Can someone please help me? It's 1:16am and its due at 7:15am. I've got school.

Answer:
2 v 13, 2 Square root 13
Step-by-step explanation:
there's 2 right triangles
a)
10^2 = 6^ + a^2
100 - 36 = a^2
64 = a ^2
a = 8
--------------------------------------------
now we know the base of the right triangle = 8
and also the base of the left one = 4
8 + 4 = 12
--------------------------------------------
now find x
b)
x^2 = 6^2 + 4^2
x^2 = 36 + 16
x^2 = 52
x = (square root) 52
x = 2 v 13 <---- "v = square root"
Instructions :- Simplify the answer if there is square root
hypotenuse² = perpendicular²+base²
First triangle:-
[tex] {h}^{2} = {p}^{2} + {b}^{2} \\ {10}^{2} = {6}^{2} + {b}^{2} \\ 100 = 36 + {b}^{2} \\ 100 - 36 = {b}^{2} \\ 64 = {b}^{2} \\ 8 = b[/tex]
Second triangle
[tex] {h}^{2} = {p}^{2} + {b}^{2} \\ {x}^{2} = {6}^{2} + {4}^{2} \\ {x}^{2} = 36 + 16 \\ {x}^{2} = 52 \\ x = \sqrt{52} = 7.2 \\ x = 7 \: \: approx[/tex]