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Suppose that y varies inversely as the square of x, and that y = 8 when x = 4. What is y when x = 3? Round your answer to two decimal places if necessary.

Respuesta :

If y varies inversely as the square of x you can say that:

[tex]x^2y=k[/tex]

Where k represents a constant.

For x=4 and y=8, then

[tex]\begin{gathered} (4)^28=k \\ k=128 \end{gathered}[/tex]

Then for x=3

[tex]\begin{gathered} x^2y=k \\ y=\frac{k}{x^2} \\ y=\frac{128}{(3)^2} \\ y=\frac{128}{9} \\ y=14.2 \end{gathered}[/tex]

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