(a) If p varies directly as (r^{2}) and p = 3.2 when r = 4, find the value of p when r = 6.5.
(b) Solve the simultaneous equations :
(frac{x}{2} + frac{y}{4} = 1 ; frac{x}{3} – frac{y}{4} = frac{-1}{6})
Explanation
(a) (p propto r^{2})
(implies p = kr^{2})
(3.2 = k(4^{2}) implies 3.2 = 16k)
(k = frac{3.2}{16} = 0.2)
(therefore p = 0.2r^{2})
When r = 6.5,
(p = 0.2 (6.5^{2}) )
= (0.2(42.25))
(p = 8.45)
(b) (frac{x}{2} + frac{y}{4} = 1 … (1))
(frac{x}{3} – frac{y}{4} = -frac{1}{6} …. (2))
Adding equation (1) and (2), we have
(frac{x}{2} + frac{x}{3} = 1 – frac{1}{6})
(frac{5x}{6} = frac{5}{6} implies x = 1)
Put x = 1 in (1),
(frac{1}{2} + frac{y}{4} = 1 implies frac{y}{4} = 1 – frac{1}{2})
(frac{y}{4} = frac{1}{2} implies y = 2)
((x, y) = (1, 2)).