JAMB Mathematics Past Questions & Answers - Page 156

\((x + 3y + 5)(2x^{2} + 5y + 2)\)

= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 6y + 10x^{2} + 25y + 10\)

= \(2x^{3} + 5xy + 2x + 6yx^{2} + 15y^{2} + 31y + 10x^{2} + 10\)

\(3\frac{7}{8} + 1\frac{1}{3} = 4\frac{21 + 8}{24}\)

= \(4\frac{29}{24}\)

\(\equiv 5\frac{5}{24}\)

\(1\frac{2}{3} - \frac{1}{8} = \frac{5}{3} - \frac{1}{8}\)

= \(\frac{40 - 3}{24}\)

= \(\frac{37}{24}\)

\(5\frac{5}{24} - \frac{37}{24} = \frac{125}{24} - \frac{37}{24}\)

= \(\frac{88}{24}\)

= 3\(\frac{2}{3}\)

Area of the rectangular picture = L x B = 8 x 6

= 48 sq.cm.

Area of the whole surface (which is gotten by adding \(\frac{1}{2}\) on every side to the original picture size) is 9 x 7 = 63 sq. cm

area of the frame is 63 - 48

= 15 sq. cm

Let x rep. the angle of depression of the foot of the tree.

tan 30^o = \(\frac{y}{100}\)

y = 100 tan 30^o

= 57.8

By Pythagoras, AC² = 300² + 58²

= 900 + 3354

tan x = \(\frac{opp}{adj}\)

= \(\frac{58}{300}\)

= 0.19

tan x = 0.19

x = tan 0.19

= 60^o

\(\sqrt{3^{\frac{1}{x}}}\) = \(\sqrt{9}\)

3\(^{\frac{1}{2x}}\) = \(9^{\frac{1}{2}}\)

3\(^{\frac{1}{2x}}\) = 3\(^{2 \times \frac{1}{2}}\)

3\(\frac{1}{2x}\) = 3\(\frac{2}{2}\) = 3

\(3^{\frac{1}{2x}}\) = \(3^{1}\)

\(\frac{1}{2x}\) = \(\frac{1}{1}\)

x = \(\frac{1}{2}\)