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Sagot :
To determine which table represents a linear function, we will examine the relationship between the [tex]\(x\)[/tex] and [tex]\(y\)[/tex] values in each table.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 7 \\ \hline 3 & 11 \\ \hline 4 & 15 \\ \hline \end{array} \][/tex]
Let's calculate the differences between consecutive [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 7 - 3 = 4 \][/tex]
[tex]\[ y_3 - y_2 = 11 - 7 = 4 \][/tex]
[tex]\[ y_4 - y_3 = 15 - 11 = 4 \][/tex]
Since the differences (slopes) are constant (4 each), Table 1 represents a linear function.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 8 \\ \hline 3 & 15 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
Let's calculate the differences between consecutive [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 8 - 3 = 5 \][/tex]
[tex]\[ y_3 - y_2 = 15 - 8 = 7 \][/tex]
[tex]\[ y_4 - y_3 = 21 - 15 = 6 \][/tex]
The differences (slopes) are not constant (5, 7, 6). Hence, Table 2 does not represent a linear function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 3 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
Let's calculate the differences between consecutive [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 9 - 3 = 6 \][/tex]
[tex]\[ y_3 - y_2 = 3 - 9 = -6 \][/tex]
[tex]\[ y_4 - y_3 = 9 - 3 = 6 \][/tex]
The differences (slopes) are not constant (6, -6, 6). Therefore, Table 3 does not represent a linear function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline \end{array} \][/tex]
Let's calculate the difference between the [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 9 - 3 = 6 \][/tex]
Since we only have two points, we can't definitively determine linearity from one difference.
### Conclusion:
After evaluating all the tables, we find that Table 1 is the one where the differences between the [tex]\(y\)[/tex]-values are constant, which means Table 1 represents a linear function.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 7 \\ \hline 3 & 11 \\ \hline 4 & 15 \\ \hline \end{array} \][/tex]
Let's calculate the differences between consecutive [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 7 - 3 = 4 \][/tex]
[tex]\[ y_3 - y_2 = 11 - 7 = 4 \][/tex]
[tex]\[ y_4 - y_3 = 15 - 11 = 4 \][/tex]
Since the differences (slopes) are constant (4 each), Table 1 represents a linear function.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 8 \\ \hline 3 & 15 \\ \hline 4 & 21 \\ \hline \end{array} \][/tex]
Let's calculate the differences between consecutive [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 8 - 3 = 5 \][/tex]
[tex]\[ y_3 - y_2 = 15 - 8 = 7 \][/tex]
[tex]\[ y_4 - y_3 = 21 - 15 = 6 \][/tex]
The differences (slopes) are not constant (5, 7, 6). Hence, Table 2 does not represent a linear function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline 3 & 3 \\ \hline 4 & 9 \\ \hline \end{array} \][/tex]
Let's calculate the differences between consecutive [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 9 - 3 = 6 \][/tex]
[tex]\[ y_3 - y_2 = 3 - 9 = -6 \][/tex]
[tex]\[ y_4 - y_3 = 9 - 3 = 6 \][/tex]
The differences (slopes) are not constant (6, -6, 6). Therefore, Table 3 does not represent a linear function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 1 & 3 \\ \hline 2 & 9 \\ \hline \end{array} \][/tex]
Let's calculate the difference between the [tex]\(y\)[/tex]-values:
[tex]\[ y_2 - y_1 = 9 - 3 = 6 \][/tex]
Since we only have two points, we can't definitively determine linearity from one difference.
### Conclusion:
After evaluating all the tables, we find that Table 1 is the one where the differences between the [tex]\(y\)[/tex]-values are constant, which means Table 1 represents a linear function.
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