Concern the cost, C, of renting a car from a company which...

Concern the cost, $C,$ of renting a car from a company which charges $\$ 40$ a day and 15 cents a mile, so $C=f(d, m)=40 d+0.15 m,$ where $d$ is the number of days, and $m$ is the number of miles. Make a table of values for $C,$ using $d=1,2,3,4$ and $m=100,200,300,400 .$ You should have 16 values in your table.

Concern the cost, $C,$ of renting a car from a company which charges $\$ 40$ a day and 15 cents a mile, so $C=f(d, m)=40 d+0.15 m,$ where $d$ is the number of days, and $m$ is the number of miles.
Make a table of values for $C,$ using $d=1,2,3,4$ and $m=100,200,300,400 .$ You should have 16 values in your table.

Key Concepts

Function with Two Variables

A function with two variables assigns a unique output to every pair of input values. In this context, it means the cost is determined by both the number of days and the number of miles, and the function can be analyzed by examining how each independent variable contributes to the final result.

Linear Functions

A linear function is one in which the dependent variable is a linear combination of the independent variables, meaning it has constant rates of change with respect to each variable. The cost function, being of the form C = 40d + 0.15m, illustrates a direct, proportional relationship between the cost, the number of days, and the number of miles.

Evaluating Functions by Substitution

Evaluating a function by substitution involves replacing the variables in the function's equation with specific values to compute the corresponding output. This is a fundamental operation in algebra that allows for the practical application of mathematical models to real-world situations.

Creating Tables of Values

Creating tables of values is a method used to systematically organize and display the outputs generated by a function when evaluated at a set of input values. It serves as a visual aid to understand the behavior of the function across different domains by correlating pairs or sets of inputs with their computed outcomes.

Transcript

00:01 So here we have the function, c equals f of d, f of d comma m equals 40d plus .15m.

00:12 So what this equation is expressing is the cost, c, of renting a car for d days and driving m of miles.

00:26 So if we want to set up a table with different values of days and miles, we can find the function values of different pairings of days and miles.

00:49 So on the top row, on the top we have miles.

00:54 On the side, we have different values of days.

00:57 And we can use this table to, show what each sort of pairing of days and miles will result in what cost.

01:13 So if we just go down, if we start with 100 miles and one day, that would be f of 1 comma 100.

01:24 And we can just substitute the values of dnm into the equation.

01:29 So that's just 40 times 1 plus 0 .15 times 100.

01:36 That's just going to result in 40 plus 15.

01:40 So that's just 55.

01:45 As we go across while keeping d equal to 1 and m is increasing by 100, what we get is actually the linear equation f of 1 comma m equals 40 plus 0 .15m.

02:10 So because we know here the slope is 0 .15, for every 100 we increase, the cost will increase by 15.

02:23 So because we're increasing by 100 miles, the cost will increase by 15.

02:29 So $15, $15, that's 70, plus 15 is 85, plus 15 is 100.

02:42 Now that d equals 2, we get f of 2 comma m, which is just 40 times 2, 80 plus 0 .15m.

02:53 So when m equals 100, that's 80 plus 15, so that's 95...

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