from the diagram, we can see that the height or line perpendicular to the parallel sides is 8.5.
likewise we can see that the parallel sides or "bases" are 24.3 and 9.7, so
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=8.5\\ a=24.3\\ b=9.7 \end{cases}\implies \begin{array}{llll} A=\cfrac{8.5(24.3+9.7)}{2}\\\\ A=\cfrac{8.5(34)}{2}\implies A=144.5~in^2 \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%20h%3D8.5%5C%5C%20a%3D24.3%5C%5C%20b%3D9.7%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B8.5%2824.3%2B9.7%29%7D%7B2%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B8.5%2834%29%7D%7B2%7D%5Cimplies%20A%3D144.5~in%5E2%20%5Cend%7Barray%7D)
The resulting equation will represent a line whose slope is 1/2 times the slope of the line
<h3>How to determine the slope of the new line?</h3>
The equation of the line is given as:
y = 3x/a + 5
The constant a is a positive constant.
So, when the value of a in the equation is doubled, we have:
y = 3x/2a + 5
A linear equation is represented as
y = mx + b
Where m represents the slope.
So, we have:
m1 = 3/a
m2 = 3/2a
Substitute m1 = 3/a in m2 = 3/2a
m2 = 1/2 * m1
Hence, the resulting equation will represent a line whose slope is 1/2 times the slope of the line
Read more about linear equation at:
brainly.com/question/14323743
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Answer:
the input number is -10
Step-by-step explanation:
output is always y
input is always x
so instead of y, put -52
-52=5x-2 add 2 to each side to ger 5x by itself
-50=5x divide both sides by 5 to get x by itself
-10=x
x is the input so the input is -10
I think it’s A. Hope it’s right. Sorry if it’s wrong.