bearing in mind that an absolute value expression is in effect a piece-wise expression, because it has a ± version.
![\bf 3|x|+7=28\implies 3|x|=21\implies |x|=\cfrac{21}{3}\implies |x|=7\implies \begin{cases} +(x)=7\\ -(x)=7 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ +(x)=7\implies \boxed{x=7}~\hfill -x=7\implies \boxed{x=-7}](https://tex.z-dn.net/?f=%20%5Cbf%203%7Cx%7C%2B7%3D28%5Cimplies%203%7Cx%7C%3D21%5Cimplies%20%7Cx%7C%3D%5Ccfrac%7B21%7D%7B3%7D%5Cimplies%20%7Cx%7C%3D7%5Cimplies%20%20%5Cbegin%7Bcases%7D%20%2B%28x%29%3D7%5C%5C%20-%28x%29%3D7%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%2B%28x%29%3D7%5Cimplies%20%5Cboxed%7Bx%3D7%7D~%5Chfill%20%20-x%3D7%5Cimplies%20%5Cboxed%7Bx%3D-7%7D%20)
Answer:
5x^2 +10x
Step-by-step explanation:
The area is the product of the height and width. For this exercise, it is convenient to compute the rectangle areas individually, then write their sum:
blue rectangle area = (5x)(x) = 5x^2
red rectangle area = (5x)(2) = 10x
Total area = 5x^2 +10x.
_____
If you start by writing an expression for the total area, expanding it requires you deal with the unlike terms separately anyway:
5x(x +2) = 5x(x) +5x(2) = 5x^2 +10x
Answer:
All integers from 2 to six inclusiv
Step-by-step explanation:
Answer:
x +0y+0z = 400
-x +y+0z = 150
-8x +0y +z = 250
Step-by-step explanation:
The last column is the solution
The rest of the columns are the coefficients of the variables
x +0y+0z = 400
-x +y+0z = 150
-8x +0y +z = 250
