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(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\(x^2\).

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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)