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Sagot :
Alright, let's tackle this step-by-step.
Given the data:
- In May, the cost was [tex]$380 for 480 miles. - In June, the cost was $[/tex]440 for 780 miles.
We need to find a linear function [tex]\( C(x) \)[/tex] that models the cost of driving [tex]\( x \)[/tex] miles per month.
(a) Finding the Linear Function [tex]\( C(x) \)[/tex]
A linear function is generally of the form:
[tex]\[ C(x) = mx + b \][/tex]
To find the values of [tex]\( m \)[/tex] (the slope) and [tex]\( b \)[/tex] (the y-intercept), we can start by finding the slope, [tex]\( m \)[/tex]. The slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
In our case, we have points:
- May: [tex]\((480, 380)\)[/tex]
- June: [tex]\((780, 440)\)[/tex]
Calculate the slope:
[tex]\[ m = \frac{440 - 380}{780 - 480} = \frac{60}{300} = 0.2 \][/tex]
Next, we use one of the points to find the y-intercept [tex]\( b \)[/tex]. We can use the May data point [tex]\((480, 380)\)[/tex]:
[tex]\[ 380 = 0.2 \cdot 480 + b \][/tex]
[tex]\[ 380 = 96 + b \][/tex]
[tex]\[ b = 380 - 96 \][/tex]
[tex]\[ b = 284 \][/tex]
Therefore, the linear function [tex]\( C(x) \)[/tex] is:
[tex]\[ C(x) = 0.2x + 284 \][/tex]
(b) Drawing the Graph of [tex]\( C(x) \)[/tex]
To draw the graph of the linear function [tex]\( C(x) = 0.2x + 284 \)[/tex], follow these steps:
1. Plot the y-intercept: Begin by plotting the point where the line crosses the y-axis. This is [tex]\( (0, 284) \)[/tex].
2. Use the slope to find another point: The slope [tex]\( m = 0.2 \)[/tex] indicates that for every 1 increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by 0.2. However, for better scaling, let's use the given points:
- For May (480 miles, [tex]$380): Plot the point \( (480, 380) \). - For June (780 miles, $[/tex]440): Plot the point [tex]\( (780, 440) \)[/tex].
3. Draw the line: Connect these points with a straight line, extending the line beyond these points as needed to represent the linear function over a range.
Your graph will show a line starting at the intercept point (0, 284) and passing through (480, 380) and (780, 440), illustrating the increase in cost as the number of miles driven increases.
By following these steps, you model and visualize how the cost [tex]\( C \)[/tex] changes in relation to the miles driven [tex]\( x \)[/tex].
So the final answers are:
(a) The linear function is:
[tex]\[ C(x) = 0.2x + 284 \][/tex]
(b) The graph of [tex]\( C(x) = 0.2x + 284 \)[/tex] will be a straight line starting at (0, 284) and passing through the points (480, 380) and (780, 440).
Given the data:
- In May, the cost was [tex]$380 for 480 miles. - In June, the cost was $[/tex]440 for 780 miles.
We need to find a linear function [tex]\( C(x) \)[/tex] that models the cost of driving [tex]\( x \)[/tex] miles per month.
(a) Finding the Linear Function [tex]\( C(x) \)[/tex]
A linear function is generally of the form:
[tex]\[ C(x) = mx + b \][/tex]
To find the values of [tex]\( m \)[/tex] (the slope) and [tex]\( b \)[/tex] (the y-intercept), we can start by finding the slope, [tex]\( m \)[/tex]. The slope [tex]\( m \)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
In our case, we have points:
- May: [tex]\((480, 380)\)[/tex]
- June: [tex]\((780, 440)\)[/tex]
Calculate the slope:
[tex]\[ m = \frac{440 - 380}{780 - 480} = \frac{60}{300} = 0.2 \][/tex]
Next, we use one of the points to find the y-intercept [tex]\( b \)[/tex]. We can use the May data point [tex]\((480, 380)\)[/tex]:
[tex]\[ 380 = 0.2 \cdot 480 + b \][/tex]
[tex]\[ 380 = 96 + b \][/tex]
[tex]\[ b = 380 - 96 \][/tex]
[tex]\[ b = 284 \][/tex]
Therefore, the linear function [tex]\( C(x) \)[/tex] is:
[tex]\[ C(x) = 0.2x + 284 \][/tex]
(b) Drawing the Graph of [tex]\( C(x) \)[/tex]
To draw the graph of the linear function [tex]\( C(x) = 0.2x + 284 \)[/tex], follow these steps:
1. Plot the y-intercept: Begin by plotting the point where the line crosses the y-axis. This is [tex]\( (0, 284) \)[/tex].
2. Use the slope to find another point: The slope [tex]\( m = 0.2 \)[/tex] indicates that for every 1 increase in [tex]\( x \)[/tex], [tex]\( y \)[/tex] increases by 0.2. However, for better scaling, let's use the given points:
- For May (480 miles, [tex]$380): Plot the point \( (480, 380) \). - For June (780 miles, $[/tex]440): Plot the point [tex]\( (780, 440) \)[/tex].
3. Draw the line: Connect these points with a straight line, extending the line beyond these points as needed to represent the linear function over a range.
Your graph will show a line starting at the intercept point (0, 284) and passing through (480, 380) and (780, 440), illustrating the increase in cost as the number of miles driven increases.
By following these steps, you model and visualize how the cost [tex]\( C \)[/tex] changes in relation to the miles driven [tex]\( x \)[/tex].
So the final answers are:
(a) The linear function is:
[tex]\[ C(x) = 0.2x + 284 \][/tex]
(b) The graph of [tex]\( C(x) = 0.2x + 284 \)[/tex] will be a straight line starting at (0, 284) and passing through the points (480, 380) and (780, 440).
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