Answer:
the answer is B
Step-by-step explanation:
<em>ik because i got it right.</em>
H(x), or the range of this function, must be y >= 0, as you can never get a negative when square rooting any number.
For the domain, x cannot exceed -2 or 2, as this would mean we'd be finding the square root of a negative number, which can't happen without getting in to imaginary numbers. Thus, the domain must be -2 <= x <= 2.
the answer is finish the fxking question nim.wit
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)
