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evablogger [386]
2 years ago
10

I need help this is the last day I can do this test!

Mathematics
2 answers:
sesenic [268]2 years ago
8 0

Answer:

B

Step-by-step explanation:

16:28 in its lowest term is 4:7

when 4 divides 16,it gives 4 and 4 divides 28 to give 7

tatiyna2 years ago
4 0

Answer:

B

Step-by-step explanation:

4:7

for every 4 boys, there are 7 girls

4, 8, 12, 16

7, 14, 21, 28

4 x 4 = 16

7 x 4 = 28

I hope this helps!

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Pavel [41]

Answer:

15. y=-4/3x+7/3

16. y=-1/3x+4

Step-by-step explanation:

5 0
3 years ago
What’s 44/24 As a mixed number
Zina [86]

Answer:

1 5/6

Step-by-step explanation:

7 0
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Plzz help, Find the volume of the figure
mezya [45]

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The answer is 896,000

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6 0
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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.
5 0
3 years ago
Find the diameter of a cone that has a volume of 56.52 cubic inches and a height of 6 inches. Use 3.14 for pi. a. 3 inches. b. 6
sergejj [24]
Volume of cone = (1/3)pi*r²h

56.52 = (1/3)*3.14*r²*6

56.52/(2*3.14) = r²

9 = r²

r² = 9

r = √9

r = 3

Diameter = 2*r = 2*3 = 6 inches.    b.
3 0
3 years ago
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