Using the equation:
\[ i = K\theta, \]
At half deflection:
\[ \frac{2}{G + R} = K\theta. \]
Rearranging:
\[ \frac{1}{\theta} = \frac{(G + R)K}{2} = RK + \frac{GK}{2}. \]
From the slope of the graph:
\[ \text{Slope} = K = 0.5 = 5 \times 10^{-1} \, \text{A/division}. \]
The largest $ n \in \mathbb{N} $ such that $ 3^n $ divides 50! is:
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to: