WAEC - Mathematics (2018 - No. 3)

Given that y varies inversely as the square of x. If x = 3 when y = 100, find the equation connecting x and y.

\(yx^2 = 300\)

\(yx^2 = 900\)

y = \(\frac{100x}{9}\)

\(y = 900x^2\)

Y \(\alpha \frac{1}{x^2} \rightarrow y = \frac{k}{x^2}\)

If x = 3 and y = 100,

then, \(\frac{100}{1} = \frac{k}{3^2}\)

\(\frac{100}{1} = \frac{k}{9}\)

k = 100 x 9 = 900

Substitute 900 for k in

y = \(\frac{k}{x^2}\); y = \(\frac{900}{x^2}\)

= \(yx^2 = 900\)