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)
So the change in between terms is 5 so you would put 5x, then for the first term, which is 2, 5*1(x)(because it's first term) then the difference between 5-2(because it's the 2nd term) So you would plug 100 in for x so 5(100)-3=497. Hope this helped!
<span>15x+12=45+4x </span>⇒15x - 4x = 45 - 12 ⇒ 11x = 33 ⇒ x = 3
Answer:
Katherine invested $12,000
Step-by-step explanation:
Use formula
where
I = interest,
P = principal,
r = rate (as decimal),
t = time (in years).
In your case,
t = 1 year,
r = 0.06 (or 6%)
P + I =$12,720, thus
pi·(7.2/2)^2·x = 2·pi·(7.2/2)^2 + 2·pi·(7.2/2)·x
x = 4.5 = 4 1/2
O = V = 1458/25·pi = 58.32·pi
