Please help due tonight!

Mathematics

1 answer:

Kamila [148]2 years ago

7 0

Just remember that vertical angles are congruent and have same degree. Straight lines add up to 180 degrees.

G = 115 degrees (vertical angle)
K = 131 degrees (vertical angle)
h = 180-115 = 65 degrees
m = 180 - 131 = 49 degrees

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I need help on this question

anyanavicka [17]

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There is a very quick way to do this. Scan the numbers. You want 75 to be the mean, so make it so. Then figure out how many up or down you go from 75. I'll make a list of the givens.

72 - 3 72 is three down from 75 so call it - 3

75 0 That is your desired average so it adds nothing to the score.

74 -1 It is one down from 75, so call it -1

76 +1 It is one up from 75

76 +1 it is one up from 75

75 adds 0.

-3 - 1 + 1 + 1 = - 2

That comes - 2/7 which means that you need a score just under 75. So 75 will give you the right average of 75

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Use the Divergence Theorem to evaluate S F · dS, where F(x, y, z) = z2xi + y3 3 + sin z j + (x2z + y2)k and S is the top half of

GenaCL600 [577]

Close off the hemisphere $S$ by attaching to it the disk $D$ of radius 3 centered at the origin in the plane $z=0$ . By the divergence theorem, we have

$\displaystyle\iint_{S\cup D}\vec F(x,y,z)\cdot\mathrm d\vec S=\iiint_R\mathrm{div}\vec F(x,y,z)\,\mathrm dV$

where $R$ is the interior of the joined surfaces $S\cup D$ .

Compute the divergence of $\vec F$ :

$\mathrm{div}\vec F(x,y,z)=\dfrac{\partial(xz^2)}{\partial x}+\dfrac{\partial\left(\frac{y^3}3+\sin z\right)}{\partial y}+\dfrac{\partial(x^2z+y^2)}{\partial k}=z^2+y^2+x^2$

Compute the integral of the divergence over $R$ . Easily done by converting to cylindrical or spherical coordinates. I'll do the latter:

$\begin{cases}x(\rho,\theta,\varphi)=\rho\cos\theta\sin\varphi\\y(\rho,\theta,\varphi)=\rho\sin\theta\sin\varphi\\z(\rho,\theta,\varphi)=\rho\cos\varphi\end{cases}\implies\begin{cases}x^2+y^2+z^2=\rho^2\\\mathrm dV=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi\end{cases}$

So the volume integral is

$\displaystyle\iiint_Rx^2+y^2+z^2\,\mathrm dV=\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi=\frac{486\pi}5$