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Sagot :
To determine if the function represents a direct variation, we need to check two main properties:
1. The function must pass through the origin (i.e., when [tex]\( \text{Time} = 0 \)[/tex], [tex]\( \text{Cost} = 0 \)[/tex]).
2. The function must have a constant rate of change.
Let's analyze the given data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (hours)} & \text{Cost (\$)} \\ \hline 0 & 0 \\ \hline 2 & 10 \\ \hline 4 & 20 \\ \hline 6 & 30 \\ \hline 8 & 40 \\ \hline \end{array} \][/tex]
### Step 1: Determine if the function passes through the origin
From the table, when [tex]\( \text{Time} = 0 \)[/tex], [tex]\( \text{Cost} = 0 \)[/tex]. This shows that the function passes through the origin.
### Step 2: Determine the rate of change
To find the rate of change, we can use any two points from the table. Let's use the first two points [tex]\((0, 0)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{\Delta \text{Cost}}{\Delta \text{Time}} = \frac{10 - 0}{2 - 0} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
Next, let's verify if this rate of change is constant for all points in the table:
- Between [tex]\((2, 10)\)[/tex] and [tex]\((4, 20)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{20 - 10}{4 - 2} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
- Between [tex]\((4, 20)\)[/tex] and [tex]\((6, 30)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{30 - 20}{6 - 4} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
- Between [tex]\((6, 30)\)[/tex] and [tex]\((8, 40)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{40 - 30}{8 - 6} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
Since the rate of change is constant at 5 \[tex]$/hour at every interval, and the function passes through the origin, we can conclude that this function represents a direct variation. ### Conclusion The function represents a direct variation because it passes through the origin and has a constant rate of change of $[/tex] 5 per hour.
The correct explanation is:
"This function represents a direct variation because it passes through the origin and has a constant rate of change of $ 5 per hour."
1. The function must pass through the origin (i.e., when [tex]\( \text{Time} = 0 \)[/tex], [tex]\( \text{Cost} = 0 \)[/tex]).
2. The function must have a constant rate of change.
Let's analyze the given data:
[tex]\[ \begin{array}{|c|c|} \hline \text{Time (hours)} & \text{Cost (\$)} \\ \hline 0 & 0 \\ \hline 2 & 10 \\ \hline 4 & 20 \\ \hline 6 & 30 \\ \hline 8 & 40 \\ \hline \end{array} \][/tex]
### Step 1: Determine if the function passes through the origin
From the table, when [tex]\( \text{Time} = 0 \)[/tex], [tex]\( \text{Cost} = 0 \)[/tex]. This shows that the function passes through the origin.
### Step 2: Determine the rate of change
To find the rate of change, we can use any two points from the table. Let's use the first two points [tex]\((0, 0)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{\Delta \text{Cost}}{\Delta \text{Time}} = \frac{10 - 0}{2 - 0} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
Next, let's verify if this rate of change is constant for all points in the table:
- Between [tex]\((2, 10)\)[/tex] and [tex]\((4, 20)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{20 - 10}{4 - 2} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
- Between [tex]\((4, 20)\)[/tex] and [tex]\((6, 30)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{30 - 20}{6 - 4} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
- Between [tex]\((6, 30)\)[/tex] and [tex]\((8, 40)\)[/tex]:
[tex]\[ \text{Rate of Change} = \frac{40 - 30}{8 - 6} = \frac{10}{2} = 5 \, \text{\$/hour} \][/tex]
Since the rate of change is constant at 5 \[tex]$/hour at every interval, and the function passes through the origin, we can conclude that this function represents a direct variation. ### Conclusion The function represents a direct variation because it passes through the origin and has a constant rate of change of $[/tex] 5 per hour.
The correct explanation is:
"This function represents a direct variation because it passes through the origin and has a constant rate of change of $ 5 per hour."
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