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If y varies directly as x^2 and y=12 when x=2, find y when x=5

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Kate200468id
[tex]y=ax^2\\\\y=12\ \ \ and\ \ \ x=2\ \ \ \Rightarrow\ \ \ 12=a\cdot 2^2\ \ \ \Rightarrow\ \ \ a= \frac{4}{12}= \frac{1}{3} \\\\y=\frac{1}{3} x^2\\\\x=5\ \ \ \Rightarrow\ \ \ y=\frac{1}{3} \cdot5^2=\frac{25}{3} =8\frac{1}{3} [/tex]

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