WAEC Mathematics Past Questions & Answers - Page 84

\( A = P \left(1 + \frac{r}{100}\right)^n \)

Where P = 126, r = 4,n = 2

A=126 \( \left(1 + \frac{4}{100}\right)^2 \text{Using LCM} \)

=126 \( \left(\frac{100+4}{100}\right)^2 = 126 \left(\frac{104}{100}\right)^2 \)


=126 \( \left(1.04^2 \right) \)

= 126 * 1.04 * 1.04

=136.28

A = 136.30 (approx.)

The Amount A, = N136.30k

The slope of the line from (0, 0) passing through (-3, -4) = \(\frac{-4 - 0}{-3 - 0}\)

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

Equation of a line is given as \(y = mx + b\), where m = slope and b = intercept.

To get the value of b, we use a point on the line, say (0, 0).

\(y = \frac{4}{3} x + b\)

\(0 = \frac{4}{3}(0) + b\)

\(b = 0\)

The equation of the line is \(y = \frac{4}{3} x\)

Given: \(x + y = 90° ... (1)\)

\((\sin x + \sin y)^{2} - 2\sin x \sin y = \sin^{2} x + \sin^{2} y + 2\sin x \sin y - 2\sin x \sin y\)

= \(\sin^{2} x + \sin^{2} y ... (2)\)

Recall: \(\sin x = \cos (90 - x) ... (a)\)

From (1), \(y = 90 - x ... (b)\)

Putting (a) and (b) in (2), we have

\(\sin^{2} x + \sin^{2} y \equiv \cos^{2} (90 - x) + \sin^{2} (90 - x)\)

= 1

The total surface area of a cylinder = 2πrl + 2πr2

= 2πr(l + r)

= 2 × 3.14 x 5(7+5)

2 × 3.14 × 12 x 5

= 377.1cm (1DP)

n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y) = 15 + 12 − 7 ∴ n(X ∪ Y) = 20