# Thread: Dividing/Multiplying fractions+square roots...

1. ## Dividing/Multiplying fractions+square roots...

Is this correct?

sq. root of 2 / 2 = 40 / sq. root of 3/2
= 80 / sq. root of 6

( refer to attachment for better understanding )

I don't know if that is proper multiplication/division?

Thanks.

2. Originally Posted by advancedfunctions2010
Is this correct?

sq. root of 2 / 2 = 40 / sq. root of 3/2
= 80 / sq. root of 6

( refer to attachment for better understanding )

I don't know if that is proper multiplication/division?

Thanks.
$\frac{\sqrt 2}{2} \neq \frac{40}{\sqrt {\frac{3}{2}}}$ Not even close. How did you get this?

But $\frac{40}{\sqrt{\frac{3}{2}}} = \frac{80}{\sqrt 6}$

3. Im not sure.

This all relates to my previous post, regarding the 2 right angle triangles..

How did you get this?
$
\frac{40}{\sqrt{\frac{3}{2}}} = \frac{80}{\sqrt 6}
$

4. Originally Posted by advancedfunctions2010
Im not sure.

This all relates to my previous post, regarding the 2 right angle triangles..

How did you get this?
$
\frac{40}{\sqrt{\frac{3}{2}}} = \frac{80}{\sqrt 6}
$
$\frac{40}{\sqrt{\frac{3}{2}}} = \frac{40 \times 2}{\frac{\sqrt 3}{\sqrt 2}\times 2}= \frac{40 \times 2}{\frac{\sqrt 3}{\sqrt 2}\times (\sqrt 2 \times \sqrt 2)}= \frac{80}{{\sqrt 3}\times{\sqrt 2}}=\frac{80}{\sqrt 6}$

5. thank you so much