Jambmaths question:
If y varies directly as the square root of x and y =3 and x =16. Calculate y when x = 64
Jamb Maths Solution:
$\begin{align} & y\propto \sqrt{x} \\ & y=k\sqrt{x} \\ & \text{when }x=16,\text{ }y=3 \\ & 3=k\sqrt{16} \\ & k=\frac{3}{4} \\ & \text{when }x=64 \\ & y=k\sqrt{x}=\frac{3}{4}\sqrt{64}=\frac{3}{4}\times 8=6 \\\end{align}$
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