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Math Help - New Inverse Variation (check this please)

  1. #1
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    New Inverse Variation (check this please)

    When a seesaw is perfectly balanced, each person’s distance, “d,” from the fulcrum varies inversely as his weight, “w.” If a 120-pound person sits 5 feet from the seesaw fulcrum, where must a 150-pound person sit to balance the seesaw?

    w = k/d

    120 = k/5

    120*5 = k

    600 = k

    ... find d when w = 150...

    d = 600/w
    d = 600/150
    d= 4

    Thanks a lot!
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    Quote Originally Posted by bobbyboy1111 View Post
    When a seesaw is perfectly balanced, each person’s distance, “d,” from the fulcrum varies inversely as his weight, “w.” If a 120-pound person sits 5 feet from the seesaw fulcrum, where must a 150-pound person sit to balance the seesaw?

    w = k/d <=== This is incorrect.

    120 = k/5

    120*5 = k

    600 = k

    ... find d when w = 150...

    d = 600/w
    d = 600/150
    d= 4


    Thanks a lot!
    Hi bobbyboy1111,

    Your original equation is incorrect. However, it didn't affect the value of k.

    It should have been d=\frac{k}{w}

    I noticed that you fixed it when you solved for the distance a 150 lb person needs to sit from the fulcrum. Your answer is correct.
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