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2 years ago

I need help with my geometry homework.

Mathematics

1 answer:

wlad13 [49]2 years ago

7 0

Step-by-step explanation:

5x + 80 = 180

x = 20

2x = 40

3x=60

7x=140

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The minute hand of a clock is 8 inches long and moves from 12 to 8 o’clock. How far does the tip of the minute hand move?

ryzh [129]

The minute hand is like an eight inch radius of a circle, whose circumference would equal 2 * PI * 8 = 50.265 inches. By moving to 8 o'clock, it travels 2/3 of the circumference or 50.265 * 2/3 = 33.510 inches

3 0

3 years ago

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How can Ari simplify the following expression? StartFraction 5 Over a minus 3 EndFraction minus 4 divided by 2 + StartFraction 1

Zielflug [23.3K]

Answer:

$-\frac{8}{3}$

Step-by-step explanation:

Given

$\frac{5}{-3} - \frac{4}{2} + \frac{1}{-3}$

Required

Simpify

The very first step is to take LCM of the given expression

$\frac{-10 -4 - 2}{6}$

Perform arithmetic operations o the numerator

$-\frac{16}{6}$

Divide the numerator and denominator by 2

$-\frac{16/2}{6/2}$

$-\frac{8}{3}$

The expression can't be further simplified;

Hence, $\frac{5}{-3} - \frac{4}{2} + \frac{1}{-3}$ = $-\frac{8}{3}$

8 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

$\dfrac{2\pi}3\left(r^3\arctan\dfrac{\sqrt{r^2-h^2}}h-\dfrac{h^3}2\left(\dfrac{r\sqrt{r^2-h^2}}{h^2}+\ln\dfrac{r+\sqrt{r^2-h^2}}h\right)\right)$