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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 .

Sagot :

okpalawalter8

Answer:

[tex]d_e =6.67ft[/tex]

Step-by-step explanation:

From the question we are told that:

Weight of Roger [tex]W_r=120\ pounds[/tex]

Distance of Roger from fulcrum [tex]d_r=6 ft[/tex]

Weight of Ellen [tex]W_e=120\ pounds[/tex]

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

  [tex]d\alpha \frac{1}{W}[/tex]

  [tex]\frac{d_1}{d_2} =\frac{W_2}{W_1}[/tex]

  [tex]\frac{d_r}{d_e} =\frac{W_e}{W_r}[/tex]

Therefore

 [tex]\frac{d_e}{d_r} =\frac{W_r}{W_e}[/tex]

 [tex]d_e =\frac{W_r*d_r}{W_e}[/tex]

 [tex]d_e =\frac{6*120}{108}[/tex]

 [tex]d_e =6.67ft[/tex]

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

[tex]d_e =6.67ft[/tex]

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