NECO PAST QUESTIONS & ANSWERS FURTHER MATH

NECO PAST QUESTIONS & ANSWERS FURTHER MATH – Past questions and answers, 100% real answers, neco expo, neco exam runz, complete further maths expo, exam paper leaked on exam day – Get more update and answers on all exams today.

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

Finally here comes the Further maths objectives Answers.

First and foremost  – 1BBEDEDABCA
Afterward  -11DCBBABDABD
Above all -21AECCBEACBA
Finally  -31DAACBDEEBD
Furthermore -41CCECBDCADA

NECO PAST QUESTIONS & ANSWERS FURTHER MATH

 

First and Foremost

NECO PAST QUESTIONS & ANSWERS FURTHER MATH

(6ii)
from y=3x+4 and y=2x-1
let y1=3x+4 nand y2=2x-1
therefore m1=3, m2=2
let tita be the angle between the
two lines.
tan titat=m1-m2/(1+m1m2)=(3-2)/
(1+3(2))
tan tita= 1/1+6 =1/7
tita= tan^-1(1/7)
Above all =tita= 8.13 degree.

 

 

4
i)seven male surgeons can not be
formed from a team consisting of
four men
Also three female surgeons cannot
be formed from two women
Hence the numbers of way is.

i)0 ways.

ii)-0 ways.

 

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

5
i)for -2<_x<_2 ie x=-2,-1,0,1,2 for x=-2,f(x)=(-2)^2+2=4+2=6 for x=-1.

f(x)=(0)^2+2=1+2=3 for x=0,

-f(x)=(0)^2+2=0+2=2 for x=1

=f(x)=1^2+2=1+2=3 for x=2,

Above all f(x)=2^2+2=4+2=6 Hence the co-domain are 2,3,6

ii) f is onto Reason since f(-2)=f(2) f(-1)=f(1) therefore f is onto iii)f(x)=x^2+2=y =>x^2+2=y
x^2=y-2
x=sqroot(y-2)
f^-1(x)=sqroot(x-2)
f^-1(11)=sqroot(11-2)
=sqroot(9)
=3 0r -3.

FURTHERMORE Answers are BELOW

NECO PAST QUESTIONS & ANSWERS FURTHER MATH

(7)
a)Given a*b=a+b+2ab where a,bER
To show wether * is commutative
a*b=a+b+2ab
b*a=b+a+2ba=a+b+2ab=a*b
Hence, the operation is
commutative/

b)a*e=a+e+2ae=a=e*a
=>a+e+2ae=a
=>e+2ae=a-a
=>e(1+2a)=0
=>e=0/1+2a
=>e=0
the identity element e=0.

Furthermore.
7c)a*a^-1
=>a+a^-1+2aa^-1=0
=a=a_1(1+2a)=0
a^1(1+2a)=0-a
=a^-1(1+2a)=-a
a^-1=-a/1+2a
the inverse element a^-1=-a/1+2a
Any element E/ -a/1+2a has no
inverse.

 

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

9

Furthermore.

Given the curve f(x)=x^3-6^2-15x-1
dy/dx=3x^2-12x-15
of the turning point dy/dx=0
=>3x^2-12x-15=0
or x^2-4x-5=0
x^2-5x+x-5=0
x(x-5)+1(x-5)=0
(x-5)(x+1)=0
x=5 or x=-1
d^2y/dx^2=6x-12.

Furthermore

i)The gradient of x=1=dy/dx x=1
=3(1^2)-12(1)-15
==3(1)-12-15
=3-12-15
=-24.

9ii)when x=5,d^2y/dx^2 x=5
=6(5)-12
=30-12
=18
when x=-1,d^2y/dx^2 x=-1
=6(-1)-12
=-6-12
=-18
since d^2y/dx^2>1 when x=5, the y
is minimum
Above all since d^2y/dx^2<1 when x=-1,then y is maximum at x=-1 when x=5,

f(x)=5^3-6(5^2)-15(5)-1 =125-150-75-1 =

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

the maximum point is (5,-101) when x=-1,f(x)= (-1)^3-6(-1)^2-15(-1)-1 =-1-6+15-1 =7

Hence the maximum point is (-1,7)t (13) diagram frictional force Fr=mg sin tita =5.6* 9.8* 0.5878 =32.26N

Reaction R= mgcos tita =5.6 * 9.8 cos^2 36 56* 9.8 * 0.8090 =44.40N..

(a) Above all the force parallel to the plane Fr==32.26N.

(b) Finally from Fr= UR =0.45* 44.40 =19.98N.

 

1
P=15kg(15N)
Efx=8cos50degrees-10cos60degrees-150cos90degrees
=8(0.6428)-10(0.5000)-150(0)
=5.142-5
=-0.1424N
Efy=8sin50degrees
+10sin60degrees-150sin90degrees
=8(0.7660)+10(0.8660)-150(1)
=6.128+8.660-150
=-135.242N.

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

i)Resultant of force R=Sqroot
(Efx^2+Efy^2)
=sqroot(-0.1424)^2+(-135.242)^2
sqroot( 0.020+18282.285)
=sqroot( 18282.305)
Finally =135.21N.

(1ii)
let titat be the direction odf the
resultant force
tan tita= Efx/Ef
=tan tita=-135.212/-0.1424
=949.5225.
tita=tan^-1(949.5225)
Finally =89.94 degree.

FINALLY _ NECO PAST QUESTIONS &ANSWERS FURTHER MATH

(1iii) acceleration
from P-R=F
i.e 150-135.21=14.74
ma=14.79
a=14.79/m=14.79/150g
Finally =0.986ms^-1.

2
a)From S=ut-1/2at^2
29=3.89*15-1/2*a*15^2
=29=58.09-0.5*a*225
29-58-58.05=-0.5*a*225
-29.05=-112.5a
a=29.05/112.5
Above all  -a=0.26m/s^2.

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

b)Total distance=21m+8m=29m
Total time=10+5=15s
from S=(v+u/2)t
v=0,s=29m,t=15s
29=(0+u/2)15
29=(u/2)15
2*29=15u
58=15u
=>u=58/15
Above all the initial velocity u=3.87m/s.

NECO PAST QUESTIONS &ANSWERS FURTHER MATH

(6i)
from 3x-y+11=0
-y=-3x-11
y=3x+11(comparing to y=mx+c)
gradient m=3,
from equation of straight line
y-y1=m(x-x1)
(x1,y1)=(1,-5) or x1=1, y1=-5
y-(-5)=3(x-1)
y+5=3(x-1)
=y=3x-3-5
finally =y=3x-8
or y-3x+8=0 or 3x-y-8=0.

 

NECO FURTHER MATHEMATICS QUESTIONS AND ANSWERS 2017

6a ) Finally – greatest height (H )= ( U ^2 sin ^2 tita )/ 2 g =956 ^ 2 sin ^2 ( 36)) / 2 (10) =( 3136 * 0. 5878 ^2 )/ 20 = 1083 .516 /20 =54 .196 =54 .18 6b ) horizontal range( R) = ( U ^ 2 sin2 tita ) /g =( 56^ 2 sin 2( 36) )/ 10 = (3136 * sin 72)/ 10 =( 3136* 0 .9511 )/ 10 =2982 .6496 /10 =298 . 26495 =298 . 26m 6c ) time of flight =2 sin tita / g =( 2 *56 sin 36) / 10 =112 * 0 .5878 =65 .8336 /10= 6 .5834 — – — — – — — – — — – — — – — — – — — 15a) capital demand=N12,000* N120 =N5,040,000 total purchasing cost =N140*12,000 =N1,680,000 holding cost amount =20% =20/100 * N5,040,000 =N1,008,000 Economic order= N1,680,000 + N1,008,000 =N2688000 ii) total value cost per n annum =N268800/12,000 =N224 15b) mean x=74 s.d=15 i) Pr(x>85)=x- xbar/ sigma
=85-74 /15
=11/15 =0.73
=0.2673

FINALLY NECO PAST QUESTIONS &ANSWERS FURTHER MATH

ii)
Finally Pr(x<62)=(62-74)/15 =-12/15=-0.8 =0.2881 iii) Pr(62 =0.5-.2673 + 0.5- 0.2881 =0.4446

Above all Use the search box to find other answers on any exam at all you need. Also like us to get first hand update on Neco Further math expo two hours before the exam starts.

 

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