ANSWER
h(-2) = 0
k(4) = 0
EXPLANATION
Given:
[tex]p(x)=\begin{cases}{2x+5\text{ if }x<0} \\ {\placeholder{⬚}} \\ {x^2-5\text{ if }x\ge0}\end{cases}[/tex]
Desired Outcome:
1. h(-2)
2. k(4)
Evaluate p(-2)
[tex]\begin{gathered} p(-2)=2(-2)+5 \\ p(-2)=-4+5 \\ p(-2)=1 \end{gathered}[/tex]
Evaluate p(4)
[tex]\begin{gathered} p(4)=4^2-5 \\ p(4)=16-5 \\ p(4)=11 \end{gathered}[/tex]
Evaluate h(-2)
h(-2) = 0 since h(x) is NOT given
Evaluate k(4)
k(4) = 0 since k(x) is NOT given
