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pashok25 [27]
3 years ago
12

Please help me !!!!!!!!

Mathematics
1 answer:
ella [17]3 years ago
5 0

Answer:

Step-by-step explanation:

a ans b

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2 years ago
#2, can someone help me with this question?
Greeley [361]

Answer:

166.5 m³

Step-by-step explanation:

Vcone = ⅓ × pi × r² × h

pi × r² × h = 3 × Vcone

Vcylinder = pi × r² × h

Vcylinder = 3 × Vcone

Vcylinder = 3 × 55.5 = 166.5 m³

8 0
3 years ago
Read 2 more answers
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
Solve the missing variable. Show your process please because I did the other three on my homework sheet but I really struggle wi
kotykmax [81]

Follow these steps:

-8.4b - 4.2 = 12.6

Add 4.2 on both sides:

-8.4b = 16.8

Divide by -8.4 on both sides to give you the final answer of:

b = -2

So you're answer is indeed correct! Hope this helps and that the steps are clear enough!

5 0
2 years ago
Read 2 more answers
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