Questions: Responses For full credit, you are required to reply to two of your classmates' main posts. Your replies should address the following: - Check how your classmates put together their equations. If it is not correct, point out why. If it is, continue to the next point. - Using your classmates' profit equations, what is the profit if you do not manufacture and sell any items? Show your work. Does this make sense? Explain why. - Using your classmates' profit equations, offer a value (number of objects sold) that gives a true profit and not a loss. How do you know this is true? Hint: compute the profit value and show your work. - Using your classmates' cost and revenue equations, use algebra to solve for the break-even point. Show your work.

To address the task, I'll provide a structured approach to responding to two hypothetical classmates' main posts regarding their profit equations. Since I don't have access to the actual posts, I'll create sample equations and demonstrate how to analyze them.

Equations:

• Cost Equation: \( C(x) = 50x + 200 \)
• Revenue Equation: \( R(x) = 70x \)
• Profit Equation: \( P(x) = R(x) - C(x) = 70x - (50x + 200) = 20x - 200 \)

1. Equation Check:

   • The profit equation is correctly derived from the cost and revenue equations. The profit equation \( P(x) = 20x - 200 \) is accurate.

2. Profit with Zero Items Sold:

   • \( P(0) = 20(0) - 200 = -200 \)
   • This result indicates a loss of $200 when no items are sold, which makes sense because the fixed cost of $200 is incurred regardless of sales.

3. True Profit (No Loss):

   • To find a value where profit is positive, solve \( 20x - 200 > 0 \).
   • \( 20x > 200 \)
   • \( x > 10 \)
   • Selling more than 10 items results in a true profit. For example, if 11 items are sold:
     • \( P(11) = 20(11) - 200 = 220 - 200 = 20 \)
     • This confirms a profit of $20.

4. Break-even Point:

   • Set \( P(x) = 0 \) to find the break-even point.
   • \( 20x - 200 = 0 \)
   • \( 20x = 200 \)
   • \( x = 10 \)
   • The break-even point is at 10 items, where costs equal revenue.

Equations:

• Cost Equation: \( C(x) = 30x + 150 \)
• Revenue Equation: \( R(x) = 50x \)
• Profit Equation: \( P(x) = R(x) - C(x) = 50x - (30x + 150) = 20x - 150 \)

1. Equation Check:

   • The profit equation is correctly derived. The profit equation \( P(x) = 20x - 150 \) is accurate.

2. Profit with Zero Items Sold:

   • \( P(0) = 20(0) - 150 = -150 \)
   • This indicates a loss of $150 when no items are sold, which is logical due to the fixed cost of $150.

3. True Profit (No Loss):

   • To find a value where profit is positive, solve \( 20x - 150 > 0 \).
   • \( 20x > 150 \)
   • \( x > 7.5 \)
   • Selling more than 7.5 items results in a true profit. For example, if 8 items are sold:
     • \( P(8) = 20(8) - 150 = 160 - 150 = 10 \)
     • This confirms a profit of $10.

4. Break-even Point:

   • Set \( P(x) = 0 \) to find the break-even point.
   • \( 20x - 150 = 0 \)
   • \( 20x = 150 \)
   • \( x = 7.5 \)
   • The break-even point is at 7.5 items, where costs equal revenue.

Both classmates have correctly formulated their profit equations. The analysis shows that the profit equations logically reflect the costs and revenues, with break-even points and conditions for true profit accurately determined.