Ethan is deciding between two parking garages. Garage A charges an initial fee of $7 to park plus $4 per hour. Garage B charges an initial fee of $10 to park plus $3 per
hour. Let A represent the amount Garage A would charge if Ethan parks for t hours,
and let B represent the amount Garage B would charge if Ethan parks for t hours.
Write an equation for each situation, in terms of t, and determine the interval of
hours parked, t, for which Garage A is cheaper than Garage B.
A=
B =
Garage A is cheaper than Garage B when t

A= 4t + 7

B= 3t+10

Garage A is cheaper than Garage B when t < 3 hours.

Step-by-step explanation:

Garage A charges $7 just to enter the garage even for t=0 hours, and will charge and additional $4 per hour parked. Therefore the amount Garage A would charge for t hours of parking is: A= 4t + 7

Garage B charges 10 just to enter the garage even for t=0 hours, and will charge and additional $3 per hour parked. Therefore the amount Garage B would charge for t hours of parking is: B= 3t+10

To find the number of hours, t, that would make the parking garages cost the same, set the two equations equal to each other and solve for t:

4t + 7 = 3t + 10

-3          -3

t + 7 =  + 10

- 7          -7   t = 3

Since both equation values are equal when t=3, we know that the two garages would charge an equal amount when parking for 3 hours.

Since Garage A has less expensive initial fee ($7 vs. $10), Garage A will continue to be less expensive until it "catches up" to Garage B at t=3 hours

When t is less than 3 hours, Garage A is cheaper.

deez nu tz

step-by-step explanation: