Answer:
23 Yards
Step-by-step explanation:
So for this question, we would have to start off with writing an equation: 19.71+0.23x=25 once we subtract 19.71 on both sides, the left over sum of money will be 5.29. Since we are dealing with BOTH Dollars and Cents, it should be remembered to use 23 CENTS instead of Dollars. 
= 23 so the finals answer is 23 yards.
AFC = FC / Quantity printed
<span>So given she prints 1,000 posters: AFC = 250.00/1000 = $0.25 </span>
<span>Given she prints 2,000 posters: AFC = 250.00/2000 = $0.125 </span>
<span>Given she prints 10,000 posters: AFC = 250.00/2000 = $0.025 </span>
<span>ATC = TC / Quantity printed </span>
<span>where TC = FC + Variable C * Quantity printed </span>
<span>If she prints 1000: TC = 250 + 2000*1000 = 2,000,250 </span>
<span>ATC = 2,000,250/1000 = 2000.25 </span>
<span>If she prints 2000: TC = 250 + 1600*2000 = 3,200,250 </span>
<span>ATC = 3,200,250/2000 = 1600.125 </span>
<span>If she prints 10000: TC = 250 + 1600*2000 + 1000*8000 ($1000 for each additional poster after 2000) = 11,200,250 </span>
<span>ATC = 11,200,250/10000 = 1120.025</span>
I can figure out this is a frefall motion.
Starting from rest => Vo = 0
Then, use the equation: d = [1/2]gt^2 => t = √(2d/g)
d = width of a black/clear stripe pair = 5cm = 0.05m
g ≈ 10 m/s^2 (the real value is about 9.81 m/s^2)
t =√(2*0.05m/10m/s^2) = 0.1 s
Answer: approximately 0.1 s
A: From the graph, car B was traveling faster.
Because the line of car B is more steep than the line of car A.
B:
The lines cross at (2,80), this means that the two cars were traveling at the same distance (80 miles) at the same time (2 hours).
C:
Since the line passes through the points (0,0) and (2,80)
So the unit rate or slope is 40 mph.
The distance which car A traveled in the first four hours = 4x40= 160 mph
