[table]
As air is pumped into an exercise ball, the surface area of the ball expands. The table shows the surface area of the ball at 2 seconds intervals for 30 seconds. Determine the average rate of change of the surface area of the ball for the interval between 10 seconds and 30 seconds. |
[/table]

Question

BlackTom AI · Accepted Answer

To tackle this problem, I first identify the given data points at 10 seconds and 30 seconds from the table. At t = 10 s, the surface area is 324.0 cm^2, and at t = 30 s, it is 2836.0 cm^2.
Next, I compute the overall change in surface area between these two times: 2836.0 − 324.0 = 2512.0 cm^2.
Then I determine the interval duration, which is 30 − 10 = 20 seconds.
With these two values, the average rate of change is the total change divided by the time interval: 2512.0 cm^2 ÷ 20 s = 125.6 cm^2 per second.
To verify consistency, I confirm that the interval used (10 s to 30 s) matches the requested range and that the table provides data at 2-second increments, ensuring no misalignment in the time endpoints.
Finally, I note that the rate is expressed in square centimeters per second, reflecting the change in surface area per unit time over the specified interval.

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