Answer :
Given the problem, we know that the cost per day to care for each dog is [tex]$ \$[/tex]6$. To model this situation, we need to verify which table matches the equation [tex]\( y = 6x \)[/tex], where [tex]\( y \)[/tex] represents the total cost of caring for a single dog for [tex]\( x \)[/tex] days.
### Step-by-Step Solution:
1. Evaluate the first table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 12 & 6 \\ \hline 15 & 9 \\ \hline 18 & 12 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (12, 6): [tex]\( 6 \neq 6 \times 12 \)[/tex] (which simplifies to [tex]\( 6 \neq 72 \)[/tex]) => False
- For (15, 9): [tex]\( 9 \neq 6 \times 15 \)[/tex] (which simplifies to [tex]\( 9 \neq 90 \)[/tex]) => False
- For (18, 12): [tex]\( 12 \neq 6 \times 18 \)[/tex] (which simplifies to [tex]\( 12 \neq 108 \)[/tex]) => False
Thus, the first table does not model the situation.
2. Evaluate the second table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 12 \\ \hline 9 & 15 \\ \hline 12 & 18 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (6, 12): [tex]\( 12 \neq 6 \times 6 \)[/tex] (which simplifies to [tex]\( 12 \neq 36 \)[/tex]) => False
- For (9, 15): [tex]\( 15 \neq 6 \times 9 \)[/tex] (which simplifies to [tex]\( 15 \neq 54 \)[/tex]) => False
- For (12, 18): [tex]\( 18 \neq 6 \times 12 \)[/tex] (which simplifies to [tex]\( 18 \neq 72 \)[/tex]) => False
Thus, the second table does not model the situation.
3. Evaluate the third table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 36 & 6 \\ \hline 54 & 9 \\ \hline 72 & 12 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (36, 6): [tex]\( 6 \neq 6 \times 36 \)[/tex] (which simplifies to [tex]\( 6 \neq 216 \)[/tex]) => False
- For (54, 9): [tex]\( 9 \neq 6 \times 54 \)[/tex] (which simplifies to [tex]\( 9 \neq 324 \)[/tex]) => False
- For (72, 12): [tex]\( 12 \neq 6 \times 72 \)[/tex] (which simplifies to [tex]\( 12 \neq 432 \)[/tex]) => False
Thus, the third table does not model the situation.
4. Evaluate the fourth table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 36 \\ \hline 9 & 54 \\ \hline 12 & 72 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (6, 36): [tex]\( 36 = 6 \times 6 \)[/tex] => True
- For (9, 54): [tex]\( 54 = 6 \times 9 \)[/tex] => True
- For (12, 72): [tex]\( 72 = 6 \times 12 \)[/tex] => True
Thus, the fourth table does model the situation.
### Conclusion:
The fourth table, which consists of the pairs [tex]\((6, 36)\)[/tex], [tex]\((9, 54)\)[/tex], and [tex]\((12, 72)\)[/tex], correctly models the relationship where [tex]\( y = 6x \)[/tex]. Therefore, the valid table is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 36 \\ \hline 9 & 54 \\ \hline 12 & 72 \\ \hline \end{array} \][/tex]
### Step-by-Step Solution:
1. Evaluate the first table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 12 & 6 \\ \hline 15 & 9 \\ \hline 18 & 12 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (12, 6): [tex]\( 6 \neq 6 \times 12 \)[/tex] (which simplifies to [tex]\( 6 \neq 72 \)[/tex]) => False
- For (15, 9): [tex]\( 9 \neq 6 \times 15 \)[/tex] (which simplifies to [tex]\( 9 \neq 90 \)[/tex]) => False
- For (18, 12): [tex]\( 12 \neq 6 \times 18 \)[/tex] (which simplifies to [tex]\( 12 \neq 108 \)[/tex]) => False
Thus, the first table does not model the situation.
2. Evaluate the second table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 12 \\ \hline 9 & 15 \\ \hline 12 & 18 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (6, 12): [tex]\( 12 \neq 6 \times 6 \)[/tex] (which simplifies to [tex]\( 12 \neq 36 \)[/tex]) => False
- For (9, 15): [tex]\( 15 \neq 6 \times 9 \)[/tex] (which simplifies to [tex]\( 15 \neq 54 \)[/tex]) => False
- For (12, 18): [tex]\( 18 \neq 6 \times 12 \)[/tex] (which simplifies to [tex]\( 18 \neq 72 \)[/tex]) => False
Thus, the second table does not model the situation.
3. Evaluate the third table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 36 & 6 \\ \hline 54 & 9 \\ \hline 72 & 12 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (36, 6): [tex]\( 6 \neq 6 \times 36 \)[/tex] (which simplifies to [tex]\( 6 \neq 216 \)[/tex]) => False
- For (54, 9): [tex]\( 9 \neq 6 \times 54 \)[/tex] (which simplifies to [tex]\( 9 \neq 324 \)[/tex]) => False
- For (72, 12): [tex]\( 12 \neq 6 \times 72 \)[/tex] (which simplifies to [tex]\( 12 \neq 432 \)[/tex]) => False
Thus, the third table does not model the situation.
4. Evaluate the fourth table:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 36 \\ \hline 9 & 54 \\ \hline 12 & 72 \\ \hline \end{array} \][/tex]
For each entry, check if [tex]\( y = 6x \)[/tex]:
- For (6, 36): [tex]\( 36 = 6 \times 6 \)[/tex] => True
- For (9, 54): [tex]\( 54 = 6 \times 9 \)[/tex] => True
- For (12, 72): [tex]\( 72 = 6 \times 12 \)[/tex] => True
Thus, the fourth table does model the situation.
### Conclusion:
The fourth table, which consists of the pairs [tex]\((6, 36)\)[/tex], [tex]\((9, 54)\)[/tex], and [tex]\((12, 72)\)[/tex], correctly models the relationship where [tex]\( y = 6x \)[/tex]. Therefore, the valid table is:
[tex]\[ \begin{array}{|c|c|} \hline x & y \\ \hline 6 & 36 \\ \hline 9 & 54 \\ \hline 12 & 72 \\ \hline \end{array} \][/tex]
