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3 years ago

HELP please and explain

Mathematics

2 answers:

jonny [76]3 years ago

7 0

(x3)(x4) is x7 because when you have a problem like that, you have to ADD your exponents.

-(m4)3 is -m12 since you're multiplying them.

(z3)5 is z15

y4 x y5 is y9

thats all i know of. hope it helps cx

Evgesh-ka [11]3 years ago

3 0

7. z^15
8. -1m^20
9. -27^6
10. x^7
11. y^9
12. -y^7

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A spherical ball just fits inside a cylindrical can that is 10 centimeters tall, with a diameter of 10 centimeters. Which expres

fgiga [73]

Step-by-step explanation:

Volume:

$volume \: of \: a\: sphere = \frac{4}{3} \pi {r}^{3} \\ = \frac{4}{3} \pi( {5)}^{3}$

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3 years ago

Find the inverse function of y=1

AveGali [126]

Answer:

To find the inverse, interchange the variables and solve for y

f^−1(x)=−7x+7

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3 years ago

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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.

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The answer would be b. 27

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Yolanda says 10 to the 5th power is the same as 50 because 10×5=50. what was yoland's mistake?

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When it's 5th to the power you do the number by itself how many times the power is so the correct way would be

10×10×10×10×10

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3 years ago

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