Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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]
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
