If y varies inversely as the square of x and y=5 when x=2.5, what is the value of y when x=9?

Question

whpalmer4 · Answer

If $y$ varies inversely with the square of $x$, that means we can write
$$y = \frac{k}{x^2}$$ or $$k=x^2y$$

whpalmer4 · Answer

From that, you should have no trouble finding the value of $k$ and finding the desired value of $y$ when $x=9$

anonymous · Answer

thank you

anonymous · Answer

i got 4.5 as my answer

anonymous · Answer

y=4.5

whpalmer4 · Answer

hmm...not what I get.  Want to show your work?

whpalmer4 · Answer

$$k = x^2y$$$$k = (2.5)^2* 5 = 6.25*5 = 31.25$$

So the variation equation is $$31.25 = x^2y$$or$$y=\frac{31.25}{x^2}$$What do you get for $y$ when you plug in $x=9$ in that equation?

anonymous · Answer

i was walking with my dad but thank you i was meaning to ask you the next step after plugging in x and y but now i got it

whpalmer4 · Answer

What do you get for your new, improved answer?

anonymous · Answer

9^2 is 81  $$31.25\div81$$ = 0.39

anonymous · Answer

i rounded it

whpalmer4 · Answer

Yes, much better.

anonymous · Answer

thank you for your help

whpalmer4 · Answer

you're welcome!