The fact that there are two length sides and two width sides.
Answer: Choice C

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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 
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
where 0 < x < 1
Let's compute the area of each general cross section.

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

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,

I used the rule that
which says swapping the limits of integration will have us swap the sign out front.
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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]](https://tex.z-dn.net/?f=%5Cdisplaystyle%200.5%5Cint_%7B-2%7D%5E%7B0%7De%5E%7Bu%7Ddu%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5Be%5Eu%2BC%5Cright%5D_%7B-2%7D%5E%7B0%7D%5C%5C%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B%28e%5E0%2BC%29-%28e%5E%7B-2%7D%2BC%29%5Cright%5D%5C%5C%5C%5C%5C%5C%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B1%20-%20%5Cfrac%7B1%7D%7Be%5E2%7D%5Cright%5D)
In short,
![\displaystyle \int_{0}^{1}e^{-2x}dx = \frac{1}{2}\left[1 - \frac{1}{e^2}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B0%7D%5E%7B1%7De%5E%7B-2x%7Ddx%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Cleft%5B1%20-%20%5Cfrac%7B1%7D%7Be%5E2%7D%5Cright%5D)
This points us to choice C as the final answer.
(A) Just because every digit has an equal chance of appearing does not mean that all will be equally represented. (See "gambler's fallacy")
(B) The experimental procedure isn't exactly clear, so assuming a table of digits refers to a table of just one-digit numbers, each with 0.1 chance of appearing (which means you can think of the digits 0-9), you should expect any given digit to appear about 0.1 or 10% of the time.
So if a table consists of 1000 digits, one could expect 7 to appear in 10% of the table, or about 100 times.
Answer: about 8 am
Step-by-step explanation:
that’s just the closest
Answer:
You add up all the sides of the shape