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#### 160
#### 0.35
\( 2 \times V_1 = 0.4 \times 250 \) → \( V_1 = (0.4 \times 250) \div 2 = 50 \) mL.
#### 0.0475
\( 0.001 = 10^{-3} \), so \( \log_{10}(0.001) = -3 \).
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$ \frac{14}{3} = \frac{28}{6} $, $ \frac{7}{2} = \frac{21}{6} $, LCM of periods $ \frac{14}{3} $ and $ \frac{7}{2} $: write as $ \frac{28}{6}, \frac{21}{6} $, GCD of 28 and 21 is 7, so period is $ \frac{14}{\gcd(14,21)} \cdot \frac{1}{\text{some factor}} $. Better: find least $ T $ such that $ \frac{3\pi}{7}T = 2\pi m $, $ \frac{4\pi}{7}T = 2\pi n $, so $ T = \frac{14m}{3} = \frac{7n}{2} \Rightarrow \frac{14m}{3} = \frac{7n}{2} \Rightarrow 28m = 21n \Rightarrow 4m = 3n $. Small