Please help me, I need this

Mathematics

1 answer:

NARA [144]3 years ago

5 0

G(-1) = 1 - 4 + 1 = -2
f(a-2) = 4(a-2)+3 = 4a - 8 + 3 = 4a - 5
g(3) = 3*3 + 4*3 + 1 = 22
g(a-2) = (a-2)^2 + 4(a-2) + 1 = a^2 - 4a + 4 + 4a - 8 + 1 = a^2 - 3

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SSSSS [86.1K]

The divergence theorem applies since $\mathcal S$ is a closed surface. We have

$\displaystyle\iint_{\mathcal S}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_{\mathcal V}\nabla\times\mathbf f(x,y,z)\,\mathrm dV$

where $\mathcal V$ is the space enclosed by the surface $\mathcal S$ . The divergence of the given vector field is

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$\displaystyle3\iiint_{\mathcal V}\,\mathrm dV=3\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=1}\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{4\pi}3$

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3 years ago

a 16 inch wire is to be cut. one piece is to be bent into the shape of a square, whereas the other piece is to be bent into the

Lyrx [107]

1. Let x be a length of square side, then the square perimeter is 4x (you need to cut 4x from 16 in. to make a square with side x in.). Then 16-4x is remained length of wire and you have to make form this piece a rectangle with sides one of which is twice bigger than another. If y is length of the smaller side, then 2y is length of the bigger side and rectangle perimeter is y+2y+y+2y=6y. You have 16-4x in. of wire left, so 6y=16-4x and $y= \frac{16-4x}{6}$ .

2.
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3. Find the derivative of the function A(x):
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4. Solve the equation A'(x)=0:
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5. Since $y= \frac{16-4x}{6}$ you have

$y= \dfrac{16-4 \frac{32}{17} }{6} = \dfrac{16\cdot17-4\cdot32}{6\cdot 17} = \dfrac{24}{17}$ .

Answer: <span>the width of the rectangle is $\frac{24}{17}$ </span>

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3 years ago

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Nikitich [7]

Answer:

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