1. **Problem 2.1: Sugar for Treats** Lebo has a five-cup bag of sugar and needs to use it for three recipes: - 1 \(\frac{3}{4}\) cups for brownies - 2 \(\frac{1}{2}\) cups for cupcakes - \(\frac{3}{8}\) cups for cookies 2. **Convert the mixed numbers to improper fractions or decimals:** - 1 \(\frac{3}{4}\) = \(1 + \frac{3}{4} = \frac{4}{4} + \frac{3}{4} = \frac{7}{4}\) cups - 2 \(\frac{1}{2}\) = \(2 + \frac{1}{2} = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}\) cups - \(\frac{3}{8}\) cups stays as it is. 3. **Find a common denominator to sum the fractions:** Denominators are 4, 2, and 8. The least common denominator is 8. - \(\frac{7}{4} = \frac{7 \times 2}{4 \times 2} = \frac{14}{8}\) - \(\frac{5}{2} = \frac{5 \times 4}{2 \times 4} = \frac{20}{8}\) - \(\frac{3}{8}\) stays \(\frac{3}{8}\) 4. **Sum the sugar amounts:** $$\frac{14}{8} + \frac{20}{8} + \frac{3}{8} = \frac{14 + 20 + 3}{8} = \frac{37}{8}$$ 5. **Convert \(\frac{37}{8}\) to a mixed number:** $$37 \div 8 = 4 \text{ remainder } 5 \implies 4 \frac{5}{8} \text{ cups}$$ 6. **Compare required sugar to amount available:** - Available sugar = 5 cups = \(5 \frac{0}{8} = \frac{40}{8}\) - Required sugar = \(\frac{37}{8}\) Since \(\frac{40}{8} > \frac{37}{8}\), Lebo has enough sugar. 7. **Calculate leftover sugar:** $$\frac{40}{8} - \frac{37}{8} = \frac{3}{8}$$ Lebo will have \(\frac{3}{8}\) cups of sugar left over. --- 8. **Problem 2.2: Simplify the expression** $$-3^{2} + 4 \times (2 + 5) - (-6)$$ 9. **Apply the order of operations (PEMDAS):** - First, evaluate exponent: $$3^{2} = 9$$ - Note the negative sign: \(-3^{2} = -(3^{2}) = -9\) 10. **Evaluate inside the parentheses:** $$2 + 5 = 7$$ 11. **Multiply:** $$4 \times 7 = 28$$ 12. **Simplify the double negative:** $$- (-6) = +6$$ 13. **Now substitute back and add: ** $$-9 + 28 + 6$$ 14. **Sum:** $$(-9 + 28) + 6 = 19 + 6 = 25$$ **Final Answers:** - Q2.1: Lebo has enough sugar and will have \(\frac{3}{8}\) cups left. - Q2.2: The simplified value is 25.