Advertisement

header ads

2017 NECO GCE Mathematics Expo Answer Obj And Essay Solution Answer – Nov/Dec Free Expo

2017  NECO GCE Mathematics Expo Answer Obj And Essay Solution Answer – Nov/Dec Free Expo


2017 NECO GCE Mathematics Expo (Obj & Essay/Theory) Questions and Answers Free Neco Gce 2017/Expo/Runs/Runz/Site


NECO GCE 2017 MATHEMATICS EXPO ANSWERS FREE | NECO GCE 2017 Maths EXPO OBJ AND THEORY ANSWERS AVAILABLE NOW


Tags::: 2017 neco gce Mathematics obj2017 neco gce Mathematics theory correct neco gce Mathematics answer expo site for neco gce Mathematics answerfree Mathematics expo runz nov/dec expo free neco gce Mathematics expo runs 2017 Mathematics 2017 neco gce answer Mathematics 2017 neco gce expo answer Mathematics answer for neco gcemathematics gce expoMathematics neco gce 2017 answer nov/decMathematics neco gce 2017 expoMathematics neco gce ansewer by examanswer Mathematics neco gce answerneco gce 2017 math runsneco gce 2017 Mathematicsneco gce 2017 Mathematics objneco gce 2017 Mathematics questionneco gce 2017 questionneco gce answer Mathematicsneco gce free math runsneco gce geo Mathematics expo wapbase.orgNeco gce geography expo 2017Neco gce geography expo answer flexygist.comneco gce Mathematics answer gidifans.comneco gce Mathematics answer with freeneco gce Mathematics expo naijapunch.comneco gce Mathematics runsneco gce Mathematics theoryneco gce nov/dec Mathematics expo 2017nov/dec neco gce Mathematics exponovember/december neco gce Mathematics expo answerreal Mathematics neco gcewapbase.org, GurusVevo, ExamWide, passing, Mathematics expo 2017where to get neco gce Mathematics expo


TO GET THE REAL 2017 NECO GCE MATHS QUESTIONS AND ANSWERS DIRECT TO YOUR PHONE CLICK HERE 




NOTE::: BELOW ANSWER IS OUR 2016 NECO GCE MATHEMATICS ANSWERS..
OUR 2016 NECO GCE MATHEMATICS WENT WELL..  


UPDATED NUMBERS: 1,2,3,4,5,7,8,9,10,11…
FULL ANSWERS COMPLETED
MATHS OBJ
1EAACBAECBE
11CDABECBCCA
21AEDCBACEDE
31ADADCCDABD
41BCEBDCEABE
51DBAECBECBE
==========
SECTION A(ANSWER ALL QUESTIONS 1-5)
1a)
(6y²)³ / 2+y^6
= 6³y^6 / 24y^6
=6³*y^6/24*y^6
= 216/24
=9
1b)
Log10(2m²+5m-2)=1
2m²+5m-2=10
2m²+5m-2-10=0
2m²+5m-12=0
Using m=-b+-√b²-4ac/2a
a=2
b=5
c=-12
m=-5+-√5²-4*2*(-12)/2*2
m=-5+-√2+96/4
m=-5+-√98/4
m=-5+-9.9/4
m=-5+9.9/4 or m=-5-9.9/4
M=1.225 or -3.725
==========
2a)
w*3√D
W=k 3√D
10=k*3√27
10/3 =k*3/3 k=10/3
W=10/3*3√64 = 10/3*4*40/3=13 1/3
2b)
X/2 + y/6 = 2/3
L.C.M = 6
Rx + y = 4 —–(1)
X/3 – y/2 = 5/3
L.C.M = 6
2x – 3y = 10 —–(2)
From eq (1)
3x + y = 4
Y= 4 – 3x ——(3)
Sub for y in eq (2)
2x – 3(4-3x) = 10
2x – 12 + 9x = 10
11x = 22
X=2
Sub for x in eq (3)
Y = 4-3(1)
Y = 4-3-
Y = 1
===========
3a)
Sn = a(r^n – 1)
a = 5
r = 3
n= 8
5(3^8 – 1)/ (3-1) = 5(6561-1)/2
= 32800/2 = 164000
3b
Cos30°=12/P
PCos30°=12
Divide through by cos 30°
PCos30°/Cos30°=12/Cos30°
P=12/cos30°
P=12/0.8660
P=13.8568
==========
4a)V=2/3Πr
1527.4=2/3*22/7*r
1527.4=44r/21
32075.4=44rb
r=32075.4/44
r=728.98
=729cm
4b)
3^3y+1 = 81^-y
3^3y= 3^-4y
3y+1=-4y
1=-4y-3y
1=-7y
Y= – 1/7
===========
5a)
Red=4
Green = 6
Black= 10
Total=20
1)p(Red) = 4/20 = 1/5
2)p(green)= 6/20 = 3/10
3) p (black)= 10/20= 1/2
5b)
( X-1)(X-8)+12=0
X²-8x-X+8+12=0
X²-9x+20=0
X= -b +√b²-4ai
X= 9+/-√-9²-4×1×20÷ 2×1
X=9+/-√81-80÷2
X=9+/-√1÷2
X= 9+1/2 or 9-1/2
X= 10/2 or 8/2
X= 5 or 4
============
+++++++++++++
SECTION B(ANSWER 5 QUESTIONS 7-11)
+++++++++++++
7a)
lets k and p be our constant
R=k+P3√5
S = speed
k + P3√64 = 182…(i)
k + P3√343 = 200…(ii)
subtitute eqn (i) from (ii)
k+4p=182
k+7p=200
3p/3 = 18/3
p=6
subtitute 6 in eqn(i)
k+4p=182
k+4(6)=182
k=182-24
k=158
recall R=k+p3√5
R=158+63√5
(a)find S when R is 188
188=158+6×3√s
188-158 = 6 3√s
30/s = 6 3√s/s
√s=30
(s) = 30^3
s = 30^3
s= 27,000k/h
7b)
find R when S = 27
R=158+6 3√27
R=158+6×3
R=158+18
R=176 newtons
============
(8a)
4a^2 b*3ab^-2/4a^3*(3ab)
=12 a^4 b^1/12a^4 b
=6^-1/b^1
=b^-1-1 =b^2
6^-2 or 1/b^2
8b)
Let the pocket money be x
2x/5 =stations
X/4=school levies
X/6= food
X-(24x/5+x/4+x/6)=55
X-(24x+15x+10x)/60 =55
X-49x/60 =55
60*x-60-49x/60 = 55*60
60x-49x=#3,300
11x/11=3300/11
X=#300
The pocket money =#300
8c)
16^x-4 = 64^x+4/8^x
24^x-16 = 2^6x+24/2^3x
2^4x-16 = 2^6x+24-3x
2^4x-16 = 2^3x+24
4x-16 = 3x + 24
4x – 3x = 16+24
x = 40
============
9a)
solve using cosine rule
m^2 = 70^2+40^2 – (2 x 70 x 40 x cos135)
m^2 = 4900 + 1600 – (5600 x cos135)
m^2 = 6500 – (5600 x cos135)
m√6500-(5600 x cos135)
m = √6500+5578.2
m = √12078.16
m = 109.9006
m == 109.9
9b)
lets find tita (θ)
using sine rule
70/sinθ = 109.9/sin135
70sin135/109.9 = 109.9sinθ/109.9
sinθ = 70 x sin135/109.9
sinθ = 0.45038
θ = sin^-1(0.45038)
θ = 26.76935
θ == 27
:- Bearing of k from R
180° + 85° + 27°
= 292°
==============
10a)
1/27^(x-2)=9^(2-x)
3^3(x-2)-1=3^2(2-x)
3-1(x-2)=4-2x
2x-4=4-2x
2x+2x=4+4
4x=8
X=8/4
X=2
10b
5x+y/x+5y= 3/1
5x+y= 3(x+5y)
5x+y=3x+15y
5x-3x=15y-y
2x=14y
X/y= 14/2 =7, y/x= 1/7
X2-y2/4xy
Divide all through by my
(X²- y²/) ÷(4xy/xy)
Xy- xy
(X/y- y/x) ÷4
(7- 1/7)÷4 ~~~49-1/7×1/4
48/7× 1/4= 24/7
= 4
============
11a)
A=(1 7,3 2) and B=(2 5,4 6)
i)A^-1=1/det(2 -7,-3 1)
Det= 2 -21=-19
A^-1=1/-19(2 -7,-3 1)
-1/19(2 -7,-3 1)
ii)B^T =(2 -4,5 6)
iii)AB =(1 7,3 2) (2 5,4 6)
(2 -28,6-8, 5+46,15+12)
(-26,-2, 51,27,)
11b)
7x^2 + 12x + 5 = 0
7x^2/7 + 12x/7 = -5/7
X^2 + 12x/7 = -5/7
X^2 12x/7 + (6/7)^2 – -5/7 + (6/7)2
(X+6/7)^2= -5/7 + 36/49
(X+6/7)^2 = -35+36/49
(X +6/7)^2 = 1/49
(X +6/7) = + and – √1/19
X+ 6/7 = + and – 1/7
X = -6/7 + and – 1/7
X = -6+1/7 or -6-1/7
X= -5/7 or -7/7
X = -5/7 or -1
============
GOODLUCK
============

Welcome: 2017 neco gce maths answer,2017 neco gce answers,neco gce answer 2017,2017 neco gce answers,2017 neco gce maths answer,2017 neco gce answers,neco gce answer 2017,2017 neco gce answers,2017 neco gce maths answer,2017 neco gce answers,neco gce answer 2017,2017 neco gce answers,2017 neco gce maths answer,2017 neco gce answers,neco gce answer 2017,2017 neco gce answers,2017 neco gce maths answer,2017 neco gce answers,neco gce answer 2017,2017 neco gce answers,2017 neco gce maths answer,2017 neco gce answers,neco gce answer 2017,2017 neco gce answers

Post a Comment

0 Comments