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log\(_2\)6 - log\(_2\)3 = log\(_2\) (\(\frac{6}{3}\)) = log\(_2\)2... A log\(_2\)8 - 2log\(_2\)\(\frac{1}{2}\) log\(_2\)8 - log\(_2\)\(\frac{1}{2}\)\(^2\) log\(_2\)8 - log\(_2\)\(\frac{1}{4}\) = log\(_2\) 32... B \(\frac{A}{B}\) = \(\frac{log_22}{log_232}\) N.B. log\(_2\)2 = 1 = \(\frac{1}{5}\)

\(\frac{2\sqrt{14} \times 3\sqrt{21}}{7 \sqrt{24} \times 2\sqrt{98}}\) = \(\frac{6\sqrt{14} \times 3 \times \sqrt{7} \times \sqrt{3}}{7 \times 2 \sqrt{6} \times \sqrt{7} \times \sqrt{14}}\) = \(\frac{3\sqrt{3}}{14\sqrt{6}}\) = \(\frac{3\sqrt{3}}{14\sqrt{2} \times \sqrt{3}}\) = \(\frac{3\sqrt{2}}{28}\)

To find the extra cost of feeding one additional person, we start with the formula for the cost \( Q \): \(Q = 1.5 + 0.5n\) Calculate the cost for \( n \) \(Q(n) = 1.5 + 0.5n\) Calculate the cost for \( n + 1 \) \(Q(n + 1) = 1.5 + 0.5(n + 1) = 1.5 + 0.5n + 0.5 = 2 + 0.5n\) Find the extra cost \(\text{Extra cost} = Q(n + 1) - Q(n) = (2 + 0.5n) - (1.5 + 0.5n) = 2 - 1.5 = 0.5\) The extra cost of feeding one additional person is \( 0.5 \) Naira. To convert this to kobo (since \( 1 \) Naira = \( 100 \) kobo): \(0.5 \text{ Naira} = 0.5 \times 100 = 50 \text{ kobo}\) Thus, the extra cost of feeding one additional person is \( 50 \) kobo.

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Nonagon = 9 Where 3 are equal and 6 other angles = 1110o (2 x 9 - 4)90o = (18 - 4)90o 14 x 90o = 1260o 9 angles = 1260^\circ; 6 angles = 110o Remaining 3 angles = 1260o - 1110o = 150o Size of one of the 3 angles \(\frac{150}{3}\) = 50o

In PNO, ONP = 180 - (35 + 86) = 180 - 121 = 59^\circ PRQ = QNP = 59^\circ(angles in the same segment of a circle are equal)

Tan \(\theta\) = \(\frac{m^2 - n^2}{2mn}\) \(\frac{\text{Opp}}{\text{Adj}}\) by pathagoras theorem = Hyp2 = Opp2 + Adj2 Hyp2 = (m2 - n2) + (2mn)2 Hyp2 = m4 - 2m2n4 - 4m2 - n2 Hyp2 = m4 + 2m2 + n2n Hyp2 = (m2 - n2)2 Hyp2 = \(\frac{m^2 + n^2}{2mn}\)

From < ZWY \(\frac{h}{a}\) = Tan60º h = a tan 60º .............................1 From < XWZ \(\frac{h}{ 8 + a }\) = tan 30º h = ( 8 + a ) tan 30º ...................2 equate 1 and 2 (8 + a ) tan30º = a tan60º \(\frac{( 8 + a )}{\sqrt3}\) = a\(\sqrt{3}\) 8 + a = 3a 2a = 8 a = 4 but h = a tan 60º = 4\(\sqrt{3}\)

ABCD is the floor, by pythagoras AC2 = 144 + 81 = \(\sqrt{225}\) AC = 15cm Height of room is 8m, diagonal of floor is 15m ∴ The cosine of the angle which a diagonal of the room makes with the floor is EC2 = 152 cosine = \(\frac{\text{adj}}{\text{hyp}}\) = \(\frac{15}{17}\)

Circumference of Latitude 0oS where R is the radius of the earth is = 2\(\pi\) R cos \(\theta\)

The locus of the mid-point of all chords of length 6cm within a circle of radius 5cm and with centre O is the perpendicular bisectors of the chords.

Period of day light is from 5.30 to 7.00p.m - 13hrs 30min. interval 24hrs make 1 day ∴ period of darkness = 24hrs - 13hrs 30mins = 10hrs 30mins. Angle of sector of daylight = \(\frac{13 \times 60 + 30 \times 360}{(24 \times 60)}\) = \(\frac{810}{1440}\) x 360o = \(\frac{291600}{1440}\) = 202.5o 202.5o = 202.30o darkness period = 360o - 202o 30' = 157o30' = 202o30' 157o 30'

A rectangle, a square and a rhombus are the only parallelogram ∴ Probability of a parallelogram = \(\frac{3}{5}\)

From the diagram, POQ is a diameter; o is the centre of the circle and TP is a tangent where RTP = 30o RTP = RQP = x(at circumference = made by tangent outside the circle) i.e x = 30o

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the nonagon = 9 where 3 are equal and 6 other angles = 1110o ( 2 x 9 - 4)90o = (18 - 4)90o = 14 x 90o = 1260o 9 angles = 12600, 6 angles = 1110o Remaining 3 angles = 1260o - 1110o = 150o size of one of the3 angles \(\frac{150}{3}\) = 50o

cos\(\theta\) = \(\frac{a}{b}\), Sin\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a}}\) Tan\(\theta\) = \(\sqrt{\frac{b^2 - a^2}{a^2}}\), Tan 2 = \(\sqrt{\frac{b^2 - a^2}{a^2}}\) 1 + tan2\(\theta\) = 1 + \(\frac{b^2 - a^2}{a^2}\) = \(\frac{a^2 + b^2 - a^2}{a^2}\) = \(\frac{b^2}{a^2}\)