Answer: [tex]6<z<18[/tex]
Step-by-step explanation:
Use the pythagorean theorem to find z first.
[tex]c^2=a^2+b^2\\c=\sqrt{a^2+b^2}[/tex]
a = 12
b = 6
c = z
[tex]c=\sqrt{(12)^2+(6)^2}\\ c=\sqrt{144+36}\\ c=\sqrt{180} \\c=13.4[/tex]
z is greater than a and b
