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2 years ago

6

In a museum there is a sculpture in the shape of a cylinder. The cylinder has a diameter of 12 feet and a height of h feet. Whic

h equation can be used to find V , the volume of the cylinder in cubic feet?

1 answer:

pashok25 [27]2 years ago

3 0

Answer:

v=π×(12÷2)^2×h=36πh ft^3

Step-by-step explanation:

v=π(d÷2)^2×h(formula)

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Evaluate the iterated integral. $$ \int\limits_0^{2\pi}\int\limits_0^y\int\limits_0^x {\color{red}9} \cos(x+y+z)\,dz\,dx\,dy $$

KengaRu [80]

$\displaystyle\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\int_{z=0}^{z=x}\cos(x+y+z)\,\mathrm dz\,\mathrm dx\,\mathrm dy=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(x+y+z)\bigg|_{z=0}^{z=x}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}\int_{x=0}^{x=y}\sin(2x+y)-\sin(x+y)\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left(\cos(2x+y)-2\cos(x+y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dx\,\mathrm dy$
$\displaystyle=\int_{y=0}^{y=2\pi}-\frac12\left((\cos3y-2\cos2y)-(\cos y-2\cos y)\right)\bigg|_{x=0}^{x=y}\,\mathrm dy$
$\displaystyle=-\frac12\int_{y=0}^{y=2\pi}(\cos3y-2\cos2y+\cos y)\,\mathrm dy$
$\displaystyle=-\frac12\left(\frac13\sin3y-\sin2y+\sin y\right)\bigg|_{y=0}^{y=2\pi}$
$=0$

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2 years ago

Quadrilateral ABCD is located at A(-2,2), B(-2,4) and D(2,2). The quadrilateral is then transformed using the rule (x+2,y-3) to

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The rule is x+2, y-3 so you apply it to each of these coordinates. Right 2, down 3.
A(-2,2) -> A'(0,-1)
B(-2,4) -> B'(0, 1)
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3 years ago

Find the Fourier series of f on the given interval. f(x) = 1, ?7 < x < 0 1 + x, 0 ? x < 7

Zolol [24]

$f(x)=\begin{cases}1&\text{for }-7$

The Fourier series expansion of $f(x)$ is given by

$\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi x}7+\sum_{n\ge1}b_n\sin\frac{n\pi x}7$

where we have

$a_0=\displaystyle\frac17\int_{-7}^7f(x)\,\mathrm dx$
$a_0=\displaystyle\frac17\left(\int_{-7}^0\mathrm dx+\int_0^7(1+x)\,\mathrm dx\right)$
$a_0=\dfrac{7+\frac{63}2}7=\dfrac{11}2$

The coefficients of the cosine series are

$a_n=\displaystyle\frac17\int_{-7}^7f(x)\cos\dfrac{n\pi x}7\,\mathrm dx$
$a_n=\displaystyle\frac17\left(\int_{-7}^0\cos\frac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\cos\frac{n\pi x}7\,\mathrm dx\right)$
$a_n=\dfrac{9\sin n\pi}{n\pi}+\dfrac{7\cos n\pi-7}{n^2\pi^2}$
$a_n=\dfrac{7(-1)^n-7}{n^2\pi^2}$

When $n$ is even, the numerator vanishes, so we consider odd $n$ , i.e. $n=2k-1$ for $k\in\mathbb N$ , leaving us with

$a_n=a_{2k-1}=\dfrac{7(-1)-7}{(2k-1)^2\pi^2}=-\dfrac{14}{(2k-1)^2\pi^2}$

Meanwhile, the coefficients of the sine series are given by

$b_n=\displaystyle\frac17\int_{-7}^7f(x)\sin\dfrac{n\pi x}7\,\mathrm dx$
$b_n=\displaystyle\frac17\left(\int_{-7}^0\sin\dfrac{n\pi x}7\,\mathrm dx+\int_0^7(1+x)\sin\dfrac{n\pi x}7\,\mathrm dx\right)$
$b_n=-\dfrac{7\cos n\pi}{n\pi}+\dfrac{7\sin n\pi}{n^2\pi^2}$
$b_n=\dfrac{7(-1)^{n+1}}{n\pi}$

So the Fourier series expansion for $f(x)$ is

$f(x)\sim\dfrac{11}4-\dfrac{14}{\pi^2}\displaystyle\sum_{n\ge1}\frac1{(2n-1)^2}\cos\frac{(2n-1)\pi x}7+\frac7\pi\sum_{n\ge1}\frac{(-1)^{n+1}}n\sin\frac{n\pi x}7$

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2 years ago

Anely and Sandra get paid per project. Anely is paid a project fee of $40 plus $11.50 per hour. Brenda is paid a project fee of

alexdok [17]

Answer:

$68 to hire and 28.25 per hour to hire both

Step-by-step explanation:

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2 years ago

What is the solution for x in the equation?

rjkz [21]

Answer:
10

Step-by-step explanation:

-3x+7x-8=34+9x-2

(Combine like terms)

4x-8=32

(Add 8 to each side)

4x=40

(Divide by 4)

X=10

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1 year ago