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Math Help - Another table of values question, use inverse proportion and find k?

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    Another table of values question, use inverse proportion and find k?

    So I have:
    10 100
    40 6
    70 2
    100 1

    That is inverse square correct?

    Now, how would I 'model that using the correct inverse proportion' and find the value of k using 10 100
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  2. #2
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    Quote Originally Posted by olen12 View Post
    So I have:
    10 100
    40 6
    70 2
    100 1

    That is inverse square correct?

    Now, how would I 'model that using the correct inverse proportion' and find the value of k using 10 100
    An inverse proportion is of the form y= k/x or xy= k. If that were correct we would have k= 10(100)= 1000 but 40(6)= 240, not 1000. If it were y= k/x^2 or x^2y= k then, depending on which number is x, we have either [tex]k= (10)(100^2)= 100000[tex] or k= (10^2)(100)= 10000. In the first case, (40)(6^2)= 1440, which is not right, and in the second, (40^2)(6)= 9600, also not right. So it is not "inverse square". If you were to try a generic y^n= k/x^m or x^my^n= k, we could write (10^m)(100^n)= k and (40^m)(6^n)= k. Dividing one equation by the other [tex]\left(\frac{10}{40}\right)^m\left(\frac{100}{6}\ri ght)^n= 1[tex] or \left(\frac{1}{4}\right)^m\left(\frac{50}{3}\right  )^n. Especially easy is the last pair: (100^m)(1^n)= 100^m= k so that \left(\frac{100}{10}\right)^m\left(\frac{1}{100}\r  ight)^n= 1. That gives you two equations to solve for m and n.
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