a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as
... c = k·s² + m·s + n
Filling in the given values gives three equations in k, m, and n.

Subtracting each equation from the one after gives

Subtracting the first of these equations from the second gives

Using the next previous equation, we can find m.

Then from the first equation
[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]
There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with
... c = 0.01s² - s + 37
b) At 150 kph, the cost is predicted to be
... c = 0.01·150² -150 +37 = 112 . . . cents/km
c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.
d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.
The answer would be 60. By rounding, 14.2=14; 15.51=16; 14.99=15; 15.8=16.
14+15+16+16=61
Rounding: 61=60
Answer:
B
Step-by-step explanation:
3000-1000 = 2000
600-200 = + 400
40 - 20 = + 20
8 - 3 = + 5
First lets remove the initial fee

150 is how much she is charging you per hour.
If each hour is 55, lets divide 150 by 55.

She charged you for 2.7 hours (almost 3) of service