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(a) Conservation- can be referred to as the intelligent use or selection of what can be changed or replaced in the community without harmfully disturbing the balance in nature. In other words, it is the preservation of natural I resources from loss or waste, harm and exploitation through wise, rational or sensible use to ensure their availability and to preserve the quality or original nature of the natural resources. (b) Reasons for conservation are: (i) To prevent destruction of natural ecosystems (ii) For medicinal purposes . (iii) To preserve naturally beautiful scenarios for their aesthetic values. (iv) For research purposes (v) To promote the recycling of some scarce mineral resources/ water (vi) To prevent destruction of natural environment (vii) To allow for continued use of natural resources for man's benefit (viii) Preservation of rare and valuable species of plant and animals for future generation or to save them from extinction. (c) (i) Methods of conserving water are: (1) Trapping or storage in tanks or wells (2) Treatment and recycling of used water (3) Damming of rivers to allow more effective management of water (4) Planting of trees to provide vegetation cover which reduces evaporation and promotes water retention in soil /rainfall. (5) Saving of water by prompt repairs of burst pipes and turning off taps immediately after use. (ii) Methods of conserving wild life are: (1) Establishment of zoological gardens (2) Establishment of game /forest reserves (3) Hunting should be controlled (4) Prohibiting bush burning (5) Prohibiting killing or poaching of animals in game reserves (6) Educating the public on the values of wildlife (7) Discouraging deforestation and encouraging afforestation. (iii) Methods of conserving Forest are: (1) Ensure cutting of trees without destroying undergrowth. (2) Reafforestation or tree planting should be encouraged. (3) Discourage bush burning. (4) Pests /diseases of plants should be controlled. (5) Establishment of forest reserve (6) Educating the public on the value of forest and importance of conservation and cutting of matured trees only. (d) Some forest or game reserves in Nigeria include: (1) Yankari game reserve in Bauchi State (2) Borgu game reserve in Niger State (3) Omo forest reserve in Ogun State (4) Sakpoba forest reserve in Edo State (5) Mamu River in Anambra State (6) Anara in Kaduna State (7) Zamfara in Zamfara State (8) Sanga River Plateau State (9) Olomu forest reserve - Kwara State (10) ljero forest reserve in Ekiti.

An Isosceles Right Triangle is a right triangle that consists of two equal length legs. i.e opp = adj hyp\(^2\) = adj\(^2\) + opp\(^2\) 2\(^2\) = x\(^2\) + x\(^2\) 4 = 2x\(^2\) 2 = x\(^2\) \sqrt2 = x

△PQR is similar to △PNM \(\frac{PQ}{PN}\) = \(\frac{PR}{PM}\) \(\frac{7 + 3 }{5}\) \(\frac{5 + NR}{3}\) cross multiply 10 x 3 = (5 x 5) + 5 x NR 5NR = 30 - 25 = 5 NR = \(\frac{5}{5}\) = 1 cm.

Area of sector = \(\frac{\theta}{360^\circ} \times \pi r^2\) = \(\frac{60}{360} \times \frac{22}{7} \times 3.5 \times 3.5\) = \(\frac{77}{12}\) = \(6\frac{5}{12}\) cm\(^2\)

where f = \(\frac{oscillations}{time}\) period = 1/ f or \(\frac{time}{oscillations}\) For pendulum x: where period = 1.5 and oscillation = 400 time = oscillations X 1.5 = 600s For pendulum y: \(\frac{time}{oscillations}\) → 600/500 period for y = 1.2s period =

E = I\(_{rms}\)\(^2\)Rt I\(_{rms}\) = \(\frac{I_o}{\sqrt{2}}\) = \(\frac{5}{\sqrt{2}}\) = 3.5355 E = 12.5 x 8 x 100 = 10,000J

a, ar, ar\(^2\), ar\(^3\), ar\(^4\), ar\(^5\) 6th term = ar\(^5\) = \(\frac{-2}{27}\) 1st term = a = 18 => ar\(^5\) = 18r\(^5\) =\(\frac{-2}{27}\) r\(^5\) = \(\frac{-2}{27 * 18}\) → \(\frac{-1}{27 * 9}\) r\(^5\) = \(\frac{-1}{243}\) r = \(^5\)\sqrt[\(\frac{-1}{243}\)] r = \(\frac{-1}{3}\)

2 89 2 44 r 1 2 22 r 0 2 11 r 0 2 5 r 1 2 2 r 1 2 1 r 0 0 r 1 89\(_{10}\) = 101100\(1_2\)

\((2\frac{1}{6} - 1\frac{2}{3}) \div 2\frac{2}{3}\) = \((\frac{13}{6} - \frac{5}{3}) \div \frac{8}{3}\) = \(\frac{3}{6} \times \frac{3}{8}\) = \(\frac{3}{16}\)

\(56x^{-4} \div 14x^{-8}\) = \(4x^{- 4 - (-8)}\) = \(4x^{4}\)

Log\(_{10}\)4 + Log\(_{10}\)25 = Log\(_{10}\)(4 + 25) = Log\(_{10}\)100 = Log\(_{10}\)10\(^2\) = 2Log\(_{10}\)10 = 2 x 1 = 2

Length = 59.6m Breath = 40.3m Perimeter = (59.6m + 40.3m) x 2 = 99.9 x 2 = 199.8m Actual measurement Perimeter = (60 + 40) x 2 = 100 x 2 = 200m Error = 200m - 199.8m = 0.2 Error =0.2/200 x 100/1 = 0.1%

18, 12, 6, 0, -6, 12, -18, 24, -30 9th term = -30

(2x - 5)(x - 3) 2x\(^2\) - 5x - 6x + 15 2x\(^2\) - 11x + 15

\(\frac{5}{x - y} - \frac{4}{y - x}\) = \(\frac{5(y - x) - 4(x - y)}{(x - y)(y - x)}\) = \(\frac{5y - 5x - 4x + 4y}{(x - y)(y - x)}\) = \(\frac{9y - 9x}{(x - y)(y - x)}\) = \(\frac{9(y - x)}{(x - y)(y - x)}\) = \(\frac{9}{x - y}\)