(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

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