1st Avenue would be more difficult because it’s rise and run is for every one foot forward it is 3 feet up. Meanwhile avenue 16th would start at (3,1) and the rise and run would be for every 3 feet it would go up 1 foot.
Answer:
the slope of line j is -1/8, the slope of lines m and n are both 8.
Explanation step by step:
Suppose j is the slope of line j, m the slope of line m and n the slope of line n. Since:
1. line j is perpendicular of line m, so j*m= -1
2. J is perpendicular of line n, so j*n=-1
3. m and n and parallel so m=n
As we want to know the slope of j we clear the equations presented to us in order to find it.
(1) m* n* j = -8, since n=m we have
* j = -8.
we clear j; 
replacing in the equation n*j = -1 we get
n*(-8/n^2) = -1. Thus n = 8. Since j = -1/n = -1/8.
Yes because each x only has one y output
Answer:
Answer is 
Step-by-step explanation:
To find the interval of x. Use our equations to equal each other.



Integrate.
![\frac{-x^3}{3}+x^2\\(\frac{-2^3}{3}+2^2)-[\frac{-0^3}{3}+0^2]\\-\frac{8}{3} +4-0\\-\frac{8}{3}+\frac{12}{3} =4/3](https://tex.z-dn.net/?f=%5Cfrac%7B-x%5E3%7D%7B3%7D%2Bx%5E2%5C%5C%28%5Cfrac%7B-2%5E3%7D%7B3%7D%2B2%5E2%29-%5B%5Cfrac%7B-0%5E3%7D%7B3%7D%2B0%5E2%5D%5C%5C-%5Cfrac%7B8%7D%7B3%7D%20%2B4-0%5C%5C-%5Cfrac%7B8%7D%7B3%7D%2B%5Cfrac%7B12%7D%7B3%7D%20%20%3D4%2F3)
Using Desmos I have Graphs of both of the equations you have provided. The problem asks us to find the shaded region between those curves/equations.
Proof Check your interval of x.