Further Mathematics Questions and Answers

\(F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j\)

\(F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i\)

\(R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j\)

\(|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}\)

= \(\sqrt{56.25 + 18.75} = \sqrt{75}\)

= 8.66 N

P(at least one green wrapper) = 1 - P(no green wrapper)

= \(1 - (\frac{2}{6} \times \frac{1}{5})\)

= \(1 - \frac{1}{15}\)

= \(\frac{14}{15}\)

\(f(x) = mx^{2} - 6x - 3\)

\(f '(x) = \frac{\mathrm d y}{\mathrm d x} = 2mx - 6\)

\(f'(1) = 2m - 6 = 12\)

\(2m = 18 \implies m = 9\)

\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3}\)

\(\frac{x^{2} - 2x - 3}{x - 3} = \frac{x^{2} - 3x + x - 3}{x - 3}\)

\(\frac{(x - 3)(x + 1)}{x - 3} = x + 1\)

\(\lim \limits_{x \to 3} \frac{x^{2} - 2x - 3}{x - 3} \equiv \lim \limits_{x \to 3} (x + 1)\) (L'Hopital rule)

\(\lim \limits_{x \to 3} (x + 1) = 3 + 1 = 4\)

\(\tan 15 = \tan (60 - 45)\)

\(\tan (x - y) = \frac{\tan x - \tan y}{1 + \tan x \tan y}\)

\(\tan (60 - 45) = \frac{\tan 60 - \tan 45}{1 + \tan 60 \tan 45}\)

= \(\frac{\sqrt{3} - 1}{1 + (\sqrt{3} \times 1)}\)

= \(\frac{\sqrt{3} - 1}{1 + \sqrt{3}}\)

Rationalizing by multiplying denominator and numerator by \(1 - \sqrt{3}\),

\(\tan 15 = 2 - \sqrt{3}\)