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3 years ago

Write an equation of the line that passes through a pair of points:

Mathematics

2 answers:

aleksklad [387]3 years ago

7 0

Answer:

y = StartFraction 7 Over 9 EndFractionx + StartFraction 17 Over 9 EndFraction

Step-by-step explanation:

Y_Kistochka [10]3 years ago

6 0

Answer:

its C

Step-by-step explanation:

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IRINA_888 [86]

Answer:

Four is the final answer (4)

Step-by-step explanation: Have a brilliant day of learning!

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3 years ago

Find the x- and y-intercepts of the graph of each equation. (set x & y equal to 0; two different

bogdanovich [222]

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3 years ago

A water pipe is 4 7/12 feet beneath the surface of the road. What is the elevation of the pipe, in feet, expressed as a decimal?

Andrej [43]

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3 years ago

Pedro fill the glass 2 4 full with orange juice write a fraction with a denominator of 6 that is greater than 2/4 what is the an

jeyben [28]

The fraction would be 3/6

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3 years ago

Read 2 more answers

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