NECO 2017 FURTHER MATHEMATICS ANSWERS | NECO RUNZ

NECO 2017 FURTHER MATHEMATICS ANSWERS – neco 2017 further mathematics answers, neco further mathematics past questions, real neco further maths expo, neco further maths, neco further maths 2015, all neco further mathematics 2015, neco further maths 2014, neco further maths answer, complete neco further mathematics answer, finally neco further mathematics answer, neco further maths expo 2015 – Get more update and answers on all exams today.

Finally here comes the Further maths objectives Answers.

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

First and Foremost

(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.

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.

(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.

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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

=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 =

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.

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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.

(6i)
from 3x-y+11=0
-y=-3x-11
y=3x+11(comparing to y=mx+c)
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.

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

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