Answer :
To determine which table represents a linear function, we need to check if the sets of values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex] exhibit a constant rate of change (i.e., the slope between points is consistent). Let's analyze each table individually.
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 9 \\ \hline 3 & 5 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, 5)\)[/tex] and [tex]\((2, 9)\)[/tex]:
[tex]\[ \text{slope} = \frac{9 - 5}{2 - 1} = \frac{4}{1} = 4 \][/tex]
- Between [tex]\((2, 9)\)[/tex] and [tex]\((3, 5)\)[/tex]:
[tex]\[ \text{slope} = \frac{5 - 9}{3 - 2} = \frac{-4}{1} = -4 \][/tex]
- Between [tex]\((3, 5)\)[/tex] and [tex]\((4, 9)\)[/tex]:
[tex]\[ \text{slope} = \frac{9 - 5}{4 - 3} = \frac{4}{1} = 4 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 10 \\ \hline 3 & -15 \\ \hline 4 & 20 \\ \hline \end{array} \][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, -5)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[ \text{slope} = \frac{10 - (-5)}{2 - 1} = \frac{15}{1} = 15 \][/tex]
- Between [tex]\((2, 10)\)[/tex] and [tex]\((3, -15)\)[/tex]:
[tex]\[ \text{slope} = \frac{-15 - 10}{3 - 2} = \frac{-25}{1} = -25 \][/tex]
- Between [tex]\((3, -15)\)[/tex] and [tex]\((4, 20)\)[/tex]:
[tex]\[ \text{slope} = \frac{20 - (-15)}{4 - 3} = \frac{35}{1} = 35 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 10 \\ \hline 3 & 20 \\ \hline 4 & 40 \\ \hline \end{array} \][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, 5)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[ \text{slope} = \frac{10 - 5}{2 - 1} = \frac{5}{1} = 5 \][/tex]
- Between [tex]\((2, 10)\)[/tex] and [tex]\((3, 20)\)[/tex]:
[tex]\[ \text{slope} = \frac{20 - 10}{3 - 2} = \frac{10}{1} = 10 \][/tex]
- Between [tex]\((3, 20)\)[/tex] and [tex]\((4, 40)\)[/tex]:
[tex]\[ \text{slope} = \frac{40 - 20}{4 - 3} = \frac{20}{1} = 20 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
Calculate the slope between the two points:
- Between [tex]\((1, -5)\)[/tex] and [tex]\((2, 0)\)[/tex]:
[tex]\[ \text{slope} = \frac{0 - (-5)}{2 - 1} = \frac{5}{1} = 5 \][/tex]
With only two points, the consistency of the slope is maintained by default. Therefore, this table represents a linear function.
### Conclusion
The table that represents a linear function is the one with [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
So, Table 4 represents a linear function.
### Table 1
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 9 \\ \hline 3 & 5 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, 5)\)[/tex] and [tex]\((2, 9)\)[/tex]:
[tex]\[ \text{slope} = \frac{9 - 5}{2 - 1} = \frac{4}{1} = 4 \][/tex]
- Between [tex]\((2, 9)\)[/tex] and [tex]\((3, 5)\)[/tex]:
[tex]\[ \text{slope} = \frac{5 - 9}{3 - 2} = \frac{-4}{1} = -4 \][/tex]
- Between [tex]\((3, 5)\)[/tex] and [tex]\((4, 9)\)[/tex]:
[tex]\[ \text{slope} = \frac{9 - 5}{4 - 3} = \frac{4}{1} = 4 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 2
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 10 \\ \hline 3 & -15 \\ \hline 4 & 20 \\ \hline \end{array} \][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, -5)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[ \text{slope} = \frac{10 - (-5)}{2 - 1} = \frac{15}{1} = 15 \][/tex]
- Between [tex]\((2, 10)\)[/tex] and [tex]\((3, -15)\)[/tex]:
[tex]\[ \text{slope} = \frac{-15 - 10}{3 - 2} = \frac{-25}{1} = -25 \][/tex]
- Between [tex]\((3, -15)\)[/tex] and [tex]\((4, 20)\)[/tex]:
[tex]\[ \text{slope} = \frac{20 - (-15)}{4 - 3} = \frac{35}{1} = 35 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 3
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 5 \\ \hline 2 & 10 \\ \hline 3 & 20 \\ \hline 4 & 40 \\ \hline \end{array} \][/tex]
Calculate the slopes between consecutive points:
- Between [tex]\((1, 5)\)[/tex] and [tex]\((2, 10)\)[/tex]:
[tex]\[ \text{slope} = \frac{10 - 5}{2 - 1} = \frac{5}{1} = 5 \][/tex]
- Between [tex]\((2, 10)\)[/tex] and [tex]\((3, 20)\)[/tex]:
[tex]\[ \text{slope} = \frac{20 - 10}{3 - 2} = \frac{10}{1} = 10 \][/tex]
- Between [tex]\((3, 20)\)[/tex] and [tex]\((4, 40)\)[/tex]:
[tex]\[ \text{slope} = \frac{40 - 20}{4 - 3} = \frac{20}{1} = 20 \][/tex]
The slopes are not consistent, so this table does not represent a linear function.
### Table 4
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
Calculate the slope between the two points:
- Between [tex]\((1, -5)\)[/tex] and [tex]\((2, 0)\)[/tex]:
[tex]\[ \text{slope} = \frac{0 - (-5)}{2 - 1} = \frac{5}{1} = 5 \][/tex]
With only two points, the consistency of the slope is maintained by default. Therefore, this table represents a linear function.
### Conclusion
The table that represents a linear function is the one with [tex]\( x \)[/tex] and [tex]\( y \)[/tex] values:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & -5 \\ \hline 2 & 0 \\ \hline \end{array} \][/tex]
So, Table 4 represents a linear function.
