+0  
 

Best Answer 

 #2
avatar+8963 
+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  smiley

 Mar 2, 2020
 #1
avatar
+1

is it 6490?????

 Mar 2, 2020
 #2
avatar+8963 
+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  smiley

hectictar Mar 2, 2020

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