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andriod09 Group Title

Joe was out on a hike in the wooks. He had a medical emergency (hangnail) and used his cellphone to call Dalene and ask her to come rescue him (i.e., bring her nail clipper to him). Joe with great difficulty was moving along toward Darlene at 2mph, whereas Darlene was goting to rescue him at 4mph. They were two miles apart when he made his call. How long until Joe is rescued?

  • one year ago
  • one year ago

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  1. andriod09 Group Title
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    @hartnn @cshalvey @ganeshie8

    • one year ago
  2. xBreakingBenjaminx Group Title
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    in the wooks?

    • one year ago
  3. andriod09 Group Title
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    woods

    • one year ago
  4. pratu043 Group Title
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    |dw:1347892389658:dw|

    • one year ago
  5. phi Group Title
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    There a few ways to solve this. One way is think like this: Joe is moving 2 mph toward Darlene. Darlene is moving 4 mph toward Joe You can add their speeds

    • one year ago
  6. pratu043 Group Title
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    So 6 mph. d = s*t t = d/s = 2/6 = 1/3 hrs is that right?

    • one year ago
  7. xBreakingBenjaminx Group Title
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    would it be 15min? because if 2mph = 1hr..then you add 2mph to make 4mph and you subtract 15min...then it would be 30min..then you add 2mph to 4mph to make 6mph and then subrtact 15min again...and you have..15 min..right? (wants to know...cause Benji is just guessing..) xD

    • one year ago
  8. phi Group Title
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    another way is use rate*time= distance we know they head toward each other, and both walk the same amount of time (from the end of the phone call until they meet). together they walk 2 miles. Joe walks rate*t or 2*t where t the time is unknown D walks 4*t together they walk 2*t+4*t= 2 simplify: 6t= 2 hours t= 2/6 hours or 1/3 hour 1/3 of an hour is also 60 mins/3 = 20 mins

    • one year ago
  9. xBreakingBenjaminx Group Title
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    CLOSE!! :D

    • one year ago
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