# Math Help - really easy cancel I can't do!

1. ## really easy cancel I can't do!

Im not sure how to times through by 3/3 on the bottom

(2/3) / (1+(2/3)^2)^0.5

trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated

2. Originally Posted by i_zz_y_ill
Im not sure how to times through by 3/3 on the bottom

(2/3) / (1+(2/3)^2)^0.5

trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated
What is $\frac{3}{3}$ squared? It is $\frac{9}{9}$.

Multiply that out through the "top" and "bottom" of the fraction.

3. Originally Posted by i_zz_y_ill
Im not sure how to times through by 3/3 on the bottom

(2/3) / (1+(2/3)^2)^0.5

trying to get 2/(9+4x^2)^0.5 If somebody could show me how that'd be much appreciated
I'm sure you mean the following:

$\frac{ \frac{2}{3} }{ \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }$

Observe the $\frac{2}{3}$ in the numerator. We can move that 3 to the denominator of the entire function.

Giving us:

$\frac{ 2 }{ 3 \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }$

But we also know that:
$3 = \sqrt{9}$

Thus:
$\frac{ 2 }{ \sqrt{9} \sqrt{1 + \left( \frac{2x}{3} \right) ^{2}} }$

Simplify the denominator.

$\frac{ 2 }{ \sqrt{9(1 + \left( \frac{2x}{3} \right) ^{2})} }$

$\frac{ 2 }{ \sqrt{9(1 + \left( \frac{4x^2}{9} \right) )} }$

$\frac{ 2 }{ \sqrt{9 + 4x^2 } }$