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Softa [21]
3 years ago
13

Help pls!! Whoever answers gets marked brainliest

Mathematics
1 answer:
Tamiku [17]3 years ago
7 0

Answer:

10. x= 10 y=sqrt 200

11. x=y=450

12. x=19 y=sqrt 1083

13. x=sqrt 48 y=sqrt 12

Step-by-step explanation:

10.

x and 10 are the same

10^2+10^2=200

sqrt 200 is not a whole number so y is just sqrt 200

11.

30^2=2x^2 because x and y are the same

x^2=450

sqrt 450 is not a whole number so x and y are just sqrt 450

12.

x= 38 divided by 2=19

38^2=19^2+y^2

1444=361+y^2

y^2=1083

sqrt 1083 is not a whole number so y is just sqrt 1083

13.

x^2=(x/2)^2+36=x^2/4+36

x^2=48

sqrt 48 is not a whole number so x is just sqrt 48

y is half of x so it is sqrt 12

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Isabella has a 22 ounce cola. She drinks 17 ounces. Enter the percentage of ounces Isabella has left of her cola. Round your ans
jenyasd209 [6]

Answer:

<u>22.73%</u> of ounces Isabella has left of her cola.

Step-by-step explanation:

Given:

Isabella has a 22 ounce cola. She drinks 17 ounces.

Now, to find the percentage of ounces Isabella has left of her cola.

Total ounces of cola = 22.

Ounces of cola Isabella drink = 17.

So, the ounces of cola she left:

<em>Total ounces of cola - ounces of cola Isabella drink.</em>

=22-17

=5\ ounces.

<u><em>Thus, 5 ounces of cola she left.</em></u>

Now, to get the percentage of cola she left:

\frac{5}{22}\times 100

=\frac{500}{22}

=22.727\%.

<u>Rounding the percentage to the nearest hundredths is 22.73%.</u>

Therefore, 22.73% of ounces Isabella has left of her cola.

7 0
3 years ago
Pls help! will give brainlist!
algol [13]

Answer:

A. (2,5)

because 3×2 is 6, 6-1 is 5

so y=5

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3 years ago
PLEASE HELP!! MATH FINAL + NO EXPLANATION NEEDED :))
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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

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