WAEC Mathematics Past Questions & Answers - Page 120

No explanation has been provided for this answer.

\(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) = \(\frac{\log 27\frac{1}{2}}{81\frac{1}{2}}\)

= \(\frac{\log 3\frac{1}{2}}{\log 3^2}\)

\(\frac{\frac{3}{2} \log 3}{2 \log 3} = \frac{3}{2} \div \frac{2}{1}\)

= \(\frac{3}{2} \times \frac{1}{2}\)

= \(\frac{3}{4}\)

x = \(\frac{1}{2}\) and x = \(\frac{-2}{3}\)

expand (x - \(\frac{1}{2}\))(x + \(\frac{2}{3}\)) = 0

x(x + \(\frac{2}{3}\)) - \(\frac{1}{2}(x + \frac{2}{3}\)) = 0

x² + \(\frac{4x - 3x}{6} - \frac{2}{6} = 0\)

\(x^2 + \frac{x}{6} - 2 = 0\)

6x² + x - 2 = 0

A prism has 3 rectangular faces and 2 triangular faces and 2 rectangular faces = 10(3 + 4 + 5) = 120

Area of triangular faces = \(\sqrt{s(s - a) (s - b) (s - c)}\)

where s = \(\frac{a + b + c}{2}\)

= \(\frac{3 + 4 + 5}{2}\)

= \(\frac{12}{2}\)

= 6

Area of \(\bigtriangleup\) = \(\sqrt{6(6 - 30(6 - 4)(6 - 5)}\)

= \(\sqrt{6 \times 3 \times 2 \times 1}\) = 6

Area of triangle faces = 2 x 6 = 12cm²

Total surface area = Area of rectangular face + Area of \(\bigtriangleup\) = 120 + 12

= 132cm²

The bearing S 20° W is equivalent to a bearing of 200°.

Moving clockwise 90° ⇒ 200° + 90°

= 290°

This is equivalent to N 70° W.