Renting a Car A rental car costs $25 plus $0.25 per mile.
(a) Write an equation that gives the cost C of driving the car x miles.
(b)When will the rental be $100 dollars? Complete numerically with a table of values.
25 + 0.25x = C
(a) To find the equation that gives the cost C of driving the car x miles, we can use the given information that the rental car costs $25 plus $0.25 per mile. Therefore, the equation can be written as:
C = 25 + 0.25x
where C represents the cost of renting the car and x represents the number of miles driven.
(b) To determine when the rental will cost $100, we can set up the equation as:
100 = 25 + 0.25x
Now, we can solve this equation numerically to find the values of x:
Let's create a table with different values of x and substitute them into the equation to check if the equation holds true:
| x | C = 25 + 0.25x |
|-------|-------------------|
| 0 | 25 + 0.25(0) = 25 |
| 100 | 25 + 0.25(100) = 50 |
| 200 | 25 + 0.25(200) = 75 |
| 300 | 25 + 0.25(300) = 100 |
| 400 | 25 + 0.25(400) = 125 |
| 500 | 25 + 0.25(500) = 150 |
From the table, we can observe that when the rental car is driven 300 miles, the cost will be $100.
(a) The equation that gives the cost C of driving the car x miles is:
C(x) = 25 + 0.25x
The $25 represents the fixed cost, and the $0.25 per mile represents the variable cost.
(b) To find when the rental will be $100, substitute C(x) = 100 into the equation and solve for x:
100 = 25 + 0.25x
Subtracting 25 from both sides:
75 = 0.25x
Dividing both sides by 0.25:
x = 75 / 0.25
x = 300
So, when the rental will be $100, it will be after driving 300 miles.
Now, let's complete numerically using a table of values:
x | C(x)
--------------
0 | $25.00
100 | $50.00
200 | $75.00
300 | $100.00
Based on the table, when driving 300 miles, the rental cost will be $100.