If $\lim\limits_{x \to \infty} \dfrac{(2x-1)^{19} \cdot (3x+2)^{11}}{(6x-5)^{30}} = 2^a \cdot 3^b$, then $a + b =$

The dual of the statement pattern $(p \wedge \sim q) \longrightarrow (q \wedge \sim p)$ is equivalent to

Negation of the statement "If an integer is greater than 4 and less than 5, then it is a multiple of 3", is

In a triangle ABC, with the usual notations, $\angle B = \dfrac{\pi}{3}$, $\angle C = \dfrac{\pi}{4}$. If D divides BC internally in the ratio 1:3, then $\dfrac{\sin \angle BAD}{\sin \angle CAD} =$