Arocket is dropping to the moons surface at a constant rate of 15 feet per second. after 2 minutes of descent the rocket is at a height of 11,400 feet above the moons surface. at what elevation in feet was the rocket when it started to drop to the moons surface?

13,200 feet

Step-by-step explanation:

So, the rocket starts descending at a given rate and time, so we first need to find the length it passed during that descent.

Length equals to speed multiplied with time:

s = v • t

since speed is given in feet per second, we need to convert minutes to seconds; if one minute has 60 seconds, then two minutes have 120 seconds.

s = 15 feet/s • 120 s

s = 1800 feet

So, the rocket descended 1800 feet, but it still has 11,400 feet to the Moon's surface. That means that the rocket was on 11,400 + 1,800 = 13,200 feet elevation when it started to drop.

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8