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Chapter 1: Problem 8

A car that gets \(m\) miles per gallon: The cost of operating a car depends on the gas mileage \(m\) that your car gets, the cost \(g\) per gallon of gasoline, and the distance \(d\) that you drive. a. How much does it cost to drive 100 miles if your car gets 25 miles per gallon and gasoline costs 289 cents per gallon? b. Find a formula that gives the cost \(C\) as a function of \(m, g\), and \(d\). Be sure to state the units of each variable. c. Use functional notation to show the cost of driving a car that gets 28 miles per gallon a distance of 138 miles if gasoline costs \(\$ 2.99\) per gallon. Use the formula from part b to calculate the cost.

Short Answer

Expert verified
Part (a) costs \$11.56. Part (c) costs \$14.74 using the formula \(C = \left(\frac{d}{m}\right) \times g\).

Step by step solution

01

Understand the Given Problem

We're tasked with finding the cost of driving 100 miles at a gas mileage of 25 miles per gallon (mpg) and a fuel cost of 289 cents per gallon. We also need to find a general formula for driving costs based on gas mileage, fuel cost, and distance, and use it to calculate a specific scenario.
02

Calculate Cost for 100 Miles (Part a)

To find the cost of driving 100 miles, first find the amount of fuel used: \[ \text{Gallons used} = \frac{\text{Distance}}{\text{Miles per gallon}} = \frac{100}{25} = 4 \text{ gallons} \]Then, calculate the cost: \[ \text{Cost} = \text{Gallons used} \times \text{Cost per gallon} = 4 \times 289 \text{ cents} = 1156 \text{ cents} \] Convert to dollars: \[ 1156 \text{ cents} = \frac{1156}{100} = \$11.56 \]
03

Derive the Formula for Cost (Part b)

The cost of operating the car can be found with the formula: \[ C = \left(\frac{d}{m}\right) \times g \] where:- \(d\) is the distance driven, in miles.- \(m\) is the mileage in miles per gallon.- \(g\) is the cost of gasoline per gallon, in cents or dollars.
04

Apply the Formula to a Specific Case (Part c)

Use the formula from Step 3 for a car that gets 28 mpg, over a distance of 138 miles, with gas at \\(2.99 per gallon:Convert \\)2.99 to cents: \[ g = 2.99 \times 100 = 299 \text{ cents} \] Calculate the cost: \[ C = \left(\frac{138}{28}\right) \times 299 \approx 4.92857 \times 299 = 1473.6713 \text{ cents} \]Convert to dollars: \[ 1473.6713 \text{ cents} \approx \$14.74 \]
05

Present the Calculated Results

The cost for part (a) is \\(11.56, and the cost for driving with the conditions in part (c) is approximately \\)14.74.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Miles per Gallon
Miles per Gallon (mpg) is a measure of how fuel-efficient your car is. It tells you how many miles your car can travel on one gallon of gasoline. Understanding this helps you assess and control your fuel expenditure. For example, if a car can travel 25 miles on a single gallon of gas, this would be expressed as 25 mpg. The higher the mpg value, the less frequently the car needs refueling, which typically translates into cost savings.To calculate the amount of fuel you鈥檇 use for a specific distance, divide the distance by miles per gallon: \[\text{Gallons used} = \frac{\text{Distance (d)}}{\text{Miles per gallon (m)}}\]Knowing how many gallons are used can help predict the overall cost when paired with the cost per gallon.
Cost per Gallon
Cost per Gallon is the price you pay for a single gallon of gasoline. Gas prices can fluctuate due to a variety of market factors, impacting the cost to drive your car.In exercises like ours, the cost per gallon might be given in cents or dollars. It鈥檚 essential to pay attention to the unit used and convert if necessary. For instance, if gas is \(2.89 per gallon, it should be represented in cents as \)289 cents for calculation purposes.The overall cost to drive a certain distance is directly proportional to the cost per gallon. If the cost per gallon rises, so will the total cost of the trip, unless other factors like miles per gallon offset this increase.
Functional Notation
Functional Notation is a way to express relationships between variables in a concise manner. In algebra, functions are used to model real-world scenarios, making it easier to calculate outcomes based on variable inputs.In this scenario, we are finding the function that expresses the cost \(C\) in terms of miles per gallon, cost per gallon, and distance driven. The formula derived is:\[C = \left( \frac{d}{m} \right) \times g\]This function allows you to input different values for \(m, g,\) and \(d\) to find the cost \(C\). Changing one variable lets you see how it impacts the overall cost, which is especially useful for budgeting and planning trips.
Distance Driven
Distance Driven refers to how far you plan to travel, typically measured in miles. Knowing this distance upfront is crucial because it determines how much fuel you will need based on your car's miles per gallon.To calculate how much fuel is required for a given distance, you divide the total distance by the mpg of your vehicle:\[\text{Gallons used} = \frac{\text{Distance}}{\text{Miles per gallon}}\]Once you know the gallons used, you can easily calculate the cost by multiplying it by the cost per gallon of gas.Taking meticulous note of the distance driven helps in planning your trip more effectively, accounting for fuel stops, and managing your travel budget efficiently.

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Most popular questions from this chapter

Inflation: During a period of high inflation, a political leader was up for re-election. Inflation had been increasing during his administration, but he announced that the rate of increase of inflation was decreasing. Draw a graph of inflation versus time that illustrates this situation. Would this announcement convince you that economic conditions were improving?

A stock market investment: A stock market investment of \(\$ 10,000\) was made in 1970 . During the decade of the \(1970 \mathrm{~s}\), the stock lost half its value. Beginning in 1980, the value increased until it reached \(\$ 35,000\) in 1990 . After that its value has remained stable. Let \(v=v(d)\) denote the value of the stock, in dollars, as a function of the date \(d\). a. What are the values of \(v(1970), v(1980)\), \(v(1990)\), and \(v(2000)\) ? b. Make a graph of \(v\) against \(d\). Label the axes appropriately. c. Estimate the time when your graph indicates that the value of the stock was most rapidly increasing.

A car: The distance \(d\), in miles, that a car travels on a 3 -hour trip is proportional to its speed \(s\) (which we assume remains the same throughout the trip), in miles per hour. a. What is the constant of proportionality in this case? b. Write a formula that expresses \(d\) as a function of \(s\).

Newton's law of cooling says that a hot object cools rapidly when the difference between its temperature and that of the surrounding air is large, but it cools more slowly when the object nears room temperature. Suppose a piece of aluminum is removed from an oven and left to cool. The following table gives the temperature \(A=A(t)\), in degrees Fahrenheit, of the aluminum \(t\) minutes after it is removed from the oven. $$ \begin{array}{|c|c|} \hline t=\text { Minutes } & A=\text { Temperature } \\ \hline 0 & 302 \\ \hline 30 & 152 \\ \hline 60 & 100 \\ \hline 90 & 81 \\ \hline 120 & 75 \\ \hline 150 & 73 \\ \hline 180 & 72 \\ \hline 210 & 72 \\ \hline \end{array} $$ a. Explain the meaning of \(A(75)\) and estimate its value. b. Find the average decrease per minute of temperature during the first half- hour of cooling. c. Find the average decrease per minute of temperature during the first half of the second hour of cooling. d. Explain how parts b and c support Newton's law of cooling. e. Use functional notation to express the temperature of the aluminum after 1 hour and 13 minutes. Estimate the temperature at that time. (Note: Your work in part c should be helpful.) f. What is the temperature of the oven? Express your answer using functional notation, and give its value. g. Explain why you would expect the function \(A\) to have a limiting value. h. What is room temperature? Explain your reasoning.

What if interest is compounded more often than monthly? Some lending institutions compound interest daily or even continuously. (The term continuous compounding is used when interest is being added as often as possible - that is, at each instant in time.) The point of this exercise is to show that, for most consumer loans, the answer you get with monthly compounding is very close to the right answer, even if the lending institution compounds more often. In part 1 of Example 1.2, we showed that if you borrow \(\$ 7800\) from an institution that compounds monthly at a monthly interest rate of \(0.67 \%\) (for an APR of \(8.04 \%\) ), then in order to pay off the note in 48 months, you have to make a monthly payment of \(\$ 190.57\). a. Would you expect your monthly payment to be higher or lower if interest were compounded daily rather than monthly? Explain why. b. Which would you expect to result in a larger monthly payment, daily compounding or continuous compounding? Explain your reasoning. c. When interest is compounded continuously, you can calculate your monthly payment \(M=\) \(M(P, r, t)\), in dollars, for a loan of \(P\) dollars to be paid off over \(t\) months using $$ M=\frac{P\left(e^{r}-1\right)}{1-e^{-r t}}, $$ where \(r=\frac{A P R}{12}\) if the APR is written in decimal form. Use this formula to calculate the monthly payment on a loan of \(\$ 7800\) to be paid off over 48 months with an APR of \(8.04 \%\). How does this answer compare with the result in Example 1.2?
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