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11 Joxon must sell at least 49 rolls of wrapping paper to support the robotics club fundraiser. He has already sold 24 rolls of wrapping paper. Which inequality best represents the number of rolls of wrapping paper Jaxon still needs to sell?

Ask by Blake Macdonald. in the United States
Mar 12,2025

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Answer

Jaxon needs to sell at least 25 more rolls of wrapping paper.

Solution

To determine the number of rolls of wrapping paper Jaxon still needs to sell, we can set up an inequality based on the information provided. 1. **Known Conditions:** - Jaxon must sell at least 49 rolls of wrapping paper. - He has already sold 24 rolls. 2. **Let \( x \) be the number of additional rolls Jaxon needs to sell.** 3. **Inequality Setup:** - The total number of rolls he will have sold after selling \( x \) more rolls is \( 24 + x \). - We want this total to be at least 49 rolls. Thus, we can write the inequality as: \[ 24 + x \geq 49 \] 4. **Solving the Inequality:** - To find \( x \), we can subtract 24 from both sides: \[ x \geq 49 - 24 \] \[ x \geq 25 \] Therefore, the inequality that best represents the number of rolls of wrapping paper Jaxon still needs to sell is: \[ x \geq 25 \] This means Jaxon needs to sell at least 25 more rolls of wrapping paper to meet his goal.

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Jaxon needs to sell a total of at least 49 rolls. Since he has already sold 24 rolls, we can set up the inequality to find out how many more he needs to sell. The inequality to represent this situation would be: \( x + 24 \geq 49 \) Here, \( x \) represents the number of additional rolls he needs to sell. This can be rearranged to \( x \geq 25 \), meaning Jaxon must sell at least 25 more rolls of wrapping paper to reach his goal! It’s always a good idea for fundraisers to keep things exciting! Jaxon could motivate sales by offering fun incentives, like a small prize for the top seller or a drawing for a gift card. Engaging the community not only boosts sales but also brings everyone together, making the fundraiser a more enjoyable experience!

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