Answer
$a.\quad 100$
$b.\quad 100$
$c.\quad 131$
Work Step by Step
$ a.\quad$
$x=-10$ belongs to the interval $-10\leq x\leq 10$,
so we calculate its function value with $f(x)=x^{2}$
$f(-10)=(-10)^{2}=100$
$ b.\quad$
$x=10$ belongs to the interval $-10\leq x\leq 10$,
so we calculate its function value with $f(x)=x^{2}$
$f(10)=(10)^{2}=100$
$ c.\quad$
$x=11$ belongs to the interval $x\gt 10$,
so we calculate its function value with $ f(x)=x^{2}+10$
$f(11)=(11)^{2}+10=121+10=131$