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

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Mar 2, 2020

#2
+8874
+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

Mar 2, 2020

#1
+1

is it 6490?????

Mar 2, 2020
#2
+8874
+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