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3 years ago

15

Jeremiah invested

1 answer:

Margaret [11]3 years ago

3 0

Answer: 1210.1

Step-by-step explanation:

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58 - 5X = 30X - 75 need help please

Tju [1.3M]

Answer:

3.8

Step-by-step explanation:

$58-5x=30x-75$

Let's add 5x to both sides of the equation...

$58=35x-75$

Then, let's add 75 to both sides of the equation...

$133=35x$

Divide by the coefficient of x, (35)...

$x=3.8$

7 0

3 years ago

How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

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:

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

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

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

6 0

3 years ago

What is the answer to this

Sphinxa [80]

X= 3/9 is the answer!!

7 0

3 years ago

55 is the product of vidya's age and 5 <br> Use the variable v to represent vidya's age

klasskru [66]

V= 11. If you divide the product 55 by 5 you get Vidya's age.
You're Welcome

5 0

3 years ago

NEED HELP, please help ASAP!!!!!!!!

nadya68 [22]

Answer: B

Step-by-step explanation:

6 0

2 years ago