Atour bus normally leaves for its destination at 5: 00p. m. for a 378 mile trip. this week however, the bus
leaves at 6: 00 p. m. to arrive on time, the driver drives 9 miles per hour faster than normal what is the normal
speed of the bus?

54 mph

Step-by-step explanation:

we know D = RT, where

D is distance

R is rate/speed

T is time

Normally:

D = RT

378 = RT

This week:

D = RT

378 = (R+9)(T-1)

These are the 2 equations we can write:

Let's mutliply out the last equation:

378 = (R+9)(T-1)

378 = RT - R + 9T - 9

387 = RT - R + 9T

we know RT = 378 and T = 378/R, we replace:

387 = 378 - R + 9(378/R)

9 = -R + 3402/R

Now solving:

R can't be negative, so we take R = 54 miles per hour

the two may have that, even a breakdown is probably offered