Question

A helium balloon is at a height of 1,200 ft. after two minutes in flight. The balloon finally pops after seven minutes in flight, at a height of 10,500 ft. What is the average rate of change in height for the balloon over this period of time?(1 point) Responses 1,860 ft./min. 1,860 ft./min. 2,340 ft./min. 2,340 ft./min. 9,300 ft./min. 9,300 ft./min. 1,033.33 ft./min.

Answer 1

To find the average rate of change in height of the helium balloon over the given time period, we can use the formula for average rate of change:

\[ \text{Average Rate of Change} = \frac{\text{Change in Height}}{\text{Change in Time}} \]

1. Determine the Change in Height:

   • Final height: 10,500 ft (after 7 minutes)
   • Initial height: 1,200 ft (after 2 minutes)
   • Change in height = Final height - Initial height
   • Change in height \( = 10,500 , \text{ft} - 1,200 , \text{ft} = 9,300 , \text{ft} \)

2. Determine the Change in Time:

   • Final time: 7 minutes
   • Initial time: 2 minutes
   • Change in time \( = 7 , \text{min} - 2 , \text{min} = 5 , \text{min} \)

3. Calculate the Average Rate of Change: \[ \text{Average Rate of Change} = \frac{9,300 , \text{ft}}{5 , \text{min}} = 1,860 , \text{ft/min} \]

Thus, the average rate of change in height for the balloon over this period of time is 1,860 ft/min.