SOLUTION: Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 mph and starts 3 hours after the first cyclist who is traveling at 6 mph. How much ti
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Question 307256: Two cyclists start biking from a trail's start 3 hours apart. The second cyclist travels at 10 mph and starts 3 hours after the first cyclist who is traveling at 6 mph. How much time will pass before the second cyclist catches up with the first from the time the second cyclist started biking?
Not sure if this is Combinatorics, but help is appreciated :) Answer by checkley77(12844) (Show Source):
You can put this solution on YOUR website! D=RT
D=6T
D=10(T-3)
THE 2 DISTANCES ARE EQUAL.
6T=10(T-3)
6T=10T-30
6T-10T=-30
-4T=-30
T=-30/-4
T=7.5 HOURS.
PROOF:
6*7.5=10(7.5-3)
45=1-*4.5
45=45
7.5-3=4.5 HOURS AFTER THE SECOND RIDER LEAVES THEY WILL MEET.
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