Question

sqrt{(2*3^2) * 2^1/2}/ 3^-2 ? answer is 54 but how do you get it?

Answer 1

like this?:
\[\frac{\sqrt{2*3^{2}*2^{\frac{1}{2}}}}{3^{-2}}\]

Answer 2

only 2* 3^2 is under the square root

Answer 3

or maybe like this:
\[\frac{\sqrt{2*3^{2}}*2^{\frac{1}{2}}}{3^{-2}}\]

Answer 4

ah yes yes. ok first lets change that 2^(1/2):
\[2^{\frac{1}{2}} = \sqrt{2}\]
Now we can stick it inside the square root with all the other numbers:
\[\sqrt{2*3^{2}}*\sqrt{2} = \sqrt{2*3^{2}*2} = \sqrt{2^{2}*3^{2}}\]

Answer 5

this can be simplified a little further:
\[\sqrt{2^{2}*3^{2}} = \sqrt{2^{2}}*\sqrt{3^{2}} = 2 * 3 = 6\]
so the top of our fraction turns out to be 6

Answer 6

so over all we have:
\[\frac{6}{3^{-2}} = 6*3^{2} = 6*9 = 54\]

Answer 7

i got 9 on the top but now i will look over
what you did so i know how its done.
thank you so much, i don''t know how to write math like that!

Answer 8

if there is anything that isnt clear just let me know :)