Author Topic: NECO (MATHEMATICS) (Read 2531 times)

JEFFREY96 · « **on:** November 14, 2014, 07:20:30 pm »

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JEFFREY96 · « **Reply #1 on:** November 15, 2014, 10:36:52 am »

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tochukwu84 · « **Reply #2 on:** November 15, 2014, 11:51:03 am »

Pls I hav nt seen d solved question, am inside d hall now n i subscribed wit u pple

benscheely · « **Reply #3 on:** November 15, 2014, 12:23:31 pm »

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benscheely · « **Reply #4 on:** November 15, 2014, 12:45:48 pm »

(1a)
R=14cm, tita=74 degree
lenght of chord= 2r sin tita/2
=2*14*sin(74/2)
=2*14*sin 37 degree
=2*14*sin37 degree
=2*14*0.6018
=16.85cm
(1b)
R=14cm, L=16.85cm
let x reps the distance
R^2=L^2 +X^2
14^2= (16.85)^2 + x^2
x^2= 87.92
x= sqr root (87.92)
=9.38
(1c)
Area of triangle POQ
=1/2 r^2 sin tita
=1/2* (14)^2 * sin 74 degree
=94.21 cm^2
(2a)
sum of inferior angle
=(n-2)180 degree
=(6-2)180 degree
=4*180 degree= 720 degree
720 degree =140 +5x
5x=720-140
5x=580
x=580/5
=116 degree
(2b)
sum is 720 degree
k+k+15+k+30+35+k+45+k+55=720 degree
6k+180= 720
6k=720-180
6k=540
k=540/6
k=90 degree

Ikole · « **Reply #5 on:** November 15, 2014, 01:10:09 pm »

(3)
f5=f4+24-----eqn1
f4=f3+8---eqn 2
f5=ar^4----eqn 3
f4=ar^3------eqn 4
f3=ar^2------eqn 5
ar^4=ar^3+24
ar^4-ar^3=24
ar^3=ar^2+8
ar^3-ar^2=8
ar^3-ar^2=8
ar^2(r^2-(r))=24
ar^2(r-1)=8
(r^2-r)/r-1=24/8
r(r-1)/r-1=3
r=3
(i) Common ratio(r)=3
(ii)
First term(a) is thus;
ar^2(r-1)=8
a(3)^2(3-1)=8
a*18=8
a=8/18=4/9
first term(a)=4/9
(iii)
sn= a(r^n-1)/(r-1)
s5=(4/9(3^5-(1))/(3-1)
s5=4/9(243-1)/2
s5=(4/9*(242))/2
s5=4*242/(9*2)
s5=(4*242)/9*2
s5=968/18
=53.7

benscheely · « **Reply #6 on:** November 15, 2014, 01:16:30 pm »

(4a) tan tita=71.5/26 tan tita= 2.75 tita= tan^-1(2.75) tita= 70 degree

(4b) adjacent = p^2 -q^2, hyp= (p+q)^2 (p+q)^2= p^2 -q^2+ opp^2 p^2 + 2pq +q^2= p^2- q^2 +opp^2 opp^2=2pq- q^2- q^2 =opp^2=2pq-2q^2 sin tita= (2pq-(2q^2))/(p+q)^2

(5a) draw the table

(5b) perfect cub=0 Pr(perfect cube)=0

(5c) divisible by 5=(5,5,5,5,10,10,10) divisible by 3=(3,3,6,6,6,6,6,9,9,9,9,12) probalitiy=19/36

Ikole · « **Reply #7 on:** November 15, 2014, 01:29:51 pm »

(6a)
Tabulate:
no:
3.267,0.483,0.385,
log:
0.5141,bar 1.3161,bar 1.4145
total bar 2.7306
0.5141
bar 2.7306
1.3835
3.267
0.43*0.385
antilog 24.18

(6b)
z=(3x-2)/(2x+3)
z(2x+3)=3x-2
2zx+3z=3x-2
2zx-3x=--3z-2
x(2z-3)=-3z-2
x(2z-3)=-3z-2
x=(-3-2)/2z-3
x=-(3z+2)/2z-3
x=(-2-3z)/(2z-3)

Ikole · « **Reply #8 on:** November 15, 2014, 01:38:30 pm »

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Ikole · « **Reply #9 on:** November 15, 2014, 01:40:22 pm »

(7a)
A:T:U= 1/3: 7/12: 1/12
let A reps Ade
T reps Tayo
U reps uche
Total ratio= 1/3 +7/12+ 1/12
=(4+7+1)/12=12/12=1
Ade=7000 + U
U=(1/12)/1*(T)
T=12U
A=(1/3)/12U
A=4U
4U=7000 + U
3U=7000
U=7000/3 =
N2333.3
T=12*(7000)/3
T=4*7000
T=N28,000
Total profit = N28,000
(7bi)
Given (A/P)^1/n -1 =r/100
(A/p)^1/n=r/100 + 1
(A/p)^1/n= (r+100)/100
A/p=((r+100)/100)^n
A=p((r+100)/100)^n)
(7bii)
given p=7808,n=3, r=5
A=7808((25+100)/100)^3
A=7808(125/100)^3
A=7800*(5/4)^3
A=7808*(125/64)
A=15250

benscheely · « **Reply #10 on:** November 15, 2014, 01:42:34 pm »

Quote from: Ikole on November 15, 2014, 01:38:30 pm

for more answers add on 2go@
EXAMASTER4

HMMMMM

JEFFREY96 · « **Reply #11 on:** November 15, 2014, 01:53:49 pm »

Quote from: tochukwu84 on November 15, 2014, 11:51:03 am

Pls I hav nt seen d solved question, am inside d hall now n i subscribed wit u pple

if u subscribe u would have received the answers

MANDINHO · « **Reply #12 on:** November 15, 2014, 01:58:34 pm »

(8ai)
area of one triangle = 1/2 r^2 sin tita
=1/2 *(10)^2* sin 45 degree
=100/2*(sqr root 2/2)
=50 sqr root 2/ 2 = (25 sqr root 2) m^2
area of the octagon =8*25 (sqr root 2) =(200 sqr root 2) m^2

(8aii)
perimeter of the octagon
=8* 27.86
=222.88 cm^2

(8b) ((3x+2)/4)-((2x+5)/3))=1/6
(3x(3x+2)-4x(2x+5))/12=1/6
(9x+6-8x-20)/12=1/6
9x-14)/12=1/6
x-14=2,x=2+14,
x= k=540/6
k=90 degree

benscheely · « **Reply #13 on:** November 15, 2014, 02:33:25 pm »

(9ai)
Lenght(L)= (p+4)m
breadth(B)=(p-1)m
diagonal(D)=(p+5)m
(p+5)^2=(p-1)^2+(p+4)^2
p^2 +10p +25= p^2 -2p +1+ p^2
+8p +16
2p^2 +6p +17 - pΛ2 -10p -25 =0
p^2-4p-8=0.
p=-(b (+-) b^2 -4ac)/2a
p=+4 (+-) sqr root ((-4^2)-4*1*-8)
/2*1
p=(4(+-) sqr root (16+32))/2
p=(4 (+-) sqr root (48))/2
p=( 4(+-)6.93)/2, p=(4-6.93)/2
p=10.93/2, p=-2.93/2
p=5.47, p=-1.47m
the value of p is 5.47m
(9aii)
area of the lawn
=l*b
=(p+4)*(p-1)
=(5.47+4)*(5.47-1)
=9,47*4.47
=42.33m^2
(9b)
Base of a pyramid is 4.5m by
2.5m
Height = 4m
volume = 1/3*area of
base*height
= 1/3*11.25*4
volume = 15m^3.
(9c)
Using SOHCAHTOA
Tan tita = 5/12
x^2=12^2-5^2
x^2=144-25
x^2=119
x=root 119
x= 10.9 = 11 apprx.
Cos tita = 11/12.

JEFFREY96 · « **Reply #14 on:** November 15, 2014, 03:30:19 pm »

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