Rom

\(12'' = \dfrac{12}{3600}^\circ = \dfrac{1}{300}^\circ\\ 6' = \dfrac{6}{60}^\circ = \dfrac{1}{10}^\circ\\ 6'12'' = \dfrac{31}{300}^\circ = 0.10\bar{3}^\circ\\ -46^\circ6'12'' = -46.10\bar{3}^\circ\\ \sec(46.10\bar{3}) \approx 1.44225\)

0.08062

\(p[k=42] = \dbinom{100}{42}(0.42)^{42}(1-0.42)^{100-42}\\ \text{I'll let you crunch the numbers}\)

\(\tan\left(-\dfrac{13\pi}{6}\right) = \tan\left(-\dfrac{\pi}{6}\right)=\\ \dfrac{\sin \left (-\frac{\pi}{ 6} \right) }{\cos\left(-\frac \pi 6 \right)} =\dfrac{-\frac 1 2}{\frac{\sqrt{3}}{2}} = -\dfrac{1}{\sqrt{3}}\)

separate the coins into pairs.

weigh one pair against another

if they balance, the 3rd pair has the heavier coin

if they don't balance the heavier pair has the heavier coin

whichever pair it is weigh the two coins against each other to determine the heavier.

2 weighings

where do you all get these questions....

\((10a+b)\left(\dfrac 7 4 \right) = 10b+a,~a, b \in \mathbb{Z}^+\\ 70a+7b = 40b + 4a\\ 66a = 33b\\ 2a=b\\ 12+24+36+48 = 120\)

\(10^{\log(x)} = x\\ \text{so yes}\\ 10^{\log(6)} = 6\)

\(\text{one real solution means the discriminant is 0}\\ \text{for the quadratic }ax^2 + b x + c \text{ the discriminant }\\ D = b^2 - 4ac\\ \text{here we have}\\ D = m^2 - 4(m+n) = 0\\ m^2 -4m - 4n = 0\\ m = 2 \left(1\pm\sqrt{n+1}\right)\)

\(\text{we're told that }m \text{ is unique, so }\\ 1+\sqrt{n+1} = 1 - \sqrt{n+1}\\ 2\sqrt{n+1} = 0\\ n=-1\)

I answered the original posting