$\sqrt{(2^2)^2+2(2^{-1})+((2^{10})^{\frac{1}{2}})}$ =

Challenger App

★

1M+ Downloads

B$\sqrt20$

C$\sqrt48$

Answer:

$2^4$ +(2x $\frac12$ )+ $2^5$ =16+1+32 = $\sqrt{49}$ =7

Related Questions:

K+ 1/K – 2 = 0, K > 0, ആയാൽ K²⁹ + 1/ K²⁹- 2 ന്റെ വില എത്ര ആകും ?

What is to be added to $2^{36}$ to get $2^{37}$ ?

47^{-5} ÷ 47^{11} × 47^{-15}