The answer is a number that can be written as a ratio of two integers. The answer is C.
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)
Answer:
Step-by-step explanation:
3v + u + 66 = 360
<em>Subtract 66 from 366</em>
3v + u = 294
<em>check images</em>
<em>for more steps</em>
Answer:
x∈R
Step-by-step explanation:
2(3x+4)=6x+8
Use the distributive property to multiply 2 by 3x+4.
6x+8=6x+8
Subtract 6x from both sides.
6x+8−6x=8
Combine 6x and −6x to get 0.
8=8
Compare 8 and 8.
True
This is true for any x.
x∈R
1 + -4
2 + -5
3 + -6
4 + -7