Answer:
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
Answer:
Here's a way this can help! If the graph it by two then any odd number will by by the middle, start by the origin and you'll be able to see it clearly.
Is that the rest of the question ?
Answer:
17
Step-by-step explanation:
both m's substitute for 4
4+4+9
8+9
17
