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anonymous

  • 5 years ago

mr. Duncan traveled to a city 180 miles from him home to attend a meeting. Due to car trouble, his average speed returning was 13 miles less than his speed going. If the total time for the road trip was 7 hours, at what rate of speed did he travel to the city? (round to nearest tenth)

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  1. anonymous
    • 5 years ago
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    It was 7 hours total round trip?

  2. anonymous
    • 5 years ago
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    yes

  3. heisenberg
    • 5 years ago
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    here is a guess:

  4. heisenberg
    • 5 years ago
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    \[\frac{miles}{hour} * hour = miles\] so\[hour = \frac{miles}{\frac{miles}{hour}}\]

  5. heisenberg
    • 5 years ago
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    so the total time is that for each leg of the journey:

  6. heisenberg
    • 5 years ago
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    \[7 = \frac{180}{x} + \frac{180}{x - 13}\]

  7. heisenberg
    • 5 years ago
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    solve for x. not sure if this makes sense, even to me :)

  8. heisenberg
    • 5 years ago
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    a better way to restate what i first said: \[average speed * time = distance\]

  9. anonymous
    • 5 years ago
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    okk. So i got 30420 when I did that. ha. I'm sure what you did makes sense. I just am lost with this problem a bit. Did you get a solution at all?

  10. heisenberg
    • 5 years ago
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    hmm you may have made a mistake with your numbers. let me try and step through it. i am checking my answer here: http://www.wolframalpha.com/input/?i=7+%3D+180%2Fx+%2B+180%2F(x-13) it gives approx 59mph. also gives approx 6mph as well, so i may be wrong.

  11. anonymous
    • 5 years ago
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    ohh hmm. I wonder which one it is. According to where you checked it the answer is 59 mph?

  12. heisenberg
    • 5 years ago
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    there are actually two solutions listed there, which seems indicative of something wrong. but I'm not too sure about that. I would guess 59 because if it was 6mph, then on the way home he would have been going 6-13 mph which is impossible.

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