y is directly proportional to \( \sqrt{x} \) \( y = \frac{1}{6} \) when \( x = 8 \) Find the value of \( y \) when \( x = 64 \) - Edexcel - GCSE Maths - Question 16 - 2017 - Paper 1

Given ( y = \frac{1}{6} ) when ( x = 8 ), substituting these values gives: [ k = \frac{y}{\sqrt{x}} = \frac{\frac{1}{6}}{\sqrt{8}} = \frac{1/6}{\sqrt{8}} = \frac{1/6}{2\sqrt{2}/2} = \frac{1}{6 \cdot 2 \sqrt{2}} = \frac{1}{12 \sqrt{2}}\ ]

Now substituting ( k ) back into the equation with ( x = 64 ) gives: [ y = k \sqrt{64} = \frac{1}{12 \sqrt{2}} \cdot 8 = \frac{8}{12 \sqrt{2}} = \frac{2}{3 \sqrt{2}}\ ] To rationalize the denominator, multiply by ( \frac{\sqrt{2}}{\sqrt{2}} ): [ y = \frac{2\sqrt{2}}{3 \cdot 2} = \frac{\sqrt{2}}{3}] Thus, ( y = \frac{\sqrt{2}}{3} ) when ( x = 64 ).