SOLUTION: I am unsure if I am submitted this question to right area but here goes. Ot os listed as Aplications: Mixture and Unform Motion Problems.
A student rode a bycycle to the re
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-> SOLUTION: I am unsure if I am submitted this question to right area but here goes. Ot os listed as Aplications: Mixture and Unform Motion Problems.
A student rode a bycycle to the re
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Question 258695: I am unsure if I am submitted this question to right area but here goes. Ot os listed as Aplications: Mixture and Unform Motion Problems.
A student rode a bycycle to the repair shop then walked home. The student averaged 14mph ridign to the shop and 3.5mph walking home. The route took 1 hour. How far is it from the student's home to the repair shop. I am unsure how to even set up the equation. All I do know is he rides 4x as fast as he walks. Answer by Fombitz(32388) (Show Source):
You can put this solution on YOUR website! Rate*Time=Distance
The distance is the same going to and coming from the shop.
It takes t1 hrs to get to the shop by bike and it takes t2 hrs to get to the shop on foot.
Since they both equal d, they equal each other.
1.
2.
Now you have 2 equations with 2 unknowns.
Please re-post if you need more help.
