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andriy [413]
3 years ago
13

I need help please and thank you

Mathematics
1 answer:
Phoenix [80]3 years ago
8 0

Answer:

28.26

Step-by-step explanation:

This is the formula for area of a circle

A = pi r^2

Substitute the given values

A = 3.14 (3)^2

A = 3.14 (9)

28.26

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67 1/2 of 32 86/2 is??
aalyn [17]

Answer:

7

Step-by-step explanation:

671/2 33 of 32 is right

7 0
3 years ago
Read 2 more answers
A local college recently reported that enrollment increased to 108% percent of last year. If enrollment last year was at 17,113,
Sever21 [200]
Answer=18,482 students

  17113        x
_______=_____
  100%      108%

Cross multiply
100x=<span>1848204
divide both sides by 100
x=18482.04
Round to the nearest whole number
x=18,482 students</span>
6 0
3 years ago
I need some help ASSAPPP please
Sedaia [141]
I cant see it sorry :(
8 0
3 years ago
Read 2 more answers
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
#19. state the possible number of imaginary numbers of f(x) = 7x3 - x2 + 10x - 4
guapka [62]
7x3=21-2=19+10=29-4=25
3 0
3 years ago
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