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Given that y varies inversely as x^2, and that y=16 when x=3, what is the value of y...

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hellook | eNoter

Posted April 10, 2012 at 4:56 PM via web

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Given that y varies inversely as x^2, and that y=16 when x=3, what is the value of y when x=4?

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mlsiasebs | College Teacher | (Level 1) Associate Educator

Posted April 10, 2012 at 5:32 PM (Answer #1)

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If y is inversely proportional to `x^(2)` then we know that

y (is proportional to) `1/(x^2)`

Since we are given a value for x and y, we can find the value for the numerator so that we are no longer looking at a proportionality but an equation

`y=z/(x^2)```

Now, we can plug in the known values

`16 = z/(3^2)`

`16 = z/9`

z = 16*9

z = 144

Now that we know the value of z, we can use the same equation to solve for the unknown value of y

`y=z/(x^2)`

`y=144/(4^2)`

`y=144/16`

`y = 9`

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