Q. 6

  1. y \, \alpha \, \frac{1}{x^2} y = 48 \, when \, x = \frac{1}{2} y = \frac{k}{x^2} 48 = \frac{k}{(\frac{1}{2})^2} k = 48 \times (\frac{1}{2})^2 k = 12 y = \frac{12}{x^2}

  2. Find x when y = 3

    y = \frac{12}{x^2} 3 = \frac{12}{x^2} x^2 = \frac{12}{3} = 4 x = \sqrt{4} x = \pm 2