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anonymous

  • one year ago

PLEASE HELP The current in a nearby river is 5 mph. John wants to be able to travel up the river for a distance of 30 miles and return to his starting point in a total of 8 hours. How fast must John be able to row his boat in order to accomplish this? (Assume that the speed that John can row refers to how fast he can row in still water.)

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  1. anonymous
    • one year ago
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    @dan815 @nincompoop

  2. mathstudent55
    • one year ago
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    He travels a total time of 8 hours. If he travels up the river for t time, how much time will he travel down the river?

  3. anonymous
    • one year ago
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    i have no idea!! please explain

  4. anonymous
    • one year ago
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    @mathstudent55

  5. mathstudent55
    • one year ago
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    Ok. The total trip is 8 hours. time going up + time going down = 8 hours If you let the time going up be called just t, then you have t + time going down = 8 hours Ok so far?

  6. anonymous
    • one year ago
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    ok

  7. mathstudent55
    • one year ago
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    Now let's see if we can get an expression for the time going down. t + time going down = 8 Subtract t from both sides: time going down = 8 - t Now we have: Time going up = t Time going down = 8 - t Ok?

  8. anonymous
    • one year ago
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    ok

  9. mathstudent55
    • one year ago
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    Ok. I put the info we have in the table below. Now we need the speed going up and the speed going down, and the distances going up and down. |dw:1436391156182:dw|

  10. mathstudent55
    • one year ago
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    Let's do the distances first bec they are very easy.

  11. mathstudent55
    • one year ago
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    How far does he row each way?

  12. anonymous
    • one year ago
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    like how am i supposed to know im so confused on this question

  13. mathstudent55
    • one year ago
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    Have you read the problem? It's written right there. There is no calculation needed.

  14. mathstudent55
    • one year ago
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    Does this help? \(\sf John ~wants ~to ~be ~able ~to ~travel ~up ~the ~river ~for ~a ~distance ~of ~\huge \color{red}{30 ~miles}\) \( \sf and ~return ~to ~his ~starting ~point ~in ~a ~total ~of ~8 ~hours. \)

  15. anonymous
    • one year ago
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    so for up its 30 ?

  16. mathstudent55
    • one year ago
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    Right, Each part of the trip is 30 miles. We can add the distances to our table.

  17. anonymous
    • one year ago
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    ok

  18. mathstudent55
    • one year ago
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    |dw:1436391741905:dw|

  19. mathstudent55
    • one year ago
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    Now we need to work on the speed going up and the speed going down. His speed in still water is an unknown. Let's call it s. When he goes up the river, does the 5 mph speed of the river speed him up or slow him down?

  20. anonymous
    • one year ago
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    slows him down?

  21. mathstudent55
    • one year ago
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    Right. That means it takes away from his speed. If the speed in still water is s, then going up the river against the 5-mph current, his speed will be s - 5 By the same token, going down the river, the 5-mph current helps him, so his speed will be s + 5 down the river. Now we add the speeds to our table.

  22. mathstudent55
    • one year ago
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    |dw:1436392136839:dw|

  23. anonymous
    • one year ago
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    ok now that we have the info how would i answer the question?

  24. mathstudent55
    • one year ago
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    Now that we have all the info we need, we can solve the problem. speed = distance/time That means distance = speed * time

  25. mathstudent55
    • one year ago
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    We write an equation for going up the river and an equation for going down the river.

  26. mathstudent55
    • one year ago
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    Up the river distance = speed * time 30 = (s - 5)t Down the river distance = speed * time 30 = (s + 5)(8 - t)

  27. anonymous
    • one year ago
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    s-5 = 30 /t for up

  28. anonymous
    • one year ago
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    oh nvm ur right

  29. mathstudent55
    • one year ago
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    Now we have a system of equations that we need to solve: 30 = (s - 5)t 30 = (s + 5)(8 - t)

  30. mathstudent55
    • one year ago
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    Solve the first equation for t and substitute in the second one: \(t = \dfrac{30}{s - 5} \) \(30 = (s + 5)(8 - \dfrac{30}{s - 5}) \) Now we have an equation in only s, so we can solve for s.

  31. mathstudent55
    • one year ago
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    Multiply both sides by s - 5 to get rid of the denominator: \(30(s - 5)= (s + 5)(s - 5)(8 - \dfrac{30}{s - 5}) \) \(30s - 150 = (s + 5)(8s - 40 - 30)\) \(30s - 150 = (s + 5)(8s - 70) \) \(30s - 150 = 8s^2 - 70s + 40s - 350\) \(30s - 150 = 8s^2 - 30s - 350 \) \(8s^2 - 60s - 200 = 0\) \(2s^2 - 15s - 50 = 0\) \((2s + 5)(s - 10) = 0\) \(s = -\dfrac{5}{2} \) or \(s = 10\) We discard the negative speed, -5/2, and the answer is the speed is 10 mph.

  32. mathstudent55
    • one year ago
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    Let's see if the answer makes sense. The speed in still water is 10 mph The speed going up is 5 mph The speed going down is 15 mph The time going up is 30/5 = 6, 6 hours The time going down is 30/15 = 2, 2 hours 6 hours + 2 hours = 8 hours. Yes, 10 mph is the correct speed he needs to row in still water.

  33. anonymous
    • one year ago
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    thank you so much!

  34. mathstudent55
    • one year ago
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    You're welcome.

  35. anonymous
    • one year ago
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    could you help me with this as well?

  36. anonymous
    • one year ago
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    @mathstudent55

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