At 9:00 a. m. a train left the station on a round trip that covered 54 miles. It returned after a 30-minute layover. The train averaged 18 miles per hour on the outgoing trip. It averaged 50% faster on the return trip. What time did it return to the station

12pm

Step-by-step explanation:

54 miles in total. I thought of a straight line & the train is travelling to the end & back & that total miles is 54 .

so we divide 54 by 2 to find out how miles is it to go & to come back

54÷2 =27

27 miles to go & 27 miles to come back.

we know on the outgoing trip the train travelled 18 mph.

theres 27 miles

27÷18=1.5

it took 1.5 hours to travel 27 miles or one hour 30 mins .

on the return trip the train travelled 50 % faster .

50 percent of 18mph =9

9+18=27 mph

the return train is travelling 27mph & we know it has to travel only 27 miles

27÷27 =1

the return train would have taken an hour.

but they said there was a 30 minute delay so we have to factor all this in .

out going train 1 hour 30 mins , Return train 1 hour, Delay 30 mins

1 hour + 1 hour +30mins +30 mins =3 hours .

The train left 9 am & returned after 3 hours which is 12pm

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.

the correct answer is 1: 2

step-by-step explanation:

hope this your welcome : )