What are the steps?

Simplify the following:
(3)/(5 sqrt(1-(3/5)^2))

 

(3/5)^2 = 3^2/5^2:
(3)/(5 sqrt(1-3^2/5^2))

 

3^2 = 9:
(3)/(5 sqrt(1-9/5^2))

 

5^2 = 25:
(3)/(5 sqrt(1-9/25))

 

Put 1-9/25 over the common denominator 25. 1-9/25  =  25/25-(9)/25:
(3)/(5  sqrt(25/25-(9)/25 ) )

 

25/25-9/25  =  (25-9)/25:
(3)/(5  sqrt((25-9)/25 ) )

 

25-9 = 16:
(3)/(5 sqrt(16/25))

 

sqrt(16/25)  =  (sqrt(16))/(sqrt(25)):
(3)/(5 (sqrt(16))/(sqrt(25)))

 

sqrt(25)  =  sqrt(5^2)  =  5:
(3)/(5 (sqrt(16))/(5))

 

sqrt(16)  =  sqrt(2^4)  =  2^2:
(3/5)/(2^2/5)

 

2^2 = 4:
(3/5)/(4/5)

 

Multiply the numerator by the reciprocal of the denominator, (3/5)/(4/5) = 3/5×5/4:
(3×5)/(5×4)

 

(3×5)/(5×4) = 5/5×3/4 = 3/4:
Answer: |  3/4

How do you solve (3/5)/Sqr(1-(3/5)^2)

 

\(\frac{3}{5}\sqrt{1-\left(\frac{3}{5}\right)^2}\\ =\frac{3}{5}\sqrt{\frac{25}{25}-\frac{9}{25}}\\ =\frac{3}{5}\sqrt{\frac{25-9}{25}}\\ =\frac{3}{5}\sqrt{\frac{16}{25}}\\ =\frac{3}{5}*\frac{4}{5}\\ =\frac{12}{25}\\ or\\ =0.48\)

 

If you have questions then just ask :)

I took your \sqrt to be just sqrt but Alan hs pointed out that you probably meant the answer that our guest gave.  :)

That is

\(\frac{3}{5}\div\sqrt{1-\left(\frac{3}{5}\right)^2}\\ =\frac{3}{5}\div\sqrt{\frac{25}{25}-\frac{9}{25}}\\ =\frac{3}{5}\div \sqrt{\frac{25-9}{25}}\\ =\frac{3}{5}\div \sqrt{\frac{16}{25}}\\ =\frac{3}{5}\div \frac{4}{5}\\ =\frac{3}{5}\times \frac{5}{4}\\=\frac{3}{4}\\\)