Final answer:
After 7 minutes, both airplanes at the airport will be at the same altitude of 18,900 feet. The time was found by equating the altitudes as functions of time and solving for the time variable.
Step-by-step explanation:
The question involves determining when two airplanes at an airport will be at the same altitude. The first airplane is at 43,400 feet and descends at a rate of 3,500 feet per minute. The second airplane ascends at 2,700 feet per minute. To find the time when they will be at the same altitude, we set up an equation that represents the altitude of each airplane as a function of time.
Let t be the time in minutes at which the airplanes will be at the same altitude. The altitude of the descending airplane can be represented as 43,400 - 3,500t, and the altitude of the ascending airplane as 2,700t.
We set these two expressions equal to each other and solve for t:
- 43,400 - 3,500t = 2,700t
- 43,400 = 3,500t + 2,700t
- 43,400 = 6,200t
- t = 43,400 / 6,200
- t = 7
Therefore, after 7 minutes, both airplanes will be at the same altitude. To find what this altitude will be, we substitute t back into either equation for altitude:
- Altitude = 43,400 - 3,500(7)
- Altitude = 43,400 - 24,500
- Altitude = 18,900 feet