Plane A is descending toward the local airport, and plane B is ascending from the same airport. Plane A is descending at a rate of 2,500 feet per minute. Plane B is ascending at a rate of 4,000 feet per minute. If plane A is currently at an altitude of 14,000 feet and plane B is at an altitude of 1,000 feet, how long will it take them to be at the same altitude? The equation representing plane A’s descent is y = -2,500x + 14,000. The equation representing plane B’s ascent is y = 4,000x + 1,000. In both equations, y represents altitude and x represents time in minutes.

2 minutes

Step-by-step explanation:

Given the following :

Plane A's descent :

y = -2,500x + 14,000

Plane B's Ascent :

y = 4,000x + 1,000

where y = altitude x = minute

Time to be at the same altitude :

Being at the same altitude means ;

Plane A's descent = Plane B's Ascent

-2,500x + 14,000 = 4,000x + 1,000

-2500x - 4000x = 1000 - 14000

-6500x = - 13000

x = 13000 / 6500

x = 2

x = 2minutes.

idk

step-by-step explanation:

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.