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.