Answer:
C (1,2)
Step-by-step explanation:
The point that is a solution must satisfy both inequalities;
The inequalities are;
[tex]y\:<\:x^2+3[/tex]
[tex]y\:>\:x^2-4[/tex]
Point A
[tex](4,0)[/tex]
[tex]0\:<\:16+3[/tex]: True
[tex]0\:>\:16-4[/tex]:False
Point B (0,4)
[tex]4\:<\:0+3[/tex]: False
[tex]4\:>\:0-4[/tex]:True
Point C (1,2)
[tex]2\:<\:1+3[/tex]: True
[tex]2\:>\:1-4[/tex]:True
Point D(-2,-4)
[tex]-4\:<\:4+3[/tex]: True
[tex]-4\:>\:4-4[/tex]:False
The correct answer is C.
