Answer:
Here's what I get.
Step-by-step explanation:
9. (6, -8)
The reference angle θ is in the fourth quadrant.
∆AOB is a right triangle.
OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100
OB = √100 = 10

10. cot θ = -(√3)/2
The reference angle θ is in the second quadrant.
∆AOB is a right triangle.
OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7
OB = √7

Acceleration is the change in velocity over time. There's are formula's for such problems which are called kinematics equations. Look em up. One of them doesn't have t (time) in them you would normally use that one to solve this problem. 16/2/20=a=0.4m/s^2
I did it for you but please look it up, understand and memorize it ;)
u is at (1,-2)
V is at (-6,-6)
using distance formula it is about 8.06 long
so Answer is C
Answer:
B. 14.075
Step-by-step explanation: