Assume that
$$\sqrt{3} = 3.605 \ (approximately)  $$
$$\sqrt{130} = 11.40 \ (approximately)  $$
Find the value of
$$\sqrt{1.3} + \sqrt{1300}+\sqrt{0.013}  $$

The smallest natural number, by which 3000 must be divided to make the quotient a perfect cube, is : 

The largest number of five digits, which is a perfect square is 

The value of $$ {\sqrt{0.6912}+{\sqrt{0.5292}}\over {\sqrt{0.6912}-{\sqrt{0.5292}}}}$$ is:

Simplify : $$\sqrt{[(12.1)^2-(8.1)^2]÷[(0.25)^2+(0.25)(19.95)]} $$

3x² = 36 then what is the value of x?

If  $$\sqrt{2}=1.414$$, the value of $${ \sqrt{2}-1\over\sqrt{2}+1}$$  is nearest to 

What mathematical operation should come at the place of ‘?’ in the equation:
2 ? 6 – 12 ÷ 4 +2 =11.