2004 - WAEC Mathematics Past Questions and Answers - page 2
3x – y = 5 ------ II\)
Add I and II;
\(4x = 12 => x = \frac{12}{4} = 3\
Y = 7 – 3 = 4, evaluate \hspace{1mm} \frac{y}{2} – 3\
\frac{4}{2} – 3 => 2-3 = -1\)
In the diagram, POQ is the diameter of the circle centre O. Calculate ∠QRS
∠PSQ = 90° (angle in a semi-circle)
(when SR is joined to SP)
∠SPQ = 180 – (90+35) = 180 – 125 = 55°
∠QRS + ∠SPQ = 180° (opposite angles in a cyclic quad is supplementary)
∠QRS = 180° - 55°
= 125°
If \(\left(\frac{1}{4}\right)^{(2-y)} = 1\), find y.
l = length h = height b = base
l = 12cm h = 10cm b = 8cm
T.S.A \(= (12 \times 8 \times 12 \times 10 + 8 \times 10)cm^\
= (96 + 120 + 80)cm^2 = 296cm^2\)
bx + ya + y = ba + b; y(a+1) = ba+b-bx\
y(a+1)=b(a+1-x); y = \frac{b(1+a-x}{a+1}\)
lf \(log_q p= r\), express p in terms of q and r
If \(log_q p\) = r convert from logarithm to indices
P = q\(^r\)
Find the next two terms of the sequence
1, 5, 14, 30, 55, ...
\(1 + 2^2 = 5 + 3^2 = 14 + 4^2 = 30 + 5^2 = 55\)
\(55 + 6^2 = 91 + 7^2 = 140\).
91, 140 are the next two terms.
Each interior angle of a regular polygon is 108°. How many sides has it?
The exterior angle of the polygon = 180 – 108 = 72°. The sum of exterior angle of the regular polygon = 360°
The number of sides \(=\frac{360}{72}=5\)
5(2 – x) + x(2 – x) = 0; (2 – x)(5 + x) = 0
2 – x = 0 or 5 + x = 0; x = 2 or x = -5