Music: P-Square – Nobody Ugly -Download Here>>>>>


Nigeria’s most famous pair, P-Square, have completely raised the bar many notches higher with this instant classic they call “Nobody Ugly”.

The track is rendered over a joyful and addictive high-life instrumental and is accompanied with a video fully packed with beautiful bursts of colour and intense, rib-cracking humour. On this song, Peter and Paul positively address some serious social issues of self esteem and the daily damage caused by the alarming “fakeness” on social media.

The extremely beautiful lyrics preach of self acceptance, financial hope, reassurance and positivity while respectfully condemning the need to be unnecessarily extra just for vain endorsement.There’s no contesting the fact that this beautiful piece will sleep at the top of both local and international charts

Download Below

Stream Below

 

Expo: Mathematics (Essay & Objective) Waec 2017 Free Expo Answers See Here>>>>>


Maths (Essay & Objective) Waec 2017 Free Expo Answers
============================
Thursday, 20th April, 2017
General Mathematics 2 (Essay) 09.30am – 12.00pm
General Mathematics 1 (Objective) 2.00pm – 3.30pm
============================

MATHS OBJ:
1-10: CBBCCACBBC
11-20: CACBAAAADB
21-30: BBADBCCACB
31-40: BBCACBDADA
41-50: DDACCCDDCB

================================
*SECTION A ANS ALL QUESTIONS*
================================

1 a)
(y -1 ) log 4 ^ 10 = ylog 16 ^ 10
log 4 ^ 10 ( y -1 )= log 16 ^ y 10
4 ^ ( y -1 )= 16 y
4 ^ y -1 = 4 ^ 2 y
y- 1 = 2 y
-1 = 2 y= y
-1 = y
y= -y

1 b )
let the actual time for 5 km / hr be t
for 4 km /hr = 30 mint + t
4 km /hr =0 . 5 + t
distance = 4 (0 . 5 + t )
= 2 * 4 t
for 5 km /hr , time = t
distance =5 t
1 + 4 t = 5 t
t= 2 hrs
actual distance = 5 * 2 = 10 km

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

2a)
2/3 (3x – 5) – 3/5 (2x – 3) = 3
(15) x 2/3 (3x – 5) – 3/5 (15) (2x – 3) = 3(15)
10(3x -5) – 9(2x – 3) = 45
30x – 50 – 18x + 27 = 45.
12x – 23 = 45
12x = 45+23
12x = 68
X = 68/12
X = 5.67

2b)
80 = n+r ( ext. < equal sum of opp int. < s )
80 = n+r ———— (i)
M = 80 + 92 – n
M = 172 – n ———– (ii)
80 + 92 – n + 180 – m = 180°‎
80 + 92 + 180 – n – m = 180°‎
352 – n – m = 180
-n – m = 180 – 352‎
– n – m = – 172°‎
m + n = 172°‎

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

3a) ‎
Tan 23.6° = h/50
Cross multiply
Tan 23.6° x h/50
h = 50 tan 23.6°
= 21.844m‎
‎aprox. 22m

3b)
Area of A = 1/2bh
45 = 1/2 x 10 x h
45 = 5h
h = 9cm
Area of < QTUS = 1/2 ( QT + US)h
= 1/2 ( 6 + 16)9
= 99cm^2

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

4 a)
T 6 =37
T 6 =a + ( 6 – 1 )d
T 6 =a + 5 d
a + 5 d = 37 – — -( eq1 )
s 6 = 147
sn = n / 2 (2 a + (n -1 ) d )
147 = 3 (2 a + 5 d )
49 = 2 a + 5 d
2 a+ 5 d = 49 — — (eq 2 )
a + 5 d = 37 – –( eq1 )
2 a+ 5 d = 49 — -( eq2 )
a =12

4 b )
S 15 = 15 /2 ( 2 (12 )+ 14 d )
S 15 = 15 /2 ( 24 + 14 d )
from(1 )
a + 5 d = 37
12 + 5 d = 37
5 d =37 -12
5 d =25
d =5
S 15 = 15 /2 ( 24 + 14 (15 )
S 15 = 15 / 2 (24 + 70 )
S 15 = 15 /2 * 94
S 15 = 15 * 42
S 15 = 630

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

5a)
Let bag=B
Shoe= S
U=120
n(BnS)=45, n(s)=x+11, n(b)=x
n(SnB’)=x+11-45
=x-34
n(BnS’) = 45

5b)
Y – 45 + 45 + Y – 34 = 120
2Y – 34 = 120
2Y = 120 – 34
2Y = 154
Y = 154/2
Y = 77
11+x=77+11
= 88
Therefor 88 bought shoes costumer

5c)
n(bag)= 77 customers
Pr. =77/120

================================
*SECTION B ANS 5 QUESTIONS ONLY*
================================

8)
In Table Form / Tabular form

X = 1,2,3,4,5
F = m+2, m-1, 2m-3, m+5, 3m-4 = 8m – 1‎
Fx = m+2, 2m-2, 6m-9, 4m+20, 15m-20 = 28m – 9‎

But x̄ ( this symbol (x̄) means X bar)
= 75/23‎
ΣFx / Σf = 75/23 = 28m – 9/8m-1
75/23 = 28m – 9/8m – 1
Cross multiply
75(8m-1) = 23(28m-9)
600m – 75 = 644m – 207
-75 + 207 = 644m – 600m
132 = 44m
M = 3

8b)
In tabular form
X = 1,2,3,4,5
F = 5,2,3,8,5
Cum Freq= 5,7,10,18,23

Q1 = (N+1/4) = (23+1/4)
= 6

Q3 = (3N + 1/4) = (3*23+1/4)
= 18

Inter quarter range = Q3 – Q1
=. 18-6
= 12

8bii)
Pr. (at least 4 mark)
= 8+3+2+5/23
= 18/23‎

10)
Sin x = 5/13
Using pythagoras rule
M^2 = 13^2 – 5^2 (^ means Raise to power)
M^2 = 169 – 25
M ^2 = 144
M = √144
M = 12
Hence:
Cos x – 2sin x / 2tan x
12/13 – 2(5/13) / 2(5/12)
‎= 12/13 – 10/23 / 5/6
FIND LCM
= 12 – 10/13 / 5/6
= 12/65

10b)
Draw a triangle LACB
in triangle LCB
Hyp^2 = Opp^2 + Adj^2
12^2 = 9.6^2 + |CB|^2
144 = 92.16 = |CB|^2
144 – 92.16 = |CB|^2
51.84 = |CB|^2
therefore, |CB| = √51.84
|CB| = 7.2m
|AC| + |CB| =|AB|
|AC| + 7.2m = 10m
|AC| = 10m – 7.2m

|AC| = 2.8m
In triangle LCA
Hyp^2 = Opp^2 + Adj^2
|LA|^2 = |AC|^2 + |LC|^2

|LA|^2 = 2.8^2 + 9.6^2

|LA|^2 = 7.84 + 92.16

|LA|^2 =100

|LA| = √100

|LA| = 10m

10bii)
in triangle LCA
sinθ = Opp/Hyp
sinθ = |LC|/|LA|
sinθ = 9.6/10
sinθ = 0.96
θ = sin^-1 (0.96)
θ = 73.74

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

13ai)
given
x(*)y=x+y/2
i)3(*)2/5=3+2/5/2
=(15+2/5)*1/2
=17/5*1/2
=17/10= 1,7/10

13aii)
8(*)y=8^1/4
=8+y/2 =33/4
32+4y=66
4y=66-32
4y=34
y=34/4
y=17/2
y=8^1/2

13b)
given DABC
AB=(^-4/6) and AC =(3/^-8)
so AP =1/2(^-4/6)
AP=(^-2/3)
hence
CP = CA + AP
CP= -(3/^8)+(^-2/3)
CP = (^-5/11)