The distance between the two points is 
The polar coordinate of A is 
The polar coordinate of B is 
Explanation:
The two points are
and 
The distance between two points is given by,

Thus, the distance between the two points is 
The polar coordinates of A can be written as 
Distance = 
Substituting
, we get,
Distance = 

To make the angle positive, let us add 360,

The polar coordinate of A is 
Similarly, The polar coordinate of B can be written as 
Distance = 
Substituting
, we get,
Distance = 

To make the angle positive, let us add 360,

The polar coordinate of B is 
The rate of change of Pete's height from 3 to 5 years can be determined as,

Thus, the required rate of change from 3 to 5 years is 4 in/year.
Answer:
1: D
2: B
3: y=-4x+22
4: y=-8x+26
Step-by-step explanation:
1: Parallel lines have the same slope, and the only one with a slope of 3x is D
2: Perpendicular lines have an opposite slope, so a line with a slope of -1/2 would be 2, so you just flip the number and add or take away a negative sign, depending on the original slope
3: Like I said before, perpendicular lines have an opposite slope, so the slope would be -4. After you've figured that out, you just plug in the numbers given to you (and remember, x is first, y is last)
Plug in: 6=-4(4)+b
You would then solve for b.
6=-16+b
22=b
Then plug that into y=mx+b, with m being the slope (-4) and b being the y intercept (22)
4: The process for finding parallel and perpendicular lines is very similar, except you don't have to change the slope.
Plug in: 10=-8(2)-b
10=-16+b
26=b
Again, plug that into the equation y=mx+b
Hope I could be of help! Sorry if it doesn't make sense, this is my first time on this website.
Answer:
C
Step-by-step explanation:
C is the only one that is an enlarged version of A.