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An air traffic controller is tracking two planes. To start, Plane A was at an altitude of 990 meters, and Plane B was at an altitude of 1476 meters. Plane A is losing altitude at 18 meters per second, and Plane B is losing altitude at 27 meters per second.Let x be the number of seconds that have passed.(a)For each plane, write an expression for the altitude of the plane after x seconds.AltitudeofPlaneA(in=meters)AltitudeofPlaneB(in=meters)(b)Write an equation to show when the two planes would be at the same altitude.

Question

An air traffic controller is tracking two planes. To start, Plane A was at an altitude of 990 meters, and Plane B was at an altitude of 1476 meters. Plane A is losing altitude at 18 meters per second, and Plane B is losing altitude at 27 meters per second.Let x be the number of seconds that have passed.(a)For each plane, write an expression for the altitude of the plane after x seconds.AltitudeofPlaneA(in=meters)AltitudeofPlaneB(in=meters)(b)Write an equation to show when the two planes would be at the same altitude.

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Solution

(a) The altitude of each plane after x seconds can be calculated by subtracting the rate of descent (in meters per second) times the number of seconds from the initial altitude.

For Plane A, the altitude after x seconds would be: 990 - 18x For Plane B, the altitude after x seconds would be: 1476 - 27x

(b) The two planes would be at the same altitude when their altitude expressions are equal to each other. Therefore, the equation to find when the two planes would be at the same altitude is:

990 - 18x = 1476 - 27x

This problem has been solved

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