Answer :
To determine which table of values matches the equation [tex]\( c = 3.5b \)[/tex] and includes only viable solutions, let's first understand what viable solutions mean in this context. Here, viable solutions should include the number of loaves of bread purchased and the corresponding cost in dollars, based on the equation provided.
### Given Tables
1. First Table
[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]
2. Second Table
[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]
3. Third Table
[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]
### Calculation Check
Let's verify which table matches the given equation [tex]\( c = 3.5b \)[/tex] using viable values.
1. First Table:
- [tex]\( b = -2 \)[/tex], [tex]\( c = 3.5 \times (-2) = -7 \)[/tex]
- [tex]\( b = 0 \)[/tex], [tex]\( c = 3.5 \times 0 = 0 \)[/tex]
- [tex]\( b = 2 \)[/tex], [tex]\( c = 3.5 \times 2 = 7 \)[/tex]
- [tex]\( b = 4 \)[/tex], [tex]\( c = 3.5 \times 4 = 14 \)[/tex]
These match the costs calculated, but [tex]\( b = -2 \)[/tex] is not a viable solution for the number of loaves of bread (as you cannot purchase negative loaves).
2. Second Table:
- [tex]\( b = 0 \)[/tex], [tex]\( c = 3.5 \times 0 = 0 \)[/tex]
- [tex]\( b = 0.5 \)[/tex], [tex]\( c = 3.5 \times 0.5 = 1.75 \)[/tex]
- [tex]\( b = 1 \)[/tex], [tex]\( c = 3.5 \times 1 = 3.5 \)[/tex]
- [tex]\( b = 1.5 \)[/tex], [tex]\( c = 3.5 \times 1.5 = 5.25 \)[/tex]
These values all match the costs calculated and include only non-negative values for [tex]\( b \)[/tex].
3. Third Table:
- [tex]\( b = 0 \)[/tex], [tex]\( c = 3.5 \times 0 = 0 \)[/tex]
- [tex]\( b = 3 \)[/tex], [tex]\( c = 3.5 \times 3 = 10.5 \)[/tex]
- [tex]\( b = 6 \)[/tex], [tex]\( c = 3.5 \times 6 = 21 \)[/tex]
- [tex]\( b = 9 \)[/tex], [tex]\( c = 3.5 \times 9 = 31.5 \)[/tex]
These values also match the costs calculated and include only non-negative values for [tex]\( b \)[/tex].
### Conclusion
While all three tables can match the cost values generated by the equation [tex]\( c = 3.5b \)[/tex], the criterion of viable solutions — typically non-negative number of loaves [tex]\( b \)[/tex] — disqualifies the first table due to the negative value.
Therefore, both the second and third tables include only viable solutions.
However, as they ask for the 'only viable solutions', it is implied that the values of loaves should not be negative. So either of the second and third tables should be correct, with second table being the primary choice for lower range of values.
### Given Tables
1. First Table
[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]
2. Second Table
[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]
3. Third Table
[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]
### Calculation Check
Let's verify which table matches the given equation [tex]\( c = 3.5b \)[/tex] using viable values.
1. First Table:
- [tex]\( b = -2 \)[/tex], [tex]\( c = 3.5 \times (-2) = -7 \)[/tex]
- [tex]\( b = 0 \)[/tex], [tex]\( c = 3.5 \times 0 = 0 \)[/tex]
- [tex]\( b = 2 \)[/tex], [tex]\( c = 3.5 \times 2 = 7 \)[/tex]
- [tex]\( b = 4 \)[/tex], [tex]\( c = 3.5 \times 4 = 14 \)[/tex]
These match the costs calculated, but [tex]\( b = -2 \)[/tex] is not a viable solution for the number of loaves of bread (as you cannot purchase negative loaves).
2. Second Table:
- [tex]\( b = 0 \)[/tex], [tex]\( c = 3.5 \times 0 = 0 \)[/tex]
- [tex]\( b = 0.5 \)[/tex], [tex]\( c = 3.5 \times 0.5 = 1.75 \)[/tex]
- [tex]\( b = 1 \)[/tex], [tex]\( c = 3.5 \times 1 = 3.5 \)[/tex]
- [tex]\( b = 1.5 \)[/tex], [tex]\( c = 3.5 \times 1.5 = 5.25 \)[/tex]
These values all match the costs calculated and include only non-negative values for [tex]\( b \)[/tex].
3. Third Table:
- [tex]\( b = 0 \)[/tex], [tex]\( c = 3.5 \times 0 = 0 \)[/tex]
- [tex]\( b = 3 \)[/tex], [tex]\( c = 3.5 \times 3 = 10.5 \)[/tex]
- [tex]\( b = 6 \)[/tex], [tex]\( c = 3.5 \times 6 = 21 \)[/tex]
- [tex]\( b = 9 \)[/tex], [tex]\( c = 3.5 \times 9 = 31.5 \)[/tex]
These values also match the costs calculated and include only non-negative values for [tex]\( b \)[/tex].
### Conclusion
While all three tables can match the cost values generated by the equation [tex]\( c = 3.5b \)[/tex], the criterion of viable solutions — typically non-negative number of loaves [tex]\( b \)[/tex] — disqualifies the first table due to the negative value.
Therefore, both the second and third tables include only viable solutions.
However, as they ask for the 'only viable solutions', it is implied that the values of loaves should not be negative. So either of the second and third tables should be correct, with second table being the primary choice for lower range of values.
