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completed

MATHS ANSWERS

1a)
Tabulate
x – 1 ,2 ,3 ,4
1 – 1 ,2 ,3 ,4
2 – 2 _ , 4 , 0 _ ,2 _
3- 3 , 0_ , 3, 0 _
4 – 4 _ , 2_ , 0_ , 4

1 b )
I = PRT /100 , p = N15000 R = 10 % and I= 3 years
A = P + I
where I = 15000 * 10 * 3 / 100 = N4500
A = 4500+ 15000 = N19500
==================================

2 a)
No2

Using sine rule
b/sin120 = 6/sin30
bsin30 = 6sin120
b= 6sin120/sin30
b=6sin30/0.5
b= 12*sin120
b=12*sin120
b= 10.4cm

2 bi )
the diagram is equivalent t triangles .
where
| AX | / | BC | = | BY | / | AC | = | XY | /| YC |
XY = 9 , BY = 7
YC = 18 -7 = 11
9 /11 = 7 / | AC |
9 | AC | = 77
| AC | = 77 /9
| AC | = 8 cm

2 bii)
XY /AB = BY /AC
9 /| AB| = 7 / 8 .6
| AB | = 9 x 8 .6 /7
| AB | = 11 cm
==================================

3)
let the son age be x
man = 5 x
son= x
4 yrs ago; the man age = 5 x – 4
the son age = x – 4
the product of their ages
(5 x – 4 )( x – 4 ) = 448
==================================
4 a)
volume of fuel = cross -sectional area of X depth of fuel rectangular
tank
30, 000litres = 7 . 5* 4 .2 * d m ^ 3
but ; 1000litres = 1 m ^ 3
therefore ; 30( M ^3 ) = 7 .5 *4 . 2* d ( M ^ 3)
30= 31.5 d
==== d = 30/31. 5 = 0 .95( 2 d .p )
4 b)
to fill the tank /volume of fuel needed
= 7 .5 * 4. 2 *1 .2
= 37.8 m ^ 3
= 37, 800 litres
addition fuel = 37, 800- 30, 000
= 7 , 800 litres
therefore , 7 , 800 more litres would be needed

5 a)
sector for building project = 48000 /144000* 360 = 120degree
sector for education = 32, 000 /144000* 360 =80degree
sector for saving = 19200 /144000* 360 = 48degree
sector for maintenance = 12000 / 144000*360 = 30degree
sector for miscellaneous = 7200/144000* 360 =18degree
sector for food items = 360 -( 120 + 80+ 48+30+ 18)
= 360- 296
= 64degree
5 b)
amount spent =144000- [48, 000 + 32000+ 19200 + 12000 +7200]
= 144000-118400
= N 25600

6)
/————————— —
/ 41.02 × √0.7124
/ ———— ———— — —
√42.87 × 0.207 × 0.0404

| No | Log |
| 41.02 | 1.6130 | 1.6130
| 0.7124 | T. 8527 ÷ 2 | T.9263/1.5393
| 42.77 | 1.6321 |
| 0.207 | T.3160 |
| 0.0404 | 2. 6064 / |
| | T.5544 | T.5544/1.9849

Antilog = 9658 ≈ 96.58

=================================

7 a)
3^ 2 n + 1 – 4 ( 3 ^ n + 1 )+ 9 = 0
3^ 2 -3 – 4 ( 3^ n – 3) + 9 = 0
(3 ^ n )^2- 3 – 4 (3 ^ n -3 )+ 9 = 0
let 3^ n = p
p ^ 2 -3 – 4 (p -3 ) + 9 = 0
3p ^ 2 /3 – 12 p /3 + 9 / 3 = 0
p ^ 2 – 4 p + 3 = 0
p ^ 2 – 3 p – p + 3 = 0
p ^ 2 p ( p – 3 ) – 1 (p – 3) = 0
(p – 1 ) (p – 3) = 0
p -1 = 0 or p – 3 = 0
p = 1 or 3
Recall 3 ^ n = p
when p = 1
3^ n = 3^ 0
n = 0
when p = 3
3^ n = 3^ 1
n = 1

7 b )
log (x ^ 2 + 4 ) = 2 + logx – log ^ 20
log (x ^ 2 + 4 ) = log ^ 100 = log ^ x – log ^ 20
(x ^ 2 + 4 ) = log ( xx )
x ^ 2 + 4 = 5 x
x ^ 2 -5 x + 4 = 0
x ^ 2 -4 x – x + 4 = 0
x (x -4 ) – 1 ( x -4 ) = 0
(x -1 ) (x -4 ) = 0
x -1 = 0 or x -4 = 0
x = 1 or 4

8 a)
| AD | ^ 2 = 13^2 -5 ^ 2
| AD | ^ 2 = 169- 25
| AD | ^ 2 = 144
AD =sqr 144
AD =12CM
| AD | = 12- r
r ^2 = ( 12-r )^2 – 5 ^ 2
r ^2 = ( 12-r ) ( 12- r ) + 25
r ^2 = 144 -24r + 25
r ^2 = 169 -24r
r ^2 + 24r -169 = 0
r ^2 + 24r = 169
r ^2 + 24r + 14^2 = 169 + 14^2
( r + 14)^2= 169 + 196
( r + 14)^2= 365
( r + 14=sqr 365
r + 14=19. 105
r = 19.105 -14
r = 5. 105
r = 5. 1 cm
8 aii )
circumfrenece of a circle = 2 pie R
C = 2 x22/ 2* ( 5 .1 )^2
C = 1144.44/ 7
C = 163 .4914cm
C = 163 .5 cm
8 b)
y2 -y1 / x2 -x1 = y- y1 /x- x1
6 -2 /2 – ( -1 )=y- 2/ x- ( -1 )
4 /2 + 1 = y- 2 /x+ 1
4 /3 = y-2 / x+ 1
3 ( y-2 )=4 ( x+ 1 )
3 y-6 = 4 x+ 4
3 y-4 x= 4 + 6
3 y-4 x= 10
y= 4 x/3 + 10/ 3

9 a)
let Xy represent the two digit number
x-y =5 — — ( i )
3 xy – ( 10x + y)=14
3 xy – 10x – y = 14 – — -( ii )
from eqn ( i )
x= 5 +y
3 y( 5 +y ) -10( 5 + y) -y= 14
15y+ 3 y^2 – 50 – 10y – y =14
3 y^2 + 4 y – 50 = 14
3 y^2 + 4 y – 50 – 14 = 0
3 y^2 + 4 y – 64 = 0
3 y^2 + 12y + 16y – 64 = 0
( 3 y^2 – 12y ) ( + 16y – 64)=0
by
( y-4 ) + 16( y- 4)= 0
( y-4 )=0
ii )
( 3 y+ 16) ( y-4 )=0
3 y+ 16=0 or y- 4 =0
3 y= -16 or y= 4
y= -16/ 3 or y= 4
when y =4!
x= 5 +y
x= 5+y
x= 5 +4
x= 9
the no is 94
9 b)
3 -2 x/ 4 + 2 x-

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