SOLUTION: Michael drove to a friend`s house a rate of 40 mph. he came back by the same route, but at a rate of 45 mph. if the round-trip took 4 hours what is the distance to his friend`s hou
Algebra.Com
Question 323432: Michael drove to a friend`s house a rate of 40 mph. he came back by the same route, but at a rate of 45 mph. if the round-trip took 4 hours what is the distance to his friend`s house?
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
D=RT
D=40T FIRST TRIP
D=45(4-T) RETURN TRIP
THE DISTANCES ARE THE SAME TEREFORE THE EQUATIONS ARE THE SAME.
40T=45(4-T)
40T=180-45T
40t+45T=180
85T=180
T=180/85
T=2.118 HRS. FOR THE FIRST TRIP.
4-2.118=1.88 HRS. FOR THE RETURN TRIP.
PROOF:
40*2.118=45*1.88
84.7~84.7
2*84.7=169.4 MILES IS THE DISTANCE OF THE ROUND TRIP.
RELATED QUESTIONS
Michael drove to a friend's house at a rate of 40 mph. He returned by the same route at a (answered by Alan3354)
michaek drove to a friends house at a rate of 40 mph. he returned the same route at a... (answered by ankor@dixie-net.com)
Michael drove to a friends house at a rate of 40 mph. He returned by the same route at a... (answered by ramkikk66,josmiceli)
Che drove to Due's house at a rate of 40 mph. He returned by the same route at a rate of... (answered by checkley77)
Denise drove to her parents' house at a rate of 70 mph. She came back by the same route,... (answered by richwmiller)
I am stumped by this question:
Michael drove to a friend's house at a rate of 40 mph. (answered by Alan3354,chessace)
Michael drove to a friend's house at a rate of 40mph. He returned by the same route at a... (answered by checkley77,richwmiller)
Michael drove to a friend's house at a rate of 40 mi/h. He returned by the same route at... (answered by Alan3354)
The problem is: Michael drove to a friend's house at a rate of 40mi/h. He returned by the (answered by Paul)