11-20: CCDCAACECA
21-30: DCDAECAEED
31-40: EAADBEEADC
41-50: BBDCCBCECD
51-60: CADBAECDEC
NEW CHANGES IN THEORY MATHS THEORY NO 1,3,4,5,6,7,8,9,10 ==================================
(1a) Tabulate: Draw both vertical and horizontal lines across the digits. I drew only verticals cos it's txt messaging.
x|1|2|3|4 1|1|2|3|4 2|2|4|0|2 3|3|0|3|0 4|4|2|0|4
(1b) S.I=PRT P=N15,000, R=10%, T=3yrs S.I=#15,000, R=10%, T=3yrss S.I=15,000* 10/100 *3 S.I=N4,500 A=P S.I A=N15,000 N4,500 A=#19,500 ============================
(3) Let x represent Donald's age and let y represent his son's age x=5y---(1) (x-4(y-4)=448---(11) substitute for x in (11) 5y^2-20y-4y 16=448 5y^2-24y 16-448=0 5y^2-24y-432=0 y=-b _sqroot(b^2-4ac)/2a a=5, b=-24 c=-432 y=24 _sqroot[(-24^2)-4*5*-432]/2*5 y=24 _sqroot(576 8640)/10 y=24 _sqroot(9216)/10 y=24 _96/10 y=24 96/10 or 24-96/10 y=120/10 or -72/10 y=12 or -7.2 since age cannot be negative y=12years but x=5y x=5*12 x=60years ================================
(4a) Convert 30000 litres to metres = 30000/1000 metres = 30metres Depth of fuel = h 7.54.2 = 30m^3 = 31.5hm^3 = 30m^3 = h = 30m^3/31.5m = h = 0.9m
(4b) Depth of the tank = l*b*h where l= 7.5, b = 4.2, h = 1.2 =(7.5 4.2 1.2)m^3 =37.8m^3 Convert metres to litres = 37.8* 1000litres = 37800litres Litres of fuel needed to fill the tank=37800litres/30800litres =7800litres ================================
(5a) sector for building project =48000/144000*360=120degree sector for education = 32,000/144000*360=80degree sector for saving = 19200/144000*360=48degree sector for maintenance = 12000/144000*360=30degree sector for miscellaneous = 7200/144000*360=18degree sector for food items = 360-(120 80 48 30 18) =360-296 =64degree THEN DRAW A PIE CHART WITH THE ANSWERS YOU GOT ABOVE.
(5b) Food = 48000 32000 19200 1200 7200 x = 144000 Food => 118400 x = 144000 Food => x = 144000 — 118400 Food = x = #25,600 =================================
(6) 3sqroot[(41.02*sqroot0.7124)/(42.87*0.207*0.0404)] Tabulate No|log 41.02|16130 *sqroot0.7124|(1^-.8527)/2 |(2^- 1.8527)/2 |1^-9264 41.02*sqroot0.7124|1.6130 | 1^-.9264 |=1.5394 42.87|1.6322 *0.207| 1^-.3160 |0.9482 *0.0404| 2^-.6064 |=1^-.5546 |1.5394 |1^-.5546 |=(1.9848)/3 antilog4.588|0.6616 Answer=4.588 ==================================
(7a) 3^2n 1 - 4(3^n 1) 9=0 3^2-3 - 4(3^n -3) 9=0 (3^n)^2-3 - 4(3^n -3) 9=0 let 3^n = p p^2 -3 - 4(p-3) 9=0 3p^2/3 - 12p/3 9/3 = 0 p^2 - 4p 3 = 0 p^2 - 3p - p 3 = 0 p^2p(p-3) - 1(p-3) = 0 (p-1)(p-3) = 0 p-1 = 0 or p-3 = 0 p = 1 or 3 Recall 3^n = p when p=1 3^n = 3^0 n = 0 when p = 3 3^n = 3^1 n = 1
(7b) log(x^2 4) = 2 logx - log^20 log(x^2 4) = log^100 = log^x - log^20 (x^2 4) = log(xx) x^2 4 = 5x x^2-5x 4 = 0 x^2-4x - x 4 = 0 x(x-4) - 1(x-4) = 0 (x-1)(x-4) = 0 x-1 = 0 or x-4 = 0 x = 1 or 4 ==========================
(8ai) In <> ABC Cos B=(a^2 c^2-b^2)/2ac Cos B=(13^3 13^2-10^2)/(2*13*13) Cos B=(338-100)/338 Cos B=238/338 Cos B=0.7041 B=cos^-0.7041 B=45.2degrees therefore =90.4degrees In <>AOC By sine rule r/sine44.8=10/sine90.4 r=10sin44.8/sin90.4 r=7.05cm
(8aii) Circumference =2*pie*R =2*22/7*7.05 =44.31 =44.3cm(1d.p)
(8b) p(-1,2) , (2,6) (y-y1)/(y2-y1)=(x-x1)/(x2-x1) (y-2)/(6-2)=(x 1)/(2 1) (y-2)/4=(x 1)/3 3(y-2)=4(x 1) 3y-6=4x 4 3y-4x=10 =============================
(9a) Let the lens digit x and unit digit be y, therefore x-y=5 ---(1) 3xy-(10x y)=14 ----(2) 3xy-10x-y=14 -----(3) frm eq(1); x=5 y --- (4) therefore, 5(5 y)(y)-10(5 y)-y=14 (15 3y)y-50-10y-y=14 3y^2 4y-50-14=0 3y^2 4y-64=0 3y^2 -12y 16y-64=0 3y(y-4) 16(y-4)=0 (3y 16)(y-4)=0 y=-16/3 or 4 therefore from eqn(1); x 4=5 x=5 4=9 the number is 94
(9b) (3-2x)/ 4 (2x-3)/3 (3(3-2x) 4(2x-3))/12 (9-6x 8x-12)/12 =(2x-3)/12 ============================
(10a) y=(2x^2 3)^5 let U=2x^2 3 Y=u^5 du/dx = 4x dy/du = 5u^4 dy/du = (2x^2 3)^4 dy/dx = du/dx dy/du dy/dx = 4x.5(2x^2 3)^4 dy/dx = 20x(2x^2 3)^4
(10b) y=3x^2 2x 5 dy/dx =6x 2 dy/dx =6(3) 2 dy/dx =18 2 dy/dx =20
(10c) R-W=Wv^2/gx Wv^2=gx(R-W) Wv^2=gRx-Wgx Wv^2 Wgx=gRx W(v^2 gx) =gRx W=gRx/V^2 gx R=2, g=10, x=3/2, V=3 W= 10*2*3/2/3^2 10*3/5 W=30/9 15 W=30/24 W=5/4
FROM PROF KEN EKWUEME
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