32.
What is the value of x for ∆DEF?
HELP! Answer if you can!!

Step-by-step explanation:
[tex]c^{2} = {a}^{2} + { b}^{2} [/tex]
Pythagoras theorem
x²=12² + 5²
x²=144+25
x²=169
[tex]x = \sqrt{169} [/tex]
x=13
Answer:
x = 13
Step-by-step explanation:
Using the Pythagorean Theorem to find the missing hypotenuse side:
[tex]a^2+b^2=c^2[/tex]
12 and 5 are a and b. 'x' would be c.
[tex]12^2+5^2=c^2\\\\12^2=144\\5^2=25\\\\144+25=c^2\\\\169=c^2\\\\\sqrt{169}=\sqrt{c^2}\\\\\boxed{13=c}[/tex]
'x' should be equal to 13.