The slope of the line that passes through points ( -10, -8 ) and ( -8, -16 ) is -4.
<u>Explanation:</u>
Points given = ( -10, -8) and ( -8, -16)
Slope = ?
( -10, -8 ) : x1 = -10 and y1 = -8
( -8, -16 ) : x2 = -8 and y2 = -16
We know,
slope = y2 - y1 / x2 - x1
Slope = -16 - ( -8) / -8 - (-10)
slope = -16 + 8 / -8 + 10
slope = -8 / 2
slope = -4
Therefore, slope of the line that passes through points ( -10, -8 ) and ( -8, -16 ) is -4.
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:
The answer should be "Outlier".
Answer:
160
Step-by-step explanation:
2+2=4
4 x 40= 160
