Question

\[3 \div 3\sqrt{9x}\]

it is supposed to look like a fraction

Answer 1

\[3/3\sqrt{9x}\]
\[\sqrt{9x}/9x\] rationalize the denominator

Answer 2

did you multiply to get that?

Answer 3

no, I put the 3 over the 3sqrt(9x)

Answer 4

that is what it is supposed to look like the 3/a cubed sq root 9x

Answer 5

a 3 and a cubed sq root of 9x, i cant type either tonight :(

Answer 6

\[\Large\frac{3}{\sqrt[3]{9x}}\]

Answer 7

how do you make it look like that ???

Answer 8

If you right click on it in my reply you should be able to click Show Math As > TeX Commands, and it will show you the code I used. Just put that between \ [ and \ ] without the spaces.

Answer 9

i so need to figure out how to do these, rationalize the denominator in simplified form,

Answer 10

Alright well you have \(3\cdot(9x)^{-1/3}\), if you multiply that by \((9x)^{2/3}(9x)^{-2/3}\) (which you can do because that is equal to one), you'll get \(3\cdot(9x)^{2/3}\cdot(9x)^{-1}\), which is \(\dfrac{3\sqrt[3]{(9x)^2}}{9x}\), or \(\dfrac{\sqrt[3]{(9x)^2}}{3x}\) if you simplify it further.

Answer 11

im just trying to figure out how you got the 2/3 so i know how to do that part