Question

If $x=3$ and $y=-2$, then the value of $x^x+y^y$ is $27\frac{1}{p}$. So, $p$ is:

Answer 1

Given, $x=3$ and $y=-2$
$x^x+y^y$

$=3^3+(-2{)}^{-2}$

$=27+\frac{1}{4}$

$=\frac{108+1}{4}$

$=\frac{109}{4}$

$=27\frac{1}{4}$

Thus, $p=4$