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 = \(\frac{100x}{9}\)
- \(y = 900x^2\)
The correct answer is: B
Explanation
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 = \(\frac{900}{x^2}\)
= \(yx^2 = 900\)