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:
Step-by-step explanation:
y=-5x-13
Since we know the value of y we can substitute it in
6x+6(-5x-13)=-6
6x-30x-78=-6
-24x=72
-x=3
x=-3
Now that we know the value of x we can solve Y
y=-5(-3)-13
y=15-13
y=2
y/5 = x
multiply by 5 on each sides
y = 5x
There you go! I really hope this helped, if there's anything just let me know! :)
Remark
The order is not quite correct.
- 7 + (-11 * 5 - - 7) ÷ - 6
The - - 7 becomes a plus
- 7 + (- 55 + 7) ÷ -6
This is where the real problem is. -55 + 7 goes down. This gives - 48
- 7 + [( - 48) ÷ - 6]
You do the division by - 6 before adding. Adding - 7 to the result of the division is your very last step. Think BedMas.
-7 + (8)
The 8 is the result of the division. -48/-6 = 8. the -7 and the 8 are added. You get 1.
1 Answer
Answer:
3 1/2 gallons.
Step-by-step explanation:
1 6/16 + 2 1/8
= 3 + 6/16 + 1/8
= 3 + 3/8 + 1/8
= 3 4/8
= 3 1/2 gallons.
