(a) \(\frac{0.45 \times 0.91}{0.0117} = \frac{45 \times 91}{117}\) = \(5 \times 7 = 35\) (b) Halfway between \(1\frac{6}{7}\) and \(2\frac{11}{28}\) = \(\frac{1\frac{6}{7} + 2\frac{11}{28}}{2}\) = \((\frac{13}{7} + \frac{67}{28}) \times \frac{1}{2}\) = \(\frac{119}{28} \times \frac{1}{2} \) = \(\frac{119}{56} = 2\frac{7}{56} = 2\frac{1}{8}\) (c) Let the interior angle = 5x^\circ and the exterior angle = x^\circ. \(\therefore 5x^\circ + x^\circ = 180^\circ\) \(6x^\circ = 180^\circ \implies x = 30^\circ\) Exterior angle = \(\frac{360^\circ}{n} \implies 30^\circ = \frac{360^\circ}{n}\) \(n = \frac{360^\circ}{30^\circ} = 12\) The polygon has 12 sides. (d) Length of pipe = 20 cm Internal diameter = 6 cm ; radius = 3cm External diameter = 8 cm ; radius = 4cm Let the external radius = R and internal radius = r \(\therefore\) Volume of material used = \(\pi R^{2} l - \pi r^{2} l\) Where l = length. \(Volume = \pi l (R^{2} - r^{2}) cm^{3}\) = \(\frac{22}{7} \times 20 (4^{2} - 3^{2}) = \frac{22}{7} \times 20 \times 7 = 440 cm^{3}\)