anonymous
  • anonymous
if you travel to work at 30mph and go home by 40 mph, what is the average rate for the entire trip?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I am totally lost, because there is no distance or time given
anonymous
  • anonymous
are you there? I appreciate the help
anonymous
  • anonymous
That's all the problem gives you?

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anonymous
  • anonymous
yes, it is why I am so frustated
regan
  • regan
(30mph*t1+40mph*t2)/(t1+t2) which reads total distance over total time thats all i can think of
anonymous
  • anonymous
yeah, regan's answer looks right
anonymous
  • anonymous
if you aren't given time or distance
regan
  • regan
and 30mph*t1=40mph*t2 this may also help
regan
  • regan
30mph*t1*2/t1+30mph*t1/40mph
regan
  • regan
t1(30*2) numerator t1(1+3/4) denominator
regan
  • regan
34.29 mph : )
anonymous
  • anonymous
how'd you get that lol :)
regan
  • regan
ok using 30mph*t1=40mph*t2 i said 30mph*t1+30mph*t1=30mph*t1+40mph*t2 in the denominator and t1+t2=t1+(30/40)*t1 in the numerator so we have (t1(30*2))/(t1(1+3/4)) numerator/denominator however they both have been written in terms of t1 therefore can cancel to get 30*2/1.75 which equals 34.29
regan
  • regan
does that make sense?
anonymous
  • anonymous
yeah it does , thank you so much

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