$$ { \sqrt{(0.798)^2+0.404×0.798+(0.202)^2}+1=?} $$

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

Find the value of   $$ {(0.75)^3\over1-.75}+[0.75+(0.75)^2+1]$$

$$ (17)^{2} + 6 × (13)^{2}-212×6$$

1-[5-{2+(-5+6-2)2}] is equal to 

The sum of the squares of 2 number is 146 and the square root of one of them is 5. The cube of the other number is 

$$\sqrt{x}÷ \sqrt{144} = 0.02, $$ then the value of x is 

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