Area of square =
where,
is the side length of the square.


s must be positive since it is a side length, so the negative root of 81 is rejected.
Perimeter=
Perimeter=
check the picture below.
namely, which of those intervals has the steepest slope, recall slope = average rate of change.
now, from the picture, notice, those two there are the steepest, the other three are leaning too much to the "ground".
so, from those two, which is the steepest anyway? let's check their slope.
![\bf \stackrel{\textit{from the 6th to the 8th hour}}{(\stackrel{x_1}{6}~,~\stackrel{y_1}{104})\qquad (\stackrel{x_2}{8}~,~\stackrel{y_2}{146})} \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{146-104}{8-2}\implies \cfrac{42}{2}\implies 21~~\bigotimes \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Bfrom%20the%206th%20to%20the%208th%20hour%7D%7D%7B%28%5Cstackrel%7Bx_1%7D%7B6%7D~%2C~%5Cstackrel%7By_1%7D%7B104%7D%29%5Cqquad%20%28%5Cstackrel%7Bx_2%7D%7B8%7D~%2C~%5Cstackrel%7By_2%7D%7B146%7D%29%7D%20%5C%5C%5C%5C%5C%5C%20slope%20%3D%20m%5Cimplies%20%5Ccfrac%7B%5Cstackrel%7Brise%7D%7B%20y_2-%20y_1%7D%7D%7B%5Cstackrel%7Brun%7D%7B%20x_2-%20x_1%7D%7D%5Cimplies%20%5Ccfrac%7B146-104%7D%7B8-2%7D%5Cimplies%20%5Ccfrac%7B42%7D%7B2%7D%5Cimplies%2021~~%5Cbigotimes%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)

Slope: (-3/2)
Y intercept: 9
The answer is C.
y = -3x + 2
y = -3x + 4
-3x + 2 = -3x + 4
2 ≠ 4
Answer: B. Subtract 28 from both sides
Explanation: You need to subtract first because the variable needs to be by itself.