(2) Resolver (a) \( 91+84: 2-(36+49)+3^{2}= \) (b) \( (1+2)^{3}: 9+(2.4-2)+\sqrt{15: 3+2^{2}} \) (c) \( \left(\frac{1}{2}+\frac{1}{3}\right): \frac{7}{6}= \)
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Beyond the Answer
First, let's tackle part (a): To solve \( 91 + 84: 2 - (36 + 49) + 3^{2} \): 1. Calculate \( 84 : 2 = 42 \). 2. Compute the sum: \( 91 + 42 = 133 \). 3. Calculate the sum inside the parentheses: \( 36 + 49 = 85 \). 4. Thus, we have \( 133 - 85 + 3^{2} \). 5. Calculate \( 3^{2} = 9 \). 6. Now plug it all back in: \( 133 - 85 + 9 = 57 \). So the answer for part (a) is \( 57 \). Now for part (b): For \( (1 + 2)^{3} : 9 + (2.4 - 2) + \sqrt{15 : 3 + 2^{2}} \): 1. Start with \( (1 + 2)^{3} = 3^{3} = 27 \). 2. Now compute \( 27 : 9 = 3 \). 3. Next, calculate \( 2.4 - 2 = 0.4 \). 4. Now calculate \( 15 : 3 = 5 \), followed by \( \sqrt{5 + 2^{2}} = \sqrt{5 + 4} = \sqrt{9} = 3 \). 5. Now gather everything: \( 3 + 0.4 + 3 = 6.4 \). Thus, the answer for part (b) is \( 6.4 \). Finally, for part (c): To solve \( \left( \frac{1}{2} + \frac{1}{3} \right) : \frac{7}{6} \): 1. First, find a common denominator for \( \frac{1}{2} + \frac{1}{3} \). The common denominator is 6. Thus, \( \frac{1}{2} = \frac{3}{6} \) and \( \frac{1}{3} = \frac{2}{6} \). 2. So, \( \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \). 3. Now divide \( \frac{5}{6} : \frac{7}{6} = \frac{5}{6} \times \frac{6}{7} = \frac{5}{7} \). Therefore, the answer for part (c) is \( \frac{5}{7} \).
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