(3/4)*{(1/3)*[8^2+(24/6*4)+10]+[(2/5)*(10^2-12+3)]}
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MathBot Answer:
\[\frac{3}{4} \cdot \left(\frac{1}{3} \cdot \left({8}^{2} + \frac{24}{6} \cdot 4 + 10\right) + \frac{2}{5} \cdot \left({10}^{2} - 12 + 3\right)\right) = \frac{249}{5} = 49.8\]
\[\begin{aligned}\frac{3}{4} \cdot \left(\frac{1}{3} \cdot \left({8}^{2} + \frac{24}{6} \cdot 4 + 10\right) + \frac{2}{5} \cdot \left({10}^{2} - 12 + 3\right)\right)& = \frac{3}{4} \cdot \left(\frac{1}{3} \cdot \left(8^{2} + \frac{24}{6} \cdot 4 + 10\right) + \frac{2}{5} \cdot \left(10^{2} - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(\frac{1}{3} \cdot \left(64 + \frac{24}{6} \cdot 4 + 10\right) + \frac{2}{5} \cdot \left(10^{2} - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(\frac{1}{3} \cdot \left(64 + 16 + 10\right) + \frac{2}{5} \cdot \left(10^{2} - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(\frac{1}{3} \cdot \left(80 + 10\right) + \frac{2}{5} \cdot \left(10^{2} - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(\frac{90}{3} + \frac{2}{5} \cdot \left(10^{2} - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(30 + \frac{2}{5} \cdot \left(10^{2} - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(30 + \frac{2}{5} \cdot \left(100 - 12 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(30 + \frac{2}{5} \cdot \left(88 + 3\right)\right)\\& = \frac{3}{4} \cdot \left(30 + \frac{2}{5} \cdot 91\right)\\& = \frac{3}{4} \cdot \left(30 + \frac{182}{5}\right)\\& = \frac{3}{4} \cdot \frac{332}{5}\\& = \frac{249}{5}\end{aligned}\]