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Pavel [41]
3 years ago
15

NEED HELP NOW 100 POINTS WILL MARK BRAINLIEST ANSWER!!!

Mathematics
2 answers:
gayaneshka [121]3 years ago
6 0
11.) Narrative 
12.) B.) 
13.) Limerick 
14.) C.)
15.) Haiku 
16.)D.)
17.) Line 1 & 2
18.) Comical 
19.) D.) 
Hope that helps!! 
Have a wonderful day!!

Art [367]3 years ago
6 0
11. C
12. C
13. B
14. B
15. B
16. A
17. D
18. B
19. D
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Can I get help with finding the Fourier cosine series of F(x) = x - x^2
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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
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(Answer coming from someone who’s stressed himself)
Upset stomach.
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A purse originally cost $45.99. This week, it is on sale for 25% off.
sertanlavr [38]

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Step-by-step explanation:

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Find the volume of a right circular cone that has a height of 5.6 m and a base with a radius of 16.6 m. Round your answer to the
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Answer:

1615.97 m^3

Step-by-step explanation:

V=πr^2 h/3

V= π (16.6)^2 (5.6/3)

V= 1615.97 m^3

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3 years ago
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Which expression is equivalent to -16 - 9? O A. 16+9 O B. -16 +(-9) O C. -16 + 9 O D. 16 + (-9)​
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Answer:

B

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because -16+(-9) is essentially the same thing because you're adding the -9 to the -16

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