11/9 is the answer of the question
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)
ANSWER
5
EXPLANATION
We want to simplify:

when x=2.
We substitute x=2 into the absolute value expression to obtain:

The about value of 5 means what is the distance of 5 from zero on the real number line.

Answer:
6 out of 12
Step-by-step explanation: there was 12 before but now there are 6 because probally somebody ate them
Just to be different we'll start with intercept intercept form which says the line through x intercept (a,0) and y intercept (0,b) is

Through (4,0) and (0,-2) that's

Multiply through by the common denominator of 4 for standard form:
x - 2y = 4
For slope intercept form we solve for y
-2 y = -x + 4
y = (1/2) x - 2
We see our slope is (1/2) and we go through (4,0) so point slope form is
y - 0 = (1/2)(x - 4)
y = (1/2)(x-4)