SOLUTION: a man commutes to work a distance of 30 miles and teturns on the same route at the end of the day. His average rate on the return trip is 15 miles per hour faster than his average
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Question 557584: a man commutes to work a distance of 30 miles and teturns on the same route at the end of the day. His average rate on the return trip is 15 miles per hour faster than his average rate on the outgoing trip. write the total time, T, in hours, devoted to his outgoing an return trips as a function of his rate on the outgoing trip, x. The find a interpret T(25). I already have 30/x + 30/x+15
I am having a problem with the interpret Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! a man commutes to work a distance of 30 miles and teturns on the same route at the end of the day.
His average rate on the return trip is 15 miles per hour faster than his average rate on the outgoing trip. write the total time, T, in hours, devoted to his outgoing an return trips as a function of his rate on the outgoing trip, x.
The find a interpret T(25). I already have 30/x + 30/x+15
:
x = the speed
t(x) = time
t(x) =
replace x with 25
t(25) = +
t(25) = 1.2 + .75
t(25) = 1.95 hrs, round trip time, when the speed is 25 mph (outgoing), is the interpretation
