An aircraft carrier traveled to dry dock an back. The trip there took 2 hours and the trip back took 3 hours. It averaged 5 miles per hour faster on the trip there than on the return trip. Find the aircraft carrier's average speed on the outbound trip

Step-by-step explanation:

Average speed = distance/time

Trip to dry dock

Let x be the velocity

Time = 2hours

Distance = average speed ×time

Distance = x×2

Distance = 2x

Return trip

Velocity = x-5 (5 miles per hour faster on the trip there than on the return trip)

Time = 3hrs

Distance = 3(x-5)

Since the distance is the same both to and fro;

2x = 3(x-5)

2x = 3x -15

2x-3x = -15

-x = 15

x = 15

Hence the aircraft carrier's average speed on the outbound trip is 15miles/hour

its definatley   this on your answer choice?

step-by-step explanation:

hjkj