Answer:
eliver mail. The table shows the truck's distance d from the post office at time t. Calculate the average rate of change over the interval from 8 to 15 minutes. t(min) d (km) 0 0 8 7 11 11 15 14 22 16 (1 point) The rate of change is about 0.8 kilometers per minute. The rate of change is 7 kilometers
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)
You can make some algebraic equations and solve it.
The first would be:

The second would be

You can then rearrange the second into

And subsitute it into the first like so:

After that, distribute the y into the parantheses.

Subtract the 21 on both sides and multiply by -1 on both sides:

You then can factor it into:

With Zero Product Property, we can determine y to be either -3 and 7. Since the variables are interchangable, you can say the same about x, just that whatever x is, y must be the other value.
Thus, the answer is 7 and -3.
Answer:
0
Step-by-step explanation:
Replace x with 2.
-(2)+2=0