SH MOBILE LOGO
Home | Waec | Neco | Jamb | Nabteb | G.c.e | Time Table | Education | Verifications | Chat | Payment | Others|

Fast & Easy Contact
examgoals: 08074125006 ||gidiclass: 08074125006

Exams QuickLink

CLICK HERE FOR OUR ANSWERS PAGE

« | »

Waec 2019 Further Maths Answer Now Available.

Posted by on May 6th, 2019


Message Us On WhatsApp (CLICK HERE)
Text Message Us SMS (CLICK HERE)

How to Get 2019 WAEC 2019 MAY/JUNE FURTHER MATHS Question Paper.

Follow the steps below to get your waec may/june further maths questions and answers 1 hours before your exam time.

*FURTHER MATHEMATICS*

_

FURTHER MATHS OBJ:
1-10: AACACBDABC
11-20: DAACCDBDAA
21-30: BCCBABABCC
31-40: BCCBCCDDDD

===

1(b)
F(x+2)-6x^2+5x-8 f(5
X+2=5
X=5-2
X=3
6x^2 +5x-8
=6(3)^2 + 5(3)-8
=54+15-8
=6!
Lord cubic
=====================
(2a)
Using y2 – y1/x2 – x1
Where y2 = 7, y1 = -5,
X2 = -2, and X1 = 7
7 – -5/-2 – 7
=7+5/-9
=12/-9
=4/-3
Coordinate points :
-4/3(3 : 2)
=-12/3 : -8/3
= -4 : -2⅔
X = (-4, 2⅔)
(2b)
2/1-√2 – 2/2+√2
=2(2+√2)-2(1-√2)/(1-√2)(2+√2)
=4+2√2 – 2+2√2/2+√2-2√2 – 2
=2 + 4√2/-√2
=(2+4√2)(-√2)/-√2(-√2)
= -2√2 – 4(2)/2
= -8 – 2√2/2
= -4 – √2
======================
(4)
Draw the right angled triangle
Using Pythagoras theorem
Third side = √(p+q)² – (p-q)²
=√(p+q+p-q)(p+q-p+q)
Difference of two squares.
=√(2p)(2q)
=√4pq
Adjacent side = 2√pq
Tanx = opp/adj = p – q/2√pq
1 – tan²X = 1-(p-q)²/4pq
=(4pq)-(p²-2pq+q²)/4pq
= -p²+6pq-q²/4pq
= -(p² – 6pq + q²)/4pq
=================
(5)
Draw the diagram
Using cosine law
Cos∅ = 16²+10²-14²/2(16)(10)
Cos∅ = 256 + 100 – 196/320
Cos∅ = 160/320
Cos∅ = 0.5
∅ = cos-¹(0.5)
∅ = 60°
Angle between 10N and 16N
= 180 – ∅ (sum of angles on a straight line)
= 180 – 60
=120°
====================
(6)
Draw the diagram
Taking moment about the pivot,
(T × 25)=(50×10)+(20×45)
25T = 500 + 900
25T = 1400
T = 1400/25
T = 56N
====================
(7)
In a tabular form
Under class interval:
1-5, 6-10, 11-15, 16-20, 21-25, 26-30
Under class mark (X):
3, 8, 13, 18, 23, 28
Under X-Xbar:
-10, -5, 0, 5, 10, 15
Under frequency:
18, 12, 25, 15, 20, 10
Ef = 100
Under f(X – XbarA):
-180, -60, 0, 75, 200, 150
f(X – XbarA) = 185
Where xA = 13
Mean = xA + Ef(X – Xbar)/Ef
=13 + 185/100
=13 + 1.85
=14.85years
=======================
9a
2x^2 ≥9x+5
2x^2+x-10x-5≥0
(2x^2+x)-(10x+5)≥0
x(2x+1)-5(2x+1)≥0
(2X+1)(x-5)≥0
2x+1≥0
2x≥-1
X≥-1/2
Or
X-5≥0
X≥5
Thus
X £ [5+∞] and x £ [-∞,-1/2]
Hence
X € [-∞,-1/2] v [5+∞]
=====================
11) u= 5ms-1 u= 6ms-1
A= 1ms-2 a= 3ms-2
p→ ←a
M—————————————N
51m
11a)t when particles are 30 meters apart
S= ut + ½ at2
Sp= st + ½ t2
Sa = 6t + 1/2 (3)t2
Where sp and sa are distances covered by particles respectively at
any time t
At time, t= 1.5s
Sp = 5(1.5) + ½(1.5)2 =8.625m
sa = 6(1.5) + ½ (3)(1.5)2 = 12.375m
sp + sa = 8.625 + 12.375
= 21m
Distance between particles = 51 – 21 = 30m
At time t = 1.5s, the particles will be 30 meters apart.
====

(10a) given 4x² – px +1 = 0
For real roots: b² – 4ac >0
(-p) ² – 4(4) (1) > 0
p² – 16 > 0
p² >16
p > ± 4

(10bi) Given: (1 +3x)⁶
Using pascal’s triangle: 1, 6, 15, 20, 15, 6, 1
(1)⁶(3x)º + 6(1)⁵ (3x)¹ + 15 (1)⁴ (3x)₂ + 20(1)³ (3x)³ + (15) (1)²(3x)⁴+6(1)¹(3x)⁵ + 1(1)º(3x)⁶
1 + 6(3x) + 15 (9x²) + 20 (27x³) + 15 (81x⁴) + 6(243 x⁵) 729x⁶
1 + 18x + 135x² + 540x³ + 1215x⁴ 1458x⁵ + 729x⁶

(ii) (1.03)⁶ = (1 + 3(0.01)]
Therefore (1.03)⁶ = 1 + 18(0.01) + 135 (0.01)² + 540(0.01)³ + 1215(0.01)⁴
+ 1458 (0.01)⁵ + 729 (0.01)⁶
+ 1 + 0.18 + 0.0135 + 0.005 + 0.00001215
+ 0.0000001458 + 0.000000000729
= 1.1940523
= 1.194 (4s.f)

MORE LOADING…..

KEEP REFRESHING

A serious Student would go vividly extreme miles to see his or her success because no one would be happy to say am going to re-write next year.

2019 WAEC FURTHER MATHS(ESSAY ; OBJ) MAY/JUNE Expo –

CLICK ANY OF THE LINK BELOW TO VIEW QUESTIONS AND ANSWERS::NECO 2019 All Questions & Answers Direct To Your Phone As SMS [CLICK HERE TO SUBSCRIBE]

GOOD LUCK IN YOUR EXAM!

Post Views: 467

RECOMMENDED POST FOR YOU

» NSUK Postgraduate Admission List, 2019/2020 Out » LASU Acceptance Fee Payment For 2019/2020 Session Has Commenced » FUGashua Supplementary Post-UTME 2019: Cut-off mark, Eligibility and Registration Details » UNN lecturer and Pastor reverses claim of #sexforgrade documentary being doctored after backlash » FUTO JUPEB Acceptance Fee Payment And Registration Procedure, 2019/2020

OR


CLICK TO DROP YOUR COMMENT

Share this post with your Friends on



No Comments Yet

Leave a Reply

Your Name:

Your Message:





Go Back To The Top


« | »


Looking for something? Search below






LIKE US ON FACEBOOK