we can pretty much split the middle part into two trapezoids. Check the picture below.
so we really have one trapezoid and one square, each twice, so simply let's get the area of the trapezoid and sum it up with the area of the square, twice, and that's the area of the shape.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{\textit{parallel sides}}{bases}\\[-0.5em] \hrulefill\\ h=5\\ a=3\\ b=7 \end{cases}\implies A=\cfrac{5(3+7)}{2}\implies A=25 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{sum of areas}}{[25+(3\cdot 3)]}\cdot \stackrel{twice}{2}\implies [34]2\implies \underset{in^2}{68}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7B%5Ctextit%7Bparallel%20sides%7D%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D5%5C%5C%20a%3D3%5C%5C%20b%3D7%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B5%283%2B7%29%7D%7B2%7D%5Cimplies%20A%3D25%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bsum%20of%20areas%7D%7D%7B%5B25%2B%283%5Ccdot%203%29%5D%7D%5Ccdot%20%5Cstackrel%7Btwice%7D%7B2%7D%5Cimplies%20%5B34%5D2%5Cimplies%20%5Cunderset%7Bin%5E2%7D%7B68%7D)
Solve the problem by doing the following
5x=75
5x/5=75/5
x=15
Answer by JKismyhusbandbae: -12 ≤ f ≤ 6
-3f - 9|-8+8 ≤ 19 + 8
|-3f-9|≤ 27
-3f-9≤ 27 and -3f - 9≥ 27
-3f-9+9≤27+9
-3f ≤ 36
(-3f) (-1) ≥ 36 (-1)3f ≥ - 36
≥ 
f ≥ - 12-3f
-9+9≥-27+9
-3f≥-18
≤ 
F ≤ 6
Answer: 100
Step-by-step explanation:
Do X+60+5x=180 then that will get you 20 Then do 5*20 and that will get you 100
Pick A that is the one I would go with.
