aFor each mile means we need to multiply just the $0.15 by m.
B
b Use the equation from Part A and substitute m=145.
C
c Use the equation from Part A. Substitute m=105, then multiply the answer by four.
D
d Use the equation from Part A. Substitute m=220, then multiply the answer by seven.
A
a 0.15m+15
B
b $36.75
C
c $123
D
d $336
Practice makes perfect
a For this exercise, we need to write a polynomial expression that represents how much it would cost to rent a car for a day given certain conditions.
$ 15 flat fee for a day
$0.15 for each mile driven
The first condition does not alter according to miles driven. The second condition tells us that the cost varies according to the number of miles driven. This is where we can place our variable, m. The following polynomial represents the cost to rent a car for mmiles.
15+ 0.15m
b This time, we need to determine how much it costs to rent a car if it is driven 145 miles. Let's substitute m= 145 into our polynomial from Part A and evalute.
If the rental car is driven 145 miles, it will cost $36.75.
c We need to find the cost of renting a car if it is driven 105 miles each day for four days. We can start like we did in Part B and substitute m= 105 into our polynomial from Part A. Then, we can multiply that answer by 4, for the four days.
Now that we have the cost for one day, we can multiply it by 4.
30.75 * 4 = 123
It will cost a total of $123 to rent a car for four days and drive it 105 miles each day.
d We now need to do the same thing we did in Part C, but with different numbers. We can start with substituting m= 220.
Now that we have the cost for one day, we can multiply it by 7 to get the cost for seven days.
48 * 7 = 336
It will cost a total of $336 to rent a car for seven days and drive it 220 miles each day.
