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3 years ago

Which equation represents this statement?

Mathematics

1 answer:

kykrilka [37]3 years ago

5 0

Answer:

3n-9=18 because the number which is times the 3 is 9

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Which<br> graph shows a dilation?

lukranit [14]

The second one from the top shows a dilatation.

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2 years ago

I need help with number 8,please.

kodGreya [7K]

Let x be equal to the number of drinks Yasmine consumed.
Jose had 2 times that drink so his number of consumed drink would be represented by 2x.
Sally had 3 fewer drink than Jose so her number of consumed drinks would be represented by 2x-3.
Altogether, the three of them consumed 72 drinks so your equation would be:
x+2x+(2x-3)=72
add like terms together:
5x-3=72
have the term with x be alone on one side of the equation, in this case by adding three to both sides:
5x=75
now divide both sides by five for the value of x and your answer is.....
x=15

7 0

3 years ago

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What is <br> -5 3/4 - 3 1/2 =

Luba_88 [7]

Answer:

-9 1/4

Step-by-step explanation:

-5 3/4 - 3 1/2

-5 3/4 - 3 2/4

-9 1/4

7 0

2 years ago

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Can I get help please someone

Kamila [148]

Answer:

Step-by-step explanation:

Im pretty sure its A

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3 years ago

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Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago