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# Easier algebra with exponents

Module by: Siyavula Uploaders. E-mail the author

## EASIER ALGEBRA WITH EXPONENTS

Easier algebra with exponents

CLASS WORK

• Do you remember how exponents work? Write down the meaning of “three to the power seven”. What is the base? What is the exponent? Can you explain clearly what a power is?
• In this section you will find many numerical examples; use your calculator to work through them to develop confidence in the methods.

1 DEFINITION

23 = 2 × 2 × 2 and a4 = a × a × a × a and b × b × b = b3

also

(a+b)3 = (a+b) × (a+b) × (a+b) and 234=23×23×23×23234=23×23×23×23 size 12{ left ( { {2} over {3} } right ) rSup { size 8{4} } = left ( { {2} over {3} } right ) times left ( { {2} over {3} } right ) times left ( { {2} over {3} } right ) times left ( { {2} over {3} } right )} {}

1.1 Write the following expressions in expanded form:

43 ; (p+2)5 ; a1 ; (0,5)7 ; b2 × b3 ;

1.2 Write these expressions as powers:

7 × 7 × 7 × 7

y × y × y × y × y

–2 × –2 × –2

(x+y) × (x+y) × (x+y) × (x+y)

1.3 Answer without calculating: Is (–7)6 the same as –76 ?

• Now use your calculator to check whether they are the same.
• Compare the following pairs, but first guess the answer before using your calculator to see how good your estimate was.

–52 and (–5)2 –125 and (–12)5 –13 and (–1)3

• By now you should have a good idea how brackets influence your calculations – write it down carefully to help you remember to use it when the problems become harder.
• The definition is:

ar = a × a × a × a × . . . (There must be r a’s, and r must be a natural number)

• It is good time to start memorising the most useful powers:

22 = 4; 23 = 8; 24 = 16; etc. 32 = 9; 33 = 27; 34 = 81; etc. 42 = 16; 43 = 64; etc.

Most problems with exponents have to be done without a calculator!

2 MULTIPLICATION

• Do you remember that g3 × g8 = g11 ? Important words: multiply; same base

2.1 Simplify: (don’t use expanded form)

77 × 77

(–2)4 × (–2)13

( ½ )1 × ( ½ )2 × ( ½ )3

(a+b)a × (a+b)b

• We multiply powers with the same base according to this rule:

ax × ay = ax+yalsoax+y=ax×ay=ay×axax+y=ax×ay=ay×ax size 12{a rSup { size 8{x+y} } =a rSup { size 8{x} } times a rSup { size 8{y} } =a rSup { size 8{y} } times a rSup { size 8{x} } } {}, e.g. 814=84×810814=84×810 size 12{8 rSup { size 8{"14"} } =8 rSup { size 8{4} } times 8 rSup { size 8{"10"} } } {}

3 DIVISION

• 4642=462=444642=462=44 size 12{ { {4 rSup { size 8{6} } } over {4 rSup { size 8{2} } } } =4 rSup { size 8{6 - 2} } =4 rSup { size 8{4} } } {} is how it works. Important words: divide; same base

3.1 Try these: a6aya6ay size 12{ { {a rSup { size 8{6} } } over {a rSup { size 8{y} } } } } {}; 323321323321 size 12{ { {3 rSup { size 8{"23"} } } over {3 rSup { size 8{"21"} } } } } {}; a+bpa+b12a+bpa+b12 size 12{ { { left (a+b right ) rSup { size 8{p} } } over { left (a+b right ) rSup { size 8{"12"} } } } } {}; a7a7a7a7 size 12{ { {a rSup { size 8{7} } } over {a rSup { size 8{7} } } } } {}

• The rule for dividing powers is: axay=axyaxay=axy size 12{ { {a rSup { size 8{x} } } over {a rSup { size 8{y} } } } =a rSup { size 8{x - y} } } {}.

Alsoaxy=axayaxy=axay size 12{a rSup { size 8{x - y} } = { {a rSup { size 8{x} } } over {a rSup { size 8{y} } } } } {}, e.g. a7=a20a13a7=a20a13 size 12{a rSup { size 8{7} } = { {a rSup { size 8{"20"} } } over {a rSup { size 8{"13"} } } } } {}

4 RAISING A POWER TO A POWER

• e.g. 324324 size 12{ left (3 rSup { size 8{2} } right ) rSup { size 8{4} } } {}= 32×432×4 size 12{3 rSup { size 8{2 times 4} } } {}= 3838 size 12{3 rSup { size 8{8} } } {}.

4.1 Do the following:

• This is the rule: axy=axyaxy=axy size 12{ left (a rSup { size 8{x} } right ) rSup { size 8{y} } =a rSup { size 8{ ital "xy"} } } {}alsoaxy=axy=ayxaxy=axy=ayx size 12{a rSup { size 8{ ital "xy"} } = left (a rSup { size 8{x} } right ) rSup { size 8{y} } = left (a rSup { size 8{y} } right ) rSup { size 8{x} } } {}, e.g. 618=663618=663 size 12{6 rSup { size 8{"18"} } = left (6 rSup { size 8{6} } right ) rSup { size 8{3} } } {}

5 THE POWER OF A PRODUCT

• This is how it works:

(2a)3 = (2a) × (2a) × (2a) = 2 × a × 2 × a × 2 × a = 2 × 2 × 2 × a × a × a = 8a3

• It is usually done in two steps, like this: (2a)3 = 23 × a3 = 8a3

5.1 Do these yourself: (4x)2 ; (ab)6 ; (3 × 2)4 ; ( ½ x)2 ; (a2 b3)2

• It must be clear to you that the exponent belongs to each factor in the brackets.
• The rule: (ab)x = ax bxalsoap×bp=abbap×bp=abb size 12{a rSup { size 8{p} } times b rSup { size 8{p} } = left ( ital "ab" right ) rSup { size 8{b} } } {} e.g. 143=2×73=2373143=2×73=2373 size 12{"14" rSup { size 8{3} } = left (2 times 7 right ) rSup { size 8{3} } =2 rSup { size 8{3} } 7 rSup { size 8{3} } } {}and32×42=3×42=12232×42=3×42=122 size 12{3 rSup { size 8{2} } times 4 rSup { size 8{2} } = left (3 times 4 right ) rSup { size 8{2} } ="12" rSup { size 8{2} } } {}

6 A POWER OF A FRACTION

• This is much the same as the power of a product. ab3=a3b3ab3=a3b3 size 12{ left ( { {a} over {b} } right ) rSup { size 8{3} } = { {a rSup { size 8{3} } } over {b rSup { size 8{3} } } } } {}

6.1 Do these, but be careful: 23p23p size 12{ left ( { {2} over {3} } right ) rSup { size 8{p} } } {}223223 size 12{ left ( { { left ( - 2 right )} over {2} } right ) rSup { size 8{3} } } {}x2y32x2y32 size 12{ left ( { {x rSup { size 8{2} } } over {y rSup { size 8{3} } } } right ) rSup { size 8{2} } } {}axby2axby2 size 12{ left ( { {a rSup { size 8{ - x} } } over {b rSup { size 8{ - y} } } } right ) rSup { size 8{ - 2} } } {}

• Again, the exponent belongs to both the numerator and the denominator.
• The rule: abm=ambmabm=ambm size 12{ left ( { {a} over {b} } right ) rSup { size 8{m} } = { {a rSup { size 8{m} } } over {b rSup { size 8{m} } } } } {}andambm=abmambm=abm size 12{ { {a rSup { size 8{m} } } over {b rSup { size 8{m} } } } = left ( { {a} over {b} } right ) rSup { size 8{m} } } {}e.g. 233=2333=827233=2333=827 size 12{ left ( { {2} over {3} } right ) rSup { size 8{3} } = { {2 rSup { size 8{3} } } over {3 rSup { size 8{3} } } } = { {8} over {"27"} } } {}anda2xbx=a2xbx=a2bxa2xbx=a2xbx=a2bx size 12{ { {a rSup { size 8{2x} } } over {b rSup { size 8{x} } } } = { { left (a rSup { size 8{2} } right ) rSup { size 8{x} } } over {b rSup { size 8{x} } } } = left ( { {a rSup { size 8{2} } } over {b} } right ) rSup { size 8{x} } } {}

end of CLASS WORK

TUTORIAL

• Apply the rules together to simplify these expressions without a calculator.

1. a5×a7a×a8a5×a7a×a8 size 12{ { {a rSup { size 8{5} } times a rSup { size 8{7} } } over {a times a rSup { size 8{8} } } } } {} 2. x3×y4×x2y5x4y8x3×y4×x2y5x4y8 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. a2b3c2×ac22×bc2a2b3c2×ac22×bc2 size 12{ left (a rSup { size 8{2} } b rSup { size 8{3} } c right ) rSup { size 8{2} } times left ( ital "ac" rSup { size 8{2} } right ) rSup { size 8{2} } times left ( ital "bc" right ) rSup { size 8{2} } } {} 4. a3×b2×a3a×b5b4×ab3a3×b2×a3a×b5b4×ab3 size 12{a rSup { size 8{3} } times b rSup { size 8{2} } times { {a rSup { size 8{3} } } over {a} } times { {b rSup { size 8{5} } } over {b rSup { size 8{4} } } } times left ( ital "ab" right ) rSup { size 8{3} } } {}

5. 2xy×2x2y42×x2y32xy32xy×2x2y42×x2y32xy3 size 12{ left (2 ital "xy" right ) times left (2x rSup { size 8{2} } y rSup { size 8{4} } right ) rSup { size 8{2} } times left ( { { left (x rSup { size 8{2} } y right ) rSup { size 8{3} } } over { left (2 ital "xy" right ) rSup { size 8{3} } } } right )} {} 6. 23×22×278×4×8×2×823×22×278×4×8×2×8 size 12{ { {2 rSup { size 8{3} } times 2 rSup { size 8{2} } times 2 rSup { size 8{7} } } over {8 times 4 times 8 times 2 times 8} } } {}

end of TUTORIAL

Some more rules

CLASS WORK

1 Consider this case: a5a3=a53=a2a5a3=a53=a2 size 12{ { {a rSup { size 8{5} } } over {a rSup { size 8{3} } } } =a rSup { size 8{5 - 3} } =a rSup { size 8{2} } } {}

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

1.1 a3a3a3a3 size 12{ { {a rSup { size 8{3} } } over {a rSup { size 8{3} } } } } {} 1.2 a3a5a3a5 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 a0 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: a0 = 1 also 1 = a0 . A few examples:

30 = 1; k0 = 1; (ab2)0 = 1 ; (n+1)0 = 1; a3bab220=1a3bab220=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?
• a3a5=a×a×aa×a×a×a×a=1a×a=1a2a3a5=a×a×aa×a×a×a×a=1a×a=1a2 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: ax=1axax=1ax 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: 1ax=ax1ax=ax 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 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 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 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"} } } } } {}
(1)

end of CLASS WORK

HOMEWORK ASSIGNMENT

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

1. x3y2×32x2y×2xy4x3y2×32x2y×2xy4 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. x43xy2×6x2x3y×2x7y3×4x2y42yx43xy2×6x2x3y×2x7y3×4x2y42y 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. 5x353x5x353x 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. 2a2b5c3d2×2abc2d3×4abcd322a2b5c3d2×2abc2d3×4abcd32 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. 6x2y2×2xy33×x43xy6x2y2×2xy33×x43xy 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. 2a23+12a308a62a23+12a308a6 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. x3y4×31x2y13×2xy32x3y4×31x2y13×2xy32 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 x3×y4×x2y5x4y8x3×y4×x2y5x4y8 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:

x3×y4×x2y5x4y8x3×y4×x2y5x4y8 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} } } } } {} = 33×24×32×2534×2833×24×32×2534×28 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} } } } } {} = 27×16×9×3281×12827×16×9×3281×128 size 12{ { {"27" times "16" times 9 times "32"} over {"81" times "128"} } } {} = 3 × 2 = 6

x3×y4×x2y5x4y8x3×y4×x2y5x4y8 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} } } } } {} = x3×x2×y4×y5x4y8x3×x2×y4×y5x4y8 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} } } } } {} = x5×y9x4×y8x5×y9x4×y8 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} } } } } {} = x54×y98x54×y98 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: 2a2b5c3d2×2abc2d3×4abcd322a2b5c3d2×2abc2d3×4abcd32 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 = xn + m 1.6.2 x n  x m = xn – m 1.6.3 x 0 = 1 1.6.4 x –n = 1xn1xn 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 × 1017

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 × 1011) × (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 3a2xy3ab2x2y33a2xy3ab2x2y3 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 a0b0c36c2ab3c52×23a2c304abc2×18b42a3c42a0b0c36c2ab3c52×23a2c304abc2×18b42a3c42 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: 2x2y3 + (xy)2 – 4x

3.2 Calculate the value of 2x2y3 + (xy)2 – 4x as x = 4 and y = –2

4. Formulae

the formula for the area of a circle is: area = π r2 (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 × 1018

1.2.2 1,3104 × 10–17

1.3.1 6,148 × 1011 × 9,23022 × 1015

= 6,148 × 9,23022 × 1011 × 1015

≈ 56,74 × 1026

= 5,674 × 1027

1.3.2 1,767×1066,553×1041,767×1066,553×104 size 12{ { {1,"767" times "10" rSup { size 8{ - 6} } } over {6,"553" times "10" rSup { size 8{ - 4} } } } } {} = 1,7676,553×106(4)1,7676,553×106(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 34a5x7y4 = 81a5x7y4

2.2 c3×2×18a6b4c86a2b6c12×4abc2c3×2×18a6b4c86a2b6c12×4abc2 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} } } } } {} = 36a6b4c1124a3b7c1436a6b4c1124a3b7c14 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"} } } } } {} = 3a32b3c33a32b3c3 size 12{ { {3a rSup { size 8{3} } } over {2b rSup { size 8{3} } c rSup { size 8{3} } } } } {}

3.1 2x2y3 + x2y2 – 4x

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 × 122 = 452,38934… ≈ 452,4 cm2

4.1.2 opp = π × 42 = 50,26548… ≈ 50 m2

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 74y5 etc.

1.3 (–7)6 = 76 , (–7)6 ≠ –76 etc.

2.1 714 (–2)17 = –217 etc.

3.1 a6–y 32 (a+b)p–12a0

4.1 a5a 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.

2. xy

3. a6b8c8

4. a8b6

5. 4x8y9

6. 1

CLASS WORK

New for most learners in grade 9.

HOMEWORK ASSIGNMENT

1. 18x6y7

2. 24x11y3

3. 0

4. 32a6b14c14d11

5. 16x10y1216x10y12 size 12{ { {"16"x rSup { size 8{"10"} } } over {y rSup { size 8{"12"} } } } } {}

6. 1

7. 108y5x108y5x 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 cm2

3.3 ≈ 1,45 × 1015

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