You are planning a movie stunt using an aircraft that moves vertically at a constant rate of -55 feet per second. You want the stunt to last 15 seconds and to end when the aircraft is only 30 feet above the ground. At what altitude should the stunt begin?

855 feet

Step-by-step explanation:

Step 1 :

Speed of the aircraft moving vertically is -55 feet / sec

Time period for which the stunt should last  =  15 seconds

Distance at which the stunt should stop = 30 feet above the ground

Step 2 :

Distance moved by the aircraft in 15 secs = 15 * 55 = 825 feet

At the end of 15 seconds the aircraft should be 30 feet above the ground

So the stunt should be started at the altitude of 15 * 55 + 30 = 855 feet

x = -3

y = -18

step-by-step explanation:

to solve by substitution, substitute one variable and solve for the other. then substitute again to find the remaining variable.

x-1/3(3x-9) = -3 since y=3x-9 substittute y.

x-1/3(3x-9)=-3

x + x +3 = -3

2x+3=-3

2x+3-3= -3-3

2x=-6

x=-3

substitute again into y=3x-9

y=3(-3)-9

y=-9-9

y=-18