To solve you would use the equation y=mx=b .
y is how far line goes up, x is how for along , m is the slope and b is where the line crosses the yaxis.
slope: -1
x-intercept ; -4
y-intercept : -4
Answer:
Step-by-step explanation:
Given two upward facing parabolas with equations
The two intersect at
=
x=
area enclosed by them is given by
A=![\int_{-\sqrt{\frac{2}{5}}}^{\sqrt{\frac{2}{5}}}\left [ \left ( x^2+2\right )-\left ( 6x^2\right ) \right ]dx](https://tex.z-dn.net/?f=%5Cint_%7B-%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5E%7B%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%7D%5Cleft%20%5B%20%5Cleft%20%28%20x%5E2%2B2%5Cright%20%29-%5Cleft%20%28%206x%5E2%5Cright%20%29%20%5Cright%20%5Ddx)
A=
A=![4\left [ \sqrt{\frac{2}{5}} \right ]-\frac{5}{3}\left [ \left ( \frac{2}{5}\right )^\frac{3}{2}-\left ( -\frac{2}{5}\right )^\frac{3}{2} \right ]](https://tex.z-dn.net/?f=4%5Cleft%20%5B%20%5Csqrt%7B%5Cfrac%7B2%7D%7B5%7D%7D%20%5Cright%20%5D-%5Cfrac%7B5%7D%7B3%7D%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D-%5Cleft%20%28%20-%5Cfrac%7B2%7D%7B5%7D%5Cright%20%29%5E%5Cfrac%7B3%7D%7B2%7D%20%5Cright%20%5D)
A=
Answer:
option B is correct. Once have a look to this solution that I have answered
Altho' I can easily guess what you're supposed to do here, I must point out that you haven't included the instructions for this problem.
I'll help you by example. Let's look at the first problem:
"Evaluate 6(z-1) at z-4."
Due to "order of operations" rules, we must do the work inside the parentheses FIRST. Replace the z inside (z-1) with "-4". We obtain
6(-4-1) = 6(-5) = -30 (answer.)
Your turn. Try the next one. If it's unclear, as questions.
Answer:
A) T'(-2,3), U'(0,5), V'(-1,0)
