A rental truck costs $50 plus $0.50 (67)
per mile.
(a) Write an equation that gives the cost C of driving
the truck x miles.
(b) Use the intersection-of-graphs method to determine
the number of miles that the truck is driven
if the rental cost is $80.
(c) Solve part (b) numerically with a table of values.
a. C = 0.50x + $50.
b C = 0.50x + 50 = 80
0.5x = 80-50 = 30
X = 60 miles.
c. Solve the Eq using several values of
x.
(a) To write an equation that gives the cost C of driving the truck x miles, we can use the given information that the truck costs $50 plus $0.50 per mile.
The cost of driving the truck depends on the number of miles driven, so we can express this relationship using the equation:
C = 50 + 0.50x
Where:
- C represents the total cost of driving the truck
- x represents the number of miles driven
(b) To determine the number of miles that the truck is driven if the rental cost is $80, we can use the intersection-of-graphs method.
First, we need to set up the equation to find the intersection point. We know that the cost C is equal to $80, so we can substitute this value into the equation we wrote in part (a):
80 = 50 + 0.50x
Now, we can solve for x:
0.50x = 80 - 50
0.50x = 30
Dividing both sides of the equation by 0.50, we get:
x = 30 / 0.50
x = 60
So, the truck is driven 60 miles if the rental cost is $80.
(c) To solve part (b) numerically with a table of values, we can create a table where we substitute different values for x into the equation we wrote in part (a) and calculate the corresponding values of C. We can then look for the value of x that gives us a rental cost of $80. Let's create a table:
| x | C = 50 + 0.50x |
|------|----------------|
| 0 | 50 |
| 10 | 55 |
| 20 | 60 |
| 30 | 65 |
| 40 | 70 |
| 50 | 75 |
| 60 | 80 |
From the table, we can see that when x is equal to 60, the rental cost C is equal to $80. Therefore, numerically, the truck is driven 60 miles if the rental cost is $80.