2014 - WAEC Mathematics Past Questions and Answers - page 2

E = \(\frac{m}{2g}\)(v² - u²)

multiply both sides by 2g

2Eg = 2g (\(\frac{M}{2g} (V^2 - U^2)\)

2Eg = m(V² - U²)

2Eg - mV² - mU²

mU² = mV² - 2Eg

divide both sides by m

\(\frac{mU^2}{m} = \frac{mV^2 - 2Eg}{m}\)

U² = \(\frac{mV^2 - 2Eg}{m}\)

= \(\frac{mV^2}{m} - \frac{2Eg}{m}\)

U² = V² - \(\frac{2Eg}{m}\)

U = \(\sqrt{V^2 - \frac{2Eg}{m}}\)

x \(\alpha \frac{1}{y}\) y \(\alpha z\) the relationship = x \(\alph \frac{1}{z}\)

(2, -3)(2, 5) gradient = \(\frac{y_ = y_1}{x_2 = x_1} = \frac{5 - (-3)}{2 - 2}\) = \(\frac{5 + 3}{0}\) = \(\frac{8}{0}\) = undefined

\(\begin{array}{c|c} height(x) & frequency(f) & fx \ \hline 2 & 2 & 4\ 3 & 4 & 12\ 4 & 5 & 20 \ 5 & 3 & 15\ 6 & 1 & 6\end{array}\) mean\(\bar{x} = \frac{\sum fx}{\sum f}\) = \(\frac{57}{15}\) = 3.8

diameter = 12cm

radius = \(\frac{12}{2}\)cm = 6cm

area of sector = 66cm²

area of sector = area of cone

66 = \(\pi rl\)

66 = \(\frac{22}{7} \times r \times 6\)

divide both sides by 6

11 = \(\frac{22}{7}r\)

22r = 11 x 7

r = \(\frac{77}{22}\)

= 3.5cm

let the chord be AB = 7cm

radius OA = 3.7cm distance of the = OM using Pythagoras theorem

OA² = AM² + OM²

3.7² = 3.5² + OM²

13.69 = 12.25 + OM²

13.69 - 12.25 = OM²

1.44 = OM²

OM = \(\sqrt{1.44}\)

OM = 1.2cm

distance of chord = 1.2cm

let the number of N50 notes = x let the number of N20 notes = y from the first statement x + y = 13.....(1) from the second statement 50x + 20y = 530 ......(2) dividing through by 10, we have 5x + 2y = 53 ....(20 solving simultaneously by eliminating x + y = 12....(1) 5x + 2y = 53.....(2) multiply eqn (1) by 2 multiply eqn (2) by 1 2x + 2y = 26 -5x + 2y = 53 -3x = -27 x = 9 substitute x = 9 into equation (1) x + y = 13 y = 13 - x = 13 - 9 = 4 y = 4 the number of N50 notes = 9