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Let the given points be: P(-3, -14) = (x\(_1\), y\(_1\)) Q(t, -5) = (x\(_2\), y\(_2\)) PQ = 9 units (given) Using the distance formula, d = \sqrt [ \((x_2 - x_1)^2 + (y_2 - y_1)^2\)] PQ = \sqrt [ \((t - (-3))^2 + (-5 + 14)^2\)] \(\implies\) \sqrt [ \((t + 3)^2 + 81\)] = 9 Squaring on both sides, ⇒ (t + 3)\(^2\) + 81 = 81 ⇒ (t + 3)\(^2\) = 0 ⇒ t + 3 = 0 ∴ t = -3

\(3x + 2y + 4 = 0\) Rearrange: \(2y = -3x - 4\) Divide both sides by 2 y = \(\frac {-3 \times - 4}{2}\) y = \(\frac {-3}{2} \times - 2\) ∴ the gradient of the line 3x + 2y + 4 = 0 is \(\frac {-3}{2}\) If two lines are perpendicular to each other ∴ \(m_1 x m_2\) = -1 Let \(m_1 = \frac {-3}{2} \therefore m_2 = \frac {-1}{m_1} = \frac {-1}{-3/2} = \frac {2}{3}\) From the equation of a line which is given as m = \(\frac {y - y_1}{x - x_1} where (x_1, y_1) = (2,3)\) \(\therefore \frac {2}{3} = \frac {y - 3}{x - 2}\) =3(y - 3) = 2(x - 2) =3y - 9 = 2 x -4 =3y = 2 x -4 + 9 ∴ 3y = 2x + 5

The line: \(3y + 6x\) = 48 Divide through by 3 ⇒ y + 2\(x\) = 16 ⇒ y = -2\(x\) + 16 ∴ The gradient of the line = -2 The points: A(-2, k) and B (4, 8) m =\(\frac{y2 - y1}{x2 - x1} = \frac{8 - k}{4 - (-2)}\) ⇒ m =\(\frac[8 - k}{4 + 2} = {8 - k}{6}\) Since the line passes through the points ∴ -2 = \(\frac{8 - k}{6}\) ⇒ \(\frac{-2}[1} = \frac{8 - k]{6}\) ⇒ 8 - k = -12 ⇒ k = 8 + 12 ∴ k = 20

The height of the second building H = h + 24 tan \theta = \(\frac {opp}{adj}\) tan 30\(^o = \frac {h}{x}\) \(\implies\frac{\sqrt 3}{3} = \frac {h}{x}\) \(\implies x = \sqrt 3 = 3h\) \(\implies x = \frac {3h}{\sqrt 3}\) ....(i) tan 60\(^o = \frac {24}{x}\) \(\implies\sqrt 3 = \frac {24}{x}\) \(\implies x\sqrt 3 = 24\) \(\implies x = \frac {24}{\sqrt 3}\) ....(ii) Equate equation (i) and (ii) \(\implies \frac {3h}{\sqrt 3} = \frac {24}{\sqrt 3}\) \(\implies\) 3h = 24 \(\implies h = \frac {24}{3}\) = 8m ∴The height of the second building = 8 + 24 = 32m

tan \theta = \(\frac {opp}{adj} = \frac {RQ}{QP} = \frac {6}{8}\) tan \theta = 0.75 \theta = tan\(^{-1} (0.75) = 36.87^o\) ∴ The bearing of R from P = 360\(^o\) - 36.87\(^o\) = 323\(^o\) (to the nearest degree)

Speed = \(\frac{Distance}{Time}\) ⇒ Time = \(\frac{Distance}{Time}\) Let D = distance between the two airports ∴ Time taken to get to the airport = \(\frac{D}{120}\) and Time taken to return =\( \frac{D}{150}\) Since total time of flight= 3hours, ⇒ \(\frac{D}{120} + \frac{D}{150}\) = 3 ⇒ \(\frac{15D + 12D}{1800}\) = 3 ⇒ \(\frac{27D}{1800}\) = 3 ⇒ \(\frac{3D}{200} = \frac{3}{1}\) ⇒ 3D = 200 x 3 ∴ D =\(\frac{ 200\times3}{3}\)= 200km

AB = \(\frac{\theta}{360}\times 2\pi Rcos\propto\) (distance on small circle) = 64 - 56 = 8\(^o\) \(\propto = 86^o\) ⇒ AB = \(\frac{8}{360}\) x 2 x 3.142 x 6370 x cos 86 ⇒ AB = \(\frac{22,338.29974}{360}\) ∴ AB = 62km (to the nearest km)

\(lim_{x\to2} \frac {x^2 + 4x - 12}{x^2 - 2x}\) = \(lim_{x\to2} \frac {(x - 2)(x + 6)}{x(x - 2)}\) \(lim_{x\to2} \frac {x + 6}{x}\) \(\frac {2 + 6}{2} = \frac {8}{2}\) = 4

y = \(\sqrt[3]{x^2(2x - x^2)} = x^{2/3} (2x - x^2)\) = \(2x^{5/3} - x^{8/3}\) Now, we can differentiate the function \(\therefore \frac {dy}{dx} = \frac {10x^{2/3}}{3} - \frac {8x^{5/3}}{3}\)

Area of a circle (A) = \(\pi r^2\) Given \(\frac{dA}{dt} = 1.5cm^2s^{-1}\) \(\frac{dr}{dt}\) = ? A = 2cm\(^2\) Now 2 = \(\pi r^2\) = r\(^2 = \frac {2}{\pi}\) r = \(\sqrt \frac {2}{\pi}\) cm = 0.798cm \(\frac {dr}{dt} = \frac {dA}{dt} \times \frac {dr}{dt}\) \(\frac {dA}{dr} = 2\pi r\) (differentiating A = \(\pi r^2)\) \(\frac {dr}{dA} = \frac {1}{2\pi r}\) \(\frac {dr}{dt} = 1.5 \times \frac {1}{2\times \pi \times 0.798} = 1.5 \times 0.199\) \(\frac {dr}{dt} = 0.299cms^{-1}\) (to 3 s.f)

\(\int_0^1 4x - 6^3\sqrt x^2\) dx =\(\int_0^1 4x - 6x^{2/3}\) dx =\((2x^2 - \frac{8x^{5/3}}{5})_0^1\) =\((2(1)^2 - \frac{18(1)^{5/3}}{5}) - (2(0)^2 - \frac{18(0)^{5/3}}{5}\) = - \(\frac{8}{5}\) - 0 - \(\frac{8}{5}\)

Class Interval Class Mark Frequency (f) fx 5 - 6 5.5 29 5.5 x 29 = 159.5 7 - 8 7.5 40 7.5 x 40 = 300 9 - 10 9.5 38 9.5 x 38 = 361 \(\sum f = 107\) \(\sum fx = 820.5\) Mean = \(\frac{\sum fx}{\sum f} = \frac {820.5}{107}\) = 7.7 (1 d.p)

The first five prime numbers are 2, 3, 5, 7 and 11 M.D = \(\frac {\sum (x - x̄)}{n}\) n = 5 \(x̄ = \frac {\sum x}{n} = \frac {2 + 3 + 5 + 7 + 11}{5} = \frac {28}{5} = 5.6\) M.D = \(\frac {(2 - 5.6) + (3 - 5.6) + (5 - 5.6) + (7 - 5.6) + (11 - 5.6)}{5}\) = \(\frac {(-3.6) + (-2.6) + (-0.6) + (1.4) + (5.4)}{5} = \frac {13.6}{5}\) Take the absolute value of all numbers in bracket = \(\frac {3.6 + 2.6 + 0.6 + 1.4 + 5.4}{5} = \frac {13.6}{5}\) \(\therefore M.D = 2.72\)

\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54) Median = \(1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})\)c \(l_m\)=lower class boundary of the median age = 6.5 \(\sum f\) = 107 \(f_b\) = sum of all frequencies before the median age = 29 \(f_m\)=frequency of the median age = 40 c = class width = 8.5 - 6.5 = 2 Median = 6.5 + \((\frac {^{107}/_2 - 29}{40})\) x 2 = 6.5 + \((\frac {53.5 - 29}{40})\) x 2 = 6.5 + \((\frac {24.5}{40})\) x 2 = 6.5 + 1.225 \(\therefore\) median age = 7.725

Number of students who scored atleast 25% = 5 + 3 + 8 = 16