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)

Answer:
m = - 2.27 or m = 16.27
Step-by-step explanation:
Given
m² - 14m - 37 = 0 ( add 37 to both sides )
m² - 14m = 37
To complete the square
add ( half the coefficient of the m- term )² to both sides
m² + 2(- 7)m + 49 = 37 + 49, that is
(m - 7)² = 86 ( take the square root of both sides )
m - 7 = ±
( add 7 to both sides )
m = 7 ±
, that is
m = 7 -
≈ - 2.27 or m = 7 +
≈ 16.27
The given expression is

We just have to divide by 2

<h2>Hence, the solution is all real numbers less than 10.</h2>
1. 46.45
2. 1.92
3. 425.04
4. 69.3
5. I can't see the symbol. Can you type it in the comments?
6. 850
7. 6,960
8. $8
I hope you find this answer the most helpful! :)