This is 16 pints.
I hope this helps
bearing in mind that, whenever we have an absolute value expression, is in effect a piece-wise function with a positive and a negative version of the expression, so
![\bf |x^2-4x-5|=7\implies \begin{cases} +(x^2-4x-5)=7\\\\ -(x^2-4x-5)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x^2-4x-5)=7\implies x^2-4x-5=7\implies x^2-4x-12=0 \\\\\\ (x-6)(x+2)=0\implies x= \begin{cases} 6\\ -2 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ -(x^2-4x-5)=7\implies x^2-4x-5=-7\implies x^2-4x+2=0 \\\\\\ (x-2)(x-2)=0\implies x = 2](https://tex.z-dn.net/?f=%5Cbf%20%7Cx%5E2-4x-5%7C%3D7%5Cimplies%20%5Cbegin%7Bcases%7D%20%2B%28x%5E2-4x-5%29%3D7%5C%5C%5C%5C%20-%28x%5E2-4x-5%29%3D7%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%2B%28x%5E2-4x-5%29%3D7%5Cimplies%20x%5E2-4x-5%3D7%5Cimplies%20x%5E2-4x-12%3D0%20%5C%5C%5C%5C%5C%5C%20%28x-6%29%28x%2B2%29%3D0%5Cimplies%20x%3D%20%5Cbegin%7Bcases%7D%206%5C%5C%20-2%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-%28x%5E2-4x-5%29%3D7%5Cimplies%20x%5E2-4x-5%3D-7%5Cimplies%20x%5E2-4x%2B2%3D0%20%5C%5C%5C%5C%5C%5C%20%28x-2%29%28x-2%29%3D0%5Cimplies%20x%20%3D%202)
#2.
a) 1/4m + 3 = 2
1/4m = -1
m = -4
b) 4y - 3(y + 8) = 12
4y - 3y + 24 = 12
y = -12
For #10, letter b. Do the same thing you did with letter a.
The equation is already given.
You have your y-intercept: -8, and your slope: 3/1.
These two create two different points you can add to your data table.
(0, -8) and (1, 3)
y:-8, 3
x: 0, 1
10.) The measure of angle 1 is 131° according to the Alternate Interior Angles Theorem. The measure of angle 2 is 49° because 180° - 131° = 49°. I can do this because of the Linear Pair Postulate.
11.) The measure of angle 1 is 65° according to the Vertical Angles Theorem. The measure of angle 2 is 65° according to the Corresponding Angles Theorem.
Hope this helps!