Question

(√0.01)^2+3^2x-(4x^y)^0+64^2/3-(√3)^4x​

Answer 1

Given:

$\bold{(\sqrt{0.01})^2+3^2x-(4x^y)^0+\frac{64^2}{3}-(\sqrt{3})^4x}$

To find:

simplify

Solution:

$\Rightarrow (\sqrt{0.01})^2+3^2x-(4x^y)^0+\frac{64^2}{3}-(\sqrt{3})^4x\\\\\Rightarrow 0.01+9x-1+\frac{4096}{3}-(9)x\\\\\Rightarrow 0.01+9x-1+1365.33-9x\\\\\Rightarrow 0.01-1+1365.33\\\\\Rightarrow 1365.34-1\\\\\Rightarrow \boxed{\bold{1364.34}}$