3=4m+ b
-1=3m+b
m=4
plug in to either equation,
3=16+ b
b=-13
Thus, C would be the answer.
Check the picture below.
so the area of the hexagon is really just the area of two isosceles trapezoids.
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ a=2\\ b=4\\ h=2 \end{cases}\implies \begin{array}{llll} A=\cfrac{2(2+4)}{2}\implies A=6 \\\\\\ \stackrel{\textit{twice that much}}{2A = 12} \end{array}](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%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2%5C%5C%20b%3D4%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B2%282%2B4%29%7D%7B2%7D%5Cimplies%20A%3D6%20%5C%5C%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Btwice%20that%20much%7D%7D%7B2A%20%3D%2012%7D%20%5Cend%7Barray%7D)
You would have to add a positive 6 to the negative 6 to get zero. Lets say you have -2 in order to get it to zero or just any positive number, you have to add a positive of the same value or higher to be able to get it there. I hope you understand that.
The functional equation of the line is f (x)=a*x+b
you have already a: -4 -->f (x)=-4 x+b
Now you can use the point (3;-3)
3 is the x-coordinate and -3 the f (x)-coordinate
--> -3 = -4*(3) + b Now you can solve it
-3 = -12+b (+12)
b = 9
ANSWER: f (x) = -4x + 9
2/3(2/8 + 4/8) = 2/3(1/4 + 1/2)
2/3(3/4) = 1/2
