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    Graph of log10T plotted against log10d
    Gradient is -1

    Q: Theory that T=k/d
    Discuss whether the value of G supports this theory

    A : T=kd^-1
    and that the power of d is equal to the gradient.

    Problem is, I don't see how this answer supports the theory whatsoever.
    So if someone could shed some light on this?
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    If T=k/d

    then taking logs gives

    Log T = Log k - Log d

    A graph of Log T against log d would have a gradient of -1 with Log k as the intercept.


    (Log T) = (-1)(Log d) + (Log k)
    (Y) = (m) (x) + (c)
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    Thanks, that clears it up.
 
 
 
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Updated: April 17, 2013
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