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    The function h is given by h(x) = k/x^n where k and n are positive integers. The points (1,100) and (2,25) lie on the graph.

    I) determine values k and n
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    (Original post by Custardcream000)
    The function h is given by h(x) = k/x^n where k and n are positive integers. The points (1,100) and (2,25) lie on the graph.

    I) determine values k and n
    Well, you know that \displaystyle \frac{k}{1^n} = 100 and \displaystyle \frac{k}{2^n} = 25

    What can you do with this information?
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    (Original post by Zacken)
    Well, you know that \displaystyle \frac{k}{1^n} = 100 and \displaystyle \frac{k}{2^n} = 25

    What can you do with this information?
    Ahh I get it do you sub in one of the values of what k=
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    (Original post by Custardcream000)
    Ahh I get it do you sub in one of the values of what k=
    You have two equations in two variables. Simultaneous equations.

    Although this one is very easy, the first equation should give you k directly and substituting it into the second gives you an equation which you can solve for n.
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    (Original post by Zacken)
    You have two equations in two variables. Simultaneous equations.

    Although this one is very easy, the first equation should give you k directly and substituting it into the second gives you an equation which you can solve for n.
    K= 100
    And n = 2 but I don't know how I show that
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    (Original post by Custardcream000)
    K= 100
    And n = 2 but I don't know how I show that
    If, \frac{k}{1^n} = 100 and we know that 1^anything = 1 then what can you say about k?

    Then, re-arrange the second equation: \frac{k}{25} = 2^n, take logs of both side or spot n by inspection.
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    (Original post by Zacken)
    If, \frac{k}{1^n} = 100 and we know that 1^anything = 1 then what can you say about k?

    Then, re-arrange the second equation: \frac{k}{25} = 2^n, take logs of both side or spot n by inspection.
    Thanks for your help, I understand it now
 
 
 
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