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)