What is the value of x in the circle to the right? (round to the nearest tenth as needed.)

Using the Pythagorean theorem a^2 + b^2 = C^2

a and b are the base and height and C is the hypotenuse of the triangle.

Base = a = 10 cm

Height = b = x

Hypotenuse = c = 7 + x

So we get 10^2 + x^2 = (7+x)^2

= 100 + x^2 = (7+x)^2

Subtract (7+x)^2 from each side:

100 + x^2 - (7+x)^2 = 0

Rewrite (7+x)^2 as (7+x) (7+x)

100 + x^2 + ((7+x) (7+x)) = 0

use FOIL on: ((7+x) (7+x)) 

((7+x) (7+x))  = (49 +14x +x^2)

Now yo have:
100 + x^2 - (49 +14x +x^2) = 0
Apply Distributive Property:
100-49 = 51
x^2 - x^2 = 0

Now you have: 
51-14x =0

Rewrite the left side to get the variable first:
-14x+51 = 0

Subtract 51 from each side:
-14x = -51

Divide each side by -14:

x = 51/14

x = 3.64

Rounded to the nearest tenth X = 3.6