F /MATH OBJ :
1 DCEBDBABAC .
11 ABDDDCCDAC .
21 CDBACAEBCB .
31 CEAEEDEBBD
41 CEBCDDCDDA
5 )
diagram
final velocity (v )= u + at
= 20 + 5 ( 4 )
= 20 + 20
= 40 m /s total distance = area of trapezium OABD
+ Area of a triagngle BCD
time between DC = t
from v = u + at
t= v -u / a = 40 - 0 /10 = 4 s
total distance = 1 /2 ( 20 + 4 ) 4 + 1 /2 ( 4 ) * 40 =
1 /2 ( 60 ) 4 + 1 / 2 (4 ) * 40
= 30 * 4 + 2 * 40
= 120 + 80
= 200 m
==========================
1 DCEBDBABAC .
11 ABDDDCCDAC .
21 CDBACAEBCB .
31 CEAEEDEBBD
41 CEBCDDCDDA
5 )
diagram
final velocity (v )= u + at
= 20 + 5 ( 4 )
= 20 + 20
= 40 m /s total distance = area of trapezium OABD
+ Area of a triagngle BCD
time between DC = t
from v = u + at
t= v -u / a = 40 - 0 /10 = 4 s
total distance = 1 /2 ( 20 + 4 ) 4 + 1 /2 ( 4 ) * 40 =
1 /2 ( 60 ) 4 + 1 / 2 (4 ) * 40
= 30 * 4 + 2 * 40
= 120 + 80
= 200 m
==========================
1 )
A = (2 / 1 -1 / 3 )
| A | =2 * 3 -1 x -1
= 6 + 1 = 7
minor( 2 )= 3
(- 1 )= 1
(1 )= - 1
(3 )= 2
minor A = (3 / -1 1 / 2 )
co- factor A (3 / 1 - 1 /2 )
adjoint (A )= ( 3 /- 1 1 / 2 )
A ^ - 1 = 1 / | A | adj (A )
= 1 / 7 (3 / -1 1 /2 )
================================
10 c )
ar ^ 3 = 54 .. .. ... (i )
ar = 6 ... .. ..( ii )
√ r^ 2 = 3
Substitute for the value or r in eqn ( ii)
3 a/ 3 = 6 / 3
a =2
Tn = 2 * 3 ^ n - 1
=========================
3a)
Pr(A passed)=4/5,pr(A failed)=1/5
Pr(B passed)=3/5,Pr(B failed)=2/5
Pr(C passed)-2/5,Pr(C failed)=3/5
Pr(atleast one passed)
=4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2/5
=24/125+9/125+4/125
=47/125
3b)
Pr(atleast one failed)
=4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2/5
=101/125
4)
(3+2y)^5 with (a+b)^5 shows
a=3,b=2y
Using pascals triangle the coefficient of (a+b)^5 are 1,5,10,10,5 and 1
(3+2y)^5
=(3)^5+5(3)^4(2y)+10(3)^3(2y)^2+10(3)^2(2y)^3+5(3)(2y)^4+(2y)^5
=243+810y+1080y^2+720y^3+240y^4+32y^5
(3.04)^5=(3+0.04)^5
=3^5+5(3)^4(0.04)+10(3)^2(0.04)^2+10(3)^2(0.04)^3+5(3)(0.04)^4+(0.04)^5
=243+5*81(0.04)+10*27(0.0016)+10*9(0.000064)+15(0.0000026)+0.0000001
=243+16.2+0.432+0.00576+0.000039+0.0000001
=259.63783
11a)
5x^2-2x+3=0
Divide both sides by 5
x^2-2/5x+3/5=0--eq1
Compare eq1 above with
x^2-(alpha+Beta)x+alphabeta=o
alpha+beta=2/5
alphabeta=3/5
11ai)
alpha^3+beta^3=(alpha+beta)(alpha+beta)^2-3alphabeta
=2/5(2/5)^2-3(3/5)
=2/5(4/25-9/5)
=2/5(-41/25)
=-82/125
11aii)
Sum of alpha +1/beta and beta+1/alpha
=(alphabeta+1)/beta +(alpha beta + 1)/alpha
=alphabeta^2+alpha^2beta+beta/alphabeta
=alphabeta(alpha+beta)+(alpha+beta)/alphabeta
product root
(alpha +1/beta)(beta+1/alpha)
=alpha(beta+1/alpha)=1/beta(beta+1/alpha)
=alphabeta+1+1+1q/alphabeta
sum=alphabeta(alpha+beta)+(alpha+beta)/alphabeta
=3/5(2/5)+(2/5)
=6/25+2/5
=(6+10)/25
=16/25
Product alphabeta+2+1/alphabeta
=3/5+2+1/(3/5)
=3/5+2/1+5/3
=64/15
Recall
x^2-(alpha+beta)+alphabeta=0
x^2-16/25x+64/15=0
11bi)
f(x)=x^3+2x-kx-6
x-2=0
x=+2
f(2)=2^3+2(2)^2-k(2)-6=0
8+8-2k-6=0
16-6-2k0
10-2k=0
2k=10
k=10/2
k=5
11bii)
(x^2+4x+3)
(x-2)rootx^3+2x^2-5x-6
=x^3-2x^2
=4x^2-5x
=4x^2-8x
=3x-6
=3x-6
=0
=>x^2+4x+3
=x^2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)
the remaining factors of f(x) are (x+1) and (x+3)
11biii)
Zero of f(x)
x-2=0,then x=2
x+1=0, then x=-1
x+3=0, then x=-3
Zeros of f(x) are 2,-1 and -3
10a)
T4=ar^4=162---(1)
T8=ar^7=4374---(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2
10ai)a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18
10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048
10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6
10c)
T2=ar=6---(1)
T4=ar^3=54---(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1
15a)
Rank correlation=1-6ED^2/n(n^2-1)
Rank correlation =1-(6*50)/8(64-1)
=1-348/504
=1-0.691
=0.309
15b)
Pr(at most 2)=Pr(0)+Pr(1)+Pr(2)
n=1000,Pr=0.05 Pr(2)=0.95
Pr(x)=e^-landa(landa^x)
Pr(0.05)<0.1 and landa=1.000*0.05=50
e^-50(50^0)/o!+e^-50(50^1)/1!+e^-50(50^2)/2!
1/e^50+50/e^50+250.2e^50
=1/e^50[1+50+25/2]
-----------------
12a)
w=weight of the plank
moment=force*perpendicular distance
clockwise moment=Anticlockwise
Make A the reference point
Pw=2.5m,
AM=(2.5-0.m
=1.7m g=10m/s
AB=3.2m,PA=0.8m
Convert 30kg to weight
=mg=30*10=300N
therefore 300*0.8=w*1.7
w=240/1.7=141.18N
Distance of its C.G from P=2.5m
12b)
For R1=A, R2=B
R1=R2=2R1
Let Q be the reference point
300x*141.18*2.5=2R1(4.8+1)
300x+352.95=11.6R1
300x-11.6R1=-352.95---(1)
Let P in the reference point
2R1(0.8+4)=141.15*2.5
9.6R1=352.95
R1=352.95/9.6
=36.77N
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A = (2 / 1 -1 / 3 )
| A | =2 * 3 -1 x -1
= 6 + 1 = 7
minor( 2 )= 3
(- 1 )= 1
(1 )= - 1
(3 )= 2
minor A = (3 / -1 1 / 2 )
co- factor A (3 / 1 - 1 /2 )
adjoint (A )= ( 3 /- 1 1 / 2 )
A ^ - 1 = 1 / | A | adj (A )
= 1 / 7 (3 / -1 1 /2 )
================================
10 c )
ar ^ 3 = 54 .. .. ... (i )
ar = 6 ... .. ..( ii )
√ r^ 2 = 3
Substitute for the value or r in eqn ( ii)
3 a/ 3 = 6 / 3
a =2
Tn = 2 * 3 ^ n - 1
=========================
3a)
Pr(A passed)=4/5,pr(A failed)=1/5
Pr(B passed)=3/5,Pr(B failed)=2/5
Pr(C passed)-2/5,Pr(C failed)=3/5
Pr(atleast one passed)
=4/5*2/5*3/5+3/5*1/5*3/5+2/5*1/5*2/5
=24/125+9/125+4/125
=47/125
3b)
Pr(atleast one failed)
=4/5*3/5*3/5+3/5*2/5*1/5+4/5*2/5*2/5
=101/125
4)
(3+2y)^5 with (a+b)^5 shows
a=3,b=2y
Using pascals triangle the coefficient of (a+b)^5 are 1,5,10,10,5 and 1
(3+2y)^5
=(3)^5+5(3)^4(2y)+10(3)^3(2y)^2+10(3)^2(2y)^3+5(3)(2y)^4+(2y)^5
=243+810y+1080y^2+720y^3+240y^4+32y^5
(3.04)^5=(3+0.04)^5
=3^5+5(3)^4(0.04)+10(3)^2(0.04)^2+10(3)^2(0.04)^3+5(3)(0.04)^4+(0.04)^5
=243+5*81(0.04)+10*27(0.0016)+10*9(0.000064)+15(0.0000026)+0.0000001
=243+16.2+0.432+0.00576+0.000039+0.0000001
=259.63783
11a)
5x^2-2x+3=0
Divide both sides by 5
x^2-2/5x+3/5=0--eq1
Compare eq1 above with
x^2-(alpha+Beta)x+alphabeta=o
alpha+beta=2/5
alphabeta=3/5
11ai)
alpha^3+beta^3=(alpha+beta)(alpha+beta)^2-3alphabeta
=2/5(2/5)^2-3(3/5)
=2/5(4/25-9/5)
=2/5(-41/25)
=-82/125
11aii)
Sum of alpha +1/beta and beta+1/alpha
=(alphabeta+1)/beta +(alpha beta + 1)/alpha
=alphabeta^2+alpha^2beta+beta/alphabeta
=alphabeta(alpha+beta)+(alpha+beta)/alphabeta
product root
(alpha +1/beta)(beta+1/alpha)
=alpha(beta+1/alpha)=1/beta(beta+1/alpha)
=alphabeta+1+1+1q/alphabeta
sum=alphabeta(alpha+beta)+(alpha+beta)/alphabeta
=3/5(2/5)+(2/5)
=6/25+2/5
=(6+10)/25
=16/25
Product alphabeta+2+1/alphabeta
=3/5+2+1/(3/5)
=3/5+2/1+5/3
=64/15
Recall
x^2-(alpha+beta)+alphabeta=0
x^2-16/25x+64/15=0
11bi)
f(x)=x^3+2x-kx-6
x-2=0
x=+2
f(2)=2^3+2(2)^2-k(2)-6=0
8+8-2k-6=0
16-6-2k0
10-2k=0
2k=10
k=10/2
k=5
11bii)
(x^2+4x+3)
(x-2)rootx^3+2x^2-5x-6
=x^3-2x^2
=4x^2-5x
=4x^2-8x
=3x-6
=3x-6
=0
=>x^2+4x+3
=x^2+3x+x+3
=x(x+3)+1(x+3)
=(x+1)(x+3)
the remaining factors of f(x) are (x+1) and (x+3)
11biii)
Zero of f(x)
x-2=0,then x=2
x+1=0, then x=-1
x+3=0, then x=-3
Zeros of f(x) are 2,-1 and -3
10a)
T4=ar^4=162---(1)
T8=ar^7=4374---(2)
Divide (2) by (1)
4374/162=ar^7/ar^4
2187/81=r^3
27=r^3
r=3root27
r=3
Recall(1)
T5=ar^4=162
162=a(3)^4
a=162/81=2
10ai)a=1st,T2=ar,T3=ar^2
a=2,T2=2*3=6,T3=2*3^2=18
therefore the three terms are 2,6 and 18
10aii)
Sn=a(r^n-1)/(r-1)
=2(3^10-1)/(3-1)
Sn=2(3^10-1)/2
=59049-1
=59048
10b)
a=6,c=60,Sn=330
Sn=n/2(a+c)
330=n/2(6+60)
300*2=n(66)
660=n(60)
n=660/66
=60
L=a+(n-1)d
60=6+(10-1)d
60=6+9d
60-6=9d
54=9d
d=54/9
d=6
10c)
T2=ar=6---(1)
T4=ar^3=54---(2)
divide 2 by 1
54/6=ar^3/ar
9=r^2
r=root9
r=3
Recall(1)
ar=6
a(3)=6
a=6/3
a=2
Tn=ar^n-1
Tn=2(3)^n-1
15a)
Rank correlation=1-6ED^2/n(n^2-1)
Rank correlation =1-(6*50)/8(64-1)
=1-348/504
=1-0.691
=0.309
15b)
Pr(at most 2)=Pr(0)+Pr(1)+Pr(2)
n=1000,Pr=0.05 Pr(2)=0.95
Pr(x)=e^-landa(landa^x)
Pr(0.05)<0.1 and landa=1.000*0.05=50
e^-50(50^0)/o!+e^-50(50^1)/1!+e^-50(50^2)/2!
1/e^50+50/e^50+250.2e^50
=1/e^50[1+50+25/2]
-----------------
12a)
w=weight of the plank
moment=force*perpendicular distance
clockwise moment=Anticlockwise
Make A the reference point
Pw=2.5m,
AM=(2.5-0.m
=1.7m g=10m/s
AB=3.2m,PA=0.8m
Convert 30kg to weight
=mg=30*10=300N
therefore 300*0.8=w*1.7
w=240/1.7=141.18N
Distance of its C.G from P=2.5m
12b)
For R1=A, R2=B
R1=R2=2R1
Let Q be the reference point
300x*141.18*2.5=2R1(4.8+1)
300x+352.95=11.6R1
300x-11.6R1=-352.95---(1)
Let P in the reference point
2R1(0.8+4)=141.15*2.5
9.6R1=352.95
R1=352.95/9.6
=36.77N
further mathematics (obj and essay) Answers Will be posted free of charge...
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Neco 2016 further mathematics essay and obj questions and answers, 2016 free Neco further mathematics obj and essay answers, neco further mathematics essay and obj answer, 100% 2016 Neco further mathematics essay and objective free expo/ runz, 2016 Neco further mathematics essay and objective free answer, further mathematics essay and objective Neco 2016 expo, Neco 2016/2017 further mathematics expo for real correct verified obj essay expo further mathematics obj and essay answer, live 2016 Neco further mathematics essay and obj answer now available, how to get 2016 Neco further mathematics question and answer, 2016/2017 Neco further mathematics essay and objective question and answers, Download 2016 Neco further mathematics essay and objective answers Neco answer, Neco 2016 further mathematics expo, Neco 2016 further mathematics essay and objective expo, Neco further mathematics essay runz/expo, 2016 Neco further mathematics essay and objective expo now available, 2016 Neco further mathematics obj and theory answer. 2016 Neco further mathematics essay and objective ANSWERS IS NOW AVAILABLE HERE,I need 2016 Neco further mathematics essay and objective answers further mathematics essay and objective answer for Neco 2016,Free 2016 Neco further mathematics essay and objective EXPO AND ANSWERS,Neco practical 2016 Neco expo for further mathematics theory/ essay and obj/ objective answers by Solutionclass Examloaded Waec Runz sureboys Waec 2016 expo Fullclub Waec 2016 expo Examfiled , Exapowap Nairabaze neco expo, Generalguru, , samzyworld, Examwap, blastexam Bombexpo, jazzyfanz, Grupdates, Wapbaze, Wapdoze, Naijashine, Naijafresh, donprosper, expoloaded, guruscamp, expocrew, codedextra.com, betavibes