Question

a rental car company charges $36.39 per day to rent a car and $0.09 for every mile driven. miguel wants to rent a car, knowing that: • he plans to drive 225 miles. • he has at most $480 to spend. which inequality can be used to determine x, the maximum number of days miguel can afford to rent for while staying within his budget? answer 480 ≤ 36.39x + 20.25 480 ≥ 36.39x + 20.25 36.39 + 20.25x ≥ 480 36.39 + 20.25x ≤ 480

Explanation:

Step1: Calculate mileage cost

The cost per mile is $0.09, and he drives 225 miles. So mileage cost = \(0.09\times225 = 20.25\) dollars.

Step2: Calculate total cost

The daily cost is $36.39, for \(x\) days, so daily cost part is \(36.39x\). Total cost is daily cost plus mileage cost, so total cost = \(36.39x + 20.25\).

Step3: Set up inequality

He has at most $480, meaning total cost LXI0 480? Wait, no: "at most" means his spending (total cost) can't exceed 480, so total cost LXI1 480? Wait, no, the inequality is about what he can afford. Wait, the total cost LXI2 must be less than or equal to 480? Wait, no, the options have LXI3, which is equivalent to LXI4. Wait, let's check the options. The total cost is LXI5, and he has at most $480, so \(36.39x + 20.25 \leq 480\), which is the same as \(480 \geq 36.39x + 20.25\).

Answer:

\(480 \geq 36.39x + 20.25\) (the second option)