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Grade 9Term 1Lesson Plans

SUBJECT: MATHEMATICS GRADE 9 WEEK 1, LESSON 1

TOPIC: Number systems and Financial Maths

GRADE 9 LEARNER NAME:DIAGNOSTIC TEST 1WHOLE NUMBERSMAX: 25TIME: 30min

QUESTION 11.1 Which numbers below belong to each group?

−2 13

9π √20 1,23 .. . −3 √16

1.1.1 Integers (1)

1.1.2 Rational numbers (2)

1.1.3 Irrational numbers (2)

1.2 List the following:1.2.1 The factors of 15 (3)

ATHEMATICS LESSON PLANS - GRADE 9 TERM 1 Page 1 of 413

1.2.2 The first four multiples of 6 (2)

1.2.3 The prime factors of 36 (1)

1.2.4 Find two rational numbers between -1,36 and -1,35 (2)

[13]QUESTION 2Find the selling price if:

2.1 Cost price is R720 with a profit of 40% (2)

2.2 Price marked at R85 with VAT included. (2)

2.3 Jamie buys 10 pens and sells them for R5 each. If the 10 pens cost him R52 calculate his profit or loss per pen.

(2)

[6]


QUESTION 33.1 A flight from Durban to Cape Town takes 1 hour if an aeroplane flies at

800km/h. At what speed will it fly if the same flight takes 2 hours?(2)

3.2 John and Jane are two brands of shoes. John sells for R7,20 for 8 pairs while Jane sells for R7,76 for 9 pairs. Which brand is the cheapest?

(4)

[6]TOTAL: 25


GRADE 9 DIAGNOSTIC TEST 1 – MemorandumNumber systems and Financial Maths MAX: 25TIME: 30min

QUESTION 1

1.1.1 −3 ; √16 (1)

1.12−2 1

3−3 √16 (2)

1.1.3 9 π √20 1 ,23. . . (2)

1.2

1.2.1 1 ; 3 ; 5 ; 15 (3)

1.2.2 6 ; 12; 18 ; 24 (2)

1.2.3 2 ; 3 (1)

1.2.4 −1 ,359 ; −1 ,358; −1 ,357 ; −1, 356 ; −1 ,355 ; −1 ,354 ; −1, 353 ; −1 ,352 ; −1 ,351 (2)

[13]


QUESTION 2

Find the selling price if

2.1 Cost price is R720 with a profit of 40% (2)

40100

×7201

=R 288Selling Pr ice =R 720+R 288=R 1066

2.2 Price marked at R85 with VAT included. (2)

14100

×851

=R 11 ,90Selling Pr ice =R 85+R 11 , 90=R 96 ,90

2.3 Jamie buys 10 pens and sells them for R5 each. If the 10 pens cost him R52 calculate his profit or loss per pen.

(2)

Cost Pr ice : R 52÷10=R 5 ,20R 5−R 5 , 20=−R 0 ,20He makes a loss of 20c .

[6]

QUESTION 33.1 A flight from Durban to Cape Town takes 1 hour if an aeroplane

flies at 800km/h. At what speed will it fly if the same flight takes 2 hours?

(2)


Speed=8002

km /h

=400 km /h

(2)

3.2 John and Jane are two brands of shoes. John sells for R7,20 for 8 pairs while Jane sells for R7,76 for 9 pairs. Which brand is the cheapest?

(4)

John=R 7 , 208

=90 c

Jane=R 7 ,769

=86 c

(4)

[6]TOTAL: 25



TOPIC: Number Patterns

GRADE 9 LEARNER NAME:DIAGNOSTIC TEST 2NUMBER PATTERNSMAX: 25 TIME: 30min

QUESTION 1

Describe the number patterns below in your own words.

1.1 3; 7; 11; 15… (2)

1.2 3; 6; 12; 24… (2)

1.3 0; 7; 26; 63… (2)

[6]


QUESTION 2

Term 1 2 3 4Value of the

term-1 2 7 14

2.1 Determine a rule for the number sequence above. (2)

2.2 Find Term 9 (T 9). (2)

2.3 Determine which term in the sequence has a value of 62. (2)

[6]

QUESTION 3

Study the following patterns formed with matches


3.1 Complete the table (2)

Pattern(n) 1 2 3 4 5Number of

matches

(2)

3.2 Describe the pattern in words (2)

3.3 Determine the nth term (general term) of the above number pattern.

(2)

3.4 How many matches would there be in the 50th diagram? (2)

[8]


QUESTION 44.1 Calculate the values of x, y and zin the table below:

Position in sequence

1 2 3 4 5 10

Term in sequence

5 7 9 x y z

(5)

[5]


GRADE 9 CLASS TEST 2 – Revision - MemorandumNUMBER PATTERNS MAX: 25TIME: 30min

QUESTION 1

1.1 Add 4 to the previous term (2)

1.2 Multiply the previous term by 2 (2)

1.3 The term position cubed minus one number added starting with 5; (2)

[6]

QUESTION 2

2.1 b) y=x2−2 (2)

2.2 y=x2

y=(9)2−2y=81−2y=79

(2)

2.3 y=x2

62=x2−262+2=x2

64=x2

x=8The 8th term.

(2)

[6]


QUESTION 3

3.1 Study the following patterns formed with matches (2)

Pattern(n) 1 2 3 4 5Number of

matches

3 6 9 12 15

3.2 The pattern number is multiplied by 3 (2)

3.3 T n=3 n (2)

3.4 T n=3 nT 50=3(50 )

=150

(2)

[8]

QUESTION 4

4.1 Calculate the values of x and y and zin the table below:

x=¿ 4 ×2+3=11 (2)

y=¿ 5×2+3=13 (2)

z=¿ 10 ×2+3=23 (1)



TOPIC: Algebraic Equations

GRADE 9 LEARNER NAME:DIAGNOSTIC TEST 3SOLVING ALGEBRAIC EQUATIONS MAX: 25TIME: 30min

QUESTION 1

Solve for the unknown variable in each of the following:

1.1 p( p+7 )=0 (2)

1.2 (c−3 )(c+2)=0 (2)

1.3 3 ( y−5 )=20 (2)

1.4 a2−15 a=0 (2)

1.5 b2−64=0 (2)


[10]

QUESTION 2

Solve the following equations:

2.1 m2+9 m−14=0 (2)

2.2 t ( t−3)=( t−2 )+2( t−3 ) (4)

2.3 a5= 3

35+ a+1

7(3)

[9]

QUESTION 3

3.1 Read the following and set up an equation for each one. Then solve for the variable, thus find the missing value.


3.1.1 James has a certain number of mangoes, and Sandy has 5 times as

many mangoes as James. They have a total of 36 mangoes.(2)

3.1.2 There are 72 people in a classroom. A certain number are males

and the other 21 are female.(2)

3.1.3 There are 6 boxes with a certain number of plates in each box. In

total there are 66 plates.(2)

[6]

TOTAL: 25


GRADE 9 DIAGNOSTIC TEST 3 - MemorandumSOLVING ALGEBRAIC EQUATIONS MAX: 30TIME: 30min

QUESTION 1

Solve for the unknown variable in each of the following:

1.1 p( p+7 )=0p=0 or p=−7

(2)

1.2 (c−3 )(c+2)=0c=3 or c=−2

(2)

1.3 3 ( y−5 )=203 y−15=20

3 y=20+153 y=35

y=353

=11235

(2)

1.4 a2−15 a=0a (a−15 )=0a=0 or a−15=0a=0 or a=15

(2)


1.5 b2−64=0(b+8)(b−8 )=0b=−8 or b=8

(2)

[10]

QUESTION 2

Solve the following equations:

2.1 m2+9 m+14=0(m+7 )(m−2)=0m+7=0 or m−2=0

m=−7 or m=2

(2)

2.2 t ( t−3)=( t−2 )+2( t−3 )

t2−3 t=t−2+2 t−6t2−3 t−t−2 t=−8t2−6 t+8=0( t−4 )( t−2 )=0t=4 or t=2

(4)

2.3 a5= 3

35+ a+1

7(3)


lcd :357 a=3+5( a+1)7 a=3+5 a+57 a−5a=82 a=8a=4

[9]

QUESTION 3

3.1.1 let James = x then Sandy = 5 x

∴ x+5 x=36

∴6 x=36

∴ 6 x6

=366

∴ x=6

(2)

3.1.2 72=m+21

∴72−21=m+21−21∴51=m

(2)


3.1.3 6 x=66

∴ 6 x6

=666

∴ x=11

(2)

[6]

TOTAL: 25



TOPIC: PROPERTIES OF NUMBERS

Remedial and Revision ExercisesMAX: 20 TIME: 30min

QUESTION 11.1

Complete the table by ticking the correct columns for each number

(10)

[10]


QUESTION 2Give the prime factors of these numbers (use either the ladder or

tree method).2.1 804 (2)

2.2 312 (2)

2.3 378 (2)

2.4 846 (2)

2.5 144 (2)

[10]TOTAL: 20


GRADE 9 DIAGNOSTIC TEST 3 -MEMORANDUMPROPERTIES OF NUMBERSMAX: 20TIME: 30min

QUESTION 1

1.1Complete the table by ticking the correct columns for each number

(10)

[10]


QUESTION 2

Give the prime factors of these numbers (use either the ladder or tree method).

2.1 (2)

2.2 (2)


2.3 (2)

2.4 (2)

2.5 144 2 72 2 36 2 18 2 9 3 3 3 1

(2)

[10]

TOTAL: 20



TOPIC: Integers, Decimals , Fractions Exponents and Surds

Remedial and Revision Integers, Decimals , Fractions Exponents and Surds MAX: 20 TIME: 30min

QUESTION 11.1 Simplify the following without using a calculator

1.1.1 −4× (8−(−3 ) )+7 (3)

1.1.2 34+5 2

7−28

36÷ 4 2

9(4)


1.1.3 0,03 ÷ 95

(2)

1.1.472+(−8)2−3√−64

27(2)

1.1.5 √ 925

+( 710 )

2

−( 15 )

3 (3)


1.1.6 −427

+8×(5+(−7 )) (2)

1.1.7 √0,25− (0,3 )3 (3)

1.1.8 20087

÷ 2 29−1 3

7× 0,42 (6)


1.1.9 0,54 ÷ 0,0009 (3)

[25]TOTAL: 25


GRADE 9 DIAGNOSTIC TEST - MEMORANDUMIntegers, Decimals , Fractions Exponents and Surds MAX: 25TIME: 30min

QUESTION 11.1 Simplify the following without using a calculator

1.1.1 −4× (8−(−3 ) )+7 (3)


1.1.2 34+5 2

7−28

36÷ 4 2

9(4)

1.1.3 0,03 ÷ 95

(2)


1.1.472+(−8)2−3√−64

27(2)

1.1.5 √ 925

+( 710 )

2

−( 15 )

3 (3)


1.1.6 −427

+8×(5+(−7 )) (2)

1.1.7 √0,25− (0,3 )3 (3)


1.1.8 20087

÷ 2 29−1 3

7× 0,42 (6)

1.1.9 0,54 ÷ 0,0009 (3)

[25]TOTAL: 25



TOPIC: Common Fractions , Decimal Fractions and Percentage

Remedial and Revision Common Fractions , Decimal Fractions and PercentageMAX: 20 TIME: 30min

QUESTION 11. Complete the table by filling in the missing values. Remember to give

your fractions in the simplest form.

(20)

[20]TOTAL: 20


GRADE 9 DIAGNOSTIC TEST 3 -MEMORANDUMCommon Fractions , Decimal Fractions and PercentageMAX: 20 TIME: 30min

QUESTION 11.1

Complete the table by filling in the missing values. Remember to give your

fractions in the simplest form.

(20)


[20]


TOPIC: Scientific Notations and Exponents

Revision and Remedial MAX: 20TIME: 30min

QUESTION 1

1.1 Write the following numbers in the correct scientific notation

1.1.1 0,00096 (2)

1.1.2 32 760 000 (2)

1.1.3 0,0000067 (2)

1.1.4 3 699 100 (2)

1.1.5 0,00004588 (2)


1.1.6 51 272 (2)

[12]

QUESTION 2

Simplify the following; leave all exponents in positive form.

2.1 (a2 b3 )2

a4 b3 ×( a2

b3 )−1 (3)

2.2 27a+9a

33a

(2)

2.3 8m2n3

(3mn2 )3÷ (4 m2 )2

9m5 n0

(4)


2.4 ( xy+ 1

x )−2 (4)

[13]

TOTAL: 25


GRADE 9 MEMORANDUMMAX: 25 TIME: 30min

QUESTION 1

1.1 Write the following numbers in the correct scientific notation

1.1.1 (2)

1.1.2 (2)

1.1.3 (2)

1.1.4 (2)

1.1.5 (2)

1.1.6 (2)

[12]

QUESTION 2

Simplify the following; leave all exponents in positive form.

2.1 (a2 b3 )2

a4 b3 ×( a2

b3 )−1 (3)


2.2 27a+9a

33a

(2)

2.3 8 m2n3

(3 mn2 )3÷ (4 m2 )2

9m5 n0

(4)

2.4 ( xy+ 1

x )−2 (4)


[13]

TOTAL: 25



TOPIC: NUMBER PATTERNS

GRADE 9 LEARNER NAME:Revision and Remedial WorkMAX: 26 TIME: 30min

QUESTION 1

Study the following patterns carefully, then for each pattern do the following:

1.1 1, 4, 9, 16….

1.1.1 Find the next 3 terms (1)

1.1.2 Write down in words how the pattern works. (1)

1.1.3 Find the 10th term of the pattern. (1)

1.1.4 Write an algebraic expression for the pattern. (1)


1.2 2, 8, 18, 32…





1.3 7; 11; 15; 19…






1.4 1; 2; 4; 8 …





[16]

QUESTION 2

Look at the following tables for each of the functions given below and complete the tables.

2.1 (2)

2.2 (2)


2.3 (2)

2.4 (2)

2.5

[10]

TOTAL: 26



TOPIC: NUMBER PATTERNS

GRADE 9 MEMORANDUMMAX: 26 TIME: 30min

QUESTION 1

Study the following patterns carefully, then for each pattern do the following:

1.1 1, 4, 9, 16….

1.1.1 Find the next 3 terms25, 36, 49

(1)

1.1.2 Write down in words how the pattern works.the position of the term squared

(1)

1.1.3 Find the 10th term of the pattern.100

(1)



1.2 2, 8, 18, 32…


(1)

1.2.2 Write down in words how the pattern works.the position of the term squared and then doubled.

(1)

1.2.3 Find the 10th term of the pattern.200

(1)


1.3 7; 11; 15; 19…


(1)

1.3.2 Write down in words how the pattern works.add 4 to the previous term

(1)

1.3.3 Find the 10th term of the pattern.4(10) +3 = 43

(1)



1.4 1; 2; 4; 8 …


(1)

1.4.2 Write down in words how the pattern works.double the previous term

(1)

1.4.3 Find the 10th term of the pattern.128, 256, 512

(1)


[16]

QUESTION 2

Look at the following tables for each of the functions given below and complete the tables.

2.1 (2)

2.2 (2)

2.3 (2)


2.4 (2)

2.5

[10]

TOTAL: 26


SUBJECT: MATHEMATICS GRADE 9 WEEK 2, LESSON 6 & 7

TOPIC: Algebraic Expressions

Revision and Remedial WorkMAX: 20 TIME: 30min

QUESTION 1If a=3 ;b=−5 ;c=10∧d=−2 find the value of the following

expressions:1.1 ( c

bd+a2)÷(−c) (2)

1.2 c2 d2−a2 b2+(cd−ab) (2)

1.3 abc+bcd−acd (2)

1.4 acb

− a2 dc−d2

(2)

1.5 7c−3 a2+5 b3−8d 4 (2)


[10]QUESTION 2Given the expression: 10 x−14 x5+2x3−8x2+11

2.1 How many terms are in the expression? (1)

2.2 What is the degree of the polynomial? (1)

2.3 What is the coefficient of the term with the highest exponent? (1)

2.4 What is the constant term? (1)

2.5 If x=−3 what is the value of the expression? (2)

[6]TOTAL: 16


GRADE 9 MEMORANDUMMAX: 16TIME: 30min

QUESTION 1

If a=3 ;b=−5 ;c=10∧d=−2 find the value of the following expressions:

1.1 ( cbd

+a2)÷(−c) (2)


1.2 c2 d2−a2 b2+(cd−ab) (2)

1.3 abc+bcd−acd (2)


1.4 acb

− a2 dc−d2

(2)

1.5 7c−3 a2+5 b3−8d 4 (2)

[10]


QUESTION 2

Given the expression: 10 x−14 x5+2x3−8x2+11

2.1 How many terms are in the expression?5 terms

(1)

2.2 What is the degree of the polynomial?5th degree

(1)

2.3 What is the coefficient of the term with the highest exponent?-14

(1)

2.4 What is the constant term?11

(1)

2.5 If x=−3 what is the value of the expression? (2)

[6]

TOTAL: 16



TOPIC: Exponents, Surds, Scientific Notations and Number Patterns

Revision and Remedial WorkMAX: 46 TIME: 30min

QUESTION 1

Without the use of a calculator find the answer for the following. Leave your answer in exponential form.

1.1 (2)

1.2 Find the HCF and LCM for the following three numbers: 868, 372 and 992.

(3)


1.3 Calculate the following without using a calculator (show all working out):

1.3.1 (3)

1.4 Write the following numbers in scientific notation:

1.4.1 1 007 996 550 (1)

1.4.2 6 302 520 (1)

1.5 Write the following scientific notation numbers as normal numbers:

1.5.1 (1)

1.6 Simplify the following


and leave with positive exponents:

1.6.1 (4)

1.7 Given the following pattern: 2; 12; 20; 30.

1.7.1 Determine the rule used to find the pattern. (2)

1.7.2 Find the value of the 9th item in the pattern. (2)

[19]

2.1 Given the following expression:


2.1.1 How many terms does the expression have? (1)

2.1.2 What is the value of the constant? (1)

2.1.3 What is the coefficient of x3 ? (2)

2.2 Simplify the following:

2.2.1 (3)

2.3 Solve for 𝑥:2.3.1 (2)

2.4 The perimeter of a rectangle is 20cm2. The one side of the rectangle is 𝑥cm and the other side is 2cm shorter than the first side. Find the lengths of the two sides.

(3)

[12]

QUESTION 3

3.1 Jamie, Anita and Thando are completing a project together. Jamie does 12% of the work, Anita does 0,3 parts of the work and Thando


does 14 of the work.

3.1.1 Who does the most work? (2)

3.1.2 How much work still needs to be done as a percentage? (3)

TOTAL: 46


GRADE 9 MEMORANDUMMAX: 46TIME: 30min

QUESTION 1

Without the use of a calculator find the answer for the following. Leave your answer in exponential form.

1.1 (2)

1.2 Find the HCF and LCM for the following three numbers: 868, 372 and 992.

(3)


1.3 Calculate the following without using a calculator (show all working out):

1.3.1 (3)

1.4 Write the following numbers in scientific notation:

1.4.1 (1)

1.4.2

(1)

1.5 Write the following scientific notation numbers as normal numbers:

1.5.1 (1)

1.6 Simplify the following and leave with positive exponents:

1.6.1 (4)


1.7 Given the following pattern: 2; 12; 20; 30.

1.7.1 Determine the rule used to find the pattern. (2)

1.7.2 Find the value of the 9th item in the pattern.90

(2)

[19]


2.1 Given the following expression:

2.1.1 How many terms does the expression have?4 terms

(1)

2.1.2 What is the value of the constant?-12

(1)

2.1.3 What is the coefficient of x3 ? (2)

2.2 Simplify the following:

2.2.1 (3)


2.3 Solve for 𝑥:2.3.1 (2)

2.4 The perimeter of a rectangle is 20cm2. The one side of the rectangle is 𝑥cm and the other side is 2cm shorter than the first side. Find the lengths of the two sides.

(3)

[12]

QUESTION 3

3.1 Jamie, Anita and Thando are completing a project together. Jamie does 12% of the work, Anita does 0,3 parts of the work and Thando

does 14 of the work.

3.1.1 Who does the most work? (2)


3.1.2 How much work still needs to be done as a percentage? (3)

TOTAL: 46



TOPIC: Whole numbers

CONCEPTS AND SKILLS TO BE ACHIEVED

By the end of the lesson, learners should know and be able todescribe the real number system by recognising, defining and distinguishing properties of:

natural numbers whole numbers integers rational numbers irrational numbers

RESOURCESDBE workbook(Book 1 : Page 2-5), Sasol-Inzalo workbook(Book 1: Page 1-6), textbook, ruler, pencil,

eraser, calculators, DVDs(Disk 1: Properties of number system)

PRIOR KNOWLEDGE1. Whole numbers2. Counting forward and backward

COMPONENTS TIME TASKS/ACTIVITIES CAPS

INTRODUCTIONMENTAL MATHS

3 min 1. List the types of number systems that you know, give examples if necessary

REVIEW AND CORRECTION OF HOMEWORK

0 minnone


LESSON PRESENTATION/ DEVELOPMENT

10 min Properties of numbers:Educator asks learners to give examples of each before the table below is

mediated with them or the table can be developed with the learner’s input.Natural Numbers(N):The most basic type of number.

N = { 1; 2; 3; 4; 5; …}

Even Numbers are a subset of Natural numbers or the numbers that are divisible by 2 or multiples of 2.

{2; 4; 6; 8; …}

Odd numbers are a subset of Natural numbers or any integer that cannot be divided exactly by 2.

{1; 3; 5; 7; 9; …}

Prime numbers are numbers greater than 1 which can only be divided by itself and 1, without a remainder.

{ 2; 3; 5; 7; 11; …)

Composite numbers can be divided by 1, itself and at least one other number without a remainder

{4; 6; 8; 9; 10; …}

1 is neither a prime number nor a composite numbers, but is a factor of every natural number

1 = 1 × 1; 2 = 1 × 2; 3 = 1 ×3;201

=20

1 is the identity element of multiplication and division.

15=1 ;√1=1 ; 3√1=1; ;etc .

Whole numbers (N0): The set of {0; 1; 2; 3; 4; 5; 6; …}

Describing the real number system by recognising, defining and distinguishing properties of natural numbers, whole numbers, integers, rational numbers and irrational numbers.


natural numbers together with 0.Whole numbers also consist of odd,

even and prime numbers, and 1 and 0

0 is neither a prime number nor composite.

4 + 0 = 4; 5 – 0 = 5;0 is the identity element of addition

and subtraction 03=0 ;0×5=0 , but 4

0isundefined

Integers (Z):The set of integers consist of the positive whole numbers, zero and the negative numbers

{…; -3; -2; -1; 0; 1; 2; 3; …}

Examples:

(i) 35 + 87 = 87 + 35 = 122(ii) 202 + 123 = 123 + 202 = 325

Use the commutative property in addition to show/make the equation equal.

(i) l+k=¿¿l+k=k+l

(ii) p+¿d+¿¿p+d=d+ p


(Activity 1)

Ask learners to present their observations and engage in a whole class discussion. Ask probing questions to assist learners to make appropriate conclusions.

Example: Why do you say 5 – 4 is not the same as 4 – 5?DO NOT provide them with answers immediately. Learners have to, through the

teachers’ assistance, discover that 5 - 4 = 1 but 4 – 5 = -1.

Tell learners that commutative property of subtraction does not apply, in subtraction we are “reducing”, “decreasing”, “minimizing”, therefore the value we subtract from matters.

For example, (Payment of debt), two scenarios:

Tebogo has an amount of R5 but he owes his friend R4. After he has paid how much will he be having?

R5 – R4 = R1 means if I have “say” R5 and required to pay R4 I will still have an amount of R1 remaining,

Lesiba has an amount of R4 but he owes his friend R5. After he has paid how much will he be having?

4 – 5 = - 1 means if I have R4 and required to pay R5, then I will have nothing left but will still owing an amount of R1 (that’s is why we write it as “- 1”).


The conclusion is that commutative property does not apply for subtraction.

ERROR ANALYSIS AND/OR MISCONCEPTIONS

The part above addresses the misconceptions around numbers: some learners see no difference between “– 1” and “1”. They need to see that

“-1” < “1”.

(Activity 3)

Rational numbers (Q): Any number

that can be written in the form ab ,

where a and b are integers but b≠ 0.

Recurring decimal : a decimal fraction in which a figure or group of figures is repeated indefinitely, as in 0.666

Integernon−zero number ,

Finite decimals: 75

100=3

4=75 % ;

−1,5=−1510

=−32

;

Recurring decimals:

0 , 3̇=13

;−0 , 8̇=−89

;

Irrational numbers (Q'): They are numbers with infinite non-terminating decimals, whose numbers goes on forever.

In practice we use rounded off values: √2≈1,414 , which is a rational approximation of the real value of √2.

Pi (π ¿ , 227

=3,141592 …


Real numbers (R): Consist of all the numbers that we use and include all the numbers

Real number system:

NON-REAL NUMBERS: These numbers are generally even powered roots of negative numbers.

Examples or non-real numbers are:√−4 ; 10√−100 ; √−2 ; 4√−20 ;

UNDEFINED NUMBERS: Division by zero is undefined.

50 ;

00 ;

CLASSWORK 10 min Activity 1

Ask learners to do the following sums, and they must present their observation subtractions.

a) 5 – 4 = b) 4 – 5 =c) 25 – 10 = d) 10 – 25 =e) 345 – 234 = f) 234 – 345 =


Activity 2

Complete the following by using <; >; + ; -

a) 79 ____ 50 ____ 50 = 79

b) 168 – 65 ____ 65 – 168

c) 5 ____ 4 ____ 4 – 5

Activity 3Educator instructs learners to do the following activities in class: Can be done

individually so that the educator can determine which learners are still experiencing any challenges. Educator needs to move and observe while this activity is in progress. Time management is important.

Complete the diagram below by providing examples :


R

Q’

NNo

Z

Q

CONSOLIDATION/CONCLUSIONAND OR HOMEWORK

3 min Complete the following table of real numbers:Number N N0 Z Q Q’ Non-

RealUndefined

15 Yes Yes Yes Yes No No No√3

0,14287

√ 14

−√1443√−1√−1

π0

√0,910

−3111

REFLECTION


Answers Week 3 Lesson 1

Mental Maths Class work Homework

Whole numbers: numbers that begin at 0 and increase by 1

Natural Numbers: numbers that begin at 1 and increase by 1.

Prime Numbers: numbers that have only two factors, 1 and itself

Even Numbers: –2 ; 4 ; 6; 8; ….Odd Numbers: - 1; 3; 5; 7; ……

Activity 1a) 5 – 4 = 1 b) 4 – 5 = -1c) 25 – 10 = 15 d) 10 – 25 = -15e) 345 – 234 =111 f) 234 – 345 = -

111g)

Activity 2a) 79 _+¿___ 50 _−¿___ 50 = 79

b) 168 – 65 __¿__ 65 – 168

c) 5 __+¿__ 4 __¿__ 4 – 5

Activity 3

Reflection / Remediation based on previous day’s work.

Since this is the 1st lesson of the year, there won’t be any homework.



1; 2; 3; 4…

π ;√5 ; 0,2356…

N 0

N0

Z…;-3; -2;

-1, ...

Q12

;√4 ;0 , 3̇ ; Q':


TOPIC: Whole numbers


By the end of the lesson, learners should know and be able to Use prime factorisation of numbers to find LCM and HCF.

RESOURCESDBE workbook (Book1 : Page 6 and 7), Sasol-Inzalo workbook(Book 1 : Page 16 and 17), textbook,

ruler, pencil, eraser, calculators, DVDs(Disk 1: Properties of number system)

PRIOR KNOWLEDGE



3 min Match Column A with Column B:

Column AColumn B

FactorsLowest Common Multiple

LCMNumbers you multiply together

to get another number

HCFThe product of a number and

another number

MultiplesReduction of a number into its

prime factors

Prime factorizationHighest Common Factor

p.119-121



5 min Reflection:Homework discussion and recap of the real number systemComplete the following table of real numbers:Number N N0 Z Q Q' Non-Real Undefined

15 Yes Yes Yes Yes No No No√3 No No No No Yes No No

0,14287 No No No Yes No No No

√ 14

No No No Yes No No No

−√144 No No Yes Yes No No No3√−1 No No Yes Yes No No No

√−1 No No No No No Yes Noπ No No No No Yes No No0 No Yes Yes Yes No No No

√0,9 No No No No Yes No No10

No No No No No No Yes

−311

No No No Yes No No No

1 Yes Yes Yes Yes No No No

Describing the real number system by recognizing, defining and distinguishing properties of natural numbers, whole numbers, integers, rational numbers and irrational numbers.

LESSONPRESENTATION/DEVELOPMENT

10 min Educator instructs learners on the chalkboard

Highest Common Factor (HCF):


Example:1. To find the highest factor of 10 and 15,

First get their prime factors

Find the HCF of 10 and 15:

F10 = 1; 2; 5; 10 . F15 = 1; 3; 5; 15

OR

∴ HCF = 5

Using prime factorization to find LCM

Example: Find the LCM of 28 and 126 by using prime factorization:


2 10 3 155 5 5 5

1 1

Step 1:2 28 2 1262 14 3 637 7 3 21

1 7 7 1

28 = 2 × 2 × 7 126 = 2 × 3 × 3 × 7Step 2: All the factors of 28 and all the extra factors of 126:

LCM = 2 × 2 × 7 × 3 × 3 = 252

CLASSWORK 10 min 1. Find the HCF of:

(a) 25 and 40(b) 18 and 30

CONSOLIDATIONCONCLUSIONAND OR HOMEWORK

3 min (1) Find the HCF and LCM by prime factorization:(a) 9 and 24(b) 15 and 40(c) 18 and 24

(2) Find the LCM by using multiples:(a) 3 and 14(b) 8 and 9

REFLECTION


Answers Term 1 Week 3 Week 3 Lesson 2


1.HCF : 25 and 40 F40: 1; 5 ; 10 ; 15 ; 20 ;25 ;30 ;35 ; 40 F25: 1; 2 ; 4 ; 5 ; 8 ; 10∴ HCF of 25 and 40 is 5

2. 18 and 30 F18: 2 ; 3 ; 6 ; 9 F30 : 2 ; 3 ; 5 ; 6 ∴ HCF of 18 and 30 = 6

1. Find the HCF and LCM by prime factorization:

a) 9 and 24F9 = 1, 3, & 9F24 = 1, 2, 3, 4, 6, 8, 12, and 24

∴ HCF of 9 and 24 = 3

M9 = 9, 18, 27, 36, 45, 54, 63, 72M24 = 24, 48, 72

∴ LCM of 9 and 24 = 72b) 15 and 40

F15 = 1, 3, 5, & 15F40 = 1, 2, 4, 8, 20, and 40

∴ HCF of 15 and 40 = 5

M15 = 15, 30, 45, 60, 75, 90, 105, 120 M40 = 40, 80, and 120

∴ LCM of 15 and 40 = 120c) 18 and 24

F18 = 1, 2, 3, 6, 9, and 18F24 = 1, 2, 3, 4, 6, 8, 12, and 24

∴ HCF of 18 and 24 = 6

M18 = 18, 36, and 72


M24 = 24, 48, and 72 ∴ LCM 18 and 24 = 72

2. Find the LCM by using multiples:

a) 3 and 14M3 = 3, 6, 9, 12, 15, …, 39, and 42M14 = 14, 28, and 42

∴ LCM of 3 and 14 = 42b) 8 and 9

M8 = 3, 6, 9, 12, 15, …, 39, and 42M9 = 9, 18, 27, 36, 45, 54, 63 and 72

∴ LCM of 8 and 9 = 72



TOPIC: Whole numbers - Solving problems involving whole numbers


By the end of the lesson, learners should know and be able toSolve problems in contexts involving Simple interest and Hire Purchase

RESOURCESDBE workbook (Book 1, pages 14-17), Sasol-Inzalo workbook (Book 1, pages 22-24), textbook, ruler,

pencil, eraser, scientific calculators.

PRIOR KNOWLEDGE

1. Substitution2. Working with percentages, fractions and decimals.3. Rounding off



3 min Find 20% of the following amounts:a) R50,00b) R120c) R109d) R2500

p.119-121


5 min 1. Find the HCF and LCM by prime factorization:a) 9 and 24

F9 = 1, 3, & 9F24 = 1, 2, 3, 4, 6, 8, 12, and 24

∴ HCF of 9 and 24 = 3M9 = 9, 18, 27, 36, 45, 54, 63, 72


M24 = 24, 48, 72 ∴ LCM of 9 and 24 = 72b) 15 and 40

F15 = 1, 3, 5, & 15F40 = 1, 2, 4, 8, 20, and 40

∴ HCF of 15 and 40 = 5

M15 = 15, 30, 45, 60, 75, 90, 105, 120 M40 = 40, 80, and 120

∴ LCM of 15 and 40 = 120c) 18 and 24

F18 = 1, 2, 3, 6, 9, and 18F24 = 1, 2, 3, 4, 6, 8, 12, and 24

∴ HCF of 18 and 24 = 6

M18 = 18, 36, and 72M24 = 24, 48, and 72

∴ LCM 18 and 24 = 72

2. Find the LCM by using multiples:a) 3 and 14

M3 = 3, 6, 9, 12, 15, …, 39, and 42M14 = 14, 28, and 42

∴ LCM of 3 and 14 = 42b) 8 and 9

M8 = 3, 6, 9, 12, 15, …, 39, and 42M9 = 9, 18, 27, 36, 45, 54, 63 and 72

∴ LCM of 8 and 9 = 72



10 min SIMPLE INTEREST

Simple Interest is calculated for a number of years on an amount (the amount stays fixed) without the interest being cumulated to the total amount each year. In other words, the interest amount added stays fixed for those number of years. Interest rates are normally expressed as percentages.

Example:Tebogo deposits R2500 with a financial institution for 3 years on simple interest. The interest rate is 10% per annum. How much interest will she have earned at the end of the three year?

Solution:10% of R2500 = R250. So, this comprises the interest amount and it is the

amount which should be added to the total amount each year.Therefore,At end of first year, the total amount will be R2500 + R250 = R2750At end of first year, the total amount will be R2750 + R250 = R3000At end of first year, the total amount will be R3000 + R250 = R3250The total interest earned over the three years is therefore R3250-R2500 = R750

It can be seen that this amount does not yield interesting results, and it will be seen in the lesson how these results can be improved when a better investment is made.

A formula can also be used for a quick calculation: A=P(1+¿), where

Per annum means “per year”.


A = accumulated amountP = Principal amounti = Interest rate (written in decimal form)n = Time period (number of years, etc.)

if we use a formula, we will then have the following: A=P (1+¿ ) ¿ 2500[1+0,1(3)] ¿ R 3250 and interest will still be the difference between the initial

amount and the final amount

Alternatively, we can still

I=Prn100

=2500 ×10 ×3100

=R 750, which is the actual interest amount.

HIRE PURCHASE (HP)

Sometimes you need an item but do not have enough money to pay the full amount at that time. One option is to buy the item on what is called hire purchase (HP). This method requires that you pay a deposit and sign an agreement which commits you to pay regular, monthly instalments until you have paid the full amount.

Therefore: HP price = deposit amount + total of instalments

The difference between the HP price and the cash price is the interest

Note the conversion from percentage to decimal.

‘r’ is substituted as given in the question.


that the dealer charges you for allowing you to pay off the item over a period of time.

Example:Sipho buys a flat screen television on hire purchase. The cash price is

R5000. He pays a deposit of R2000 and allowed to pay the balance over 24 months on equal, monthly instalments of R300.

(a) Calculate the total HP price.(b) How much interest does she pay?Solution:(a) The total HP price would be deposit amount + total of instalments.

= R2000 + (24 × R200) = R2000 + R4800 = R6800

2. The interest that Sipho would pay is: R6800 – R5000 = R1800

It is a good idea to know the total amount that will have been paid at the end of the agreement period - the law insists on this.

Learners should be made aware of possible, additional costs such insurance, etc.


CLASSWORK 10 min

1. Money is invested for 1 year at an interest rate of 8% per annum. Complete the table of equivalent rates.

Invested amount 1000 2500 8000 20000 90000 x

Interest earned

2. Mapula bought a kitchen unit for R15000 on HP. She paid a deposit of R5000 and agreed to pay the balance over 24 months. The interest rate charged on this HP transaction is 20%. Calculate the total amount that Mapula would have to pay for the kitchen unit, and also what her monthly would be.


3 min Sebolelo buys a car on Hire Purchase. The car costs R130 000. She pays a 10% deposit on the cash price and will have to pay monthly instalments of R4 600 for a period of three years. Dikgang buys the same car, but chooses another option where he has to pay a 35% deposit on the cash price and monthly instalments of R3 950 for two years.

REFLECTION


Answers Term 1 Grade 9 Week 3 Lesson 3


20% of:a) R50,00 = R10,00b) R120 = R24,00c) R109 = R21,80d) R2500 = R500,00

1.Invested

amount

1000 2500 8000 20000 90000 x

Interest earned

80 200 640 1600 72000.08x or (

x× 8%)

2. Deposit = R5 000.Instalment would be on the balance of R15 000 – R5000= R10 000 .20% of R10 000 = 0,02 × R10 000 = R2 000 (Deposit is subtracted).Therefore, the total amount Mapula would pay is R15 000 + R2 000 = R17 000.However, her monthly instalments would be based on the total amount less the deposit amount = R12 000.

∴ Monthly instalment = 12000

24=¿ R500.

(a) HP price for bothSebolelo: R13 000 + (36 × 4 600) =

R178 600Dikgang: (35% × R130 000) + (24 ×

R3 950) = R45 500 + R94 800 = R140 300

(b) The difference between the two

R178 600 − R140 300 = R38 300(c) Interest that both have to paySebolelo: R178 600 − R130 000 =

R48 600.R48 600 ÷ R130 000 × 100% = 37,38

…% ≈ 37,4%Dikgang: R140 300 − R130 000 =

R10 300.R10 300 ÷ R130 000 × 100% = 7,92

…% ≈ 7,9%.



TOPIC: Whole numbers - Solve problems in financial context


By the end of the lesson, learners should know and be able toSolve problems that involve whole numbers, percentages and decimal fractions in financial contexts such as:

simple interest and compound interest


pencil, eraser, scientific calculators.

PRIOR KNOWLEDGE1. Simple Interest2. Substitution3. Rounding off

COMPONENTS TIME TASKS/ACTIVITIES CAPSINTRODUCTIONMENTAL MATHS

4 min Calculate the monthly instalment if R48 000 is paid over:(a) 12 months(b) 24 months(c) 64 months

p.119-121


5 min (a) HP price for bothSebolelo: R13 000 + (36 × 4 600) = R178 600Dikgang: (35% × R130 000) + (24 × R3 950) = R45 500 + R94 800 =

R140 300(b) The difference between the twoR178 600 − R140 300 = R38 300(c) Interest that both have to pay


Sebolelo: R178 600 − R130 000 = R48 600.R48 600 ÷ R130 000 × 100% = 37,38 …% ≈ 37,4%Dikgang: R140 300 − R130 000 = R10 300.R10 300 ÷ R130 000 × 100% = 7,92 …% ≈ 7,9%.


10 min Simple interest is often used on short-term payments. For long term transactions, compound interest is used. Compound interest is always calculated from the balance remaining at any time, unlike simple interest which is only calculated from the original amount.

e.g. calculating how much you would get if you invested an amount of R60 000 over three years and interest is compounded yearly at 5% per annum, gives:

Year 1: R60 000 + (0,05 of R 60 000 ) = R63 000Year 2: R63 000 + (0,05 of R 63 000 ) = R66 150Year 3: R66 150 + (0,05 o f R 66 150 ) = R69 457,50

A formula used for compound interest is A = P(1+i)n

Using this formula for the same example done above would yield:

A = P(1+i)n

= 60 000 (1+0,05 )3

= R69 457,50

If interest was compounded monthly, then we would have 36 conversion periods (n-value) since there are 12 months in one year.

Remember that per annum means per year.

Notice how interest is calculated from the amount available at the time.

Note the letter ‘n’ in the exponent, positioned differently from the simple interest formula.

There are 48 months in four years.

The final answer is rounded to the nearest cent since context is money.


CLASSWORK 10min Andrew and Zinzi are arguing about interest on money that they have been given for Christmas. They each received R750. Andrew wants to invest his money in ABC Building Society for 2 years at a compound interest rate of 14% per annum, while Zinzi claims that she will do better at Bonus Bank, earning 15% p.a. simple interest over 2 years. Who is correct?


1 min Calculate the interest generated by an investment (P) of R5 000 at 10% (r) compound interest over a period (n) of 3 years. A is the

final amount. Use the formula A P(1+ r

100)n

to calculate the interest.

A = R5 000(1 + 0,1)3 = R6 655R6 655 − R5 000 = R1 655 interest.

REFLECTION




Monthly instalment of R48 000 paid over:

(d) 12 months is R4 000(e) 24 months is R 2 000(f) 64 months is R 750

Compound interest: R750 + (14% × R750) = R855Second year: R855 + (14% × R855) = R974,70Simple interest: R750 + 2 × (R750 × 15%) = R975∴ Zinzi is correct.

A = R5 000(1 + 0,1)3 = R6 655R6 655 − R5 000 = R1 655 interest.



TOPIC: Whole numbers - Solving problems involving whole numbers


By the end of the lesson, learners should know and be able toSolve problems in contexts involving

Ratio

RESOURCES DBE workbook (Book 1, pages 8-13), Sasol-Inzalo workbook (Book 1, pages 18-20), textbook, ruler, pencil, eraser, calculators, DVD (Term 1, Disc 1: Ratio)

PRIOR KNOWLEDGE

Basic mathematics operations.



4 min Simplify the following terms:

(a)1020 (b) 10% : 50%

c) 6 : 3 (d) 10 mm : 30 cm

p.119-121


5min A = R5 000(1 + 0,1)3 = R6 655R6 655 − R5 000 = R1 655 interest.


10 min Educator asks learners to explain the meaning of ratio and to provide appropriate examples. Educator can then show them the 4 blocks below and ask them to determine the ratio of the two colours and to explain the meaning of the ratio.

Whole numbersConcepts and skills:Solving problems

involving whole


Ratio: A ratio is a way of comparing two or more quantities of the same kind

It is written in the form a : b. We say a is to b.a and b are called the quantities of the ratio

Example: 3 : 1

Ratios have no units. The order in which we express a ratio is important.

Example: If there are 2 boys and three girls in a group then the ratio of boys to girls is 2 : 3 and the ratio of girls to boys is 3 : 2 .

Educator shows learners the 6 blocks below and asks them to first write down the ratio and then simplify it. Ratios are always expressed in their simplest form

Example: 6 : 2 = 3 : 1 A ratio with two quantities can be expressed as a fraction

Example: The ratio of 3 blocks is to 1 block is 3 : 1 or 31 .

Educator asks learners to express the ratio below as a percentage.

A ratio can be expressed as a percentage.

Example: Express the ratio 4 : 5 as a percentage:

numbers percentages, and decimal fractions in financial context:

Compound interest Percentages

Interest, time, principal amount

Accounts & loansCompound interest


4 : 5 = 45

45

× 2020

= 80100

=80 %

Educator tasks learners to discuss the examples below in groups(1) Express ratio with quantities given in different units:Example: What ratio is 125 ml of milk to 1 litre of milk?

Write down the ratioWrite the ratio as a

fraction

Convert to the same unit

Simplify

Answer

125 ml: 1 l125 ml

1l125 ml

1000 ml

18

1 : 8

Educator explains the following three examples to the class:(2) Calculations with ratios:

Examples:Divide 300 apples amongst A, B and C in the ratio 1 : 5 : 9. 1 + 5 + 9 = 15


CLASSWORK 10min 1) Simplify the following ratio:a) 6 : 12b) 9 : 12 : 15c) 1,5 : 3

2) Express the ratio 3 : 2 as a percentage3) Which ratio is the smallest: 1 : 5 or 1 : 6 ?4) There are 450 learners in a small rural school. The ratio of the number

of boys to the number of girls is 7 : 8. How many boys and how many girls are there in the school?

5) Increase a mark of 28% in the ratio 3 : 2.


1min (1) Simplify the ratio 55 : 121(2) Express the ratio R50 : R500 as a percentage(3) Divide R1 200 between partners Jimmy and Thabang, according to the

ratio 65% : 35%.(4) Decrease 60 minutes in the ratio 2 : 3.(5) Increase 28 kg in the ratio 13 : 7(6) Decrease 45 minutes in the ratio 2 : 3

REFLECTION


Answers Term 1 Week 3 Grade 9 Lesson 5


(a) 1020

=12

(b) 10% : 50% = 1 : 5(c) 6 : 3 = 2 : 1(d) 10 mm : 30 cm = 10

mm : 300 mm = 1 : 30

1.a) 6 : 12

612

=12

b) 9 : 12 : 15

LCM is 3: ∴93

: 123

: 153

=3 :4 :5

c) 1,5 : 3

LCM is 1,5 ∴1,51,5

: 31,5

=1: 2

2. 3 : 2 as a percentage

¿32

× 5050

=150100

=150 %

3. Which ratio is the smallest: 1 : 5 or 1 : 6 ?

LCM is 30 ∴ 15

× 66=

630

∧1

6× 5

5= 5

30 ∴

16 is the smallest

4. There are 450 learners in a small rural school. The ratio of the number of boys to the number of girls is 7 : 8.How many boys and how many girls are there in the school?

7 + 8 = 15 ∴7

15× 450=210 boys and

(1) Simplify the ratio 55 : 121

∴55

121÷ 11

11= 5

11(2) Express the ratio R50 : R500 as

a percentage R50 : R500 = 1 : 10

∴ 110

∴ 110

× 1010

= 10100

=10 %

(3) Divide R1 200 between part-ners Jimmy and Thabang, ac-cording to the ratio65% : 35%.

65% : 35% = 65 : 35 = 13 : 7 13 + 7 = 20

∴1320

× R 1200=R 780

7

20× R 1 200=R 420

(4) Decrease 60 minutes in the ra-tio 2 : 3.

∴ 23

× 60=40


∴ 815

× 450=240 girls

5. Increase a mark of 28% in the ratio 3 : 2.

∴32

× 28 %=42 %


TOPIC: Integers – Properties of integers



Ratio and rate Direct and indirect proportion

FINANACIAL MATHS Exchange rates

RESOURCESDBE workbook(Book 1: Pages 8-13), Sasol-Inzalo workbook(Book 1: Pages 18,19 and 26), textbook, ruler,

pencil, eraser, calculators, DVDs (Term 1 Disk 1: Ratio; Direct and indirect proportion)

PRIOR KNOWLEDGE

1. Concepts of time, speed, and distance.2. Foreign currencies3. Conversions4. Basic mathematical operations




3 min Simplify:

(a) 30 km

2h(b) (b) 20 km/h × 3 h(c) Convert to minutes: 3 h 20 min(d) Convert to hours: 3 h 20 min

p.119-121


5 min (5) Simplify the ratio 55 : 121

∴55

121÷ 11

11= 5

11(6) Express the ratio R50 : R500 as a percentage R50 : R500 = 1 : 10

∴1

10∴ 1

10× 10

10= 10

100=10 %

(7) Divide R1 200 between partners Jimmy and Thabang, according to the ratio 65% : 35%.

65% : 35% = 65 : 35 = 13 : 7 13 + 7 = 20

∴1320

× R 1200=R 780

720

× R 1200=R 420

(8) Decrease 60 minutes in the ratio 2 : 3.

∴ 23

× 60=40

LESSONPRESENTATION/

10 min Educator asks learners to explain the meaning of rate and to provide real life examples:



DEVELOPMENT Rate:The comparison of two amounts with different units.

Notice that any rate uses the word per to indicate that it compares one quantity to one unit of another quantity.

Examples:Cost of an SMS: R1,50 per SMS

Speed limit: 20 km per hourCricket score: 30 runs per overEducator can use fake money as props and learners can role play

using the example below. This can be done as group work. Distance, time and speed:

The speed of a moving object is the rate of its motion, or the distance covered per unit of time.

Speed formula: speed=distancetime

(s=dt)

Distance formula: speed × time (d = s × t)

Time formula: time=distancespeed (t=d

s )Example:1. What is the average speed of a car that travels a distance of 352

km in 3 hours? Round off your answer off to two decimal places. What distance will a truck cover if it travels for 4 h 20 min at an average speed of 75 km/h?

4 h 20 min ¿4 13

h

involving whole numbers percentages, and decimal fractions in financial context:





d = s × t ∴d=75 kmh

× 4 13

h=325 km

2. In what time will a motorbike travel a distance of 240 km at an average speed of 100 km/h?

t=ds∴t= 240 km

100 km /h = 2,4 hours = 2 h and 0,4 × 60 min

= 2 hours and 24 minutes

s=dt∴ s= 352 km

3hours=117,33km /h

Exchange rate: Example: Mary works in a Foreign Exchange bank at OR Tambo

airport. The various exchange rates are shown on the electronic notice board below.

Currency Buy (R) Sell (R)US Dollar $ 1 8,1090 8,4586

British Pound ￡ 1 12,7850 13,4767

Euro 1 11,2780 11,9060Botswana Pula P 1 0,9320 0, 8080

Mary buys $500 from a customer: How much did she pay in rand?

500 × 8,1090 = R4 054,50Mary sells ￡400 to a customer. How much did she receive in rand?


400 × 13,4767= R5390,68CLASSWORK 10 min 1. Make use of the exchange rate table below to calculate each of

the given transactions

Currency Buy (R) Sell (R)US Dollar $ 1 8,1090 8,4586British Pound￡1 12,7850 13,4767Euro €1 11,2780 11,9060Botswana Pula P 1 0,9320 0, 8080

(a) Mr. D from Germany bought rands with 4500 euros from Mary.(b) Mr. H from South Africa bought Pula with R1 000 from Mary.(c) Mary sells R 4 000 worth of dollars to Ms. B

2. A car traveling at a constant speed travels 54 km in 18 minutes. How far, traveling at the same constant speed, will the car travel in 1 hour 12 minutes?


3 min (1) A car traveling at an average speed of 100 km/h covers a certain distance in 3 hours 20 minutes. At what constant speed must the car travel to cover the same distance in 2 hours 40 minutes?

(2) John travelled 10 km by taxi to his home. It took 12 minutes to cover the distance. The driver charged John R25.a) At what rate per kilometre did the taxi driver charge John?b) The driver stopped for petrol. He paid R390, 50 for 55 litre.

Calculate the price rate of the petrol.REFLECTION




(a) 30 km

2 h =15km/h

(b) 20 km/h × 3 h = 60km (c) 3 h 20 min = 200 minutes

(b) 3 h 20 min = 3 13 h

1.a) If Mr. D bought Rands with 4500 Euros, it

means that Mary bought 4500 Euros.4500 × 11,2780 = R50751,00

b) If Mr. H bought Pula with R1000, it means that Mary sold Pula.

1000 ÷ 0,8080 = 1237,62 P

c) 4000 ÷ 8,4586 = $ 472,89

2. Convert 1 h 12 minutes to minutes: 1 × 60 + 12 = 72 minutes.

7218

×54=¿216km

1.Convert 3 h 20 minutes to minutes: 3 ×

60 + 20 = 200 minutes. Convert 2 h 40 minutes to minutes: 2 × 60 + 40 = 160 minutes.

Distance is the same: 100 x 200 = 160 x New Speed (D=SxT)

Constant speed = distance

time=200

160× 100 =

125 km/h

2.(a) Taxi rate per kilometre: R25 ÷ 10 km = R2,50 /km(b) The driver stopped for petrol. He

paid R390,50 for 55 litres.(c) Rate of petrol: R390,50 ÷ 55 =

R7,10/litre



TOPIC: Whole Numbers – Solving problems involving whole numbers



Ratio and rate Direct and indirect proportion

RESOURCESDBE workbook(Book1 : Pages 8-13), Sasol-Inzalo workbook 1: (Pages 18,19 and 26), textbook, ruler,

pencil, eraser, calculators, DVDs(Term 1 Disk 1 : Ratio, Direct and indirect proportion)

PRIOR KNOWLEDGE Basic mathematical operations: addition, multiplication, division, multiplication.



4 min 1.(a) 12 : 24

(b)3020

2.

(a)12= 5

10(b) 4 : 10 = 8 : 20

p.119-121

REVIEW AND CORRECTION OF

5 min 1.Convert 3 h 20 minutes to minutes: 3 × 60 + 20 = 200 minutes.


HOMEWORK Convert 2 h 40 minutes to minutes: 2 × 60 + 40 = 160 minutes.Distance is the same: 100 x 200 = 160 x New Speed (D=SxT)

Constant speed = distance

time=200

160× 100 = 125 km/h

2. (a) Taxi rate per kilometre: R25 ÷ 10 km = R2,50 /km

(b) The driver stopped for petrol. He paid R390,50 for 55 litres.(c) Rate of petrol: R390,50 ÷ 55 = R7,10/litre


10 min Educator asks learners what proportion is and that they must come up with everyday examples of direct proportion.

Proportion:When quantities are presented in the form:

a : b = c : d or ab= c

d In other words, when two ratios are equal.

Direct propor-tion:

When two quantities in-crease or de-crease by the same ratio, we







say that they are in direct proportion to each other.

x is directly proportional to y if xy = constant.

x and y are directly proportional if, as the value of x in-creases the value of y increases, and as the value of x de-creases the value of y decreases in the same proportion.

The direct proportional relationship is represented by a straight line (linear graph).Example time and distance .


Educator asks learners what the difference is between direct and indirect proportion and that they must give examples.

Indirect proportion: Two quantities x and y are inversely proportional or in-

directly proportional to each other if, as the value of x increases the value of y decreases, and as the value of x decreases the value of y increases .The product of the values is constant: x× y= a constant.

An indirect proportional relationship is represented by a non-linear curve. Example pressure and volume


As time increases the distance covered also increases

Example: A farmer hires 12 men to construct a building. The job is completed in 18 days. If he had hired 24 men, the job would have been completed in half the time. If he had hired 6 men, the job would take twice as long to complete.

Mathematically: 1224

× 18=9 days 126

×18=36 days

× 24) 12 × 18 = 9 × 24 × 6) 12 × 18 = 36 × 6

1224

= 918

126

=3618

Proportion: 12 : 24 = 9 : 18 and 12 : 6 = 36 : 18 ∴12 :24=x :18 12: 6=x :18


CLASSWORK 10 min 1. The table shows the costs of various amounts of petrol sold at a filling station.

Volume of petrol in litres 1 25 40 50

Cost amount in rand 1 300 480 600

Check whether the cost per litre (the price rate) is the same for each of the four volumes of petrol.

2. Ten men build a wall in 30 days. How long will it take 5 men to build the same wall?



1min (1) Tau wanted to build a doll’s bed for his sister, Miriam. He measured his parent’s bed and found it was 2 m long and 1,5 m wide. He wanted to make the doll’s bed 20 cm long. How wide should he make it so that the ratios are the same as his par-ent’s bed?

(2) It takes 5 pipes 18 hours to fill a dam. Determine how long 9 pipes will take to fill the dam

REFLECTION


http://www.google.co.za/url?sa=i&source=images&cd=&cad=rja&docid=U3t4JJDnoQEbuM&tbnid=GHhGJTDJyeimfM:&ved=0CAgQjRwwAA&url=http://www.sse.com/images/&ei=RJNzUv7uMs-VhQeZzIHQAw&psig=AFQjCNEU1cXSVK7eQnnZoW9tFOS4L5Docg&ust=1383392452936479



(1) Simplify:

(a) 12 : 24 = 1224

=12=1:2

(b ) 3020

=32=3 :2

(2) True or false:

(a)12= 5

10 true

(b) 4 : 10 = 8 : 20 true

1.Cost

volume=180

15=R 12 per litre ;

30025

=R 12 per litre ;

48040

=R 12 per litre ;

60050

=12 per litre

2. 10 :5=x :30

∴ 105

= x30

∴ x=10 ×305

=60 days

1. 2 m = 200 cm and 1,5 m = 150 cm The proportion is : ratio of

parent’s bed : ratio of doll’s bed ∴200 :150 ¿20: x200150

=20x

∴ x=150×20200

∴ x=15cm2.

5 :3=x :18

∴ 53= x

18

∴ x=5×183

∴ x=30 hrs 5 :9=x :18

∴59= x

18

∴ x=5 ×189

∴ x=10 hrs



TOPIC: Whole Numbers



FINANCIAL MATHS profit, loss, discount and VAT budgets accounts and loans rentals commission


pencil, eraser, calculators

PRIOR KNOWLEDGE Working with decimals and percentages; basic knowledge of finances.



3 min (1) Choose the correct answer:a) Write 12% as a common fraction

i. 0,12 ii. 12

100 iii. 0,012 iv. 1,2100

b) 100% has the same value as

p.119-121


i. 0,1 ii. 1 iii. 10

100 iv. 0,100

c) R36 expressed as a percentage of R100

(i) 0,36 (ii) 36% (iii) 3,6% (iv) 13


5 min 1. 2 m = 200 cm and 1,5 m = 150 cm The proportion is : ratio of parent’s bed : ratio of doll’s bed ∴200 :150 ¿20 : x200150

=20x

∴ x=150 ×20200

∴ x=15 cm2.

5 :3=x :18

∴ 53= x

18

∴ x=5 ×183

∴ x=30 hrs 5 :9=x :18

∴ 59= x

18

∴ x=5 ×189

∴ x=10 hrs



10 min Educator asks learners what the word percentage means and the use of percentages in everyday life.

Percentages: Percentage means “out of a hundred”

Example:

(a) Calculate 10% of R83 ∴10

100× R 83=R 8,30

(a) Peter earns a salary of R9 400 a month. He receives an in-crease of 6%.Calculate his new salary.

6100

× R 9 400=R 564,00

∴new salary : R 9 400+R 564=R 9 964Educator can use examples of existing invoices to show learners

where VAT appears on these documents. Learners must also explain what VAT means and why it exists.

Value-Added tax (VAT): When we purchase a product or receive service for something we have to pay VAT.Example: Milly has a contract for her landline telephone. She pays R122, 45 for calls, and a monthly fee of R19,90 for use of the line. VAT of 14% is also added to her monthly bill. Calculate her total monthly bill.







Educator asks learners what profit, loss and discount means. Educator can show learners advertisements from pamphlets, etc. of sales.

Profit, loss and discount:

Profit is when there is a surplus because the selling price is greater than the cost price.Loss is when there is a deficit because the cost price is greater than the selling price.Discount is the amount by which the selling price is reduced .Cost price is the price it costs to make an item before the profit is added.Selling price is the price at which an item is sold .

Profit = selling price – cost priceLoss = cost price – selling price

Example:A shopkeeper bought a tin of coffee for R35 and sold it for R50.

(a) Did he make a profit or loss? Profit = R50 – R35 = R15(b) Express the profit as a percentage of the cost price, correct to

2 decimal places:


Calculate total cost of calls and service fee

Calculate VAT at 14%

Add VAT to the total cost

R122,45 + R19,90 = R142,35

14100

× R 142,35=R 19,93

R142,35 + R19,93 = R162,28

R 15R 35

×100 %=42,86 %

Educator asks learners what a budget is and the importance of having a budget to control spending and savings.

Budgets and accounts: A budget is a statement showing income and expenses.

Gross income is the income earned before deductions are made

Net income is the income received after deductions have been made.

Most common deductions are tax, medical aid, pen-sion, UIF, etc.

Net income = gross income – deductions

Example: Joan’s net income is R9 000 per month.

Net income R9 000Expenses

Home loan R3 200 Food R2 000Transport R1 000 Life insurance R 500Electricity R 200 Clothing store R 500Medical Aid R1 000 Entertainment R 750

Total expenses: R9 150


(a) Is Joan living within her means? No, because R9 000 – R9 150 = -R150

(b) Suggest a solution: Certain items are important such as food, electricity, transport, etc. She could spend less on clothing and entertainment

Educator can start a discussion on various types of employment conditions of which earning a commission is earned, etc. Learners must also give real life examples of what rentals are.

Commission: When employees earn commission, they earn a cer-tain percentage of the value of their sales.

Rentals: The amount of money you pay for the use of a venue or equipment

Example: Jerome has a job selling clothing. He earns a commission of 19, 5% on all weekly sales in excess (more than) of R5 000. How much commission does he earn on sales of R4 800 and R10 450?

On R4800:Jerome does not earn

any commission because

R4 800 is less thanR5 000

On R10 450 Jerome earns:R10 450 – R5 000 = R5 45019,5100

× R 5450=R 1062,75

CLASSWORK 10 min (1) Calculate the VAT on R89,79(2) Lerato bought a bicycle for R600. Two years later she sold it at

a loss of 25%. How much money did Lerato get for the bicycle?



1 min (1) Mr Dlamini rents out two of the bedrooms in his house to make extra money. He charges R1 750 per room per month. He has to pay a rental agency 6% per month, as they found him the tenants. How much commission, in rand, is he paying the agency?

(2) Mary sells and buys burgers. She pays for one burger R6 and sells it for R9.

(a) How much profit does she make?(b) What is her percentage profit?

(3) Penny is a teacher. Her gross income is R20 000 per month. Her deductions are: Income tax = R2 681,63; Medical Aid = R1 800; UIF = R124,78; Pension fund = R1 200.

(a) Calculate her net income (b) Penny has the following expenses,

Home loan R2 000Food: R2 200

Transport R520 Life insurance R700

Electricity R350Clothing store R600Entertainment: R425TOTAL = R6 795Use the given expenses to draw up a budget

(c) Is she living above or below her means?REFLECTION


Answers Term 1 Grade 9 Week 3 Lesson 8Mental Maths Class work Homework

1. (ii) 12

100

2. (ii) 1

3. (ii) 36%

1.14100

× R 89,79=R12,57

2. Loss: 25100

× R 600 = R150

Lerato got: R600 – R150 = R450

1. Commission = R1052. Mary’s burgers:

(a) Profit = R9 – R6 = R3(b) Percentage profit = 50%

3. Penny:(a) Her net income = R14 193,59(b) Penny’s budget:

Item Expenditure

Balance

Income salary: R20 000Income tax R2 681,63Medical Aid R1 800UIF R124,78Pension fund R1 200TOTAL R5806,41 R14 193,59Home loan R2 000Food R2 200Transport R520Life insurance R700Electricity R350Clothing store R600Entertainment R425TOTAL R6 795 R7 398,59

(c) She is living below her means.



TOPIC: Whole Numbers – Weekly Assessment

WEEKLY ASSESSMENTTERM 1 WEEK 2MAX: 20

QUESTION 1

Choose the correct answer

1.1 Which number is missing?

12 ;

14 ;

18 ; … ;

132

(2)

A. 16

B. 164

C. 464

D. 216


1.2 Which statement is incorrect? (2)

A. The only prime factors of 24 are 2 and 3 .

B. The difference between - 8 and 3 is 11.

C. All integers are natural numbers.

D. 3√−27 is a rational number.

1.3 Which of the following numbers is a rational number? (2)

A. √3

B. √16

C. √−¿9¿

D. √13

1.4 A painter is paid by the hour. If he is paid R360 for 12 hours work. How much will he be paid for 9 hours work?

A. R120

B. R180

C. R270 (2)

D. R720 (2)

[10]


WEEKLY ASSESSMENT: MEMORANDUMTERM 1 WEEK 3MAX: 20

QUESTION 1

1.1 C. 4

64(2)

1.2 C. All integers are natural numbers (2)

1.3 B. √16 (2)

1.4 C. R270 (2)

TOTAL [08]



TOPIC: Integers – Simple interest



Simple interest

RESOURCES DBE workbook( Page 70 – 77), Sasol-Inzalo workbook ( Page 25 – 38), textbook, ruler, pencil, eraser, calculators, DVDS ( Disc 1 : GDE 16 01 2014, GDE 28 01 2014, GDE 30 01 2014 & GDE 13 02 2014)

PRIOR KNOWLEDGE Working with decimals and percentages; basic knowledge of finances.



3 min Calculate:

1. 20% of R4000.002. 15% of R2500.003. 18% of R3000.00

p.119-121


5 min New week

LESSON 10 min What happens when you borrow money: You pay back with interest.


PRESENTATION/DEVELOPMENT Why do we take loans? To buy houses ,cars and so onWe are going to discuss how simple interest is calculated

Example: I borrowed a loan of R8000 to buy a television .the simple interest was 10 % per annum for two years.

QUESTION(I) What is the rate? 10% per annum(II)What is the time I should spend paying the loan: 2 years(III) How is simple interest calculated?

A formula is used simple interest (si) =prt100 tell me what these words

stand for

P=amount borrowed t= time r = rate

QUESTION 2 If we look at the story what is the principal, the rate and the time

Questions to be answered:

i) what is the simple interest chargedii) what is the final amount to be paid

Si =prt100 =

8000× 10 ×2100 = 80 × 10 × 2=R1600


Amount to be paid its p+ni= amount =8000+1600= R9600

CLASSWORK 10 min My friend took a loan to buy a car for R 6 500 and has to pay back a loan in 2 years at 12% interest.

Calculate1. Simple interest2. Final amount to be paid.


3 min Tony wants to buy a flat screen. He takes a loan of R12500 @ 12% per annum for 2 years.

a)Calculate Sib) Final amount to be paid.

REFLECTION


Week 4 Lesson 1 Answers

Mental Maths Classwork Home work

1. 20% of R4000.00 ¿20

100 × R4000

¿ R800.00

2. 15% of R2500.00¿ 15100

× R2500

¿ R375.00

3. 18% of R3000.00¿ 18100

× R3000

¿ R540.00

My friend took a loan to buy a car forR 6 500 and has to pay back a loan in 2

years at 12% interest.Calculate1. Simple interest

Si =prt100 =

6500× 12× 2100 = R1560

2. Final amount to be paid.Amount ¿p+si ¿6500+1560 = R8060

Tony wants to buy a flat screen. He takes a loan of R12500 @ 12% per annum for 2 years.

a)Calculate Si

Si =prt100 =

12500× 12× 2100 = R3000

b) Final amount to be paid.Amount ¿p+si ¿12500+3000 = R15500.00



TOPIC: Integers – Solve problems in financial context



simple interest and higher purchase compound interest

RESOURCES DBE workbook(Page 70 – 77), Sasol-Inzalo workbook 1 (25 – 38), textbook, ruler, pencil, eraser, calculators, DVDS( Disc 1 : GDE 16 01 2014, GDE 28 01 2014, GDE 30 01 2014 & GDE 13 02 2014)

PRIOR KNOWLEDGE

Working with decimals and percentages; basic knowledge of finances.



3 min Choose the correct answer:R8 000 is taken out as a loan for five years at a simple interest rate of

9,5% p.a. How much money must be repaid at the end of five years?(a) R118,00(b) R8 000(c) R11 800(d) 9,5%

p.119-121

REVIEW AND 5 m 1. 18 monthly payments = 18 × R135,60 = R2 440,80


CORRECTION OF HOMEWORK

Total cost = R229 + R2 440,80 = R2 669,80

2. Saving = R2 669,80 – R2 298 = R371.80

3. SI=P ×r × n100

∴SI=200× 4× 1100

=R 8 Final amount = R200 +

R8 = R208

4. SI = final amount – starting amount = R4 260 – R 3 000 = R1 260

i= r100

= 6100

=0,06

n=SIPi

∴i= 1260R 3000× 0,06

=7 years

LESSONPRESENTATION/

DEVELOPMENT

10 min

Compound interest: If the interest over a fixed period is added to the principal amount, and the interest for the next period is calculated on this combined amount, we call this interest compound interest.

To calculate compound interest we work out the amount of money in the bank account at the end of each year.

Example:Penny invested R2 000 at 5% per annum compound interest for a period

of 3 years.Year 1: P = R2 000

Interest = 2000× 5× 1

100=R 100

= R2 000 + R100 = R2 100

Principal (initial) value

I=Pnr100 interest earned in year

1 value of investment at end of year 1

Year 2: P = R2 100 Principal (initial) value







Interest = 2100 ×5 ×1

100=R 105

= R2 100 + R105 = R2 205



Year 3: P = R2 205 Interest = 2205 ×5 × 1

100=R 110,25

= R2 205 + R110,25 = R2 315,25




For simple interest it would have been:

SI=Pnr100

∴SI=2000×5× 3100

¿ R 300Final amount = R2 000 + R300

= R2 300 A difference of R15,25Formula for compound interest:Remember that A = P(1 + ni) where A is total amount, P is principal

(initial) amount; n is time in years and i=r

100 where r is interest

rate.

Year 1: A = P(1 + ni)Principal (P) amount invested for

n years at a rate of r% p.aYear 2: A = P(1 + ni) + P( 1 + ni) ×

IThis amount now becomes the

principal amount for the


A =P(1 + i)(1 + i)A = P(1 + i)2

second year

Year 3:A = P(1 + i)2 + P(1 + i)2 × IA = P(1 + i)2 (1 + i)A = P(1 + i)3

At the end of year 3

Year n:A = P(1 + i)n At the end of year nExample: Lets revisit the first example using the formula A = P(1 + i)n. Penny invested R2 000 at 5% per annum compound interest for a period

of 3 years. A = final amount

P = R2 000r = 5% ∴ i = 5 ÷ 100 = 0,05n = 3 years

A = P(1 + i)n

A = 2 000(1 + 0,05)3

A = R2 315,25 Calculate Principal amount:

A=P¿P ¿

∴P= A¿¿

CLASSWORK 1. Sam invests R5 000, compounded annually at a rate of 6% per year. Calculate the future value (Final amount) after 3 years.

2. You have paid R11 800 on a 5 year loan which was compounded yearly at 9,5% interest. What was your initial amount?



3 min (1) Suppose you have R10 000 to invest in a savings account over a period of 20 years at an interest rate of 11% p.a. Complete the table to compare the growth of the savings with simple and com-pound interest

years 0 5 10 15 20A = P(1 + ni) R10 000A = P(1 + i)n R10 000

(2) Mr. martin invests R12 750 for 3 years compounded annually at 5,3% p.a. How much is it worth after 3 years?

Rx is invested at 8% p.a. for 6 years compounded annually. An amount of R1 586,87 was paid out after 6 years. Determine Rx.

REFLECTION




A=P(1+ni)

=8000(1+5. 9,5100

)

=R 11 800 aAnswer: c)

1.

i= r100

= 6100

=0,06

A=P¿∴ A=5 000 ¿¿ R 5 955,08

2.

i= r100

= 9,5100

=0,095

P= A¿¿

∴P=11800¿¿

¿ R 7 495,69

1. 18 monthly payments = 18 × R135,60 = R2 440,80 Total cost = R229 + R2 440,80 = R2 669,80

2. Saving = R2 669,80 – R2 298 = R371.80

3. SI=P ×r × n100

∴SI=200× 4× 1100

=R 8

Final amount = R200 + R8 = R208

4. SI = final amount – starting amount = R4 260 – R 3 000 = R1 260

i= r100

= 6100

=0,06

n=SIPi

∴i= 1260R 3000 ×0,06¿7 years


Compound Interest worksheetExample: Penny invested R2 000 at 5% per annum compound interest for a period of 3 years.Year Interest earned New amount1 5

100× R 2 000=R 100 R 100+R 2000=R 2100

2 5100

× R 2100=R 105 R 105+R 2100=R 2205

3 5100

× R 2 205=R 110,25 110,25+R 2 205=R 2 315,25

Remember that A = P(1 + ni) where A is total amount, P is principal (initial) amount; n is time in years and i=r

100 where r is

interest rate.Use the example above to do the following:

1. Maxin invests R14 500 at 12% interest rate compounded annually. Calculate the value of the investment for one year up to five years.

2. Temoso borrowed R500 from the bank for 3 years at 8% p.a. compound interest. Calculate how much he owes the bank at the end of three years.

3. Sam invests R5 000, compounded annually at a rate of 6% per year. Calculate the future value (Final amount) after 3 years.

4. Mr. Martin invests R12 750 for 3 years compounded annually at 5,3% p.a. How much is it worth after 3 years?


5. Calculate the compound interest on R120 at 10% p.a. for 2 years.Year 1: P = R2 000

Interest = 2000 × 5× 1

100=R 100

= R2 000 + R100 = R2 100


I=Pnr100 interest earned in year 1 value of investment

at end of year 1Year 2: P = R2 100

Interest = 2100 ×5×1

100=R 105

= R2 100 + R105 = R2 205



at end of year 2Year 3: P = R2 205

Interest = 2205 ×5× 1

100=R 110,25 = R2 205 +

R110,25 = R2 315,25



at end of year 3

6. Calculate the interest if R6 500 is invested for 3 years at 7,5% per annum compound interest.

7. Calculate what R10 000 will amount to if it is invested at 10% per annum compound interest for 3 years.

8. Use the formula A = P (1+ i) n to calculate the compound interest at 7% per annum on a loan of R 5 600 for 4 years. Round your answer to the nearest cents.

9. Bongiwe invests R12 000 in a savings account at 6,5% per annum compound interest. Calculate how much there will be in the savings account after 5 years.

10.Use the formula A=P(1+i)n or A=P(1+¿) to calculate the compound interest at 7% per annum on a loan of R 5 600 for 4 years. Round your answer to the nearest cents.



TOPIC: Integers – Properties of integers


By the end of the lesson, learners should know and be able to Perform calculations involving all four operations with integers. Use additive and multiplicative inverses operations for integers

RESOURCES DBE workbook ( Page 12), Sasol-Inzalo workbook (Page 30 -38), textbook, ruler, pencil, eraser, calculators, DVDS( Disc 1 : GDE 16 01 2014, GDE 28 01 2014, GDE 30 01 2014 & GDE 13 02 2014)

PRIOR KNOWLEDGE Basic Mathematics operations



3 min Simplify without using a calculator:

(1) 3(3 + 2)(2) 3 – (5 + 11) + 13(3) (3 + 4) + 6 = 3 + (….. +……)(4) -5 × 3 = 3 × ……

p.121


5 min 1. 18 monthly payments = 18 × R135,60 = R2 440,80 Total cost = R229 + R2 440,80 = R2 669,80

2. Saving = R2 669,80 – R2 298 = R371.80

LESSON 10 min Learners must explain what the three properties are and how they In Grade 9 learners


PRESENTATION/DEVELOPMENT

differ with examples. Under which operations are they true and false?

Commutative propertyThis law applies to addition and multiplication of numbers; it tells us that even if we change the order of the numbers we still get the same answer.

Examples: Simplify without using a calculator:

(1) 4 + (-2) = (-2) + 4 (2) (- 5) × 3 = 3 × (- 5) (3) a + b = b + a

Note: Subtraction and division is not commutative

Associative propertyThis rule also applies to addition and multiplication; it allows us to group numbers when adding or multiplying and still get the same answer.Examples:

(1) 3 + (-5) + 6 = [3 + (-5)] + 6 = 3 + [(-5) + 6] (2) 3 × (-5) × 6 = [ 3 × (-5)] × 6 = 3 × [(-5) × 6] (3) (a + b) + c = a + (b + c)

Note: Subtraction and division is not associative

Distributive propertyWhen multiplying across addition or subtraction this property allows us to redistribute the numbers and still get the same answer.Examples:

consolidate number knowledge and calculation techniques for integers, developed in Grade 8.

In Grade 9, learners work with integers mostly as coefficients in algebraic expressions and equations. They are expected to be competent in performing all four operations with integers and using the properties of integers appropriately where necessary.


(1) 7(-5 + 2) = [7 × (-5)] + [7 × 2] (2) -4[(-2) – (-5)] = [(-4) × (-2)] – [(-4) × (-5)] (3) a(b + c) = ab + ac

Learners must explain what additive inverse means and examples where it is applicable. They must also explain the order of operations, namely: BODMAS Additive inverse: Is the number that is added to it to get 0

Examples: (1) 6 and -6, 6 + (-6) = 0 (2) –a and a, a + (-a) = 0 (3) 4b and -4b, 4b + (-4b) = 0

Order of operation (BODMAS):B – Brackets, o – other(exponents and surds), D – Division, M – multiplication, A – addition,S – Subtraction Example: Simplify without using a calculator: (1) 2 + 7 × 5 = 2 + 35 = 37 (2) 72 ÷ 9 + 7 = 8 + 7 = 15 (3) 20 ÷ (4 – {10 – 5}) = 20 ÷ (4 - 5) = 20 ÷ -1 = -20

CLASSWORK 10 min Educator instructs learners to do the 4 examples individually with feedback(1) Rearrange and calculate after filling in the missing numbers: (-16) × 7 = … × … = …

What property did you apply?


(2) Calculate by grouping numbers and filling in the missing numbers: -21 + (-11) + (-20) = (-21 + … ) + … = … + … = …. What property did you apply?

(3) Calculate by filling in the missing numbers: -5[10 + (-3)] = [-5 × 10] + [-5 × … ] = … + … = …. What property did you apply?

CONSOLIDATION/CONCLUSION AND OR HOMEWORK

3 min Simplify without using a calculator:(1) What is the additive inverse of 999 and -2(2) (6 + 25 – 7 ) ÷ 6(3) Check whether the answers In each pair are equal:(a) – 3 + (5 – 11) and (- 3 + 5) -11(b) – 25 + 72 and 72 – 25(4) Simplify by using the distributive law:(a) – 3(-11 + 5)(b) 4 (−6 x−4)

REFLECTION




(1) 3(3 + 2) = 3(5) = 15(2) 3 – (5 + 11) + 13 = 3 – 16 +

13 = 0(3) (3 + 4) + 6 = 3 + (4 + 6)(4) -5 × 3 = 3 × -5

(1) Rearrange and calculate after filling in the missing numbers: (-16) × 7 = … × … = …. 7 × (-16) = -112

What property did you apply? Commutative property

(2) Calculate by grouping numbers and filling in the missing numbers: -21 + (-11) + (-20) = (-21 + … ) + … = … + … = ….-21 + (-11)] + (-20) = -21 + [-11 + (-20)] = -52

What property did you apply? Associative property

(3) Calculate by filling in the missing numbers:-5[10 + (-3)] = [-5 × 10] + [-5 × … ] = … + … =-50 + 15 = - 35

What property did you apply? Distributive property

(1) (5 + 16) ÷ 7 – 2 = 21 ÷ 7 – 2 = 3 – 2 = 1

Simplify without using a calculator:

1. For 999 it is -999 and for – 2 it is + 2

2. (6 + 25 – 7 ) ÷ 6 = 24 ÷ 6 = 43.a) – 3 + (5 – 11) and (- 3 + 5) -11

yes the answer is – 9 in both cases

(b) – 25 + 72 and 72 – 25 yes because – 25 + 72 = 47 and 72 – 25 = 44.

a) – 3(-11 + 5) = - 3 × - 11 + - 3 × 5 = 33 +(-15)= 18b) 4 (−6 x−4 ) ¿−24 x−16



TOPIC: Integers


By the end of the lesson, learners should know and be able to Perform calculations involving all four operations with numbers that involve the squares, cubes, square roots and cube

roots of integers

RESOURCES DBE workbook (Page 70 -77), Sasol-Inzalo workbook, textbook (Page 25 – 38), ruler, pencil, eraser, calculators, DVDS( Disc 1 : GDE 16 01 2014, GDE 28 01 2014, GDE 30 01 2014 & GDE 13 02 2014)

PRIOR KNOWLEDGESquares and square roots,Exponents



3 min (1) Simplify without using a calculator:(a) (-3)2

(b) (-3)3

(c) –(3)2

(d) 53 × 2 ÷ (2 ×5)(2) What does BODMAS stand for?(3) Apply it in the following example: ( 3 + 12) – 5 × 2 ÷ 5

p.121


5 min Simplify without a calculator:(a) -9 – (21 – 4) = - 26(b) 3 – (-9) = 12(c) 80 – 0 = 18


(d) 4 × - 5 = - 20(e) 30 ÷ -5 = - 6


10 min Educator asks learners what squares and square roots are? Learners must provide examples of each. Squares and square roots: Squares: A square number is the result of multiplying a

number by itself.Example1 5 × 5 = 52 = 25

Perfect squares: Numbers like 1; 4; 9; 16; … because 1 = 12 ; 4 = 22; 9 = 32 ; 16 = 42; …

Example 2 Simplify without using a calculator:(1) (7−4 )2 = (3)2=9 and not 72−42=49−16=33(2) x2+ x2=2x2∧not x 4

Square roots: Finding the square roots and finding the square of a number are inverse operations.Example: √16=4 because 4 × 4 = 16.This means that √16=√(4)2=4when finding the square root of monomial expression, find the square root of the coefficient and divide the exponent by 2.

Example 1 Simplify without using a calculator:

(1) √16+9 = √25 = 5 and not √16+ √9 = 4 + 3 = 7(2) √90−9=√81=9

Whole numbersConcepts and

skills:Solving problems






Example 2

√25 x6=5 x3

Educator instructs learners to do the 4 examples in their respective groups. Cubes and cube roots: A cube number is the result of

multiplying a number by itself three times:Example 1 4 ×4 × 4 = 43 = 64

Perfect cubes: Numbers like 1; 8; 27; 64; …,Because 13 = 1; 23 = 8; 33 = 27; 43 = 64; …Example 2 Simplify without using a calculator:

(1) (7−4 )3 = (3)3=27∧not 73−43=343−64=279 (2) x3+ x3=2x3∧not x6

Educator asks learners what cube and cube roots are and examples of each. Cube roots: A cube root is the opposite of a cube num-

ber. Cube root of a negative number gives a negative number: Example: Simplify without using a calculator: (1) 3√18+9+ 3√8=3√27+ 3√8=3+2=5

(2) 3√−8=−2

(3)

3√27 x6

¿3 x2


4 44

When finding the cube roots of monomial expressions, find the cube root of the coefficient and divined the exponent by 3.

Educator instructs learners to do the 3 examples in their respective groups with feedback late

CLASSWORK 10 min 1. Simplify without using a calculator:(1) (4+6)2

(2). √132+12+√25(3) √0(4) 2. Simplify without using a calculator:

(1) 33+ 3√−27 × 2 (2) 3√37−10 −√100

(3) 3√−64 y12 z15


3 min Simplify without using a calculator:(1) –(6)2

(2) (-5)3

(3) 3√27+(−3√64)(4) √81−√16−12× 3√125

(5) √4 x 4−3√27 x6+(3 x )2

REFLECTION




1.a) (-3)2 = 9b) (-3)3 = -27c) –(3)2 = -9d) 53 × 2 ÷ (2 ×5) = 25

2. Brackets other Division Multiplication Addition Subtraction

3. ( 3 + 12) – 5 × 2 ÷ 5 = 15 – 5 × 2 ÷ 5

= 15 – 5 × 25

= 15 – 2 = 13

(1) (4+6)2=(10)2=100(2) √132+12+√25=√144+√25=12+5=17(3) √0=0(4)

=2 x2−3 x2+9 x2

¿8 x2

Simplify without using a calculator: (1) 33+ 3√−27 × 2 = 27 – 3 × 2 = 21

(2) 3√37−10 −√100 = 3√27 −√100 = 3 – 10 = -7

(3) 3√−64 y12 z15

=−4 y 4 z5

Simplify without a calculator:a) -9 – (21 – 4) = - 26b) 3 – (-9) = 12c) 80 – 0 = 18d) 4 × - 5 = - 20e) 30 ÷ -5 = - 6



TOPIC: Integers – Problem Solving


By the end of the lesson, learners should know and be able to Perform multiplication and division with integers

RESOURCES DBE workbook (Page 70 – 77), Sasol-Inzalo workbook (Page 25 – 38), textbook, ruler, pencil, eraser, calculators, DVDS( Disc 1 : GDE 16 01 2014, GDE 28 01 2014, GDE 30 01 2014 & GDE 13 02 2014)

PRIOR KNOWLEDGE Basic Mathematics operations



3 min Write the following words into symbols:(1) Sum(2) Difference(3) Product(4) Quotient(5) Multiply(6) Add

p.122


5 min Simplify without using a calculator:1. For 999 it is -999 and for – 2 it is + 22. (6 + 25 – 7 ) ÷ 6 = 24 ÷ 6 = 43.

(a) – 3 + (5 – 11) and (- 3 + 5) -11 yes the answer is – 9 in both cases

(b) – 25 + 72 and 72 – 25 yes because – 25 + 72 = 47


and 72 – 25 = 44.

(a) – 3(-11 + 5) = - 3 × - 11 + - 3 × 5 = 33 +(-15)= 18

(b) 4 (−6 x−4 ) ¿−24 x−16

LESSONPRESENTATION/

DEVELOPMENT

10 min Educator asks learners what steps should be followed when dealing with problem solving. Problem Solving:

Steps to follow: Step 1: Identify what you have been asked to solve. Step 2: Identify what you have been given. Step 3: Write a number sentence: Step4: Substitute your values and solve/calculate. Step 5: Write your answer with the correct SI units.

Educator asks learners to interpret the given table and must answer the questions that follow. Example with temperature: The minimum and maximum

temperatures in Sutherland , Western Cape are shown in the table:

January August OctoberMin 0C 120 -160 50

Max 0C 290 00 220

Solving problems in context involving multiple operations with integers


(a) What was the coldest temperature in Sutherland? -16 0C in August(b) What was the warmest temperature in Sutherland?29 0C in January(c) What is the difference between the two temperatures? 29 – (-16) = 45 0C(d) What is the difference between the coldest in January and the

coldest in August? 12 – (-16) = 28 0CEducator gives learners the following two examples to interpret and to

solve. Example with money: Donny has R49,64 in his wallet. He has to

pay R100 for transport to work Write the equation and calculate how much money he still needs.

Money needed = money for transport – money in wallet = R100 – R49,64 = R50,36

Example with area: A builder builds a hotel. There are 65 bedrooms. Each room is a square and has walls 5 m long.(a) What is the area of one room?

Area = length × length A = l2 A = (5 m)2 A = 25 m2

(b) What is the area of all the rooms? Area of all the rooms = total number of rooms × area of one rooms Total area = 65 × 25 m2 = 1 625 m2

CLASSWORK 10 min Educator instructs learners to do the 3 examples in their respective


groups with feedback later. (1) If Anton loses weight at an average of 2,5 kg per month, how much

would he lose in 7 months?

(2) Four students rent a house and share expenses whilst they are at university. Their expense account for November has a balance of –R588. How much does each student owe?

(3) Mrs Dlamini runs a spaza. She pays R14 per square metre per month for rent. The store has an area of 58 square metres. What rent does she pay per year?


3 min (2) The weather forecast for the world on TV shows the following temperatures:

Minimum MaximumNew York -4 0C 15 0CMoscow -10 0C -1 0CLondon -3 0C 14 0C(a) What is the difference between the minimum temperatures in

New York and the minimum in Moscow?(b) What is the rise in temperature from minimum to maximum in

New York?(c) What is the rise in temperature from minimum to maximum in

Moscow?(d) What is the rise in temperature from minimum to maximum in

London?(3) At a restaurant where she is working during the holidays, Maya

earned the following tips in one week:


R95 MondayR88 WednesdayR398 FridayR243 Sunday

What is the average amount she earns in tips per day?(4) A chef cooks breakfast in a restaurant. He has four dozen eggs.

54 people order one egg each. Write down the equation for the following and calculate the answers:

(a) How many eggs does he have?(b) How many eggs will he have left over or be short?

REFLECTION




(1) Sum ( + )(2) Difference ( - )(3) Product ( × )(4) Quotient ( ÷ )(5) Multiply ( × )(6) Add ( + )

(1) Rearrange and calculate after filling in the missing numbers: (-16) × 7 = … × … = …. 7 × (-16) = -112

What property did you apply? Commutative property

(2) Calculate by grouping numbers and filling in the missing numbers: -21 + (-11) + (-20) = (-21 + … ) + … = … + … = ….

-21 + (-11)] + (-20) = -21 + [-11 + (-20)] = -52What property did you apply? Associative property

(3) Calculate by filling in the missing numbers:-5[10 + (-3)] = [-5 × 10] + [-5 × … ] = … + … =-50 + 15 = - 35What property did you apply? Distributive property

(5) (5 + 16) ÷ 7 – 2 = 21 ÷ 7 – 2 = 3 – 2 = 1

1.(a) - 40C – (- 100C) = 6 0C(b) 150C – (-40C) = 19 0C(c) -10C – ( -100C) = 9 0C(d) 140C – (-30C) = 17 0C

2. Average amount = Total divided by 4 days

Average amount = (95 + 88 + 398 + 243) ÷ 4

= 824 ÷ 4 = R2063. a) Four dozen eggs = 4 × 12

= 48 eggs b) Difference between

eggs ordered and eggs available

= 48 – 54 = - 6 thus short 6 eggs



TOPIC: Common Fractions - Calculations using fractions


By the end of the lesson, learners should know and be able to Calculations using fractions involving all four operations with common fractions and mixed numbers

RESOURCES DBE workbook ( Page 14-19,68-84), Sasol-Inzalo workbook ( 45- 56), textbook, ruler, pencil, eraser, calculators, DVDS( Disc 1 : GDE 04 02 2014, GDE 06 02 2014, GDE 11 02 2014)

PRIOR KNOWLEDGE Basic Mathematics operations, Multiplicative inverse, Additive inverses



3 min Simplify without a calculator:

(1)12+ 1

2

(2)34−1

4

(3)202

p. 122


5 min 1.(a) - 40C – (- 100C) = 6 0C(b) 150C – (-40C) = 19 0C(c) -10C – ( -100C) = 9 0C(d) 140C – (-30C) = 17 0C



Average amount = (95 + 88 + 398 + 243) ÷ 4 = 824 ÷ 4 = R2063. a) Four dozen eggs = 4 × 12 = 48 eggs b) Difference between eggs ordered and eggs available = 48 – 54 = - 6 thus short 6 eggs

LESSONPRESENTATION/

DEVELOPMENT

10 min Learners must explain what they think common fractions are, the different types of fractions and the components of common fractions

Common fractions: A fraction is a part of a whole, for example if a pie is cut half; each part is half of one whole pie.

The denominator is the bottom number of a fraction. It tells us how many equal parts something can be divided into.

The numerator is the top number of a fraction. It tells us how many equal parts we have.

Proper fraction: When the numerator is smaller than the







denominator, example 35

Improper fraction: When the numerator is bigger than the

denominator 1, example 134

Mixed number: Consist of a whole number and a fraction, example 314

Educator asks learners the meaning of equivalent and what are equivalent fractions with examples.

Equivalent forms: Fractions can have the same value, even though they may look different

Example 12 and

24 are equivalent, because they are both half of

the whole

12 =

24

If you multiply both numerator and denominator of 12 by 2 the

result is 24


=

If you divide both numerator and denominator of 24 by 2 the res-

ult is 12

× 2 × 2

1 =

2 =

42 4 8

× 2 × 2

Educator tests learners whether they know the rules involving addition and subtraction of fraction. Addition and subtraction of common fractions: When adding and subtracting fractions, you must have a common

denominator. To add and subtract fractions, all the fractions must be

expressed in the same unit.

Adding and subtracting fractions with the same denominator and different denominators:

Examples: Simplify without using a calculator:

(a) Same denominator: 14 +

14 =

24 =

12 and

34−1

4:


¿ 3−14

=24=1

2

(b) Different denominators: Calculate 35+3 1

6

Find the LCD (Lowest common denominator) of the denominators: LCD = 30

Convert the mixed number to an improper fraction: 316=19

6

Convert both fractions to equivalent fractions: 35

× 66+ 19

6× 5

5

Add the fractions: 1830

+ 9530 =

11330 ¿3 23

30

(CLASSWORK)

Educator tests learners whether they know the rules involving multiplication and division of fraction. Multiplication and division of common fractions:

Multiplication is repeated addition: 5 ×3=5+5+5=15. Division is splitting into equal parts, example”.

5×3 12=5+5+5+(half of 5 )=17 1

2 When you multiply: you must change the whole numbers into

fractions and mixed numbers into fractions, also leave the


answer in mixed numbers unless instructed otherwise Division is the opposite of multiplying: 3 × 4 = 12 and

123

=4∨124

=3

Example: Finding a fraction of a whole number involves multiplying a whole number by a fraction.

We multiply the whole number by the numerator and then divide

by the denominator, example: 57

of 9=57

× 91=45

7=6 3

7 When we multiply fractions we multiply the numerators and the

denominators. We do not need an LCD, example:

23

× 910

=1830

1830

÷ 66=3

5

¿22124

=9 524

CLASSWORK 10 min Educator instructs learners to do the 3 examples individually with feedback.Simplify without using a calculator

a)24+ 4

4

b) 45+1 7

10


c) 356−2 1

3CONSOLIDATION/CONCLUSION AND OR HOMEWORK


(1) 112−1

6+5

3

(2) (3 12+3 1

2 )÷ 1 34

(3) 337

×(4 23−12

5)

REFLECTION

Answers Week 4 Lesson 6Mental Maths Class work Homework

Simplify without a calculator:

(1 ) 12+1

2=2

2=1

(2) 34−1

4=2

4=1

2

(3) 1+ 35=5

5+ 3

5=8

5=1 3

5

(4) 12

×2=22=1

Simplify without using a calculator

(a)24+ 4

4=2+4

4

¿ 64=3

2

¿1 12

(b)45+1 7

10= 4

5+17

10

¿8+17

10=25

10

1.(e) - 40C – (- 100C) = 6 0C(f) 150C – (-40C) = 19 0C(g) -10C – ( -100C) = 9 0C(h) 140C – (-30C) = 17 0C


Average amount = (95 + 88 + 398 + 243) ÷ 4

= 824 ÷ 4 = R2063. a) Four dozen eggs = 4 × 12 = 48 eggs


¿52=2 1

2

(c) 3 56−2 1

3 ¿236

−73

¿236

−7 × 23 × 2

¿236

−146

¿96=3

2=1

b) Difference between eggs ordered and eggs available

= 48 – 54 = - 6 thus short 6 eggs


SUBJECT: MATHEMATICS GRADE 9 WEEK 4, LESSON 7 &8

TOPIC: Common Fractions - solving problems


By the end of the lesson, learners should know and be able to Solve problems in contexts involving common fractions, mixed numbers and percentages

RESOURCESDBE workbook ( Page 14-19,68-84), Sasol-Inzalo workbook ( 45- 56), textbook, ruler, pencil, eraser,

calculators, DVDS( Disc 1 : GDE 04 02 2014, GDE 06 02 2014, GDE 11 02 2014)



INTRODUCTION MENTAL MATHS

3 min Calculate:(1) 20% of R2 million

(2) 5

16+ 2

16− 3

16

(3)516

× 8

(4) 210

÷ 210

×1

p. 122


5 min 1. (a) 15x

= 320


∴3 x=15 ×20∴ x=15 ×203

=100

(b) x

144= 1

12

∴12 x=144 ×1∴ x=14412

=12

(c) 132100

=33x

∴132 x=100 ×33∴ x=100 ×33132

=25

2.Common fraction Decimal fraction Percentage %

25

0,4 40%

60100

=35

0,6 60%

23100

0,23 23%

LESSONPRESENTATION/

DEVELOPMENT

10 min Educator asks learners to write down the steps to follow when doing problem solving. Learners must also do the examples in their respective groups.

Problem Solving: Steps to follow: Step 1: Identify what you have been asked to solve. Step 2: Identify what you have been given. Step 3: Write a number sentence: Step4: Substitute your values and solve/calculate. Step 5: Write your answer with the correct SI units (unit in the




question, unless instructed otherwise).Step 6: Remember to conclude with the sentence at the end.

Word problems can be written as mathematical statements so that they can be solved mathematically.

Example: If John uses three quarters of his savings of R10 234 to buy a bicycle, you will be able to calculate the cost of the bicycle in rand.Solution:

Steps to follow

Explanation

Step 1 Determine the cost of the bicycle in randStep 2 Savings of R10 234 and the fraction

34

Step 3 & 4 Let the cost of the bicycle be x

∴ x=34

of R 10 234

Step 5 ∴ x=34

× R 10 234=R 7675,50

Step 6 The cost of the bicycle is R7675.50

Problems involving rectangles (measurement): Your kitchen is a

rectangle: The width is 34 of the length, which is 10½ metres.

Calculate the area of the kitchen.Area = length × width We need to find the width (w)A = 10½ × w





w = 34

of 10

w=¿ 34

× 212

=638

A=212

× 638

A=82 1116

m2

The area of the kitchen is 82.7m2

Problems involving time:Jerome, Donny and Bernard swam the 200 metre freestyle event in a gala. Jerome finished in a time of 3 minutes 40 seconds. Bernard swam 5 % faster than Jerome and Donny swam 10% slower than Jerome.

(a) Calculate the time that Bernard took for his race(b) Calculate the time that Donny took for his race(a) Bernard swam 5% faster

than 3 min 40 sec

¿ 5100

×3 min 40 sec

¿ 5100

×220 sec

¿11secondsBernard swam 11 sec faster than Jerome. So we need to subtract

Find 5% of 3 min 40 sec

Remember that 5% = 5

100

Convert minutes to seconds:Remember 1 min = 60 sec, so3 min 40 sec = 3 × 60 + 40 = 220 sec


the timeBernard’s time = 3 min 40 sec – 11 sec

= 3 min 29 sec

(b) Danny swam 10% slower than 3 min 40 sec

Donny’s time¿10

100×220 sec

¿22 secDonny swam slower than Jerome,

so we need toad the time.Donny’s time¿220 sec+22 sec ¿242 sec ¿4 min 2 sec

Find 10% of 3 min 40 sec

CLASSWORK 10 min (1) Find the area of a triangle with a base of 212 centimetres and a

height of 135 centimetres

(2) Peter paints a seventh of his house on Monday, a sixth on Tuesday and a quarter on Wednesday. What fraction of the house still needs to be painted?


3 min (1) Thato works 5 days per week for 7 hours per day at a rate of one and a quarter times of what Mary earns when she works overtime. Mary earns R15 per hour normal pay. Overtime is paid at 2½ times normal pay. How much does Thato earn per day?


(2) The Sulliman family has a holiday flat in Durban. The flat cost R3½ million. Maya owns a 3 eights, Hamzah owns a quarter and Lamees owns one sixteenth of the house. Pappa owns the rest.

(a) What fraction does Pappa own(b) What is the value of Pappas portion in rand?

REFLECTION



(1) 20% of R2 million

¿ 20100

×2000 000=R 400 000

(2 ) 516

+ 216

− 316 =

5+2−316

= 416

=14

(3) 5

16× 8=40

16=5

2

(4 ) 210

÷ 210

× 1= 210

× 102

× 1 = 1

1. Area of a triangle = half times the base times(x) the perpendicular height A = ½ b h

A = ½ ×2 12 cm×1 3

5cm

A=12

× 52

× 85=2 cm2

2. Sum of the painted parts = 17+ 1

6+ 1

4=1

7× 12

12+ 1

6× 14

14+ 1

4× 21

21=12

84+ 14

84+ 21

84=47

84 Remaining fraction to be painted

= 1−4784

=8484

−4784

=3784

1. (a) 15x

= 320

∴3 x=15 ×20∴ x=15 ×203

=100

(b) x

144= 1

12

∴12 x=144 ×1∴ x=14412

=12

(c) 132100

=33x

∴132 x=100 ×33∴ x=100 ×33132

=25

2.



TOPIC: Whole Numbers – Weekly Assessment

TERM 1 WEEK 2MAX: 20

QUESTION 1 - Choose the correct answer

1.1 Which number is missing?

12 ;

14 ;

18 ; … ;

132

(2)

A. 16

B. 164

C. 464

D. 216

1.2 Which statement is incorrect? (2)

A. The only prime factors of 24 are 2 and 3 .

B. The difference between - 8 and 3 is 11.

C. All integers are natural numbers.

D. 3√−27 is a rational number.


1.3 Which of the following numbers is a rational number? (2)

A. √3

B. √16

C. √−¿9¿

D. √13

1.4 A painter is paid by the hour. If he is paid R360 for 12 hours work. How much will he be paid for 9 hours work? (2)

A. R120

B. R180

C. R270

D. R480

[8]

QUESTION 2

2.1 Identify the following numbers (3)

2.2 You have paid R11 800 on a 5 year loan which was compounded yearly at 9,5% interest. What was your initial amount?

(3)


[12]

TOTAL : 20



QUESTION 1

1.1 C (2)

1.2 C (2)

1.3 B (2)

1.4 C (2)

[8]

QUESTION 2

2.1 Number N N0 Z Q Q '3√−27

√25

3,14 0,4444..

π

(9)

2.2 i= r100

= 9,5100

=0,095

P= A¿¿

∴P=11800¿¿

¿ R 7 495,69

(3)

[12]

TOTAL : 20



TOPIC: Common fractions- Solving problems ; squares; cubes; square roots ; cube roots


By the end of the lesson, learners should know and be able to Perform multiple operations with common fractions, using a calculator where appropriate. Perform multiple operations with or without brackets, with numbers that involve the squares, cubes, square roots and

cube roots of common fractions.





3 min Simplify without using a calculator;(1) 202

(2) (1 + 4)2

(3) 13

(4) √16+9(5) 3√27(6) √1(7) 3√−8

p. 122

REVIEW AND CORRECTION OF 5 min (1) Mary’s overtime earnings are R15 × 2½ =


HOMEWORK R37, 50 an hour.Thato earns 1¼ times Mary’s overtime earnings per hour = 1¼ ×R37, 50 = R46,88 per hourThato’s earnings per day =

R 46 78

× 7=R 328 18=R 328,13

(2) a) Fraction owned by Maya, Lamees and Hamzah

= 38+ 1

16+ 1

4=6+1+4

16=11

16

Fraction owned by Pappa:

1−1116

=1616

−1116

= 516

b)Value of Pappa’s portion ¿5/16 of R3½ million

= 5/16 ×7/2million

¿ R 1093750 million


10 min Educator asks learners what square and square roots are and they must give examples. Learners are asked to do the examples below in their respective groups.

Square and square roots:

Calculations using Fractions:

All four operations, with numbers that involves the squares, cubes,


To find the square of a fraction we: multiply it by itself. find the square of its numerator and

denominator separately.Example 1: Simplify without a calculator:

( 25 )

2

=25

× 25= 4

25∨22

52 =4

25

Example 2: Simplify without a calculator: (3 15 )

2

Convert 315 to an

improper fraction

Square the numerator and denominator separately

Square

Write as a mixed number

( 165 )

2

162

52

25625

10 625

To find the square root of a fraction we: write the fraction as a square and take the

base as the answer find the square root of the numerator and

denominators separately

square roots and cube roots of common fractions


Example 1: Simplify without a calculator:

√ 964

=√( 38 )

2

=38∨ √9

√64=3

8

Example 2: Simplify without a calculator: √2 14

Convert to an improper fraction

Find the square roots of 9 and 4 separately

Write as a mixed number

√ 94

√9√4

=32

1 12

Educator instructs learners to do the following 4 examples in pairs with feedback.

Educator asks learners what cube and cube roots are and they must give examples. Learners are asked to do the examples below in their respective groups. Cube and cube root: To find the cube of a

fraction we: Multiply it three times by itself Find the cube of its numerator and de-

nominator separately.Example 1: Simplify without a calculator:


( 25 )

3

=25

× 25

× 25= 8

125∨23

53 =8

125Example 2: Simplify without a calculator:

(−3 23 )

3

=(−113 )

3

=−113

33 =−1 33127

=−49 827

To find the cube root of a fraction we: Write the fraction as a cube and take the

base as the answer Find the cube root of the numerator and de-

nominator separatelyExample 1: Simplify without a calculator:

3√ 2764

=3√( 34 )

3

= 34

❑

or 3√ 2764

=3√273√64

=34

Example 2: Simplify without using a calculator: 3√15 5

8= 3√ 125

8=

3√1253√8

=52=2 1

2Educator instructs learners to do the 4 examples in

their respective groups with feedback.

CLASSWORK 10 min Activity 1Simplify without a calculator:

(a) ( 94 )

2

(b) (−2 23 )

2

(c) √ 144100

(d) √4 16

Activity 2


Simplify without a calculator:

(a) (−2−3 )

3

(b )(5 12 )

3

(c) 3√−864

(d )− 3√ 6425



(1) (−45

)2

(2) (2 23)

3

(3) (4 16 )

2

+(−2 23 )

3

(4)√ 19

×32

(5) 16 3√ 64−64

(6) √ 11 000 000

÷ 3√ 11 000 000

REFLECTION




(1) 202 = 400(2) (1 + 4)2 = (5)2 = 25(3) 13 = 1(4) √16+9=√25=5(5) 3√27=3(6) √1=1(7) 3√−8=−2

Activity 1

(a) 814

=20 14

(b) (−83 )

2

=649

=7 19

(c) √144√100

=1210

=1 15

(d) √ 256

=√25√6

=5√6

Activity 2

(a) ( 23 )

3

= 827

(b) ( 112 )

3

=13318

=166 38

(c)3√−83√64

=−24

=−12

(d) −3√64

3√25= −4

3√25

1.Mary’s overtime earnings are R15 × 2½ = R37, 50 an hour.Thato earns 1¼ times Mary’s overtime earnings per

hour = 1¼ ×R37, 50 = R46,88 per hour

Thato’s earnings per day = R 46 78

× 7=R 328 18=R 328,13

2. a) Fraction owned by Maya, Lamees and Hamzah

= 38+ 1

16+ 1

4=6+1+4

16=11

16

Fraction owned by Pappa: 1−1116

=1616

−1116

= 516

b)Value of Pappa’s portion ¿5/16 of R3½ million = 5/16 ×7/2million

¿ R 1 093750 million



TOPIC: Decimal fractions- Calculations using decimal fractions


By the end of the lesson, learners should know and be able to: Perform multiple operations with decimal fractions, using a calculator where appropriate.





3 min Simplify by first making a rough estimate of the answer before using a calculator:(1) 0,25 + 0,75(2) 0,5 – 0,24(3) 10,1 × 2,1

(4) 0,750,5








5 min(1) (−4

5)

2

¿1625

(2)(2 23)

3

=( 83)

3

=51227

=18 2627

(3)

( 256 )

2

+(−83 )

3

=62536

−51227

=1875−2048108

=−175108

=92 14

(4) √ 19

×32

¿√ 19

× 9=√1=1

(5) 16 3√ 64−64

¿16 3√−1=16 ×−1=−16

(6) √ 11000 000

÷ 3√ 11000 000

¿ 11000

÷ 1100

= 11000

× 1001

= 110


10 min Educator asks learners to explain what decimal fractions are as well as the place values of the different digits.

Decimal fractions: The word decimal really means “based on 10” or a tenth part. A decimal number usually means there is a decimal comma

(point).Example: The decimal fraction 45,6: Which is forty-five and six-

tenths. (45 610 )

Place value: When we write numbers, the position or place value


of each digit is important.Example: In the number 327:

The 7 is in the ones position, meaning 7 ones (seven)The 2 is in the tens position meaning 2 tens (Twenty)The 3 is in the hundreds position meaning 3 hundreds

Educator asks learners what rounding means and how it is done with examples.

Rounding off: “Rounding off” means making a number simpler but keeping its

value close to what it was. The result is less accurate but easier to use. To round off any decimal place, look at the next decimal place: If this number is less than 5 round down

Example: 6,419 ≈ 6,4 (to 1 decimal place) If this number is 5 or more than 5, round up.

Example: 6,5 ≈ 7 (to nearest whole number) Rounding to tenths means to leave one number after the decimal

comma. Example: 1,2735 ≈ 1,3 (to 1 decimal place)

Rounding to hundredths means to leave two numbers after the decimal comma. Example: 3,1416 ≈ 3,14 ( to 2 decimal places)

Estimation: To solve a problem, we start by estimating the answer. Rounding off to the nearest whole number.


Look at the digit after the comma. if this digit is: 5 or more, the digit showing units increases by 1; less than 5, the digit showing units stays the same

Example: 125,56 ≈ 126 and 125,46 ≈ 125Educator asks learners to the examples below without using a calculator. They must explain the importance of place values.

Addition and subtraction:When you add or subtract decimal numbers the answer must have the same number of decimal places;Example 1: 25,792 + 7,195Estimated answer: 26 + 7 = 33

10s Units

, 10ths

100ths

1000ths

2 5 , 7 9 27 , 1 9 5

3 2 , 9 8 7

Example 2: 72,6 – 18,73Estimated answer: 73 – 19 = 54

10s Units

, 10ths

100ths

1000ths

7 2 , 6 01 8 , 7 35 3 , 8 7


Educator asks learners to the examples below without using a calculator initially and then. Calculator can then be used to calculate the actual answer. They must explain the importance of place values.

Multiplication and division:(8) Example: Calculate and round off to two decimal places:

6,23× 9,8919,5

Estimate the answer: Round each number up or down to the nearest whole number

Use a calculator to work out the actual answer.

6,23× 9,8919,5

≈ 6 ×1020

≈ 3

6,23× 9,8919,5

=3,1597 …=3,16

CLASSWORK 10 min Activity 1Simplify by first estimating the answer:

(1) 27,8 + 14,3(2) 61,7 – 33, 8(3) Peter sells meat. On Monday he sells 28,6 kg; on Tuesday he sells 33,38 kg ; on

Wednesday he sells 22,75 kg. Calculate the total mass.Activity 2Simplify by first estimating the answer than calculate the answer accurately to two decimal places by using a calculator:

(1) 4,98 × 6,12

(2)5,799× 3,1

8,86CONSOLIDATION/CONCLUSIONAND OR HOMEWORK

3 min Simplify by first estimating the answer before using a calculator for the accurate answer to 2 decimal places:(1) 17,89 + 21,91(2) 34,8183 – 9,8


(3)32,91× 4,8

3,1REFLECTION


Answers: Week 5 Lesson 2


(1) 0,25 + 0,75Estimation: ≈0 + 1 = 1 Answer: 1

(2) 0,5 – 0,24Estimation: ≈ 1 – 0 = 1

Answer: 0,26(3) 10,1 × 2,1

Estimation: ≈ 10 × 2 = 20 Answer: 21,21

(4)0,750,5

Estimation:≈ 11=1

Answer: 1,5

1. 27,8 + 14,3 ≈28 + 14 = 42 =42,12.61,7 – 33, 8 ≈ 62 – 34 = 28 = 27,9

2.28,6 + 33,38 + 22,75 = 84,73 kg≈29 + 33 + 23 = 85

Activity 21.4,98 × 6,12 ≈5 × 6 = 30

And 4,98 × 6,12 = 30,4776 = 30,48

3. 5,799× 3,1

8,86

≈ 6 ×39

=2

(1) (−45

)2

¿1625

(2)(2 23)

3

=( 83)

3

=51227

=18 2627

(3)

( 256 )

2

+(−83 )

3

=62536

−51227

=1875−2048108

=−175108

=92 14

(4) √ 19

×32

¿√ 19

× 9

¿√1¿1

(5) 16 3√ 64−64

¿16 3√−1¿16×−1¿−16

(6) √ 11 000 000

÷ 3√ 11 000000

¿ 11000

÷ 1100

¿ 11000

× 1001


¿ 110


TOPIC: Decimal fractions- Calculations using decimal fractions


By the end of the lesson, learners should know and be able to Perform multiple operations with or without brackets, with numbers that involve the squares, cubes, square roots and

cube roots of decimal fractions

RESOURCES DBE workbook ( Page 14-19,68-84), Sasol-Inzalo workbook 1 ( 45- 56), textbook, ruler, pencil, eraser, calculators, DVDS( Disc 1 : GDE 04 02 2014, GDE 06 02 2014, GDE 11 02 2014)

PRIOR KNOWLEDGE Basic Mathematics operations, Multiplicative inverse, Additive inverses Common fractions



3 min Simplify without using a calculator:(1) √36(2) (−2)2

(3) (−4 )4 × (−3 )3

(4) 3√−8(5) √9+16−3√18+9

p.124-126



5 min (1) 17,89 + 21,91Solution: ≈18 + 22 = 40 and 17,89 + 21,91 = 39,8

(2) 34,8183 – 9,8Solution: ≈ 35 – 10 = 25 and 34,8183 – 9,8 = 25,0183 = 25,02

(3) 32,91× 4,8

3,1

(4) Solution: ≈33× 5

3=55∧32,91 ×4,8

3,1=50,9574 …=50,96

LESSONPRESENTATION/

DEVELOPMENT

10 min Educator asks learners what square numbers and square root are and that they must provide a few examples. What are the rules that are applicable when dealing with multiplying decimal squares and square roots

A square number has two like factors

Example: 25 = 5 × 5 = 52 and ¿ 32

42=9

16 Squares of whole numbers 1, 2, 3… are called perfect

squares. To find the square root of a number means to find the factor

that was squared Example: √36=√6× 6=6 and

√ 364

=√ 62

22 =62=3

We can apply the same knowledge to finding the square and square root of decimals

Calculations using Decimal fractions:

All four operations, with numbers that involves the squares, cubes, square roots and cube roots of decimal fractions


Product Rule2 × 2 = 4,0 In the product there is no decimal places

after the comma0,2 × 0,2 = 0, 04 In the product there are two decimal

places after the comma0,02 × 0,02 = 0,0004 In the product there are four decimal

places after the comma0.002 × 0,002 = 0,000004

In the product there are 6 decimal places after the comma

Example: 42 = 4 × 4 = 16 so (0,4)2 = (0,4) × (0,4) = 0,16Example: √36=6 so √0,36=√0,6× 0,6=0,6

Educator instructs learners to do the 4 examples in their respective groups with feedback.(Activity 1)

Educator asks learners what cube numbers and cube root are and that they must provide a few examples. What are the rules that are applicable when dealing with multiplying decimal cubes and cube roots

A cube number has three like factors,Example: 73 = 7 ×7 ×7 = 343 and (0,7)3 = (0,7) × (0,7) × (0,7) = 0,342 (Remember the rule of counting the decimal places after the comma in multiplication of decimal fractions)

To find the cube root of a number means to find the factor that was cubed


Example: 3√8= 3√2×2×2=3 and 3√0,008=¿ 3√0,2× 0,2 ×0,2=0,2CLASSWORK 10 min Activity 1

Calculate without using a calculator:(a) (0,1)2

(b) (1,2)2

(c) √0,16(d) √1,44

Activity 2Calculate without using a calculator:

(a) 3√1(b) (-0,4)3

(c) 3√−0,064


3 min Simplify without using a calculator:(1) (1,1)2 - √0,81(2) (0,5)3 + (0,5)2

(3) 3√0,000125(4) √1,44 ×0,36(5) 3√−0,027 ×0,008(6) ¿0,4)3

(7) √0,0001REFLECTION



(1 ) √36 = √6 ×6=¿6(2 )(−2)2 =(−2)×(−2) ¿4(3 ) (−4 )4 × (−3 )3

¿ (−4 ) × (−4 ) × (−4 )× (−4 )× (−3 ×−3×−3 )=256 ×−27¿−6 912(4 ) 3√−8 =3√−2 ×−2×−2=¿ -2(5 ) √9+16−3√18+9¿√25− 3√27

¿√5× 5− 3√3 ×3 × 3¿5−3=2

Activity 1(a) (0,1)2 = 0,01(b) (1,2)2 = 1,44(c) √0,16 = 0,4(d) √1,44 =1.2

Activity 2

(a) 3√1 = 1(b) (-0,4)3 = -0,064(c) 3√−0,064 = -0,4

1. 17,89 + 21,91Solution: ≈18 + 22 = 40 and 17,89 +

21,91 = 39,82. 34,8183 – 9,8Solution: ≈ 35 – 10 = 25 and

34,8183 – 9,8 = 25,0183 = 25,02

3. 32,91× 4,8

3,1

Solution: ≈33× 5

3

¿ 55∧32,91 ×4,83,1

¿50,9574 …=50,96



TOPIC: Decimal fractions- Equivalent forms


By the end of the lesson, learners should know and be able to Convert common fraction and decimal fraction forms of the same number Convert common fraction, decimal fraction and percentage forms of the same number.


PRIOR KNOWLEDGE Basic Mathematics operations, Multiplicative inverse, Additive inverses Common fractions



3 min (1) Convert 12 and

14 into:

(a) A common fraction of hundredths(b) A percentage(c) A decimal fraction

p.124-126


5 min 1. Calculate without using a calculator(a) (0,1)2 = 0,01(b) (1,2)2 = 1,44(c) √0,16 = 0,4


(d) √1,44 1.2

2. Calculate without using a calculator:(a) 3√1 = 1(b) (-0,4)3 = -0,064(c) 3√−0,064 = -0,4


10 min Educator asks learners what common and decimal fractions are and to list examples of each.

Educator also asks learners what equivalent forms of fractions mean and to list examples.

A common fraction can be changed into an equivalent common fraction with a different denominator which is a multiple of the original denominator.

Example: Write three equivalent fractions of 38 .

38

× 22= 6

16; 3

8× x

x=3 x

8 x; 38

× 55=15

40 Common fractions can be converted to decimal fractions in the

form of tenths, hundredths, thousandths, etc.

Educator asks learners to convert common fractions into decimal fractions

Example: Change 310

;

12100

∧5

1000 into a decimal fraction without a

calculator.

Convert common fraction and decimal fraction forms of the same number

Convert common fraction, decimal fraction and percentage forms of the same number.

Equivalent formsCommon fractionsDecimal fractionsPercentage


310

=0,3 ; 12100

=0 , 12; 5

1000=0,005

Educator asks learners what percentages are and how percentages can be expressed as common and decimal fractions. They must also provide examples

Common fractions, decimal fractions and percentages represent the same numbers.

Percentages means “out of a hundred”.Example: Express 50% as a common and decimal fraction.

50 %= 50100

=12=0,5

CLASSWORK 10 min Educator assigns the following exercise to be completed in groups and feedback must be given from the groups.

(1) Express:

(a)12 as tenths

(b)75100 as quarters

(2) Change 17

100 into decimal fraction.

(3) Complete the following table:

Common fraction Decimal fraction Percentage %0,75


25%

24

CONSOLIDATION/CONCLUSION

AND OR HOMEWORK

3 min (1) Express:

(a) 14 as hundredths

(b) 25100 as 200 hundredths

(2) Change 170

1000 into decimal fraction.

(3) Complete the following table:

REFLECTION


Common fraction Decimal fraction Percentage %0,25

75%

13


Convert 12 and

14 into:

a) 12

× 5050

=

50100

∧1

4× 25

25= 25

100

b) 12=50 %∧1

4=25 %

c) 12=0∧1

4=0,25

1.

a) 12

× 55= 5

10= 75

100

b)75÷ 25

100÷ 25=3

4

2.17

100=0,17

3. Complete the following table:

Common fraction

Decimal fraction

Percentage %

75100

=34

0,75 75%

25100

=14

0,25 25%

24=1

20,5 50%

1.Calculate without using a calculator:a) (0,1)2 = 0,01b) (1,2)2 = 1,44c) √0,16 = 0,4d) √1,44 1.2

2. Calculate without using a calculator:

a) 3√1 = 1b) (-0,4)3 = -0,064c) 3√−0,064 = -0,4



TOPIC: Decimal fractions- Equivalent forms


By the end of the lesson, learners should know and be able to Solve problems in context involving decimal fractions.


PRIOR KNOWLEDGE



3 min Simplify, using a calculator, by rounding and estimation first:(a) 75,435 + 111,595(b) 17 857 – 50 – 0,88(c) 11,02 × 131,29(d) 17 857,15375 ÷ 89,23

p.124-126


5min 1. a) 14

× 2525

= 25100

b) 25

100× 2

2= 50

200

2. 170

1000=0,17

3.Complete the following table:


Common fraction Decimal fraction Percentage %14

0,25 25%

34

0,75 75%

13

0,3333.. 33 13

%


10 min Educator asks learners the use of decimal fractions in everyday life. They must try to provide examples:

Word problems with decimals:

When measuring buildings you may have a room with length of 4,21 metres and a width of 3,21 metres. Decimals are also used in finance, for example dealing with money like R23,45. More examples are with time.

Educator asks learners what steps must be followed when solving problems involving decimal fractions.

Steps to be followed in problem solving:Step 1: Identify what you have been asked to solve. Step 2: Identify what you have been given. Step 3: Write a number sentence: Step4: Substitute your values and solve/calculate. Step 5: Write your answer with the correct SI units and be sure to round off as requested and avoid premature rounding.

Word problems can be written as mathematical statements so

Solve problems in context involving decimal fractions

Common fraction and decimal fraction and percentage forms of the same number


that they can be solved mathematically.Educator asks learners to discuss the following examples in their different groups or pairs.

Example 1: As a beginning writer, Thando earns 40 cents a word.(a) If she submits an article with 1 533 words, how much will she be

paid in rand?(b) If she contracted to write a 90 thousand word book but has

written only 32 310 words, what percentage has she completed?Solution: (a) 1 533 × 0,40 = R613,20

(b) 32 31090 000

×100 %=35,9 %

Example 2: An empty bottle weighs 55,505 grams and when it is full of cooldrink it weighs 1 kilogram. Calculate the mass of the cooldrink.Solution: Convert: 1 kg = 1 000 gMass of cooldrink: 1 000g – 55,505 g = 944,495 g

CLASSWORK 10 min (1) Determine the area of a triangle with height of 45,525 cm and a base of 7,897 cm. Round off to the second decimal.

(2) A cartographer used a scale of 1 mm for 10 km. On the map the distance between two towns is 23,45 cm. Calculate the actual distance between the towns.


AND OR HOMEWORK

3 min (1) They dig a path for a walkway 30,75 m long, 0,6 m wide and 5 cm deep. How many cubic metres of concrete must they or-der? Give your answer to the nearest amount of concrete to be ordered.


(2) The electricity charge for Mrs Ruka for the month was R417,59. The first 10 kWh are free. For the next 100kWh the charge is R1,13 per kWh, and thereafter for each kWh the charge is R1,42.

(3) How much electricity did the Ruka household use?REFLECTION



(a) 75,435 + 111,595Solution: Rounding: ≈ 75 + 112 = 186.

Accurately: = 187,03(b) 17 857 – 50 – 0,88

Solution: Rounding: ≈ 18 000 – 50 – 1 = 17 949; Accurately: = 17 806,12(c) 11,02 × 131,29

Solution: Rounding: ≈ 11 × 131 = 1 441. Accurately: = 1 446,81(d) 17 857,15375 ÷ 89,23

Solution: Rounding: ≈ 17 857 ÷ 89 = 200,64; Accurately: = 200,13

1.Area = ½ base × height = 0,5 × 7,897 × 45,525 = 179,7554625 ≈ 179,76 cm2

2.Convert: 23,45 cm = 234,5 mm Thus if 1 mm on the map is equal 10 km on land then, Distance between the two towns:234,5 × 10 = 2345 km

1. They dig a path for a walkway 30,75 m long, 0,6 m wide and 5 cm deep. How many cubic metres of concrete must they order? Give your answer to the nearest amount of concrete to be ordered.

2. The electricity charge for Mrs Ruka for the month was R417,59. The first 10 kWh are free. For the next 100kWh the charge is R1,13 per kWh, and thereafter for each kWh the charge is R1,42.

3. How much electricity did the Ruka household use?



TOPIC: Exponents - Comparing and representing numbers in exponential form


By the end of the lesson, learners should know and be able to Compare and represent integers in exponential form

RESOURCES DBE workbook (p.8-11, p.88-109), Sasol-Inzalo Workbook (p.73-79), ruler, pencil, eraser, calculators, DVD( Grade 9 Term 1 Disc 2 GDE 13 03 2014)

PRIOR KNOWLEDGE Basic Mathematics operations, Calculations with integers,



3 min Simplify without using a calculator:1. 24

2. 22

3. 24

22

4. 33 ÷ 33

5. 42× 22

6. 23+22

p.124


4min 1. Convert: 5 cm = 0,05 m Volume of a cuboid = l × b × h = 30,75 × 0,6 × 0,05 = 0,9225 m3 ≈ 1 m3

2. Cost of 100 kWh electricity @ R1,13: (100×1,13)=R 113,00 Amount of electricity @ R1,42/kWh: ( R417,59 – R113,00) ÷ 1,42 = 214,5 kWh


∴ Electricity used: 10 kWh(Free) + 100kWh(first 100kWh @R1,13) +

214,5 kWh = 324,5 kWh


10 min Educator asks learners to identify the different components of the term given below and to give their own examples.

Compare and represent Whole numbers (N0)in exponential form:

a (base): the number which is multiplied by itself or raised to a power.b (index): the number of times the base is multiplied by itself or the power to which the base is raised.2 (coefficient): The value in front of the base

Educator asks learners to write 27 as a product of its prime factors and exponential form.

A number can be written in as a product of its prime factors, e.g.27 = 3 × 3 × 3 = 33

Or you can write in exponential form, e.g.

Representing Whole numbers and integers in exponential form.

Calculating exponents, not using the rules.


51 = 5 ; any number to the power of 1, is itself50 = 1 ; any number to the power of 0, is 152 = 5 × 5 = 25

Educator asks learners to compere two exponents 43 and 34, etc.

If the basis are the same, the higher the power, the larger the number, e.g. 24>22

If the basis are not the same, we calculate the value of the number of powers, and then compare them, e.g. 43 and 34. 43 = 64 and 34 = 81, thus ∴ 34>¿ 43 .Educator asks learners to express negative numbers to a power and they must give examples.

Compare and represent Integers (Ζ) in exponential form:

A negative number raised to a power must be inserted in brackets, e.g.

(−52 ¿ = (-5) × (-5) = 25, as appose to –(5¿¿2 = -( 5 ×5 ) = -25, where the base is positive.

When a negative integer is raised to an even power the answer is positive, e.g.

(−2¿¿4 = −2 ×−2×−2×−2 = +16

When a negative integer is raised to an odd power, the answer is negative, e.g.

(-35 ¿ = (-3) × (-3) × (-3) × (-3) = -243


Educator asks learners how they will go about solving the following problems:

Power of a power:Example: (2¿¿2)3=(22 ) × (22 ) × ( 22)=(26 )=64¿ ¿ (2 ×2 ) × (2 ×2 ) × (2×2 )=4 × 4 × 4=64

Power of a product:Example: (5×3)3=153=3375 ¿53 ×33=5× 5× 5×3×3×3 ∴125 ×27=3375

Power of a quotient:

Example: ( 23 )

3

=(23

× 23

× 23 ) ¿ 8

27

= 23

33 =8

27

CLASSWORK 10min Activity 1

1. Write the following numbers in exponential form:1.1 1 × 1 ×1 × 1 × 11.2 6 × 6 × 6 × 6 × 6 × 6

2. Write the following in expanded form:2.1 15

2.2 53

2.3 77

3.Arrange in ascending order: 33 ;62 ;50;161; 45


4.Compare the following integers and insert, ¿ ;<¿=¿

4.1 (-675 ¿¿0 …… (-4 ¿¿6

4.2 (-5) × (-5) × (-5) ….. -125 4.3 −40 ……….¿


AND OR HOMEWORK

3 min DBE workbook 1 Page 54 No2

REFLECTION



(1) 24 = 16 and 22 = 4

(2) 24

22 =164

=4=22

(3 ) 33 ÷33=30=1(4) 42× 22=24 × 22=26=64(5) 23+22=8+4=12

Activity 11. 1.1 15

1.2 66

2.2.1 1 × 1 × 1 × 1 × 12.2 5 ×5×52.3 7 × 7 × 7 × 7 × 7 × 7 × 73. 33=27 ;62=36 ;50=1;161=16 ;45=1024 1 ; 16 ; 27 ; 36 ; 1 024 ∴50;161 ;33 ;62; 45

4.4.1 (-675 ¿¿0 …… (-4 ¿¿6

1 …. 4 096∴¿ < (-4 ¿¿6

4.2 (-5) × (-5) × (-5) ….. -125 (-5)3 = -125 …… -125 ∴−125=−1254.3 −40 ……… ..(−4)0

−1<1 ∴−40<(−4 )0

1. Convert: 5 cm = 0,05 m Volume of a cuboid = l × b × h = 30,75 × 0,6 × 0,05 = 0,9225 m3 ≈ 1 m3

2. Cost of 100 kWh electricity @ R1,13: (100×1,13)=R 113,00

Amount of electricity @ R1,42/kWh:

( R417,59 – R113,00) ÷ 1,42 = 214,5 kWh

∴ Electricity used: 10 kWh(Free) + 100kWh(first 100kWh @R1,13) +

214,5 kWh = 324,5 kWh



TOPIC: Exponents - Comparing and representing numbers in scientific notation


By the end of the lesson, learners should know and be able to Compare and represent numbers in scientific notation Extend scientific notation to include negative exponents

RESOURCES DBE workbook (p.8-11, p.88-109), Sasol-Inzalo textbook (p.73-79), ruler, pencil, eraser, calculators, DVD( Grade 9 Term 1 Disc 2 GDE 13 03 2014)




3 min Write the following example as base 10.(1) 1(2) 10(3) 100(4) 0,1(5) 0,000 001

p.124


4min Fill in <, > or =.

a.–(10)² < (–10)² b. –(6) > (–6)³c. (–9)³ = –(9)³ d. (–8)³ < (8)³e. (–6)² > –(6)² f. (–4)³ = –(4)³



10 min Educator asks learners to have a look at the number given below.What is the decimal number system? They must dissect the number into the different place values, namely thousands, hundreds, etc.

The decimal number system is a system based on powers of 10 as place values. In a decimal table the place value of any column is one tenth of the place value of a column to its left and 10 times the place value of a column to its right.

Example: Have a look at the number 1 245,367Thousands

Hundreds

tens units comma tenths hundredths

thousandths

1 2 4 5 , 3 6 7

1 000 200 40 5 3

10 6

100

71000

1 × 1000 2 × 100 4 × 10 5 × 1 3 × 110 6 × 1

100

7× 11000

1×103 2 ×102 4 ×101

Large numbers in words:

Compare and represent integers in exponential form

Compare and represent numbers in scientific notation


101=10 : ten; 102=100: hundred; 103=1 000 : thousand; 106=1000 000 : million109=1000 000 000 : billion; 1012=1000 000 000 000:trillion

( Activity 1 )

Educator asks learners whether there are easier ways to express very large or very small numbers.

It is not easy to read or write very large numbers. Large numbers have many digits so they are not easy to read and take a long time to write down accurately. These large numbers are also difficult to compare, to add, subtract, multiply and divide.

To solve this problem, scientist and mathematicians have developed a method of representing very large numbers that makes them easier to work with. This method is called scientific notation.

A number written in scientific notation is written as a significant multiplied by a power of 10.

To convert from decimal notation to scientific notation:General form: significant S × 10n power of 10

Very big numbers:Step 1: If there is no decimal comma. Imagine there is one at the end of the numberStep 2: Count how many place to the left you need to move the decimal


comma until there is only one digit on the left of the decimal comma.Step 3: The number of places you have moved the decimal comma gives the power of ten by which you need to multiply the significant.

Example: Write 35 million into scientific notation: 35 = 3,5 × 10 350 = 3,5 × 102

3 500 = 3,5 × 103

35 million (35 000 000) = 3,5 × 107

Very small numbers:

Step 1: If there is no decimal comma. Imagine there is one at the end of the numberStep 2: Count how many place to the left you need to move the decimal comma until there is only one digit on the left of the decimal comma.Step 3: The number of places you have moved the decimal comma gives the negative power of ten by which you need to multiply the significant.

Example: Write 0,000 012 into scientific notation: 0,12 = 1,2 × 10-1

0,012 = 1,2 × 10-2

0,0012 = 1,2 × 10-3

0,000 012 = 1,2 × 10-5

(Activity 2 – classwork)Educator demonstrates to learners how to use the scientific calculator when


dealing with scientific notation. Use of the calculator:

It depends whether they are using the known brands like Sharp or Casio!!

Educator asks learners to convert the following scientific examples to normal notation

Expanded form:Example 1: Write 3,94×104in normal notation

Multiply by 104 means multiply by 10 000The number will become bigger; therefore the decimal comma will move 4 places to the right

3,94 ×10 000

39 400

Example 2: Write 6,73 ×10−5 in normal distributionMultiply by 10−5 means divide by 100 000The number will become smaller; therefore the decimal comma will move 5 places to the left

3,94 ×10 000

39 400

Example 34,32 × 104

4,32 × 104 = 4,32 × 10 000 = 43 200


43 2004,32 × 10-4

4,32 × 10-4 = 4,32 × 0,0001 = 0,0004320,0004333. Convert an ordinary number to scientific notation, or scientificNotation to an ordinary number.

Fill in <, > or =

43 200 > 0,000432CLASSWORK 10 min Activity 1

Complete the table by filling in the number 2 357,254

Thousands Hundreds tens units comma tenths hundredths thousandths

_____ _____ _____ _____ _____ _____ _____

_____ _____ _____ _____ _____ _____ _____

Activity 21. The weight of the moon is 7 350 000 000 000 000 000 000 kg. Write this

weight in scientific notation.2. Write 0,000 057 in scientific notation.


3 min (1) Write in scientific notation”(a) 5 000 000


AND OR HOMEWORK

(b) 0,000 000 235(2) Write in normal notation:2,96 ×106

(3) DBE workbook 1 Page 55 No 4REFLECTION



(1) 1 = 100

(2) 10 = 101

(3) 100 = 102

(4) 0,1 = 10−1

(5) 0,000 001 = 10−6

2 × 1 000 3 × 100 5 × 10 7 × 1

2 × 110

5 × 1100

4 × 11000

2 × 103 3 × 102 5 × 101

2 ×10−1 5 ×10−2 4 × 10−3

Activity 21. 7,350 × 1021

2. 5,7 × 10-5

1) 22 × 24

Solution: 22× 24= 4 ×16=¿642) 34 ÷ 32

Solution: 34÷ 32=81÷ 9=93) 30 + 34

Solution: 1 + 34 = 1 + 81 = 82

4) (22)3

Solution: (2¿¿2)3=(4)3=64¿5) (22 × 7)3

Solution: (4 ×7¿¿3=(28)3= 21 952

6) ( 12 )

2

Solution: = ½ x ½ = ¼



TOPIC: Exponents - Calculations using numbers in exponential form


By the end of the lesson, learners should know and be able toApply the following general laws of exponents:

am× an=am+n

am÷ an=am−n , if m>n ¿ (a × t )n=an ×t n

a0=¿ 1RESOURCES DBE workbook (p.8-11, p.88-109), Sasol-Inzalo textbook (p.73-79), ruler, pencil, eraser, calculators,

DVD( Grade 9 Term 1 Disc 2 GDE 13 03 2014)


COMPONENTS TIME TASKS/ACTIVITIES CAPSINTRODUCTIONMENTAL MATHS

3 min Simplify:(1) 23 ×24

(2) 35 ÷ 32

(3) (2¿¿3)2¿(4) (5 b¿¿3)3 ¿(5) (2 x)0

p.124


4min (1) Write in scientific notation”(a) 5 000 000

Solution: 5×106

(b) 0,000 000 235Solution: 2,35 ×10−7


(2) Write in normal notation:(a )2,96 ×106

Solution := 2 960 0003.a. 2,24 × 104 >0,25 × 10-4 b. 2,5 × 103 > 2,5 × 10-3

c. 1,75 × 10-6 < 1,75 × 106 d. 1,95 × 10-5 < 1,95 × 105

e. 0,75 × 10-5 = 0,75 × 10-5 f. 0,5 × 102 >0,5 × 10-2


10 min Educator asks learners to name the General Laws of Exponents they know. They must also give examples

1. Product of powers: am× an=am+n , a > 0

Examples: Simplify the following1.2 32 ×34=32+4=36

1.3 y5 × yo= y5+0= y5

1.4 x2+4=x6∨x3+3=x6∨x1+5=x6

2. Quotient of powers: am÷ an=am−n , if m>n

Examples: Simplify the following2.1 56 ÷ 51=56−1=55

2.2 a10 ÷ a6=a10−6=a4

Apply the following general laws of exponents:

am× an=am+n

am÷ an=am−n , if m>n a0=¿ 1 ¿ (a × t )n=an ×t n


2.3 x3=x4−1

Examples: Simplify the following3.1 ¿

3.2 x3× x3=¿ x6

3. Exponent zero a0=1 am÷ am=am−m=a0 ,a>0

am

am =1

Examples: Simplify the following

5.1 40 × 41 ×44=1× 45=1 024 5.2 (3 × 4)0=(12)0=1 5.3 ¿ 5.4 ¿ Classwork

Then :Educator asks learners to simplify the following examples in their respective groups.Examples: Simplify without using a calculator:

1. (−2 x2)¿−2x2

4 x2 =−24

=−12

Educator asks learners to name the General Laws of Exponents they know. They must also give examples


4. Power of a power: ¿, a > 0


3.2 x3× x3=¿ x6

5. Power of a product: ¿

Example: Simplify the following

4.1 ¿ 4.2 ¿ 4.3 (2 ×6)2 ¿122=144


am

am =1


5.1 ¿ 5.2 ¿


Do ClassworkEducator asks learners to simplify the following examples in their respective groups.Examples: Simplify without using a calculator:

2. (−2 x2)¿−2x2

4 x2 =−24

=−12

3. 2n 8n+2

43 n2−2 n

¿2n (23 )n+2

(22 )3n2−2n= 2n23n+6

26n 2−2n ( all the basis of the exponents need to

be the same)

¿ 2n+3 n+6

26n−2 n =24 n+6

24 n

¿24 n+6−4 n=26=64

CLASSWORK 11 in Simplify without using a calculator:(1) 4 m5×3m7 ×−2m3

(2) a7 ×a6

a8 ×a3

(3) ¿


AND OR HOMEWORK

3 min Simplify:

(1) 4 x2 y3

2x y5


(2) 2 p8 q4

8 p3qr÷ 16 p4

4 r

(3)2x+1

3.2x−1

REFLECTION



(1 ) 23× 24 = 23+ 4=27

(2) 35 ÷ 32=35−2=33

(3 ) (2 x)0=1

(1 ) 4 m5 ×3 m7×−2m3

Solution: −24m15

(2 ) a7× a6

a8× a3

Solution: a13

a11 =a13−11=a2

(3 ) ¿Solution : 1 – 6 = -5

1 5 000 000Solution: 5 ×106

1. 0,000 000 235Solution: 2,35 ×10−7

2. :(a )2,96 ×106

Solution := 2 960 000 (b) 28,72 ×10−5

Solution: 0,0002872



TOPIC: WEEKLY ASSESSMENT


1. Which of the numbers below is a mixed number?

0 ; 0,2 ; 18 ; 2

14

A. 2 14

B. 0,2C. 0

D.18

2. Which of the following numbers lie between 0,07 and 0,08 on a number line?A. 0,00075B. 0,0075C. 0,075D. 0,75

3. Complete: 34

of 1 13+1=¿

A. 3B. 5C. 2


D. 12

4. Which group of fractions is in descending order?

5. Write 0,25 as a percentageA. 0,25%B. 25%

C.25

100%

D.14

6. 3-2×3−3=¿

A. 36

B. 9-5

C. 3-5

D. 36


7. 0.012 in scientific notation is

A. 1.2×10−2

B. 1.2×102

C. 1.2×10−3

D. 12×10−2



(1) Which of the numbers below is a mixed number?

0 ; 0,2 ; 18 ; 2

14

A. 2 14

B. 0,2C. 0

D.18

(2) Which of the following numbers lie between 0,07 and 0,08 on a number line?A. 0,00075B. 0,0075C. 0,075D. 0,75

(3) Complete: 34

of 1 13+1=¿

A. 3B. 5C. 2D. 12


4. Which group of fractions is in descending order?

5. Write 0,25 as a percentageA. 0,25%B. 25%

C.25

100%

D.14

6. 3-2×3−3=¿

A. 36

B. 9-5

C. 3-5

D. 36


7. 0.012 in scientific notation is

A. 1.2×10−2

B. 1.2×102

C. 1.2×10−3

D. 12×10−2


SUBJECT: MATHEMATICS GRADE 9 WEEK 6, LESSONS 1 and 2




Compare and represent integers in exponential form Compare and represent numbers in scientific notation Extend scientific notation to include negative exponents

RESOURCES DBE workbook (p.8-11, p.88-109), Sasol-Inzalo textbook (p.73-79), ruler, pencil, eraser, calculators, DVD( Grade 9 Term 1 Disc 2 GDE 13 03 2014)




4 min Simplify the following expressions:

(1) 53

5−4

(2) 35

37

(3) 2−3 ×25+(3 )3 ÷ 3−1

p.124


5 min(1) 4 x2 y3

2x y5


Solution: 2x2−1

y5−3 =2 xy2

(2) 2 p8 q4

8 p3qr÷ 16 p4

4 r

Solution: p8−3q4−1

4 r× 4 r

16 p4

¿ p5q3

4 r× r

4 p4

¿ p5 q3 r16 p4 r

= p5− 4 q3

16= pq3

16

(3) 2x+1

3.2x−1

Solution: ¿ 2x+1− x+1

3

22

3= 4

3


10 min Activity 1: Which of the following are true? Please correct any false statements.

1. 3−1=−3

2. ( 13 )

−1

=3

3. ( 23 )

−2

=( 32 )

2

4. 3 x−2= 13x2


5. 3−5 × 92=3

Solution

1. False. 3−1=13

2. True3. True

4. False. 3 x−2= 3x2

5. 3−5 × 92=3−5 ×34=3−1=13

Scientific notationScientific notation is used to express very large numbers in shortened form: a×10n, where a is a decimal number between 1 and 10 and n is an integer, e.g. 2,3 ×10−2 is an equivalent form of 0,023.Other examples:Decimal notation Scientific notation6 130 000 6,13 ×106

0,00001234 1,234 ×10−5

Activity 2Express the following numbers in scientific notation:1. 134,562. 0,00000056783. 0,00014. 1000


Examples:Calculate: 27

2 y x 7 ¿= 128

First key in the base 2, then the y x key, then the exponent 7, and then the ¿ key. The answer 128 appears on the screen

Calculate: 6 ×53

5 y x 3 ¿ × 6 ¿= 750

First calculate 53 and then multiply it with 6. Note that this sequence might differ depending on which brand of calculator you are using

CLASSWORK 10 min 1. Are the following statements true or false? Correct any false statements:

a. 10−3=0,001

b. ( 15 )

−2

=52

c. 252 ×10−6× 26=5

2. Write in scientific notation:

a. 876 500 000b. 89 100 000 000 000c. 0,006789d. 0,0000000000321

CONSOLIDATION/ 1 min 1. Use ¿, ¿, or ¿ to make the following statements true:


CONCLUSION AND OR HOMEWORK a. −(2 )5 _____ (−2)5

b. t 3× t8_____ t 24 for t >1c. (−6 a )× (−6 a ) × (−6a ) ¿ (−6 a)3

d. (5 x)2 ¿ 5 x2 for x>1

2. Calculate each of the following and leave the answer in scientific notation:

a. 135 000 ×246 000 000b. 987 654 × 123 456c. 0,000065 ×0,000216

REFLECTION


Answers Week 6 Lessons 1 & 2


Simplify the following expressions:

(1) 53

5−4 =57

(2) 35

37 =3−2=19

(3) 2−3 ×25+(3 )3 ÷ 3−1

¿4+27÷ 13

¿313

1. Are the following statements true or false? Correct any false statements:

a. 10−3=0,001 True

b. ( 15 )

−2

=52 True

c. 252 ×10−6× 26=5 False, correct answer

is 1

25

2. Write in scientific notation:

a. 876 500 000 = 8,765 ×108

b. 89 100 000 000 000 = 8,91 ×1013

c. 0,006789 = 6,789 ×10−3

d. 0,0000000000321 = 3,21×10−11

1. Use ¿, ¿, or ¿ to make the following statements true:

a. −(2 )5 ¿ (−2)5

b. t 3× t8 ¿ t 24

c. (−6a )× (−6a ) × (−6 a )_____ (−6 a)3

d. (5 x)2 _____ 5 x2

2. Calculate each of the following and leave the answer in scientific nota-tion:

a. 135 000 ×246 000 000= 1,35 ×105 ×2,46 ×108

= 3,321 ×1013

b. 987 654 × 123 456= 9,87654 × 105× 1,23456× 105

= 12,19318122× 1010

= 1,219318122× 1011

c. 0,000065 ×0,000216


SUBJECT: MATHEMATICS GRADE 9 WEEK 6, LESSONS 3 & 4




am× an=am+n

am÷ an=am−n

¿ (a × t )n=an ×t n

a0=¿ 1RESOURCES DBE workbook (p.8-11, p.88-109), Sasol-Inzalo textbook (p.73-79), ruler, pencil, eraser, calculators,





3 min Simplify:(6) 23 ×24

(7) 35 ÷ 32

(8) (2¿¿3)2¿(9) (5b¿¿3)3 ¿(10) (2 x)0

p.124


4min 3. Use ¿, ¿, or ¿ to make the following statements true:


HOMEWORK a. −(2 )5 ¿ (−2)5

b. t 3× t8 ¿ t 24

c. (−6 a )× (−6a ) × (−6 a )_____ (−6 a)3

d. (5 x)2 _____ 5 x2

4. Calculate each of the following and leave the answer in scientific notation:

a. 135 000 ×246 000 000= 1,35 ×105 ×2,46 ×108

= 3,321 ×1013

b. 987 654 × 123 456= 9,87654 × 105× 1,23456× 105

= 12,19318122× 1010

= 1,219318122× 1011

c. 0,000065 ×0,000216


10 min Educator asks learners to name the General Laws of Exponents they know. They must also give examples

1. Product of powers: am× an=am+n , a > 0

Examples: Simplify the following1.1 32 ×34=32+4=36

1.2 y5 × yo= y5+0= y5

Apply the following general laws of exponents:

am× an=am+n

am÷ an=am−n , if m>n a0=¿ 1 ¿ (a × t )n=an ×t n


1.3 x2+4=x6∨x3+3=x6∨x1+5=x6

2. Quotient of powers: am÷ an=am−n, if m>n

Examples: Simplify the following2.1 56 ÷ 51=56−1=55

2.2 a10÷ a6=a10−6=a4

2.3 x3=x4−1


2.2 x3× x3=¿ x6


am

am =1


3.1 40 × 41 ×44=1× 45=1024 3.2 (3× 4)0=(12)0=1 3.3 ¿ 3.4 ¿


Classwork

Then :Educator asks learners to simplify the following examples in their respective groups.Examples: Simplify without using a calculator:

4. (−2 x2)¿−2 x2

4 x2 =−24

=−12

Educator asks learners to name the General Laws of Exponents they know. They must also give examples

5. Power of a power: ¿, a > 0


5.2 x3× x3=¿ x6

6. Power of a product: ¿

Example: Simplify the following


6.1 ¿ 6.2 ¿ 6.3 (2×6)2 ¿122=144


am

am =1


7.1 ¿ 7.2 ¿

Educator asks learners to simplify the following examples in their respective groups.Examples: Simplify without using a calculator:

1. (−2 x2)¿−2 x2

4 x2 =−24

=−12

2. 2n 8n+2

43 n2−2 n

¿2n (23 )n+2

(22 )3n2−2n= 2n23n+6

26n 2−2n

All the bases of the exponents need to be the same when simplifying.


¿ 2n+3 n+6

26 n−2 n =24 n+6

24 n

¿24n+6−4n=26=64

CLASSWORK 10min Simplify without using a calculator:(1) 4 m5×3m7 ×−2 m3

(2) a7 ×a6

a8 ×a3

(3) ¿


3 min Simplify:

(1) 4 x2 y3

2x y5

(2) 2 p8 q4

8 p3qr÷ 16 p4

4 r

(3)2x+1

3.2x−1

REFLECTION


Answers Week 6 Lessons 3 & 4Mental Maths Class work Homework

(1 ) 23× 24 = 23+ 4=27

(2) 35 ÷ 32=35−2=33

(3 ) (2 x)0=1

(1 ) 4 m5 ×3 m7×−2m3

Solution: −24m15

(2 ) a7× a6

a8× a3

Solution: a13

a11 =a13−11=a2

(3 ) ¿Solution : 1 – 6 = -5

2 5 000 000Solution: 5 ×106

3. 0,000 000 235Solution: 2,35 ×10−7

4. :(a )2,96 ×106

Solution := 2 960 000 (b) 28,72 ×10−5

Solution: 0,0002872


SUBJECT: MATHEMATICS GRADE 9 WEEK 6, LESSONS 5 and 6




am× an=am+n

am÷ an=am−n

¿ (a × t )n=an ×t n

a0=¿ 1RESOURCES DBE workbook (p.8-11, p.88-109), Sasol-Inzalo Workbook 1 (p.73-79), ruler, pencil, eraser, calculators,




INTRODUCTION &MENTAL MATHS

4 min Simplify:

(1) 53

5−4

(2) 35

37

(3) 2−3 ×25+(3 )3 ÷ 3−1

p.124


5 min(1) 4 x2 y3

2x y5


HOMEWORK

Solution: 2x2−1

y5−3 =2 xy2

(2) 2 p8 q4

8 p3qr÷ 16 p4

4 r

Solution: p8−3q4−1

4 r× 4 r

16 p4

¿ p5q3

4 r× r

4 p4

¿ p5 q3 r16 p4 r

= p5− 4 q3

16= pq3

16

(3) 2x+1

3.2x−1

Solution: ¿ 2x+1− x+1

3

22

3= 4

3


10 min Educator asks learners how they will solve the following example and that they must supply examples:

1. Negative exponents:

a−x= 1ax and ax= 1

a−x

Example: 32

35 =32−5=3−3= 133 =

127

Extend the general laws of exponents to include:

Negative exponents

a−m= 1

am


Simplify: 16 b5

−8b7 =−2b5−7=−2 b−2=−2b2

Educator instructs learners to do the following examples individually with feedback.

Educator asks learners to apply substitution in the following examples:

2. Substitution: We can determine the values of exponential expressions, if we are provided with a possible value. We substitute the given value in for the variable and calculate:

Example: Determine the value of: 3 x2−5 x−1if x=−2¿3¿

¿3 (4 )−5(−12 )

¿12+52=14 1

2Educator instructs learners to practice exponents using a calculator.

3. Using a calculator for exponents:If the power also has to be multiplied or divided by another number, then it is better to first calculate the power and then multiply or divide it by the other number.

Educator might need to teach learners how to use different brands of calculators.


Examples:Calculate: 27:

2 y x 7 ¿= 128

First key in the base 2, then the y x key, then the exponent 7, and then the ¿ key. The answer 128 appears on the screen

Calculate: 6 ×53

5 y x 3 ¿ × 6 ¿= 750

First calculate 53 and then multiply it with 6. Note that this sequence might differ depending on which brand of calculator you are using

CLASSWORK 10 min Activity 1:

(1) −10 x3

x8 × 5 x2

x

(2) ( 34 )

5

÷( 43 )

−3

Activity 2:

(1) Determine the value of 12¿

(2) First simplify and then calculate the value of ¿¿CONSOLIDATION/CONCLUSION AND OR HOMEWORK

1 min (4) Calculate the following using substitution:

a. ( 3 ba )

3

if a=2∧b=−1

b. (−2 x2)( 1(−2x )2 )


c. 2n . 8n+2

43 n . 2−2 n

2. Use a calculator to calculate:

(a) ( 12 )

5

(b) (−0,9)4

(c) 43

πr 2if r=2,5 cm

REFLECTION

Answers Week 6 Lessons 5 and 6Mental Maths Class work Homework

(1) 53

5−4

¿53 .5−4

¿53−4

¿5−1

¿ 15

(2) 35

37

¿ 137−5

Activity 1

Simplify: (1) −10 x3

x8 × 5 x2

x

¿ −50 x5

x9

¿−50 x5−9

¿−50 x−4

¿−50x4

(2)(34 )

5

÷( 43 )

−3

1) ( 3 ba )

3

if a=2 an d b=−1

¿3(−1)

2

3[❑❑ ]3

2) (−2 x2)( 1(−2 x )2 )

3) 2n. 8n+2

43n .2−2 n

Use a calculator to calculate:


¿ 132

¿19

(3) 2−3 ×25+(−3)3÷ 3−1

¿123 ×25+(3)3 ×3

¿ 25

23 +34

¿22+34

¿4+81¿85

(4) 102 n−3×106−n

10n+3

¿ 102n−3+6−n

10n+3

¿10n+3−n−3

¿100

¿1

¿ 35

45 × 43

33

¿ 35× 43

45× 33

¿35−3 ×43−5

¿32 ×4−2

¿ 916

Activity 2

1. 12 [(−1)−2(2)3 ]2=1

2¿¿

12

[ 1× 8 ]2=12

×64=32

2. 43 x6 y9

x4 y=64 x6−4 y9−1

¿64 x2 y8

∴64 ¿¿64 × (4 ) (1 )

¿256

(a) ( 12 )

5

(b) (−0,9)4

(c) 43

π r 2if r=2,5 cm


SUBJECT: MATHEMATICS GRADE 9 WEEK 6 , LESSON 7 & 8

TOPIC: FUNCTIONS AND PATTERNS


By the end of the lesson, learners should know and be able toInvestigate and extend numeric and geometric patterns looking for relationships between numbers, including patterns:

not limited to sequences involving a constant difference or ratio. of learner’s own creation. Describe and justify the general rules for observed relationships between numbers in own words or in algebraic language.

RESOURCES DBE workbook (p.20-23, p.112-113); Sasol-Inzalo Workbook 1 (p.87-99), ruler, pencil, eraser, calculators, notebook. Tablets and DVDs(Grade 9 Term 1 Disc 2GDE 25 02 2014)

PRIOR KNOWLEDGE Basic Mathematics operations, Terminology such as sum, difference, ratio, etc.



INTRODUCTION &MENTAL MATHS

3 min (1) Can this flow diagram be used to calculate the number of squares?

(2) Complete the table:

Topic: 2.2 CAPS page 129

Functions and relationships

Input and output values

HOMEWORK 0 min Flow diagramsInputOutputVariableFormulaeEquations


LESSON DEVELOPMENT

20 min Educator asks learners what flow diagrams, input and output values mean.

Flow diagrams

Flow diagrams show how input numbers are changed into output numbers. We use rules to guide us which operation to apply.A flow diagram is the same as an equation.

Example: y=x+30 equation x→+30 → y flow diagram

What is the rule to get an output of 50 from an input of 20?20→+x→ 50 or 20→× x→ 50 flow diagram20+x=30 or 20 × x=50 equation

It is important to identify the set of numbers that you are working with as being natural numbers (N), whole numbers (N 0), integers (Z) or rational numbers (Q).

When two numbers are mathematically related one of the numbers is the input value and the other number is the resulting output value.


(1). Predict what happened to the 4 to become an 8 in the example:

The 4 became an 8. It can be 4 × 2 = 8 or 4 + 4 = 8

(2) Predict what happened to the 17 to become 34.

The 17 became a 34: It can be 17 × 2 = 34 or 17 + 17 = 34


Educator instructs learners to do the following example on their own

Classwork: 4 11

5 a ×3−1 9 b

12 35

What are the values of a∧b?What is the domain of the numbers used in this machine?

Educator asks learners what variables are and how formulae and equations can be used to find the value of the variable

Formulae and equations:Determine input and output values using formulae and equations”Variables or unknown, are expressed by using letters of the alphabet. Variables can also be in the rule box of a flow diagram.Example:

1 a 2 b 1 2 n+1 c 15 d 20 eThe rule is 2n+1, this means we have to substitute in each input value for n.To find the value of a: 2 (1 )+1=3=aTo find the value of b: 2 (2 )+1=5=b


Educator asks learners to discuss the two examples below within their respective groups with feedback.

Using tables:

Example: Use the equation y=3 x to determine the output values in the table: Describe the domain of the input values.

Input (x) -3 -2 -1 0 1 2 3Output (y)

3(-3)=-9 3(-2)=-6 -3 0 3 6

The domain is Integers (Z) between -3 and 3.

Example: Describe the relationship between the values in the top and bottom row in the given table. Write down the values of m∧n:

x -3 -2 -1 0 1 5 ny 4 5 6 7 8 m 27

The equation is: 4 = -3 +7 ; 5 = -2 + 7 ; 6 = -1 + 7 ; ∴ y=x+7∴m=5+7=12 and 27=n+7 ∴n=20

Educator instructs learners to do the following example individually

Classwork:


HOMEWORK ACTIVITIES

(1) Find the missing values and describe the domain:

1 6 2 11 3 ×5+1 a 4 21

(2) 2 -6 3 b 4 ×(−3) -12 4 a

(3) Determine the variables: 25 u x 6 y √n−1 9 144 w

(4) Use the equation y=2 x+1 to complete the table and describe the domain of the input values:Input (x) 0 1 2 4 5Output ( y )

13

LESSON REFLECTION:


Answers Week 6 Lessons 7 and 8Mental Maths Class work

1. Yes2.Input

number

5 10 15 20 25 30

Output number

65 115 165 215 265 315

1. The first input is 4. So 4 ×3−1=11 The last input is 12 ×3−1=35 The second input is 5 ×3−1=14=a The third input is 9×3−1=26=b

The domain is natural numbers (N )

2. Let’s identify the pattern:First term: -5 = -2 - 3Second term: -4 = -1 - 3The third term: -3 = 0 - 3Formula: y=x−3Value of a: 34=x−3=37Value of b: y=12−3=9


WEEKLY ASSESSMENT

1.1 23 ×24

A. 212

B. 27

C. 2 x 107

D. 21

1.2 Simplify: 4 x2 y3

2 x y5

A. 2xy-2

B. 4xy-2

C. 2xy2

D. 4x2y2

1.3 Write 876 500 000 in scientific notation:

A. 87,6 x 10-7

B. 8,76 x 108

C. 8,76 x 10-8

B. D. 87,6 x 107


1.4 Solve for x in : 2x−1=18

A. -3B. -2C. 2D. 3

1.5 Find the missing values and describe the domain: 1 6 2 11 3 ×5+1 a 4 21

A. 6B. 18C. 15D. 16


Memo1.1 1.2 1.3 1.4 1.5B A C B D





By the end of the lesson, learners should know and be able toInvestigate and extend numeric and geometric patterns looking for relationships between numbers, including patterns:

not limited to sequences involving a constant difference or ratio. of learner’s own creation. Describe and justify the general rules for observed relationships between numbers in own words or in algebraic language.

RESOURCES DBE workbook (p.20-23, p.112-113); Sasol-Inzalo textbook (p.87-99), ruler, pencil, eraser, calculators, notebook. Tablets and DVDs(Grade 9 Term 1 Disc 2GDE 25 02 2014)


COMPONENTS TIME TASKS/ACTIVITIES CAPSMENTAL MATHSWarmer

3 minutes Extend the following pattern:(1) 100 ; 90; 81; 73; 66; …(2) 0; 3; 7; 12; 18; …(3) 5; 15; 25; 35; …(4) 1; 2; 3; 4; 5; …(5) 1; 4; 9; 16; …

CAPS page 126Numeric patterns


HOMEWORK 5 minutes Look at these three arrangements. They consist of black, grey and white squares.

(a) Draw the next arrangement.

(b) Describe what is constant in these arrangements. Answer: The number of black squares

(c) What are the variables in these arrangements? Answer: The number grey squares and the number white squares are both variables

Common differenceCommon ratioTables rulenth term


LESSONDEVELOPMENT

20 minutes Educator asks learners what numeric (number) patterns are with examples

Numeric (number) patterns and the general rule:They are patterns that consist of numbers.

Examples from previous lesson:

Natural numbers: 1; 2; 3; 4; 5; 6; … the pattern is add one to the previous term Even numbers: 2; 4; 6; 8; 10; … the pattern is 2 times each term (2n) Square numbers: 0;1; 4; 9; 16; 25; … the pattern is square the position of the term

Educator asks learners to use tables in order to find the patterns in the given examples.

The use of tables:To see the pattern and use it to find the next two terms in a sequence of numbers is good but we want to do better than that. We want to be able to work out the value of any term in the sequence.

Examples: Describe the pattern(1)

Answer: Term value = 15 × term number - 5(2)

Answer: Term value = 10 ×2term number

(3)

Answer: Term value = 10 ×2term number−1


Term number 1 2 3 4Term value 10 25 40 55



Let’s look at the following example: Find the 20th term in the sequence 8; 14; 20; 26; …

Step 1: Organize the number sequence in a table. Note that n stands for the position of the number in the sequence and T stands for the value of the term.

Step 2: Find the constant gap between the terms in the sequence. In this case it is 6. Each term is 6 more than the previous term.

The difference between any term and the term before it is 6. We say there is a constant difference between the terms.

Step 3: Add an extra row to your table and write in the 6 times table.

Look carefully at every term; the six times table and its position.Can you see that the first term is 6 + 2; the second term is 12 + 2 and the third term is 18

+ 2

Thus: First term = 1(position) × 6(constant difference) + 2 = 8 T 1=1× 6+2=8 T 2=2 ×6+2=14 T 3=3 ×6+2=20So the general rule for finding terms in this sequence is:Nth term = n(position)×6(constant difference) + 2. ∴T n=6 n+2

Step 4: So to find the 20th term in the sequence we use the general rule and substitute n=20.So the 20th term in the sequence is (20 × 6) + 2 = 122.


Position in the sequence (n¿

1 2 3 4

Term (T n¿ 8 14 20 26

Position in the sequence (n¿

1 2 3 4 5

Term (T n¿ 8 14 20 26 32Six times table 6 12 18 24 30

Educator instructs learners to do the following example individually:

Classwork:(1) Describe the pattern:

(2) In the numeric pattern 5; 8; 11; 14; …, the first term is 5 because it is the first number in the

sequence.We write T 1=5. The second term is 8 because it is the second number in the sequence. We write T 2=8.

(a) Write these values in a table

(b) Find the rule that describes the relationship between the numbers.



HOMEWORK ACTIVITIES

Complete by filling in the gaps:Consider the pattern: 9; 14; 19; 24; …

T 1 T 2 T 3 T 4 Position in the sequence

9 14 19 24 ………

+5 +5 +5 ……………………….

In this pattern, the constant ………….. is …... We can use equations to describe the rule as follows: T 1=5 (1 )+¿…… = 9 T 2=5(….)+4 =14 T …..=5 (3 )+4 = 19 T 4=… (4 )+4 = ….. Notice how the position of the term corresponds to the number in the brackets: ∴T 15=5(…..)+4 = …….

∴n thTerm of the pattern is T n=5 (…. ) ….where we substitute the term number and the number in brackets with n.

REFLECTION


HHOMEWORKComplete by filling in the gaps:

Consider the pattern: 9; 14; 19; 24; …

T 1 T 2 T 3 T 4 Position in the sequence

9 14 19 24 Terms of the sequence

+5 +5 +5 Common difference of 5

In this pattern, the constant difference is 5 We can use equations to describe the rule as follows: T 1=5 (1 )+¿4 = 9

T 2=5(2)+4 =14

T …..=5 (3 )+4 = 19

T 4=5 ( 4 )+4 = 24

Notice how the position of the term corresponds to the number in the brackets:

∴T 15=5(15)+4 = 79

∴n thTerm of the pattern is T n=5 n+4where we substitute the term number and the number in brackets with n.


Answers Week 7 Lessons 1 and 2Mental Maths Class work

(1) 100 ; 90; 81; 73; 66; 60; 55; 51; …

(2) 0; 3; 7; 12; 18; 25; 33; …(3) 5; 15; 25; 35; 45; 55; …(4) 1; 2; 3; 4; 5; 6; 7; …(5) 1; 4; 9; 16; 25; 36; …

(1)Describe the pattern:


Solution: Term value = 5 × term value + 10(2) In the numeric pattern 5; 8; 11; 14; …, the first term is 5 because it is the first number in the sequence.We write T 1=5. The second term is 8 because it is the second number in the sequence. We write T 2=8.

(a) Write these values in a table

We can write these values in a table:Term number (n¿

1 2 3 4

Answer (T n) 5 8 11 14

(b) Find the rule that describes the relationship between the numbers.

The rule that describes the relationship between the numbers in this sequence is:Add three to each number to get the next number in the sequence.To find each term, we do the following:T 1=3 (1 )+2=5T 2=3 (2 )+2=8T 3=3 (3 )+2=11T 4=3 ( 4 )+2=14The pattern is thus: T n=3 (n )+2





By the end of the lesson, learners should know and be able to Investigate and extend numeric and geometric patterns looking for relationships between numbers, including patterns:

not limited to sequences involving a constant difference or ratio. of learner’s own creation.

Describe and justify the general rules for observed relationships between numbers in own words or in algebraic language.

RESOURCES DBE workbook (p.20-23, p.112-113); Sasol-Inzalo textbook (p.87-99), ruler, pencil, eraser, calculators, notebook. Tablets and DVDs(Grade 9 Term 1 Disc 2GDE 25 02 2014)




MENTAL MATHSWarmer

3 min Complete the table: Topic: 2.2 CAPS page 129Functions and

relationshipsEquivalent forms


CLASSWORK

5 m

inut

es

(1) Find the missing values:

1 6 2 11 3 ×5+1 a 4 21Solution: a=3×5+1=16

(2) 2 -6 3 b 4 ×(−3) -12 4 a

Solution: b=3× (−3 )=−9

(3) Determine the variables where n represents the given values:

25 u x 6 y √n−1 9 144 w

Solution: u=√25−1=4 x=√6−1=1,45 y=√9−1=1 w=√144−1 = 11

(4) Use the equation y=2x+1 to complete the table:Input (x)

0 1 2 4 5 6

Output ( y )

1 3 5 9 11 13

Flow diagramsInputOutputWordsVariableFormulaeEquations


Educator asks learners what equivalence means and how relationships can be expressed in different equivalent forms

Equivalent forms:

We can represent relationships and functions in five different ways: In words In flow diagrams In tables Using formulae Using equations

Example:

The function describing the relationship between input and output values can be presented in a number of different ways.

Here is an example of a relationship between two quantities:

In each arrangement there are some red dots and some blue dots.


Different ways to describe the relationship between the red and blue dots: In words: The number of blue dots is four times the number of red dots plus two

OR to calculate the number of blue dots you multiply the number of red dots by two, add one and multiply the answer by two.

In a flow diagram: number of red dots ¿¿ × 4 ¿¿ +2 → number of blue dotsOR

In a table:

Formulae/Equations:Number of blue dots = 2× the number of red dots + 4

OR

Number of blue dots = 2 ×¿ the number of red dots + 1)


Educator instructs learners to complete the given example in their respective groups

Classwork:

The second quantity is always three times the first the first quantity plus eight.The first quantity varies between one and five, and it is always a whole number.

(1.1). Complete the flow diagram:

(1.2). Describe the relationship between the two quantities with this table:

(1.3). Describe the relationship between the two quantities with a word formula:

Output number = ……………. × ….. + ………:


×3

Input number 1

11

HOMEWORK ACTIVITIES

(1) Complete the flow diagram

(2) Explain the relationship in words and give a verbal formula(3) Give an algebraic formula(4) Complete the table

Input numbers -1 -2 -3 -4 -5

Function values

LESSON REFLECTION:

Challenges:Recommendations:


SOLUTIONSMental MathsComplete the table

Words Flow diagram ExpressionMultiply a number by two and add six

to the answer2× x+6

Add three to a number and then multiply the answer by two

2×( x+3)

Multiply a number by five and then subtract one from the answer

5 × x−1

Multiply a number by four and then add seven to the answer

7+4× x

Multiply a number by negative five and then add ten to the answer

10−5× x


CLASSWORK:The second quantity is always three times the first the first quantity plus eight.The first quantity varies between one and five, and it is always a whole number.

(1.1). Complete the flow diagram:

(1.2). Describe the relationship between the two quantities with this table:

(1.3). Describe the relationship between the two quantities with a word formula:

Output number = … Input number × 3 + 8


HOMEWORK(1) Complete the flow diagram

(2) Explain the relationship in words and give a verbal formula

Solution: Each input number is multiplied by 5, then 20 is added, to produce the output numbers.

Output number = 5 × input number + 20

(3) Give an algebraic formula

Solution: y=5 x+20

(4) Complete the table

Input numbers -1 -2 -3 -4 -5Function values 15 10 5 0 -5



TOPIC: FUNCTIONSAND PATTERNS


By the end of the lesson, learners should know and be able to Investigate and extend numeric and geometric patterns looking for relationships between numbers, including patterns:

not limited to sequences involving a constant difference or ratio. of learner’s own creation.

Describe and justify the general rules for observed relationships between numbers in own words or in algebraic language.

RESOURCES DBE workbook (p.20-23, p.112-113); Sasol-Inzalo Workbook 1 (p.87-99), ruler, pencil, eraser, calculators, notebook. Tablets and DVDs(Grade 9 Term 1 Disc 2GDE 25 02 2014)

PRIOR KNOWLEDGE

Basic Mathematics operations, Terminology such as sum, difference, ratio, etc.


MENTAL MATHSWarmer

3 min Extend the following pattern:(1) 2; 5; 8; 11; …(2) 4; 5; 8; 13; …(3) 1; 2; 4; 16; …(4) 3 ;5 ;7 ; 9; …(5) 1; 8; 27; 64; …

Topic CAPS page 126Patterns


HOMEWORK 5 min HOMEWORK(1) Complete the flow diagram





Solution: y=5 x+20

Common differenceCommon ratioFibonacci, Golden

pattern, Pascal


15

10

5

0

-5

LESSONDEVELOPMENT

Educator asks learners to give examples of number patterns.

Examples: Patterns in numbers:

The most common patterns in the number system : Natural numbers: 1; 2; 3; 4; 5; 6; … the pattern is add one to the previous term Even numbers: 2; 4; 6; 8; 10; … the pattern is 2 times each term (2n) Square numbers: 0;1; 4; 9; 16; 25; … the pattern is square the position of the term

0 = 0 × 01 = 1 ×14 = 2 ×2

9 = 3 ×3Thus n2

Cube numbers: 1; 8; 27; 64; … the pattern is cube the position of the term1=1×1×18=2× 2× 227=3 ×3 ×3

Thus n3

Fibonacci numbers: 1; 1; 2; 3; 5; 8; 13; … the pattern is the sum of the previous two numbers give you the next number.

Pascal’s Triangle1 = 1 = 20

1 + 1 = 2 = 21

1 + 2 + 1 = 4 = 22

1 + 3 + 3 + 1 = 8 = 23

1 + 4 + 6 + 4 + 1 =16 = 24


Educator instructs learners to do the following examples in pairs:

Classwork:Write down the next two terms of each sequence and identify the pattern:(1). 4; 7; 10; 13; ……;…….. the pattern is …………………………………(2). 1; 1; 2; 3; 5; …..;……… the pattern is …………………………………(3). 3; 6; 9; 12; ……; …….. the pattern is ………………………………….(4). 2; 6; 18; …..;…….. the pattern is …………………………………

Educator asks learners to give everyday examples of patterns in physical or diagram form.

Patterns in physical or diagram form:Patterns can be presented in physical and diagrammatic form, e.g.

Sierpinski’s Triangle:Repetition of triangles within triangles.Description:Pattern: Cut out triangle in the centre.For each of the remaining triangles (second figure), perform this same act (third figure). Iterate indefinitely (final figure)

Matchsticks


Tiling:Blue and yellow square tiles arte combined to form the arrangements below.

(a) How many yellow tiles are there in each arrangement? Answer: There are 1, 2, 3 and 4 yellow tiles in arrangements 1, 2, 3 and 4 respectively.

(b) How many blue tiles are there in each arrangement? Answer: There are 8, 10, 12 and 14 blue tiles in arrangements 1, 2, 3 and 4 respectively

(c) How many yellow and blue tiles will there be in arrangement 5? Answer: 5 yellow and 16 blue tiles.(d) Complete the table:

Number of yellow tiles

1 2 3 4 5 8

Number of blue tiles

8 10 12 14 16 22


Educator instructs learners to do the following example in their respective groups:

Classwork:(a) Make two more arrangements of black and grey squares so that a pattern is formed.

(b) Is there a constant inn your pattern and what is this value?(c) Is there a variable in your pattern? If yes, give the values of the variables

HOMEWORK ACTIVITIES

Look at these three arrangements. They consist of black, grey and white squares.

(a) Draw the next arrangement(b) Describe what is constant in these arrangements(c) What are the variables in these arrangements?

LESSON REFLECTION:



SOLUTIONSMental Maths: Extend the following patter

(1) 2; 5; 8; 11; 14; 17; … add 3(2) 4; 5; 8; 13; 20; 29; … add consecutive odd numbers: 1 then 3 then 5, etc.(3) 1; 2; 4; 16; 32; 64; … Multiply by 2(4) 3 ;5 ;7 ; 9; 11; 13; … add 2 each time or count in odd numbers(5) 1; 8; 27; 64; 125; 216; … cube numbers

CLASSWORK:Write down the next two terms of each sequence and identify the pattern:(1). 4; 7; 10; 13; 16; 19 the pattern is a common difference(2). 1; 1; 2; 3; 5; 8; 13 the pattern is Fibonacci(3). 3; 6; 9; 12; 15; 18 the pattern is multiples of 3(4). 2; 6; 18; 54; 162 the pattern is common ratio of 3

(a) Make two more arrangements of black and grey squares so that a pattern is formed.

(b) Is there a constant inn your pattern and what is this value? Answer: Yes, the number of black squares is constant. It is 1 in all the arrangements


(c) Is there a variable in your pattern? If yes, give the values of the variables. Answer: Yes, the number of grey squares is a variable: 4, 8, 12, 16, 20

HOMEWORKLook at these three arrangements. They consist of black, grey and white squares.

(d) Draw the next arrangement.

(e) Describe what is constant in these arrangements. Answer: The number of black squares

(f) What are the variables in these arrangements? Answer: The number grey squares and the number white squares are both variables

SUBJECT: MATHEMATICS GRADE 9 WEEK 7 , LESSON 7


TOPIC: FUNCTIONS AND RELATIONSHIPS


By the end of the lesson, learners should know and be able toDetermine, interpret and justify equivalence of different descriptions of the same relationship or rule presented:

by formula by graphs on the Cartesian Plane

RESOURCES Sasol-Inzalo Workbook 1 (p. 101-113), ruler, pencil, eraser, calculators, notebook. Tablets and DVD (Grade 9 Term 1 Disc 2 –GDE 04 03 2014)

PRIOR KNOWLEDGE Substitution in algebraic expressions, To find input or output values in flow diagrams, tables, formulae and equations.

COMPONENTS TIME TASKS/ACTIVITIES CAPSMENTAL MATHSWarmer

3 m

inut

es

Complete the table

x -2 -1 0 1 2 3y=2 x−5

Topic: 2.2CAPS page 129Functions and

relationshipsEquivalent formsBy formula and graphs


HOMEWORK

5 m

inut

es

(1) Complete the flow diagram





Solution: y=5 x+20




Educator asks learners what equivalence means and how relationships can be expressed in different equivalent forms

Equivalent forms:We can represent relationships and functions in five different ways:

In words In flow diagrams In tables Using formulae Using equations By a graph on a Cartesian plane

Educator asks learners what is needed to represent a function on a graph.

Functions on a graph To present functions on a graph you need to use a set of ordered pairs. An ordered pair is made up of the input value (x) and its corresponding

output value ( y ) . Ordered pairs are always expressed in the same way with the input value (x)

given first, followed by the output value ( y ) in the form (x ; y). On a graph the x value is expressed on the horizontal axis and the y value is

expressed on the vertical axis.

To find output values, learners should be given the rule/formula as well as the domain of the input values.


Educator refers to previous lesson and asks learners how they are going to present the homework graphically

Homework from previous lesson:(1) Complete the flow diagram





Solution: y=5 x+20


Educator asks learners to do the following example individually.Classwork: Represent the function y=−3x+4 with

(a) A flow diagram(b) A table of values for the set of integers from -3 to 3(c) A graph

HOMEWORK ACTIVITIES

Represent the function y=12

x+2 with

(a) A flow diagram(b) A table of values for the set of integers from -5 to 5(c) A graph

LESSON REFLECTION:



SOLUTIONS:

MENTAL MATHSComplete the table

x -2 -1 0 1 2 3y=2 x−5 -9 -7 -5 -3 -1 1

CLASSWORK: Represent the function y=−3x+4 with(a) A flow diagram

Solution:

(b) A table of values for the set of integers from -3 to 3


Solution:x -3 -2 -1 0 1 2 3

y=−3x+4 13 10 7 4 1 -2 -5

(c) A graph

Solution:

(a) A graph

Solution:4

1.33


HOMEWORK: Represent the function y=12

x+2 with

(a) A flow diagram

12½


x -5 -4 -3 -2 -1 0 1 2 3 4 5

y=12

x+2 -½ 0 ½ 1 1½ 2 2½ 3 3½ 4 4½

(c) A graph

2

-4



TOPIC: FUNCTIONS AND RELATIONSHIPS


By the end of the lesson, learners should know and be able to Determine, interpret and justify equivalence of different descriptions of the same relationship or rule presented:

by formula by graphs on the Cartesian Plane

RESOURCES Sasol-Inzalo textbook (p. 101-113), ruler, pencil, eraser, calculators, notebook. Tablets and DVD (Grade 9 Term 1 Disc 2 –GDE 04 03 2014)

PRIOR KNOWLEDGE Substitution in algebraic expressions, To find input or output values in flow diagrams, tables, formulae and equations.


MENTAL MATHS

3 min Describe the rule that determines the successive terms of the patterns:(1) 2; 8; 32; 128; …(2) 6; 3; 1½; ¾; …

Topic: 2.1 CAPS page 126Geometric

patterns


HOMEWORK 5 min HOMEWORK: Represent the function y=12

x+2 with

(a) A flow diagram

12½


x -5 -4 -3 -2 -1 0 1 2 3 4 5

y=12

x

+2

-½ 0 ½ 1 1½ 2 2½ 3 3½ 4 4½

(c) A graph

2

-4

Common difference

Common ratioFibonacci,

Golden pattern, Pascal

Tables Rulenth term


y=12

x+2

LESSON DEVELOPMENT

20 min Educator asks learners to distinguish between numeric and geometric patterns and give everyday examples, etc.

Geometric patterns:Triangular numbers:Geometric patterns are number patterns that can be represented by a diagram.A table will help us to count and organize the numbers we see in the pattern we notice in the diagram.Always try to draw or determine what the next diagram in the pattern looks like. This helps you make sure you understand the pattern.

Example: Mary uses circles to form a pattern of triangular shapes:

If the pattern continued, how many circles must Mary have(a) In the bottom row of picture 5? Answer: 5(b) In the second row from the bottom of picture 5? Answer: 4(c) In the third row from the bottom of picture 5? Answer: 3(d) In the second row from the top of picture 5? Answer: 2(e) In the top row of picture 5? Answer: 1(f) In total in picture 5? Answer: 15(g) Complete the table:

Picture 1 2 3 4 5 6 12 15Circles 1 3 6 10 15 21 78 120


Example:A factory makes window frames, Type 1 has one windowpane, type 2 has four windowpanes, type 3 has nine windowpanes, and so on.

(1.1) How many windowpanes will there be in type 5? 25 windowpanes(1.2). Draw type 5

(1.3). How many windowpanes will there be in type 6? 36 windowpanes(1.4). How many windowpanes will there be in type 7? 49 windowpanes(1.5). How many windowpanes will there be in type 12? ExplainThere should be 144, because this is the patterns of squares.(1.6). Complete the table now.

(1.7). Describe the rule: T n=n2

(1.8). Name the number: Square numbers


Educator asks learners to discuss the following example: Determining the rule (formula)

Example: Hamzah is making triangles using sticks as shown in the diagram. Write a formula to determine the number of sticks needed to make any number of triangles.

Sticks: 3 5 7 9Number of triangles (n¿

1 2 3 4 5

Number of sticks (T)

3 5 7 9

n×2=2n 2 4 6 8 10Number of sticks required = 2 × number of triangles + 1T n=2 n+1Number of sticks to make 5 triangles: T 5=2 (5 )+1=11

Educator instructs learners to do the following example in pairs with feedback:

Classwork: Determine how many matches are required to produce a row of 30 squares.


Recap

The concepts discussed in the lesson must be consolidated with the learners

Min 2 - Geometric patterns are number patterns that can be represented by a diagram.- A table will help us to count and organize the numbers we see in the pattern we notice in the diagram.- Always try to draw or determine what the next diagram in the pattern looks like. This helps you make sure

you understand the pattern.

HOMEWORK ACTIVITIES

Use the given sketch to draw the next pattern in the sequence.Then use equation to describe the rule.Complete the table.

Term(pattern) 1 2 3 4 5 20 nNo of match sticks

6 11 16

Complete the table: T 4=5 ( 4 )+1=¿ 21 T 5=¿……………. T 20=¿………………..

The rule: T 1=5 (1 )+1 = 6 T 2=5 (2 )+1 = 11 T 3=5(…)+1 = 16 T n=5 (… ) …. The formula: T n=5 ………….

LESSON REFLECTION:



SOLUTIONS Mental Maths Extend the following pattern:Describe the rule that determines the successive terms of the patterns:(1) 2; 8; 32; 128; …

Solution:Term position 1 2 3 4Term 2 8 32 128

× 4 × 4 × 4 Multiply the previous term by 4 to get the next term

(2) 6; 3; 1½; ¾; …

Solution:Term position 1 2 3 4Term 6 3 1½ ¾

× ½ × ½ × ½ Multiply the previous term by ½ to get the next term.


Weekly Assessment

Lesson 9Informal Assessment 1 Choose the correct answer:

(1) The pattern for the sequence below is…

12

; 14 ;

116 ; …

A. CubicB. SquareC. Common difference of 2D. × 2

(2) The rule for the table below is…Term

number

1 2 3 4

Term value

15 20 25 30

A. n+10

B. 5 n+10

C. 5 n

D. n+5


(3) How many yellow tiles in will there be in arrangement 5?A. 5B. 6C. 16D. 18

(4) The rule for the pattern below is…A. 2 nB. n2

C. 16 nD. 2 n2


A. 29 and 39B. 39 and 79C. 29 and 79D. 19 and 39


Memorandum

(1) B(2) B(3) A(4) B(5) B



TOPIC: Algebraic Expressions – Algebraic Language


By the end of the lesson, learners should know and be able to Recognize and identify conventions for writing algebraic expressions, Identify and classify like and unlike terms in algebraic expressions, Recognize and identify coefficients and exponents in algebraic expressions, Recognize and differentiate between monomials, binomials and trinomials

RESOURCES DBE workbook (p. 114-141), Sasol-Inzalo textbook (p.117-142), textbook, ruler, pencil, eraser, calculators, DVDS (Disc 2 : GDE 11 03 2014 & GDE 13 02 2014 )

PRIOR KNOWLEDGE - Patterns



4 min Simplify by collecting like terms:(a) 2 x+3 y+5 x+ y(b) 3 x2 y2 z+5 xyz−x2 y2 z(c) 3 y−1−x

p.130


5 min N/A

LESSONPRESENTATION/

10 min Educator asks learners whether algebra does have its own language and what is an algebraic expression.

Recognize and identify


DEVELOPMENT Algebraic expression:

An algebraic expression is a representation of the operations that are done on numbers in that expression, for example the sentence “the sum of a number and 2, we can replace:

“the number “with x “sum” with +

Now the expression will be x+2

We use symbols to write words and numbers algebraically, for example in the sentence “28 added to a certain number gives 42”, we can replace :

“a certain number” with x “added to” with + “gives” with =.

Now the problem will be 28+x=42(Activity 1)Educator asks learners to discuss what variables and constants are?

Variables and constants

A variable is a letter that represents an unknown number.It is called a variable because its value in the equation or

expression can change, for example. 7; -12; π; 34 ; ..

A constant is a specific number whose value in the equation or

conventions for writing algebraic expressions,

Identify and classify like and unlike terms in algebraic expressions,

Recognize and identify coefficients and exponents in algebraic expressions


expression is fixed for example. x ; y ;a ;n ;θ ;…Example:Let us identify variables and constants in the following term:

−12+ xy2

-12 → constantx → variabley → variable 2 → exponent

(Activity 2)Educator asks learners to discuss the differences between like and unlike terms:

Like and unlike terms

Like terms contain the same letters of the alphabet and the same exponents.Examples of like terms are:

3 x ; 4 x ;−6 x ; 12

x ;0,34 x; (3+4 ) x ; … (All the terms have

the variable x)

12 ab2 ;−3 a b2; 34

a b2;… (All the terms have the variable

a b2) Note that constants do not effect whether terms are like

or unlike


Unlike terms contain different letters of the alphabet and/or different exponents.Examples of unlike terms are:

11 x2 y2 and 11 xy ( x2 y2 and xy do not have the same exponents)

3 p ;3 and p3(p , 3and p3 are not the same)

All four operations can be performed on like terms, namely addition(+), subtraction(-), multiplication(×) and division(÷ ) .

Addition and subtraction cannot be performed on unlike terms, only multiplication and division

Addition(+) and subtraction separate terms and division, multiplication and brackets do not separate terms, for example:

- p+q has 2 terms separated by the + sign- abc has one term

-x+1x−2 has one term

- (k+l )+(m−p) has two terms- −3 ( pq )−5 (p+1) has two terms

(Activity 3)CLASSWORK 10 min Activity 1

(1) Write the following in algebraic language:(1.1) A certain number is decreased by 16


(1.2) A number that is 5 more than a(1.3) Twice a number(1.4) The product of two and a number

Activity 2Say in each case whether the value is a constant or a variable:

(a) -25(b) p(c) m n6

(d) 0(e) 1

Activity 3Which term does not fit in with the rest?

(a) xyz(b) 3 yxz(c) 2 xy(d) zyx


1 min (1) Write the following in algebraic language:(a) A number that is half the sum of two and three(b) The quotient if x is divided by two

(2) Classify the following terms as either like or unlike:(a) 4 a+3 b(b) 3 x3 y+7 x3 y

(3) Simplify by collecting like terms: 6 a+10b−2 a−5 b


REFLECTION



(a) 2 x+3 y+5 x+ y¿7 x+4 y

(b) 3 x2 y2 z+5 xyz−x2 y2 z¿2 x2 y2+5xyz

(c) 3 y−1−x¿3 y−1−x

Activity 1(1)(1.1) x−16(1.2) a+5(1.3) 2 x(1.4) 2 x

Activity 2a) constantb) variablec) Variablesd) Constante) constant

Activity 3

Answer : C - 2 xy

1.a)Term(pattern)

1 2 3 4 5 20 n

No of match sticks

6 11

16 21 26 101

5n + 1

b) T 4=5 ( 4 )+1=¿ 21 T 5=¿ 5(5) + 1 = 26 T 20=¿5(20) + 1 =101

c) The rule: T 1=5 (1 )+1 = 6 T 2=5 (2 )+1 = 11 T 3=5(3)+1 = 16 T n=5 (4 )+1=21

d) The formula: T n=5n+1



TOPIC: Algebraic Expressions – Expand and simplify algebraic expressionsCONCEPTS AND SKILLS TO BE ACHIEVED

By the end of the lesson, learners should know and be able toUse the commutative, associative and distributive laws for rational numbers and laws of exponents to:

Add and subtract like terms in algebraic expressionsMultiply integers and monomials by:

monomials binomial trinomials

Divide the following by integers or monomials: monomials binomials trinomials Simplify algebraic expressions involving the above operations


PRIOR KNOWLEDGE



4 min Calculate the following without a calculator:(a) 5(3 + 4)(b) 5 × 3 + 5 × 4


(c) 6 × 3 + (4 + 6)(d) (6 + 4) + 3 × 6(e) 3 × (4 × 5)(f) (3 × 4) × 5


5 min 1.(a) 2 x+6+x2

x2+2 x+6 ; 2nd degree

(b) 3 x4 y2−5 x2 y3+2 x3 y−x6 y5+8xy+1 −x6 y5+3 x4 y2+2 x3 y−5 x2 y3+8 xy+1; 6th degree

2.(a) 39 z2−14 y2 z+19 y z2−22 xyz

yes

(b) 6 x2 y6−x−5

no


10 min Educator asks learners to discuss: The use of commutative, associative and distributive laws for rational numbers and the laws of exponents.

Expand and simplify Algebraic expressions:

Properties: Commutative property

This law applies to addition and multiplication of numbers; it

Expand and simplify algebraic expressions

Use the commutative, associative and distributive laws for rational numbers and


tells us that even if we change the order of the numbers we still get the same answer.

4 x+ (−2 x )=−2 x+4 x −5 x× 3 y=3 y×−5 x

Associative propertyThis rule also applies to addition and multiplication; it allows us to group numbers when adding or multiplying and still get the same answer. 3 x+ (−5 x )+6 x= [3 x+(−5 x ) ]+6 x=3x+[−5x+6 x ]

3 x× (−5 x ) ×6 x= [3 x× (−5 x ) ]× 6 x=3x ×[−5 x ×6 x ] Distributive property

When multiplying across addition or subtraction this property allows us to redistribute the numbers and still get the same answer.

7 (−5 x+2 )=[ 7×−5 x ]+ [ 7×2 ]=−35 x+14

Educator asks learners to discuss the rules regarding addition of like terms

Addition of like terms.

Example: Simplify the following algebraic expressions: (a2+3a−1 )+(2a2−5a+7) and (7a2+5 a−4 )−(2a2−7 a+3)

NOTE: We can add and subtract only like terms.

Example: Addition

laws of exponents to:

Add and subtract like terms in algebraic expressions


Method 1: HorizontallyThink DoWrite down the problemMultiply out the bracketsCollect the like termsAdd like terms separately and write answer in descending order of x

(a2+3 a−1 )+(2a2−5 a+7)a2+3 a−1+2 a2−5 a+7a2+2a2+3 a−5 a−1+73a2−2 a+6

Method 2: Add like terms vertically

a2+3 a−12a2−5 a+73 a2−2 a+6(Activity 1)Educator asks learners to discuss the rules regarding subtraction of like terms

Subtraction of like terms

Example: SubtractionMethod 1: Horizontally


Think DoWrite down the problemMultiply out the bracketsCollect the like termsAdd like terms separately and write answer in descending order of x

(7a2+5 a−4 )−(2a2−7a+3)7a2+5 a−4−2a2+7 a−37a2−2a2+5 a+7a−4−35a2+12 a−7

Method 2: Vertically7 a2+5a−42 a2−7 a+35a2+2a−7

(Activity 2)

CLASSWORK 10 min Activity 1Simplify:

(5 p2 q+ p q2−6 )+(−p2q−7 ) + (-pq2+14 ¿

Activity 2Simplify:

( 4b2+9b−1 )−(11b2−8b+19)CONSOLIDATION/ 1 min Simplify the following expressions:


CONCLUSION AND OR HOMEWORK

(a) 3a2−2a−7+3 a+4−2a2+9 a(b) x2+6 xy+5 y2−6 xy+2 y2

(c) Add 2a2+5 ab−7b2

3a3+6 a2−7 ab+8 b2

¿¿

(d) Subtract 2a2+5 ab−7b2

3a3+6 a2−7 ab+8 b2

¿¿

REFLECTION




Calculate the following without a calculator:

(a) 5(3 + 4)= 5 × 3 + 5 × 4

(b) 6 × 3 + (4 + 6)= (6 + 4) + 3 × 6

(c) 3 × (4 × 5)= (3 × 4) × 5

1.5 p2 q+ p q2−p2q−7−p q2+14

¿4 p2q+72.

5 p2 q+ p q2−p2q−7−p q2+14−6 ¿4 p2q+73.

b2+9b−1−11b2+8b−19 ¿−7b2+17 b−20

1.a)2 x+6+x2

x2+2 x+6 ; 2nd degree

b) 3 x4 y2−5 x2 y3+2 x3 y−x6 y5+8xy+1 −x6 y5+3 x4 y2+2 x3 y−5 x2 y3+8 xy+1; 6th degree

2.a)39 z2−14 y2 z+19 y z2−22 xyz

yes

b)6 x2 y6−x−5

no





By the end of the lesson, learners should know and be able toUse the commutative, associative and distributive laws for rational numbers and laws of exponents to: Add and subtract like terms in algebraic expressions Multiply integers and monomials by:

monomials binomial trinomials

Divide the following by integers or monomials: monomials binomials trinomialsSimplify algebraic expressions involving the above operations


PRIOR KNOWLEDGE



4 min Simplify without a calculator(1) (-9)[(-1) + (-3)](2) -7[9 – (-2)](3) 12 + (13 – 1)(4) 2 (5 x+8 )−2 x−3


involving whole numbers percentages, and


(5) (2 a+5 )−¿) decimal fractions in financial context:





5 min Simplify the following expressions:(a) 3a2−2 a−7+3 a+4−2 a2+9 a=a2+10 a−3

(b) x2+6 xy+5 y2−6 xy+2 y2=x2+7 y2

(c) Add 2 a2+5ab−7 b2

3 a3+6 a2−7ab+8 b2

3 a3+8 a2−2ab+b2

(d) Subtract 2 a2+5ab−7 b2

a3+6 a2−7 ab+8 b2

−3 a3−4 a2+12 ab−15 b2


10 min Educator asks learners to discuss some of the important conventions in Algebra and the laws of exponents.


Important conventions:

x+x is written as 2 x Repeated addition can be shortened

x× x× x × x can be written as x4

Exponential form instead

x ÷ y is written as xy

Fractional form preferably

4 x instead of x 4 Numerical coefficient firstx and not 1 x ;− y and not −1 y

No need to show 1

6 xyz and not 6 zxy Arrange variables alphabeticallya+b=a+b and not ab Unlike terms and cannot be

added(Activity 1)

Laws of exponents Product of powers: am× an=am+n , a > 0

x6=x2+ 4∨x3+3∨x1+5

Quotient of powers: am÷ an=am−n , if m>n a10÷ a6=a10−6=a4

Raising a power to a power: ¿, a > 0 x6=x3× x3

Raising a product to a power: ¿ ¿


Educator asks learners to discuss multiplication of integers and monomials by monomials, binomials and trinomials. They must use the following examples in their discussions:

Simplify the following Expressions:5 x2 y ×−2x5 y2 ; −5(a+2 b) and 5a(3a+8 b−c)

Multiplication:

Multiply monomialsSimplify: 5 x2 y ×−2x5 y2

¿−10x7 y3

Multiply integer with a binomialSimplify: −5(a+2b) ¿−5a−10 b

Multiply monomial with a trinomialSimplify: 5a(3a+8 b−c) ¿15a2+40 ab−5 ac

Educator asks learners to discuss division of monomials, binomials and trinomials by integers or monomials and how will they simplify the following example?

Divide monomials: 15 a÷ 3b

Divide binomials: 14 x2+7 x14 x

Divide trinomials: 4 x2 y4−6 x y3+ y2

4 xy


Division:

Divide monomialsSimplify: 15a ÷ 3b

= 15 a3 b

¿5ab

Divide binomials

Simplify: 14 x2+7 x14 x

¿ 14 x2

14 x+ 7 x

14 x

¿ x+12

Divide trinomials

Simplify: 4 x2 y4−6 x y3+ y2

4 xy

¿ 4 x2 y4

4 xy−6 x y3

4 xy+ y2

4 xy

¿ x y3−3 y2

2+ y

4 xCLASSWORK 10 min Activity 1

Choose the correct form: (1) 5 y∨ y 5 (2) x2+ x2=2x2 or x2+ x2=2x4

(3) −x (3 x+1 )=−3 x2+1 or −x (3 x+1 )=−3 x2−x


Activity 2(1) Simplify:(1) −3 r3 ×−r ×5 r2

(2) −3 x (5 x−11)(3) 10(6 x3 y−7 x2 y−12 x )

Activity 3(1) Simplify:

(1.1) −10 p12q8

40 p7q

(1.2) 12 x12 y10−3 x ( 2 x9 y12)8 x3 y10

(1.3) 3x20 y2−9 x4 y4+x y 9

9 x y2


1 min (1) 5( p2+3 pq−6 q2)(2) −2 x (x2+3 x−6)

(3) 12 x2 y−6 x y2

3 xy

(4)14 a+21 b−35

7

(5) 9 p4 q5−6 p3q+3 pq3 pq

REFLECTION




1.Simplify without a calculator(1) (-9)[(-1) + (-3)] = 9 + 27 = 36 or

(-9)[-4] = 36(2) -7[9 – (-2)] = -63 – 14 = -77 or

-7[11] = -77(3) 12 + (13 – 1) = 12 + 12 = 24(4)2 (5 x+8 )−2 x−3=10 x+16−2 x−3=8 x+13(5)(2a+5 )−(4 a−5 )=2a+5−4 a+20=−2a+25

Activity 1Choose the correct form:

(1) 5 y∨ y 5 Answer: 5 y

(2) x2+ x2=2x2 or x2+ x2=2 x4 Answer: x2+ x2=2x2

(3) −x (3 x+1 )=−3 x2+1 or −x=−3 x2−xAnswer: −x (3 x+1 )=−3 x2−x

Activity 2Simplify:(1) −3 r3 ×−r ×5 r2=15r6

(2) −3 x (5 x−11 )=−15 x2+33 x(3) 10 (6 x3 y−7 x2 y−12 x )=60 x3 y−70 x2 y−120 x

Activity 3Simplify:

(1) −10 p12q8

40 p7q

Solution: −p12−7 q8−1

4=−p5 q7

4

(1) 5( p2+3 pq−6q2)(2) −2 x (x2+3 x−6)

(3) 12 x2 y−6 x y2

3 xy

(4)14 a+21 b−35

7

(5) 9 p4 q5−6 p3q+3 pq3 pq


(2)12 x12 y10−3 x (2 x9 y12)

8 x3 y10

Solution: 12 x12 y10

8 x3 y10 −6 x10 y12

8 x3 y10 =3x9

2−3 x7 y2

4=6 x9−3 x7 y2

4

(3) 3 x20 y2−9 x4 y4+x y 9

9 x y2

Solution: 3 x20 y2

9 x y2 −9 x4 y4

9 x y2 + x y9

9x y2 =x19

3−x3 y2+ y7

9





By the end of the lesson, learners should know and be able to Determine the squares, cubes, square roots and cube roots of single algebraic terms or like algebraic terms, Determine the numerical value of algebraic expressions by substitution, Multiply integers and monomials by polynomials,

RESOURCES DBE workbook (Page 72 – 92), Sasol-Inzalo workbook (Page 152 – 143), textbook, ruler, pencil, eraser, calculators, DVDS ((Disc 2 : GDE 11 03 2014 & GDE 13 02 2014 )

PRIOR KNOWLEDGE



4 min (1)Write in exponential form:(a) 2 × 2 (b) 2 × 2 × 2(2) Simplify without a calculator:(a) √4 (b) 3√27 (c) √16+9(3) Complete:(a) ( ym )n

=… (b) ( xy )m=…REVIEW AND CORRECTION OF HOMEWORK

5 min (1) 5( p2+3 pq−6 q2)Solution: 5 p2+15 pq−30 q2

(2) −2 x (x2+3 x−6)


Solution: −2 x3−6 x2+12 x

(3) 12 x2 y−6 x y2

3 xy

Solution: 12 x2 y3 xy

−6 x y2

3xy=4 x−2 y

(4)14 a+21b−35

7

Solution: 14 a

7+21 b

7−35

7=2 a+3b−5

(5) 9 p4 q5−6 p3q+3 pq3 pq

Solution: 9 p4 q5

3 pq−6 p3 q

3 pq+ 3 pq

3 pq=3 p3 q4−2 p2+1


10 min Educator asks learners to discuss squares and square roots, cube and cube roots with examples

Squares and cubes:

A square is a number with two identical factorsExample 9 = 3×3 = 32

A cube is a number with three identical factorsExample 27 = 3×3×3 = 33

Determine the squares, cubes, square roots and cube roots of single algebraic terms or like algebraic terms,


Square roots and cube roots:

Finding square roots means finding the two identical factors that multiply together to form the number, which is the square of the factor, for example:

Example: √36 = 6 because 6 × 6 = 36

Finding cube roots means finding the three identical factors that multiply together to form the number, which is the cube of the factor, for example:

Example: 3√64 = 4 because 4 × 4 × 4 = 64Educator asks learners to discuss: Squares, cubes, square roots and cube roots of single algebraic termsHow will you simplify the following examples?

3√−27 x6 y12 ; √49 x6 y2 ; √25 x2−9 x2 ; 3√ 81 x5 y 8

3 x2 y2 25 ;

( x y3 )2× 2 ( xy )3

Examples: 3√−27 x6 y12

¿−3 x2 y4

√49 x6 y2

¿7 x3 y


√25 x2−9 x2

¿√16 x2

¿4 x

3√ 81 x5 y 8

3 x2 y2 25

¿ 3√27 x3 y6

¿3 x y2

( x y3 )2× 2 ( xy )3

¿ x2 y6 ×2 x3 y3=2 x5 y 9

CLASSWORK 10 min Simplify:a) √36 x4

b) (2) 3√−8 x12 y15

c) √ 162 x7 y5

2xy

d) √(2 x+3)2

e) 3√ x3

64


f)(2x2 y )3

(4 x2 y2 )2


AND OR HOMEWORK

1 min Simplify:a) 3 (5 x2)2

b) ( 3 x2

6 y )2

c) √144 x4 ×9 y2

d) √16 x4+9 x4

e) 3√27 x6

REFLECTION




(1)(a) 2 × 2 (b) 2 × 2 × 2 (a) 22 (b) 23

(2)(a) √4 (b) 3√27 (c) √16+9(a) 2 (b) 3 (c) √25=5

(3)(a) ( ym )n

= ymn (b) ( xy )m=xm ym

(1) √36 x4=6 x2

(2) 3√−8 x12 y15=−2 x4 y5

(3) √ 162 x7 y5

2xy=√ 81 x6 y4

1=9 x3 y2

(4) √ (2 x+3 )2=2 x+3

(5) 3√ x3

64= x

4

(6)(2 x2 y )3

(4 x2 y2 )2= 8 x6 y3

16 x4 y4 =x2

2 y

(1) 5( p2+3 pq−6 q2)Solution: 5 p2+15 pq−30 q2

(2)−2 x (x2+3 x−6)Solution: −2 x3−6 x2+12 x

(3)12 x2 y−6 x y2

3 xySolution: 12 x2 y3 xy

−6 x y2

3xy=4 x−2 y

(4)14 a+21b−35

7Solution: 14 a

7+21 b

7−35

7=2a+3 b−5

(5)9 p4 q5−6 p3q+3 pq3 pq

Solution: 9 p4 q5

3 pq−6 p3 q

3 pq+ 3 pq

3 pq¿3 p3 q4−2 p2+1





By the end of the lesson, learners should know and be able to Determine the squares, cubes, square roots and cube roots of single algebraic terms or like algebraic terms, Determine the numerical value of algebraic expressions by substitution,

Multiply integers and monomials by polynomials,RESOURCES DBE workbook (Page 72 – 92), Sasol-Inzalo c textbook, ruler, pencil, eraser, calculators, DVDS (Disc 2 :

GDE 11 03 2014 & GDE 13 02 2014 )PRIOR KNOWLEDGE



3 min Complete the table by substituting the valuesx -2 5 -1y 4 -3 2

x+ y7+x− y

(2 y ) ( 4 x )+5


5 min Simplify:(1) 3 (5 x2)2

=3 (25 x4 )=75 x4

(2) ( 3 x2

6 y )2

= 9 x4

36 y2 =x4

4 y2


(3) √144 x4 ×9 y2=12 x2×3 y=36 x2 y

(4) √16 x4+9 x4=√25 x4=5x2

(5) 3√27 x6=3 x2


10 min Educator asks learners to discuss what substitution is with examples

Substitution:

When we determine the numerical value of an expression we replace the variable with a constant.

Examples:(1) If x=−3∧ y=5 , find the values of:(1.1) x+ y ¿−3+5=2

(1.2) ( y− x)(x− y ) ¿ [5− (−3 ) ] [−3−(5 ) ]=[ 8 ] [−8 ]=−64

(1.3)5 x4 y

¿5(−3)4(5)

=−1520

=−34

(2) If x=2 and y=5 x−7, find the value of y


∴ y=5 (2 )−7∴ y=3

(3) Calculate the area of the rectangle with sides 2 x and ( x+1 ) given that x=¿6

A=l× b∴ A=2 x × ( x+1 )∴ A=2 x2+2 x

(Activity 1)Educator asks learners to discuss the following substitutions in their respective groups

Substitution in square and cube rootsExample: If x=3∧ y=4 , calculate the value of each of the following

(a) √ x2+ y2=√32+42=√25=5(b) 3√ x3=3√33=3

Substitution in formulaeExample: Calculate the perimeter of the following shapes if x=4

(a) x+3 (b)

x+1 5 x7 x

5 xP=2 ( l+b )∴P=2 ( x+3+x+1 ) P=5 x+5 x+7 x∴P=2 (2 x+4 )∴P=17 x∴P = 2[2(4) + 4] = 24 P = 17(4) = 68


CLASSWORK 10 min Activity 1:

(1) Given y=x2−4, calculate y when x=−7(2) Given that x=3, find the perimeter of a triangle with sides

5 x;7 x∧3 x(3) If a=4 and b=2 a, find the value of b . Use the formula of b to

find c, if c=2b−2 a


3min (1) Evaluate the following, if a=3∧b=−2(a) 2a2+3b(b) 2ab−a2

(c) b2+2ab(2) If x=2 find the values of:(a) √ x6

(b) 3√ x12

REFLECTION


Answers Week 8 Lesson 5 & 6

Mental Maths Class work

(1) Given y=x2−4, calculate y when x=−7Solution: y=(−7)2−4=49−4=45

(2) Given that x=3, find the perimeter of a triangle with sides 5 x ;7 x∧3 x

Solution: P=5 x+7 x+3 x=12 x=12 (3 )=36(3) If a=4 and b=2a, find the value of b . Use the

formula of b to find c, if c=2b−2 aSolution: b=2 (4 )=8∧c=2 (8 )−2 ( 4 )=8

(1) If x=−3∧ y=−2, calculate the numerical value of:(a) √ x2.√ y2

Solution: √(−3)2 .√(−2)2=√9 .√4=3 × 2=6(b) 3√−3 x2−√x2+ y4

Solution: 3√−3¿¿¿ 3√−27−√9+16=−3−√25

¿−3+5=2


HomeworkSimplify:

(1) 3 (5 x2)2=3 (25 x4 )=75 x4

(2) ( 3 x2

6 y )2

= 9 x4

36 y2 =x4

4 y2

(3) √144 x4 ×9 y2=12 x2×3 y=36 x2 y

(4) √16 x4+9 x4=√25 x4=5 x2

(5) 3√27 x6=3 x2





By the end of the lesson, learners should know and be able to Divide polynomials by integers or monomials Determine the square of a binomial,

RESOURCES DBE workbook (Page 72 – 92), Sasol-Inzalo workbook (115 – 143), textbook, ruler, pencil, eraser, calculators, DVDS ((Disc 2 : GDE 11 03 2014 & GDE 13 02 2014 )

PRIOR KNOWLEDGE



3 min Check whether ( x− y ) ( x+3 y )=x2+2 xy−3 y2 by substitution and completing the table

x -1 0 2y 0 1 0

( x− y ) ( x+3 y )x2+2 xy−3 y2








5 min (1) Evaluate the following, if a=3∧b=−2(a) 2a2+3

Solution: 2 (3 )+3=9

(b) 2ab−a2

Solution: 2 (3 ) (−2 )−(3 )2=−21

(c) b2+2abSolution: (−2)2+2 (3 ) (−2 )=−8

(2) If x=2 find the values of:

(a) √ x6

Solution: ¿ x3=23=8

(b) 3√ x12

Solution: = x4=24=16


10 min Educator asks learners what binomials are and give examples

BinomialsA binomial is an expression containing two terms.All the variables have positive whole numbers as exponents

Product of two binomials

Example: Simplify (a+b)(x+ y)


Using a diagram:

We can compare this to the area of a rectangle:

Area of ABCD = l ×b=(a+b)(x+ y)Area of ABCD consists of: Area 1 + Area 2 + Area 3 + Area 4 ax+ay+bx+by ∴ ( a+b ) ( x+ y )=ax+ay+bx+by

Using multiplication

(a+b ) ( x+ y )=ax+ay+bx+by


Steps:1 the two first terms: a× x F2 the two outer terms: a × y O3 the two inner terms: b× x I4 the two last terms: b× y L

Use the acronym FOIL, to help you remember this order

Example: Simplify (a+4)(a−2) ¿a2−2 a+4 a−8 ¿a2+2a−8

(Activity 1)

Educator asks learners what the square of a binomial means

Square of a binomialWhen two identical binomials are multiplied by each other, the process

is called squaring a binomial.

Example: Simplify ( x+ y ) (x+ y ) can also be written as ( x+ y )2

¿ x2+ xy+xy+ y2=x2+2 xy+ y2


(Activity 2)CLASSWORK 10 min Activity 1

Simplify:


(1) (x+4 )( x+3)(2) (x+3)(x−3)(3) ( y+½)( y−½)(4) (3 p+q)( p−q)

Activity 2(1)

(2) (x+3)(x+3)

(3) (x+ 12 )

2

(4) ( x−5 )2

(5) ( x+ y )2−( x− y )2


3 min Simplify:(1) (8−6 c )(8+6 c )(2) (2 x−6)(8 x−5)(3) (2 a+b)(c+d)(4) (2 x−7 )2

(5) 3 ( x−5 )2+( x−3)(x+3)REFLECTION

Answers Week 8 Lesson 7 & 8



x -1 0 2y 0 1 0

( x− y ) ( x+3 y ) 1 -3 4x2+2 xy−3 y2 1 -3 4

Conclusion:Yes ( x− y ) ( x+3 y )=x2+2 xy−3 y2

Activity 1Simplify:

(1) (x+4 )( x+3)Solution: x2+3 x+4 x+12=x2+7 x+12

(2) (x+3)(x−3)Solution: x2−3 x+3x−9=x2−9

(3) ( y+½)( y−½)

Solution: y2−12

y+ 12

y−14= y2−1

4(4) (3 p+q)( p−q)

Solution: 3 p2−3 pq+ pq−q2=3 p2−2 pq−q2

Activity 21. ∴ ( p+q ) ( p+q )=( p+q )2=p2+ pq+ pq+q2=p2+2 pq+q2

(1) (x+3)(x+3)Solution: x2+6 x+9

(2) (x+ 12 )

2

Solution: x2+ x+ 14

(3) ( x−5 )2

Solution: x2−10 x+25(4) ( x+ y )2−( x− y )2

Solution: (x¿¿2+2xy+ y2)−(x2−2xy+ y2)¿ ¿ x2+2 xy+ y2−x2+2 xy− y2=4 xy

1.a) 2 a2+3

Solution: 2 (3 )+3=9

b) 2ab−a2

Solution: 2 (3 ) (−2 )−(3 )2=−21

c) b2+2 abSolution: (−2)2+2 (3 ) (−2 )=−82.If x=2 find the values of:

a) √ x6

Solution: ¿ x3=23=8

b) 3√ x12

Solution: = x4=24=16





QUESTION 11.1 Simplify the following expression:

3 a2−2 a−7+3 a+4−2a2+9 a(2)

1.2 Simplify:1.2.

15 x2 y ×−2 x5 y2 (2)

1.2.2

√144 x4 ×9 y2 (2)


1.3 If x=−3∧ y=5 , find the values of:1.3.

1x+ y (1)

1.3.2

5x4 y

(3)

[10]QUESTION 2

2.1 A certain number is decreased by 16 (2)2.2 A number that is 5 more than a (2)2.3 Solve the equations by means of inspection:

2.3.1

10+a=4 (2)

2.3.2

x−33=50 (2)

2.4 In two years, Laaiqah will be twice as old as Hamzah. Set up an equation that expresses this situation.

(2)


[10]

[20]TOTAL: 20



QUESTION 1

1.1 3 a2−2 a−7+3 a+4−2a2+9 a (2)

¿a2+10a−3 (2)

1.2

1.2.1 5 x2 y ×−2 x5 y2 (2)

¿−10x7 y3

1.2.2 √144 x4 ×9 y2 (2)

¿12x2 ×3 y=36 x2 y (2)

1.3

1.3.1

−3+5=2 (1)

1.3.2

5(−3)4(5)

=−1520

=−34 (3)

[10]

QUESTION 2

2.1 x−16 (2)

2.2 a+5 (2)

2.3


2.3.1

10−6=4∴a=−6 (2)

2.3.2

83−33=50∴ x=83 (2)

2.4 x+2=2(x−5) (2)

[10]

[20]

TOTAL: 20





By the end of the lesson, learners should know and be able to Solve equations by using exponential laws

RESOURCES DBE workbook(p.24-26; p.136-141), Sasol-Inzalo Textbook(p.148-156), textbook, ruler, pencil, eraser, calculators, DVD(Grade 9 Term 1 Disc 2-GDE 18 03 2014; GDE 20 03 2014)

PRIOR KNOWLEDGE Basic Mathematics operations, Substitution, Number sentences, Algebraic expressions



4 min Simplify:(a) a2+a2=…(b) a2× a3=…(c) a6÷ a4=…

p.132


5 min (1) Solve by inspection:

(a) x−6=3Solution: 9−6=3∴ x=9


(b)3x2

=6

Solution: 3(4 )

2=12

2=6∴ x=4

(2) Solve for x :

(a) 8 x+3=27Solution: 8 x+3−3=27−3 ∴8 x=24

∴8 x8

=248

∴ x=3

(b) 2 ( x+4 )=x+10Solution: 2 x+8=x+10 ∴2 x−x+8−8=x−x+10−8 ∴ x=2

(c)x2+ x+1

3=17

Solution: LCM: 6

∴× 6 ¿ x2+ x+1

3=17

1

∴ x2

×6+ x+13

×6=171

× 6

∴3 x+2 (x+1 )=102∴3 x+2x+2=102∴5 x+2=102∴5 x=100÷ 5¿∴ x=20



10 min Educator solves the following equations with the learners.

1. Solve for x :(a) x2+4=20

Solution: x2+4−4=20−4 ∴ x2=16∴ x2=24∴ x=4

(b) x4

x2 −12=24

Solve equations by using exponential laws

CLASSWORK 10 min Solve for x :

a. 5 x2−x0=80

b. 3√27 x6=12


1 min Solve the following equations:(a) x3=64(b) 4 x+2=64

(c) 4−x= 116

REFLECTION




Simplify:(a) a2+a2=2 a2

(b) a2× a3=a5

(c) a6÷ a4=a2

Solve for x :

(a) 5 x2−x0=80Solution: 5 x2−1=80 ∴5 x2−1+1=80+1 ∴5 x2=81

∴ 5 x2

5=81

5 ∴ x2=16,2 ∴√x2=√16,2 ∴ x=4

(b) 3√27 x6=12Solution: ∴3 x2=12 ∴ x2=4∴ x2=22 ∴ x=2OR √ x2=√4∴ x=2





By the end of the lesson, learners should know and be able to Solve equations by using exponential laws.





3 min Simplify:(d) (a2 )3=…(e) √ x6=…(f) 3√ x12=…

p. 132


5 min (a) x3=64Solution: x3=43 ∴ x=3OR 3√ x3=3√64∴ x=4

(b) 4 x+2=64Solution: 4 x+2=43 ∴ x+2=3∴ x=1

OR by inspection 41+2=43=64


(a) 4−x= 116

Solution: 4−x=4−2 ∴−x=−2∴ x=2


10 min Educator asks learners to discuss: Solving equations involving laws of exponents

Solving equations with a variable in the base

Example: Solve for x : x3=23

∴ x=2 The powers on either side are equal and the exponents are equal it follows that the basis are equal as well. Test your answer by substitution!

Example: Solve for x : 3+x4=24+3

Method ApplicationConsider the equationWe isolate the variables by subtracting 3 from both sidesIn order for the equation to be true, if the exponents are the same, then the bases have to be equalTest your answer

3+x4=24+3∴3+x4−3=24+3 – 3

∴ x4=24

∴ x=2

Solve equations by using exponential laws


Example: Solve for x : x3−6=2 ∴ x3−6+6=2+6 ∴ x3=8 ∴ x3=23 ∴ x=2Example: Solve for x : √4 x2=8 ∴2 x=8

∴ x=4

Educator asks learners to discuss equations where the base is raised to a power containing the variable.

Solving equations with a variable as an exponent.

Example: Solve for x : 3x−3=24

3x−3=24∴3x−3+3=24+3∴3x=27∴3x=33

∴ x=3

Add 3 to both sides of the equationSimplifyExpress the RHS as a power of 3Same basis

Example: Solve for x: 5x−1=125 ∴5x−1=53


∴ x−1=3∴ x=4CLASSWORK 10 min Solve for x :

(a) x0=1(b) 2x=16(c) 4 ×5x=100

(d) 3x= 127


3 min Solve the following equations:(a) 2 x3=−16(b) √49 x4=28

REFLECTION




a) (a2 )3=a6

b) √ x6=x3

c) 3√ x12=x4

(a) x0=1Solution: x can be any number except 0

(b) 2x=16Solution: ∴2x=24 ∴ x=4

(c) 4 ×5x=100

Solution: 4× 5x

4=100

4 ∴5x=25∴5x=52 ∴ x=2

(d) 3x= 127

Solution: 3x= 1

33

3x=3−1 ∴ x=−1

a) x3=64Solution: x3=43 ∴ x=3OR 3√ x3=3√64∴ x=4

b) 4 x+2=64Solution: 4 x+2=43 ∴ x+2=3∴ x=1

OR by inspection 41+2=43=64

c) 4−x= 116

Solution: 4−x=4−2 ∴−x=−2∴ x=2





By the end of the lesson, learners should know and be able to Substitute into equations to solve for a variable





3 min Solve the following equations(a) 4−3 x=−2(b) 4 (2x−1 )=−8

p.132


5 min (a) 2 x3=−16

Solution: 2x3

2=−16

2 ∴ x3=−8∴ 3√x3= 3√−8∴ x=−2

(b) √49 x4=28Solution: ∴7 x2=28 ∴ x2=4∴ x2=22 ∴ x=2


Or √ x2 = √4 ∴ ᵡ = 2LESSONPRESENTATION/DEVELOPMENT

10 min Educator asks learners to use substitution to solve the following problems

Substitution into equationsExample:

Determine the value of y, in 2 y=2 x+35

,if x=−6

∴2 y=2(−6)+35

∴2 y=−95

∴10 y=−9

∴ y=−910

Substitute the value of -6 for x

Simplify RHS

× LCM: 5Divide by 10

Example: If y=2x2+4 x+3 , calculate y when x=−2

∴ y=2¿∴ y=2 ( 4 )−8+3∴ y=8−8+3∴ y=3

Solve equations by using substitution.

CLASSWORK 10 min Determine the values of the following equations, for the given values:

(1) y= 5t−6

−7 t if t=−1

(2) t=√16 x2+1−3

if x=−2

(3) m=3 x2−42

if x=3

CONSOLIDATION/CONCLUSION AND

3 min (1) Determine the values of the following equations, for the given


OR HOMEWORKvalues: y=7 (−x )+7

−2 xif x=7

REFLECTION



1.Solve the following equations(a) 4−3 x=−2

Solution: ∴−4+4−3 x=−2−4 ∴−3 x=−6∴ x=2

(b) 4 (2x−1 )=−8

Solution: 4 (2x−1)

4=−8

4 ∴2 x−1=−2 ∴2 x−1+1=−2+1 ∴2 x=−1

∴ x=−12

OR remove brackets by applying the distributive law

Determine the values of the following equations, for the given values:

(1) y= 5t−6

−7 t if t=−1

Solution: ∴ y=5 (−1 )−6

−7 (−1)

∴ y=56+7

∴ y=7 56

(2) t=√16 x2+1−3

if x=−2

Solution: ∴t=4 x+1−3

∴t=4 (−2 )+1−3

∴t=−7−3

∴t=2 13

(a) 2 x3=−16

Solution: 2 x3

2=−16

2 ∴ x3=−8

∴ 3√x3=3√−8 ∴ x=−2

(b) √49 x4=28Solution: ∴7 x2=28 ∴ x2=4 ∴ x2=22

∴ x=2

Or √ x2 = √4 ∴ ᵡ = 2


(1) m=3 x2−42

if x=3

Solution :∴m=3 ¿¿

∴m=232

∴m=11,5




By the end of the lesson, learners should know and be able to Solve equations with fractions.





5 min Solve the following equations(c) 2 x+1=3(2 x−1)(d) ( x+2 ) (x−4 )=x2+5 x−1

p.132



2min1. y=7 (−7 )+7

−(7)

∴ y=−42−7

∴ y=6


10 min Educator asks learners how they will go about solving equations with fractions Equations with fractions

The first step is to find the lowest common denominator (LCD)/(LCM)Change the fraction to fractions with the same denominatorOnce the denominators are the same, we can simplify both sides and

work only with the numerators

Example: Solve for x : x4+ x+2

4=6

4

x+x+24

=64

∴ 2x+24

=64

∴2 x+2=6∴2 x=4∴ x=2

Simplify the LHS

Simplify

× 4 on both sides

Example: Solve for x : 3 x−15

=2x7

∴ 3 x−15

× 77=2 x

7× 5

5 [ multiply the RHS with 77 and the LHS with

55 ]

Solve equations with fractions.


∴7(3 x−1)

35=10 x

35×35¿∴21 x−7=10 x∴11 x=7

∴ x= 711


CLASSWORK 10 min Solve for x :

a) 3 x+2

5= x−1

4

b)13

x−53=−2

3x+ 1

3


1 min (1) Solve the following equations:

(a)x2=4

(b)3x−4

4−3 x−2

2=2

REFLECTION




(a) 2 x+1=3(2 x−1)Solution: 2 x+1=6 x−3 ∴2 x−6 x+1−1=6x−6 x−3−1 ∴−4 x=−4 ∴ x=1

(b) ( x+2 ) (x−4 )=x2+5x−1Solution: x2−2 x−8=x2+5 x−1 ∴ x2−x2−2 x−5 x−8+8=x2−x2+5 x−5 x+8 ∴−7 x=8

∴ x=−87

Solve for x :

a)3 x+2

5= x−1

4

∴ 3 x+25

× 44= x−1

4× 5

5

∴ 4 (3x+2 )20

=5 (x−1 )20

×20¿∴4 (3 x+2 )=5 ( x−1 )∴12 x+8=5 x−5∴7 x=−13

∴ x=−137

b)13

x−53=−2

3x+ 1

3

Solution: 3 ×13

x−53

×3

¿3×−23

x+ 13

×3

∴ x−5=2 x+1∴−x=6∴ x=−6

1. y=7 (−7 )+7−(7)

∴ y=−42−7

∴ y=6




CONCEPTS AND SKILLS TO BE ACHIEVEDBy the end of the lesson, learners should know and be able to

Analyse and interpret equations that describe a given situation





4 min Complete the tableVerbal description Algebraic languageThe sum of a number and two x+2A number that is five more than aThe difference between two and a numberA number increased by seven

p.132


5 min (a) 2 x3=−16

Solution: 2 x3

2=−16

2 ∴ x3=−8∴ 3√x3= 3√−8∴ x=−2


(b) √49 x4=28Solution: ∴7 x2=28

∴ x2=4∴ x2=22 ∴ x=2LESSONPRESENTATION/DEVELOPMENT

10 min Educator asks learners the steps to follow when dealing with problem solving

Problem solving stepsWord problems can be written as mathematical statements so that they can be solved mathematically.Steps to follow: Step 1: Identify what you have been asked to solve.Step 2: Let this value be x or any other variable Step 3: Identify what you have been given. Step 4: Write a number sentence: Step 5: Substitute your values and solve/calculate. Step 6: Write your answer with the correct SI units.

Example: On weekends Donny works in a restaurant. He earns a basic wage of R105 plus commission for every table he serves.

(a) Write an equation to show the amount y, that he can earn.(b) Find out how many tables he served on Friday night if he

earned R177.

xy=8x+105177=8 x+105∴8 x=177−105∴8 x=72∴ x=9

The number of tables he servedEquation to calculate his earningsEquation for the number of tablesSolving this equation



He served 9 tablesExample: Translate the following problem into an equation and then

solve it.Think of a number, multiply it by 2, and then add 3. If the answer is

15, find the original number.

x2 x+3=152 x+3−3=15−3∴2 x=12∴ x=6

The numberMultiply by 2, add 3, the answer is 15Solving this equation

The number is 6(Activity 1)

Problems involving ageExample: If we add the ages of a mother and her son, the total of

their ages is 60. The mother is currently 4 times as old as her son. Find their ages.

Let the son be x years old.The mother is 4 xyears old.∴ x+4 x=60∴5 x=60∴ x=12The son is 12 years old and the mother is 4 × 12 = 48 years old.Educator asks learners to do the following example in pairs(Activity 2)

CLASSWORK 10 min Activity 1


Two numbers added together equal 20. The difference between the numbers is 4. Find the numbers.

Activity 2A Father is currently 5 times as old as his son. If the difference

between their ages is 36 years, how old is the son now?


1 min 1.The sum of the ages of two brothers is 70 years. In 10 years’ time, Thabo will be twice as old as Tshepo was 8 years ago. How old are Thabo and Tshepo now?

REFLECTION




Verbal description

Algebraic language

The sum of a number and two

x+2

A number that is five more than a

a+5

The difference between two and a number

x−2

A number increased by seven

b+7

Activity 1The second number is: x−4. Add the numbers: x+x−4=20.Solve the equation: ∴2 x−4=20∴2 x=24∴ x=12The number and the second number are:

12 and (12 – 4) = 8

Activity 2Solution: Let the son’s age be x.The father’s age is 5x.

∴5 x−x=36∴4 x=36∴ x=9The son is 9 years old

a)2 x3=−16

Solution: 2x3

2=−16

2 ∴ x3=−8 ∴ 3√x3=3√−8 ∴ x=−2

b) √49 x4=28Solution: ∴7 x2=28

∴ x2=4 ∴ x2=22

∴ x=2





By the end of the lesson, learners should know and be able to Analyse and interpret equations that describe a given situation





4 min Complete the tableVerbal description Algebraic languageThe product of two and a numberA number that is half the sum of 2 and 3The quotient if x is divided by twoA number that is decreased by nine

p.132



5 min x+10=2 (62−x )∴ x+10=124−2 x∴ x+2 x=124−10∴3 x=114∴ x=38

The brothers are 32 and 38 years oldLESSONPRESENTATION/DEVELOPMENT

10 min Educator asks learners to REVISE the steps to follow when dealing with problem solving


Problems involving cost:ExampleThe cost of travelling by private car is R10 plus R1,50 per kilometre. If Thando paid R16 for the ride, use an equation to find the distance that he travelled.Solution:Let the number of kilometres be x

Total Cost =10+1, 50× x



16=10+1 ,50 x1 ,50 x=6x=4

He travelled 4 kilometersCLASSWORK 10 min 1.Tickets for the Gr12 dinner and dance costs R 250 per couple and

R 150 for a single person. Twice as many couple tickets were sold as single ones. Let x be the single tickets sold. The total income for

the ticket sales was R 65 000 .

a) What equation would you use to find x?b) Calculate how many single tickets were sold.c) How many people attended the dance altogether?

2. Henry has a certain amount of pocket money. His sister, Lucy, gets 23

of the amount of money that Henry gets. Altogether they get R600 pocket money.


1 min 1. Thirty five shots were fired at a target. Each time the target is hit the keeper has to pay R2,50. For every missed target he has to pay R1,50. If he makes a profit of R35, how many times did he hit the target?

Number of times missed 30−x


25 x − 15(30−x ) =35025 x−450+15 x=35025 x+15 x=350+45040 x=800

x=20He makes 20 hits

2. To rent a room in a certain building, you have to pay a deposit of R400 and then R80 per day.a) How much money do you need to rent the room for 10 days?b) How much money do you need to rent the room for 15 days?

a) R80 x 10 + R400 = R800 + R400 = R1 200b) R80 x 15 + R400 = R1 200 + R400 = R1 600

REFLECTION




Verbal description

Algebraic language

The product of two and a number

2 x

A number that is half the sum of 2 and 3

b=2+32

The quotient if x is divided by two

x2

A number that is decreased by nine

y−9

1.Tickets for the Gr12 dinner and dance costs R 250 per couple and R 150 for a single person. Twice as many couple tickets were sold as single ones. Let x be the single tickets sold. The total income for the ticket sales was R 65 000 .

a) What equation would you use to find x?

b) Calculate how many single tickets were sold.

c) How many people attended the dance altogether?

2. Henry has a certain amount of pocket

money. His sister, Lucy, gets 23 of the

amount of money that Henry gets. Altogether they get R600 pocket money.

x+10=2 (62−x )∴ x+10=124−2 x∴ x+2 x=124−10∴3 x=114∴ x=38

The brothers are 32 and 38 years old










4 min Solve the following:

1. 2( x−9 )=4

2. 3( x−1 )+2=4

p. 132


5 min 1.Number of times missed 30−x


25 x − 15(30−x ) =35025 x−450+15 x=35025 x+15 x=350+45040 x=800


2.a) R80 x 10 + R400 = R800 + R400 = R1 200b) R80 x 15 + R400 = R1 200 + R400 = R1 600


10 min Educator asks learners to REVISE AGAIN the steps to follow when dealing with problem solving


Problems involving perimeter and area:ExampleThe length of a rectangle is 10cm longer than its breadth. The perimeter is 48cm. Determine the area of the rectangle.



Solution:Let the breadth be x

The length =x+10

2×l+2×b=perimeter2 x+2( x+10 )=484 x=28x=7

Breadth = 7cm and length =17cm

Area

=l×b=17×7=119 cm2

CLASSWORK 10 min 1.The perimeter of a rectangle is 30cm. If the breadth is doubled and the length is not changed, the rectangle will be a square. Determine the dimensions of the rectangle.


1 min 1. If one side of a square is extended by 8m and the adjacent side is reduced by 6 m, a rectangle with perimeter of 100m is ob-tained. Determine the area of the original square.

REFLECTION




Solve the following:

1. 2( x−9 )=42 x−18=4

2 x=4+182 x=22x=11

2. 3( x−1 )+2=43 x−3+1=43 x=4+23 x=6 xx=2

Let the breadth be xIf the perimeter is 30m, AB + BC =15m

The length =15−x15−x=2 x15=2x+x15=3xx=5The original rectangle is 5m x 10m

1.Number of times missed 30−x

25 x − 15(30−x ) =35025 x−450+15 x=35025 x+15 x=350+45040 x=800


2.c) R80 x 10 + R400 = R800 + R400

= R1 200d) R80 x 15 + R400 = R1 200 +

R400 = R1 600






RESOURCES DBE workbook(p.24-26; p.136-141), Sasol-Inzalo Workbook 1 (p.148-156), textbook, ruler, pencil, eraser, calculators, DVD(Grade 9 Term 1 Disc 2-GDE 18 03 2014; GDE 20 03 2014)




4 min Give the formula for the following:1. Area of a rectangle

A=l×b2. Perimeter of a rectangle

2( l+b )3. Area of a triangle

A=12

b×h

4. Perimeter of a squareP=4 s

5. Area of a square

p.132


A=s2A


5 min Length =x+8

Breadth = x−6P=2 l+2 b100 =2( x+8 )+2( x−6 )100 =2 x+16+2 x−124 x=100−16+124 x=96x=24

Area =24 m×24 m=576 m2


10 min Educator now asks learners repeat the problem solving steps in a drill fashion




Problems involving distance, speed and time:A roller coaster rides against the wind for 36km at a speed of 12km/h in

order to complete the journey on time. What was his original speed?

Against wind

With wind

d 36km 36kms 12km/h 18km/ht 3 hours 2 hours

Average speed

=total dis tancetotal time

=725

=14 ,4 km /hCLASSWORK 10 min (1) The average speed (s) at which a car travels is given by s=

dt ,

where d = distance traveledand t = time taken to travel the distance.

(a) At what average speed does Jerome drive if he covered 327 km in 2 hours and 45 minutes?

(b) How long did Jerome take to travel 234 km at an average speed of 104km/h?


1 min Learners prepare for the weekly assessment by revising the weeks work


REFLECTION



Give the formula for the following:1. Area of a rectangleA=l×b2. Perimeter of a rectangle

2( l+b )3. Area of a triangle

A=12

b×h

4. Perimeter of a squareP=4 s5. Area of a square

A=s2A

1.

a) s=dt=327 km

2,75 h=118,9km /h

b)

t=ds= 234 km

104 km /h=2,25 hours∴2 hours∧15 minutes

Length =x+8

Breadth = x−6P=2 l+2 b100 =2( x+8 )+2( x−6 )100 =2 x+16+2 x−124 x=100−16+124 x=96x=24

Area =24 m×24 m=576 m2



QUESTION 11.1 Simplify1.1.

18 x+3=27 (3)

1.1.2

2 ( x+4 )=x+10 (3)

1.1.3

x4

x2 −12=24

1.1.4

If y=2 x2+4 x+3 , calculate y when x=−2 (2)

[10]QUESTION 2

2.1 The cost of travelling by private car is R10 plus R1,50 per kilometre. If Wendy paid R16 for the ride, use an equation to find the distance that he travelled.

2.2 The length of a rectangle is 10cm longer than its breadth. The perimeter is 48cm. Determine the area of the rectangle.

(6)


QUESTION 11.1 Simplify1.1.

18 x+3=27 (3)

8 x+3−3=27−3 ∴8 x=24

∴8 x8

=248

∴ x=31.1.

22 ( x+4 )=x+10 (3)

2 x+8=x+10 ∴2 x−x+8−8=x−x+10−8 ∴ x=2

1.1.3

x4

x2 −12=24

x4−2−12+12=24+12 ∴ x2=36

∴√x2=√36 ∴ x=6OR x2=62 ∴ x=6

(2)

1.1.4

If y=2 x2+4 x+3 , calculate y when x=−2 (2)

∴ y=2¿


∴ y=2 ( 4 )−8+3∴ y=8−8+3∴ y=3

[10]QUESTION 2

2.1 Let the number of kilometres be x

Total Cost =10+1,50× x

16=10+1 ,50 x1 ,50 x=6x=4

He travelled 4 kilometers

(4)

2.2 Let the breadth be x

The length =x+10

2×l+2×b=perimeter2 x+2( x+10 )=484 x=28x=7

Breadth = 7cm and length =17cm

Area

=l×b=17×7=119 cm2

(6)

[10]TOTAL: 20


CONSOLIDATION WORKSHEETS


WORKSHEET SOLUTION


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