(a) Without using tables, find the value of \(\frac{0.45 \times 0.91}{0.0117}\) (b) Find the number which is exactly halfway between \(1\frac{6}{7}\) and \(...

Answer Details

(a) Multiply numerator and denominator by \(10^4\) to remove decimals:

\[ \frac{0.45 \times 0.91}{0.0117} = \frac{4500 \times 91}{117} = \frac{409500}{11700}. \]

Since \(0.45 \times 0.91 = 0.4095\) and \(0.0117 \times 35 = 0.4095\),

\[ \frac{0.4095}{0.0117} = 35. \]

(b) The number halfway between two values is their average. Write both as improper fractions over 28:

\[ 1\tfrac{6}{7} = \frac{13}{7} = \frac{52}{28}, \qquad 2\tfrac{11}{28} = \frac{67}{28}. \] \[ \text{Halfway} = \frac{1}{2}\left(\frac{52}{28} + \frac{67}{28}\right) = \frac{1}{2}\times\frac{119}{28} = \frac{119}{56} = \frac{17}{8} = 2\tfrac{1}{8}. \]

(c) Each interior angle \(= 5 \times\) exterior angle, and interior \(+\) exterior \(= 180^\circ\):

\[ 5e + e = 180^\circ \implies 6e = 180^\circ \implies e = 30^\circ. \] \[ n = \frac{360^\circ}{e} = \frac{360^\circ}{30^\circ} = 12\ \text{sides}. \]

(d) Pipe length 20 cm, internal radius \(r = 3\) cm, external radius \(R = 4\) cm, \(\pi = \tfrac{22}{7}\). Volume of material (hollow cylinder):

\[ V = \pi(R^2 - r^2)\times \text{length} = \frac{22}{7}(4^2 - 3^2)(20) = \frac{22}{7}(7)(20) = 22 \times 20 = 440\ \text{cm}^3. \]