Question

suqre root of 64 over squre root of 3 ?
my only choices are
8 square root of 3 over 3
8 over 3
8 over square root of 3
or square root of 3 over 3

i don't understand how to work it out ?

Answer 1

Hint:

\[\Large \frac{\sqrt{64}}{\sqrt{3}}\]


\[\Large \frac{\sqrt{64}*\sqrt{3}}{\sqrt{3}*\sqrt{3}}\]


\[\Large \frac{\sqrt{64*3}}{\sqrt{3*3}}\]

Answer 2

\[\sqrt{64 * 3}\] is 24 right ?

Answer 3

Oh I should have done it this way


\[\Large \frac{\sqrt{64}}{\sqrt{3}}\]


\[\Large \frac{8}{\sqrt{3}}\]


\[\Large \frac{8*\sqrt{3}}{\sqrt{3}*\sqrt{3}}\]


\[\Large \frac{8*\sqrt{3}}{\sqrt{3*3}}\]


Hopefully that's a bit clearer

The first way isn't wrong, it will just take more steps (than needed)

Answer 4

Okay, thank you.
is that the right answer ^

Answer 5

no,

\[\Large \sqrt{3*3} =\sqrt{9} = 3\]

this is all for the denominator

Answer 6

Oh, okay. So are possible answers ?