The rate constant $K$ for a first-order reaction is given by: \[ K = \frac{1}{t} \ln\left(\frac{100}{100 - \text{completion percentage}}\right) \]
For 99.9% completion:
\[ t_{99.9\%} = \frac{\ln(1000)}{K} \]
For 90% completion:
\[ t_{90\%} = \frac{\ln(10)}{K} \]
Ratio of times:
\[ \frac{t_{99.9\%}}{t_{90\%}} = \frac{\ln(1000)}{\ln(10)} = \frac{3 \ln(10)}{\ln(10)} = 3 \]
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: