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Zanzabum
3 years ago
12

I need the answer and the work (it’s geometry)

Mathematics
1 answer:
Kisachek [45]3 years ago
6 0

Answer:

D

Step-by-step explanation:

The starting point is B and the ray extends towards A.

So the asnwer is D.

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A sphere of radius r is cut by a plane h units above the equator, where
Anika [276]
Consider the top half of a sphere centered at the origin with radius r, which can be described by the equation

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\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx

Converting to polar coordinates will help make this computation easier. Set

\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi

Now, the volume can be computed with the integral

\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta

You should get

\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)
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3 years ago
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mixas84 [53]

Answer:

a.   s = .9*r

b. Yes, because the regular price is only being multiplied by a number without any additional operations.

Step-by-step explanation:

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3 years ago
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