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3 years ago

What's the equation of the line that passes through the points (-7,0) and (-7,8)

Mathematics

2 answers:

Elis [28]3 years ago

7 0

Answer:

x = -7

Step-by-step explanation:

This equation produces a vertical line that will go through (-7,0) and (-7,8) along with all other real y values.

serious [3.7K]3 years ago

4 0

Answer: x = -7

Step-by-step explanation: use the slope formula and slope intercept form y = mx + b to find the equation

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A gym employee is paid monthly. After working for three months, he earned $4,500. To determine how much money he will make over

Bogdan [553]

I don’t know your question but he would get 1500 each month so he gets 9000 over sixth months

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3 years ago

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Which of the following are reasons used in the proof that the angle-bisector construction can be used to bisect any angle? Check

Bad White [126]

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All of the radii of a circle are congruent

CPCTC

SSS triangle congruence postulate

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3 years ago

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Helppp meeeeeeee plsssssss you got till nine

damaskus [11]

Answer:

3/20 of the cover page is text about endangered species.

Step-by-step explanation:

We know that 3/8th of the page is text, and 2/5ths of that text is about endangered species.

So all we have to do is find out how much 2/5ths of 3/8ths are, which can be done simply via fraction multiplication.

3/8 * 2/5 = ?

For multiplying fractions, we multiply the top parts together, and then the bottom parts together. So 3/ 8 * 2/5 would be equal to (3*2) / (8*5). This gives us 6 / 40 which can be simplified to 3/20.

So 3/20ths of the cover page is about endangered animals.

3 0

3 years ago

You will write a 5-paragraph essay explaining how you would solve the following equation:

sattari [20]

Answer:

$x=0$

Step-by-step explanation:

$\frac{7}{3}(2x+3)+\frac{3}{4}(\frac{x}{5}-\frac{15}{2})=\frac{11}{8}$ <-- Given

$\frac{14}{3}x+7+\frac{3}{20}x-\frac{45}{8}=\frac{11}{8}$ <-- Distributive Property

$\frac{280}{60}x+7+\frac{9}{60}x-\frac{45}{8}=\frac{11}{8}$ <-- Find LCD of x-terms

$\frac{289}{60}x+7-\frac{45}{8}=\frac{11}{8}$ <-- Combine Like Terms

$\frac{289}{60}x+7=\frac{56}{8}$ <-- Add 45/8 to both sides

$\frac{289}{60}x+7}=7$ <-- Simplify Right Side

$\frac{289}{60}x=0$ <-- Subtract 7 on both sides

$x=0$ <-- Divide both sides by 289/60

3 0

2 years ago

Solve using Fourier series.

Olin [163]

With $2L=\pi$ , the Fourier series expansion of $f(x)$ is

$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos\dfrac{n\pi x}L+\sum_{n\ge1}b_n\sin\dfrac{n\pi x}L$
$\displaystyle f(x)\sim\frac{a_0}2+\sum_{n\ge1}a_n\cos2nx+\sum_{n\ge1}b_n\sin2nx$

where the coefficients are obtained by computing

$\displaystyle a_0=\frac1L\int_0^{2L}f(x)\,\mathrm dx$
$\displaystyle a_0=\frac2\pi\int_0^\pi f(x)\,\mathrm dx$

$\displaystyle a_n=\frac1L\int_0^{2L}f(x)\cos\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle a_n=\frac2\pi\int_0^\pi f(x)\cos2nx\,\mathrm dx$

$\displaystyle b_n=\frac1L\int_0^{2L}f(x)\sin\dfrac{n\pi x}L\,\mathrm dx$
$\displaystyle b_n=\frac2\pi\int_0^\pi f(x)\sin2nx\,\mathrm dx$

You should end up with

$a_0=0$
$a_n=0$
(both due to the fact that $f(x)$ is odd)
$b_n=\dfrac1{3n}\left(2-\cos\dfrac{2n\pi}3-\cos\dfrac{4n\pi}3\right)$

Now the problem is that this expansion does not match the given one. As a matter of fact, since $f(x)$ is odd, there is no cosine series. So I'm starting to think this question is missing some initial details.

One possibility is that you're actually supposed to use the even extension of $f(x)$ , which is to say we're actually considering the function

$\varphi(x)=\begin{cases}\frac\pi3&\text{for }|x|\le\frac\pi3\\0&\text{for }\frac\pi3$

and enforcing a period of $2L=2\pi$ . Now, you should find that

$\varphi(x)\sim\dfrac2{\sqrt3}\left(\cos x-\dfrac{\cos5x}5+\dfrac{\cos7x}7-\dfrac{\cos11x}{11}+\cdots\right)$

The value of the sum can then be verified by choosing $x=0$ , which gives

$\varphi(0)=\dfrac\pi3=\dfrac2{\sqrt3}\left(1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots\right)$
$\implies\dfrac\pi{2\sqrt3}=1-\dfrac15+\dfrac17-\dfrac1{11}+\cdots$

as required.

5 0

3 years ago