#2**+1 **

average rate of change over the interval [9, 10] = \(\frac{f(10)-f(9)}{10-9}\ =\ \frac{11014-4052}{1}\ =\ \frac{6962}{1}\ =\ 6962\)

average rate of change over the interval [5, 8] = \(\frac{f(8)-f(5)}{8-5}\ =\ \frac{1491-75}{8-5}\ =\ \frac{1416}{3}\ =\ 472\)

How much greater is the average rate of change over the interval [9, 10] than the interval [5, 8] ?

In other words...

How much greater is 6962 than 472 ?

6490 seems to be correct

hectictar Mar 2, 2020

#2**+1 **

Best Answer

average rate of change over the interval [9, 10] = \(\frac{f(10)-f(9)}{10-9}\ =\ \frac{11014-4052}{1}\ =\ \frac{6962}{1}\ =\ 6962\)

average rate of change over the interval [5, 8] = \(\frac{f(8)-f(5)}{8-5}\ =\ \frac{1491-75}{8-5}\ =\ \frac{1416}{3}\ =\ 472\)

How much greater is the average rate of change over the interval [9, 10] than the interval [5, 8] ?

In other words...

How much greater is 6962 than 472 ?

6490 seems to be correct

hectictar Mar 2, 2020