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* FRESHREALS * FRESHREALS (1a )
6whole 1 / 2 - 3 whole 2/ 5 / 2whole 1 / 2 - 1 whole 3 / 5
25/4 - 17/5 / 5/2 - 8/5
125-68 / 20 / 25-16 / 10
==> 57/20 / 9/10
=57/20 / 9/10
=57/20 x 10/9
=57/20 x 10/9
=57/18
= 3whole 1/6
=============================
(3a)
c= - 1, y= -3 , z=- 4 and w= -7
x^2 - y^2/2w-z
(-1)^3 - (-3)^2 / 2(-7) -(-4)
= -1-9 /+ 4+4 = -10/- 10
=1
(3b)
< MNQ = 90 degree
< MQN + < NQO = 90 degree
< MNQ + 46 = 90 degree
< MNQ = 90 - 46 degree
< MNQ = 44 degree
Considering < MNQ
y + n + MQN = 180 degree
y + 44 + 44 = 180
y = 180 - 88
Y = 92 degree
===============================
(4a)
2/ 3(1 -4 x) -1 /2 (5- 3x ) less /equal to 1/ 4(7 + 9x )- 1/ 3
multiply through by 12
8(1 -4 x) -6 (5- 3x ) less /equal to 3(7 + 9x )- 4
8- 32 x- 30 + 18 x less/ equal to 21 + 27 x -4
-32 x + 18 x - 27 x less /equal to 21 -4 + 30 -8
-41 x less /equal to 39
x less/ equal to -39 / 41
(4b)
DRAW THE ANGLE
from DTMR /
Tan 65 degree = TR/3
TR = 3Tan 65
= 3 x 22.1445
From DRMB
Tan 20 = RB / 3
RB = 3Tan 20
= 3 x 0.3640
= 1.092 m
H = TR + RB
= 6.4335 + 1 . 092
= 7.5255
= 7.53m
====================================
(5)
Area of shed segment = Area of Sector - Area of Triangle
=tita/360 *pie r^2 - (1/2 abSinc )
=90/ 360 * 22/7*7^2) - 1/2 * 7 *7*sin90degree
=154/4 - 42/2
=(754-98)/4
=56/4
=14cm^2
Cost of painting it = 14*750
=N10500
==============================
(6a )
Taxable income = x
25 /100 x x /1
= 14, 000
25 x= 1400000
X =1400000 / 25
X = 560000
Total income =
56 ,000 / 4 x 5
=N70,000
(6b)
Education = 2 /5
Clothes = 1 /6
Food = 3/ 8
Expenditure = 2 /5 + 1/ 6 + 3 /8
48 + 20 + 45 / 120
= 113 /120 * 36, 000 / 1
Le 39 , 000
Savings annually = Le 21 ,000
To save Le 63,000.00
To save Le 63,000 / 2100 yrs
= 30 years
================================
(8a)
Drawing
(8bi)
Using Pythagoras theory
| xz|^ =550^2 320^2
= 302500 102400 =404900
xz=√ 404999
=636 km( 3 s.f )
Total distance = 320 550 636
= 1506km
(8bii )
Using SOHCAHTOA
Tan z = 320 /550
Z = Tan ^-1(320 /550)
Z = Tan ^-1(0.5818)
Z = 30 degree
From the diagram in 8 a
Bearing of X from Z = 55 30
= 085 degree or N85 degree E
====================================
(7a)
diagram
(7b)
Angula difference in long(tita)=42-12
tita=30 degree
(7bi)
lenght of chord Xy=2Rsin tita/2
XY=2*6400*Sin 30/2
=2*6400*sin15 degree
=12800*0.2588
=3.312.64km
=3312.64km
=3310km(to the nearest 10km)
(7bii)
let the angle that the chord xy substends at the centre of
the earth be alpha degree
diagram
sin alpha/2= opp/hyp=/NY/ /6400
/NY/= 1/2 /XY/= 1/2 * 3312.64
=1656.32km
sin alpha/2= 1656.32/6400
sin alpha/2= 0.25888
alpha/2= sin^-1(0.2588)
alpha/2=14.999
alpha=14.999 *2
alpha= 29.998
alpha= 30.0 degree (to 1 dp)
(7biii)
XY bar= tita/360 * 2pie R cos lat
=30 degeree/360 * 2*3.142 * 6400* cos60
XY bar= 30/360 * 2* 3.142* 6400* 0.5
XY bar=1675.73 km
(9a)
3log 10^2 - 2log10^3= 1+ log 10(1/x)
log 10^6 - log10^6= log 10 + log 10(1/x)
log 10^6 / 10^6= log 10* 10(1/x)
1=10^1+1/x
1+10(x+1 /x)
x+1 /x =0
x+1=0
x=1
(9b)
h=8m
speed =3km/hr
time= 3mins=3/60 =1/20hr
(i)
Distance= 2 pie r= speed * time
2* 22/7 *r= 3* 1/20
44r/7 = 3/20
44r= 7*3 /20
r=7*3/ 44*20
r=0.02386km
r=23.86km
r=24m(nearest whole n)
(ii)
vol= pie r^2 h
=22/7 * 24^2* 8
=14,482.29m^3
vol= 14,482m^2(nearest whole number)
2a)
m/m-y+2 = r/y+r-1
r(m-y+2)=m(y+r-1)
rm-ry+2r = my+mr-m
rm+2r-m+mr=my+ry
y(m+r)=rm+2r-m+mr
y=2mr+2r-m/m+r
2b)
p[one of them owns a bicycle]
[60/100+70/100] – [40/100 + 30/100]
Because % of boys that don’t have bicycle = 40/100
% of girls without bicycle = 30/100
=130/100 – 70/100
=60/100
=0.6
1b)
let the number of students be X therefore total age of
students be 15x
15x + 45/x + 1 = 18
15x + 45 = 18 (x + 1)
15x + 45 = 18x + 18
18x – 15x = 45 – 18
3x = 27
X = 27/3
X = 9 students
9b)
Distance= speed * time
Distance= 3km/h * 3mins
= 3000m/60min * 3min
=50*3 =150min.
9bi)
Circumference= 150m
2π(r+1) = 150
22/1 * 22/7 * (r+1)= 150
r+1= 150*7/44
=1050/44
r+1 = 23.9 ≈ 24 to the nearest whole number.
r+1= 24
r=24-1 = 23min
9bii)
Volume of cylinder=πr²h
Volume= 22/7 * (23)² * 8
22/7 * 529/1 * 8/1
= 93104/7 = 13,300.57
13301m³ to the nearest whole number
=========
10a)
GIVEN: y = 2X^2- 3X – 1
When x = -3
Y= 2(-3)^2- 3(-3) – 1 = 18 + 9 – 1 = 26
When x= -1
Y = 2(-1)^2 – 3(-1)-1 = 2+3-1 = 4
When x = 1
Y = 2(-1)^2 – 3(1)-1 = 2 – 3 – 1 = -2
When x = 2
Y=2(2)^2-3(2)-1= 8 – 6 – 1 =1
When x = 3
Y =2(3)3-3(3)-1 = 18-9-1 = 8
When x = 4
Y = 2(4)^3 – 3(4)-1 = 32-12-1 =19
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