We have been given that fuel efficiency for a 2007 passenger car was 31.2 mi/gal and the same model of car, the fuel efficiency increased to 35.6 mi/gal in 2012. Also, the gas tank for this car holds 16 gallons of gas.
We need to write a function and graph a linear function that models the distance that each car can travel for a given amount of gas up to one tankful.
Let represent the functions as
and
where
and
represent the distances traveled by car in years 2007 and 2012 and x represents the number of gallons. Therefore, we can express the required functions as:

Domain of both these functions are [0,16] and ranges are [0,499.2] and [0,569.6] respectively for years 2007 and 2012.
The difference function will be:


Domain of this function is [0,16] and range is [0,67.2].
The graphs are shown below.
1.54 because 1.5 is the same thing as 1.50
Answer:
4y^2-y+21
Step-by-step explanation:
(4y-5y)+(5y^2-y^2)+(-6+27)
-y+4y^2+21
put it in standard form
4y^2-y+21
KMN = 90° (tangent perpendicular to radius)
Using Pythagoras Theorem,
√(8^2 + 6^2) = √(100) = 10
Therefore KN = 10km
Answer:
-5
Step-by-step explanation: