g) \( \frac{1}{\left(\frac{16}{3} \div \frac{16}{5}\right) * \frac{4}{7}} \quad \) h) \( \left[\frac{\left(\frac{4}{3}+\frac{8}{3}\right) * \frac{7}{3}}{\frac{14}{4} \div \frac{6}{7}}\right] \)
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For the first expression, we simplify the division in the parentheses first. When you divide \(\frac{16}{3}\) by \(\frac{16}{5}\), it becomes \(\frac{16}{3} \times \frac{5}{16} = \frac{5}{3}\). Now, multiply it by \(\frac{4}{7}\): \(\frac{5}{3} \times \frac{4}{7} = \frac{20}{21}\). Therefore, the entire expression is \(\frac{1}{\frac{20}{21}} = \frac{21}{20}\). In the second expression, start by simplifying the inside. \(\frac{4}{3} + \frac{8}{3} = \frac{12}{3} = 4\). Then, multiply by \(\frac{7}{3}\), giving \(4 \times \frac{7}{3} = \frac{28}{3}\). Next, handle the division at the bottom: \(\frac{14}{4} \div \frac{6}{7} = \frac{14}{4} \times \frac{7}{6} = \frac{98}{24} = \frac{49}{12}\). Now, divide the numerator by the denominator: \(\frac{\frac{28}{3}}{\frac{49}{12}} = \frac{28}{3} \times \frac{12}{49} = \frac{336}{147} = \frac{112}{49}\). Thus, the final result is \(\frac{112}{49}\).
