For the first question, simply find a point that is on the line segment. For the second question, knowing that in quadrant iii the x values are negative and the y values are also positive using this fact find the point that has x negative and y positive.
First you need to know how many pounds of carbon dioxide you get when burning 1 gallon of gasoline. That number is 19.64 pounds
To get the number of pounds released when entire tank is burned you need to multiply gas tank volume and number of pounds you get when 1 gallon is burned.
17.2 * 19.64 = 337.808
ADC should measure to 115 degrees if both measurements are equal to 180 degrees.
If not, ADC might be 25 degrees if both measurements are equal to 90 degrees.
Hope this helps.
The area of a circle with radius of 13 millimeters is A≈530.93A=πr2=π·132≈530.92916
Answer:
Instantaneous Velocity at t = 1 is 20 feet per second
Step-by-step explanation:
We are given he following information in the question:

B) Instantaneous Velocity at t = 1

A) Formula:
Average velocity =

1) 0.01

2) 0.005 s

3) 0.002 s

4) 0.001 s
