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operties of Lincar Question 25, F.3.117 WW Secre: 24.05\%, 6.01 al 25 points Pental3 Points 0 of 1 Save polications A truck rental company rerts a moving fruck for one day by charging 524 plus 50.12 per mile. Whate a linear ocuration that relates the cost \( C \), in dollars, of renting the fruck to the number \( x \) of miles diven. What is the cost of rerfing the truck ithe truck is driven 200 milles? 489 milles? Type the linear equation hant relates the cost \( C \), in dollars, of reating mis frick to the rumber of x milles divent \[ c=21+012 x \] (Use integers or decimals for any mumbers in the expression. De not indude the \$ symbot in your answer) \[ c=5 \square \]

Ask by Chen Spencer. in the United States
Mar 13,2025

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**Linear Equation:** \[ C = 524 + 50.12x \] **Cost Calculations:** - For 200 miles: \( C = 524 + 50.12 \times 200 = 10548 \) dollars - For 489 miles: \( C = 524 + 50.12 \times 489 = 25032.68 \) dollars

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To represent the cost \( C \) of renting the truck as a linear equation, we start with a base fee of $524 and then add the cost per mile, which is $0.12 per mile. Thus, the equation that describes this situation is: \[ C = 524 + 0.12x \] Now, for the number of miles driven, plugging in 200 miles gives us: \[ C = 524 + 0.12(200) = 524 + 24 = 548 \] And for 489 miles: \[ C = 524 + 0.12(489) = 524 + 58.68 = 582.68 \] So the costs would be $548 for 200 miles and approximately $582.68 for 489 miles. --- Understanding the significance of linear equations can be fascinating! They model various real-world scenarios, be it budgeting for a road trip or calculating the cost of goods based on quantity. Whenever you have a starting point (y-intercept) and a consistent change (slope), you're likely dealing with a linear function. If you’d like to delve deeper into linear equations and their applications, think about exploring real-life situations, such as how to graph data or analyze trends. You can check out "Algebra Unplugged" by Kenn Amdahl for a fun and enlightening take on the topic!

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