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.

constant acceleration- a

constant velocity- c

no velocity or acceleration- b

explanation:

for body moving with a constant acceleration, the rate of change of displacement changes over time. this is proven by the increasing gradient of the curve (a) against time.

a body that is moving at constant velocity covers equal distances over equal periods of time, thus the straight line with a constant gradient c.

a stationary body   covers no displacement as time elapses. thus the flat line with   zero gradient.

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation: