Answer :
To determine which relations represent a function, we need to check if each input value maps to only one unique output value. Here are the detailed steps and results for each of the provided tables:
First table:
[tex]\[ \begin{tabular}{c|cccc} Input & -2 & 9 & -5 & 9 \\ \hline Output & 7 & 0 & 9 & 3 \\ \end{tabular} \][/tex]
- Input [tex]\(-2\)[/tex] maps to output [tex]\(7\)[/tex].
- Input [tex]\(9\)[/tex] maps to output [tex]\(0\)[/tex].
- Input [tex]\(-5\)[/tex] maps to output [tex]\(9\)[/tex].
- Input [tex]\(9\)[/tex] maps to output [tex]\(3\)[/tex].
In this table, the input [tex]\(9\)[/tex] maps to two different outputs, [tex]\(0\)[/tex] and [tex]\(3\)[/tex]. Therefore, this relation is not a function.
Second table:
[tex]\[ \begin{tabular}{c|cccc} Input & -5 & 8 & -2 & 2 \\ \hline Output & 3 & 0 & -4 & -4 \\ \end{tabular} \][/tex]
- Input [tex]\(-5\)[/tex] maps to output [tex]\(3\)[/tex].
- Input [tex]\(8\)[/tex] maps to output [tex]\(0\)[/tex].
- Input [tex]\(-2\)[/tex] maps to output [tex]\(-4\)[/tex].
- Input [tex]\(2\)[/tex] maps to output [tex]\(-4\)[/tex].
In this table, each input maps to a unique output. Therefore, this relation is a function.
Third table:
[tex]\[ \begin{tabular}{c|cccc} Input & -4 & -2 & 3 & 6 \\ \hline Output & 6 & 5 & 6 & 8 \\ \end{tabular} \][/tex]
- Input [tex]\(-4\)[/tex] maps to output [tex]\(6\)[/tex].
- Input [tex]\(-2\)[/tex] maps to output [tex]\(5\)[/tex].
- Input [tex]\(3\)[/tex] maps to output [tex]\(6\)[/tex].
- Input [tex]\(6\)[/tex] maps to output [tex]\(8\)[/tex].
In this table, each input maps to a unique output. Therefore, this relation is a function.
Conclusion:
Based on the analysis, the relations that are functions are:
1. [tex]\[ \begin{tabular}{c|cccc} Input & -5 & 8 & -2 & 2 \\ \hline Output & 3 & 0 & -4 & -4 \\ \end{tabular} \][/tex]
2. [tex]\[ \begin{tabular}{c|cccc} Input & -4 & -2 & 3 & 6 \\ \hline Output & 6 & 5 & 6 & 8 \\ \end{tabular} \][/tex]
Hence, the correct selections are the second and third tables.
First table:
[tex]\[ \begin{tabular}{c|cccc} Input & -2 & 9 & -5 & 9 \\ \hline Output & 7 & 0 & 9 & 3 \\ \end{tabular} \][/tex]
- Input [tex]\(-2\)[/tex] maps to output [tex]\(7\)[/tex].
- Input [tex]\(9\)[/tex] maps to output [tex]\(0\)[/tex].
- Input [tex]\(-5\)[/tex] maps to output [tex]\(9\)[/tex].
- Input [tex]\(9\)[/tex] maps to output [tex]\(3\)[/tex].
In this table, the input [tex]\(9\)[/tex] maps to two different outputs, [tex]\(0\)[/tex] and [tex]\(3\)[/tex]. Therefore, this relation is not a function.
Second table:
[tex]\[ \begin{tabular}{c|cccc} Input & -5 & 8 & -2 & 2 \\ \hline Output & 3 & 0 & -4 & -4 \\ \end{tabular} \][/tex]
- Input [tex]\(-5\)[/tex] maps to output [tex]\(3\)[/tex].
- Input [tex]\(8\)[/tex] maps to output [tex]\(0\)[/tex].
- Input [tex]\(-2\)[/tex] maps to output [tex]\(-4\)[/tex].
- Input [tex]\(2\)[/tex] maps to output [tex]\(-4\)[/tex].
In this table, each input maps to a unique output. Therefore, this relation is a function.
Third table:
[tex]\[ \begin{tabular}{c|cccc} Input & -4 & -2 & 3 & 6 \\ \hline Output & 6 & 5 & 6 & 8 \\ \end{tabular} \][/tex]
- Input [tex]\(-4\)[/tex] maps to output [tex]\(6\)[/tex].
- Input [tex]\(-2\)[/tex] maps to output [tex]\(5\)[/tex].
- Input [tex]\(3\)[/tex] maps to output [tex]\(6\)[/tex].
- Input [tex]\(6\)[/tex] maps to output [tex]\(8\)[/tex].
In this table, each input maps to a unique output. Therefore, this relation is a function.
Conclusion:
Based on the analysis, the relations that are functions are:
1. [tex]\[ \begin{tabular}{c|cccc} Input & -5 & 8 & -2 & 2 \\ \hline Output & 3 & 0 & -4 & -4 \\ \end{tabular} \][/tex]
2. [tex]\[ \begin{tabular}{c|cccc} Input & -4 & -2 & 3 & 6 \\ \hline Output & 6 & 5 & 6 & 8 \\ \end{tabular} \][/tex]
Hence, the correct selections are the second and third tables.
