Answer :
Answer:
a. [1,2]
[tex] m= \frac{9-0}{2-1}=9[/tex]
b. [1,3.5]
[tex] m =\frac{17-0}{3.5-1}=6.8[/tex]
c. [1,5]
[tex] m =\frac{0-0}{5-1}=0[/tex]
d. [0,3.5]
[tex] m =\frac{17-(-5)}{3.5-0}=6.29[/tex]
So then we can conclude that the highest slope is for the interval [1,2] and that would be our solution for the fastest average rate.
a. [1,2]
[tex] m= \frac{9-0}{2-1}=9[/tex]
Step-by-step explanation:
Assuming that we have the figure attached for the function. For this case we just need to quantify the slope given by:
[tex] m = \frac{\Delta y}{\Delta x}[/tex]
For each interval and the greatest slope would be the interval on which the volume of the box is changing at the fastest average rate
a. [1,2]
[tex] m= \frac{9-0}{2-1}=9[/tex]
b. [1,3.5]
[tex] m =\frac{17-0}{3.5-1}=6.8[/tex]
c. [1,5]
[tex] m =\frac{0-0}{5-1}=0[/tex]
d. [0,3.5]
[tex] m =\frac{17-(-5)}{3.5-0}=6.29[/tex]
So then we can conclude that the highest slope is for the interval [1,2] and that would be our solution for the fastest average rate.
a. [1,2]
[tex] m= \frac{9-0}{2-1}=9[/tex]
