2000 - WAEC Mathematics Past Questions and Answers - page 4

(\frac{1}{x-3}-\frac{3(x-1)}{x^2 - 9}<br /> \frac{1}{x-3}-\frac{3(x-1)}{(x-3)(x+3)}<br /> \frac{x+3-3x+3}{(x-3)(x+3)};\frac{-2x+6}{(x-3)(x+3)}<br /> \frac{-2(x-3)}{(x-3)(x+3)}=\frac{-2}{x+3})

Equation with roots -8 and 5: (x + 8)(x - 5) = 0

\(x^2 - 5x + 8x - 40 = 0\)

\(x^2 + 3x - 40 = 0\)

\(k = m\sqrt{\frac{t - p}{r}}\)

\(\frac{k}{m} = \sqrt{\frac{t - p}{r}}\)

\((\frac{k}{m})^2 = \frac{t - p}{r}\)

\(rk^2 = m^2 (t - p)\)

\(\therefore m^2 t = rk^2 + m^2 p\)

\(t = \frac{rk^2 + m^2 p}{m^2}\)

(3y^2 = 27y<br /> 3y^2 - 27y = 0<br /> 3y(y - 9) = 0<br /> 3y = 0 or y - 9 = 0<br /> y = 0 or y = 9<br /> y = 0 or 9)

\(\frac{1}{x} + \frac{4}{3x} - \frac{5}{6x} + 1 = 0\)

\(\frac{6 + 8 - 5 + 6x}{6x} = 0\)

\(\frac{9 + 6x}{6x} = 0 \implies 9 + 6x = 0\)

\(6x = -9 \implies x = \frac{-3}{2}\)

circumference of the bicycle wheel
(= 2\pi r = 2 \times \frac{22}{7} \times 42 - 264cm = 2.64m\)
Number of revolution
(=\frac{66}{2.64} = 25)

Volume of pyramid with square base = \(\frac{\text{base area} \times height}{3}\)

\(80 = \frac{\text{base area} \times 15}{3}\)

\(80 = 5 \times \text{base area}\)

\(\text{Base area} = 16 cm^2\)