Answer:
[tex]\sqrt{11}[/tex]
Step-by-step explanation:
well first you have to plug in a and c in the Pythagorean theorem.
[tex]a^2+b^2=c^2\\(2\sqrt{2})^2+b^2 =(\sqrt{15})^2[/tex]
then squaring the square roots cancels them out sooo it becomes
[tex]4 +b^2=15[/tex]
and then isolate b
[tex]b^2 = -4 +15\\\\\\b=\sqrt{-4+15} \\b=\sqrt{11}[/tex]
and there you have it. b is [tex]\sqrt{11}[/tex]
