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\(\begin{array}{c|c} x & f \\\hline 1 & 3\\ 2 & 13\\3 & 15\\ 4 & 28\\ 5 & 21\\ 6 & 10\\ 7 & 7\\ 8 & 4\end{array}\) Median is half of total frequency = 50th term 4 falls in the range = 4

Area of the paper = area of square = L x B or S2 where s = S x k Area of the paper = (2.524)2 area of the diagram = (0.524)2 area not covered = (2.524)2 - (0.524)2 = 6.370576 - 0.274576 = 6.096 = 6.10cm2 (2 s.f)

\(\sqrt[3]{\frac{27a ^{-9}}{8}}\) = \(\frac{3a^{-3}}{2}\) = \(\frac{3}{2}\) x \(\frac{1}{a^3}\) = \(\frac{3}{2a^3}\)

Circumference of PRS = \(\frac{\pi d}{2}\) = \(\frac{22}{7}\) x \(\frac{7}{1}\) x \(\frac{1}{2}\) = 11cm Side PT = 7cm, Side TR = 7cm Perimeter(PTRS) = 11cm + 7cm + 7cm = 25cm

By using cosine formula, p2 = Q2 + R2 - 2QR cos p Cos P = \(\frac{Q^2 + R^2 - p^2}{2 QR}\) = \(\frac{(3)^2 + 2(\sqrt{3})^2 - 3^2}{2\sqrt{3}}\) = \(\frac{3 + 12 - 9}{12}\) = \(\frac{6}{12}\) = \(\frac{1}{2}\) = 0.5 Cos P = 0.5 p = cos-1 0.5 = 60o = < P = 60o If < P = 60o and < Q = 30 < R = 180o - 90o angle P = 60o and angle R is 90o

\(\frac{24.57 + 25.63 + 24.32 + 26.01 + 25.77}{5}\) mean = \(\frac{126.3}{5}\) = 25.26

Speed = \(\frac{distance}{time}\) let x represent the speed, d represent distance x = \(\frac{d}{4}\) d = 4x 2x = \(\frac{600 - d}{3}\) 6x = 600 - d 6x = 600 - 4x 10x = 600 x = \(\frac{600}{10}\) = 60km/hr Since speed was doubled for the last three hours = 2 x 60km\hr = 120km\hr.

(2x - 24)^\circ + (3x - 18)^\circ + (2x + 12)^\circ + (x + 12)^\circ + x^\circ = 360^\circ 9x = 360^\circ + 18^\circ x = \(\frac{378}{9}\) = 42^\circ, if x = 42^\circ, then add maths = \(\frac{2x - 42}{360}\) x 60 = \(\frac{2 \times 42 - 24}{360}\) x 60 = \(\frac{84 - 24}{6}\) = 10

Angle subtended at any part of the circumference of the circle \(\frac{125^o}{2}\) at centre = 360o - 235o = 125o \(\bar{PQR}\) = \(\frac{125}{2}\) = 62\(\frac{1}{2}\)o

The roots of the function are -1 and \(\frac{1}{3}\) sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\) product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\) \(x^2\) - (sum of roots)x + (product of roots) = 0 \(x^2\) - (-\(\frac{2}{3}\))x + (-\(\frac{1}{3}\)) = 0 \(x^2\) + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0 3\(x^2\) + 2x - 1 = 0( multiple through by -1 since the graph of the quadratic equation is maximum) 1 - 2x - 3\(x^2\).

o, o o

tan\(\theta\) = \(\frac{0.8}{2}\) = (0.4) \(\theta\) = tan-1(0.4) From the diagram, the inclination of the diagonal PR to the horizontal is 10o 42'

From the diagram; The value of x = 360o - 2(130o) = 360 - 260 = 100o

< PSQ = < QPT = 50º In \(\triangle\) PQS, PS = PQ (given) base angles of an isosceles triangle. < QRS = 180º - < SPQ (opp

Source: Myschool Past Questions Archive (Record #32771)