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To determine which table represents a linear function, we need to check if the relation between [tex]\(x\)[/tex] and [tex]\(y\)[/tex] follows a constant rate of change, known as the slope. For a function to be linear, the differences between consecutive [tex]\(x\)[/tex] values and their corresponding [tex]\(y\)[/tex] values should result in the same slope when calculated.
Let's examine each table in detail:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 6 \\ \hline 3 & 12 \\ \hline 4 & 24 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = 3, 6, 12\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are 3, 6, and 12)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{3}{1}, \frac{6}{1}, \frac{12}{1} \][/tex]
The slopes are 3, 6, and 12, which are not consistent. Thus, Table 1 does not represent a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 2 \\ \hline 2 & 5 \\ \hline 3 & 9 \\ \hline 4 & 14 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = 3, 4, 5\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are 3, 4, and 5)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{3}{1}, \frac{4}{1}, \frac{5}{1} \][/tex]
The slopes are 3, 4, and 5, which are not consistent. Thus, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -3 \\ \hline 2 & -5 \\ \hline 3 & -7 \\ \hline 4 & -9 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = -2, -2, -2\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are all -2)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{-2}{1}, \frac{-2}{1}, \frac{-2}{1} \][/tex]
The slopes are all -2, which are consistent. Therefore, Table 3 represents a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -4 \\ \hline 3 & -2 \\ \hline 4 & 0 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = -2, 2, 2\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are -2, 2, and 2)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{-2}{1}, \frac{2}{1}, \frac{2}{1} \][/tex]
The slopes are -2, 2, and 2, which are not consistent. Thus, Table 4 does not represent a linear function.
### Conclusion
Only Table 3 represents a linear function.
Let's examine each table in detail:
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 6 \\ \hline 3 & 12 \\ \hline 4 & 24 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = 3, 6, 12\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are 3, 6, and 12)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{3}{1}, \frac{6}{1}, \frac{12}{1} \][/tex]
The slopes are 3, 6, and 12, which are not consistent. Thus, Table 1 does not represent a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 2 \\ \hline 2 & 5 \\ \hline 3 & 9 \\ \hline 4 & 14 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = 3, 4, 5\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are 3, 4, and 5)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{3}{1}, \frac{4}{1}, \frac{5}{1} \][/tex]
The slopes are 3, 4, and 5, which are not consistent. Thus, Table 2 does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -3 \\ \hline 2 & -5 \\ \hline 3 & -7 \\ \hline 4 & -9 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = -2, -2, -2\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are all -2)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{-2}{1}, \frac{-2}{1}, \frac{-2}{1} \][/tex]
The slopes are all -2, which are consistent. Therefore, Table 3 represents a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -2 \\ \hline 2 & -4 \\ \hline 3 & -2 \\ \hline 4 & 0 \\ \hline \end{array} \][/tex]
Calculate the differences in [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values:
- [tex]\(\Delta x = 1, 1, 1\)[/tex] (differences between consecutive [tex]\(x\)[/tex] values are all 1)
- [tex]\(\Delta y = -2, 2, 2\)[/tex] (differences between consecutive [tex]\(y\)[/tex] values are -2, 2, and 2)
Calculate the slopes:
[tex]\[ \frac{\Delta y}{\Delta x} = \frac{-2}{1}, \frac{2}{1}, \frac{2}{1} \][/tex]
The slopes are -2, 2, and 2, which are not consistent. Thus, Table 4 does not represent a linear function.
### Conclusion
Only Table 3 represents a linear function.
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