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- anonymous

1/2 / (Square root of 3)/2

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- anonymous

1/2 / (Square root of 3)/2

- schrodinger

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- anonymous

Like this?
\[\Large \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\]

- anonymous

Yeah!

- anonymous

Okay...
\[\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}\]

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- anonymous

I keep getting the answer 1/squr root 3

- anonymous

and its supposed to be 3/ sqr root 3

- anonymous

So you can re-write it as:
\[\frac{1}{2} \div \frac{\sqrt{3}}{2}\]
which equals
\[\frac{1}{2} \times \frac{2}{\sqrt{3}}\]

- anonymous

1/sqrt 3 would be correct, except you have to get rid of the squareroot in the denominator.

- anonymous

|dw:1354415263802:dw|

- anonymous

Why cant it be in the denominator?

- anonymous

The answer is not \(\Large \frac{3}{\sqrt{3}}\).

- anonymous

Oops squr root 3/3 right?

- anonymous

There's nothing wrong with leaving it as \(\Large \frac{1}{\sqrt{3}}\), although if you understand negative exponents, there's a neater way you could write it.

- anonymous

Is it \[\frac{ \sqrt{3} }{ 3 }\]?

- anonymous

No. You were right with your answer. I was responding to your post that the answer is "supposed to be" 3/sqrt 3.

- anonymous

How though

- anonymous

It's not. You're right with 1/sqrt 3.

- anonymous

You're the one that said it's "supposed to be."

- anonymous

|dw:1354415644279:dw|

- anonymous

how is square root of 3 the square root of 3 =3? i thought it is 1

- anonymous

\[\sqrt{3} \times \sqrt{3} = 3\]
I think he was trying to show how you could get the answer \( \Large \frac{\sqrt{3}}{3}\).

- anonymous

|dw:1354415812944:dw|

- anonymous

He's showing how to get rid of the radical. \(\Large \frac{1}{\sqrt{3}}\) is the same as \(\Large \frac{\sqrt{3}}{3}\).

- anonymous

ok thanks!

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