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:
x
−
5
=
0
and
x
+
8
=
0
.
x
=
5
,
−
8
idk why it looks like that xD
:3
Answer:
2x + 2x + x - 8 = 4 (x + 2)
x = 16
Step-by-step explanation:
Because that's a square, the perimeter would be 4s where s is the measure of the side:
2x + 2x + x - 8 = 4 (x + 2)
To solve, refer below:
2x + 2x + x - 8 = 4x + 8
5x - 8 = 4x + 8
5x = 4x + 8 + 8
x = 16
Opposite the 60: 9 square roots of 3 and the hypotenuse is 18
Answer:
84 seconds
Step-by-step explanation:
Because if you divide all of that you would get that answer