An aircraft carrier made a trip. the trip there took 5 hours and the trip back took 6 hours. it averaged 3 mph faster on the trip there then on the return trip. find the aircraft carriers speed on the outbound trip.

The outbound trip is 18 miles per hour

Step-by-step explanation:

Let x represent the speed of the plane.

An aircraft carrier made a trip. Let us assume that this trip was outbound. The trip there took 5 hours.

Distance = speed × time. Therefore

Distance = 5x

The trip back took 6 hours. It averaged 3 mph faster on the trip there then on the return trip. This means that the speed on the trip back is x - 3 mph.

Distance = 6(x-3) = 6x - 18

Since the distance is the same,

5x = 6x - 18

6x - 5x = 18

x = 18

The speed on return or inbound trip would be 18 - 3 = 15 mph

answer: 40392

step-by-step explanation:

10098 x 4 = 40392

answer:   expression= p+p*d%

price reduction + sale price = $625

sale price: $p+$p*d%

where d% is the percentage of discount

step-by-step explanation:

original price = p

price went down by $150 price reduction $p*d%

sale price= $ 475

therefore the price before sale is: $625

and the expression that shows the price before sale is: p+p*d%