Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To determine which table represents a function, we need to examine if every [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value in each table. A relation is a function if no [tex]\( x \)[/tex]-value is repeated with different [tex]\( y \)[/tex]-values. Let's analyze each table.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & -1 \\ \hline 0 & 0 \\ \hline -2 & -1 \\ \hline 8 & 1 \\ \hline \end{array} \][/tex]
In this table:
- Each [tex]\( x \)[/tex] value (-3, 0, -2, 8) is unique and maps to exactly one [tex]\( y \)[/tex] value.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -5 & -5 \\ \hline 0 & 0 \\ \hline -5 & 5 \\ \hline 6 & -6 \\ \hline \end{array} \][/tex]
In this table:
- The [tex]\( x \)[/tex]-value -5 maps to two different [tex]\( y \)[/tex]-values (-5 and 5), which disqualifies it from being a function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 8 \\ \hline -2 & 2 \\ \hline -2 & 4 \\ \hline 0 & 2 \\ \hline \end{array} \][/tex]
In this table:
- The [tex]\( x \)[/tex]-value -2 maps to two different [tex]\( y \)[/tex]-values (2 and 4), which disqualifies it from being a function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 2 \\ \hline 3 & 5 \\ \hline 1 & 3 \\ \hline -4 & 0 \\ \hline \end{array} \][/tex]
In this table:
- The [tex]\( x \)[/tex]-value -4 maps to two different [tex]\( y \)[/tex]-values (2 and 0), which disqualifies it from being a function.
### Conclusion:
After examining all the tables, only Table 1 ensures that each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value. Therefore, Table 1 represents a function.
### Table 1:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -3 & -1 \\ \hline 0 & 0 \\ \hline -2 & -1 \\ \hline 8 & 1 \\ \hline \end{array} \][/tex]
In this table:
- Each [tex]\( x \)[/tex] value (-3, 0, -2, 8) is unique and maps to exactly one [tex]\( y \)[/tex] value.
### Table 2:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -5 & -5 \\ \hline 0 & 0 \\ \hline -5 & 5 \\ \hline 6 & -6 \\ \hline \end{array} \][/tex]
In this table:
- The [tex]\( x \)[/tex]-value -5 maps to two different [tex]\( y \)[/tex]-values (-5 and 5), which disqualifies it from being a function.
### Table 3:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 8 \\ \hline -2 & 2 \\ \hline -2 & 4 \\ \hline 0 & 2 \\ \hline \end{array} \][/tex]
In this table:
- The [tex]\( x \)[/tex]-value -2 maps to two different [tex]\( y \)[/tex]-values (2 and 4), which disqualifies it from being a function.
### Table 4:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline -4 & 2 \\ \hline 3 & 5 \\ \hline 1 & 3 \\ \hline -4 & 0 \\ \hline \end{array} \][/tex]
In this table:
- The [tex]\( x \)[/tex]-value -4 maps to two different [tex]\( y \)[/tex]-values (2 and 0), which disqualifies it from being a function.
### Conclusion:
After examining all the tables, only Table 1 ensures that each [tex]\( x \)[/tex]-value maps to exactly one [tex]\( y \)[/tex]-value. Therefore, Table 1 represents a function.
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.