Uhh I think something like y=x+5
Answer:
28.35 lb. since it has 2 decimal places
Answer:
![A = [\frac{32}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B32%7D%7B3%7D%5D)
Step-by-step explanation:
Given


Required
Determine the area bounded by the curves
First, we need to determine their points of intersection

Subtract 2x from both sides

Multiply through by -1

Take square root of both sides
or 
This Area is then calculated as thus
![A = \int\limits^a_b {[y_1 - y_2]} \, dx](https://tex.z-dn.net/?f=A%20%3D%20%5Cint%5Climits%5Ea_b%20%7B%5By_1%20-%20y_2%5D%7D%20%5C%2C%20dx)
<em>Where a = 2 and b = -2</em>
Substitute values for
and 

Open Brackets

Collect Like Terms


Integrate
](https://tex.z-dn.net/?f=A%20%3D%20%5B-%5Cfrac%7Bx%5E%7B3%7D%7D%7B3%7D%20%2B4x%5D%282%2C-2%29)
![A = [-\frac{2^{3}}{3} +4(2)] - [-\frac{-2^{3}}{3} +4(-2)]](https://tex.z-dn.net/?f=A%20%3D%20%5B-%5Cfrac%7B2%5E%7B3%7D%7D%7B3%7D%20%2B4%282%29%5D%20-%20%5B-%5Cfrac%7B-2%5E%7B3%7D%7D%7B3%7D%20%2B4%28-2%29%5D)
![A = [-\frac{8}{3} +8] - [-\frac{-8}{3} -8]](https://tex.z-dn.net/?f=A%20%3D%20%5B-%5Cfrac%7B8%7D%7B3%7D%20%2B8%5D%20-%20%5B-%5Cfrac%7B-8%7D%7B3%7D%20-8%5D)
![A = [\frac{-8+ 24}{3}] - [\frac{8}{3} -8]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B-8%2B%2024%7D%7B3%7D%5D%20-%20%5B%5Cfrac%7B8%7D%7B3%7D%20-8%5D)
![A = [\frac{-8+ 24}{3}] - [\frac{8-24}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B-8%2B%2024%7D%7B3%7D%5D%20-%20%5B%5Cfrac%7B8-24%7D%7B3%7D%5D)
![A = [\frac{16}{3}] - [\frac{-16}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B16%7D%7B3%7D%5D%20-%20%5B%5Cfrac%7B-16%7D%7B3%7D%5D)
![A = [\frac{16}{3}] + [\frac{16}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B16%7D%7B3%7D%5D%20%2B%20%5B%5Cfrac%7B16%7D%7B3%7D%5D)
![A = [\frac{16 + 16}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B16%20%2B%2016%7D%7B3%7D%5D)
![A = [\frac{32}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B32%7D%7B3%7D%5D)
Hence, the Area is:
![A = [\frac{32}{3}]](https://tex.z-dn.net/?f=A%20%3D%20%5B%5Cfrac%7B32%7D%7B3%7D%5D)
Answer:
87.4%
Step-by-step explanation:
The radius to the top of the atmosphere as a fraction of the moon's radius is ...
(2575 +600)/2575 ≈ 1.23301
The ratio of the moon's volume with atmosphere to the moon's volume without is the cube of this, or 1.87456
Then the ratio of the atmosphere's volume to the moon's volume is ...
(1.87456 -1)/1 = 0.87456
Atmospheric haze is about 87.4% of the moon's volume.
_____
We have assumed that the desired ratio is to the solid moon's volume, not the volume of moon and atmosphere together. The latter ratio would be 0.875 to 1.875 or about 46.7%.