2011 - WAEC Mathematics Past Questions and Answers - page 2

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

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

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

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

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

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

Perimeter of sector = 2r + \(\frac{\theta}{360^o} \times 2\pi r\)

\(\pi + 8 = 2 \times 4 + \frac{\theta}{3360^o} \times 2 \pi \times 4\)

\(\pi + 8 + \frac{\theta}{360^o} \times 8 \pi\)

P + 8 - 8 = \(\frac{\theta \pi}{456o}\)

\(\pi = \frac{\theta \pi}{45^o}\)

\(\theta \pi = 45^o\)

Error = 1.80m - 1.75m = 0.05m

%error = \(\frac{\text{error}}{\text{true measurement}}\) x 100%

volume of cylinder = \(\pi r^2\)h

0.216\(\pi m^3 = \pi \times r^2 \times 1m\)

assumed that h = 1m

0.216 = r²

r² = 0.216

r = \(\sqrt{0.216}\)

= 0.46

volume of cylinder = \(\pi r^2\)h

0.216\(\pi m^3 = \pi \times r^2 \times 1m\)

assumed that h = 1m

0.216 = r²

r² = 0.216

r = \(\sqrt{0.216}\)

= 0.46

am + bn - an - bm

am - an - bm + bn

a(m - n) - b(m - n)

(a - b)(m - n)