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Problem Situation:
Bernie spends \(\$6.50\) on ingredients and cups for his lemonade stand.
He charges \(\$1.50\) for each cup of lemonade.
He will only sell whole cups of lemonade (not 0.75 cups, 1.5 cups, etc.).
How many cups, \(x\), will Bernie need to sell to make a profit of at least \(\$20\)?

Inequality that represents this situation:
[tex]\[
20 \leq 1.50x - 6.50
\][/tex]

Drag each number to show if it is a solution to both the inequality and the problem situation or to the inequality only, or if it is not a solution.

Numbers to consider:
- 18
- 23
- 10
- 17.7
- -4
- 35.5

Categories:
- Solution to both the inequality and the situation
- Solution to the inequality only
- Not a solution

Sagot :

GinnyAnswer
To solve this problem, we need to determine whether the given numbers satisfy the inequality, and if they also represent whole non-negative numbers of cups sold by Bernie.

Let's first analyze the inequality we are given:
[tex]\[ 20 \leq 1.50x - 6.50 \][/tex]
To check if each number satisfies the inequality, we'll substitute each number into \( x \) and see if the inequality holds true.

Step-by-Step Analysis:

1. Check \( x = 18 \):
[tex]\[ 1.50 \times 18 - 6.50 = 27 - 6.50 = 20.50 \][/tex]
Since \( 20 \leq 20.50 \), \( x = 18 \) satisfies the inequality.
- It is a non-negative integer, hence it is a solution to both the inequality and the situation.

2. Check \( x = 23 \):
[tex]\[ 1.50 \times 23 - 6.50 = 34.50 - 6.50 = 28 \][/tex]
Since \( 20 \leq 28 \), \( x = 23 \) satisfies the inequality.
- It is a non-negative integer, hence it is a solution to both the inequality and the situation.

3. Check \( x = 10 \):
[tex]\[ 1.50 \times 10 - 6.50 = 15 - 6.50 = 8.50 \][/tex]
Since \( 20 \leq 8.50 \) is false, \( x = 10 \) does not satisfy the inequality.
- It is not a solution.

4. Check \( x = 17.7 \):
[tex]\[ 1.50 \times 17.7 - 6.50 = 26.55 - 6.50 = 20.05 \][/tex]
Since \( 20 \leq 20.05 \), \( x = 17.7 \) satisfies the inequality.
- It is not a whole number (cups sold must be integer), so it is a solution to the inequality only.

5. Check \( x = -4 \):
[tex]\[ 1.50 \times -4 - 6.50 = -6 - 6.50 = -12.50 \][/tex]
Since \( 20 \leq -12.50 \) is false, \( x = -4 \) does not satisfy the inequality.
- It is not a solution.

6. Check \( x = 35.5 \):
[tex]\[ 1.50 \times 35.5 - 6.50 = 53.25 - 6.50 = 46.75 \][/tex]
Since \( 20 \leq 46.75 \), \( x = 35.5 \) satisfies the inequality.
- It is not a whole number (cups sold must be integer), so it is a solution to the inequality only.

Summary of Classification:

- Solution to both the inequality and the situation:
- 18, 23

- Solution to the inequality only:
- 17.7, 35.5

- Not a solution:
- 10, -4

Thus, we can classify each number as follows:

- Solution to both the inequality and the problem situation: 18, 23
- Solution to the inequality only: 17.7, 35.5
- Not a solution: 10, -4
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