1). Make your equation equal to 0
2). Then plug in your numbers to the quadratic equation
2). Then plug in your numbers to the quadratic equation
Im going to give 15 points for this question please answer it
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Answer:
A. 3 possible combinations
B. 8 4-ounce's bags and 3 3-ounce's bags
C. 2 4-ounce's bags and 11 3-ounce's bags
D. 8 4-ounce's bags and 3 3-ounce's bags
E. All solutions offer the same revenue.
Step-by-step explanation:
You have been tasked with filling 4 ounce and 3 ounce bags from a 41 ounce container of candy. Let x be the number of 4 ounce bags and y be the number of 3 ounce bags. Then
A. Find all integer solutions:
- When x=0, then 3y=41 - impossible, because 41 is not divisible by 3.
- When x=1, then 3y=37 - impossible, because 37 is not divisible by 3.
- When x=2, then 3y=33, y=11 - possible.
- When x=3, then 3y=29 - impossible, because 29 is not divisible by 3.
- When x=4, then 3y=25 - impossible, because 25 is not divisible by 3.
- When x=5, then 3y=21, y=7 - possible.
- When x=6, then 3y=17 - impossible, because 17 is not divisible by 3.
- When x=7, then 3y=13 - impossible, because 13 is not divisible by 3.
- When x=8, then 3y=9, y=3 - possible.
- When x=9, then 3y=5 - impossible, because 5 is not divisible by 3.
- When x=10, then 3y=1 - impossible, because 1 is not divisible by 3.
You get 3 possible combinations.
B. 1. 2 + 11 = 13,
2. 5 + 7 = 12,
3. 8 + 3 = 11.
The minimal number of bags is 11.
C. 1. 2·7+11·5=69 cents
2. 5·7+7·5=70 cents
3. 8·7+3·5=71 cents
The cheapest is 1st solution.
D. 1. 2·6+11·5=67 cents
2. 5·6+7·5=65 cents
3. 8·6+3·5=63 cents
The cheapest is 3rd solution.
E. 1. 2·2+11·1.50=$20.50
2. 5·2+7·1.50=$20.50
3. 8·2+3·1.50=$20.50
All solutions offer the same revenue.