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zello Group Title

Nicole hiked up a mountain trail and camped overnight at the top. The next day she returned down the same trail. Her average rate traveling uphill was 2.6 kilometers per hour and her average rate downhill was 3.9 kilometers per hour. If she spent a total of 12 hours hiking, how long was the trail?

  • one year ago
  • one year ago

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  1. geoffb Group Title
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    You know the trail is the same length going up as it is going down. Therefore, her total distance travelled is \(2d\). \[ v = \frac{d}{t}\] \[t_{\text{up}} = 12 - t_{\text{down}}\] This should get you started.

    • one year ago
  2. zello Group Title
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    |dw:1354392763687:dw|

    • one year ago
  3. zello Group Title
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    I didn't get it =(

    • one year ago
  4. geoffb Group Title
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    Yeah, that looks right. Now you just need to make it into an equation.

    • one year ago
  5. geoffb Group Title
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    Well, if \(d_{total} = (vt)_{total}\), and \(d_{total} = d_{up} + d_{down}\), then \(d_{total} = (vt)_{up} + (vt)_{down}\)

    • one year ago
  6. zello Group Title
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    So, the answer is 36 ?

    • one year ago
  7. geoffb Group Title
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    Said another way: \[d_{up} = (vt)_{up}\] \[d_{down} = (vt)_{down}\] \[d_{up} + d_{down} = 2d = (vt)_{up} + (vt)_{down}\]

    • one year ago
  8. geoffb Group Title
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    I haven't calculated it, so I'm not sure. One sec.

    • one year ago
  9. zello Group Title
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    2.6 X +3.9(12-X)= the answer ?

    • one year ago
  10. geoffb Group Title
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    No, the answer is not 36.

    • one year ago
  11. zello Group Title
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    >.<

    • one year ago
  12. geoffb Group Title
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    Okay, let's start over. First, you know that the distance going up is the same as it is going down. Therefore, you can make \(d_{up}\) = \(d_{down}\).

    • one year ago
  13. geoffb Group Title
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    Let's call time \(t\) instead of \(x\).

    • one year ago
  14. geoffb Group Title
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    So, since \(d_{up} = d_{down}\), \((vt)_{up} = (vt)_{down}\)

    • one year ago
  15. zello Group Title
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    what is the (vt) ?

    • one year ago
  16. geoffb Group Title
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    \(vt\) is velocity times time.

    • one year ago
  17. geoffb Group Title
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    \(d = vt\)

    • one year ago
  18. geoffb Group Title
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    We're just plugging in \(vt\) for \(d\).

    • one year ago
  19. geoffb Group Title
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    Let's use the formula \((vt)_{up} = (vt)_{down}\). Can you plug in your values? Let's say that \(t\) represents time going up. Therefore, \(12 - t\) would be time going down.

    • one year ago
  20. geoffb Group Title
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    Also, remember that \((vt)_{up}\) is the same as saying \(v_{up} \times t_{up}\). Same goes for down.

    • one year ago
  21. zello Group Title
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    oh okay ,so you're saying that the Distance is the Rate*Time which is \[d=vt\] ,right?

    • one year ago
  22. geoffb Group Title
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    Yes, exactly.

    • one year ago
  23. geoffb Group Title
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    We can equate rate*time up to rate*time down, because they are both the same trail (i.e., same distance).

    • one year ago
  24. geoffb Group Title
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    That's what makes it possible to solve the equation.

    • one year ago
  25. zello Group Title
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    \[2.6x+3.9(12-x) \]

    • one year ago
  26. zello Group Title
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    ?

    • one year ago
  27. geoffb Group Title
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    You're adding them together. You need to make them equal to each other.

    • one year ago
  28. zello Group Title
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    Okay

    • one year ago
  29. zello Group Title
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    I think i got it

    • one year ago
  30. geoffb Group Title
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    Because \(2.6x = d = 3.9(12-x)\)

    • one year ago
  31. geoffb Group Title
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    Solve for \(x\) (time up the hill), then plug it into your original formula to solve for \(d\).

    • one year ago
  32. zello Group Title
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    \[2.6x=3.9(12-x) \] \[2.6x=46.8-3.9x \] \[46.8=6.5x \] \[x=7.2\]

    • one year ago
  33. geoffb Group Title
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    Good! So what does \(x\) represent?

    • one year ago
  34. zello Group Title
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    time up the hill ?

    • one year ago
  35. geoffb Group Title
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    Yes, exactly!

    • one year ago
  36. geoffb Group Title
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    So now you can plug it into your distance formula (remember, \(d = vt\)) and solve for \(d\).

    • one year ago
  37. geoffb Group Title
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    You also know that time down the hill is \(12 - x\), which you can calculate now to 4.8. You can plug *that* into your downhill formula, and you should get the same answer for \(d\). That will help prove your answer is right.

    • one year ago
  38. zello Group Title
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    \[d=vt\]\[3.9*4.8=18.72\] \[d=18.72\]

    • one year ago
  39. geoffb Group Title
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    Excellent. And, conversely, \(2.6 \times 7.2 = 18.72 \text{ km}\), so you know your answer is right.

    • one year ago
  40. zello Group Title
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    Thank you so much !

    • one year ago
  41. geoffb Group Title
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    No problem. :)

    • one year ago
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