Acargo plane flew roundtrip from smithtown to johnsville. it took one hour less time to get there than it did to get back. the average speed on the trip was 280 mph. the average speed on the way back was 240 mph. how many hours did the trip take?

13 hours

Step-by-step explanation:

Average speed = Distance traveled / time taken

                   ⇒ Distance = Average speed × Time

                                     d = s × t

For the first trip;

Average speed = 280 mph

                          d₁ = 280t₁                    (1)

where;

d₁ is the distance covered to get to the destination

t₁ is the time taken to get to the destination

For the second trip;

Average speed = 240 mph

                           d₂= 240t₂                     (2)

where;

d₂ is the distance covered on the way back

t₂ is the time taken on the way back

The trip is the same distance to and fro. Therefore,

                           d₁ = d₂                              

Substituting the equation for d₁ and d₂

                          280t₁ = 240t₂              (3)

It took one hour less time to get there than it did to get back, then,

                                 t₁ = t₂ - 1

                                 t₂ = t₁ + 1                (4)

Substituting equation (4) into  equation (3)

                               280t₁ = 240(t₁ + 1)

                               280t₁ = 240t₁ + 240

                               280t₁ - 240t₁ = 240

                               40t₁ = 240

                                   t₁ = 240/40

                                   t₁ = 6 hours

From equation (4)

                                   t₂ = t₁ + 1

                                   t₂ = 6 + 1  

                                   t₂ = 7 hours

The total time for the trip is t₁ + t₂ = 6 + 7

                                       = 13 hours    

answer: (1,3) (3,-1) (5,-5) ?

step-by-step explanation:

i can you with this answer

step-by-step explanation: