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y varies directly as the square of x. When x = 5, y = 5. Find y when x = 10.
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PLEASE HELP ME
First, try to find 'k' \(\large y=kx^2\)
did you get 20?
yes? no?
Let's substitute x and y into our \(\large \begin{align} y&=kx^2\\ 5&=k*5^2 \\ 5 &= 25k\\ k& =5\div 25\\ k&=\frac{1}{5} \end{align}\)
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y = 10 y = 20 y = 50 y = 100 are my choices
im pretty sure im right
Okay, now you found k, Find y when x = 10. \(\large y=\frac{1}{5}10^2\) y=20 Yeah, you are right :)
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