JAMB Mathematics Past Questions & Answers - Page 317

The first poster has 7 ways to be arranges, the second poster can be arranged in 6 ways and the third poster in 5 ways.

= 7 x 6 x 5

= 210 ways

or $\frac{7}{P_3}$ = $\frac{7!}{(7 - 3)!}$ = $\frac{7!}{4!}$

= $\frac{7 \times 6 \times 5 \times 4!}{4!}$

= 210 ways

($\sqrt2 + \frac{1}{\sqrt3})(\sqrt2 - \frac{1}{\sqrt3}$)

$\sqrt4 - \frac {\sqrt2}{\sqrt3} + \frac {\sqrt2}{\sqrt3} - \frac {1}{\sqrt9}$

= 2 - $\frac {1}{3}$

= $\frac {16 - 1}{3}$

= $\frac{5}{3}$

$\frac{2 - \sqrt5}{3 - \sqrt5}$ x $\frac{3 + \sqrt5}{3 + \sqrt5}$

$\frac{(2 - \sqrt5)(3 + \sqrt5)}{(3 - \sqrt5)(3 + \sqrt5)}$ = $\frac{6 +2\sqrt5 - 3\sqrt5 - \sqrt25}{9 + 3\sqrt5 - 3\sqrt5 - \sqrt25}$

= $\frac{6 - \sqrt5 - 5}{9 - 5}$

= $\frac{1 - \sqrt5}{4}$

log$_{3}^{18}$ + log$_{3}^{3}$ - log$_{3}^{x}$ = 3

log$_{3}^{18}$ + log$_{3}^{3}$ - log$_{3}^{x}$ = 3log₃3

log$_{3}^{18}$ + log$_{3}^{3}$ - log$_{3}^{x}$ = log₃3³

log₃($\frac{18 \times 3}{X}$) = log₃3³

$\frac{18 \times 3}{X}$ = 3³

18 x 3 = 27 x X

x = $\frac{18 \times 3}{27}$

= 2

$(\frac{16}{81})^{\frac{1}{4}} \div (\frac{9}{16})^{-\frac{1}{2}}$

$(\frac{16}{81})^{\frac{1}{4}} \div (\frac{16}{9})^{\frac{1}{2}}$

$(\frac{2^4}{3^4})^{\frac{1}{4}} \div (\frac{4^2}{3^2})^{\frac{1}{2}}$

$\frac{2^{4 \times \frac{1}{4}}}{3^{4 \times \frac{1}{4}}} \div \frac{4^{2 \times \frac{1}{2}}}{3^{2 \times \frac{1}{2}}}$

$\frac{2}{3} \div \frac{4}{3}$

$\frac{2}{3} \times \frac{3}{4}$

$\frac{2}{4}$

$\frac{1}{2}$