SOLUTION: according to recent study, the gas mileage of an automobile decreases by ten percent for every 5 miles per hour over 55 mph that is driven. The car gets 30 miles per gallon at 55

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Question 199878: according to recent study, the gas mileage of an automobile decreases by ten percent for every 5 miles per hour over 55 mph that is driven. The car gets 30 miles per gallon at 55 mph.
A. Create a function that models the mileage of the car as a function of miles per hour over 55. Graph the function on the axes and label ana then use the graph to answer the following questions
B. What is the cars mileage at 65 mph?
C. When the car is driven at 70 miles an hour how many miles per gallon is the car getting.
D. What is the fastest that the car can drive and still get 15 miles per gallon
E. When there is only 1.5 gallons of gas left in the car and 54 miles to drive home what is the fastest that the car can drive and still have enough gas to get home?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
according to recent study, the gas mileage of an automobile decreases by ten
percent for every 5 miles per hour over 55 mph that is driven.
The car gets 30 miles per gallon at 55 mph.
:
A. Create a function that models the mileage of the car as a function of miles
per hour over 55. Graph the function on the axes and label and then use the
graph to answer the following questions
:
Let x = mph over 55 mph; y = mpg
It says: 10% of 30 = 3 mpg decrease for 5 mph increase passed 55 mph
We can create table to represent this:
x | y
-------
0 | 30
5 | 27
10 | 24
15 | 21
:
Assign the values as follows:
x1=0; y1=30
x2=15; y2=21
find the slope:
m = %2821-30%29%2F%2815-0%29 = %28-9%29%2F15 = -3%2F5 = -.6
:
Find the equation using the point/slope formula: y - y1= m(x - x1)
y - 30 = -.6(x - 0)
y = -.6x + 30 is the equation

B. What is the cars mileage at 65 mph?
x = 65 - 55 = 10
y = -.6(10) + 30
y = -6 + 30
y = 24 mpg at 65 mph
:
C. When the car is driven at 70 miles an hour how many miles per gallon is the car getting.
x = 70 - 55 = 15
y = -.6(15) + 30
y = -9 + 30
y = 21 mpg at 70 mph
:
D. What is the fastest that the car can drive and still get 15 miles per gallon
y = 15 mpg
-.6x + 30 = 15
-.6x = 15 - 30
-.6x = -15
x = %28-15%29%2F%28-.6%29
x = 25
25 + 55mph = 80 mph for 15 mpg
:
E. When there is only 1.5 gallons of gas left in the car and 54 miles to drive home what is the fastest that the car can drive and still have enough gas to get home?
Find the no. of mpg required to travel 54 mi with 1.5 gal
54%2F1.5 = 36 mpg
y = 36, find x
-.6x + 30 = 36
-.6x = 36 - 30
-.6x = 6
x = 6%2F%28-.6%29
x = -10 mph
55 - 10 = 45 mph to get 54 mi out of 1.5 gal of gas
:
Did all this make sense to you? Any questions?