String.sub` `sm` `(i`+``1`)` `(n`-``1`))` `l)::q` `` `` `` `` `` `` `` `` `else` `(Lettre(c`,`b`,`l))::(ajoute` `sm` `q)` `` `` `in` `aux` `m` ``0`` `(String.length` `m);;`val ajoute : string -> arbre_lex -> arbre_lex = <fun>`# let` `rec` `verifie` `l` `d` ``=`` `match` `l` `with` `` `` `` `[]` `->` `[]` ``|`` `t::q` `->` `if` `existe` `t` `d` `then` `t::(verifie` `q` `d)` `` `` `` `` `` `` `` `` `` `` `` `else` `verifie` `q` `d` `;;`val verifie : string list -> arbre_lex -> string list = <fun>`# let` `string_of_char` `c` ``=`` `String.make` ``1`` `c;;`val string_of_char : char -> string = <fun>`# let` `rec` `filter` `p` `l` ``=`` `match` `l` `with` `` `` `` `[]` `->` `[]` ``|`` `t::q` `->` `if` `p` `t` `then` `t::(filter` `p` `q)` `` `` `` `` `` `` `` `` `` `` `else` `filter` `p` `q;;`val filter : ('a -> bool) -> 'a list -> 'a list = <fun>`# let` `rec` `selecte` `n` `d` ``=`` `` `` `` `match` `d` `with` `` `` `` `` `` `[]` `->` `[]` `` `` ``|`` `(Lettre(c`,`b`,`l))::q` `->` `` `` `` `` `` `` `` `` `if` `n` ``=`` ``1`` `then` `` `` `` `` `` `` `` `` `` `filter` `(function` `x` `->` `x` ``<>`` ``"!"`)` `` `` `` `` `` `` `` `` `` `` `` `` `(List.map` `` `(function` `(Lettre(c`,`b`,_`))` `->` `` `if` `b` `then` `string_of_char` `c` `else` ``"!"`)` `d)` `` `` `` `` `` `` `` `else` `` `` `` `` `` `` `` `` `` `` `let` `r` ``=`` `selecte` `(n`-``1`)` `l` `` `` `` `` `` `` `` `` `` `` `and` `r2` ``=`` `selecte` `n` `q` `` `in` `` `` `` `` `` `` `` `` `` `` `` `let` `pr` ``=`` `List.map` `(function` `s` `->` `(string_of_char` `c)`^`s)` `r` `` `` `` `` `` `` `` `` `` `` `` `` `in` `pr`@`r2;;`val selecte : int -> arbre_lex -> string list = <fun>`
```
```# let` `lire_fichier` `nom_fichier` ``=`` `` `` `` `let` `dico` ``=`` `ref` `[]` `` `` `` `and` `canal` ``=`` `open_in` `nom_fichier` `in` `` `` `try` `` `` `` `` `` `while` `true` `do` `dico` ``:=`` `ajoute` `(input_line` `canal)` ``!`dico` `done` `;` `` `` `` `` `failwith` ``"cas impossible"`` `` `` `` `with` `` `` `` `` `` `` `` `` `End_of_file` `->` `close_in` `canal` `;` ``!`dico` `` `` `` `` `` ``|`` `` `x` `->` `close_in` `canal` `;` `raise` `x` `` `;;`val lire_fichier : string -> arbre_lex = <fun>`
```
2. Write a function words that takes a character string and constructs the list of words in this string. The word separators are space, tab, apostrophe, and quotation marks.
```# let` `mots` `s` ``=`` `` `` `` `let` `est_sep` ``=`` `function` ``' '``|``'\t'``|``'\''``|``'"'`` `->` `true` `` ``|`` `` ``_`` `->` `false` `in` `` `` `let` `res` ``=`` `ref` `[]` `and` `p` ``=`` `ref` `((String.length` `s)`-``1`)` `in` `` `` `` `let` `n` ``=`` `ref` ``!`p` `in` `` `` `while` ``!`p`>=``0`` ``&&`` `est_sep` `s`.[!`p`]`` `do` `decr` `p` `done` `;` `` `` `n` ``:=`` ``!`p` `;` `` `` `while` `(`!`n`>=``0`)` `do` `` `` `` `` `while` ``!`n`>=``0`` ``&&`` `not` `(est_sep` `s`.[!`n`]`)` `do` `decr` `n` `` `done` `;` `` `` `` `` `res` ``:=`` `String.sub` `s` `(` ``!`n` ``+``1`)` `(`!`p` ``-`` ``!`n)` `::` ``!`res` `;` `` `` `` `` `while` ``!`n`>=``0`` ``&&`` `est_sep` `s`.[!`n`]`` `do` `decr` `n` `done` `;` `` `` `` `` `p` ``:=`` ``!`n` `` `` `` `done` `;` `` `` ``!`res` `;;`val mots : string -> string list = <fun>`
```
3. Write a function verify that takes a dictionary and a list of words, and returns the list of words that do not occur in the dictionary.
```# let` `rec` `verifie` `dico` ``=`` `function` `` `` `` `` `` `[]` `->` `[]` `` `` ``|`` `m::l` `->` `if` `existe` `m` `dico` `then` `verifie` `dico` `l` `else` `m::(verifie` `dico` `l)` `;;`val verifie : arbre_lex -> string list -> string list = <fun>`
```
4. Write a function occurrences that takes a list of words and returns a list of pairs associating each word with the number of its occurrences.
```# let` `rec` `ajoute` `x` ``=`` `function` `` `` `` `` `` `[]` `->` ``[`(x`,``1`)`]`` `` `` ``|`` `((y`,`n)` `as` `p)::l` `->` `if` `x`=`y` `then` `(y`,`n`+``1`)::l` `else` `p::(ajoute` `x` `l)` `;;`val ajoute : 'a -> ('a * int) list -> ('a * int) list = <fun>`# let` `rec` `ajoute_liste` `ld` ``=`` `function` `` `` `` `` `` `[]` `->` `ld` `` `` ``|`` `n::l` `->` `let` `res` ``=`` `ajoute_liste` `ld` `l` `in` `ajoute` `n` `res` `;;`val ajoute_liste : ('a * int) list -> 'a list -> ('a * int) list = <fun>`# let` `occurences` `l` ``=`` `ajoute_liste` `[]` `l` `;;`val occurences : 'a list -> ('a * int) list = <fun>`
```
5. Write a function spellcheck that takes a dictionary and the name of a file containing the text to analyze. It should return the list of incorrect words, together with their number of occurrences.
```# let` `orthographe` `dico` `nom` ``=`` `` `` `` `let` `f` ``=`` `open_in` `nom` `` `and` `res` ``=`` `ref` `[]` `in` `` `` `try` `` `` `` `` `` `` `while` `true` `do` `` `` `` `` `` `` `` `let` `s` ``=`` `input_line` `f` `in` `` `` `` `` `` `` `` `let` `ls` ``=`` `mots` `s` `in` `` `` `` `` `` `` `` `let` `lv` ``=`` `verifie` `dico` `ls` `in` `` `` `` `` `` `` `` `res` ``:=`` `ajoute_liste` ``!`res` `lv` `` `` `` `` `` `done` `;` `` `` `` `` `failwith` ``"cas impossible"`` `` `` `with` `` `` `` `` `` `` `` `End_of_file` `->` `close_in` `f` `;` ``!`res` `` `` `` `` ``|`` `x` `->` `close_in` `f` `;` `raise` `x` `` `;;`val orthographe : arbre_lex -> string -> (string * int) list = <fun>`
```
### Set of Prime Numbers
We would like now to construct the infinite set of prime numbers (without calculating it completely) using lazy data structures.
1. Define the predicate divisible which takes an integer and an initial list of prime numbers, and determines whether the number is divisible by one of the integers on the list. On sait que si un nombre x possède un diviseur supérieur à x alors il en possède inférieur à x. Nous nous contentons donc de ne tester que les éléments de la liste qui sont inférieurs à la racine carrée de l'argument.
```# let` `rec` `est_divisible` `x` ``=`` `function` `` `` `` `` `` `[]` `->` `false` `` `` ``|`` `n::l` `->` `` `(x` `mod` `n)`=``0`` ``||`` `(` `(n`*`n`<=`x)` ``&&`` `(est_divisible` `x` `l))` `;;`val est_divisible : int -> int list -> bool = <fun>`
```
2. Given an initial list of prime numbers, write the function next that returns the smallest number not on the list. On calcule le dernier élément de la liste.
```# let` `rec` `dernier` ``=`` `function` `` `` `` `` `` `[]` `->` `failwith` ``"liste vide"`` `` `` ``|`` ``[`x`]`` `->` `x` `` `` `` ``|`` ``_::`l` `->` `dernier` `l` `;;`val dernier : 'a list -> 'a = <fun>`
```
On cherche le premier nombre premier à partir d'un certain entier en les testant de deux en deux.
```# let` `rec` `plus_petit_premier` `l` `n` ``=`` `` `` `` `if` `est_divisible` `n` `l` `then` `plus_petit_premier` `l` `(n`+``2`)` `else` `n` `;;`val plus_petit_premier : int list -> int -> int = <fun>`
```
Et on assemble.
```# let` `suivant` ``=`` `function` `` `` `` `` `` `[]` `` `->` ``2`` `` `` ``|`` ``[``2``]`` `->` ``3`` `` `` ``|`` `` `l` `` `->` `let` `pg` ``=`` `dernier` `l` `in` `plus_petit_premier` `l` `(pg`+``2`)` `;;`val suivant : int list -> int = <fun>`
```
3. Define the value setprime representing the set of prime numbers, in the style of the type 'a enum on page ??. It will be useful for this set to retain the integers already found to be prime.
```# type` `'a` `ens` ``=`` `{mutable` `i`:`'a` `;` `f` ``:`` `'a` `->` `'a` `}` `;;`type 'a ens = { mutable i: 'a; f: 'a -> 'a }`# let` `next` `e` ``=`` `let` `x` ``=`` `e`.`i` `in` `e`.`i` ``<-`` `(e`.`f` `e`.`i)` `;` `x` `;;`val next : 'a ens -> 'a = <fun>`
```
```# let` `ensprem` ``=`` `` `` `` `let` `prec` ``=`` `ref` ``[``2``]`` `in` `` `` `` `let` `fonct` ``_`` ``=`` `let` `n` ``=`` `suivant` ``!`prec` `in` `prec` ``:=`` ``!`prec` ``@`` ``[`n`]`` `;` `` `n` `` `` `` `in` `{` `i` ``=`` ``2`` `;` `f` ``=`` `fonct` `}` `;;`val ensprem : int ens = {i=2; f=<fun>}`
``` | 5,221 | 13,235 | {"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-2018-26 | latest | en | 0.678076 |
https://www.coursehero.com/file/p3nbl7sq/b-Because-is-the-inverse-function-of-the-graph-of-g-is-obtained-by-plotting-the/ | 1,600,920,988,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400213454.52/warc/CC-MAIN-20200924034208-20200924064208-00222.warc.gz | 792,156,947 | 41,397 | b Because is the inverse function of the graph of g is obtained by plotting the
# B because is the inverse function of the graph of g
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b.Because is the inverse function of the graph of gisobtained by plotting the points and connecting them with a smoothcurve. The graph of is a reflection of the graph of fin the line asshown in Figure 3.18.Now try Exercise 43.Before you can confirm the result of Example 4 using a graphing utility, youneed to know how to enter You will learn how to do this using the change-of-base formuladiscussed in Section 3.3.log2 x.yx,gf x, xf x2x,g xlog2 xf x2x,g xlog2xf x2xSTUDY TIPIn Example 5, you can alsosketch the graph of by evaluating the inverse function of f,forseveral values of x. Plot thepoints, sketch the graph of g, andthen reflect the graph in the lineto obtain the graph of f.yxg x10x,f xlog10 xExample 5Sketching the Graph of a Logarithmic FunctionSketch the graph of the common logarithmic function by hand.SolutionBegin by constructing a table of values. Note that some of the values can beobtained without a calculator by using the Inverse Property of Logarithms. Othersrequire a calculator. Next, plot the points and connect them with a smooth curve,as shown in Figure 3.19.Now try Exercise 47.The nature of the graph in Figure 3.19 is typical of functions of the formThey have one x-intercept and one vertical asymptote.Notice how slowly the graph rises for x>1.f xlogax, a>1.f xlog10xx210123f x2x14121248Without calculatorWith calculator110258010.3010.6990.90312f xlog10x1101100xFigure 3.19Figure 3.18
Section 3.2Logarithmic Functions and Their Graphs199Example 6Transformations of Graphs of Logarithmic FunctionsEach of the following functions is a transformation of the graph of a.Because the graph of can be obtained byshifting the graph ofone unit to the right, as shown in Figure 3.20.b.Because the graph of can be obtained byshifting the graph oftwo units upward, as shown in Figure 3.21.Figure 3.20Figure 3.21Notice that the transformation in Figure 3.21 keeps the y-axis as a vertical asymp-tote, but the transformation in Figure 3.20 yields the new vertical asymptoteNow try Exercise 57.x1.-15-1(1, 2)(1, 0)3f(x) = log10xh(x) = 2 + log10x(2, 0)(1, 0)0.5421x= 1f(x) = log10xg(x) = log10(x1)fhh x2log10x2f x,fgg xlog10x1f x1 ,f xlog10x.Library of Parent Functions: Logarithmic FunctionThe logarithmic functionis the inverse function of the exponential function. Its domain is the set ofpositive real numbers and its range is the set of all real numbers. This is theopposite of the exponential function. Moreover, the logarithmic function hasthe y-axis as a vertical asymptote, whereas the exponential function has thex-axis as a horizontal asymptote. Many real-life phenomena with a slow rateof growth can be modeled by logarithmic functions. The basic characteris-tics of the logarithmic function are summarized below. A review of logarith-mic functions can be found in the Study Capsules.
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Answers in as fast as 15 minutes | 1,067 | 4,184 | {"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-2020-40 | latest | en | 0.841067 |
http://coertvonk.com/math/pre-calc/complex-numbers-23086 | 1,719,133,732,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198862464.38/warc/CC-MAIN-20240623064523-20240623094523-00784.warc.gz | 7,962,833 | 35,916 | # Complex numbers
Instead of projecting the future merits of complex numbers, we will introduce them in an intuitive way. We draw a parallel to negative numbers that have been universally accepted around the same time.
We start this writing with a review of concepts that should be evident. Nevertheless, I encourage you to read through them, as we build on these concepts while introducing complex numbers.
## Review
Arithmetic gives us tools to manipulate numbers. It allows us to transform one number into another using transformations such as negation, addition, subtraction, multiplication and division.
### Positive numbers
In first grade, we learned the concept of the number line and how numbers can be represented by vectors starting at $$0$$. We visualized addition by putting these vectors head to tail, where the net length and direction is the answer.
Soon thereafter, we learned how to subtract numbers by rotating the subtrahend (the value that you subtract) vector and then putting the head to tail.
We will expand on this as we discuss negative numbers and imaginary numbers. Before we introduce such numbers, let’s also refresh on the concept of equations with squares and square roots.
### Square and square root
When we solve the equation $$2x^2=8$$, we look for a transformation ($$\times x$$) that, when applied twice, turns the number $$2$$ into $$8$$.
$$2 \times x \times x = 8$$
As shown in the animation below, the two solutions $$x=2$$ and $$x=-2$$, both satisfy the equation $$2x^2=8$$.
## Negative numbers
Negative numbers have lingered around since 200 BC, but with mathematics based on geometrical ideas such as length and count, there was little place for negative numbers. After all, how can a pillar be less than nothing in height? How could you own something less than nothing?
Even a hundred years after the invention of algebra in 1637, the answer to $$4=4x+20$$ would be thought of as absurd as illustrated by the quotes:
Negative numbers darken the very whole doctrines of the equations and make dark of the things which are in their nature excessively obvious and simple. Francis Maseres, British mathematician (1757)
To really obtain an isolated negative quantity, it would be necessary to cut off an effective quantity from zero, to remove something of nothing: impossible operation. Lazare Carnot, French mathematician (1803)
### Acceptance
Some mathematicians in the 17th century discovered that negative numbers did have their use in solving cubic and quadratic equations, provided they didn’t worry about the meaning of these negative numbers. While the intermediate steps of their calculations may involve negative numbers, the solutions were typically real positive numbers.
Only in the 19th century were negative numbers truly accepted when mathematicians started to approach mathematics in terms of logical definitions.
Physical meaning has given way to algebraic use.
The English mathematician, John Wallis (1616 – 1703) is credited with giving meaning to negative numbers by inventing the number line. Even today, the number line helps students as they actively construct mathematical meaning, number sense and an understanding of number relationships. We learned to use negative numbers without a thinking about the thousands of years it took to develop the principle.
You were probably introduced to the concepts of absolute value and direction through the geometric representation on this number line. With negative numbers so embedded in our mathematics, we accept the solution to $$3-5$$ without a second thought.
In general, we learned that the negative symbol represents the “opposite” of a number. To change the sign of a number, we rotate its vector 180 degrees ($$\pi$$) around the point $$0$$. We will extend on this important concept as we introduce imaginary numbers.
## Imaginary numbers
First off, imaginary numbers are called such because for a long time they were poorly understood and regarded by some as fictitious. Even Euler, who used them extensively, once wrote:
Because all conceivable numbers are either greater than zero or less than 0 or equal to 0, then it is clear that the square roots of negative numbers cannot be included among the possible numbers [real numbers]. Consequently we must say that these are impossible numbers. And this circumstance leads us to the concept of such number, which by their nature are impossible, and ordinarily are called imaginary or fancied numbers, because they exist only in imagination. Leonhard Euler, Swiss mathematician and physicist, Introduction to Algebra pg.594
### History
Around the same time that mathematicians were struggling with the concept of negative numbers, they also came across square roots of negative numbers.
Girolamo Cardano, an Italian mathematician who published the solution to cubic and quartic equations, studied a problem that in modern algebra would be expressed as $$x^2-10x+40=0\,\land\,y=10-x$$
Find two numbers whose sum is equal to 10 and whose product is equal to 40. Translation of Ars Magna chapter 37, pg 219 (1545) by Girolamo Cardano
Cardano states that it is impossible to find the two numbers. Nonetheless, he says, we will proceed. He goes on to provide two objects that satisfy the given condition. Cardano found a sensible answer by working through his algorithm, but he called these numbers “ficticious” because not only did they disappear during the calculation, but they did not seem to have any real meaning.
This subtlety results from arithmetic of which this final point is as I have said as subtle as it is useless. Translation of Ars Magna chapter 37, pg 219 (1545) by Girolamo Cardano
### Acceptance
In 1849, Carl Friedrich Gauß, produced a rigorous proof for complex numbers what gave a big boost to the acceptance of these numbers in the mathematical community.
With this historic perspective, let’s see how these imaginary numbers fit into what we know about number theory.
We learned that the negative symbol represents the “opposite” of a number. On the number line we can represent this by rotating the vector that represents the number 180° ($$\pi$$) around the origin $$0$$.
Let’s dive straight in and consider the equation
\begin{align} x^2&=-1\\ \Rightarrow\quad 1\times x\times x&=-1\label{eq:1tomin1} \end{align}
What multiplication with $$x$$, when applied twice, turn $$1$$ into $$-1$$? Multiplying twice by a positive number gives a positive result. Same for a negative number.
Time to take a step back: we said that a negation represents a rotation of $$\pi$$ around the origin. What if we rotate the vector $$1$$ by $$\frac{\pi}{2}$$ twice, and worry about its meaning later.
Indeed, twice rotating the vector $$1$$ around the origin by $$\frac{\pi}{2}$$ gives us $$-1$$. All that we have left to do, is to find a name for the vector where $$1$$ is rotated by $$\frac{\pi}{2}$$. To credit Euler’s “imaginary or fancied numbers”, we call it $$i$$. The first multiplication turns $$1$$ into $$i$$, and the second multiplication turns $$i$$ into $$-1$$.
More over, rotating in the opposite direction works as well. The first multiplication turns $$1$$ into $$-i$$, and the second multiplication turns $$-i$$ into $$-1$$. So there are two square roots of $$-1$$: $$i$$ and $$-i$$. We have two solutions to $$x^2=-1$$!
### Interpretation of “$$i$$”
What is the meaning of this mysterious value $$i$$? The first rotation turned $$1$$ into $$i$$, so the rotation is a visualization of multiplying with $$i$$. Rotating $$i$$ once more turns $$i$$ into $$-1$$.
Substituting $$x=i$$ in $$\eqref{eq:1tomin1}$$ implies that $$i\times i=-1$$ or written as
$$\shaded{ i^2=-1 }$$
By introducing an axis perpendicular to the number line, we have extended or number space to a two dimensional plane called $$\mathbb{C}$$. This $$\mathbb{C}$$-plane includes the real numbers from the real number line, along with imaginary numbers on the $$i$$-axis and every combination thereof. We call this new number set “Complex Numbers“.
### Notation
By introducing $$i$$, we have added a dimension to the number line. With that come three new notation forms that each have their own use case. We will express the complex number $$z$$ as shown below in these forms.
#### Cartesian form
Each complex number has a real part $$x$$ and an imaginary part $$y$$, where $$x$$ and $$y$$ are real numbers. Using $$i$$ as the imaginary unit, we can denote a complex number $$z$$ as
$$\shaded{z=x+iy}$$
The point $$z$$ can be specified by its rectangular coordinates $$(x,y)$$, where $$x$$ and $$y$$ are the signed distances to the imaginary $$y$$ and real $$x$$-axis. This $$xy$$-plane is commonly called the complex plane $$\mathbb{C}$$.
This cartesian form is a logical extension of the number line and will prove useful when adding or subtracting complex numbers.
#### Polar form
The same point $$z$$ can be specified by its polar coordinates $$(r,\varphi)$$, where $$r$$ is the distance to the origin and $$\varphi$$ is the angle of the vector, in radians, with the positive $$x$$-axis. With $$r\in\mathbb{R}^+$$ and $$\varphi\in\mathbb{R}$$, we can describe point $$z$$ as
$$\require{enclose} \shaded{ r\enclose{phasorangle}{\small\varphi}=r\,(\cos\varphi+i\sin\varphi) }$$
here $$r$$ corresponds to modulus $$|z|$$, and $$\varphi$$ is called the argument. The value $$z=0$$ was excluded because the angle $$\varphi$$ is not defined that that point.
The polar form simplifies the arithmetic when used in multiplication or powers of complex numbers.
From the illustration, it is clear how to convert from cartesian to polar form
\shaded{ \begin{align} x&=r\cos\varphi\nonumber\\ y&=r\sin\varphi\nonumber \end{align}} and back \shaded{\begin{align} r&=\sqrt{x^2+y^2}\nonumber\\[6mu] \varphi&=\mathrm{atan2}(y,x)\nonumber\\[10mu] \end{align} }
Here $$\mathrm{atan2(y,x)}$$ prevents negative signs in $$\arctan\frac{y}{x}$$ from canceling each other out. Otherwise, we would not be able to distinguish $$\varphi$$ in the 1st from that in the 3rd quadrant, or $$\varphi$$ in the 2nd from that in the 4th.
\begin{align} \mathrm{atan2}(y,x) &= \begin{cases} \arctan\left(\frac{y}{x}\right) & x\gt0\nonumber\\ \arctan\left(\frac{y}{x}\right)+\pi & x\lt0 \land y\geq0\nonumber\\ \arctan\left(\frac{y}{x}\right)-\pi & x\lt0 \land y\lt0\nonumber\\ \frac{\pi}{2} & x= 0 \land y\gt0\nonumber\\ -\frac{\pi}{2} & x= 0 \land y\lt0\nonumber\\ \text{undefined} & x= 0 \land y = 0\nonumber \end{cases} \end{align}
Consider complex number $$z$$ with angle $$\varphi_0$$. If you make any integer number of rotation rotate around the origin, you will be back the your initial starting point. Since a full rotation corresponds to an angle of $$2\pi$$, the same point $$z$$ can be described as
$$\require{enclose} r\enclose{phasorangle}{\small\varphi+2k\pi}=r\,(\cos(\varphi+2k\pi)+i\sin(\varphi+2k\pi)),\quad k\in\mathbb{Z}$$
We will the effects of this further when discussing multi-valued functions such as square root.
#### Exponential form
Euler’s formula was introduced in a separate write-up as
$$\mathrm{e}^{i\varphi} = \cos\varphi+j\sin\varphi \nonumber$$
Using Euler’s formula we can rewrite the polar form of a complex number into its exponential form
$$\shaded{ z=r\,\mathrm{e}^{i\varphi} }$$
Similar to the polar form, the angle can be expressed in infinite different ways
\begin{align} z &= r\,\mathrm{e}^{i(\varphi+2k\pi)}, & k\in\mathbb{Z} \end{align}
This exponential form is often preferred over the polar form, because it reduces the need for trigonometry.
## What next?
My follow-up article Complex Functions introduces functions that operate on complex numbers. Such functions include addition, subtraction, multiplication to the most obscure trigonometry functions.
We will use the $$\mathbb{C}$$-plane extensively as we explore electronics and domain transforms. From here on
we will refer to the imaginary unit as “$$j$$”, to avoid confusion with electronics where the variable $$i$$ is already used for electrical current.
The same Leonard Euler that once said these numbers “only exist in our imagination” also used imaginary numbers to unite trigonometry and analysis in his most beautiful formula.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 3,040 | 12,320 | {"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": 2, "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": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2024-26 | latest | en | 0.935745 |
https://www.physicsforums.com/threads/physics-3-images-interference-and-difraction.744129/ | 1,521,548,439,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257647406.46/warc/CC-MAIN-20180320111412-20180320131412-00739.warc.gz | 847,050,001 | 15,845 | # Physics 3: Images, Interference and Difraction
1. Mar 19, 2014
### Kiwithepike
So three problems im stuck on.
1.) You have a converging lens with n=1.5,for a symmetric lens so the two lens have same magnitude, what should the radius of curvature be so the focal length is 10cm. I know 1/f = (n-1)[1/r1-1/r2] so that is 1/0.10m =(1.5-1)[1/r1-1/r2] where im lost is in the r1 and r2. would r2 be -r1 so we get 2r?
2)A 3cm tall object is located 10cm in front of a converging lense with a focal length of 15cm. What is the magnification of the imaged formed? |m|= h'/h and m =-i/p h is 3cm and f=15cm and p=10cm which gives a i =-30cm so would the image m be 3cm? or am i missing something?
3.) The minimum width of a slit for single diffraction to produce an interference pattern is? no min, lamda, lambda/2, or 2 lamda. since its sin=lambda/a where a is the slit width as long as a>0 theres no limit?
2. Mar 20, 2014
### Simon Bridge
1. both sides have the same radius - so there are only two possibilities.
Try both and see which makes sense.
2. check by sketching the ray diagram
3. this one will depend on your course.
strictly - there is no maximum, you are right. You also get diffraction for the limit of zero width.
diffraction can occur around an edge - so the "slit width" there is infinite.
however - your course may be talking about a specific kind of diffraction ... so check your notes. | 420 | 1,410 | {"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-2018-13 | latest | en | 0.908724 |
http://www.varsitytutors.com/high_school_math-help/finding-derivative-at-a-point | 1,484,636,939,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560279489.14/warc/CC-MAIN-20170116095119-00282-ip-10-171-10-70.ec2.internal.warc.gz | 756,508,164 | 24,840 | # High School Math : Finding Derivative at a Point
## Example Questions
### Example Question #1 : Finding Derivatives
Find if the function is given by
Explanation:
To find the derivative at , we first take the derivative of . By the derivative rule for logarithms,
Plugging in , we get
### Example Question #1 : Finding Derivative At A Point
Find the derivative of the following function at the point .
Explanation:
Use the power rule on each term of the polynomial to get the derivative,
Now we plug in
### Example Question #2 : Finding Derivatives
Let . What is ?
Explanation:
We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:
In order to find the derivative of , we will need to employ the Chain Rule.
We can factor out a 2x to make this a little nicer to look at.
Now we must evaluate the derivative when x = . | 215 | 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} | 3.8125 | 4 | CC-MAIN-2017-04 | latest | en | 0.845026 |
https://www.clutchprep.com/chemistry/practice-problems/100439/what-is-the-mass-in-grams-of-6-33-mol-of-nahco3 | 1,611,329,155,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703530835.37/warc/CC-MAIN-20210122144404-20210122174404-00088.warc.gz | 724,760,904 | 31,350 | # Problem: What is the mass, in grams, of 6.33 mol of NaHCO3?
###### FREE Expert Solution
Molar Mass:
NaHCO3: 1 Na x 23.00 g/mol = 23.00 g/mol
1 H x 1.01 g/mol = 1.01 g/mol
1 C x12.01 g/mol = 12.01 g/mol
3 O x 16.00 g/mol = 48.00 g/mol
_____________________________
Sum =
84.02 g/mol
Solving for mass of 6.33 mol NaHCO3:
92% (355 ratings)
###### Problem Details
What is the mass, in grams, of 6.33 mol of NaHCO3? | 174 | 427 | {"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-2021-04 | latest | en | 0.595716 |
https://stats.stackexchange.com/revisions/145764/1 | 1,579,671,698,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250606696.26/warc/CC-MAIN-20200122042145-20200122071145-00268.warc.gz | 684,878,798 | 18,043 | A message from our CEO about the future of Stack Overflow and Stack Exchange. Read now.
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# What is the probability that a sequence of events occurs within a given time interval?
If an event has probability p of occurring in some time interval, then the probability the event does not occur q is: $$q=1-p$$
The probability of the event not occurring by time t will be:
$$P(q_1 \cap q_2 \cap...q_t) =q_1q_2...q_t = q^t$$
So the probability it did occur by is one minus this value ($$q^t$$), which is equal to the cumulative sum of p times the probability it had not occurred up to that point ($$q^{t-1}$$): $$p_t=1-q^t=\sum\limits_{i=1}^t pq^{t-1}$$
If we are concerned with n independent events occurring with probabilities $$p_1, p_2, ... p_n$$ by time t, then: $$P(p_{1t} \cap p_{2t} \cap...p_{nt}) =p_{1t}p_{2t}...p_{nt}$$
If $$p_1=p_2= ... p_n$$ then the above will be simply $$p_t^n$$. So the probability that all n events have occurred by time t will be:
$$P(t_{Allevents}\leq t)=(1-q^t)^n=\sum\limits_{i=1}^t (pq^{t-1})^n$$
If there is only one sequence of these events that results in the outcome of interest (e.g. $$t_1, where $$t_i$$ refers to time of occurrence), the probability it is the observed sequence will be one over the total number of permutations ($$1/n!$$). So:
$$P(t_{Sequence}\leq t)=\frac{(1-q^t)^n}{n!}=\sum\limits_{i=1}^t \frac{(pq^{t-1})^n}{n!}$$
That gives us the CDF. To get the PDF we take the first derivative which is: $$P(t\geq t_{Sequence}\leq t+1)=\frac{-nq^tln(q)(1-q^t)^{n-1}}{n!}$$
Given the above assumptions, we would expect the probability that the sequence events occurs at any given time interval to follow the PDF, shown in the lower row of plots:
t=1:100; p=.025; q=1-p
par(mfrow=c(2,4))
for(n in c(1,2,4,6)){
plot(t,(cumsum((p*q^(t-1)))^n)/factorial(n), xlab="Time",
ylab="P(t.Seq <= t)",main=paste(n, "Events"))
lines(t,((1-q^(t))^n)/factorial(n))
}
for(n in c(1,2,4,6)){
plot(t,(-n*(q^t)*log(q)*(1-q^t)^(n-1))/factorial(n), log="xy", xlab="Time",
ylab="P(t<= t.Seq<= t+1)",main=paste(n, "Events"))
lines(t[-1]-.5,diff(((1-q^(t))^n)/factorial(n)), col="Red")
}
I can find no flaw with the above reasoning. So my question is what did Armitage and Dodd calculate here: I have an epidemiology question with logs ? | 774 | 2,281 | {"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": 15, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2020-05 | latest | en | 0.797299 |
https://empireessays.com/electricity-and-energy/ | 1,708,684,928,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474377.60/warc/CC-MAIN-20240223085439-20240223115439-00047.warc.gz | 244,319,435 | 18,810 | # Electricity and Energy
### Energy and Climate simple quiz
SPU 25 Energy and Climate: Vision for the Future, Spring 2018
Homework Set #3 Electricity and Energy. DUE Monday 3/12/18 (1:00 pm in lecture)
Name: _____________________________________ Section: __________ TF: _____________________ Score: ______/_______
Collaborators: _________________________________________________________________________________________________
Electricity consumption is a large fraction of our overall energy use. Electricity is generated using both fossil fuels and renewable energy sources. In this assignment, you’ll review the basics of electricity, some energy sources, and examine electricity flow and efficiency. [40 points total]
Conceptual and quantitative problems. You will only receive credit if you show all your work and write legibly.
1 .A circuit includes a 9 V battery and a copper wire of a given length and diameter. You measure the current in the wire to be 1.5 Amperes.
a. What is the resistance of the wire? Your answer should be in Ohms. [1.5 pts]
b. With this battery and wire, how much power can be provided? Report your answer in Watts. [1.5 pts]
c. What changes could you make to the wire to increase the power? [1.5 pts]
d. If you half the resistance of the wire, what is the power output? [1.5 pts]
1
e. If you double the voltage of the battery, what is the power output? [1.5 pts]
2. You have been using a 75W traditional incandescent light bulb in your desk lamp. However, in a SPU 25 section, you have learned about LED light bulbs and decided to replace your inefficient light bulb with a 12W LED that provides the same amount of light.
a. Assuming that you are using your lamp for 5 hours per day and the electricity rate is 15 cents per kilowatt-hour, what is your annual energy cost savings? [2.5 pts]
b. The incandescent bulb cost about \$0.75, and the LED bulb cost about \$6.00. Based on the energy savings, does the LED bulb save you money in the long run? For more information, visit: http://energy.gov/energysaver/lighting-choices-save-you-money. [2 pt]
3.At the large scale, electricity cannot be effectively stored with current technology. In other words, the electricity generated and transmitted over the power lines needs to match the electricity that is being used. ISO New England is an organization that carefully predicts how much energy will be used in the New England power grid during every hour of every day. They then communicate with energy generating companies (nuclear, solar, gas-burning power plants, etc.) to supply the most economic electricity to the grid at a certain time. Visit the ISO New England website: http://www.iso-ne.com.
a. What is the current actual demand (MW) and fuel mix (percentages)? [1 pt]
b. Describe the daily pattern of system demand (i.e. what time of day is energy use high or low). [1 pt]
c. At the top of the page, click on the “About Us” tab. Under “Our Three Critical Roles,” click on “Operating the Power System.” Read the section “Forecasting New England’s Electricity Use.”
What three factors largely determine the hourly demand for electricity? [2 pt]
d. At the top of the page, click on the “About Us” tab again. Under “Key Grid and Market Stats,” click on “Electricity Use.” Scroll down and read the section “Long-Term Changes in Annual and
Peak Demand.” What is the biggest determinant of electricity demand in any given year? What are three additional factors that have played a role in the decreased energy demand since 2005? [2 pts]
3
e. Read the report from ISO New England on air emissions from electricity generation: https://www.iso-ne.com/about/key-stats/air-emissions. What are the top 4 factors that have contributed to a reduction in emissions from electricity generators? [2 pts]
BI. Energy Flow
4.The Lawrence Livermore National Laboratory publishes flow charts for energy for multiple countries and states. Visit https://flowcharts.llnl.gov/commodities/energy.
a. If the going rate for electricity is \$0.15/kWh, what is the “cost” of the wasted energy from electricity? Report your answers in trillions of dollars. [2 pts]
c. What is the fate of most of this “rejected energy?” Energy cannot be created or destroyed, so where does it go? [1 pt]
5.The Lawrence Livermore National Laboratory published flow charts for both energy and carbon (https://flowcharts.llnl.gov/commodities).
2010 2014 Energy Use CO2 Emissions Energy Use CO2 Emissions (Quads) (million metric (Quads) (million metric tons) tons) Solar 0.130 0 0.427 0 Nuclear 8.40 0 8.33 0 Hydro 2.50 0 2.47 0 Wind 0.920 0 1.73 0 Geothermal 0.210 0.4 0.202 0.400 Natural Gas 25.0 1286 27.5 1440
4
Coal 21 1985 17.9 1710 Biomass 3.5 0 4.78 0 Petroleum 36 2362 34.8 2260 Total 98 5632 98.3 5410
a. Note that between 2010 and 2014, CO2 emissions decreased even though energy use increased. Look closely at which energy sources and emissions increased and which decreased. Explain the decrease in CO2 emissions based on these data and what you have learned in class. [2 pts]
IV. Energy Efficiency
6.According to energycalculator.com, an average 14-15 inch laptop runs at about 60 W.
a. Assuming that every undergraduate in Harvard College (about 6,700) uses their laptop for six hours per day, how much energy is consumed per day? Report your answer in kWh. [1 pt]
b. How much of each fuel below is needed to supply this amount of energy each day? Assume 100% efficiency. Show your work and report your answers in kilograms in the table below. [2 pts]
c. How much CO2 results from the needed amount of fuel each day? Show your work and report your answer in kilograms in the table below. [2 pts]
5
Fuel Energy Content Carbon Footprint Daily Harvard College laptop use (kWh/kg) (g CO2 eq/kWh)* Fuel (kg) CO2 Footprint (kg) Coal 8.3 900 Oil 13.0 650 Natural Gas 15.4 420 Uranium (breeder) 2.2*107 5
*http://www.geni.org/globalenergy/library/technical-articles/carbon-capture/parliamentary-office-of-science-and-technology/carbon-footprint-of-electricity-generation/file_9270.pdf
7. What did you find most surprising or interesting from this homework assignment? Why? [3 points]
8.BONUS: Can you identify immigrants who had a major influence on US technology? If you have a candidate or candidates, give a brief summary paragraph on why you think they belong in this technology Hall of Fame. [4 points] We already mentioned:
a. Alexander Graham Bell (1847-1922)
b. Nikola Tesla (1856-1943)
c. Elon Musk (1971- )
6 | 1,624 | 6,513 | {"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-2024-10 | latest | en | 0.876183 |
https://www.shabupc.com/what-is-a-multiple-sided-shape-called/ | 1,679,673,937,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296945287.43/warc/CC-MAIN-20230324144746-20230324174746-00160.warc.gz | 1,105,744,646 | 10,192 | Shabupc.com
Discover the world with our lifehacks
# What is a multiple sided shape called?
## What is a multiple sided shape called?
The ‘poly-‘ prefix simply means ‘multiple’, so a polygon is a shape with multiple sides, in the same way that ‘polygamy’ means multiple spouses. There are names for many different types of polygons, and usually the number of sides is more important than the name of the shape.
### What are all the 5 sided shapes?
A pentagon is a shape with 5 sides and 5 angles.
What are all the shapes with 3 sides?
A 3-sided shape is called a triangle. Triangles are polygons with three sides, so any polygon with three sides is called a triangle.
What is 13 sided shape called?
Tridecagon
A 13-sided polygon, sometimes also called the triskaidecagon.
## What shape has 3 sides with different lengths?
triangle
### What shape has 3 sides besides a triangle?
2D Shapes
Triangle – 3 Sides Square – 4 Sides
Pentagon – 5 Sides Hexagon – 6 sides
Heptagon – 7 Sides Octagon – 8 Sides
Nonagon – 9 Sides Decagon – 10 Sides
More …
What shape has 4 sides 4 angles and all sides different lengths?
Quadrilaterals are polygons with four sides (hence the beginning “quad”, which means “four”). A polygon with non-equal sides is called irregular, so the figure that you are describing is an irregular quadrilateral. This figure has side lengths of 1, 2, 3, and 4 respectively, so it is an irregular quadrilateral.
What has 4 sides but is not a square?
The Rhombus A rhombus is a four-sided shape where all sides have equal length (marked “s”). Also opposite sides are parallel and opposite angles are equal. Another interesting thing is that the diagonals (dashed lines) meet in the middle at a right angle.
## What is a 8 sided polygon?
octagon
In simple words, the octagon is an 8-sided polygon, also called 8-gon, in a two-dimensional plane. A regular octagon will have all its sides equal in length. Each interior angle of a regular octagon is equal to 135°. Therefore, the measure of exterior angle becomes 180° – 135° = 45°.
### What is a 9 sided polygon?
All of these polygons have different properties based on the number of sides and angles they have. A nonagon is a nine sided polygon and it is also called an Enneagon or 9-gon. In other words, a nonagon is a nine sided closed two dimensional figure. | 587 | 2,339 | {"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-2023-14 | longest | en | 0.938341 |
https://engineering.stackexchange.com/questions/12315/magnetic-decoupling-torque | 1,718,885,423,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861940.83/warc/CC-MAIN-20240620105805-20240620135805-00371.warc.gz | 199,985,076 | 39,339 | # Magnetic decoupling torque
I have a system consisting of 4 magnets rotating at speeds between 200 and 6,000rpm, which are facing 4 other magnets with a 0.5mm gap between the two sets. The set-up looks like this:
Both sets of magnets have their noth pole facing out so that they repel each other (the 0.5mm gap is maintained by other means). The resulting effect is that when the bottom set starts spinning, the top set follows because the top magnets naturally fall (radially) between the gaps of the bottom ones.
What I'm trying to do is work out what the decoupling torque provided by this set up is. In other words, if the bottom part is spinning, how much torque do I need to provide to stop the top part from rotating?
The diameter between the centres of the magnets is 17mm, they are 90 degrees apart, and I have the following (very basic) information on the magnets themselves:
• Diameter: 6mm
• Performance: 5,900 Gauss
• Vertical pull: 1.4 kg
• Slide resistance: 0.28kg
Can anybody point me in the right direction? I have made the following attempt, but I don't think that's right:
T = 4 * (8.5e-3 * 0.28 * 9.81) = 0.093 Nm
• My hunch would be to go look at motor slip, especially in the case of a stepper motor. I would imagine whatever equations are used there would be very similar to the scenario you describe, as stepper motors work by a very similar principle, but they use electromagnets instead of permanent magnets (so they can alter which sets are "active" and thus drive/step the output shaft). Commented Nov 15, 2016 at 20:30
• Are you sure it is kg mass, not kg force? As "pull" suggest it is a force between two objects. Now, the magnate is capable of resisting a torque produced by rotation and sliding, T (Nm)= Slide Resistance*Diameter*2 = 9.54*P(kW)/Speed(RPM).
– r13
Commented Dec 8, 2021 at 1:26 | 480 | 1,834 | {"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} | 3.671875 | 4 | CC-MAIN-2024-26 | latest | en | 0.951605 |
https://matheducators.stackexchange.com/questions/995/unusual-applications-of-integration/1015#1015 | 1,638,252,676,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358953.29/warc/CC-MAIN-20211130050047-20211130080047-00373.warc.gz | 430,732,295 | 46,101 | # Unusual applications of integration
I am trying to teach my calculus students to apply integration by thinking about what they are integrating rather than just applying formulas. Calculus books are full of formulas like "to find the volume of a solid of revolution obtained by rotating region A around axis B, write down integral C". I don't want my students to just memorize these formulas, but to understand where they come from and to be able to apply integration in contexts where these formulas don't necessarily apply. What are some good examples of applications of integration that don't fit into the textbook formulas?
Here's a simple example: find the area of a circle by dividing it into concentric rings and integrating the area $2\pi x\, dx$ of each ring from 0 to $r$. Of course you can do the same finding the volume of a sphere.
• Are you just looking for in terms of geometry or any interesting applications of integrals? Mar 29 '14 at 4:40
• @JyrkiLahtonen I think, your comment (with more details) might be better off as an answer. Mar 29 '14 at 7:36
• Give the students an $I$-$t$-graph of e.g. a capacitor loading process, the formula $I=\dot Q$ and ask them for the quantum of electrical charge in the capacitor. Mar 29 '14 at 11:25
• I remember there was a question about how many people at a concert given the people allowed in at any given time on the AP test a while ago. Mar 29 '14 at 18:49
• There's the classic question of, cut at two randomly selected points on a unit segment; what is the probability that the resulting three pieces can be assembled to form a triangle? As I recall, among the many ways to do this is one for which an integral is carefully set up (possibly corresponding to the standard graphical approach of representing the two breaks $a$ and $b$ as the point $(a,b)$ in the unit square, and then coloring in the part of the square that admits the assembly of a triangle). Jun 26 '17 at 23:39
An example that illustrates a slightly different integral is the following.
An observer at point $O$ is exposed to traffic noise from a long straight highway (the line $AP$). The task is to show that the total noise level at $O$ is inversely proportional to the distance $r=|OA|$ from the highway.
Use the following approximations/simplifications.
• The traffic density is constant along the road (a kind of an average).
• The level of "atomic noise intensity" at $O$ follows the inverse square law (this is probably a good approximation you can easily justify).
• The total noise intensity is the sum of the infinitesimal/atomic noise intensities (i.e. the quantity measured in Watts per square meter is additive. You only switch to the logarithmic decibel scale afterwards if ever, as things are not additive on the decibel scale).
So if you place $x$-axis along the road, and $P$ is at $x$, then the noise generated along a segment of the road of length $dx$ around $P$ has (at a unit distance from $P$) intensity $K\,dx$ for some constant $K$ proportional to traffic density (= number of passing vehicles per hour). When we take the distance $|OP|=\sqrt{r^2+x^2}$ into account we get $$dI=K\,\frac{dx}{r^2+x^2}$$ for the contribution of this atomic noise to the total observed noise level at point $O$. The total noise intensity is thus $$I=\int dI=\int_{-\infty}^{\infty}K\,\frac{dx}{r^2+x^2}.$$
So you get to illustrate/discuss:
• improper integrals
• indefinite integral of a simple rational function
• discuss what goes wrong when $r\to0+$ (the approximations fail, and you cannot be at distance zero from a continuous source of the noise)
• point out that a similar (but also different) integral emerges when you calculate the field generated by an electric current along a long straight wire
I have tried this both as an in class and as a homework example. As a HW problem it did create some discussion as the physics majors protested at first (assuming that I was adding decibels together). I guess the non-physics majors may have been bored. You know your customers (and if somebody can add an example from chemistry I will be very grateful).
• This looks excellent to me, thanks! (But then I do tend to like physics examples best myself.) Mar 30 '14 at 20:38
• Have you used any real-world data to confirm your theoretical results? For example, taking noise-meter measurements along a road that is perpendicular to such a highway? Or looking up average noise levels on a map like maps.bts.dot.gov/arcgis/apps/webappviewer/… Jun 28 '17 at 6:07
• Afraid not, @Jasper. It may well be that soundwaves don't propagate the way I thought. I have seen at least one exercise in some calculus book use this model. There the problem was to find the most quiet spot between two highways the other with twice as much traffic, and an inverse distance law was assumed. Jun 28 '17 at 6:57
Chemical reaction rates and how they relate to changes in temperature.
I use this example when teaching chemical engineering students about integrals. The Boltzmann distribution describes the velocity distribution of molecules in a gas.
$$f(v) = \sqrt{\frac{2}{\pi} \left( \frac{m}{k_B \cdot T} \right)^3} \cdot v^2 \cdot \exp\left( \frac{-m v^2}{2 k_B T} \right)$$
The integral $\int_a^b f(v) dv$ is then the fraction of the gas molecules with a velocity in the range $[a, b]$. For a certain reaction to occur there must be a minimal amount of kinetic energy present (the activation energy, $E_a$), from that the lowest possible velocity of a gas molecule with this kinetic energy can be determined by
$$v_\textrm{min} = \sqrt{\frac{2 E_a}{m}}$$
The fraction of gas molecules that can participate in the reaction is then given by
$$\int_{v_\textrm{min}}^{\infty} f(v) \, dv$$
This can be used to investigate how changes in temperature alters the fraction of gas molecules that have sufficient kinetic energy to overcome the activation energy.
One application is to test the rule of thumb, which states that an increase in temperature of 10 kelvin will roughly double the reaction rate.
Below is Mathematica code for solving this
assump = Assumptions -> {k > 0, T > 0,
m > 0}; MaxwellBoltzmannSpeedDistribution =
Sqrt[2/Pi (m/(k T))^3] v^2 Exp[-m v^2/(2 k T)];
FractionOfParticlesWithEnergyAbove =
Integrate[
MaxwellBoltzmannSpeedDistribution, {v, minVelocity, Infinity},
assump];
TemperatureDependency =
FractionOfParticlesWithEnergyAbove //. {minVelocity ->
Sqrt[2 10^-19/m], k -> 1.3806488*10^(-23), m -> 1.6726 10^-27}
res = Table[{T, TemperatureDependency} /. {T -> Tval}, {Tval, 293,
343, 10}];
res // TableForm
The above code gives the following output
293 1.05132*10^-10
303 2.33909*10^-10
313 4.94258*10^-10
323 9.96656*10^-10
333 1.926*10^-9
343 3.58024*10^-9
• Amazing! As a math educator who has never seen this equation before, what would be some good values of those constants m and k_B for a real-world example? Apr 2 '14 at 15:05
• The missing values are: $m = 1.67\cdot 10^{-27} kg$ (atom mass of hydrogen) and $k_B = 1.38\cdot 10^{-23} \cdot m^2 \cdot kg \cdot s^{-2} \cdot K^{-1}$ (Boltzmanns constant). Apr 3 '14 at 6:04
Physics provides with a lot of examples, whose main disadvantage is that they might be considered too much physics by some, i.e., require too much thinking or knowledge about something other than mathematics. (Also engineers and physicists might rather consider them the usual applications of integration.) I kept the examples simple, and as a result they might still be solved without understanding by guessing what to integrate, but they can easily made more difficult to avoid this.
Lengthening of an elastic rope under its own weight
Q: An elastic rope of length $l$ has a weight per length of $ρ$ (e.g., $ρ=1\,\text{kg}/\text{m}$) – both if unstretched. If a total of mass $m$ is pulling on a piece of rope, it extends to $f(m)$ of its unburdened length (typically $f(m)=1+Em$). To what length does the rope extend under its own weight?
A: A infinitesimally short piece of rope of length $\mathrm{d}x$ located at point $x$ (measured from the bottom of the rope in the unstretched case) is burdened with a weight of $ρx$ and thus its burdened length is $f(ρx)\mathrm{d}x$. The burdened length of the entire rope thus is the integral $\int_0^l f(ρx) \mathrm{d}x$.
A difficulty lies in considering the rope unstretched for most purposes (but this can be given as a hint). To “spice up” this problem, one could attach weights to the ropes (at the end or other positions) or have the rope vary in thickness (i.e., make $ρ$ as well as $f$ depend on $l$).
Uniformly accelerated object
Q: An apple starts falling from a tree at $t=0$. Its speed $v(t)$ at time $t$ is $gt$ (with $g$ being the local gravity of Earth). How far has it fallen after time $T$?
A: The distance covered by the apple during an infinitesimally short time $\mathrm{d}t$ is $v(t)\mathrm{d}t$ and thus the total distance covered up to time $T$ is $\int_0^T v(t)\mathrm{d}t = \int_0^T at\mathrm{d}t = \tfrac{1}{2}aT^2$.
To “spice up” the problem, one could consider more complicated acceleration functions (e.g., considering air resistance), initial speeds and so on.
• I think most of my students have memorized that distance is the integral of velocity, that being a common example in the book, so I'm not sure that example will be what I want. Mar 30 '14 at 14:27
• The first one looks good, but I'm confused about what it means to talk about a rope burdened with a mass m but not with its own weight, to make sense of the function f. Mar 30 '14 at 14:32
• @MikeShulman: I am not certain whether I understand your problem. The usual way to burden a rope is by attaching a piece of weight to it. Also, using forces may be clearer, but adds another physical concept, which is why I did not do that. Finally, I rephrased, what I thought to be the phrase in question. Mar 30 '14 at 18:37
• The rephrasing helps a bit. Maybe my problem is that the very phrase "If a total of mass m is pulling on a piece of rope, it extends to f(m) of its unburdened length" is only literally true when the "piece of rope" is infinitesimal. Right? Mar 30 '14 at 20:37
• @MikeShulman: Yes. Unless the rope is not oriented vertically, but horizontally (and the force of the mass pulling is mediated by a weigth-less rope and redirected via a wheel, e.g.). Mar 30 '14 at 20:42
Let them guess the mass of a mountain by showing them some pictures of it. They then can then model the mountain in different ways:
• cone
• rotational body
• irregular pyramid
Additionally they can model the tip of the mountain in different ways
• flat
• flat with inverse semi-ball for the caldera
• flat with inverse cone for the caldera
You can also give them elevation data and let them model the mountain as a stack of layers.
Newton's law of cooling: The rate at which the cup of tea cools is proportional to it's temperature difference to the environment.
Compute the areas/volumes of curved figures/solids. I.e., the area under a parabola, the area of a circle, the volume of a cone, a cylinder, a sphere. Can even introduce/prove Cavalieri's principle.
• Doesn’t Newton’s law of cooling (or any cooling) lead to a differential equation? Mar 29 '14 at 23:36
• @Wrzlprmft, yes, but one easily solved (it is just $y'(t) = -a(y - y_\infty)$, with $y(0) = y_0$) Mar 29 '14 at 23:59
• Your examples of areas and volumes are all easily computed with the standard textbook formulas. Mar 30 '14 at 14:17
• @MikeShulman, I understood the question was on ways to show how integration is used, in a setting different from "compute $\int_a^b f(x) \, \mathrm{d} x$, where $a$, $b$, $f$ are given." To get the point of what it is used for across, it is best if the computation of the resulting integral is simple. You don't want a series of complex maneuvers to obscure that. Mar 30 '14 at 23:53
• It has nothing to do with simplicity or complexity of the resulting integral. I don't care whether the resulting integral is even doable by elementary means or not; mainly I want them to be able to set up integrals correctly in situations that don't fit neatly into formaulas that they've memorized. Mar 31 '14 at 21:00
This example is actually from a textbook I am using.
You are in a submarine and want to know how long you have gone in some time (there are no km marks to read under water!). You have a speedometer, so you can just write down the speedometer writings, and integrate them numerically with respect to time.
(You can do that in a car too, but ther there is no point in doing it).
• This sort of example is good, but for me it belongs to a different part of calculus, when we are learning the meaning of integrals and how they relate to Riemann sums. Mar 31 '14 at 21:02
Integration is everywhere there is superposition. One very nice class of examples comes from electrostatics. In particular, if we set the zero for voltage at $\infty$ then the tiny voltage $dV$ at $\vec{r}$ due to a tiny charge $dQ$ at $\vec{r}'$ is given by the Coulomb Potential $$dV = \frac{dQ}{4\pi \epsilon_o \| \vec{r} - \vec{r}' \|}$$ If we're given a bunch of $dQ$'s over some reasonably nice geometric object then we can use integration to find the net voltage for such a collection.
Example: ring of charge if we have total charge $Q$ around ring of radius $R$ in the $xy$-plane (aka $z=0$ plane) then at the origin it's not bad to calculate $V$. We have $\vec{r} = (0,0,z)$ and using $s = \sqrt{x^2+y^2}$ and $\theta$ for the usual standard angle, the ring of charge is at $s=R$ and $z=0$ and $$dV =\frac{dQ}{4\pi \epsilon_o\| \vec{r} - \vec{r}'\|} = \frac{dQ}{4\pi \epsilon_o \sqrt{R^2+z^2}}$$ since every $dQ$ is distance $\sqrt{R^2+z^2}$ from $(0,0,z)$. Thus adding up the bits of voltage due to each $dQ$ distributed over the loop which I have labeled $C$, $$V = \int dV = \int_C \frac{dQ}{4\pi \epsilon_o \sqrt{R^2+z^2}} = \frac{1}{4\pi \epsilon_o \sqrt{R^2+z^2}} \int_C dQ = \frac{Q}{4\pi \epsilon_o \sqrt{R^2+z^2}}$$ Then, by symmetry the electric field must point along the $z$-axis and we can derive its magnitude from the voltage just calculated: $E_z = -\frac{dV}{dz}$; $$E_z = -\frac{d}{dz}\frac{Q}{4\pi \epsilon_o \sqrt{R^2+z^2}} = \frac{zQ}{4\pi \epsilon_o (R^2+z^2)^{3/2}}$$ As $z \rightarrow \infty$ we obtain $E_z \sim \frac{Q}{4\pi \epsilon_o z^2}$. Intuitively this makes sense, from very far away a ring of charge with charge $Q$ is the same as a point charge $Q$ at the origin. We find the Coulomb field in the appropriate limit.
Example: plane of charge with uniform density $\sigma = \frac{dQ}{dA}$. Building off what we just did, if we divide the plane into rings of radius $s$ with thickness $ds$ then each such ring has charge $dQ = \sigma dA = 2\pi s \sigma ds$ hence the voltage at $(0,0,z)$ due to the ring of charge is (using my first example with $R=s$ and $Q$ replaced with appropriate $dQ$): $$dV = \frac{2\pi s \sigma ds}{4\pi \epsilon_o \sqrt{s^2+z^2}}$$ The voltage due to a disk of radius $s=R$ is thus, assuming $z>0$, \begin{align} V &= \int_{0}^R \frac{\pi s \sigma ds}{2\pi \epsilon_o \sqrt{s^2+z^2}} \\ &= \frac{ \sigma }{2 \epsilon_o}\int_{0}^R \frac{ s ds}{\sqrt{s^2+z^2}} \\ &= \frac{ \sigma }{2 \epsilon_o}\sqrt{s^2+z^2} \bigg{|}_0^R \\ &= \frac{ \sigma }{2 \epsilon_o} \left( \sqrt{R^2+z^2} - z \right) \end{align} Thus, again by symmetry the electric field must be of the form $\vec{E} = \langle 0,0, E_z \rangle$ at $(0,0,z)$ and we calculate: for $z >0$, $$E_z = -\frac{dV}{dz} = -\frac{ \sigma }{2 \epsilon_o}\left( \frac{z}{(R^2+z^2)^{3/2}} - 1\right)$$ As $R \rightarrow \infty$ for fixed $z>0$ the expression above simply yields $E_z \sim \frac{ \sigma }{2 \epsilon_o}$. Now, we usually derive this result using a little cylinder and Gauss' Law, but, it's nice to see it can be derived directly by potential methods. Also, we can stop before taking $R \rightarrow \infty$ and we have formulas for the potential due to a disk of charge.
Typically we teach how to find electric fields by direct superposition of the Coulomb fields due to the point charges in the source distribution. However, that is harder than what I show here in that the integration is over vectors. In contrast, the potential is a scalar function. In both calculation schemes, if we diverge from standard geometries then the actual integrals get really tough. For example, if we get away from the axis in my example then I think Bessel functions are needed. Much of the study of electrostatics is centered around how to set up these sort of integrals.
• These are okay examples, but they really fit much better into a multivariable calculus course, since you're really doing double and triple integrals. Jun 26 '17 at 6:51
• But, we cover area and volume integrals in calculus I (or maybe II depending on the curriculum). Those take $dA$ or $dV$ and relate them a single integral depending on the coordinate. Here I do the same with $dV$. Volumes by method of washers or shells really fit your question, but, only as much as you free yourself from a blackbox calculus book. The same could be said about the chain rule in some regard, you've probably noticed calculus texts with 10 different chain rules, replete with their boxes for each function. Instead, we ought to teach one chain rule and illustrate those 10 as cases.. Jun 26 '17 at 12:17
• Ha, now that I look at my comment, maybe I've made the case for using $\varphi$ instead of $V$ for voltage. Jun 26 '17 at 12:18
• Seems pretty advanced. Jun 28 '17 at 2:12
You can do something graphical where they use "box counting" to determine the integral. For instance an electrical current that has a set of triangles on top of a base current. (for every t unit increment, I goes from 1 to 2 in a line and then drops immediately to 1. [the average current is 1.5 per t.] Obviously they won't know a formula for that weird sawtooth graph.
If you want something "different" to the normal applications, do some box counting with a peak oil bell curve shape. See the 1955 or maybe 1952 paper by Hubbard for a nice discussion. (And FWIW, I'm not pushing any peak oil slant. If you want to show failure of peak oil predictions give them the US natural gas prediction and then production to date. But the nice thing about the Hubbard paper is the explication of the math. Even someone who does not know calculus can follow it super easy. And just the boxes are super well drawn...I know it sounds silly, but it makes it seem simple.) The one thing to watch out for with this sort of answer though is you need to get an estimate versus the triangle graph where it can be exact. Maybe tell the kids "to the closest TCF" (or whatever the unit is). Also knowing the range can help your grading later! Also, this is a little more sophisticated as you ask them to estimate, show a Gaussian and they think a little more about needing a formula. Than the sawtooth, where it is a quicker insight to count boxes.
Comment:
I do think having the kids learn formulas, bag of tricks, etc. is also useful. I would not replace the "toolkit" approach with an emphasis on integration fundamental. Instead make the sort of thing that you want additive and enriching. But the bag of tricks and knowing how to find the hammer for the nail and screwdriver for the screw and even to combine tools when needed is also powerful. And important for engineers and physicists.
• Once you have a graph and are looking for the area under it, then that's what I call a textbook formula. Whether you compute the integral by counting boxes or antidifferentiating is irrelevant for the question I'm asking. Jun 26 '17 at 6:50 | 5,257 | 19,613 | {"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": 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": 1, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2021-49 | latest | en | 0.935406 |
https://justaaa.com/physics/331959-struggling-almost-at-the-end-of-my-class-and-im | 1,725,809,514,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651013.49/warc/CC-MAIN-20240908150334-20240908180334-00012.warc.gz | 310,239,534 | 10,495 | Question
# Struggling, almost at the end of my class and I'm lost. 1. An ice chest at...
Struggling, almost at the end of my class and I'm lost.
1. An ice chest at a beach party contains 12 cans of soda at 4.38 °C. Each can of soda has a mass of 0.35 kg and a specific heat capacity of 3800 J/(kg C°). Someone adds a 8.54-kg watermelon at 29.8 °C to the chest. The specific heat capacity of watermelon is nearly the same as that of water. Ignore the specific heat capacity of the chest and determine the final temperature T of the soda and watermelon in degrees Celsius.
2. A piece of glass has a temperature of 86.0 °C. Liquid that has a temperature of 45.0 °C is poured over the glass, completely covering it, and the temperature at equilibrium is 53.0 °C. The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.
3. Two bars of identical mass are at 31 °C. One is made from glass and the other from another substance. The specific heat capacity of glass is 840 J/(kg · C°). When identical amounts of heat are supplied to each, the glass bar reaches a temperature of 96 °C, while the other bar reaches 289.0 °C. What is the specific heat capacity of the other substance?
1.
n = number of cans = 12
Ts = initial Temperature of soda = 4.38
Tw = Initial temperature of watermelon = 29.8
Te = final equilibrium temperature = ?
ms = mass of each can of soda = 0.35 kg
Ms = Total mass of soda = n ms = 12 x 0.35 = 4.2 kg
Mw = mass of water melon = 8.54 kg
Cw = specific heat of watermelon = 4.2 x 103
Using conservation of energy :
Heat lost water melon = Heat gained by soda
Heat is given as
Q = m c T
so Qw = Qs
Mw Cw (Tw - Te) = Ms Cs (Te - Ts)
(8.54) (3800) (29.8 - Te) = (4.2) (4186) (Te - 4.38)
Te = 20.85
#### Earn Coins
Coins can be redeemed for fabulous gifts. | 558 | 1,971 | {"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-2024-38 | latest | en | 0.9012 |
https://www.tutorialspoint.com/compute-the-hyperbolic-sine-of-the-array-elements-in-python | 1,680,320,822,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949701.0/warc/CC-MAIN-20230401032604-20230401062604-00744.warc.gz | 1,155,204,324 | 9,810 | Compute the Hyperbolic sine of the array elements in Python
To compute the Hyperbolic sine of the array elements, use the numpy.sinh() method in Python Numpy. The method is equivalent to 1/2 * (np.exp(x) - np.exp(-x)) or -1j * np.sin(1j*x). Returns the corresponding hyperbolic sine values. This is a scalar if x is a scalar. The 1st parameter, x is input array. The 2nd and 3rd parameters are optional.
The 2nd parameter is an ndarray, A location into which the result is stored. If provided, it must have a shape that the inputs broadcast to. If not provided or None, a freshly-allocated array is returned.
The 3rd parameter is the condition is broadcast over the input. At locations where the condition is True, the out array will be set to the ufunc result. Elsewhere, the out array will retain its original value.
Steps
At first, import the required library −
import numpy as np
Get the Trigonometric Hyperbolic sine of the array elements. Create an array using the array() method in Numpy −
arr = np.array((0., 30., 45., 60., 90., 180., np.pi*1j/2, np.pi*1j))
Displaying our array −
print("Array...\n",arr)
Get the datatype −
print("\nArray datatype...\n",arr.dtype)
Get the dimensions of the Array −
print("\nArray Dimensions...\n",arr.ndim)
Get the number of elements of the Array −
print("\nNumber of elements in the Array...\n",arr.size)
To find the hyperbolic sines of the array elements, use the numpy.sinh() method in Python Numpy −
print("\nResult...",np.sinh(arr))
Example
import numpy as np
# To compute the Hyperbolic sine of the array elements, use the numpy.sinh() method in Python Numpy
# The method is equivalent to 1/2 * (np.exp(x) - np.exp(-x)) or -1j * np.sin(1j*x).
print("Get the Trigonometric Hyperbolic sine of the array elements...")
# Create an array using the array() method in Numpy
arr = np.array((0., 30., 45., 60., 90., 180., np.pi*1j/2, np.pi*1j))
# Display the array
print("Array...\n", arr)
# Get the type of the array
print("\nOur Array type...\n", arr.dtype)
# Get the dimensions of the Array
print("\nOur Array Dimensions...\n",arr.ndim)
# Get the number of elements in the Array
print("\nNumber of elements...\n", arr.size)
# To find the hyperbolic sines of the array elements, use the numpy.sinh() method in Python Numpy
print("\nResult...",np.sinh(arr))
Output
Get the Trigonometric Hyperbolic sine of the array elements...
Array...
[ 0.+0.j 30.+0.j 45.+0.j 60.+0.j
90.+0.j 180.+0.j 0.+1.57079633j 0.+3.14159265j]
Our Array type...
complex128
Our Array Dimensions...
1
Number of elements...
8
Result... [ 0.00000000e+00+0.0000000e+00j 5.34323729e+12+0.0000000e+00j
1.74671355e+19+0.0000000e+00j 5.71003695e+25+0.0000000e+00j
6.10201647e+38+0.0000000e+00j 7.44692100e+77+0.0000000e+00j
0.00000000e+00+1.0000000e+00j -0.00000000e+00+1.2246468e-16j] | 887 | 2,849 | {"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} | 3.734375 | 4 | CC-MAIN-2023-14 | latest | en | 0.598208 |
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Leap year Functional Maths challenge
A hastily written resource (I didn’t want to have to wait another 4 years!) on Feb 29th 2020 (but can be used throughout the leap year). Covers Functional Skills (Measures) content descriptors relating to using dates and units of time. Number topics such as estimation and checking, multiplication, division, odd and even numbers, and sequences are also included. There is an emphasis is on using non-calculator written methods to convert between units of time.
Level
E2
E3
L1
L2
Maths
FM Contextualised underpinning
FM Simple one step problem(s)
FM Straightforward problem(s) with more than 1 step
FM L1.4 Use multiplication facts and make connections with division facts
L2.2 Carry out calculations with numbers up to one million including strategies to check answers including estimation and approximation
E2.13 Read and record time in common date formats, and read time displayed on analogue clocks in hours, half hours and quarter hours, and understand hours from a 24-hour digital clock
FM L1.20 Convert between units of length, weight, capacity, money and time, in the same system
General
Generic resources for literacy, numeracy and beyond
Spring Festivals / the Equinox
An ESOL lesson embedding numeracy, diversity and British values, centred on the Spring Equinox and three festivals which happen around / on this day - Purim, Holi and Shunbun No Hi. Learners collect unfamiliar words on a vocabulary sheet and ask and answer questions to elicit meaning / explanation. There is a powerpoint to introduce the topic, using acronyms eg 7DIAW - Seven Days In A Week to elicit key words, and lead to an explanation of what the Vernal Equinox is and when it is.
Level
E1
E2
E3
Maths
FM E3.4 Multiply two-digit whole numbers by single and double digit whole numbers
E1.7 Know the number of days in a week, months, and seasons in a year. Be able to name and sequence
E2.7 Know the number of hours in a day and weeks in a year.
ESOL
General ESOL
General
Generic resources for literacy, numeracy and beyond
Context
Faith & Religion
Winter Storms underpinning for phonics, spelling, alphabetical order
A set of Entry 1 and Entry 2 English tasks that focus on two short texts about the alphabetical list of winter storm names released by the Met Office. The first Functional English resource I have written based on the revised Functional Skills English content (2018) which will be used by awarding organisations from September 2019 [1].
Level
E1
E2
English
FE E1.1 Say the names of the letters of the alphabet
FE E1.10 Understand a short piece of text on a simple subject
FE E2.8 Understand the main points in texts
FE E2.9 Understand organisational markers in short, straightforward texts
E1.14 Write the letters of the alphabet in sequence and in both upper and lower case
E2.15 Use the first and second letters to sequence words in alphabetical order
FE E1.15 Spell correctly words designated for Entry Level 1
FE E2.16 Spell correctly words designated for Entry Level 2
General
Staff development
Context
Science, Nature & Weather
Word Wheel Template (editable)
This is a basic editable word wheel which you can print, cut out and attach together with a brass paper fastener.
Learners can concentrate on one word at a time. Word wheels can be used for building vocabulary, spelling practice and much more. Suitable for all levels of Functional Skills and ESOL.
Instructions and ideas are included within the PPT. For example:
Stretchy:
• Write positive or negative words in the table
• Look up words that are unfamiliar
Stretchier:
Level
E2
E3
L1
L2
English
FE WRITING Spelling
FE WRITING Sentence structure and paragraphs
AL Apply strategies to spell correctly
AL Recognise and understand a range of words
ESOL
ESOL Writing: word focus (spelling and handwriting)
ESOL Reading: word focus (vocabulary, word recognition, phonics)
General
Generic resources for literacy, numeracy and beyond
Literacy Starter Parts of Speech Verbs
A two sided A4 sheet that builds on the identification of verbs and the spelling rules for past tense verbs. This is followed by a simple knowledge check.
It is designed to be used as a whole group lesson starter. It is differentiated in the sense that more able learners will be expected to complete the more complex tasks.
Editor’s note
The vocabulary in the example sentences is aimed at Painting & Decorating students, but the resource is suitable for all.
Level
E3
L1
L2
English
AL Apply strategies to spell correctly
Apply grammar
ESOL
ESOL Writing: word focus (spelling and handwriting)
ESOL Writing: sentence focus (punctuation and grammar)
General
Generic resources for literacy, numeracy and beyond
Context
Painting Decorating & DIY
Ice Falling from Planes
A simplified version of a newspaper article. The reading is followed by comprehension questions, a meaning match, a word find, and a scanning for correct spelling exercise. Developed for an adult student with Asperger’s, as well as dyslexia, who needs lots of motivation and exercises easy enough for him to be able to be successful without too much help.
Adapted from a Level 1 resource What happens to Ice Falling from planes
from Dave Norgate (2010).
Level
E2
E1
English
General
Dyslexia support
Autistic spectrum support
Context
Motor vehicles & Transport
L1 Functional Maths self assessment activity
A comprehensive list of Level 1 maths topics and descriptions.
Students to match the correct description to each topic. Then RAG (red, amber, green) rate their skills for each one using colour stickers. You can then use results to work on “Top 3 trickiest” skills, etc.
Editor’s note Ideal for start of the term or year, and also for final revision.
Level
L1
Maths
Functional Maths
General
Aims & objectives, ILPs, reflections
What have I learnt today? Self-reflection tool
This is a simple resource that has proven to be a very effective motivational tool with pre entry to level 2 learners.
Level
M8
E1
E2
E3
L1
L2
Pre-entry
pre-Entry General
General
Aims & objectives, ILPs, reflections
Generic resources for literacy, numeracy and beyond
A reading comprehension and writing activity based on a Cadbury’s Easter Brand Fact Sheet. Ideal for Level 1 Functional English.
Editor’s note
Fully mapped with answer sheet and S&L idea.
Level
L2
L1
English
Writing: text focus (composition)
General literacy / English
Functional Skills English
General
Generic resources for literacy, numeracy and beyond
Context
Catering Food Nutrition
Identifying BIAS - tips & tasks L2 English
Tips and four tasks designed to help Level 2 Functional English learners think about how they can identify BIAS in texts.
Level
L2
English
Rt/L2.5
Rt/L2.4 | 1,578 | 6,758 | {"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-2020-16 | latest | en | 0.896284 |
https://kidsjoybox.com/12-divided-by-4-using-long-division/ | 1,723,336,100,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640826253.62/warc/CC-MAIN-20240810221853-20240811011853-00563.warc.gz | 267,171,154 | 21,535 | # What is 12 divided by 4 using long division?
How to calculate 12 divided by 4 using the long division method? Here is a simple method.
The method of long division is essential for various mathematical problems and so you should acquire the knowledge of how to divide a given number by the method of long division.
Here we will take the help of this given example 12 and divide it 4 through the method of long division and teach you the steps in which the sum can be completed.
What is 12 divided by 4: The Quotient is 3 and the remainder is 0
## Basic concepts of long division method
1) Here the number 12 is the dividend. A dividend is a number that is placed on the right of the
divisor and it is divided into a smaller value during the process of division.
2) Here, in this case, the dividend is 12.
3) A divisor is a number or the value that is placed on the left of the dividend and it is used in the process of long division to divide the given dividend and yield the results.
4) Number 4 is the divisor in this case and we shall learn how the process is initiated.
5) Now we also need to know what a quotient is. A quotient is a number that is the final result that is obtained after the process of long division is completed.
6) Remainder refers to the value that is left behind after you finish the entire process.
Also read: What is 1/2 as a decimal?
## Calculations to show 12 divided by 4 using long division
Step 1
The very first step of long division is to arrange the divisor and the dividend in the proper positions. If the placement is wrong then the entire answer will be different.
Step 2
Now you need to check with the first digit 1 and see if one feature in the multiplicative table of 4. In other words, you have to check how many times four can go in 1 or the number one is divisible by 4.
Step 3
Now you need to subtract zero from 1 and in the place of the quotient, it becomes 0 zero because in the previous step one did not show up in the multiplication table for 4.
Then you will see the following arrangement.
Step 4
This time you need to bring down 2 such that you now have the entire value of the dividend as 12 and check for the next step.
Step 5
The number is now 12 in the dividend column and you need to check how many times 4 can divide 12. For this, you may use the multiplication table where you will see that 4×1= 4,4×2=8,4×3=12 and so on.
Hence 4 will go thrice in 12 and Thu,s in the quotient column you will write 3. The arrangement will look something like this.
Step 6
Now again you need to subtract for the remainder and you will obtain the value of the final remainder. Here is what the final arrangement of long division for 12 by 4 will look like.
The final results for the division of 12 by 4 in the process of long division are as follows
## Extra information for you
Here is some of the other information that might prove to be helpful for you.
1) You can also use a calculator to solve sums like this where you need to type the number 12 first followed by the division sign and then 4 and then the results will be reflected as 3. It will look like this mathematical expression: 12÷4=3.
2) You can also express the value of the quotient and the remainder in the form of a mixed fraction as your answer. The bigger number will be 3 and the smaller numerator will be 0 while the denominator will remain the same as the given divisor 4.
3) Hence the obtained mixed fraction after arranging will be 3 0/4.
4) Now as the numerator is 0 the entire fraction becomes zero and thus we can simply write it as 3 as the final result which is a whole number.
The final answer to question 12 divided by 4 in the long division will be 3. | 879 | 3,710 | {"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 | 5 | CC-MAIN-2024-33 | latest | en | 0.942315 |
https://byjus.com/question-answer/figure-shows-a-relation-between-set-p-and-q-such-that-x-in-p-y-3/ | 1,702,189,824,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679101282.74/warc/CC-MAIN-20231210060949-20231210090949-00211.warc.gz | 179,733,710 | 27,834 | Question
# Figure shows a relation between set P and Q such that x ∈ P, y ∈ Q. What is the relation, domain and range?
A
R = {(x, y): y is the square of x}, Domain = {9, 4, 25, 36}, Range = {-2, 2, -3, 3, -4, 4, -5, 5}
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B
R = {(x, y): x is the square of y}, Domain = {9, 4, 25, 36}, Range = {-2, 2, -3, 3, -5, 5, -6, 6}
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C
R = {(x, y): y is the square of x}, Domain = {9, 4, 25, 36}, Range = {-2, 2, -3, 3, -5, 5, -6, 6, -4, 4}
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D
R = {(x, y): x is the square of y}, Domain = {9, 4, 25}, Range = {1, -2, 2, -3, 3, -4, 4}
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Solution
## The correct option is B R = {(x, y): x is the square of y}, Domain = {9, 4, 25, 36}, Range = {-2, 2, -3, 3, -5, 5, -6, 6} The Relation R is "x is the square of y” x = y2 R = {(x, y): x is the square of y, x ∈ P, y ∈ Q} Domain = {9, 4, 25, 36} Range = {-2, 2, -3, 3, -5, 5, -6, 6}
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Join BYJU'S Learning Program | 522 | 1,258 | {"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} | 3.96875 | 4 | CC-MAIN-2023-50 | latest | en | 0.86108 |
https://stats.stackexchange.com/questions/343456/derivation-of-standard-error-of-sample-standard-deviation | 1,721,358,299,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514866.33/warc/CC-MAIN-20240719012903-20240719042903-00161.warc.gz | 467,994,169 | 38,104 | # Derivation of standard error of sample standard deviation
Given $x \sim N(\mu ,\sigma ^{2})$, we know $\frac{(n-1)s^{2}}{\sigma ^{2}} \sim x^{2}_{n-1}$, where $s$ is sample standard deviation.
But to find sample error of $\sigma$, I took the standard deviation of both sides, treating $\sigma$ and $n-1$ as constants, and got $Var(s^{2}) = Var(\frac{x^{2}_{n-1}} {\sigma ^{2} (n-1)}）$. Since Variance of chi-squared with $n-1$ degrees of freedom is $2(n-1)$, the right hand side expression reduces to $Var(\frac {2} {\sigma ^{2}})$, which is not the correct answer of $\frac {\sigma ^{4}} {n-1}$. Where did my math break down?
Thanks!
You made an algebra error. When you move $(n-1)/\sigma^2$ to the RHS you should get: $$s^2=\frac{\sigma^2}{n-1}\chi^2_{n-1}.$$ For the next step, use the fact that $$\operatorname{Var}(cX)=c^2\operatorname{Var}(X)$$ for any constant $c$.
Note: The quantity $2\sigma^4/(n-1)$ is the variance of the sample variance, not the variance of the sample standard deviation (which is quite difficult to obtain: see here) | 331 | 1,052 | {"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} | 3.578125 | 4 | CC-MAIN-2024-30 | latest | en | 0.802874 |
https://en.motivationactions.com/exterior-angle-what-it-is-definition-and-concept/amp/ | 1,660,118,083,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882571150.88/warc/CC-MAIN-20220810070501-20220810100501-00279.warc.gz | 249,414,818 | 24,423 | # Exterior angle – What it is, definition and concept
The exterior angle of a polygon is that formed by one side of the figure and the prolongation of its continuous side. Thus, the angle is formed outside the polygon.
To understand it in another way, the exterior angle is one that shares the same vertex with an interior angle, being supplementary to it. That is, the exterior and interior angles of the same vertex add up to 180º or form a straight angle.
As we can see in the image above, the exterior angle of vertex D measures 56.3º, which corresponds to an interior angle of 123.7º.
The following equality can then be taken for granted, where x is the exterior angle and Ɵ is the interior angle of the respective vertex
## Sum of exterior angles
The sum of the exterior angles of a polygon is equal to a complete angle, that is, 360º or 2π radians. This, regardless of the number of sides of the polygon.
We must specify that this calculation is taking into account only one external angle for each vertex. On the other hand, if we consider two, the total sum of the exterior angles of the polygon would be 720º or 4π radians.
That said, in the case of a regular polygon (where all the sides and interior angles measure the same), the exterior angle of all vertices are identical to each other and could be calculated with the following equation:
In the formula presented, x is the measure of the exterior angle and n, the number of sides of the regular polygon.
## Exterior angle example
Suppose that the interior angle of a regular polygon is greater than its exterior angle by 90º. What shape is it and how large is its exterior angle?
First, we remember that the exterior and interior angle are supplementary. So if x is the exterior angle and Ɵ the interior angle:
Then, to know which polygon it is, we must remember that the sum of all exterior angles is 360º:
Therefore, we are facing a regular octagon. | 424 | 1,931 | {"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} | 4.5625 | 5 | CC-MAIN-2022-33 | latest | en | 0.920216 |
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