Further Mathematics Questions and Answers

The equation of a circle with centre coordinate (a, b) and radius r is :

$(x - a)^{2} + (y - b)^{2} = r^{2}$

Given centre = (2, -3) and radius r = 2 units

Equation = $(x - 2)^{2} + (y - (-3))^{2} = 2^{2}$

$x^{2} - 4x + 4 + y^{2} + 6y + 9 = 4$

$x^{2} + y^{2} - 4x + 6y + 4 + 9 - 4 = 0 \implies x^{2} + y^{2} - 4x + 6y + 9 = 0$

The sum of deviations from the mean of a set of numbers equals 0.

$(k+3)^{2} + (k+7) + (-2) + k + (k+2)^{2} = 0$

$(k^2 + 6k + 9) + (k+7) - 2 + k + (k^2 + 4k + 4) = 0$

$2k^{2} + 12k + 18 = 0$

$2k^{2} + 6k + 6k + 18 = 2k(k + 3) + 6(k + 3) = 0$

$k = -3 (twice)$

$F = F\cos\theta + F\sin\theta$

$\implies 90N = 90\cos 120° + 90\sin 120°$

$120N = 120 \cos 240° + 120 \sin 240°$

$R = F_{1} + F_{2}$

= $(90 \cos 120 + 120 \cos 240)i + (90\sin 120 + 120 \sin 240)j$

= $90(-0.5) + 120(-0.5))i + (90(\frac{\sqrt{3}}{2}) + (120(-\frac{\sqrt{3}}{2}))j$

= $-105i - 15\sqrt{3}j = -15(7i + \sqrt{3}j)$

P(at least one head) = 1 - P(4 tails)

Let $p = \frac{1}{2}$ = probability of head and $q = \frac{1}{2}$ = probability of tail.

$(p + q)^{4} = p^{4} + 4p^{3}q + 6p^{2}q^{2} + 4pq^{3} + q^{4}$

P(4 tails) = $q^{4} = (\frac{1}{2})^{4} = \frac{1}{16}$

P(at least one head) = $1 - \frac{1}{16} = \frac{15}{16}$

$(3x + 4)^{4} = ^{4}C_{0}(3x)^{0}(4)^{4} + ^{4}C_{1}(3x)^{1}(4)^{3} + ^{4}C_{2}(3x)^{2}(4)^{2} + ^{4}C_{3}(3x)^{3}(4)^{1} + ^{4}C_{4}(3x)^{4}(4)^{0}$

$x^{3} = ^{4}C_{3}(3x)^{3}(4) = \frac{4!}{3!1!} \times 3^{3} \times 4$

= $432x^{3}$