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)
the answer will be $7401.22
Answer:
k<0 or k>4
Step-by-step explanation:
In the generic equation ax² + bx + c = 0, there are two solutions only if discriminant D>0 where D = b²-4ac.
Mapping your equation on the generic means that:
a = k
b = k
c = 1
So when we have to solve D>0 we solve:
k² - 4k > 0
k(k-4) > 0
roots are k=0 and k=4
k² - 4k > 0 when k<0 or k>4
Answer: Y = 7x/2 -2
Step-by-step explanation:
Y intercept of -2 means there's a point at (0,-2).
Given the two points (4,12) and (0,-2), we can now find the slope
12-(-2)/4-0 = 14/4 = 7/2 which is the slope
so the equation is Y = 7x/2 -2
Answer:
<h2>30 mm</h2>
Step-by-step explanation:
The formula of an area of a circle:
<em>r</em><em> - radius</em>
<em />
We have
Substitute:
<em>divide both sides by π</em>
<em>use π ≈ 3.14</em>
The diameter <em>d = 2r</em>.
Therefore
