Answer :
Let's analyze each of the given tables and determine which one matches the equation [tex]\(c = 3.5 \cdot b\)[/tex] while including only viable solutions. The equation [tex]\(c = 3.5 \cdot b\)[/tex] means that the cost [tex]\(c\)[/tex] is directly proportional to the number of loaves [tex]\(b\)[/tex] with a proportionality constant of 3.5.
### Table 1
[tex]\[ \begin{tabular}{|c|c|} \hline Loaves (b) & Cost $(c)$ \\ \hline -2 & -7 \\ \hline 0 & 0 \\ \hline 2 & 7 \\ \hline 4 & 14 \\ \hline \end{tabular} \][/tex]
Let's check each pair:
- For [tex]\(b = -2\)[/tex], [tex]\(c = 3.5 \times (-2) = -7\)[/tex]. The value matches.
- For [tex]\(b = 0\)[/tex], [tex]\(c = 3.5 \times 0 = 0\)[/tex]. The value matches.
- For [tex]\(b = 2\)[/tex], [tex]\(c = 3.5 \times 2 = 7\)[/tex]. The value matches.
- For [tex]\(b = 4\)[/tex], [tex]\(c = 3.5 \times 4 = 14\)[/tex]. The value matches.
All the values in this table match the equation.
### Table 2
[tex]\[ \begin{tabular}{|c|c|} \hline Loaves $(b)$ & Cost $(c)$ \\ \hline 0 & 0 \\ \hline 0.5 & 1.75 \\ \hline 1 & 3.5 \\ \hline 1.5 & 5.25 \\ \hline \end{tabular} \][/tex]
Let's check each pair:
- For [tex]\(b = 0\)[/tex], [tex]\(c = 3.5 \times 0 = 0\)[/tex]. The value matches.
- For [tex]\(b = 0.5\)[/tex], [tex]\(c = 3.5 \times 0.5 = 1.75\)[/tex]. The value matches.
- For [tex]\(b = 1\)[/tex], [tex]\(c = 3.5 \times 1 = 3.5\)[/tex]. The value matches.
- For [tex]\(b = 1.5\)[/tex], [tex]\(c = 3.5 \times 1.5 = 5.25\)[/tex]. The value matches.
All the values in this table match the equation.
### Table 3
[tex]\[ \begin{tabular}{|c|c|} \hline Loaves $(b)$ & Cost $(c)$ \\ \hline 0 & 0 \\ \hline 3 & 10.5 \\ \hline 6 & 21 \\ \hline 9 & 31.5 \\ \hline \end{tabular} \][/tex]
Let's check each pair:
- For [tex]\(b = 0\)[/tex], [tex]\(c = 3.5 \times 0 = 0\)[/tex]. The value matches.
- For [tex]\(b = 3\)[/tex], [tex]\(c = 3.5 \times 3 = 10.5\)[/tex]. The value matches.
- For [tex]\(b = 6\)[/tex], [tex]\(c = 3.5 \times 6 = 21\)[/tex]. The value matches.
- For [tex]\(b = 9\)[/tex], [tex]\(c = 3.5 \times 9 = 31.5\)[/tex]. The value matches.
All the values in this table match the equation.
### Conclusion
All three tables (Table 1, Table 2, and Table 3) have values that match the equation [tex]\(c = 3.5 \cdot b\)[/tex]. However, we need to choose the table with only viable solutions. Typically, for such problems, viable solutions refer to non-negative values, as negative values for number of loaves or cost don't make practical sense.
- Table 1 includes a negative value for [tex]\(b\)[/tex] and [tex]\(c\)[/tex] (-2, -7), making it not entirely viable.
- Table 2 includes non-negative values that are fractions and whole numbers of loaves, which are viable.
- Table 3 also includes non-negative values that are whole numbers of loaves, which are viable.
Both Table 2 and Table 3 include only viable solutions. But considering standard practice and the need to align with realistic whole numbers for purchasing loaves of bread, Table 3 is often considered more practical with whole numbers.
Thus, Table 3 is the table of values that matches the equation [tex]\(c = 3.5 \cdot b\)[/tex] and includes only viable solutions.
### Table 1
[tex]\[ \begin{tabular}{|c|c|} \hline Loaves (b) & Cost $(c)$ \\ \hline -2 & -7 \\ \hline 0 & 0 \\ \hline 2 & 7 \\ \hline 4 & 14 \\ \hline \end{tabular} \][/tex]
Let's check each pair:
- For [tex]\(b = -2\)[/tex], [tex]\(c = 3.5 \times (-2) = -7\)[/tex]. The value matches.
- For [tex]\(b = 0\)[/tex], [tex]\(c = 3.5 \times 0 = 0\)[/tex]. The value matches.
- For [tex]\(b = 2\)[/tex], [tex]\(c = 3.5 \times 2 = 7\)[/tex]. The value matches.
- For [tex]\(b = 4\)[/tex], [tex]\(c = 3.5 \times 4 = 14\)[/tex]. The value matches.
All the values in this table match the equation.
### Table 2
[tex]\[ \begin{tabular}{|c|c|} \hline Loaves $(b)$ & Cost $(c)$ \\ \hline 0 & 0 \\ \hline 0.5 & 1.75 \\ \hline 1 & 3.5 \\ \hline 1.5 & 5.25 \\ \hline \end{tabular} \][/tex]
Let's check each pair:
- For [tex]\(b = 0\)[/tex], [tex]\(c = 3.5 \times 0 = 0\)[/tex]. The value matches.
- For [tex]\(b = 0.5\)[/tex], [tex]\(c = 3.5 \times 0.5 = 1.75\)[/tex]. The value matches.
- For [tex]\(b = 1\)[/tex], [tex]\(c = 3.5 \times 1 = 3.5\)[/tex]. The value matches.
- For [tex]\(b = 1.5\)[/tex], [tex]\(c = 3.5 \times 1.5 = 5.25\)[/tex]. The value matches.
All the values in this table match the equation.
### Table 3
[tex]\[ \begin{tabular}{|c|c|} \hline Loaves $(b)$ & Cost $(c)$ \\ \hline 0 & 0 \\ \hline 3 & 10.5 \\ \hline 6 & 21 \\ \hline 9 & 31.5 \\ \hline \end{tabular} \][/tex]
Let's check each pair:
- For [tex]\(b = 0\)[/tex], [tex]\(c = 3.5 \times 0 = 0\)[/tex]. The value matches.
- For [tex]\(b = 3\)[/tex], [tex]\(c = 3.5 \times 3 = 10.5\)[/tex]. The value matches.
- For [tex]\(b = 6\)[/tex], [tex]\(c = 3.5 \times 6 = 21\)[/tex]. The value matches.
- For [tex]\(b = 9\)[/tex], [tex]\(c = 3.5 \times 9 = 31.5\)[/tex]. The value matches.
All the values in this table match the equation.
### Conclusion
All three tables (Table 1, Table 2, and Table 3) have values that match the equation [tex]\(c = 3.5 \cdot b\)[/tex]. However, we need to choose the table with only viable solutions. Typically, for such problems, viable solutions refer to non-negative values, as negative values for number of loaves or cost don't make practical sense.
- Table 1 includes a negative value for [tex]\(b\)[/tex] and [tex]\(c\)[/tex] (-2, -7), making it not entirely viable.
- Table 2 includes non-negative values that are fractions and whole numbers of loaves, which are viable.
- Table 3 also includes non-negative values that are whole numbers of loaves, which are viable.
Both Table 2 and Table 3 include only viable solutions. But considering standard practice and the need to align with realistic whole numbers for purchasing loaves of bread, Table 3 is often considered more practical with whole numbers.
Thus, Table 3 is the table of values that matches the equation [tex]\(c = 3.5 \cdot b\)[/tex] and includes only viable solutions.
