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Math Help - Inverse Variation

  1. #1
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    Inverse Variation

    The gravitational attraction between two masses varies inversely as the square of the distance between them. The force of attraction is 4 lbs. when the masses are 3 ft apart, what is the attraction when the masses are 6 ft. apart?

    My Work:

    Let g = gravitational attraction force in terms of pounds between two masses.

    My equation is: g = k/d^2, where k is the constant of proportionality and d is the distance squared.

    I know that g = 4.
    I also know that 3 = d.
    I first need to find k.
    4 = k/(3)^2
    4 = k/9
    4*9 = k
    36 = k
    Now that I know k, I need to find g.
    g = 36/(6)^2
    g = 36/36
    g = 1 pound.


    Is this correct?


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  2. #2
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    Quote Originally Posted by sharkman View Post
    The gravitational attraction between two masses varies inversely as the square of the distance between them. The force of attraction is 4 lbs. when the masses are 3 ft apart, what is the attraction when the masses are 6 ft. apart?

    My Work:

    Let g = gravitational attraction force in terms of pounds between two masses.

    My equation is: g = k/d^2, where k is the constant of proportionality and d is the distance squared.

    I know that g = 4.
    I also know that 3 = d.
    I first need to find k.
    4 = k/(3)^2
    4 = k/9
    4*9 = k
    36 = k
    Now that I know k, I need to find g.
    g = 36/(6)^2
    g = 36/36
    g = 1 pound.

    Is this correct?

    correct
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