Knowing how much space an object takes up would help minimize the amount of material because they know how much they have to order and would not have to order any extra
Answer:
I think it would be 33+27q
tell me if this helps! :)
Sorry if it doesn't though :(
Answer:
Statement Reasoning
LM=NO Given
∠LMO≅∠NOM. Given
OM=OM Reflexive Property of Equality
ΔLMO=ΔNOM SAS Congruence Postulate
Answer:
The first one I believe
Step-by-step explanation:
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.