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kari74 [83]
3 years ago
6

Need the answer ASAP!! 20 points

Mathematics
1 answer:
aniked [119]3 years ago
8 0

Answer:

I think C. is it

Step-by-step explanation:

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5+[30-(8-1)² ÷ 11-2²
Xelga [282]

Answer:

26.5454545455

Step-by-step explanation:

6 0
2 years ago
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

z=\sqrt{r^2-x^2-y^2}

and consider a plane

z=h

with 0. Call the region between the two surfaces R. The volume of R is given by the triple integral

\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)
5 0
3 years ago
6.17 greater or less 61 87/100
Mrrafil [7]

Answer:

Less

Step-by-step explanation:

6.17 < 6.187

3 0
3 years ago
26 INCHES AS A FRACTION
ohaa [14]
Yes I think so if not am sorry
8 0
3 years ago
Read 2 more answers
Which statements about a rhombus are always true? Its diagonals bisect each other. All 4 angles are 90°. Its diagonals are congr
GREYUIT [131]
Hello,


The answer is Its diagonals are congruent. Its diagonals are ⊥ to each other.


I believe you can pick more than 1.

Hope this helps
7 0
3 years ago
Read 2 more answers
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