Question
The number of loaves of bread purchased and the total cost of the bread in dollars can be modeled by the equation c = 3.5b. Which table of values matches the equation and includes only viable solutions?
A 2-column table with 4 rows. The first column is labeled loaves (b) with entries negative 2, 0, 2, 4. The second column is labeled cost (c) with entries negative 7, 0, 7, 14.
A 2-column table with 4 rows. The first column is labeled loaves (b) with entries 0, 0.5, 1, 1.5. The second column is labeled cost (c) with entries 0, 1.75, 3.5, 5.25.
A 2-column table with 4 rows. The first column is labeled loaves (b) with entries 0, 3, 6, 9. The second column is labeled cost (c) with entries 0, 10.5, 21, 31.5.
Answers
GPT-4o mini
To determine which table of values matches the equation \( c = 3.5b \) and includes only viable solutions, we will validate each table by calculating the cost \( c \) for the given number of loaves \( b \).
1. **First Table**
- Loaves (b): \(-2, 0, 2, 4\)
- Cost (c): \(-7, 0, 7, 14\)
Calculations:
- For \( b = -2 \): \( c = 3.5 \times (-2) = -7 \) (matches)
- For \( b = 0 \): \( c = 3.5 \times 0 = 0 \) (matches)
- For \( b = 2 \): \( c = 3.5 \times 2 = 7 \) (matches)
- For \( b = 4 \): \( c = 3.5 \times 4 = 14 \) (matches)
Choices: This table includes negative loaves, which are not viable solutions.
2. **Second Table**
- Loaves (b): \(0, 0.5, 1, 1.5\)
- Cost (c): \(0, 1.75, 3.5, 5.25\)
Calculations:
- For \( b = 0 \): \( c = 3.5 \times 0 = 0 \) (matches)
- For \( b = 0.5 \): \( c = 3.5 \times 0.5 = 1.75 \) (matches)
- For \( b = 1 \): \( c = 3.5 \times 1 = 3.5 \) (matches)
- For \( b = 1.5 \): \( c = 3.5 \times 1.5 = 5.25 \) (matches)
Choices: All entries are viable.
3. **Third Table**
- Loaves (b): \(0, 3, 6, 9\)
- Cost (c): \(0, 10.5, 21, 31.5\)
Calculations:
- For \( b = 0 \): \( c = 3.5 \times 0 = 0 \) (matches)
- For \( b = 3 \): \( c = 3.5 \times 3 = 10.5 \) (matches)
- For \( b = 6 \): \( c = 3.5 \times 6 = 21 \) (matches)
- For \( b = 9 \): \( c = 3.5 \times 9 = 31.5 \) (matches)
Choices: All entries are viable.
After evaluating all three tables, the second table and the third table consist of only viable solutions with valid costs. However, the third table seems to be better as it consists purely of whole loaves.
The best answer is:
**Third Table** (loaves: \(0, 3, 6, 9\) and cost: \(0, 10.5, 21, 31.5\)).
1. **First Table**
- Loaves (b): \(-2, 0, 2, 4\)
- Cost (c): \(-7, 0, 7, 14\)
Calculations:
- For \( b = -2 \): \( c = 3.5 \times (-2) = -7 \) (matches)
- For \( b = 0 \): \( c = 3.5 \times 0 = 0 \) (matches)
- For \( b = 2 \): \( c = 3.5 \times 2 = 7 \) (matches)
- For \( b = 4 \): \( c = 3.5 \times 4 = 14 \) (matches)
Choices: This table includes negative loaves, which are not viable solutions.
2. **Second Table**
- Loaves (b): \(0, 0.5, 1, 1.5\)
- Cost (c): \(0, 1.75, 3.5, 5.25\)
Calculations:
- For \( b = 0 \): \( c = 3.5 \times 0 = 0 \) (matches)
- For \( b = 0.5 \): \( c = 3.5 \times 0.5 = 1.75 \) (matches)
- For \( b = 1 \): \( c = 3.5 \times 1 = 3.5 \) (matches)
- For \( b = 1.5 \): \( c = 3.5 \times 1.5 = 5.25 \) (matches)
Choices: All entries are viable.
3. **Third Table**
- Loaves (b): \(0, 3, 6, 9\)
- Cost (c): \(0, 10.5, 21, 31.5\)
Calculations:
- For \( b = 0 \): \( c = 3.5 \times 0 = 0 \) (matches)
- For \( b = 3 \): \( c = 3.5 \times 3 = 10.5 \) (matches)
- For \( b = 6 \): \( c = 3.5 \times 6 = 21 \) (matches)
- For \( b = 9 \): \( c = 3.5 \times 9 = 31.5 \) (matches)
Choices: All entries are viable.
After evaluating all three tables, the second table and the third table consist of only viable solutions with valid costs. However, the third table seems to be better as it consists purely of whole loaves.
The best answer is:
**Third Table** (loaves: \(0, 3, 6, 9\) and cost: \(0, 10.5, 21, 31.5\)).