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Sav [38]
3 years ago
6

For geometry:( i’ll give a brainist!!!

Mathematics
1 answer:
kvasek [131]3 years ago
4 0
<h3>Answer: Angle Q = 133 degrees</h3>

========================================================

Work Shown:

Recall that for any triangle, the three angles always add to 180

P+Q+R = 180

(x+13) + (10x+13) + (2x-2) = 180

(x+10x+2x) + (13+13-2) = 180

13x+24 = 180

13x = 180-24

13x = 156

x = 156/13

x = 12

Now that we know x, we can find the angle measures

  • angle P = x+13 = 12+13 = 25 degrees
  • angle Q = 10x+13 = 10*12+13 = 120+13 = 133 degrees
  • angle R = 2x-2 = 2*12-2 = 24-2 = 22 degrees

As a way to check if we have the right answer or not, we see that,

P+Q+R = 25+133+22 = 180

So the answer is confirmed.

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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

From this we need to subtract the contribution of

\displaystyle\iint_D\vec F(x,y,z)\cdot\mathrm d\vec S

that is, the integral of \vec F over the disk, oriented downward. Since z=0 in D, we have

\vec F(x,y,0)=\dfrac{y^3}3\,\vec\jmath+y^2\,\vec k

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\displaystyle\int_0^{2\pi}\int_0^3-u^3\sin^2v\,\mathrm du\,\mathrm dv=-\frac{81\pi}4

and the value of the integral we want is

(integral of divergence of <em>F</em>) - (integral over <em>D</em>) = integral over <em>S</em>

==>  486π/5 - (-81π/4) = 2349π/20

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