Mathematics lesson note for SS2 First Term is now available for free. The State and Federal Ministry of Education has recommended unified lesson notes for all secondary schools in Nigeria, in other words, all private secondary schools in Nigeria must operate with the same lesson notes based on the scheme of work for Mathematics.
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For prospective school owners, teachers, and assistant teachers, Mathematics lesson note is defined as a guideline that defines the contents and structure of Mathematics as a subject offered at SS level. The lesson note for Mathematics for SS stage maps out in clear terms, how the topics and subtopics for a particular subject, group works and practical, discussions and assessment strategies, tests, and homework ought to be structured in order to fit in perfectly, the approved academic activities for the session.
To further emphasize the importance of this document, the curriculum for Mathematics spells out the complete guide on all academic subjects in theory and practical. It is used to ensure that the learning purposes, aims, and objectives of the subject meant for that class are successfully achieved.
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This post is quite a lengthy one as it provides in full detail, the Mathematicsapproved lesson note for all topics and subtopics in Mathematics as a subject offered in SS2.
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SS2 Mathematics Lesson Note (First Term) 2024
FIRST TERM: ELEARNING NOTES
SS 2 MATHEMATICS
SCHEME FIRST TERM
WEEK  TOPIC  CONTENT 
1  LOGARITHM 1  (a) Revision of logarithm of numbers greater than 1. (b) Comparison of characteristics of logarithms and standard form of numbers. 
2  LOGARITHM 2  (a) Logarithm of numbers less than one, involving: Multiplication, Division, Powers and roots. (b) Solution of simple logarithmic equations. 
3 
SEQUENCE AND SERIES 1  (a) Meaning and types of sequence. (b) Example of an A. P. (c) Calculation of: (i) first term (ii) common difference (iii) nth term (iv) Arithmetic mean (v) sum of an A. P. (d) Practical problems involving real life situations. 
4 
SEQUENCE AND SERIES 2  (a) Examples of geometric progression. (b) Calculation of; (i) First term (ii) Common ratio (iii) nth term, (iv) Geometric menu (v) sum of terms of geometric progression. (vi) Sum to infinity. (c) Practical problems involving real life situation. 
5 
QUADRATIC EQUATION  (a) Revision of factorization of perfect squares. (b) Making quadratic expression perfect squares by adding a constant K. (c) Solution of quadratic equation by the method of completing the square. (d) Deducing the quadratic formula from completing the square. (e) Construction of quadratic equation from sum and product of roots. (f) Word problems leading to quadratic equations. 
6 
SIMULTANEOUS LINEAR AND QUADRATIC EQUATIONS  (a) Simultaneous linear equations (Revision). (b) Solution to linear and quadratic equations. (c) Graphical solution of linear and quadratic equations. (d) Word problems leading to simultaneous equations (capital market). (e) Gradient of curve. 
7  MIDTERM BREAK  
8 
*COORDINATES GEOMETRY OF STRAIGHT LINES  (a) Distance between two points. (b) Midpoint of line joining two points. (c) Gradients and intercept of a straight line. (d) Determination of equation of a straight line. (e) Angle between two intersecting straight lines. (f) Application of linear graphs to real life situation. 
9 
APPROXIMATIONS  (a) Revision of approximation. (b) Accuracy of results using logarithm table and calculators. (c) Percentage error. (d) Application of approximation to everyday life. 
10  REVISION  
11  EXAMINATION 
WEEK 1:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: logarithm 1
Content:
 Comparison of characteristics of logarithms and standard form of numbers.
 Revision of logarithm numbers greater than 1.
Comparison of characteristics of logarithms and standard form of numbers
There is a relationship between the standard form and the logarithm of a number.
For instance, (a) 189.7 =
This shows that the logarithm of a number is the power to which the base 10 is raised. Hence, Logarithm of 189.7 = 2.2781, where 189.7 =
(b) 850.9 = 8.509 (standard form)
Log850.9 = 2.9299
The integer (characteristics) is the same with the power 10
CLASS ACTIVITY
Show how the characteristics following Logarithms are related to standard form
 82000
 9
 6895
 8
Revision of logarithm numbers greater than 1
Logarithm of numbers is the power to which 10 is raised to give that number. Logarithms used in calculations are normally expressed in base 10.
 Rules for the use of Logarithms
 Multiplication: find the logarithms of the numbers andadd them together
 Division: find the logarithm of each number. Then subtract the logarithm of the denominator from that of the numerator
 Powers: find the logarithm of the number and then multiply it by the power or the index
 Roots:find the logarithm of the number and then divide it by the root
Example1: Evaluate using logarithm tables
Numbers  Log 
19.28  1.2851 
2.987  0.4752 
195.8  2.2918 
11270  4.0521 
Example 2: Evaluate using logarithm tables
Numbers  Log  
173.8  2.2400  2.2400  
1.1673  2.3346  
Numerator  4.5746  4.5746  
0.4166  1.2498  1.2498  
3.3248  
3.3248  
45.96  1.6624 
CLASS ACTIVITY
Use logarithm tables to evaluate correct to 4 s.f.
PRACTICE EXERCISE
Use log tables to find the value of
(SSCE 1991)
ASSIGNMENT
Use log tables to find the value of
KEYWORDS: base, logarithm, integer, antilogarithm, mantissa, characteristics etc.
WEEK 2:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: logarithm 2
Content:
 Logarithm of numbers less than one, involving: Multiplication, Division, Powers and roots.
 Solution of simple logarithmic equations.
Logarithm of numbers less than one
Simple logarithm operations
To find the logarithms of numbers less than 1, (i.e. numbers between 0 and 1), we use negative powers of 10.
For example, 0.08356 = 8.356 (standard form)
0.08356 = (from log tables)
=
So Log0.08356 = 2+0.9220
Characteristics (i.e. power of 10) = 2
Mantissa = 0.9220
Note: 2 is called bar 2 i.e.
Example 1: Work out the following giving the answers in bar notation
SOLUTION
(a)
(b)
⇒
Example 2: Work out the following giving the answers in bar notation
SOLUTION
CLASS ACTIVITY
Work out the following in bar notation form
 ii. iii. iv. v.
 + 5.6 – – –
(b)
Logarithm of numbers less than one, involving: Multiplication, Division, Powers and roots.
Example 1: a. Evaluate, using logarithm tables 0.9807 x 0.007692
Solution: 0.9807 x 0.007692
Number  Log 
0.9807  
0.007692  
0.007543 
antilog of = 0.00754 to 3s.f
Evaluate the following using logarithm tables
 00889 204.6
Numbers  Log 
0.00889  
204.6
 2.3109 
0.00004345 
antilog of = 0.0000435 to 3s.f
Note: In Logarithm, powers take multiplication while roots take division.
Example 2: a. Evaluate
Numbers  Log 
0.00001188 
Solution:
Numbers  Log 
0.3355 
Numbers  Log  
Numerator  
Denominator  
0.000315 
0.000315
CLASS ACTIVITY
Evaluate the following using Logarithm tables
(1)i.
(2) i.
SOLUTION OF SIMPLE LOGARITHMIC EQUATIONS
In this lesson, the equations have to be solved first, then tables used to evaluate
If
Then
RULES OF LOGARITHM
 note that any logarithm to the same base is 1, that is, .
e.g
e.g , e.t.c
Example 1: Solve the logarithmic equations
Solution:
from the formula above, we have that
this implies that because the powers will cancel each other
EXAMPLE 2: Solve the logarithmic equations
CLASS ACTIVITY
 Solve for
 Evaluate the following
PRACTICE EXERCISE
 Use logarithm tables to evaluate
 Evaluate using logarithm tables, correct to 3 significant figures
 Evaluate using tables
(SSCE 1993)
 Evaluate using tables, leaving your answer in standard form
Where P =3.6 × 10^{3} and Q = 2.25 × 10^{6}
 Evaluate using logarithm table, correct to 1 decimal place
ASSIGNMENT
 Evaluate using logarithm table, leaving your answer in standard form
 Use logarithm tables to evaluate
(SSCE 1991)
 Use logarithm table to Evaluate, correct to 3 significant figures;
(SSCE 1994)
 Use logarithm tables to Evaluate, correct to 3 significant figures;
(SSCE 1997)
 Evaluate using logarithm tables
 Evaluate using logarithm tables
 Given that log_{10}p = 1, find the value of p.
 B. 3 C. 10 D. 100 E.1000
 Evaluate log_{5}0.04
 0.008 B. – 1.4 C. – 2 D. 1
KEYWORDS: base, logarithm, integer, antilogarithm, mantissa, characteristics etc.
WEEK 3:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: Sequence and Series 1
Content:
 Meaning and types of sequence.
 Example of an A. P.
 Calculation of: (i) first term (ii) common difference (iii) nth term (iv) Arithmetic mean (v) sum of an A. P.
 Practical problems involving real life situations.
Meaning and types of sequence
SEQUENCES:
A sequence is an ordered list of numbers whose subsequent values are formed based on a definite rule. The numbers in the sequence are called terms and these terms are normally separated from each other by commas.
Examples:
2, 4, 6, 8, 10,……
Rule: Addition of 2 for subsequent terms.
70, 66, 62, 58, 54,……
Rule: Subtraction of 4 for subsequent terms.
3, 6, 12, 24,……
Rule: Multiply each term by –2.
There are many types of sequences. We shall be considering the Arithmetic Progression and Geometric progression.
Finite and Infinite Sequences
A finite sequence is a sequence whose terms can be counted. i.e. it has an end. These types of sequences are usually terminated with a full stop. e.g. (i) 3,5,7,9,11,13. (ii) 7,10,13,16,19,21.
If however, the terms in the sequence have no end, the sequence is said to be infinite. These types of sequences are usually ended with three dots, showing that it is continuous. e.g. (i) 5,8,11,14,17,20… (ii) 35,33,31,29,27,…
ARITHMETIC PROGRESSION (AP) {LINEAR SEQUENCE}
If in a sequence of terms T_{1}, T_{2}, T_{3, }…T_{n1}, T_{n} the difference between any term and the one preceding it is constant, then the sequence is said to be in arithmetic progression (A.P) and the difference is known as the common difference, denoted by d.
\d = T_{n} – T_{n1}, where n = 1, 2, 3, 4, …
i.e d = T_{2} – T_{1} = T_{3} – T_{2} = T_{4} – T_{3} and so on.
Examples of A.P
(i) 1, 3, 5,7, 9, …
T_{n} – T_{n1} Þ 5 – 3 = 2
7 – 5 = 2
9 – 7 = 2
 d = 2
The difference is common, hence it is an A.P.
(ii) 2, 4, 8, 16, 32, …
T_{n} – T_{n1} Þ 4 – 2 = 2
8 – 4 = 4
16 – 8 = 8
32 – 16 = 16
The difference is NOT common; therefore it is not an A.P.
(iii) 70, 66, 62, 58, 54, …
T_{n} – T_{n1} Þ 66 – 70 = 4
62 – 66 = 4
58 – 62 = 4
\ d = 4
The difference is common; hence it is an A.P.
(iv) –2, 5, 8, 11, …
T_{n} – T_{n1} Þ (5) – (2) = 5 + 2 = 3
(8) – (5) = 8 + 5 = 3
(11) – (8) = 11 + 8 = 3.
The difference is common; hence it is an A.P.
CLASS ACTIVITY
Which of the following are arithmetic progressing sequence?
 4,6,8,10,…
 3,7,9,11,..
 1,6,11,16,21,26…
 100,96,92,88,84,…
 20,17,15,11,…
 45,42,39,36,…
THE nth TERM OF AN A.P
If the first term of an A.P is 3 and the common difference is 2. The terms of the sequence are formed as follows.
1^{st} term = 3
2nd term = 3+2 = 3 + (1)2
3rd term = 3+2+2 = 3 + (2)2
4th term = 3+2+2+2 = 3 + (3)2
5th term = 3+2+2+2+2 = 3 + (4)2
nth term = 3+2+2+2+ … = 3 + (n – 1)2
Hence, the nth term (T_{n}) of an A.P whose first term is “a” and the common difference is “d” is given as
Example 1:
a.Find the 21^{st} term of the A.P 3, 5, 7, 9, …
Solution
a = 3
d = 2
n = 21
T_{n} = a + (n – 1)d
T_{21} = 3 + (21 – 1)2
= 3 + 20 x 2
= 3 + 40
= 43.
b.Find the 27th term of the A.P
100, 96, 92, 88, …
Solution
a = 100
d = 4
n = 27
T_{n} = a + (n – 1)d
T_{27} = 100 + (27 – 1)(4)
= 100 + 26 x –4
= 100 – 104
= 4.
Example 2:
a.Find the value of n given that 77 is the nth term of an A.P 3½, 7, 10½, …
Solution:
a = 3½
d = 7 – 3½
\ d = 3½
T_{n }= 77
T_{n} = a + (n – 1)d
77 = 3½ + (n – 1)3½
77 = 3½ + 3½n – 3½
77 = 3½n
77 = ^{7}/_{2}n
7n = 77 x 2
n = 77 x 2
7
n = 11 x 2
\n = 22.
 What is the first term of an A.P whose 21^{st}term is 43 and the common difference is 2 ?
Solution:
T_{21} = 43
n = 21
d = 2
T_{n} = a + (n –1)d
43 = a + 20 x 2
43 = a + 40
a = 43 – 40
\a = 3
 Find the common difference of an A.P given that 43is the21^{st} term of the sequence and the first term is 3.
Solution:
a = 3
T_{21} = 43
n = 21
T_{n} = a + (n – 1)d
43 = 3 + (21 – 1) d
43 = 3 + 20d
43 –3 = 20d
20d = 40
d = 40
20
\ d = 2.
CLASS ACTIVITY
 Find the 31^{st}term of the sequence –7, 10, 13, 16, …
(2) What is the 26th term of the A.P 5, 10, 15, 20, …?
FURTHER Example 1:
The first three terms of an A.P are
x, 3x + 1, and (7x – 4). Find the
(i) Value of x
(ii) 10th term
Solution:
(i) Recall that given an A.P T_{1}, T_{2}, T_{3}
T_{2} – T_{1} = T_{3} – T_{2}
Hence for, x, (3x + 1), (7x – 4)
(3x + 1) – x = (7x – 4) – (3x + 1)
3x+1 – x = 7x – 4 – 3x – 1
2x + 1 = 4x – 5
1+ 5 = 4x – 2x
2x = 6
x = 6
2
\ x = 3.
The sequence x, (3x + 1), (7x – 4) is
= 3, (3×3 + 1), (7×3 – 4)
= 3, 10, 17.
(ii) a = 3
n = 10
d = 7
T_{n} = a + (n – 1)d
T_{10} = 3 + (10 – 1)7
= 3 + 9×7
= 3 + 63
= 66.
Example 2:
The 6th term of an A.P is –10 and the 9th term is –28.
Find the (i) Common difference
(ii) First term
(iii) 26th term of the sequence.
Solution:
(i) T_{6} = 10 T_{n} = a + (n – 1)d
n = 6 10 = a + (6 – 1)d
10 = a + 5d ———– (1)
T_{9} = 28 28 = a + (9 – 1)d
n = 9 28 = a + 8d ——— (2)
Solve equation (1) and (2) simultaneously.
Eqn. (1): 10 = a + 5d
Eqn. (2): 28 = a + 8d
18 = 3d
d = 18
3
\ d = 6.
(ii) Put d = 6 in equation (1)
10 = a + 5(6)
10 = a – 30
10+30 = a
\ a = 20.
(iii) To find the 26th term of the sequence.
a = 20
d = 6
n = 26
T_{n }= a + (n – 1)d
T_{26} = 20 + (26 – 1)(6)
= 20 + 25(6)
= 20 – 150
= 130.
CLASS ACTIVITY
 The 6th term of an AP is –10 and the 9th term is 18 less than the 6th term. Find the
(a) common difference (b) first term
(c) 26th term of the sequence.
(2) The 7th term of an AP is 17 and the 13th term is 12 more than the 7th term. Find the (i) common difference (ii) first term (iii) 21st term of the AP.
Arithmetic Series:
These are series formed from an arithmetic progression. e.g.
1 + 4 + 7 + 10 + …
In general, if S_{n} is the sum of n terms of an arithmetic series then
S_{n} = a + (a + d) + (a + 2d) + … + (l – d) +
l — (1)
Where l is the nth term, a is the first term and d is the common difference.
Rewriting the series above starting with the nth term, we have.
S_{n} = l +(l – d) + (l – 2d) +… + (a + d) + a —– (2)
Adding equation (1) and (2) we have
2S_{n} = (a + l) + (a + l) + … + (a + l) + (a + l) in n places
2S_{n} = n(a + l)
\S_{n} = ^{n}/_{2}(a + l)
But l is the nth term i.e a + (n – 1) d
S_{n} = ^{n}/_{2}{a +a + (n – 1)d}
\ S_{n} = ^{n}/_{2} {2a + (n – 1)d}
Example 1 :
Find the sum of the first 20 terms of the series 3+5+7+9+ …
Solution:
^{ }a = 3
d = 2
n = 20
S_{n} = ^{n}/_{2} {2a +(n – 1)d}
S_{20} = ^{20}/_{2} {2×3 +(20 – 1)2}
S_{20} = 10{6 + 19 x 2}
= 10{6 + 38}
= 10{44}
\ S_{20} = 440
Example 2:
Find the sum of the first 28 terms of the series –17 + (14) + (11) + (8) + …
Solution:
a = 17
d = 3
n = 28
S_{n} = ^{n}/_{2}{2a + (n – 1)d}
S_{28} = ^{28}/_{2} {2 (17) + (28 – 1) 3}
= 14 {34 + 27 x 3}
= 14 { 34 + 81}
= 14 {47}
\ S_{28} = 658
CLASS ACTIVITY
(1) Find the sum of the numbers from 1 to 100.
(2) Find the sum of the first 26 terms of the A.P –18, 15, 12, 9, …
FURTHER Example :
The sum of the first 9 terms of an A.P is 117 and the sum of the next 4 terms is 104.
Find the(i) Common difference
(ii) First term
(iii)25th term of the A.P
(WAEC)
Solution:
T_{1}, T_{2}, T_{3}, T_{4} … T_{8}, T_{9}, T_{10}, T_{11}, T_{12}, T_{13}
117 104
S_{9} = 117 ————————–(*)
n = 9
since a sequence is normally summed from the first term
S_{9 + 4} = 117 + 104
\ S_{13} = 221 —————(**)
n = 13
S_{n} = ^{n}/_{2} {2a + (n – 1) d}
From (*) above;
117 = ^{9}/_{2} {2a + (9 – 1) d}
117 = ^{9}/_{2} x 2a + ^{9}/_{2} x 8d
117 = 9a + 36d
Divide through by 9 to have;
13 = a + 4d ———————(1)
From (**) above;
221 = ^{13}/_{2} {2a + (13 – 1) d}
221 = ^{13}/_{2} x 2a + ^{13}/_{2} x 12d
221 = 13a + 78d
Divide through by 13 to have;
17 = a + 6d ———————(2)
From equation (1) and (2)
Eqn. (1): 13 = a + 4d
Eqn. (2): 17 = a + 6d
4 = 2d
d = 4
2
\ d = 2.
(ii) From equation (1) we have
13 = a + 4 x 2
13 = a + 8
a = 13 – 8
\ a = 5
(iii)a = 5
d = 2
n = 25
T_{n} = a + (n – 1) d
T_{25} = 5 + (25 – 1) 2
= 5 + 24 x 2
= 5 + 48
T_{25} = 53
CLASS ACTIVITY
(1) The sum of the first 9 terms of an A.P is 171 and the sum of the next 5 terms is 235.Find the (a) Common difference
(b) First term
(c) Sequence
(2) The sum of the first 8 terms of an A.P is 172 and the sum of the next three terms is 15. Find the
(a) Common difference
(b) First term
(c) 21^{st} term of the A.P
PRACTICAL PROBLEMS INVOLVING REAL LIFE SITUATION
Example 1:
A clerk employed by a private establishment on an initial salary of N5000 per annum. If his annual increment in salary is N300. Find the total salary earned by the clerk in 20years.
Solution:
a = 5000
d = 300
n = 20
S_{n} = ^{n}/_{2} {2a + (n – 1) d}
S_{20} = ^{20}/_{2} {2 x 5000 + (20 – 1) 300}
= 10 {10000 + 19 x 300}
= 10 {10000 + 5700}
= 10 {15700}
\ S_{20} = 157000
\ The total amount earned in 20years is N157000.00
EXAMPLE 2
A sum of money is shared among nine people so that the first gets N75, the next N150, the next N225, and so on.
 How much money does the ninth person get?
 How much money is shared altogether?
Solution:
a=75 d=15075=75 n=9
T_{n}=a + (n1)d
T_{9}=75 + (91)75
=75+600
=N675
S_{n}
S_{9 }
=4.5 × 750
=N3375
CLASS ACTIVITY
 The value of a machine depreciates each year by 5% of its value at the beginning of that year. If its value when new on 1^{st}January 1980 was N10,250.00, what was its value in January 1989 when it was 9years old? Give your answer correct to three significant figures.
(WAEC) 1989
(2) The houses on one side of a particular street are assigned odd numbers, starting from 11. If the sum of the numbers is 551, how many houses are there? (SSCE 1999)
PRACTICE EXERCISE
(1) The 6th term of an A.P is 26 and the 11th term is 46. Find the
(i) Common difference
(ii) First term
(iii) 25th term of the A.P
 The 5th term of an A.P is 11 and the 9th term is 19. Find the
(i)
common difference
(ii) First term
(iii) 21^{st} term of the A.P
(3) The fourth term of an A.P is 37 and the 6th term is 12 more than the
fourth term. Find the first and seventh terms.
SSCE, June 1994, No 11a (WAEC)
(4) The first three terms of an A.P are (x+2) ,(2x5) and (4x+1). Find the
(i) Value of x
(ii) 7th term.
(5) The first three terms of an A.P are x, (2x5) and (x+6). Find the
(i) Value of x
(ii) 21^{st} term.
ASSIGNMENT
 If the first three terms of an A.P are (4x+1), (2x5) and (x+3). Find the
(i) Value of x
(ii) Sequence
(iii) 11th term of the sequence
 Given that 9, x, y, 24 are in A.P, find the values of x and y.
(10) If –5, a, b, 16 are in A.P, find the values of a and b.
 The 8th term of an arithmetic progression (A.P) is 5 times the third term while the 7th term is 9 times greater than the 4th term. Write the first five terms of the A.P. (SSCE 2009)
 If 3, x, y, 18 are the arithmetic progression (A.P). Find the values of x and y. (SSCE 2008)
 If are successive terms of an arithmetic progression (A.P), show that (SSCE 2007)
 The 3rd and 8th terms of an arithmetic progression (A.P)are 9 and 26 respectively. Find the:
 Common difference
 First term (SSCE 2007)
 The 2nd, 3rd and 4th terms of an A.P. are Calculate the value of
(SSCE 2006)
 An arithmetic progression (A.P) has 3 as its first term and 4 as the common difference.
 Write an expression, in its simplest form for the nth term.
 Find the least term of the A.P that is greater than 100 (SSCE 2003)
 The first term of an arithmetic progression (A.P)is 3 and the common difference is 4. Find the sum of the first 28 terms. (SSCE 2002)
 The first term of an arithmetic progression (A.P) is 8. The ratio of the 7th term to the 9th term is 5:8. Calculate the common difference of the progression. (SSCE 2000)
 The 6th term of an A.P. is 35 and the 13th term is 77. Find the 20th term. (SSCE 1997)
KEYWORDS: sequence, series, first term, common difference, last term, arithmetic progression, finite sequence etc.
WEEK 4:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: Sequence and Series II
Content:
 Examples of geometric progression
 Calculation of; (i) First term (ii) Common ratio (iii) nth term, (iv) Geometric progression (v) sum of terms of geometric progression. (vi) Sum to infinity
 Practical problems involving real life situation.
GEOMETRIC PROGRESSION (G.P) OR EXPONENTIAL SEQUENCE
Given any sequence of terms T_{1}, T_{2}, T_{3}, T_{4}, … T_{n1}, T_{n}. If the ratio between any term and the one preceding it is constant then the sequence is said to be in geometric progression (G.P). The ratio is called the common ratio denoted by r.
i.e.
r =
where n = 1, 2, 3, 4, …
Example1: find out which of the sequence is a GP
(i) 1, 2, 4, 8, 16, …
Þ = 2
1
= 2
= 2
The ratio is common, hence the sequence is a G.P \ r = 2
(ii) 16, 8, 4, 2, …
Þ =
=
=
The ratio is common, hence the sequence is a G.P \ r =
Example 2: find out which of the sequence is a GP
(i) 3, 7, 9, 12, …
Þ =
3
=
The ratio is NOT common, hence not a G.P.
(ii) 2, 10, +50, 250, …
T_{n} Þ = 5
T_{n1}
= 5
= 5
The ratio is common, hence the sequence is a G.P \ r = 5
CLASS ACTIVITY
Which of the following sequences is G .P.?
 3,6,12,24…
 2,4,6,8,…
 5,15,45,…
 1,4,16,…
THE NTH TERM OF A G.P
Let 5 be the first term of a G.P whose common ratio is 2. Then
The 2nd term is 5×2 = 5(2)^{1}
The 3rd term is 5x2x2 = 5(2)^{2}
The 4th term is 5x2x2x2 = 5(2)^{3}
The 5th term is 5x2x2x2x2 = 5(2)^{4}
The nth term is 5x2x2 … 2 = 5(2)^{n – 1}
In general, the nth term of a G.P denoted by T_{n}, whose first term is “a” and whose common ratio is “r” is ar^{n1}. i.e.
T_{n} = ar^{n – 1}^{ }where n = 1, 2, 3, 4, …
NOTE
The four examples below show how the formula can be used to find the nth term, n, r and a.
Example 1:
 Find the 8th term of the G.P
3, 6, 12, 24, …
Solution
a = 3
r = 2
n = 8
T_{n} = ar^{n – 1}
T_{21} = 3(2)^{8 – 1}
T_{21} = 3×2^{7}
T_{21} = 3×128
\ T_{21} = 384
 Find the value of n given that the nth term of a G.P is 2916 and the first term and common ratio are 4and 3
Solution:
T_{n} = 2916
r = 3
a = 4
T_{n} = ar^{n – 1}
2916 = 4(3)^{ n – 1}
2916 = 3^{n – 1}
4
729 = 3^{n – 1}
3^{6} = 3^{n – 1}
n1 = 6
n = 6+1
\ n = 7.
EXAMPLE 2:
 Find the common ratio of an exponential sequence whose 10th term is –512and the first term is
Solution
a = 1
T_{10} = 512
n = 10
T_{n} = ar^{n – 1}
512 = 1(r)^{10 – 1}
512 = r^{9}
(512)^{1/9} = r
r = (2^{9})^{1/9}
\ r = 2
 Find the first term of an exponential sequencewhose 7th term is 4096 and the common ratio is
Solution
T_{7} = 4096
n = 7
r = 4
T_{n} = ar^{n – 1}
4096 = a(4)^{71}
4096 = a4^{6}
4096 = 4096a
a = 4096
4096
\a = 1
CLASS ACTIVITY
(1) The 3rd and 6th term of a geometric progression (G.P) are 48 and 14^{2}/_{9 }respectively. Write down the first four terms of the G.P.
SSCE, June. 1993, No9 (WAEC)
(2) The first and third terms of a G.P are 5 and 80 respectively. What is the 4th term?
SSCE, Nov. 1993, No 11b (WAEC)
GEOMETRIC SERIES
The general expression for a geometric series is given as
S_{n} = a + ar + ar^{2} + ar^{3} + … + ar^{n1} —– (1)
Where S_{n} represents the sum of n terms of the series
Multiply both sides of equation (1) by r to have
rS_{n} = ar + ar^{2} + ar^{3} + ar^{4} + — + ar^{n} — (2)
Subtract (2) from (1) to have
S_{n} – rS_{n} = a – ar^{n}
S_{n} (1 – r) = a(1 – r^{n})
S_{n}= —————— (3)
If the numerator and denominator of equation (3) is multiplied by –1, we have
S_{n}= ——————– (4)
If r < 1, formula (3) is more convenient
If r >1, formula (4) is more convenient
Example 1:
Find the sum of the first 8 terms of the G.P 3, 6, 12, 24, …
Solution:
a = 3
n = 8
r = 2
S_{n} = a(r^{n} – 1) r > 1
r – 1
S_{8} = 3(2^{8} – 1)
2 – 1
= 3(256 – 1)
1
= 3 (255)
S_{8} = 765.
Example 2:
Find the sum of the first 10 terms of the G.P 2, 6, 18, 54, …
Solution:
a = 2
n = 10
r = 3
S_{n}= r < 1
S_{10} = 2(1(3)^{10})
1(3)
= 2(159049)
1+3
= 2(59048)
4
= 59048
2
S_{10} = 29524
CLASS ACTIVITY
(1) Find the sum of the first 9 terms of the sequence 84, 42, 21, 10½, …
(2) Find the sum of the first 10 terms of the G.P 4, 8, 16, 32, …
SUM TO INFINITY
In general, if the common ratio, r, is a fraction such that 1<x<1, the value of r^{n} approaches zero as n increases towards infinity. It then follows that the sum
S= becomes
This formula gives the sum to infinity of a geometric progression.
EXAMPLE 1:
Find the sum to infinity of the series S_{n}=36+24+6+4+….
Solution
a=36 r=
EXAMPLE 2:
The sum to infinity of a GP series is 15/7 and its second term is 6/5. Find the common ratio.
Solution
Now, ar=6/5
a=6/5r
15 (1r)=7 (6/5r)
15 15r =42/5r
5r(1515r)=42
75r75r^{2}= 42
25r – 25r^{2} = 14
25r^{2}25r14=0
25r^{2}35r+10r14=0
5r(5r7)+2(5r7)=0
(5r+2)(5r7)=0
5r+2=0 or 5r7=0
r=2/5 or 7/5
Note that r=7/5 is not a true value since (7/5)^{n} will not approach zero(0) as n increases towards infinity. Hence the valid common ratio of the series is 2/5.
CLASS ACTIVITY
 Find the sum to infinity of S_{n}=200+120+72+43 +……..
 the first term of a GP series is 4/9 and the sum to infinity is . find the common ratio.
Practical problems involving real life situation
Example 1:
 A ball was dropped from a height 80m above a concrete floor. It rebounded to the height of ½ of its previous height at each rebound. After how many bounces is the ball 2.5m high? NECO 2001
Solution
The rebounds forms a sequence of the order
80, 40, 20, …,2.5.
a = 80
r = ½
T_{n} = 2.5 T_{n }= ar^{n1 }
2.5 = 80(½)^{n1}
2.5 = (½)^{n1}
80
= (½)^{n1}
5 = (½)^{n1}
2×80
^{1}/_{32 }= (½)^{n1}
^{ }(½)^{5 }= (½)^{n1}
^{ } n – 1 = 5
n = 5 + 1
\ n = 6
 The 3rd term of a G.P is 54 and the 5th term is 486. Find the
 Common ratio
 First term
 7th term of the G.P
Solution
(a) T_{3} = 54 T_{n} = ar^{n1}
n = 3 54 = ar^{31}
54 = ar^{2} ————— (1)
T_{5} = 486 486 = ar^{51}
n = 5 486 = ar^{4 } ————– (2)
Equation (2) ¸ equation (1)
486 = ar^{4}
54 ar^{2}
9 = r^{2}
r = ±Ö9
r = ±3
(b) Substitute in equation (1)
54 = a(±3)^{2}
54 = 9a
a = ^{54}/_{9}
\ a = 6
(c) a = 6
r = ±3
n = 7
T_{n} = ar^{n – 1}
T_{7} = 6(±3)^{7 – 1}
= 6×3^{6}
= 6×729
= 4373
Example 2:
 If 2, x, y, 54 are in G.P, find x and y.
Solution:
a = 2
n = 4
T_{4} = 54
T_{n} = ar^{n – 1}
54 = 2(r)^{4 1}
54 = 2r^{3}
^{54}/_{2} = r^{3}
r^{3} = 27
r^{ 3} = 3^{3}
\r = 3
a = 2
r = 3
T_{2} = x^{ }
n = 2
T_{n} = ar^{n – 1}
x = 2(3)^{2 – 1}
x = 2×3
\ x = 6
a = 2
r = 3
T_{3} = y
n = 3
y = 2(3)^{3 – 1}
y = 2(3)^{2}
y = 2×9
\ y = 18
ii.Given that x, (x2), (2x1) are in G.P. Find the
 Value(s) of x
 Sequences
 Possible value(s) of the 8th term of the G.P
Solution:
Since x, (x2), (2x1) are in G.P, the ratio between any of the terms and the one preceding must be common.
i.e (x2) = (2x1)
x (x2)
Crossmultiplying
(x – 2)(x – 2) = x(2x – 1)
x^{2} – 2x –2x + 4 = 2x^{2} – x
0 = 2x^{2} – x – x^{2} + 2x + 2x – 4
0 = x^{2 }+ 3x – 4
i.e x^{2 }+ 3x – 4 = 0 (Factorize the equation)
4x^{2}
x^{2 }+ 4x – x – 4 = 0
x(x + 4) – 1(x + 4) = 0
(x + 4)(x – 1) = 0
x + 4 = 0 or x – 1 = 0
x = 4 or x = 1
(b) Since x = 4 and 1, we shall have two sequences
For x = 4
x, (x2), (2x1)
= 4, (4 – 2), (2(4)1)
= 4, 6, 9
For x = 1
x, (x2), (2x1)
= 1, (12), (2×11)
= 1, 1, 1
(c) For x = 4 the sequence = 4, 6, 9
a = 4
n = 8
r = 3
2
T_{8} = 4(^{3}/_{2}) ^{8 – 1}
= 4(^{3}/_{2})^{7}
= 4 x 2187
128
\T_{8 } = – 2187
32
For x =1, the sequence = 1, 1, 1
a = 1
n = 8
r = 1
T_{8} = 1(1)^{8 – 1}
^{ }= 1(1) ^{7}
= 1
Class activity
 Three consecutive terms of a geometric progression are given as n2,n and n+3. Find the common ratio.
 The 3^{rd}and 6^{th} terms of a geometric progression are 48 and 14 Write down the first four terms of the GP.
PRACTICE EXERCISE
(1) Find the sum of the first 7 terms of the G.P 3, 9, 27, 81, …
(2) Find the sum of the first 11 terms of the G.P 5, 10, 20, 40, …
(3) If the 3rd and 7th terms of a G.P are 12 and 192 respectively, find the sum of the first 6 terms of the sequence.
(4) The third and fifth terms of a geometric progression are ^{9}/_{2 }and ^{81}/_{8 } respectively.
Find the (i) Common ratio
(ii) First term.
(5) If 2, x, y, 250, … is a geometric progression, find x and y.
(JAMB)
ASSIGNMENT
 Given that 2, a, b, 686 are in G.P, find the value of a and b.
 The first three terms of a G.P are x+1, (x+4) and 2x. Find the
 Value (s) of x
(ii) Sequence(s)
(iii) 6th term of sequences
 What is the 25^{th}term of 5, 9, 13? The 3^{rd} and 6^{th} terms of a Geometric progression are 48 and 14 .Write down the first four terms of the G.P. (SSCE 1993)
 The third term of a Geometric Progression is 360 and the sixth term is 1215.Find the
(a) common ratio (b) first term (c) sum of the first four terms. (SSCE 1998)
 The 1^{st}and 3^{rd} terms of a geometric Progression are 2 and Find
(i) the common difference (ii) the 5^{th} term. (SSCE 1999)
 Write down the 15^{th}term of the sequence , , , …, (SSCE 2003)
 The sum of the second and third terms of a geometric progression is six times the fourth term. Find the two possible values of the common ratio.(ii) If the second term is 8 and the common ratio is positive, find the first six terms (SSCE 2008)
 The third term of a Geometric Progression (G.P) is 24 and its seventh term is 4.
Find its first term. (SSCE 2010)
 Given the Geometric Progression 6, 12, 24, 48, …
Find (i) its common ratio (ii) the 24^{th} and 40^{th} term of the G.P.
 The sum of the Geometric Progression 2 + 6 + 18 + 54 +…+ 1458 is 2186.
Find the nth term of the G.P.
 Find the sum of the G.P 3 + 9 + 27 + …. up to the 7^{th}
 Given the Geometric Progression (G.P) 4, 16, 64…
Find the 6^{th} and 7^{th} term of the G.P respectively.
 The sum to infinity of a geometric progression is 1/10 and the first term is 1/8. Find the common ratio of the progression. JAMB 2012
 Find the sum to infinity of the series ½ ,1/6, 1/18,…. JAMB 2004
KEYWORDS:
sequence, series, first term, common ratio,last term, geometric progression, infinite series etc.
WEEK 5:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: Quadratic Equations
Content:
 Revision of factorization of perfect squares
 Making quadratic expression perfect squares by adding a constant K.
 Solution of quadratic equation by the method of completing the square
 Deducing the quadratic formula from completing the square.
 Construction of quadratic equation from sum and product of roots.
 Word problems leading to quadratic equations.
Revision of factorization of perfect squares
A quadratic expression is a perfect square if it can be expressed as the product of two linear factors that are identical. For example,x^{2}+4x+4=(x+2)(x+2) ,x^{2}+6x+9=(x+3)(x+3) are perfect squares.
Example 1: Factorize
The expression is a perfect square if the first term and the constant terms are both perfect squares. The sign of the middle term of the quadratic expression can be put between the terms of the linear factors.
For , the first term is the square of and the constant term is 121 which is the square of 11.
Notice that the middle term is twice the product of the terms of the linear factor. In this case
Example 2: Factorize
The leading term (term containing the highest power of the variable) is a perfect square. Also the constant term, 400 is a perfect square.
So,
Example 3: Factorize
Solution;
Notice that the middle term of the quadratic expression is twice the product of the terms of the linear factors.
CLASS ACTIVITY
Factorize the following quadratic expressions
MAKING QUADRATIC EXPRESSION PERFECT SQUARES BY ADDING A CONSTANT K
In this section, we shall consider the constant K to be added to a quadratic expression to make it a perfect square.
Example 1: What must be added to to make the expression a perfect square?
Solution:
Let the constant term be k, is a perfect square if
i.e.
equating the coefficients of m, we have
Equating the constant terms,
Therefore 16 must be added to the expression to make it a perfect square.
In general, the quantity to be added is the square of half of the coefficient of m (or whatever letter is involved)
Example 2: What must be added to to make it a perfect square?
Solution:
What is to be added is the square of half of the coefficient of w
The coefficient of w is
Half of the coefficient is
The square of half of the coefficient is , so must be added to to make it a perfect square. i.e
CLASS ACTIVITY
What must be added to …
 X^{2}+4x
 M^{2}3m
SOLUTION OF QUADRATIC EQUATION BY THE METHOD OF COMPLETING THE SQUARE
In this section, we shall solve quadratic equations using method of completing the square
Example 1:
Solve the quadratic equation using completing the square method:
Solution
the L.H.S of the equation does not factorize. So, one can rearrange the equation to make the L.H.S a perfect square
We then find half of the coefficient of , square it and add it to both sides of the equation. The reason for that is to make the L.H.S a perfect square.
i.e
Take square root of both sides,
EXAMPLE 2:
Solve using completing the square method.
Solution
L.H.S of cannot be factorized,
Make the coefficient of one by dividing through by 3 to get
CLASS ACTIVITY
Solve the following using completing the square method
DEDUCING THE QUADRATIC FORMULA FROM COMPLETING THE SQUARE
The general form of a quadratic equation is .
We shall now derive the formula for solving the equation by method of completing the square
Given that,
Since we can divide through by to get
We shall now make the left hand side a perfect square by adding to both sides i.e half of the coefficient of all squared.
Take square root of both sides;
This is known as quadratic formula
Example 1: Use quadratic formula to solve,
We shall write the equation in the form
So,
Use the formula,
EXAMPLE 2: i. Solve using the quadratic formula,
Solution
Considering the coefficients of
 Use quadratic formula to solve;
Solution; first is to clear the fractions by multiplying each term of both sides of the equation by the LCM of the denominators which
On rearranging, we have;
CLASS ACTIVITY
Use quadratic formula to solve the following equations:
 i.
CONSTRUCTION OF QUADRATIC EQUATION FROM SUM AND PRODUCT OF ROOTS
We shall now consider how to construct quadratic equations from sum and product of roots. The root on an equation is the value that satisfies the equation.
Let be roots of a quadratic equation then
The general form of a quadratic equation is
Divide through by
Comparing equations (i) and (ii); equating coefficients of x, we have,
Equating their constant terms; we have
Consequently, a quadratic equation can be expressed or written as;
Example 1: Find the quadratic equation whose roots are
Solution
The equation is
Clear the fraction by multiplying through by the LCM of the denominator i.e. 6
⇒
EXAMPLE 2:
Construct the quadratic equation whose sum and product of roots are respectively .
Solution
⇒
CLASS ACTIVITY
 Find the sum and product of the roots of the following quadratic equations:
 Find the quadratic equation whose roots are;
WORD PROBLEMS LEADING TO QUADRATIC EQUATIONS
In this section, we shall consider some word problems leading to quadratic equation.
Example 1: If 63 is subtracted from twice the square of a number, the result is the same as five times of the number. Find the number
Solution
Let the number,
Twice the square of the number is
The equation required is
On rearranging, we have
This is then solved using any of the known methods (completing the square method or quadratic formula or factorization method)
Example 2: Find two consecutive odd numbers whose product is 399
Solution:
Let the smaller of the odd number be the other one will be (note that this approach also goes for consecutive even numbers)
Solving this equation, we obtain,
If then the other will be
If then the other will be
Therefore, the numbers are 19 & 21 or 19 & 21.
CLASS ACTIVITY
 A mother is 36 years old and her son is 6 years. When will the product of their ages be 451
 The hypotenuse of a right angled triangle is one unit more than twice the shortest side. The third side is one unit less than twice the shortest side. Find the; (a) shortest side (b) hypotenus
 A man is 3 times as old as his son, 8 years ago, the product of their ages was 112. Find their present ages.
PRACTICE EXERCISE
 What value of k makes the given expression a perfect square?^{2}
 2 4 C. 8 D. 16 E. 64(SSCE 1988)
 Factorize: ^{2}^{2}
 (5ya)(y+3a) (5y+a)(y3) C. (5y^{2}+a)(2y3a)
 (ya)(5y+3a) E. (y+a)(5y3a) (SSCE 1988)
 Factorize the following expression: 2x^{2}+x – 15.
 (2x+5)(x3) B.(2x5)(x3) C. (2x5)(x3) D.(2x3)(x+5)
 (2x+5)(x+3) (SSCE 1989)
 If 4x^{2}12x+c is a perfect square, find the value of c.
 36 B. 9 C. 9/4 D. 9/4 E. 24 (SSCE 1990)
 Factorize completely 4a^{3}a
 a(4a^{2}1) B. (2a1)(2a+1) C. a(2a1) D. 2a^{2}(a1) E. (2a+1)(2a1)
(SSCE 1990)
 Factorize 3a^{2}11a+6
 (3a2)(a – 3) B. (2a2)(a3) C. (3a – 2)(a – 3) D. (3a + 2)(a 3)
 (2a3)(a+2) (SSCE 1991)
 Factorize a^{2}– 3a – 10
 (a+5)(a+2) B. (a – 5)(a – 2) C. (a + 5)(a – 2) D.(a – 5)(a + 2)
 Factorize the expressions: ^{2}^{2}
 2()3^{2} B.(2)(2) C. ()()
 E.^{2}
 What must be added to the expression ^{2}18 to make it a perfect square?
 3 B. 9 C.36 D.72 E. 81 (SSCE 1992)
 Factorize the expression 2s^{2}3st2t^{2}
 (2st)(s+2t) B. (2s+t)(s2t) C. (s+t)(2st) D. (2s+t)(st)
 (2s+t)(s+2t) (SSCE 1993)
 Factorize ^{2}
 B. C.
 E. (SSCE 1993)
 Factorize ^{2}^{2}
 B. C.
 E. (SSCE 1994)
ASSIGNMENT
 Factorize ^{2} (SSCE 1990)
 Factorize ^{2} (SSCE 1990)
 Find the value of m which makes ^{2}a perfect square. (SSCE 1990)
 Given that ^{2}factor (SSCE 2000)
 Derive an equation whose coefficients are integers and which has roots of ^{1}/_{2} and 7.
 Three years ago a father was four times as old as his daughter is now. The product of their present ages is 430. Calculate present ages of the daughter and the father.(SSCE 1989)
 Solve the equation correct to two decimal places: ^{2}. (SSCE 1991)
 Using the method of completing the square, find the roots of the equation ^{2}
Correct to 1 decimal place. (SSCE 1995)
 The product of two consecutive positive odd numbers is 195. By constructing a quadratic equation and solving it, find the two numbers. (SSCE 1995)
 The area of a rectangular floor is ^{2}. One side is longer than the other. Calculate
(a) The dimensions of the floor
(b) If it costs per square metre to carpet the floor and only is
available, what area of the floor can be covered with the carpet? (SSCE 1998)
 A rectangular lawn of length metres is metres wide. If the diagonal is
metres, find;
(i) the value of
(ii) the area of lawn.
 The sum of the ages of a woman and her daughter is 46 years. In 4 years’ time, the ratio of their ages will be 7:2. Find their present ages. (SSCE 2001)
 The sides of a rectangular floor are The diagonal is .
Calculate in metres;
(a) the value of
(b) the area of the floor. (SSCE 2001)
 Solve correct to two decimal places the equation ^{2} (SSCE2008)
 The lengths, in cm of the sides of a right angledtriangle are where Find, in cm, the length of its hypotenuse.
KEYWORDS:
Quadratic equation, Factorize, perfect square, coefficient, roots, quadratic formula, etc.
WEEK 6:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: Simultaneous Linear and Quadratic Equation
Content:
 Simultaneous linear equations (Revision).
 Solution to linear and quadratic equations.
 Graphical solution of linear and quadratic equations.
 Word problems leading to simultaneous equations.
 Gradient of curve.
SIMULTANEOUS LINEAR EQUATIONS (REVISION)
Simultaneous linear equations (revision)
Recall: Simultaneous means happening or done at the same time i.e. following each other. It can be solved either graphically or algebraically. Algebraically involves using either substitution or elimination methods
Example 1: Solve the following pairs of simultaneous equation
Solution:
ELIMINATION METHOD
(a) ,
On adding both equations, we have
Substituting for in equation (1)
⇒
Using substitution method;
 ………….(1)
, …………(2)
In (2),
Substituting in equation (1), we have
In (3),
EXAMPLE 2: Solve the pairs of equations;
and
Solution;
Recall;
⇒
Comparing the powers of 3, we have,
Solving equations (i) and (ii) we obtain values for
i.e
CLASS ACTIVITY
 Solve the following pairs of simultaneous equations
,
 ,
 Solve the questions below; (WAEC)
,
SOLUTION OF LINEAR AND QUADRATIC EQUATIONS
When solving a simultaneous equation involving one linear and one quadratic such as (or ) graphical or substitution not elimination method is frequently used.
EXAMPLE 1: Solve the pair of equation; ………………. (i)
…………………(ii)
Solution
from equation (ii),
Substituting for in (i),
Collecting like terms and rearranging the equation; we have
Using any of the methods learnt previously for solving quadratic equations, we have that
When When
EXAMPLE 2: Solve completely;
Solution:
Note:
……….(ii)
Substituting for m+n in (ii),
Multiply equation (ii) by 7
Add (iii) & (iv)
Substitute for m in (ii),
CLASS ACTIVITY
 Solve: (a)
(b.)
 (a)A man is years old while his son is years old. The sum of their ages is equal to twice the difference of their ages. The product of their ages is 675. Write down the equations connecting their ages and solve the equations in order to find the ages of the man and his son. (WAEC)
(b)Solve:
GRAPHICAL SOLUTION OF LINEAR AND QUADRATIC EQUATIONS
Note: To use graphical method, plot the graphs of each equation on the same axes. Then read off the x & y coordinates of the points where both lines cross to obtain the required solution.
Example 1:
Solve the following simultaneous equations graphically:
Solution
Tables of values ()
3  2  1  0  1  2  
18  8  2  0  2  8  
3  2  1  0  1  2  
5  5  5  5  5  5  
10  1  4  5  2  5 
2  0  2  
5  3  1 
From the graph, x≃1.6 and y≃1.5 or x≃ 1.6 and y=5.6
EXAMPLE 2:
Draw the graphs of
 Use your graph to solve the following equations;
 Find the minimum value of
Solution
Tables of values for
2  1  0  1  2  3  4  
12  3  0  3  12  27  48  
10  5  0  5  10  15  20  
 8  8  8  8  8  8  8 
14  0  8  10  6  4  20 
 From the graph,
 For we have to make it look like the first graphical equation given. So we add 12 to both sides;
The line is drawn on the graph
 For this, we add to both sides ;
Then subtract 2 from both sides as well,
The line has tables of values as follows
2  0  2  
8  2  4 
From the graph above,
(b) The minimum value of is where the turning point of the curve is, i.e at and it occurs at
CLASS ACTIVITY
 Solve the following simultaneous equations graphically;
 (a) Copy and complete the following table of values for the relation;
2  1.5  1  0.5  0  0.5  1  1.5  
0  1.25 
.(b) Draw the graph of the relation using a scale of 2cm to 1 unit on each axis.
 Using the same axes, draw the graph of
 From your graphs determine the roots of the equation (WAEC)
WORD PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS
Note: Read the question carefully so as to understand what you need to find. Choose letters for the unknowns. Solve the equations formed from the information.
Example 1:
The diagram below is a rectangle. Find the perimeter and the area of the rectangle
Opposite sides of a rectangle are equal, so we equate each of them and we obtain a simultaneous equation;
With the previous knowledge on simultaneous equation, solving equations (i) & (ii)
We obtain
Perimeter = 2L + 2B or 2(L+B)
= 2(25) + 2(17)
= 50 + 34
= 84cm
Area = LB
= 25cm x 17cm
= 425
EXAMPLE 2:
A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the number. (WAEC)
Solution:
Let the digits be , where p is the tens digit and q is the unit part, so the number is 10p+q. when the digits are interchanged, the tens digit becomes q and the unit digit becomes p. hence the number is 10q+p
But ,
Divide through by 9
We were told that the sum of the digits 11, i.e
Adding (i) & (ii), 2q = 12
q = 6
substituting for q in (ii)
p + 6 = 11
p = 5
the number =
CLASS ACTIVITY
 The sum of two numbers is 110 and their difference is 20. Find the two numbers
 The perimeter of a rectangular lawn is 24m. if the area of the lawn is how wide is the lawn? (JAMB)
LINEAR GRAPH (REVISION)
Recall that any equation whose highest power of the unknown is one is a linear equation. The expression y = ax + b where a & b are constants, represents a general linear function. Its graph is a straight line. To plot the graph of a linear function, two distinct points are sufficient to draw the straight line graph.
Example 1:
Draw the graph of y = 3x – 2
Solution: A table of three points is prepared below
1  2  5  
5  4  13 
Scale: 1cm to 1 unit on xaxis
1cm to 5 units on yaxis
GRADIENT OF A STRAIGHT LINE X & Y INTERCEPTS
The yintercept of a line is the point () where the line intersects the yaxis. To find b, substitute for x in the equation of the line and solve for y.
The xintercept of a line is the point (a,) where the line intersects the xaxis. To find a, substitute for y in the equation of the line and solve for x.
Example 1:
Solve the equation using the intercept method
Solution: To find the xintercept, we let y be and solve for x,
4x – 2(0) = 8
4x = 8
X = 2, the intercept is (2,0)
To find the yintercept, we let x be 0 and solve for y
4(0) – 2y = 8
Y = 4, the intercept is (0,4)
The graph of the given equation should then be plotted.
Gradient of a straight line: The gradient of a nonvertical line is a number that measures the line’s steepness. The gradient of a straight line is the rate of change of y compared with x. For example, if the gradient is 3, then for any increase in x, y increases three times as much. We can calculate gradient by picking two points on the line and writing the ratio of vertical change (change in y) to the corresponding horizontal change (change in x).
The gradient formula: The gradient of a straight line is generally represented by m.
gradient of any line (m) =
=
=
Consider moving from point P to another point Q on a straight line PQ, as in the diagram below
Yaxis
Increase in y
Increase in x
Xaxis
The gradient of line PQ passing through points is;
Example 2:
Find the gradient of the lines joining the following pairs of points(3,2) , (4,4)
Solution: from the points
Gradient of AB =
=
= =
=
CLASS ACTIVITY
Find the gradients of the lines joining the following pairs of points
 (9,7) , (2,5)
 (4,4) , (1,5)
 (7,2) , (1,2)
GRADIENT OF A CURVE
The gradient of any point on a curve is the gradient of the tangent to the curve at that point. The tangent must be produced at equidistant to the point.
Note: there are two major types of quadratic curves from the general form
Where is +ve
i.e
where is –ve
i.e
There are two types of gradients namely: positive and negative gradients
Positive Gradient Negative Gradient
() ()
() ()
Note: The gradient of a straight line is the same at any point on the line, but the gradient of a curve changes from point to point.
Example 1:
Copy and complete the following tables of values for
1  0  1  2  3  4  5  6  
1  8  17 
Using a scale of 1cm to 5units on yaxis and 1cm to 1unit on xaxis, draw the graph of
Use the graph to find the gradient of the curve at
Solution
1  0  1  2  3  4  5  6  
2  0  2  8  18  32  50  72  
9  0  9  18  27  36  45  54  
1  1  1  1  1  1  1  1  
10  1  8  11  10  5  4  17 
(b) At point x = 3 traced to the curve, we draw an equidistant line as shown, then take the slope.
Slope = = = 2.78
Example 2: The table below is for the curve
4  3  2  1  0  1  2  3  
3  2  5  6  5  2  3  10 
 Find the value of the constant m
 Using a scale of 1cm to represent 1unit on the xaxis and 1cm to represent 2units on the yaxis, draw the graph of for
 Use the graph to obtain the gradient of the curve at point
Solution
 Using any two values of and we can solve for ‘m’, let us make use of (2, 3) i.e
⇒
(c.) Gradient = = = = =
CLASS ACTIVITY
 (a) Copy and complete the following table of values for the relation.
x  2  1  0  1  2  3  4  5 
y  19  3  9 
(b) Using 2cm to 1 unit on the xaxis and 2cm to 5 units on the yaxis,
draw the graph of for 2
 (a)from your graph, find the:
 minimum value of y
 gradient of the curve at x=1
(b) By drawing a suitable straight line, find the values of x for which
(SSCE 2004)
PRACTICE EXERCISE
 Find a two digit number such that three times the tens digit is 2 less than twice the units digit and twice the number is 20 greater than the number obtained by reversing the digits.(JAMB)
 Find the perimeter and the area of the equilateral triangle in the diagram below.
 Five years ago, a father was 3 times as old as his son, now their combined ages amount to 110 years, how old are they?
 Solve for (x,y) in the equations;
 Draw the graph of for values of x from 4 to 4. Using your graph, find
 The roots of (i) (ii) , giving your answer correct to one decimal place
 The least value of and the corresponding value of (WAEC)
ASSIGNMENT
OBJECTIVE QUESTIONS
The graph above represents the relation. Use it to answer questions 1 and 2.
 Find the values of for which
 1.55, 4.55 B. 1.55, 4.55 C. 1.55, 4.55 D. 1.55, 4.55 (SSCE 2011)
 What is the equation of line of symmetry of the graph?
 C. D. (SSCE 2011)
The following is a graph of a quadratic function. Use it to answer question 3 and 4.
 Find the coordinates of point
 (0, 4) (1, 4) C. (0, 4) D. (4, 0) (SSCE2009)
 Find the values of when
 1, 3 B. 1, 4 C. 2, 3 D.1, 6 (SSCE 2009)
Use the graph below to answer question below
 What is the equation of the curve?
 C. D. (SSCE 2006)
THEORY QUESTIONS
 Draw the graph of 25 and for.
Use the graph(s) to:
 find the roots of the equation 25 and
 determine the line of symmetry of the curve25.
 (a) Copy and complete the table of values for the relation for.
X  3  2  1  0  1  2  3 
Y  4  2  4 
(b)Using scales of 2cm to 1 unit on the xaxis and 2cm to 2 units on the yaxis.
Draw a graph of the relation
(c) From the graph find the:
 minimum value of y
 roots of the equation
iii. gradient of the curve at (SSCE 2010)
 (a) Copy and complete the table
.
X  4  3  2  1  0  1  2  3 
Y  22  2  1  6 
(b) Using a scale of 2cm of 1 unit on the xaxis and 2cm to 5 units on the yaxis,
draw the graph of
(c) Use your graph to find:
 the roots of the equation
 the values of x for which
iii. the equation of the line of symmetry of the curve. (SSCE 2005)
 (a) Copy and complete the following table of values for the relation.
x  2  1  0  1  2  3  4  5 
y  19  3  9 
(b) Using 2cm to 1 unit on the xaxis and 2cm to 5 units on the yaxis,
draw the graph of for 2
(c)from your graph, find the:
 minimum value of y
 gradient of the curve at x=1
(d) By drawing a suitable straight line, find the values of x for which
(SSCE 2004)
 The table shows the values of the relationfor 4
x  4  3  2  1  0  1  2  3 
y  13  11 
(a) Copy and complete the table
(b) Using 2cm to 1 unit on the xaxis and 2cm to 5 units on the yaxis,
draw the graph of
(c) Use your graph to find:
 the roots of the equation
 the value for which
iii. the gradient of the curve at x=1. (SSCE 2003)
 a. Draw the table of values for the relation for the interval
 Using a scale of 2cm to 1 on the xaxis and 2cm to 2 units on the yaxis, draw the graphs of:
 for
 Use your graph to find:
 The roots of the equation
 The gradient of at
 a. Draw the table of values for the relation y=for the interval
 Using a scale of 2cm to 1 on the xaxis and 2cm to 2 units on the yaxis, draw the graphs of:
 for
 Use your graph to find:
 The roots of the equation
 The gradient of at (SSCE 2001)
 a. Copy and complete and the following table of values for the relation .
x  3  2  1  0  1  2  3  4 
y  2  6  2  3  10 
 Draw the graph of the relation ; using a scale of 2cm to 1 unit on the xaxis,
and 2cm to 2 units on the yaxis
 Using the same axes, draw the graph of
 Obtain in the form where a, b and c are integers, the equation which is
satisfied by the xcoordinate of the points of intersection of the two graphs.
 from your graphs, determine the roots of the equation obtained in d. above.(SSCE 2000)
 (a) Copy and complete the values for the relation for
x  3  2  1  0  0.5  0  1  2 
y  28  6  5 
(b) Using a scale of 2cm to 1 unit on the xaxis and 2cm to 5 units on the yaxis, draw the:
 Graph of
 line y=3 on the same axis
(c) Use your graph to find the
 roots of the equation
 Maximum value of (SSCE 1998)
 a. Copy and complete the following table of values for
x  1  0  1  2  3  4  5  6 
y  1  8  11 
 Using a scale of 2cm to represent 1 unit on the x axis and 2cm to represent 5 units on
the yaxis, draw the graph of .
 Use your graph to find the:
 Roots of the equation , correct to one decimal place
 Gradient of the curve at x=3 (SSCE 1994)
KEYWORDS: gradient, line of symmetry, linear graph, quadratic graph, coordinator, xaxis, yaxis, table of values etc.
WEEK 7:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC:
MID TERM BREAK
WEEK 8:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: Coordinates geometry of straight lines
Content:
 Distance between two points.
 Midpoint of line joining two points.
 Gradients and intercept of a straight line.
 Determination of equation of a straight line.
 Angle between two intersecting straight lines.
 Application of linear graphs to real life situation.
DISTANCE BETWEEN TWO POINTS
Let be two distinct points. The distance between them can thus be calculated.
Applying Pythagoras theorem to right angled triangle in the graph above,
^{ }
Example 1:
Calculated the distance between the points, (4,1) and (3,2)
Solution:
The distance between the points (4,1) and (3, 2) is
EXAMPLE 2:
Find the length of the line segment with end points (2,8) and (6,5).
Solution
X_{1}=2
Y_{1}=8
X_{2}=6
Y_{2}=5
CLASS ACTIVITY
Find the distances between the given points:
MIDPOINT OF A LINE SEGMENT
In the Cartesian plane above, let be the midpoint of the line segment with the coordinate and .
As triangles are similar;
Since R is the midpoint,
:.
Similarly,
Hence, the coordinates of the midpoint of the line joining and are:
Example 1: Find the midpoint ‘R’ of the line segment where and .
Solution:
and
GRADIENT OF A STRAIGHT LINE
The gradient of a line is defined as the ratio, increase in in going from one point to another on a line.
Is the change in x as the variable x increases or decreases from x_{1} to x_{2} and is the change in y with respect to y_{1} and y_{2} .
The slope (gradient) m of a straight line L is defined as
If is the angle of inclination to the slope of L, then ; is called the angle of slope of the line.
Example 2: Find the slope and the angle of inclination of the L through points and
Solution:
The slope m of points P_{1} and P_{2} on L is
The slope of the points P1 and P3 on L is
Therefore, implies that the slope of the line L is 3.
Since
Therefore, the angle of inclination is 71.57
It can therefore be concluded from the example above that any given line has one and only one slope.
CLASS ACTIVITY:
 Find the angle between lines L_{1}, with slope 7 and L2 which passes through (2,1) and (5,3)
 Find the gradients of the lines joining the following pairs of points :
DETERMINATION OF EQUATION OF A STRAIGHT LINE
ONE POINT FORM OF A LINE
The equation of a line passing through any point(x_{1},y_{1}) and with gradient m is given by
LINE THROUGH TWO POINTS
The equation of a line joining the twopoints (x_{1},y_{1}) and (x_{2},y_{2}) is given by
EXAMPLE 1:
Find the equation of the straight line which passes through the following pairs of points: (3, 4) and (5,3)
SOLUTION
X_{1}=3 y_{1}= 4 x_{2}=5 y_{2}= 3
or 2yx+11=0
EXAMPLE 2:
Find the equation of the straight line with gradient 5 and passing through the points(3,5)
SOLUTION
X_{1}=3 and y_{1}=5 m=5
PERPENDICULAR AND PARALLEL LINES
Two lines are said to be perpendicular to each other if the product of their gradient is equal to 1. If m_{1} and_{ }m_{2 }are gradients of two perpendicular lines then m_{1}m_{2}=1.
Two lines are said to be parallel to each other if their gradients are equal. If m_{1 }and m_{2 }are gradients of two parallel lines, then m_{1}=m_{2}
EXAMPLE 1:
Find the equation of the straight line which is perpendicular to the line 5x2y=3 and passing through (3, 5).
SOLUTION
m_{1}= 5/2
m_{1} × m_{2} = 1
=
Using
EXAMPLE 2:
Find the equation of the straight line which is parallel to 4x5=12y and is passing through the point (4,7).
SOLUTION
4x5=12y
4x+y=17
Y=4x+17
M_{1}=4 and m_{2}= 4 since they are parallel
CLASS ACTIVITY
 Find the equation of the line which is perpendicular to the line y=4x, passing through the points(7,1)
 Find the equation of the line parallel to 3x+5y=1 and passing through (4,2)
ANGLE OF SLOPE
Y B
A C X
From the diagram above, line AB makes an angle with positive xaxis. is called the angle of slope of the line. The gradient of the line AB = = , therefore the gradient of the line is equal to tangent of the angle the line makes with the positive xaxis.
Example 1: Find the gradient of the line joining (2,4) and (1,3); find also the angle of slope of the line.
Solution: Let and
But, ;
⇒
Example 2: Find the gradient of the line joining (6,2) and (3, 2), also the angle of slope.
Solution:
Let and
Also,
or
CLASS ACTIVITY
 Find the gradient of the line joining Find the angle of slope of the line.
 Find the gradient of the line joining Find the angle of slope of the line.
ANGLE BETWEEN TWO LINES
The angle between the lines
Y=xtanα+c_{1 }and
Y=xtan c_{2}
is given by .
We can also use the formula
Tanθ=
where M_{2} and m_{1} are the gradients of the straight lines.
EXAMPLE 1:
Find the angle between the lines
Y=3x4 and 2xy+1=0
SOLUTION
y=3x4
tan
Y=2x+1
tan

=71.6^{0}63.5^{0}
=8.1^{0}
2^{ND} METHOD
Using tanθ=
tanθ=
tanθ=
tanθ= 0.1429
θ=
θ=8.1^{0}
EXAMPLE 2:
Find the acute angle between the following pair of lines
2x3y+5=0 and
4yx+2=0
SOLUTION
Tanθ=
2x3y+5=0
3y=2x+5
Y= x + ,m_{1}=2/3
4yx+2=0
4y=x2
Y= ,m_{2}= ¼
Tanθ= 
Tanθ=
Tanθ=
Tanθ= 0.3571
Θ≃ 19.7^{0}
CLASS ACTIVITY
 Find the acute angle between the pair of lines
3x+4y=2 and
2x+y=5
 Find the acute angle between the following pairs of lines:
Y=5x1 and y3x+2=0
APPLICATION OF LINEAR GRAPHS TO REAL LIFE SITUATION
EXAMPLE 1:
P(6,1) and Q(6,6) are the two ends of the diameter of a given circle. Calculate the radius
SOLUTION
X_{1}=6 Y_{1 }=1 X_{2}=6 Y_{2}=6
units
EXAMPLE 2:
SOLUTION
X_{1}=1 y_{1}=α x_{2}= y_{2}=1
9=
9 2 +
7+2(
7+2 ×2=
CLASS ACTIVITY
 What is the value of r if the distance between the points (4,2) and (1,r) is 3 units.
 Find the distance between the point Q(4,3) and the point common to the lines . JAMB
PRACTICE EXERCISE
 Find the acute angle between the lines 3x+2y=1 and 7x+4y=5.
 Find the coordinates of the midpoint of the x and y intercepts on the line 2y=4x8
 Find the equation of the line which is parallel to the line 5x+4y=18 and makes an intercept of 2 units on the xaxis. WAEC
 Find the equation of the line which passes through the point P(4,3) and is perpendicular to the line 2x+5y+1=0 WAEC
 Find the coordinates of the midpoint of the x and y intercepts on the line 2y=4x8 JAMB
ASSIGNMENT
 What is the value of p if the gradient of the line joining (1,P) and (p,4) is ? JAMB
 Find the value of p if the line that passes through (1,p) and (2p,2) is parallel to the line 8x+2y17=0 JAMB
 If the lines 3y=4x1 and qy=x+3 are parallel to each other, find the value of q.
 PQ and RS are two parallel lines. If the coordinates of P,Q,R and S are (1,q), (2,3),(3,4) and (5,2q) respectively, find the value of q. JAMB
 Find the midpoints of the line segments with the following end points:
 (3,0) and (4,4)
 (4,5) and (3,9)
 (1/4 ,0) and (1, ½ )
 Find the value of α if the lines 2yαx+4=0 is perpendicular to the line y+ ¼ x7=0 JAMB
 Find the point of intersection of the two lines 3x2y+5=0 and y4x+3=0
KEYWORDS:
Coordinates, parallel, perpendicular, intercept, gradient, equation, midpoint, intersect etc.
WEEK 9:
DATE……………………….
Subject: Mathematics
Class: SS 2
TOPIC: Approximations
Content:
 Revision of approximation.
 Accuracy of results using logarithm table and calculators.
 Percentage error.
 Application of approximation to every day life.
REVISION OF APPROXIMATION
An approximation helps us to see and understand easily the size of a number. It is a number taken as close as possible to the actual value of the number.
In order to come close to the actual value, the number must be rounded off. This implies digits 1 to 4 are rounded down while those from 5 to 9 are rounded up. Other methods could be approximating to decimal places or significant figures.
Decimal places mean the number of places after the decimal point while significant figure is the first nonzero digit from the left.
Example 1: The distance between the earth and the sun is 148729440km. Round this number to the nearest (a) million ( b) 2s.f (c) 3s.f
SOLUTION
 148729440km = 149000000km (i.e round 7 up)
 148729440km = 150000000km (i.e round 8 up)
 148729440km = 149000000km (i.e round 7 up)
EXAMPLE 2: (a) Write 78.45831kg to (i) 2d.p, (ii) 3d.p
SOLUTION
(i) 78.45831kg = 78.46kg (i.e round 8 up)
(ii) 78.45831kg = 78.458kg (round down 3)
(b) Round 0.0004996 to (i) 1s.f (ii) 2s.f
SOLUTION
(i) 0.0004996 = 0.0005 (round the first 9 up)
(ii) 0.0004996 = 0.00050 (round the second 9 up)
CLASS ACTIVITY
 Round off 586.5764 to (i) 1d.p (ii) 3d.p
 Approximate 964572183665 to (i) 3s.f (ii) 6s.f
ACCURACY OF RESULTS USING LOGARITHM TABLE AND CALCULATORS
All measurements are approximations, so they cannot be exact. Any stated measurement has been rounded off to some degree of accuracy.
Example 1:
A plot of land measuring 4532m by 431m.Calculate the area of the plot, using the calculator and the Logarithms table.
Solution: (i) 4532m x 431m = 1953292
(ii) 4532m x 431m
Number  Log 
4532  3.6563 
431  2.6345 
1954  6.2908 
Antilog of 6.2908 = 1954000
EXAMPLE 2: Calculate the difference due to using two different methods of calculation by evaluating
(a) 2321 x 4122
(b) 12204 x 2123
SOLUTION
(a) i. 2321 x 4122 = 9567162
 2321 x 4122
Number  Log 
2321  3.3657 
4122  3.6151 
9568  6.9808 
Antilog of 6.9808= 9568000
Difference = 9568000 – 9567000
= 838
.(b) (i) 12204 x 2123 = 25909092
(ii) 12204 x 2123
Numbers  Log 
12204  4.0864 
2123  3.3269 
2590  7.4133 
Antilog 0f 7.4133= 25900000
Difference = 25909092 – 25900000
= 9092
It can be seen that the result from the calculator is more accurate. The logarithm table results are higher because of premature approximation. The last example was different because the table deals with fourfigures only while one is 5figures
CLASS ACTIVITY
Solve this problems, using the calculator and the logarithm table. Calculate the difference from your answers and state why?
 153 x 5234
 12536 x 2413
 213564 x 1215
 1211015 x 123143
PERCENTAGE ERROR
Every measurement no matter how carefully carried out is an approximation, not exact. If the height of a wall is 25m, then the true or actual height is between 24.5m and 25.5m i.e (range difference). The error involved cannot exceed , the maximum absolute error.
Note: The maximum absolute error is an allowance within which the actual measurement falls.
Relative error is the ratio of the maximum absolute error (precision) and the true measurement value.
In measurements,
Percentage error =
If the true value is known,
Percentage error =
Example1:
A sales girl gave a balance of #1.15 to a customer instead of #1.25. Calculate her percentage error.
SOLUTION
Balance given = #1.15
Real or actual bal. = #1.25
Error = #1.25 – #1.15 = #0.10
Percentage error =
= 8
EXAMPLE 2:
A student who was asked to correct 0.02539 to two significant figures gave its value to two decimal places. His percentage error is ………..
SOLUTION:
The correct value in significant figure is 0.025
His answer is 0.03
% error = %
=
CLASS ACTIVITY
 A Man made a table with a rectangular top of dimension 36cm by 44cm instead of 37cm by 41cm. what is the percentage error in the perimeter of the table correct to 1 decimal place?
 A boy measures the length and breadth of a rectangular lawn as 59.6m amd 40.3m respectively instead of 60m and 40m. what is the percentage error in his calculation of the perimeter of the lawn?
APPLICATION OF APPROXIMATION TO EVERYDAY LIFE
This aspect deals with real life situations.
Example 1:
A car travels a distance of 100km for 1h 39mins. Calculate the speed of the car. If the time is rounded up to the nearest 1hr.Find the difference between the actual speed and when the time was rounded off.
SOLUTION
Speed = =
=
= 66.67km/hr
1hr 39mins to the nearest one hour is 2hrs, speed = = 50km/hr
Difference = 66.67km/hr – 50km/hr
= 16.67km/hr
EXAMPLE 2:
Find the sum of 34.25 (to 4s.f), 26 (to 3s.f) and 10 (to 2s.f) and leave your answer to a reasonable degree of accuracy.
Solution: the actual values
34.25 0.005
26 0.5
10 0.5
Sum = 70.25 1.105
the actual sum lies between 69.145 and 71.355
Hence, the sum of the values to a reasonable degree of accuracy is 69 to 2s.f
Note: Maximum error can be derived thus;
Whole numbers = = 0.5
1 d.p numbers = 0.05
2 d.p numbers = 0.005 etc
PRACTICE EXERCISE
 If it takes a proton to move in seconds, find the speed of the proton in metre per second correct to a suitable degree of accuracy
 Instead of writing as a decimal correct to 3s.f, a student wrote it correct to 3d.p, find his error in standard form.
 Find the value of each of the following and the degree of accuracy
 3 + 200.5 + 670.3 + 504.4
 56 – 9.062 – 4.147 – 10.20 – 5.108
 A string is 4.8m . a boy measured it to be 4.95m. Find the percentage error
 A sales boy gave a change of N68 instead of N72.calculate his percentage error.
ASSIGNMENT
 The length of a piece of stick is 1.75m. A girl measured it as 1.80m. Find the percentage error. NECO 2011
 A man estimated his transport fare a journey as N210.00 instead of N220.00. Find the percentage error in his estimate, correct to 3 s.f. NECO 2012
 A boy when rounding up a number wrote 98 instead of 980(to 2 significant figures).What is the percentage error? NECO 2007
 Approximate 0.0033780 to 3 significant figures
 Express 302.10495 correct to 5 s. f.
 Express the product of 0.007 and 0.057 to 2 s. f. NECO 2004
 Evaluate (0.13)^{3}correct to 3 s. f. NECO 2006
 A rectangular room has sides 5m by 4m, measured to the nearest metre.
 Write down the limits of accuracy for each length
 Find the greatest area the room could have
 Find the smallest perimeter the room could have
 Evaluate: , correct to 2 s. f. NECO 2008
KEYWORDS: error, percentage error, approximate, limits of accuracy, actual value,etc.
WEEK 10 REVISION
WEEK 11 EXAMINATION
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