To balance a seesaw, the distance a person is from the fulcrum is inversely proportional to his or her weight . Roger , who weighs 120 pound

Question

To balance a seesaw, the distance a person is from the fulcrum is inversely proportional to his or her weight . Roger , who weighs 120 pounds is sitting 6 feet from the fulcrum . Ellen weighs 108 pounds . How far from the fulcrum must she sit to balance the seesaw ? Round to the nearest hundredth of a foot .

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Thanh Hà 1 year 2021-08-14T15:13:58+00:00 1 Answers 56 views 0

Answers ( )

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    2021-08-14T15:15:12+00:00

    Answer:

    d_e =6.67ft

    Step-by-step explanation:

    From the question we are told that:

    Weight of Roger W_r=120\ pounds

    Distance of Roger from fulcrum d_r=6 ft

    Weight of Ellen W_e=120\ pounds

    Generally the equation for distance-weight relationship is mathematically given by

      d\alpha \frac{1}{W}

      \frac{d_1}{d_2} =\frac{W_2}{W_1}

      \frac{d_r}{d_e} =\frac{W_e}{W_r}

    Therefore

     \frac{d_e}{d_r} =\frac{W_r}{W_e}

     d_e =\frac{W_r*d_r}{W_e}

     d_e =\frac{6*120}{108}

     d_e =6.67ft

    Therefore the distance from the fulcrum she must sit to balance the seesaw is given as

    d_e =6.67ft

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