If $3A + 4B^{t} = \left( \begin{array}{cc} 7 & -10 \\ 0 & 6 \end{array} \right) $ and $ 2B - 3A^{t} = \left( \begin{array}{cc} -1 & 18 \\ 4 & -6 \\ -5 & -7 \end{array} \right) $, then find $ (5B)^{t}$:
If \(P(B) = \frac{3}{5}\), \(P(A/B) = \frac{1}{2}\), and \(P(A \cup B) = \frac{4}{5}\), then \(P(A \cup B)' + P(A')\) is :
\(\int x^x(1 + \log x) \, dx\), \(\text{ is equal to:}\)