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  • Discuss the following two problems, and make two more rules to cover these cases.

1.1 a 3 a 3 size 12{ { {a rSup { size 8{3} } } over {a rSup { size 8{3} } } } } {} 1.2 a 3 a 5 size 12{ { {a rSup { size 8{3} } } over {a rSup { size 8{5} } } } } {}

2 WHEN THE EXPONENT IS ZERO

  • The answer to 1.1 is a 0 when we apply the rule for division.
  • But we know that the answer must be 1, because the numerator and denominator are the same.
  • So, we can say that anything with a zero as exponent must be equal to 1.
  • The rule is now: a 0 = 1 also 1 = a 0 . A few examples:

3 0 = 1; k 0 = 1; (ab 2 ) 0 = 1 ; (n+1) 0 = 1; a 3 b ab 2 2 0 = 1 size 12{ left ( { {a rSup { size 8{3} } b} over { left ( ital "ab" rSup { size 8{2} } right ) rSup { size 8{2} } } } right ) rSup { size 8{0} } =1} {} and

1 = (anything) 0 , in other words, we can change a 1 to anything that suits us, if necessary!

3 WHEN THE EXPONENT IS NEGATIVE

  • Look again at 1.2. According to the rule, the answer is a –2 . But what does it mean?
  • a 3 a 5 = a × a × a a × a × a × a × a = 1 a × a = 1 a 2 size 12{ { {a rSup { size 8{3} } } over {a rSup { size 8{5} } } } = { {a times a times a} over {a times a times a times a times a} } = { {1} over {a times a} } = { {1} over {a rSup { size 8{2} } } } } {} .
  • So the rule is: a x = 1 a x size 12{a rSup { size 8{ - x} } = { {1} over {a rSup { size 8{x} } } } } {} and vice versa.
  • From now on we always try to write answers with positive exponents, where possible.
  • The rule also means: 1 a x = a x size 12{ { {1} over {a rSup { size 8{ - x} } } } =a rSup { size 8{x} } } {} and vice versa. These examples are important:

ab 2 c 3 = ab 2 c 3 size 12{ ital "ab" rSup { size 8{2} } c rSup { size 8{ - 3} } = { { ital "ab" rSup { size 8{2} } } over {c rSup { size 8{3} } } } } {}

2x m y = 2y x m size 12{2x rSup { size 8{ - m} } y= { {2y} over {x rSup { size 8{m} } } } } {}

a 2 b 5 a 3 b 5 = a 2 a 3 b 5 b 5 = a 5 b 10 size 12{ { {a rSup { size 8{2} } b rSup { size 8{ - 5} } } over {a rSup { size 8{ - 3} } b rSup { size 8{5} } } } = { {a rSup { size 8{2} } a rSup { size 8{3} } } over {b rSup { size 8{5} } b rSup { size 8{5} } } } = { {a rSup { size 8{5} } } over {b rSup { size 8{"10"} } } } } {}

end of CLASS WORK

HOMEWORK ASSIGNMENT

  • Simplify without a calculator and leave answers without negative exponents.

1. x 3 y 2 × 3 2 x 2 y × 2 xy 4 size 12{x rSup { size 8{3} } y rSup { size 8{2} } times 3 rSup { size 8{2} } x rSup { size 8{2} } y times 2 ital "xy" rSup { size 8{4} } } {}

2. x 4 3 xy 2 × 6x 2 x 3 y × 2x 7 y 3 × 4x 2 y 4 2y size 12{ { {x rSup { size 8{4} } } over {3 ital "xy" rSup { size 8{2} } } } times { {6x rSup { size 8{2} } } over {x rSup { size 8{3} } y} } times 2x rSup { size 8{7} } y rSup { size 8{3} } times { {4x rSup { size 8{2} } y rSup { size 8{4} } } over {2y} } } {}

3. 5 x 3 5 3 x size 12{ left (5 rSup { size 8{x} } right ) rSup { size 8{3} } - left (5 rSup { size 8{3} } right ) rSup { size 8{x} } } {}

4. 2a 2 b 5 c 3 d 2 × 2a bc 2 d 3 × 4 ab cd 3 2 size 12{ left (2a rSup { size 8{2} } b rSup { size 8{5} } c rSup { size 8{3} } d right ) rSup { size 8{2} } times 2a left ( ital "bc" rSup { size 8{2} } d right ) rSup { size 8{3} } times 4 ital "ab" left ( ital "cd" rSup { size 8{3} } right ) rSup { size 8{2} } } {}

5. 6 x 2 y 2 × 2x y 3 3 × x 4 3 xy size 12{6 left ( { {x rSup { size 8{2} } } over {y} } right ) rSup { size 8{2} } times left ( { {2x} over {y rSup { size 8{3} } } } right ) rSup { size 8{3} } times { {x rSup { size 8{4} } } over {3 ital "xy"} } } {}

6. 2a 2 3 + 12 a 3 0 8a 6 size 12{ left (2a rSup { size 8{2} } right ) rSup { size 8{3} } + left ("12"a rSup { size 8{3} } right ) rSup { size 8{0} } - 8a rSup { size 8{6} } } {}

7. x 3 y 4 × 3 1 x 2 y 1 3 × 2 xy 3 2 size 12{x rSup { size 8{3} } y rSup { size 8{ - 4} } times left (3 rSup { size 8{ - 1} } x rSup { size 8{2} } y rSup { size 8{ - 1} } right ) rSup { size 8{ - 3} } times left (2 ital "xy" rSup { size 8{3} } right ) rSup { size 8{2} } } {}

end of HOMEWORK ASSIGNMENT

CLASS WORK

  • Let us make sure that we can replace variables with numerical values properly.

1 To calculate the perimeter of a rectangle with side lengths 17cm and 13,5 cm, we use normal formula:

  • Perimeter = 2 [ length + breadth ]
  • Put brackets in place of the variables: = 2 [ ( ) + ( ) ]
  • Fill in the values: = 2 [ (17) + (13,5) ]
  • Remove brackets and simplify = 2 [ 17 + 13,5 ]according to the rules: = 2 × 20,5
  • Remember the units (if any): = 41 cm

2 What is the value of x 3 × y 4 × x 2 y 5 x 4 y 8 size 12{ { {x rSup { size 8{3} } times y rSup { size 8{4} } times x rSup { size 8{2} } y rSup { size 8{5} } } over {x rSup { size 8{4} } y rSup { size 8{8} } } } } {} if x = 3 and y = 2 ?

  • There are two possibilities: first substitute and then simplify or simplify first and then substitute. Here are both methods:

x 3 × y 4 × x 2 y 5 x 4 y 8 size 12{ { {x rSup { size 8{3} } times y rSup { size 8{4} } times x rSup { size 8{2} } y rSup { size 8{5} } } over {x rSup { size 8{4} } y rSup { size 8{8} } } } } {} = 3 3 × 2 4 × 3 2 × 2 5 3 4 × 2 8 size 12{ { { left (3 right ) rSup { size 8{3} } times left (2 right ) rSup { size 8{4} } times left (3 right ) rSup { size 8{2} } times left (2 right ) rSup { size 8{5} } } over { left (3 right ) rSup { size 8{4} } times left (2 right ) rSup { size 8{8} } } } } {} = 27 × 16 × 9 × 32 81 × 128 size 12{ { {"27" times "16" times 9 times "32"} over {"81" times "128"} } } {} = 3 × 2 = 6

x 3 × y 4 × x 2 y 5 x 4 y 8 size 12{ { {x rSup { size 8{3} } times y rSup { size 8{4} } times x rSup { size 8{2} } y rSup { size 8{5} } } over {x rSup { size 8{4} } y rSup { size 8{8} } } } } {} = x 3 × x 2 × y 4 × y 5 x 4 y 8 size 12{ { {x rSup { size 8{3} } times x rSup { size 8{2} } times y rSup { size 8{4} } times y rSup { size 8{5} } } over {x rSup { size 8{4} } y rSup { size 8{8} } } } } {} = x 5 × y 9 x 4 × y 8 size 12{ { {x rSup { size 8{5} } times y rSup { size 8{9} } } over {x rSup { size 8{4} } times y rSup { size 8{8} } } } } {} = x 5 4 × y 9 8 size 12{x rSup { size 8{5 - 4} } times y rSup { size 8{9 - 8} } } {} = x × y = (3) × (2) = 6

  • Without errors, the answers will be the same.

3.1 Calculate the perimeter of the square with side length 6,5 cm

3.2 Calculate the area of the rectangle with side lengths 17 cm and 13,5 cm.

3.3 If a = 5 and b = 1 and c = 2 and d = 3, calculate the value of: 2a 2 b 5 c 3 d 2 × 2a bc 2 d 3 × 4 ab cd 3 2 size 12{ left (2a rSup { size 8{2} } b rSup { size 8{5} } c rSup { size 8{3} } d right ) rSup { size 8{2} } times 2a left ( ital "bc" rSup { size 8{2} } d right ) rSup { size 8{3} } times 4 ital "ab" left ( ital "cd" rSup { size 8{3} } right ) rSup { size 8{2} } } {} .

end of CLASS WORK

Assessment

Exponents ω

I can . . . ASs Now I have to . . .
Recognise which rules to use 1.6 <
Simplify expressions with exponents 2.8
Give answers with positive exponents 1.6.4
Use scientific notation 1.6.1
Do substitutions 1.6
Apply formulae 1.6 >

good average not so good

For this learning unit I . . .
Worked very hard yes no
Neglected my work yes no
Did very little work yes no Date :
Learner can . . . ASs 1 2 3 4 Comments
Recognise which rules to use 1.6
Simplify expressions with exponents 2.8
Give answers with positive exponents 1.6.4
Use scientific notation 1.6.1
Do substitutions 1.6
Apply formulae 1.6
Critical outcomes 1 2 3 4
Cooperative group work
Communication using symbols properly
Accuracy
Understanding of Maths in everyday life
Educator:
Signature: Date :
Feedback from parents:
Signature: Date :

Assessment

Learning outcomes(LOs)
LO 1
Numbers, Operations and RelationshipsThe learner will be able to recognise, describe and represent numbers and their relationships, and to count, estimate, calculate and check with competence and confidence in solving problems.
Assessment standards(ASs)
We know this when the learner :
1.1 describes and illustrates the historical development of number systems in a variety of historical and cultural contexts (including local);
1.2 recognises, uses and represents rational num­bers (including very small numbers written in scientific notation), moving flexibly between equivalent forms in appropriate contexts;
1.3 solves problems in context including contexts that may be used to build awareness of other learning areas, as well as human rights, social, economic and environmental issues such as:
1.3.1 financial (including profit and loss, budgets, accounts, loans, simple and compound interest, hire purchase, exchange rates, commission, rentals and banking);
1.3.2 measurements in Natural Sciences and Technology contexts;
1.4 solves problems that involve ratio, rate and proportion (direct and indirect);
1.5 estimates and calculates by selecting and using operations appropriate to solving problems and judging the reasonableness of results (including measurement problems that involve rational approximations of irrational numbers);
1.6 uses a range of techniques and tools (including technology) to perform calculations efficiently and to the required degree of accuracy, including the following laws and meanings of exponents (the expectation being that learners should be able to use these laws and meanings in calculations only):
1.6.1 x n × x m = x n + m
1.6.2 x n  x m = x n – m
1.6.3 x 0 = 1
1.6.4 x –n = 1 x n size 12{ { {1} over {x rSup { size 8{n} } } } } {}
1.7 recognises, describes and uses the properties of rational numbers.
LO 2
Patterns, Functions and AlgebraThe learner will be able to recognise, describe and represent patterns and relationships, as well as to solve problems, using algebraic language and skills.
We know this when the learner :
2.8 uses the laws of exponents to simplify expressions and solve equations.

Memorandum

TEST 2

1. Scientific Notation

1.1 Write the following values as ordinary numbers:

1.1.1 2,405 × 10 17

1.1.2 6,55 × 10 –9

1.2 Write the following numbers in scientific notation:

1.2.1 5 330 110 000 000 000 000

1.2.2 0,000 000 000 000 000 013 104

1.3 Do the following calculations and give your answer in scientific notation:

1.3.1 (6,148 × 10 11 ) × (9 230 220 000 000 000)

1.3.2 (1,767 × 10 –6 )  (6,553 × 10 –4 )

2. Exponents

Simplify and leave answers without negative exponents. (Do not use a calculator.)

2.1 3a 2 xy 3 ab 2 x 2 y 3 size 12{3a rSup { size 8{2} } ital "xy" left (3 ital "ab" rSup { size 8{2} } x rSup { size 8{2} } y right ) rSup { size 8{3} } } {}

2.2 a 0 b 0 c 3 6c 2 ab 3 c 5 2 × 2 3a 2 c 3 0 4 abc 2 × 18 b 4 2a 3 c 4 2 size 12{ { { left (a rSup { size 8{0} } b rSup { size 8{0} } c right ) rSup { size 8{3} } } over {6c rSup { size 8{2} } left ( ital "ab" rSup { size 8{3} } c rSup { size 8{5} } right ) rSup { size 8{2} } } } times { {2 left (3a rSup { size 8{2} } c rSup { size 8{3} } right ) rSup { size 8{0} } } over {4 ital "abc" rSup { size 8{2} } } } times "18"b rSup { size 8{4} } left (2a rSup { size 8{3} } c rSup { size 8{4} } right ) rSup { size 8{2} } } {}

3. Substitution

3.1 Simplify: 2 x 2 y 3 + ( xy ) 2 – 4 x

3.2 Calculate the value of 2 x 2 y 3 + ( xy ) 2 – 4 x as x = 4 and y = –2

4. Formulae

the formula for the area of a circle is: area = π r 2 (r is the radius).

4.1 Calculate the areas of the following circles:

4.1.1 A circle with radius = 12 cm ; round answer to 1 decimal place.

4.1.2 A circle with a diameter 8 m ; approximate to the nearest metre.

TEST 2

1.1.1 240 500 000 000 000 000

1.1.2 0,000 000 006 55

1.2.1 5,330 110 × 10 18

1.2.2 1,3104 × 10 –17

1.3.1 6,148 × 10 11 × 9,23022 × 10 15

= 6,148 × 9,23022 × 10 11 × 10 15

≈ 56,74 × 10 26

= 5,674 × 10 27

1.3.2 1, 767 × 10 6 6, 553 × 10 4 size 12{ { {1,"767" times "10" rSup { size 8{ - 6} } } over {6,"553" times "10" rSup { size 8{ - 4} } } } } {} = 1, 767 6, 553 × 10 6 ( 4 ) size 12{ { {1,"767"} over {6,"553"} } times "10" rSup { size 8{ - 6 - \( - 4 \) } } } {} ≈ 0,26 × 10 –2 = 2,6 × 10 –1

2.1 3 4 a 5 x 7 y 4 = 81 a 5 x 7 y 4

2.2 c 3 × 2 × 18 a 6 b 4 c 8 6a 2 b 6 c 12 × 4 abc 2 size 12{ { {c rSup { size 8{3} } times 2 times "18"a rSup { size 8{6} } b rSup { size 8{4} } c rSup { size 8{8} } } over {6a rSup { size 8{2} } b rSup { size 8{6} } c rSup { size 8{"12"} } times 4 ital "abc" rSup { size 8{2} } } } } {} = 36 a 6 b 4 c 11 24 a 3 b 7 c 14 size 12{ { {"36"a rSup { size 8{6} } b rSup { size 8{4} } c rSup { size 8{"11"} } } over {"24"a rSup { size 8{3} } b rSup { size 8{7} } c rSup { size 8{"14"} } } } } {} = 3a 3 2b 3 c 3 size 12{ { {3a rSup { size 8{3} } } over {2b rSup { size 8{3} } c rSup { size 8{3} } } } } {}

3.1 2 x 2 y 3 + x 2 y 2 – 4 x

3.2 2(4) 2 (–2) 3 + (4) 2 (–2) 2 – 4(4) = 2(16)(–8) + (16)(4) – 16 = –256 + 64 – 16 = – 208

4.1.1 opp = π × 12 2 = 452,38934… ≈ 452,4 cm 2

4.1.2 opp = π × 4 2 = 50,26548… ≈ 50 m 2

CLASS WORK

The learners are likely to know the work in the first part already. Those who have not mastered the simplest laws of exponents now have an opportunity to catch up. For the rest it serves as revision in preparation for the new work in the second part.

1.1 4 × 4 × 4 ( p +2) × ( p +2) × ( p +2) × ( p +2) × ( p +2) etc.

1.2 7 4 y 5 etc.

1.3 (–7) 6 = 7 6 , (–7) 6 ≠ –7 6 etc.

2.1 7 14 (–2) 17 = –2 17 etc.

3.1 a 6– y 3 2 ( a + b ) p–12 a 0

4.1 a 5 a etc.

TUTORIAL

The tutorial should be done in silence in class in a fixed time. Recommendation: Mark it immediately – maybe the learners can mark one another’s work.

Answers: 1. a 3

2. xy

3. a 6 b 8 c 8

4. a 8 b 6

5. 4 x 8 y 9

6. 1

CLASS WORK

New for most learners in grade 9.

HOMEWORK ASSIGNMENT

Answers:

1. 18 x 6 y 7

2. 24 x 11 y 3

3. 0

4. 32 a 6 b 14 c 14 d 11

5. 16 x 10 y 12 size 12{ { {"16"x rSup { size 8{"10"} } } over {y rSup { size 8{"12"} } } } } {}

6. 1

7. 108 y 5 x size 12{ { {"108"y rSup { size 8{5} } } over {x} } } {}

CLASS WORK

Substitution gives lots of trouble because it looks so easy. Learners who leave out steps (or don’t write them down) often make careless mistakes. Force them to use brackets.

2. They should conclude that simplification should come first – after all, this is why we simplify!

3.1 26 cm

3.2 229,5 cm 2

3.3 ≈ 1,45 × 10 15

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Source:  OpenStax, Mathematics grade 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11056/1.1
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