Question

Evaluate the value of the following:

${\left(6^2\right)}^0+{\left(\frac{-4}{7}\right)}^0$

Answer 1

Given number: ${\left(6^2\right)}^0+{\left(\frac{-4}{7}\right)}^0$

Using the above law of exponents:

$$\begin{gathered} {\left(6^2\right)}^0+{\left(\frac{-4}{7}\right)}^0={\left(6^2\right)}^0+{\left(\frac{-4}{7}\right)}^0[\because U\sin g{\left(x^b\right)}^c=x^{b\times c}] \\ =6^0+\frac{{(-4)}^0}{{\left(7\right)}^0}[\because U\sin g{\left(\frac{x}{y}\right)}^a=\frac{x^a}{y^b}] \\ =1+1[\because U\sin g{x}^0=1] \\ =2 \end{gathered}$$

Hence, $2$, This is the required value.