(a) Given that P = (\(\frac{rk}{Q} – ms\))\(^{\frac{2}{3}}\)
(i) Make Q the subject of the relation;
(ii) find, correct to two decimal places, the value of Q when P = 3, m = 15, s = 0.2, k = 4 and r = 10.
(b) Given that \(\frac{x + 2y}{5}\) = x – 2y, find x : y
Explanation
P = (\(\frac{rk}{Q} - ms\))\(^{\frac{2}{3}}\)
P\(^{\frac{3}{2}}\) = \(\frac{rk}{Q}\) - ms
P\(^{\frac{3}{2}}\) + ms = \(\frac{rk}{Q}\)
Q = \(\frac{rk}{p^{\frac{3}{2}}ms}\)
(a)(ii)
Q = \(\frac{10 \times 4}{3\frac{3}{2} + 15 (0.2)}\)
Q = \(\frac{40}{5.1958 + 3}\)
Q = \(\frac{40}{8.1958}\)
= 4.88 to 2 d.p
(b) \(\frac{x + 2y}{5} = x - 2y\)
x + 2y = 5(x - 2y)
x + 2y = 5x - 10y
x - 5x = -10y - 2y
-4x = -12y
\(\frac{x}{y} = \frac{12}{4}\)
x : y = 3 : 1