Assume that the lines that appear to be tangent are tangent p is the center if each circle find x.

Step-by-step explanation:

Problem One

All quadrilaterals have angles that add up to 360 degrees.

Tangents touch the circle in such a way that the radius and the tangent form a  right angle at the point of contact.

Solution

x + 115 + 90 + 90 = 360

x + 295 = 360

x + 295 - 295 = 360 - 295

x = 65

Problem Two

From the previous problem, you know that where the 6 and 8 meet is a right angle.

Therefore you can use a^2 + b^2 = c^2

a = 6

b =8

c = ?

6^2 + 8^2 = c^2

c^2 = 36 + 64

c^2 = 100

sqrt(c^2) = sqrt(100)

c = 10

x = 10

Problem 3

No guarantees on this one. I'm not sure how the diagram is set up. I take the 4 to be the length from the bottom of the line marked 10 to the intersect point of the tangent with the circle.

That means that the measurement left is 10 - 4 = 6

x and 6 are both tangents from the upper point of the line marked 10.

Therefore x = 6

2/3x + 7 = 29

step-by-step explanation:

two thirds = 2/3

plus a number = x

at least 29 = 29

there for you should get 2/3x + 7 = 29

5/20 so 25%

step-by-step explanation: