Welcome, Guest: Register On Nairaland / LOGIN! / Trending / Recent / NewStats: 3,194,672 members, 7,955,463 topics. Date: Sunday, 22 September 2024 at 06:38 AM |
Nairaland Forum / Nairaland / General / Education / Nabteb Maths (563 Views)
Schools To Teach Maths, Science In Indigenous Languages-Onu / NABTEB Result 2016/2017 - Eworld.nabtebnigeria.org | NABTEB May/june Result Chec / Cre In My NABTEB Result.. (2) (3) (4)
(1) (Reply)
Nabteb Maths by Benosky1(m): 9:29am On May 25, 2015 |
(MATHEMATICS ESSAY AND OBJ) verified Nabteb math 1-10.ABAAACBBAB 11-20.ABBCCBBCAC 21-30.CBBBADAABB 31-40.ABCDAADBAC 41-50.DACACABCCC 6a 3^x+1/9^-1 = 1 3^x+1 3^-2= 3^0 3^x+1+2 = 3^0 3^x+3 = 3^0 X + 3 = 0 X = -3 6b 3x+2y=7 ……(1) 7x-3y = 1 ……(2) Multiply (1) by 3 and (2) by 2 we have. 9x +6y = 21 14x – 6y = 2 23x = 23 X= 1 From (1) 3x+2y=7 3(1)+2y=7 3+2y=7 2y=7-3 2y=4 =y=2 X=1,and y=2 6c 18x^2 – 8/6x^2–11x–10 =2 (9x^2 - 4)/(6x2-15x) + (4x-10) =(3x-2)(3x+2)^2/3x(2x-5)+(2x-5)^2 =(3x-2)(3x+2)^2/(3x+2) (2x -5) = (3x-2)^2 2x -5= 6x–4/2x – 5 Or (3x-2)^2/2x-5 7a _/3(2_/2_/3)+_/2(_/3-_/2) clearing the bracket =2_/6+3+_/6-2 =3_/6+1 = 1+3_/6 7b 11,a,b,21 1/2,.. 1st term=a=11--(i) 2nd term=a+d=a..(ii) 3rd term=a+2d=b..(iii) 4th term=a+3d=21 1/2..(iv) put a=11 in Equa(iv) to get i.e a+3d=21 1/2 11+3d=21 1/2 3d=21 1/2-11 3d=10 1/2 d=10 1/2/3 d=7/2 =3 1/2 since a&d are 11 and 3 a=a+d=11+3.5=14.5 b=a+2d=11+2(7/2)=11 clearly, a=14 1/2 or 14.5 and b=18 7c Q = P(1+r/100)^2 Given Q=625 and P=225 substitute the value of P and Q 625=225(1+r/100)^2 625/225=(1+r/100)^2 25/9=(1+r/100)^2 (1+r/100)^2=25/9 Taking the square root of both sides [(1+r/100)^2]^2=_/25/9 =1+r/100=5/3 =1+r/100=5/3 r/100 =5/3-1/1 r/100=5-3/3 r/100 =2/3 r= 2x100/3=200/3 r=66.6667 r=66.7(3s.f) 11. DRAW THE LATITUDE CIRCLE where M is the midpoint Angular difference between A and B=45-15=30¤ 11a. Distance between A and B along the parallel of latitude i.e |AB|=@/360*2pieRCOS& where @=30¤=angular difference &=common latitude=40¤ |AB|=30¤/360*2*22/7*6400*COS40 =30*2*22*6400*COS40/360*7 =30*44*6400*0.7660/2520 =6471168/2520 =2567.9238 =3000km (to the nearest km) 11b. Distance between B and C along of longitude is, since the distance between |BC|=6700KM and the distance betwen A and B is AB=2567.9238KM Hence,Total distance covered by the aircraft is =|AB|+|BC| =(2567.9238+6700)km =9267.9238km using the relation i.e speed= Total distance covered/total time taken but speed=850km/hr 850=9267.9238/Total time taken :.total time taken=9267.9238/50 =10.9034hr total time taken=11hr(to the nearest hr) 11bii. Distance between B and C along of longitude is given by the |BC|=@/360*2pieR Where |BC|=6700Km 6700=@/360*2*22/7*6400 cross multiply i.e @=6700*360*7/2*22*6400 =16884000/281600 @=59.9574¤ hence,the latitude of C=@=59.95 hence,the latitude of C=20 1. Log8/log12-log3 =log8/log12/3-log3 =3log2/2log2=3/2 =1 1/2 1b. 81^0.25/32^0.2 =81^1/4/32^1/5 =3/2 =1 1/2 2a. T5=a+4d=9 a+4d=9---(i) T8=a+7d=27--(ii) a+7d=27--(i) -1*a+4d=27 substitute equation i frm ii -9-4d=-9 a+7d=27 3d=18 d=18/3 d=6 substitute d=6 into equa(i) to get a a+4d=9 a+4(6)=9 a+24=9 a=-15 hence the first term is -15 2bi. TI=a=x T2=ar^2=y a=x--(i) ar^2=y--(ii) divide eq(ii) by eq(i) ar^2/a=y/x r^2=y/x r=_+_/y/x hence the common ratio r=_+_/y/x 2bii. T1=a=8--(i) T3=ar^2=18--(ii) divide eq (ii) by (i) ar^2/a=18/8 r^2=18/8=9/4 r=_+_/9/4=_+3/2=_+1 1/2 since the common ratio is p common ratio is 1 1/2 3a. 4x-3/3-4x+2/5=1 taking the L.c.m 5(4x-3)-3(4x+2)/5=1 20x-15-12x-6/15=1 20x-12x-15-6/15=1 8x-21/15=1 cross multiply i.e 8x-21=15 8x=15+21 8x=36 x=36 x=9/2 =4 1/2 b. Universal set £={1,2,3,4,5} Elements; X=(2,3,5,7,9) Y=(1,2,3,4,5,6) Z=(2,4,6,8,10) i. X=(2,3,5,7,9) ii.Y=(1,2,3,4,5,6) iii. Z=(2,4,6,8,10) 3cii. X'=(1,4,6,8,9,10) Z'=(1,3,5,7,9) Y=(1,2,3,4,5,6) X'nZ'nY=(1) 4. Let the probability of Musa=M """ of Isa=I "" of Umaru=U Pr(M)=2/3 Pr(I)=5/8 Pr(U)=3/4 Will pass 4a. Pr(all three)=Pr(M)*Pr(I)*Pr(U) =2/3*5/8*3/4 =5/16=0.3125 4b. Pr(none of them pas the exam) Pr(M')*Pr(I')Pr(U') 1/3*3/8*1/4 =1/32=0.03125 4c.Pr(of Musa and Isa only pass) =Pr(M*)Pr(I) =2/3*5/8 =5/12 =0.4167 5. |AD|=|DC| Draw consider right angle ^ ABC i.e tantan68¤=8m/|BC| =>tan68¤/1=8m/|BC| |BC|*tan68¤=8m |BC|=8m/2.4751 |BC|=3.2322 Since D is the midpoint hence,|BD|=3.2322/2 consider right angle ^ ABD using triangle tantanhenceconsider ^ABD i.e Draw triangle 90¤+78.5792+=11.4208¤ also,consider right angle _^ABC i.e Draw 90¤+68¤+=22¤ Therefore=22¤-11.42 =10.5792 verified Nabteb math 1-10.ABAAACBBAB 11-20.ABBCCBBCAC 21-30.CBBBADAABB 31-40.ABCDAADBAC 41-50.DACACABCCC for more 07068397314 and ur neco answers |
(1) (Reply)
Student Graduate With 3 First Class Law Degrees From 3 Institutions / Unilorin 2015/16 Aspirants, This Is For You / YOBE STATE 2015/16 School Of Nursing And Midwifery Form Is On Sale Now, For More
(Go Up)
Sections: politics (1) business autos (1) jobs (1) career education (1) romance computers phones travel sports fashion health religion celebs tv-movies music-radio literature webmasters programming techmarket Links: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) Nairaland - Copyright © 2005 - 2024 Oluwaseun Osewa. All rights reserved. See How To Advertise. 16 |