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son4ous [18]
3 years ago
10

Find the volume round to the nearest tenth

Mathematics
2 answers:
AleksAgata [21]3 years ago
7 0

Answer: 4. 1436.8 ft^3     5. 113.1  cm^3

Step-by-step explanation:

4.  V= 4/3* 7^3 * \pi

  V= 4/3 * 343 *\pi

    V=  1436.755 ft^3  rounded to the nearest tenth is 1436.8 cubic feet.

5. V= 4/3* 3^3 * \pi

   V= 4/3 * 27 * \pi

   V=  113.097  cm^3   rounded to the nearest tenth is 113.1 cubic centimeters.

juin [17]3 years ago
4 0
4/3*兀*r^3

#4)4/3*3.14*7^3=1436.8
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Without a [roblem to go along with x=4, this would be underfined

Step-by-step explanation:

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Can someone give me the answer :( and then explain how to get the answer..
Pani-rosa [81]

x=months

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make them equal:

200+75x=350+50x

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WILL GIVE BRAINLIEST!!!!! 100 POINTS
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Answer:the new line will be 1/4 of the original line.

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3 years ago
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Need help with AP CAL
anzhelika [568]

Answer: Choice C

\displaystyle \frac{1}{2}\left(1 - \frac{1}{e^2}\right)

============================================================

Explanation:

The graph is shown below. The base of the 3D solid is the blue region. It spans from x = 0 to x = 1. It's also above the x axis, and below the curve y = e^{-x}

Think of the blue region as the floor of this weirdly shaped 3D room.

We're told that the cross sections are perpendicular to the x axis and each cross section is a square. The side length of each square is e^{-x} where 0 < x < 1

Let's compute the area of each general cross section.

\text{area} = (\text{side})^2\\\\\text{area} = (e^{-x})^2\\\\\text{area} = e^{-2x}\\\\

We'll be integrating infinitely many of these infinitely thin square slabs to find the volume of the 3D shape. Think of it like stacking concrete blocks together, except the blocks are side by side (instead of on top of each other). Or you can think of it like a row of square books of varying sizes. The books are very very thin.

This is what we want to compute

\displaystyle \int_{0}^{1}e^{-2x}dx\\\\

Apply a u-substitution

u = -2x

du/dx = -2

du = -2dx

dx = du/(-2)

dx = -0.5du

Also, don't forget to change the limits of integration

  • If x = 0, then u = -2x = -2(0) = 0
  • If x = 1, then u = -2x = -2(1) = -2

This means,

\displaystyle \int_{0}^{1}e^{-2x}dx = \int_{0}^{-2}e^{u}(-0.5du) = 0.5\int_{-2}^{0}e^{u}du\\\\\\

I used the rule that \displaystyle \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx which says swapping the limits of integration will have us swap the sign out front.

--------

Furthermore,

\displaystyle 0.5\int_{-2}^{0}e^{u}du = \frac{1}{2}\left[e^u+C\right]_{-2}^{0}\\\\\\= \frac{1}{2}\left[(e^0+C)-(e^{-2}+C)\right]\\\\\\= \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

In short,

\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]

This points us to choice C as the final answer.

5 0
1 year ago
Don't be childish I will report thanks ​
Semenov [28]

Answer:

what do you mean childish?.. anyway your answer is

-2| 3 - 6 |

= -6

happy to help!

5 0
2 years ago
Read 2 more answers
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