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Sagot :
To determine which of the given tables represents a function, we need to ensure that each [tex]\( x \)[/tex]-value is paired with exactly one [tex]\( y \)[/tex]-value. A relation is a function if each input value from the domain (all possible [tex]\( x \)[/tex]-values) is associated with one and only one output value from the range (all possible [tex]\( y \)[/tex]-values).
Let's examine each table:
### Table [tex]\( W \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 15 \\ \hline -2 & 6 \\ \hline 0 & 15 \\ \hline 4 & 6 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\(-1, -2, 0, 4\)[/tex], which are all unique.
- Each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value.
Since each [tex]\( x \)[/tex]-value corresponds to one and only one [tex]\( y \)[/tex]-value, Table [tex]\( W \)[/tex] is a function.
### Table [tex]\( X \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 4 \\ \hline 6 & 15 \\ \hline 0 & 6 \\ \hline -2 & -1 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\( 0, 6, 0, -2\)[/tex].
- The [tex]\( x = 0 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (4 and 6).
Since the [tex]\( x \)[/tex]-value [tex]\( 0 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( X \)[/tex] is not a function.
### Table [tex]\( Y \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 4 & 0 \\ \hline -1 & -2 \\ \hline 6 & -2 \\ \hline 6 & 15 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\( 4, -1, 6, 6 \)[/tex].
- The [tex]\( x = 6 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (-2 and 15).
Since the [tex]\( x \)[/tex]-value [tex]\( 6 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( Y \)[/tex] is not a function.
### Table [tex]\( Z \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & -2 \\ \hline 4 & 0 \\ \hline 15 & -1 \\ \hline 4 & 3 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\( 6, 4, 15, 4 \)[/tex].
- The [tex]\( x = 4 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (0 and 3).
Since the [tex]\( x \)[/tex]-value [tex]\( 4 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( Z \)[/tex] is not a function.
Therefore, the correct answer is:
[tex]\[ \boxed{A. \, W} \][/tex]
Let's examine each table:
### Table [tex]\( W \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -1 & 15 \\ \hline -2 & 6 \\ \hline 0 & 15 \\ \hline 4 & 6 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\(-1, -2, 0, 4\)[/tex], which are all unique.
- Each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value.
Since each [tex]\( x \)[/tex]-value corresponds to one and only one [tex]\( y \)[/tex]-value, Table [tex]\( W \)[/tex] is a function.
### Table [tex]\( X \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 0 & 4 \\ \hline 6 & 15 \\ \hline 0 & 6 \\ \hline -2 & -1 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\( 0, 6, 0, -2\)[/tex].
- The [tex]\( x = 0 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (4 and 6).
Since the [tex]\( x \)[/tex]-value [tex]\( 0 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( X \)[/tex] is not a function.
### Table [tex]\( Y \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 4 & 0 \\ \hline -1 & -2 \\ \hline 6 & -2 \\ \hline 6 & 15 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\( 4, -1, 6, 6 \)[/tex].
- The [tex]\( x = 6 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (-2 and 15).
Since the [tex]\( x \)[/tex]-value [tex]\( 6 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( Y \)[/tex] is not a function.
### Table [tex]\( Z \)[/tex]
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & -2 \\ \hline 4 & 0 \\ \hline 15 & -1 \\ \hline 4 & 3 \\ \hline \end{array} \][/tex]
- The x-values are [tex]\( 6, 4, 15, 4 \)[/tex].
- The [tex]\( x = 4 \)[/tex] appears twice with different [tex]\( y \)[/tex]-values (0 and 3).
Since the [tex]\( x \)[/tex]-value [tex]\( 4 \)[/tex] maps to two different [tex]\( y \)[/tex]-values, Table [tex]\( Z \)[/tex] is not a function.
Therefore, the correct answer is:
[tex]\[ \boxed{A. \, W} \][/tex]
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