Line through (-4,2) with slope of 3/4 in point-slope form

20 POINTS!!! I GIVE BRAINLIEST PLEASE HELP!!!!A circle with a circumference of approximately 15.71 inches drawn inside a square,

Oksi-84 [34.3K]

answer:

5.002 inches

step-by-step explanation:

find the diameter of the circle
the diameter is the side length of the square since the sides of it are touching the circle

c = 2πr

c stands for circumference
π stands for pi (or 3.14)
r stands for radius

- plug in the values given to find the diameter

c = 2πr

15.71 = 2(3.14)r

15.71 = 6.28r

r = 2.501 (rounded or estimated)

now that we have the radius, we can find the diameter by multiplying by 2, since diameter is two times bigger than the radius

diameter = 5.002 (estimated)

now, that will be the side length because the diameter = the side length

3 0

2 years ago

David has a hose that is 27 ft long. He wants to cut it into two pieces so that the longer piece is 3 ft more than 2 times the l

evablogger [386]

One piece is 8ft and the other is 19ft

4 0

3 years ago

Read 2 more answers

Please help ASAP please I’ll mark as brainlister

faust18 [17]

Problems

The bottom of the red figure should be long; similar to the top of the blue figure.

The top of the red figure should be the same as the bottom of the blue figure.

Red are: (1, -1), (4, -1), (1, -5), (3, -5), (2, -2), (2, -4), (3, -4), (4, -2), (4, -2)

red points should be:

(1, -1), (3, -1), (1, -5), (4, -5), (2, -2), (2, -4), (4, -4), (3, -2), (3, -2),

7 0

2 years ago

Can you help?

Ber [7]

-5 1/3, -2 1/2, 3.75, 4.2 ( this is the order from least to greatest)

5 0

3 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