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1 year ago

11

Find the area of a circle that a has a radius of 10 in. a Use 3.14 for it. 10 in. Hint: A = ttr2 Area = [?] square inches Enter

the number that goes in the green box. Do not round your answer.

2 answers:

damaskus [11]1 year ago

3 0

Answer:

A≈314.16in²

A=πr2=π·102≈314.15927in²\\

Step-by-step explanation:

Alona [7]1 year ago

3 0

<u>Answer:</u>

The area of the circle is 314 in.²

<u>Step-by-step explanation:</u>

<u>We know that:</u>

Area of circle: πr²
r = 10 in.

<u>Work:</u>

πr²
=> 3.14 x 10 x 10
=> 3.14 x 100
=> 314

Hence, <u>the area of the circle is 314 in.²</u>

Hoped this helped.

$BrainiacUser1357$

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Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.

We can find the area like this:

$A = \pi r^2 = \pi (2-\sqrt{\frac{y}{2}})^2 = \pi (4 -2y +\frac{y^2}{4})$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \pi \int_{0}^8 2-2y +\frac{y^2}{4} dy$

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

The figure 3 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (8-2x^2)^2 = \pi (64 -32x^2 +4x^4)$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= \pi \int_{0}^2 64-32x^2 +4x^4 dx$

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

The figure 4 attached show the radius. We can find the area like this:

$A = \pi r^2 = \pi (\sqrt{\frac{y}{2}})^2 = \pi\frac{y}{2}$

And we can find the volume with this formula:

$V = \int_{a}^b A(y) dy$

$V= \frac{\pi}{2} \int_{0}^8 y dy$

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Suppose that a school has 20 classes: 16 with 25 students in each, three with 100 students in each, and one with 300 students, f

astraxan [27]

Answer

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100 | 0.15

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300 | $\dfrac{1\times 300}{1000}=0.3$

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E(y) = 25 × 0.4 + 100 × 0.3 + 300 × 0.3

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2 years ago

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sergey [27]

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