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Indices and Surds

6

167

What a greatspace saver!

Chapter Contents

6:01

Indices and the index laws

NS5·1·1, PAS5·1·1

Investigation: Exploring index notationChallenge: Family trees

6:02

Negative indices

NS5·1·1, PAS5·2·1

Investigation: Zero and negative indices6:03

Fractional indices

NS5·1·1, PAS5·2·1

Fun Spot: Why is a room full of marriedpeople always empty?Investigation: Reasoning with fractional indices

6:04

Scientific (or standard) notation

NS5·1·1

Investigation: Multiplying and dividing by powers of 10

6:05

Scientific notation and the calculator

NS5·1·1

Investigation: Using scientific notation

6:06

The real number system

NS5·3·1

Reading Maths: Proof that is irrationalChallenge: f-stops and

6:07

Surds

NS5·3·1

6:08

Addition and subtraction of surds

NS5·3·1

6:09

Multiplication and division of surds

NS5·3·1

Investigation: Iteration to find square roots6:10

Binomial products

NS5·3·1

6:11

Rationalising the denominator

NS5·3·1

Fun Spot: What do Eskimos sing at birthday parties?Challenge: Rationalising binomial denominators

Maths Terms, Diagnostic Test, Revision Assignment, Working Mathematically

22

Learning Outcomes

PAS5·1·1 Applies the index laws to simplify algebraic expressions.PAS5·2·1 Simplifies, expands and factorises algebraic expressions involving fractions and

negative and fractional indices.NS5·1·1 Applies index laws to simplify and evaluate arithmetic expressions and uses

scientific notation to write small and large numbers.NS5·3·1 Performs operations with surds and indices.Working Mathematically Stages 5·3·1–5.1 Questioning, 2 Applying Strategies, 3 Communicating, 4 Reasoning, 5 Reflecting.

168

NEW SIGNPOST MATHEMATICS

9

STAGE

5.1–5.3

6:01

|

Indices and the

Outcomes NS5·1·1, PAS5·1·1

Index Laws

Multiplication using indices34

× 32

= (3

× 3

× 3

× 3)

× (3

× 3) x5

× x3

= (x

× x

× x

× x

× x)

× (x

× x

× x)

= 36 [

= 34 + 2]

= x8 [

= x5 + 3]

Division using indices

35

÷ 32

= x4

÷ x3

=

= 33 [

= 35

− 2]

= x1 [

= x4

− 3]

Powers of indices(33)2

= 33

× 33 (x5)4

= x5 × x5 × x5 × x5

= 33 + 3 [Using Law 1] = x5 + 5 + 5 + 5 [Using Law 1]= 36 [= 33 × 2] = x20 [= x5 × 4]

If we simplify the division xn ÷ xn, using the second law above:xn ÷ xn = xn − n

= x0

But any expression divided by itself must equal 1.xn ÷ xn = 1

Therefore x0 must be equal to 1.x0 = 1

■ 2 × 2 × 2 × 2 = 16 • 2 is called the base.• 4 is called the index.

24 = 16 • 16 is called the basic numeral.

xn = x × x × x × . . . × x × x (where n is a positive integer)

n factors

For: x is the basexnn is the index.

⎧ ⎨ ⎩■ 24 is calleda power of 2.

24 is the 'index form'(base 2 ) of 16

Law 1 When multiplying terms, add the indices: xm × xn = xm + n

You should

learn these

laws.

3 3× 3× 3× 3×3 3×

---------------------------------------- x x× x× x×x x× x×

-----------------------------

Law 2 When dividing terms, subtract the indices: xm ÷ xn = xm − n

Law 3 For powers of a power, multiply the indices: (xm)n = xmn

Law 4 x0 = 1

CHAPTER 6 INDICES AND SURDS 169

worked examples

1 Simplify:a 43 b 135 c (−4)2

2 Simplify:a 32 × 35 b x3 × x2 c 6m2n × mn4

3 Simplify:a x7 ÷ x2 b 15a5 ÷ 3a2 c 20a3b2 ÷ 10ab

4 Simplify:a (a4)2 b (2a4)3 c (p4)3 ÷ (p2)4

5 Simplify:a 70 b 18x3 ÷ 6x3 c (2y3)4 ÷ (4y6)2

Solutions1 a 43 = 4 × 4 × 4

= 64b 135 = 13 × 13 × 13 × 13 × 13

Using the calculator to evaluate 135

PRESS 13 5

135 = 371 293c (−4)2 = −4 × −4

= 162 Using index law 1:

a 32 × 35 = 32 + 5

25= 37

b x3 × x2 = x3 + 2

32 = x5

c 6m2n × mn4 = 6 × m2 + 1 × n1 + 4

24 = 6m3n5

3 Using index law 2:



■ Remember the xy button. Enter the base x first,

press

then enter the index y.

xy

■ Another name for an INDEX is an EXPONENT.

xy =

a x7 ÷ x2 = x7 − 2

= x5b 15a5 ÷ 3a2 =

=

= 5 × a5 − 2

= 5a3

15a5

3a2-----------

153

------ a5

a2-----×

c 20a3b2 ÷ 10ab

=

=

= 2a2b

20a3b2

10ab-----------------

2010------ a3

a-----× b2

b-----×

a (a4)2 = a4 × 2

= a8b (2a4)3 = 23 × (a4)3

= 8 × a4 × 3

= 8a12

c (p4)3 ÷ (p2)4 = p12 ÷ p8

= p4

a 70 = 1 b 18x3 ÷ 6x3 =

= 3 × x3 − 3

= 3x0

= 3 × 1= 3

18x3

6x3----------- c (2y3)4 ÷ (4y6)2 =

=

= 1 × y0

= 1

24 y3( )4×42 y6( )2×------------------------

1616------ y12

y12-------×■ With practice, many of

the steps in the above solutions can be left out.

Note: n = n1

170 NEW SIGNPOST MATHEMATICS 9 STAGE 5.1–5.3

Write each expression in index form.a 2 × 2 × 2 × 2 b 3 × 3c 5 × 5 × 5 d 7 × 7 × 7 × 7 × 7e 10 × 10 × 10 f 9 × 9 × 9 × 9g x × x h a × a × a × ai n × n × n j m × m × m × m × mk p × p × p × p × p × p l y × ym 4 × 4 × 4 n t × t × t × to x × x × x × x × x

Rewrite in expanded form.a 23 b 42 c 65 d 104

e 73 f 31 g a3 h x4

i y2 j m5 k n7 l p3

Determine the basic numeral for:a 25 b 34 c 72

d 112 e 104 f 63

g 28 h 57 i 85

j 95 k 412 l 315

m 26 × 53 n 34 × 73 o 93 × 54

p 123 × 73

Simplify these products, writing answers in index form.a 102 × 103 b 10 × 102 c 103 × 103

d 52 × 54 e 23 × 22 f 75 × 7g 32 × 33 × 34 h 2 × 22 × 25 i 107 × 10 × 102

Simplify these quotients, writing answers in index form.a 102 ÷ 10 b 106 ÷ 103 c 105 ÷ 104

d 58 ÷ 54 e 75 ÷ 72 f 310 ÷ 3g 23 ÷ 22 h 53 ÷ 53 i 28 ÷ 22

Simplify these powers, writing answers in index form.a (102)3 b (103)3 c (106)2 d (23)4

e (22)2 f (27)5 g (34)2 h (53)5

i (72)4 j (23 × 32)2 k (73 × 114)2 l (3 × 22)4

Simplify:a x3 × x2 b y4 × y2 c m3 × m3

d m × m4 e p5 × p5 f a × ag y3 × y4 h x2 × x i m2 × m5

j 3y4 × y k m6 × 3 l 5x3 × 3

Simplify:a x6 ÷ x3 b x6 ÷ x2 c x6 ÷ xd m3 ÷ m e y5 ÷ y2 f m4 ÷ m2

g y6 ÷ y6 h x5 ÷ x5 i y8 ÷ y2

j 6m6 ÷ 3 k 4y8 ÷ 2 l 20x5 ÷ 5

Exercise 6:01The index laws NS5·1·11 Evaluate:

a 23 b 34 c 53

2 Use your calculator to evaluate:a 63 b 45 c 84

Foundation Worksheet 6:01

This sure ispowerfulstuff!

1

2

3

4

5

6

7

8


Simplify:a (x2)3 b (y4)2 c (a3)5

d (m2)0 e (x0)3 f (a7)0

g (y3)2 h (a6)3 i (x3)3

j (2x)3 k (3x2)2 l (5m2)4

Simplify:a 8x4 × x3 b 5a2 × a c 4m6 × m4

d 8x4 ÷ x3 e 5a2 ÷ a f 4m6 ÷ m4

g 10y3 × 5y h 16m2 × 2m2 i 8a5 × 4a4

j 10y3 ÷ 5y k 16m2 ÷ 2m2 l 8a5 ÷ 4a4

m 12x5 × 6x3 n 9a7 × 3a2 o 18y6 × 6y

p q r

s t u

Simplify:a 6a0 b 6(a3)0 c (6a3)0

d ab0 e x0y3 f m5n0

g (2m2)3 h (4n3)2 i (2p3)4

j x2y3 × x3 k a2b5 × b4 l xy3 × x4

m x2y2 × xy2 n a3b × a2b o m2n × mn3

p (x2y3)2 q (abc)2 r ( pq3)3

s 5x2y × 2xy t 4a2b2 × 7ab3 u 11a3 × 4a2b2

v a × a × a × 3a w 3a × 2a × −4a x 5c4 × 4c2 ÷ 10c5

y 12x2 − 5x3 + 10x2 z 4x(3x + 2) − (x − 1)

Simplify:a 5x2 × 2x3 × 3x b 5 × 2a × 4a2 c 5y × y2 × xyd 4x4 ÷ 8x e 7y3 ÷ 49y2 f 100x3 ÷ 10x4

g (x2)3 × x2 h (a4)2 × a5 i (y7)3 × y5

j (a2)3 ÷ a4 k (m3)4 ÷ m10 l n 8 ÷ (n2)3

m (y5)2 × (y3)3 n (2a4)3 × (a3)2 o (b4)3 ÷ (b2)5

p (x4 × x7) ÷ x9 q (4a3 × 5a4) ÷ 10a5 r 7p7q5 ÷ ( p2q)3

s t u

Expand and simplify:a x2(x2 − 1) b a3(5 − a2) c a2(5a − a3)d x(x2 + y) e m(7 − m2) f y(y2 − xy)g 3a2(2a3 + 3a) h 5x(3x2 − x) i 2m3(n2 − m2)j x(5x2 − 3x + 7) k x2(2x2 + 7x − 14) l y(y2 − 7y − 1)m x2(x − 7) − x3 n y(4y3 + 2) − 2y o x(x2 − 7x + 1) − (x3 − x2)

Simplify:a 3x × 3x + 1 b 52y ÷ 5y + 1 c (2x)2 ÷ (21 − x)2

d e2x + 1 × ex e e2x + 1 ÷ ex f (ex + 2)2 ÷ ex + 1

9

10

12x5

6x3----------- 9a7

3a2-------- 18y6

6y-----------

3a5

a3-------- 10x5

5----------- 42a7

21a6-----------

11

12

5x3 4x7×10x5

----------------------- 3x3( )2 4x5×6x4 x×

------------------------------- x2 xy( )3×2x( )4

------------------------

13

14


Investigation 6:01 | Exploring index notationSearching for patterns is part of mathematics and being able to explain concepts is important if our ideas are to be shared.• Find pairs of terms that can be multiplied to give 27. Explain the relationship between the

members of any pair.• Find pairs of terms that can be divided to give 3x2. Explain the difference between the

members of any pair.• Explain the difference between 7x0 and (7x)0.• Explain why 32 × 34 ≠ 96.• List all the pairs of expressions that could be multiplied together to give 10xy2.

(Use only whole numbers.)• Fold a sheet of A4 paper in half as many times as you can. How does the number of regions

increase with each new fold. Write a formula for the number of regions (R) for n folds.• As a reward for service, a man asked his king for the amount of rice that would be needed to

place one grain of rice on the first square of a chessboard, 2 on the second, 4 on the third, 8 on the fourth, doubling each time, until the 64th square is considered. How many grainsof rice would be needed for the 32nd square?

Challenge 6:01 | Family treesA part of Alan’s family tree is drawn below.

1 How many great-grandparents would Alan have had?

2 Estimate the number of generations you would need to go back to before more than one million boxes would be required to show that generation. Would this mean that in that generation there would be over 1 000 000 different ancestors?

3 Estimate Alan’s total number of ancestors in the previous 20 generations.

It is much harder to estimate the number of descendants Alan will have in any one generation as a lot of assumptions will need to be made. Will all of his five children marry? Will there be wars, diseases or population control in the future?

4 Estimate how many generations would be needed before Alan has a total number of descendants in excess of 1 000 000.

inve

stigation

6:01

ch

allenge

6:01Robert McSeveny

Thomas McSeveny

Alan McSeveny

Sarah McNaughton Stephen Newby

Edna Newby

Ruby Dann


6:02 | Negative Indices Outcomes NS5·1·1, PAS5·2·1

All the indices seen so far have been positive integers or zero.If we had 23 ÷ 25, the answer, according to the second index law, should be 23 − 5,

ie 23 ÷ 25 = 2−2.

But this could also be written in this way:

=

=

=

So: 2−2 =

Also 23 ÷ 25 = 8 ÷ 32

2-2

= 4

= 221

1

What happens if theindex is negative?

23

25----- 21 21× 21×

21 21× 21× 2× 2×------------------------------------------------

12 2×------------

122-----

122-----

In general, the meaning of a negative index can be x−m = , (x ≠ 0)summarised by the rules:

x−m is the reciprocal of xm, since xm × x−m = 1

1xm------

■ Examples

x−3 =

x3 × x−3 = x0

= 1

1x3-----

worked examples

Note: Since x-m is thereciprocal of xm. and

=1 –24

41

2

=2 –33

32

3

1 Simplify the following:a 3−2 b 5−1 c x7 × x−3 d 6x2 ÷ 3x4 e ( )−2 f ( )−3

2 Evaluate, using the calculator:a 2−3 b ( )−2

Solutions1 a 3−2 = b 5−1 = c x7 × x−3 = x7 + (−3) d 6x2 ÷ 3x4 = 2x2 − 4

= x4 = 2x−2

= = =

e ( )−2 = f ( )−3 =

= =

= 16 =

= 3

14--- 2

3---

13---

132----- 1

51-----

19--- 1

5--- 2

x2-----

14---

1

14---⎝ ⎠

⎛ ⎞ 2----------- 2

3---

1

23---⎝ ⎠

⎛ ⎞ 3-----------

1

116------⎝ ⎠

⎛ ⎞----------- 1

827------⎝ ⎠

⎛ ⎞-----------

278

------

38---

■1

116------⎝ ⎠

⎛ ⎞----------- 1 1

16------÷ 1 16

1------× 16= = =

continued ➜➜➜


Write down the value of each of the following.a 3−1 b 5−1 c 2−1

d 6−2 e 4−2 f 10−3

g 2−4 h 10−4 i 5−2

Write each with a negative index.

a b c d

e f g h

i j k l

Write true or false for:

a 1024 = 210 b 8 = 24 c 3−2 = d 2(3)2 = 36

e 2(3)−1 = f 4−1 = g 2−1 < 1 h −28 = (−2)8

Simplify, leaving answers as powers of ten.

a 102 ÷ 105 b 10−2 × 103 c 10 ÷ 106 d 10−1 × 10−2

e f g h

Write each without a negative index.a a−1 b x−1 c m−1 d y−1

e x−3 f y−2 g x−4 h m−6

i 2x−1 j 5a−3 k 10y−2 l 36q−4

Rewrite each using a negative index to avoid having a fraction.

a b c d

e f g h

i j k l

2 The key can also be used for negative indices by entering y as a negative number.

Examine the following:

a Press: 2 3 answer: 0·125, ie or 2−3

b Press: 1 3 2 answer: 9, ie ( )−2

xy

xy +/− = 18---

÷ = xy +/− = 13---

Exercise 6:02Negative indices NS5·1·11 Write down the value of:

a 4-1 b 2-3 c 3-2

2 Write with a negative index.

a b c14--- 1

23----- 1

32-----


1

21

11------ 1

3--- 1

5--- 1

7---

133----- 1

54----- 1

28----- 1

72-----

1102-------- 1

103-------- 1

106-------- 1

105--------

319---

16--- 1

2---

4

102 103×106

----------------------- 10 2–( )2

102------------------ 10

102 107÷----------------------- 104 10 2–×

103( )2-------------------------

5

61x--- 1

x2----- 1

x3----- 1

x4-----

5y2----- 3

a--- 10

m4------ 75

x3------

xy2----- m

a3----- 3a

b2------ 4x

y------


Write each as an integer, fraction or mixed number.

a b c d

e f g h

Rewrite each expression with a positive index.a x−2 b a−5 c 3x−1 d 5m−2

e (x + 1)−2 f (3 + a)−1 g (6x)−2 h 4(x + 2)−1

Evaluate the following, using your calculator. Leave your answers in decimal form.a 2−3 b 4−2 c 5−2 d 8−2

e 2−6 f 4−3 g (0·5)−3 h (0·2)−2

i (0·05)−2 j (2·5)−3 k (0·1)−5 l (0·625)−2

Simplify, writing your answers without negative indices.a x3 × x−2 b a−2 × a5 c m4 × m−1 d n5 × n−5

e 3a2 × a−1 f 6x−2 × 5x4 g a−2 × 5a3 h 15m−1 × 2m3

i x−4 × x j 2a−2 × a−3 k 4y × 2y−2 l 15m−4 × 2m−1

Simplify, writing your answers without negative indices.a m4 ÷ m−1 b x2 ÷ x−2 c y−6 ÷ y−8 d x3 ÷ x−1

e a−2 ÷ a2 f y−1 ÷ y3 g y−2 ÷ y h x−3 ÷ x−1

i 6x2 ÷ 2x−1 j 10a3 ÷ 5a7 k 24a−2 ÷ a3 l 18n−1 ÷ 9n−2

Simplify, writing your answers without negative indices.a (a−3)−2 b (x2)−1 c (y−3)2 d (m−2)−2

e (2x2)−1 f (3x)−2 g (5x−1)2 h (7x−2)2

i (abc)−1 j (a2b2c2)−1 k (2a2b)−1 l 2(a2b)−1

If x = 2, y = 3 and z = , evaluate:

a x−1 + y−1 b (xy)−1 c (xz)−1 d x−1y−1z

Simplify:a 3x ÷ 3−x b 5y ÷ 52 − y c ex + 1 ÷ e1 − x d (ex)2 × e−(x − 1)

7

12---⎝ ⎠

⎛ ⎞ 1– 13---⎝ ⎠

⎛ ⎞ 1– 23---⎝ ⎠

⎛ ⎞ 1– 110------⎝ ⎠

⎛ ⎞ 1–

12---⎝ ⎠

⎛ ⎞ 2– 13---⎝ ⎠

⎛ ⎞ 2– 23---⎝ ⎠

⎛ ⎞ 2– 310------⎝ ⎠

⎛ ⎞ 2–

8

9

10

11

12

13 12---

14

• The formula for the volume of a sphere is:

V = πr3

where π � 3·142 and r is the radius of the sphere.

43--


Investigation 6:02 | Zero and negative indices1 What does it mean to have a zero or negative index?

Complete these tables writing answers less than 1 as fractions.

Divide each answer by 10 to reduce the power.

• As the power of 10 decreases, does the answer decrease?

Divide each answer by 5 to reduce the power.

• As the power of 5 decreases, does the answer decrease?

2 Use the tables above to write true or false for:

a 10−1 = b 50 = 1 c 5−1 =

d 10−2 = e 5−3 = f 100 = 1

g 10−4 = h 5−2 = i 10−3 =

inve

stigation

6:02

Power of 10 104 103 102 101 100 10−1 10−2 10−3 10−4

Answer 10 000 1000

÷10

■ 104 ÷ 10 = 103

103 ÷ 10 = 102

102 ÷ 10 = 101

101 ÷ 10 = 100

100 ÷ 10 = 10−1

. . . . . . = . . .

Power of 5 54 53 52 51 50 5−1 5−2 5−3 5−4

Answer 625

÷5

■ 54 ÷ 5 = 53

53 ÷ 5 = 52

52 ÷ 5 = 51

51 ÷ 5 = 50

50 ÷ 5 = 5−1

. . . . . = . . .

= 10a

= -mma

1a

110------ 1

5---

1102-------- 1

53-----

1104-------- 1

52----- 1

103--------


6:03 | Fractional Indices Outcomes NS5·1·1, PAS5·2·1

What is the meaning of a fractional index?The meaning is shown in the examples below.

1 = 3 × 3 = 9

= 91 = 9= 9

multiplied by itself gives 9 and multiplied by itself gives 9.

So is the square root of 9.

∴ =

2 == 51

= 5

Now = 5

So =

3 Similarly:

== 81

= 8

So = , (the cube root of 8)

∴ = 2

Complete:

1 52 = 25 2 72 = 49 3 23 = 8 4 53 = 125

= . . . = . . . = . . . = . . .

Consider the following:

5 5a × 5a = 52a 6 8a × 8a = 82a

If 5n × 5n = 51, If 8n × 8n = 81,what is the value of n? what is the value of n?

7 xa × xa = x2a 8 ya × ya = y2a

If xn × xn = x1, If yn × yn = y1,what is the value of n? what is the value of n?

9 5a × 5a × 5a = 53a 10 xa × xa × xa = x3a

If 5n × 5n × 5n = 51, If xn × xn × xn = x1,what is the value of n? what is the value of n?

prep quiz

6:03

25 49 83 1253

912---

912---

× 912--- 1

2---+⎝ ⎠

⎛ ⎞

9 9×

912---

9

912---

912---

9 That's neat!

That means thatis the squareroot of 5.

= 512

2

5

125

The number that multiplies itself to give 5

(ie ) is the square root of 5.512---

⎧⎨⎩

512---

512---

× 512--- 1

2---+⎝ ⎠

⎛ ⎞

5 5×

512---

5

2 × 2 × 2 = 8

= 8

Two is the cube root of 8.

83 83× 83×

■ Since ,x3( )3x=

x3 x13---

=

813---

813---

× 813---

× 813--- 1

3--- 1

3---+ +⎝ ⎠

⎛ ⎞

813---

83

813---


Write each of the following using a square root sign.

a b c

d e f

Use a fractional index to write:a b c d

, , = nth root of x

is the number that, when used three times in a product, gives x.

x12---

x= x13---

x3= x1n---

x13---

worked examples

1 Simplify the following:

a b c d e f

2 Evaluate using your calculator:

a b c d

Solutions1 a = b = c =

= 5 = 3 = 12x

d = e = f == = == 7m3 = 22

= 4=

=Note from examples e and f the rule:

2 aUsing the square root key

b To evaluate

Press: 32 5

Answer:

c To evaluate d To evaluate

Press: 256 4 Press 125 3 4

Answer: Answer: 6

2512---

2713---

3x12---

4x12---

× 49m6( )12---

823---

932---–

19612---

3212---

25614---–

12543---–

2512---

25 2713---

273 3x12---

4x12---

× 12x12--- 1

2---+

49m6( )12---

4912---

m6 1

2---×

× 823---

813---

⎝ ⎠⎛ ⎞

2

932---–

912---–

⎝ ⎠⎛ ⎞ 3–

49 m3× 83( )29( ) 3–

133-----

127------

(f) is pretty tricky!

For roots higher than a square root ( )

[or cube root ( ), if your calculator has a

key], the x1/y or key can be used.

You may need to use the inverse button.

12---

13---

3 x

or xp

q--

xpq= xq( )p

19612---

196=

196 14=

3215---

x1/y =

3215---

2=

25614---–

12543---–

x1/y +/− = x1/y xy +/− =

25614---–

0·25= 12543---–

0·001=

Exercise 6:03Fractional indices NS5·1·11 Simplify:

a b c

2 Evaluate:

a b c

36 64 121

3612---

6412---

12112---


1

512---

1012---

212---

3 212---

× 4 312---

× 7 612---

×

23 3 2 113 7 3


Find the value of the following:

a b c d

e f g h

i j k l

Assuming that all pronumerals used are positive, simplify:

a b c

d e f

g h i

j k l

Evaluate:

a b c

d e f

g h i

j k l

Evaluate using your calculator, leaving answers asdecimal numerals:

a b c

d e f

g h i

j k l

If a = 4, b = 8 and c = 9, evaluate the following:

a b c d

e f g h

Use the fact that , to simplify:

a b c

d e f

3

412---

4912---

813---

1614---

1612---

10012---

14412---

112---

12112---

3215---

8112---

8114---

126 34a 4 a

12

138 1000

8000

13

13

■ ab( )13---

a13---b

13---

=

■ xa( )b xab=

4

x12---

x12---

× a13---

a23---

× m12---

m12---

×

6x12---

2x12---

× 3y14---

2y34---

× 9n23---

2n13---

×

x2( )12---

y6( )13---

4a6( )12---

a2b4( )12---

9x4y6( )12---

8x3y3( )13---

5

912---–

2512---–

813---–

932---

432---

452---

1634---

12523---

823---–

952---

3245---

1634---–

6

22512---

78412---

102412---

72913---

337513---

800013---

22532---

72923---

800053---

0·125( )23---–

0·25( )52---–

0·01( )32---–

7

a12---

b13---

+ ab( )1

5---

2c32--- 1

2--- ac( )

12---

a12---–

b13---–

+ 2b( )14---–

2ab( )16---–

2bc( )12---

a32---

–

8■ As ,

stands for the positive square root of x.

Note .

x12---

x=

x12---

x2 x=

xpq--

xpq=

27a3( )23---

x6y12( )23---

8m9( )23---

a3

b3-----⎝ ⎠

⎛ ⎞23--- 16

x4------⎝ ⎠

⎛ ⎞34--- y6

25------⎝ ⎠

⎛ ⎞32---

6:03 Who wants to be a millionaire?

Challenge worksheet 6:03 Algebraic expressions and indices


Fun Spot 6:03 | Why is a room full of married people always empty?Work out the answer to each part and put the letter for that part in the box that is above the correct answer.

Write in index form:E 10 × 10 E 8 × 8 × 8 E yyyyaa

Find the value of:E 32 E 23 E 52 − 5A 103 A 23 − 32

Find the value of x in:I 2x = 16 I 3x = 9

Write as a basic numeral:I 1⋅7 × 102 I 6⋅04 ÷ 10

Evaluate:O x0 O 70 + 9 U 11y0

N 2−2 N 10−3 N ( )−1

N B C 7a × a−1

H To fill a jar in 6 minutes, Jan doubled the number of peanuts in the jar every minute. After how many minutes was the jar half full?

Simplify:

S 3x − x S 3x × x T 5 × 5x R 10x ÷ 10

T x2 × x3 T x10 ÷ x2 S (x2)3 L 60x3 ÷ 5x3

S R x10 × x−3 P x3 ÷ x−1 G

Investigation 6:03 | Reasoning with fractional indices• Write , , , , , . . . as expressions with fractional indices and describe the

pattern that emerges.

• Find the value of b if (xb)3 = x

• Explain why .

• Find some values that x, p and q could take if .

fun spot

6:03

23---

3612---

2713---

4aabab

------------ 1102--------

3

102 7 -1 11 2x 9 x5 5 83 x 8 2

3x2 1

25x

1000 4a

0·60

4

1

0·01 12

y4 a2 x4

20 x7 x6

10 6 4

170 x8

1 4---

1 2---

110

00----

--------

inve

stigation

6:03x x2 x3 x4 x5

8 232---

2 2 2( )3= = =

xpq--

2=


6:04 | Scientific (or Standard) Outcome NS5·1·1

Notation

The investigation above should have reminded you that:1 when we multiply a decimal by 10, 100 or 1000, we move the decimal point 1, 2 or 3 places to

the right2 when we divide a decimal by 10, 100 or 1000, we move the decimal point 1, 2 or 3 places to the left.

• This number is written in scientific notation(or standard form).

6·1 × 105 • The first part is between 1 and 10.

• The second part is a power of 10.

Scientific notation is useful when writing very large or very small numbers.

Numbers greater than 1

5 9 7 0 · = 5·97 × 103

To write 5970 in standard form:• put a decimal point after the first digit• count the number of places you have to move the

decimal point to the left from its original position.This will be the power needed.

Investigation 6.04 | Multiplying and dividing by powers of 10• Use the xy button on your calculator to answer these questions.• Look for a connection between questions and answers and then fill in the

rules at the end of the investigation.

Exercise1 a 1⋅8 × 101 b 1⋅8 × 102 c 1⋅8 × 103

d 4⋅05 × 101 e 4⋅05 × 102 f 4⋅05 × 103

g 6⋅2 × 104 h 6⋅2 × 105 i 6⋅2 × 106

j 3⋅1416 × 102 k 3⋅1416 × 103 l 3⋅1416 × 104

2 a 1⋅8 ÷ 101 b 1⋅8 ÷ 102 c 1⋅8 ÷ 103

d 968⋅5 ÷ 102 e 968⋅5 ÷ 103 f 968⋅5 ÷ 104

noitagitsevni

6:04

To multiply by 10n move the decimal point ______ places to the ______.

To divide by 10n move the decimal point ______ places to the ______.

When expressing numbers in scientific (or standard) notation each number is written as the product of a number between 1 and 10, and a power of 10.

■ ‘Scientific notation’is sometimes called ‘standard notation’ or ‘standard form’.

To multiply 5.9 by 103, wemove the decimal point 3 placesto the right - which gives 59 0.


Numbers less than 10·005 97 = 5·97 × 10−3

To write 0·005 97 in scientific notation:• put a decimal point after the first non-zero digit• count the number of places you have moved the

decimal point to the right from its original position.This will show the negative number needed asthe power of 10.

worked examples

1 Express the following in scientific notation.a 243 b 60 000 c 93 800 000

2 Write the following as a basic numeral.a 1·3 × 102 b 2·431 × 102 c 4·63 × 107

Solutions1 a 243 = 2·43 × 100

= 2·43 × 102

b 60 000 = 6 × 10 000= 6 × 104

c 93 800 000 = 9·38 × 10 000 000= 9·38 × 107

2 a 1·3 × 102 = 1·30 × 100= 130

b 2·431 × 102 = 2·431 × 100= 243·1

c 4·63 × 107 = 4·630 000 0 × 10 000 000= 46 300 000

If end zeros are significant, write them in your answer.eg 60 000 (to nearest 100) = 6⋅00 × 104

■ We have moved the decimal point7 places from its original position.

■ To multiply by 107, move thedecimal point 7 places right.

5.9 × 10–3

Multiplying by 10–3 is the

same as dividing by 103 so we

would move the decimal point

3 places left to get 0.005 9 .

■ Short-cut method:0⋅043

• How many places must we move the decimal point for scientific notation?Answer = 2

• Is 0⋅043 bigger or smaller than 1?Answer = smaller

• So the power of 10 is ‘−2’.∴ 0⋅043 = 4⋅3 × 10−2

5·97 × 10−3 is the same as 5·97 ÷ 103.

worked examples

1 Express each number in scientific notation.a 0·043 b 0·000 059 7c 0·004

2 Write the basic numeral for:a 2·9 × 10−2 b 9·38 × 10−5

c 1·004 × 10−3


a Explain the difference between 2 × 104 and 24.b Explain the difference between 5 × 10−2 and 5−2.c How many seconds are in 50 000 years?d Have you lived 8·2 × 104 hours?e Order the following, from smallest to largest.

3·24 × 103 6 9·8 × 10−5 5·6 × 10−2

1·2 × 104 2·04 5·499 × 102 0·0034f Write the thickness of a sheet of paper in scientific notation if 500 sheets of paper have a

thickness of 3·8 cm.g Estimate the thickness of the cover of this book. Write your estimate in scientific notation.

Write the basic numeral for:a 2·1 × 101 b 2·1 ÷ 101 c 2·1 × 10−1

d 7·04 × 102 e 7·04 ÷ 102 f 7·04 × 10−2

g 1·375 × 103 h 1·375 ÷ 103 i 1·375 × 10−3

Express in scientific notation.(Assume that final zeros are not significant.)a 470 b 2600 c 53 000d 700 e 50 000 f 700 000g 65 h 342 i 90j 4970 k 63 500 l 2 941 000m 297·1 n 69·3 o 4976·5p 9 310 000 q 67 000 000 r 190 100s 600 000 t 501 700 u 100 000

Express in scientific notation.a 0·075 b 0·0063 c 0·59d 0·08 e 0·0003 f 0·009g 0·3 h 0·0301 i 0·000 529j 0·426 k 0·001 l 0·000 009 7m 0·000 06 n 0·000 907 o 0·000 000 004

Solutions1 a 0·043 = 4·3 ÷ 100 b 0·000 059 7 = 5·97 ÷ 100 000 c 0·004 = 4 ÷ 1000

= 4·3 × 10−2 = 5·97 × 10−5 = 4 × 10−3

2 a 2·9 × 10−2 = 002·9 ÷ 100 b 9·38 × 10−5 = 000009·38 ÷ 100 000= 0·029 = 0·000 093 8

c 1·004 × 10−3 = 0001·004 ÷ 1000= 0·001 004

Exercise 6:04Scientific notation NS5·1·11 Evaluate:

a 6 × 101 b 6 × 102 c 6 × 103

2 Write in scientific notation.a 430 b 4300 c 43 000


1

2

If you’re stuck with thisexercise, think back toInvestigation 6:02 . . .

3

4


Write the basic numeral for:a 2·3 × 102 b 9·4 × 104 c 3·7 × 103 d 2·95 × 102

e 8·74 × 101 f 7·63 × 105 g 1·075 × 103 h 2·0 × 104

i 8 × 101 j 2·9 × 10−2 k 1·9 × 10−3 l 9·5 × 10−1

m 3·76 × 10−3 n 4·63 × 10−4 o 1·07 × 10−2 p 7 × 10−2

q 8·0 × 10−1 r 5 × 10−6 s 9·73 × 105 t 6·3 × 10−3

u 4·7 × 107 v 9·142 × 102 w 1·032 × 10−2 x 1·0 × 108

6:05 | Scientific Notation and Outcome NS5·1·1

the Calculator

• On a calculator:

5·517 × 1012 is shown as

3·841 × 10−6 is shown as

This is the calculator’s way of showing scientific notation.

• To enter scientific notation, press:

5·517 12, to enter 5·517 × 1012, and

3·841 6 , to enter 3·841 × 10−6.

Write in scientific notation:

1 690 2 4000 3 963·2 4 0·073 5 0·0003

Rewrite as basic numerals:

6 2·9 × 103 7 8·0 × 105 8 4·6 × 10−2 9 5 × 10−7 10 8·14 × 10−1

5

pr

ep quiz

6:05

Some calculators arecalled ‘Scientific

Calculators’ becausethey can give answers inscientific (or standard)

notation.

5.517 12

3.841 −06

Exp

Exp +/−

■ To convert calculator answers into decimal form:

1 Locate the decimal point in the first part of the number (the part betwen 1 and 10).

2 Look at the sign of the second part. This tells you in which direction to move the decimal point. If it is negative the point moves to the left. If it is positive the point moves to the right.

3 Look at the size of the second part. This tells you how many places the decimal point has to be moved.

4 Move the decimal point to its new position, filling in any gaps, where necessary, with zeros.

2·16 −03

First part

Second part

0·002 16

worked examples

Use a calculator to find the answers for:

1 630 000 × (47 000)2

2 45 ÷ (8614)3

3 (8·4 × 106) + (3·8 × 107)

4 1·44 10 6–×

A calculator will give ananswer in scientific notationif the number is too large orsmall to fit on the screen.


Enter each of these on your calculator using the key, and copy the calculator readout.a 6·3 × 1015 b 1·4 × 10−12 c 9·2 × 1011

Rewrite these calculator readouts in scientific notation using powers of 10.a 3.02 05 b 4.631 09 c 1.37 15d 1.31 −04 e 6.9 −08 f 4.6327 −10g 7.6514 08 h 1.03124 −12 i 6.9333 −05

Explain why a calculator readout of has adifferent value to 24.

Place the nine numbers in question 2 in order of size from smallest to largest.

Give the answers to these in scientific notation,correct to 5 significant figures.a 38144 b 0·0004 ÷ 84002

c (0·000 7)5 d 93 000 000 ÷ 0·000 13e (65 × 847)3 f (0·0045)3 × (0·0038)2

g h

Use a calculator to answer correct to 4 significant figures, then use the index laws to check your answer.a 13·85 × (2·3 × 104) b (8·14 × 10−2)2 c (2·1 × 108) ÷ (8·6 × 108)d (3·8 × 10−3)2 e 468 × (1·8 × 10−5) f (9·1 × 104) + (6⋅8 × 105)

g h i

a An American reported that the diameter of the sun is approximately 8·656 × 105 miles. Write this in kilometres, using scientific notation written correct to four significant figures. There are 1·609 km in a mile. If the sun’s diameter is 109 times that of the earth, what is the earth’s diameter, correct to three significant figures?

b The distance to the sun varies from 1·47 × 108 km in January to 1·52 × 108 km in July. This is because the earth’s orbit is an ellipse. What is the difference between these distances?

Solutions1 630 000 × (47 000)2 2 45 ÷ (8614)3

= == 1·39167 × 1015 = 7·040409359 × 10−11

= 1 391 670 000 000 000 � 0·000 000 000 070 404 093 59

3 (8·4 × 106) + (3·8 × 107) 4

Press: 8·4 6 3·8 7 Press: 1·44 6

= 46 400 000 = 0·0012

1·39167 15 7·040409359 −11

1·44 10 6–×

Exp + Exp Exp +/− =

■ Note: Not all calculators work the same way.

The answers to 1 and 2are too long to fit on the screen.

Exercise 6:05

1 Exp

2

■

has 4 significant figures, as four figures are used in the decimal part.

1·402 × 107

■ Use index lawsto check the sizeof your answer.

2.04

3

4

9865 8380×0·000 021

------------------------------- 68000·0007( )5

-------------------------

5

7·45 109× 9·1 10 8–×3 6·714 10 12–×3

6

186

NEW SIGNPOST MATHEMATICS

9

STAGE

5.1–5.3

c If we use the average distance to the sun (1·50

× 108 km), how long would it take light travelling at 3·0

× 108 m/s to reach the earth? (Answer correct to the nearest minute.)

d The mass of the earth is approximately 6 × 1021 tonnes. The sun’s mass is about 333 400 times greater than the mass of the earth. What is the mass of the sun correct to one significant figure?

e We belong to the galaxy known as the Milky Way. It contains about 1 × 1011 stars. If the sun is taken to have average mass [see part d], what is the total mass, correct to 1 significant figure, of the stars in the Milky Way?

Investigation 6:05 | Using scientific notation1 The speed of light is 3·0 × 108 m/s. Use reference books and your calculator to complete this

table for five stars of your choice (eg Vega, Dog Star, Pole Star, Sirius).

Order the distances of the five stars from the earth, from smallest to largest.

2 Research nanotechnology, which involves the use of very small machine parts. Parts are often measured in micrometres. Make comparisons between the sizes of components.

inve

stigation

6:05

Name of star Distance from the earth

Time taken for light to travel to the earth

The sun 1·5 × 108 km

Alpha Centauri 4·2 × 1013 km

■ 106 is 1 million.109 is 1 billion.

• Distances in astronomy are measured in light years, which is the distance that light travels in a year. A light year is approximately 9.6 × 1012 km.


6:06 | The Real Number System Outcome NS5·3·1

The real number system is made up of two groups of numbers: rational and irrational numbers.

Rational numbersAny number that can be written as a fraction, where a and b are whole numbers and b ≠ 0, is a rational number. These include integers, fractions, mixed numbers, terminating decimals and recurring decimals.

eg , 6 , 1·25, 0·07, ,

These examples can all be written as fractions.

, , , , ,

Note: An integer is a rational number whose denominator is 1.

Irrational numbersIt follows that irrational numbers cannot be written as a fraction, where a and b are whole numbers. We havemet a few numbers like this in our study of the circle and Pythagoras’ theorem.

eg π, , ,

The calculator can only give approximations for these numbers. The decimals continue without terminatingor repeating.

3·141 592 65 . . ., 1·414 213 56 . . ., 1·587 401 05 . . ., 3·732 050 80 . . .

Irrational numbers on the number lineAlthough irrational numbers cannot be given an exact decimal value, their positions can still be plotted on the number line. As can be seen from exercises on Pythagoras’ theorem, a number such as can correspond to the length of a side of a triangle, and so this length can be shown on a number line.

Examine the diagram on the right.

If the length of the hypotenuse of this triangle is transferred, using compasses, to the number line as shown, we have the position of on the number line. (This agrees with the decimal approximation from the calculator of 1·414213 56.)

The previous construction can be extended to give the position of other square roots on thenumber line.

Another irrational number you have met beforeis π. You should know that it has an approximate value of 3·142, so it would lie on the number line in the position shown.

ab--

78--- 3

5--- 0·4̇ 81

78--- 33

5------ 5

4--- 7

100--------- 4

9--- 9

1---

ab--

2 43 3 2+

2

1

1

2

2 2102

210 4 5 3 6 3

–1 0 1 2 3 4 5ππ


For each number write rational or irrational. (A calculator might help you to decide.)

a b c 0·3 d 1

e 1·6 f π g h

i j 0·99 k l 9

m n 0·666 o 70 p

q 1 r s t

u 0·0005 v w x

y z

Use your calculator to find an approximation correct to 1 decimal place for the following. Also use these values to show the position of each number on the number line.

a b c d e

f g h i j π

Between which two consecutive integers does each number below lie?

a b c d e

f g h i j

Arrange each set of numbers below in order, from smallest to largest.

a b c

d e f

g h i

j k l

This diagram shows another constructionfor locating square roots on the number line.

Use a set square to draw the triangles on graph paper, then use compasses to drawthe arcs on the number line.

Extend your diagram to show . Check the accuracy of your constructionswith your calculator.

Exercise 6:06

1

That looks

irrational.

710------ 2 1

2---

5 34---

16 227

------

625 13

13--- 1 3+ 1 4 9+

1 9– 3– 7 2+

273 103

2

2 3 5 6 7

8 10 12 20

3

11 18 41 78 95

125 180 250 390 901

4

5 2 3, , 8 3 π, , 10 12 3, ,

7 40 50 6·5, , , π 2 2·1 12, , , 5·6 26 6 30, , ,

8·1 65 7·9 60, , , 98 10 102 10·1, , , 3·1 π 9 3·2, , ,

20 4·1 4·5 21, , , 20 390 21 420, , , 600 610 24 25, , ,

1 1

1

12

345

1 2 3 2 50

5

6


To show multiples of a square root on the number line we can use a pair of compasses to mark off equal intervals.a Repeat the instructions in question 5 to

find the position of on the number line. Then use a pair of compasses to mark the position of and .

b Draw a diagram and show the position of, and on a number line.

The position of π on the number line can be shown by doing the following. Use the diameter of a twenty cent coin to mark off unitson a number line.

Then mark a point on the circumference of the coin, align it with zero on the number line, then roll it along the line carefully until the mark meets the number line again. This will show the position of π on the number line.

Reading maths 6:06 | Proof that is irrationalLet us suppose that is rational and,therefore, can be written as a fraction in

the form where p and q are positive

integers with no common factor.(This assumption is essential.)

so =

then 2 = (squaring both sides)

and 2q2 = p2

6

1

21022 23

2

2 2 3 2

3 2 3 3–

Why is it so?

7

0 1 2 3 4

0 1 2 3 4

reading maths

6:06

2

These numbers

are absurd.

They can’t

be written as

an exact

decimal.

2

pq--

2 pq--

p2

q2-----


This last step implies that p2 must be divisible by 2 (since 2 is prime). Therefore 2 must divide into p exactly.

∴ p can be expressed in the form 2k for some integer k.

∴ 2q2 = (2k)2

2q2 = 4k2

q2 = 2k2

Now, as for p above, it can be argued from this last step that q must be divisible by 2.But p and q were said to have no common factor, hence a contradiction exists. So our original assumption was wrong.

Therefore p and q cannot be found so that . Hence must be irrational.

• Try to use the method above to prove that these are irrational.

1 2 3

Challenge 6:06 | f-stops and Professional photographers have cameras that can alter shutter time and aperture settings using whatare called f-stops.

The f-stops 2, 4, 8 and 16 are accurate. More accurate readings for the rest are given below the scale.• Find the pattern in the accurate f-stops.

* Try squaring each accurate f-stop number.* Try dividing each f-stop number by the one before it.

• Try to discover how f-stops are used.

Perhaps

that’s

why they

called them

SURDS.

2 pq--= 2

3 5 11

ch

allenge

6:06

2

5·642·8

1·42

811

1622

1·4142136

2·82842715·65685425

11·313708

22·627417


6:07 | Surds Outcome NS5·3·1

Surds obey the following rules, which aresuggested by Prep Quiz 6:07.

Find the value of:

1 2 3 4 5

6 7 8 9 10

prep quiz

6:07

16 9 36 16 9+ 16 9+

16 9× 16 9× 369

------ 36

9---------- 16( )2

Surds are numerical expressions that involveirrational roots. They are irrational numbers.

Rule 1 xy x y×=

worked examples

1 = 2 = 3 =

= = =

= 10 (which is true) =

100 4 25× 27 9 3× 5 7× 5 7×

2 5× 3 3× 35

3 3

So √5, 3√7, 2+√3 and

√11 – √10 are all surds.

Rule 2 xy--- x

y-------=

■ Note: means the positive square root of x when x > 0.

when x = 0.

x

x 0=

worked examples

1 = 2 = 3 =

= =ie =

= 5= 2 (which is true)

164

------ 16

4---------- 125 5÷ 125 5÷ 30 5÷ 30 5÷

25 64 4

2---

Rule 3 x( )2 x= ■ Note: For to exist, x cannot be negative.x

worked examples

1 = (5)2 2 = 7 3 == 25 = 9 × 2

= 18

25( )2 7( )2 3 2( )2 32 2( )2×


A surd is in its simplest form when the number under the square root sign is as small as possible. To simplify a surd we make use of Rule 1 by expressing the square root as the product of two smaller square roots, one being the root of a square number. Examine the examples below.

Simplify:

a b c d

e f g h

i j k l

m n o p

q r s t

Square each of the surds below (Rule 3).

a b c d

e f g h

i j k l

m n o p

q r s t

Simplify each of these surds.

a b c d

e f g h

i j k l

m n o p

q r s t

u v w x

worked examplesSimplify the following surds.

1 = 2 = 3 =

= = =

= = =

18 9 2× 75 25 3× 5 48 5 16× 3×

3 2× 5 3× 5 4× 3×

3 2 5 3 20 3

Exercise 6:07Surds NS5·3·11 Simplify:

a b

2 Simplify:

a b

6 5× 7 3×

18 3 8


1

3 5× 5 3× 7 6× 6 7×

10 3× 23 2× 13 5× 11 3×

5 2× 7 2× 11 10× 13 7×

26 2÷ 55 5÷ 77 11÷ 34 17÷

38

2---------- 57

3---------- 60

10---------- 22

11----------

2

16 9 1 100

5 8 15 73

2 2 3 5 2 3 5 3

7 3 2 7 9 11 6 5

10 10 9 20 6 50 15 15

3

8 20 12 50

24 32 45 54

28 90 56 63

44 52 108 40

99 60 96 76

68 126 200 162


Simplify each surd and then, taking and , give a decimal approximation correct to 1 decimal place.

a b c d

e f g h

Write each surd below in its simplest form.

a b c d

e f g h

i j k l

m n o p

q r s t

Simplified surds can be written as an entire root by reversing the above process. For example,

. Express the following as entire square roots.

a b c d

e f g h

i j k l

m n o p

q r s t

6:08 | Addition and Subtraction Outcome NS5·3·1

of Surds

As can be seen from the Prep Quiz, if x and y are two positive numbers,

does not equal and

does not equal

For example:

Simplify: 1 2 3 4

Evaluate: 5 6 (to 1 decimal place)

7 8 (to 1 decimal place)

9 10 (to 1 decimal place)

4 2 1·41= 3 1·73=

18 27 8 12

32 48 50 162

5

2 12 3 8 2 50 4 18

5 20 2 75 10 27 3 56

2 125 4 45 3 24 2 54

7 56 3 72 3 44 5 90

6 200 5 98 9 108 5 68

6

4 3 16 3× 48= =

2 3 3 2 2 5 3 6

4 2 5 3 3 7 5 2

6 2 5 6 3 10 4 7

6 7 5 10 7 2 10 2

9 3 8 4 7 9 12 3

prep quiz

6:08

12 20 32 50

4 9+ 4 9+

16 4– 16 4–

25 36+ 25 36+

x y+ x y+

x y– x y–Surds are

irrational.

■ does not equal .

does not equal .

5 8+ 13

10 7– 3


Only ‘like’ surds can be added or subtracted. For example:

and

However, we can only tell whether surds are like or unlike if each is expressed in its simplest form.

Examine the following examples.

Simplify:

a b c

d e f

g h i

j k l

m n o

Simplify by collecting like surds.

a b c

d e f

g h i

j

■ Remember how in algebra only like terms could be added or subtracted.eg 5x + 3x = 8x

7a − 6a = a

I think I’m beginning

to like this.2 3 4 3+ 6 3= 5 2 4 2– 2=

When adding or subtracting surds• write each surd in its simplest form• add or subtract like surds only.

worked examples

Simplify each of the following.

1 2

= =

3 4

= =

= =

5 6

= =

= =

= =

4 3 7 3 2 3–+ 8 2 5 2– 2 5+ +

9 3 7 2 3 5+

8 18+ 75 27 2 3–+

2 2 3 2+ 5 3 3 3 2 3–+

5 2 6 3

2 12 3 48+ 2 75 45– 2 20+

2 2 3( ) 3 4 3( )+ 2 5 3( ) 3 5– 2 2 5( )+

4 3 12 3+ 10 3 3 5– 4 5+

16 3 10 3 5+

Exercise 6:08Addition and subtraction of surds NS5·3·11 Simplify.

a b

2 Collect like surds.

a

b

3 5 5+ 2 7 7–

5 3 2 3+ +

6 5 4 3– 7 3+


1

3 2 2 2+ 4 3 7 3+ 5 6 2 6+

10 3 7 3– 9 5 6 5– 4 2 3 2–

7 4 7+ 5 3 3+ 9 6 6–

9 5 2 5 3 5+ + 4 10 7 10 2 10–+ 4 3 3 3– 5 3+

10 2 2 2– 5 2– 3 7 3 5 3–+ 2 7 3 7 5 7–+

2

2 5 3 7 4 5+ + 7 3 7 2 5+ + 2 3 3 5 2 3 5 5+ + +

9 3 2 2 7 3– 3 2+ + 6 10 5 7 5 10– 7–+ 3 3 6 5 3– 4 5–+

6 11 2 7– 2 11 5 7+ + 9 2 3 3 9 3 8 2–+ + 5 2 4 5 3 5– 6 2–+

10 7 2 5– 8 7– 3 5–


Simplify completely:

a b c d

e f g h

i j k l

m n o p

q r

Simplify:

a b c

d e f

g h

6:09 | Multiplication and Outcome NS5·3·1

Division of Surds

3

8 2+ 12 2 3+ 2 2 18+ 2 5 20+

27 2 3+ 3 6 24+ 2 8 2– 3 5 20–

32 3 2– 18 32+ 20 45+ 2 27 48–

75 2 12– 98 3 50+ 3 50 2 32+ 5 28 2 63+

4 45 2 20– 2 75 3 48–

4

2 8 18– 3 2+ 2 5 45 20–+ 27 2 48 5 3–+

5 7 63– 2 28+ 5 3 50 12–+ 2 45 20 32+ +

9 8 3 12 27–+ 5 18 72 75–+

worked examples

Simplify the following.

1 2

3 4

5 6

Solutions1 = 2 = ,

= == 30

3 4 == == ===

5 = 6 =

==

=

=

=

=

7 3× 3 2 5 2×5 8 3 6× 96 12÷4 3 18×

12--------------------------- 3 2 3 5–( )

7 3× 7 3× 3 2 5 2× 3 5× 2× 2× 2 2× 2=( )21 15 2×

5 8 3 6× 96 12÷ 96 12÷5 3× 8× 6× 8

15 48× 2 2

15 4 3×60 3

■ At this point we could cancel like this:

42

3× 31

× 2×2

131

-------------------------------------------- 6 2=

4 3 18×12

--------------------------- 4 3 3 2×2 3

--------------------------- 3 2 3 5–( ) 3 2 3 3 5×–×

4 3× 3 2××2 3

---------------------------------------2 3 3 5×–×

12 6

2 3-------------

6 15–

6 63---

6 2

■ 1

2

x y× xy=

x y÷ x y÷=

Using the surd

rules, these are

easy!


Simplify these products:

a b c

d e f

g h i

j k l

m n o

p q r

s t u

v w x

Simplify:

a b c

d e f

g h i

j k l

m n o

p q r

s t u

Simplify fully:

a b c

d e f

g h i

Expand and simplify:

a b c

d e f

g h i

j k l

m n o

p q r

Exercise 6:09Multiplication and division of surds NS5·3·11 Simplify:

a b

2 Simplify:

a b

7 5× 2 3 3 2×

20 5÷ 30 6÷


1

2 3× 5 7× 3 11×

5 2× 3 7× 10 7×

2 8× 3 12× 5 20×

3 6× 5 10× 2 10×

2 3 5× 3 2 4 5× 7 2 2 3×

5 2 2 2× 3 3 2 3× 4 5 5×

4 2 3 10× 2 6 8× 4 3 2 15×

2 x 3 x× 8x 2× a x a x×

2

10 2÷ 12 4÷ 6 3÷

27 3÷ 32 8÷ 45 5÷

5 2 2÷ 6 5 5÷ 10 3 5 3÷

16 8 8÷ 30 5 10÷ 24 7 24÷

12 10 2 5÷ 9 12 3 6÷ 10 15 5 5÷

20 2÷ 75 5÷ 5 8 10÷

2x 2÷ 5a a÷ 20p 2 p÷

3

2 3 2 6×4

--------------------------- 4 5 2 6×10

--------------------------- 2 5 3 8×6 20

---------------------------

15 3×3 5

------------------------ 2 3 6×12

------------------------ 3 7 2 6×21

---------------------------

2 6 5 5×10 15

--------------------------- 6 2 6×4 3

------------------------ 12 27×8 2 6×

---------------------------

4

2 3 2+( ) 5 5 2+( ) 7 2 7 2–( )

3 5 3–( ) 2 2 3 1–( ) 10 5 2 4–( )

2 2 2 1+( ) 3 5 5 2+( ) 4 3 2 3–( )

3 6 5 6+( ) 2 7 7 2–( ) 3 7 3 3–( )

2 2 3 2 2+( ) 4 5 2 2 5–( ) 5 6 2 6 3 2–( )

a a 1+( ) x 2 x 3+( ) 2 y 3 y 2 x+( )

6:09 Golden ratio investigations


Investigation 6:09 | Iteration to find square rootsIteration is the repetition of a process.We can use a simple process to find square roots.

ExampleFind correct to 4 decimal places without using a calculator.

Step 1Estimate .

Let E = 1·6(We want E2 to be close to 3.)

Step 2Divide 3 by your estimate.

3 ÷ 1·6 = 1·875Since 1·6 × 1·875 = 3, the correct answermust lie between 1·6 and 1·875.

Step 3Average these two numbers to get a betterestimate.

∴ � 1·7375

• Use 1·7375 as the new estimate andrepeat the steps above (iterate).

E = 1·73753 ÷ 1 7375 � 1·72662

= 1·73206

∴ � 1·73206

• If we use 1·73206 as our next estimate we get abetter approximation (ie � 1·732051). Since lies between 1·73206 and 1·732051 then = 1·7321 correct to 4 decimal places.

1 Use iteration to find, correct to 4 decimal places:

a b c d

2 Investigate finding the square root of a number n, using iteration of the formula:

New estimate =

where x is your last estimate and we wish to find .

noitagitsevni

6:09To iterate, repeat the

process over and over

again.

To iterate, repeat the

process over and

over again.

Can you find

other uses for

iteration?

3

3

1·6 1·875+2

----------------------------- 1·7375=

3

1·7375 1·72662+2

---------------------------------------------

3

3 33

2 5 70 110

x2 n+2x

--------------

n


6:10 | Binomial Products Outcome NS5·3·1

In Chapter 3, you saw how to expand a binomial product.(a + b)(c + d) = a(c + d) + b(c + d)

= ac + ad + bc + bd

The same procedure is used if some of the terms are notpronumerals, but surds. Examine the following examples.

Simplify the following:

1 2 3 4 5

6 7

Expand and simplify where possible:

8 9 10

pr

ep quiz

6:10

3 5× 18 2× 5( )2 3 2( )2 3 2 7 2+

9 3 3– 3 3 3 3–

5 7 4–( ) 10 10 3+( ) 2 3 3 2 3–( )

ees

3:06

I

remember!

worked examples

Expand and simplify:

1 a b

2 a b

3 a b

Solutions1 a b

= =

= =

= =

=

2 a b

= =

= =

= =

3 a == 5 − 2

= 3

b == 50 − 7

= 43

2 3+( ) 2 5–( ) 3 2 5–( ) 2 2 3 5+( )

2 3 5+( )2 7 3–( )2

5 2–( ) 5 2+( ) 5 2 7–( ) 5 2 7+( )

2 3+( ) 2 5–( ) 3 2 5–( ) 2 2 3 5+( )

2 2 5–( ) 3 2 5–( )+ 3 2 2 2 3 5+( ) 5 2 2 3 5+( )–

2( )2 5 2 3 2 15–+– 12 9 10 2 10– 15–+

2 2 2– 15– 7 10 3–

13– 2 2–

2 3 5+( )2 7 3–( )2

2 3( )2 2 2 3 5 5( )2+××+ 7( )2 2 7 3 3( )2+××–

12 20 3 25+ + 7 2 21– 3+

37 20 3+ 10 2 21–

5 2–( ) 5 2+( ) 5( )2 2( )2–

5 2 7–( ) 5 2 7+( ) 5 2( )2 7( )2–

These give

‘the difference

of two squares’.

■ Remember!(a + b)(a − b) = a2 − b2

These are

‘perfect

squares’.

■ Remember!(a + b)2 = a2 + 2ab + b2

(a − b)2 = a2 − 2ab + b2

ees

3:07A

ees

3:07B


Expand and simplify the following:

a b

c d

e f

g h

i j

k l

m n

o p

q r

s t

u v

w x

a b c

d e f

g h i

j k l

m n o

p q r

a b

c d

e f

g h

i j

k

l

m

n

o

p

Exercise 6:10Binomial products — surds NS5·3·11 Expand and simplify:

a (x + 2)(x + 3)

b

2 Simplify:a (x + 2)2 b

5 2+( ) 5 3+( )

3 2+( )2


1 2 3+( ) 2 1+( ) 3 5+( ) 3 1–( )

7 2–( ) 7 5–( ) 2 3+( ) 2 5+( )

7 2–( ) 5 2–( ) 10 2+( ) 5 3+( )

5 2+( ) 3 2+( ) 2 3–( ) 2 5–( )

6 2–( ) 5 6–( ) 2 3+( ) 2 2 1+( )

3 2 5+( ) 3 5+( ) 2 3 2+( ) 3 2 2+( )

5 2 7–( ) 7 2 2–( ) 2 5 3–( ) 5 3 5+( )

7 5 2–( ) 2 5 7–( ) 5 3 2 7+( ) 2 3 5 7–( )

10 10 7+( ) 2 7 10+( ) 5 3 7 2+( ) 7 3 2 2–( )

9 2 3–( ) 4 2 2 3+( ) 5 7 4+( ) 2 7 7–( )

6 3 5+( ) 7 2 3–( ) x 3+( ) x 2+( )

m n+( ) 2 m n+( ) 3 a 2 b–( ) 2 a 3 b+( )

2 2 1+( )2 3 5–( )2 5 2+( )2

3 2+( )2 5 2–( )2 3 10+( )2

2 3 1+( )2 3 2 4–( )2 5 2 5+( )2

2 2 7+( )2 3 5 10+( )2 7 3 5–( )2

5 3 2 2+( )2 7 3 2 5–( )2 5 10 10 3–( )2

x y+( )2 2 m 5+( )2 3 p 2 q–( )2

3 2 1+( ) 2 1–( ) 5 3+( ) 5 3–( )

10 7–( ) 10 7+( ) 4 2–( ) 4 2+( )

2 3+( ) 2 3–( ) 7 5–( ) 7 5+( )

10 8–( ) 10 8+( ) 11 7+( ) 11 7–( )

2 3 5–( ) 2 3 5+( ) 6 3 2–( ) 6 3 2+( )

7 2 3+( ) 7 2 3–( )

3 5 3–( ) 3 5 3+( )

5 2 2 3–( ) 5 2 2 3+( )

2 3 3 2+( ) 2 3 3 2–( )

x y+( ) x y–( )

■ Important notice!The two binomials in each part of question 3 are said to be conjugatesurds.Note that when a binomial surd is multiplied by its ‘conjugate’, the answer is always a rational number.

2 a 3 b+( ) 2 a 3 b–( )


6:11 | Rationalising the Outcome NS5·3·1

Denominator

If a fraction has a surd (ie an irrational number) in its denominator, we generally rewrite the fraction with a ‘rational’ denominator by using the method shown below.

Rationalise the denominator for each of the following:

a b c d e f

g h i j k l

m n o p q r

s t u v w x

Simplify the following:

1 2 3 4 5

6 7 8

9 10

pr

ep quiz

6:11

5 5× 10 10× 2 3 3× 5 2 2× 2 6 6×

2 1+( ) 2 1–( ) 3 2–( ) 3 2+( ) 5 2–( ) 5 2+( )

2 3 2–( ) 2 3 2+( ) 5 2 3 3–( ) 5 2 3 3+( )

worked examples

For these fractions, we multiply top

and bottom by the square root in

the denominator.

Rewrite with rational denominators:

1 2 3 4

Solutions

1 = 2 =

= =

= =

3 = 4 =

= =

= =

3

3------- 1

5 2---------- 5

12---------- 2 3+

2 3----------------

3

3------- 3

3------- 3

3-------× 1

5 2---------- 1

5 2---------- 2

2-------×

3 33

---------- 25 2×------------

3 210-------

5

12---------- 5

2 3---------- 3

3-------× 2 3+

2 3---------------- 2 3+

2 3---------------- 3

3-------×

152 3×------------ 3 2 3+( )

2 3×-----------------------------

156

---------- 2 3 3+6

--------------------

■ Note:Multiplying by is the

same as multiplying by 1.

3

3-------

Exercise 6:11

1

1

2------- 1

5------- 2

3------- 5

10---------- 3

2------- 6

3-------

10

5------- 2

11---------- 2

3------- 3

5------- 5

10---------- 3

15----------

1

2 2---------- 2

5 3---------- 7

2 5---------- 10

2 3---------- 6

2 3---------- 5

5 2----------

2 3

3 2---------- 5 7

3 5---------- 2 3+

3---------------- 1 5+

2---------------- 7 3+

2 7--------------------- 10 5–

5 10-----------------------


Evaluate each fraction correct to 3 significant figures (using your calculator). Then rationalise the denominator and evaluate the fraction again. Compare this answer with your first calculation.

a b c d

Rationalise each denominator, then express as a single fraction.

a b c d e

f g h i

Fun Spot 6:11 | What do Eskimos sing at birthday parties?Answer each question and put the letterfor that question in the box above thecorrect answer.

A E E

L E O

Simplify:

L O Y

E L O

F G D

E F G H

I J L N

O R S T

Solve:

W Y Z

2

2

5------- 3

7------- 3

2 2---------- 7

3 5----------

3

1

2------- 1

3-------+ 1

5------- 1

2-------– 1

6------- 1

5-------+ 2

10---------- 3

5-------– 3

8------- 5

2-------+

2

2 3---------- 1

3 2----------– 2

5 2---------- 5

10----------+ 5

2------- 5

3-------+ 2

5------- 3

2-------–

fun spot

6:11

x2--- 5= x

5--- 3= x

10------ 1·86=

x7--- 0·185= x

5--- 0·5= 0·560 x

5---=

5x x– 5x x÷ 5x x×2x 3y× x3 x5× x6 x2÷2x5( )3 5x0 4× 35 10 2–×169 7 7× 3 3( )2 2 2+

50 3 6× 3 7 7– 50 18+

5 2 3 2× 32 2÷ 6 2÷ 3 2 3+

5x 3 21=+ 14x

------ 1= 2x 5 7 6x–=–

20 7 4 13

33

22

x18

·6= x

14=

3

52

82

x15

=

x1

·5=

x2

·5=

x10

=

30 x8

4x 5x2

27 x4

0·35

8x15

6xy 5

32

x2

·8=

x1

·295

=

27

x3

·6=


Challenge 6:11 | Rationalising binomial denominatorsExamplesRationalise the denominators foreach expression.

1 2

3

Solutions

1 = 2 =

= =

= =

3 =

=

=

Now try these exercises!

1 Express with a rational denominator.

a b c d

e f g h

i j k l

m n o p

2 Rationalise each denominator, then express as a single fraction.

a b c

ch

allenge

6:11

For these fractions,

we multiply top

and bottom by

the conjugate

of the denominator.5

5 2–---------------- 1

3 2+---------------------

2 3 5+

2 3 5–------------------------

5

5 2–---------------- 5

5 2–---------------- 5 2+

5 2+----------------× 1

3 2+--------------------- 1

3 2+--------------------- 3 2–

3 2–--------------------×

5 5 2+( )25 2–

------------------------- 3 2–3 2–

--------------------

25 5 2+23

----------------------- 3 2–

■ Note:The product of a binomial surd and its conjugate is always rational.

2 3 5+

2 3 5–------------------------ 2 3 5+

2 3 5–------------------------ 2 3 5+

2 3 5+------------------------×

2 3 5+( )2

12 5–-------------------------------

17 4 15+7

--------------------------

1

1 2+---------------- 1

3 1–---------------- 1

7 5–-------------------- 1

10 2+------------------------

3

3 2+---------------- 5

5 2–---------------- 10

5 2–-------------------- 12

7 3–--------------------

1

2 3 5+-------------------- 2

5 2 2–------------------- 3

3 2 2 3+--------------------------- 1

4 3 3 2–---------------------------

5 2+

5 2–---------------- 4 3+

4 3–---------------- 5 3–

5 3+--------------------- 3 2 3–

3 2 3+------------------------

1

2 3–---------------- 1

2 3+----------------+ 1

5 3–-------------------- 1

7 5+---------------------– 5

6 3–---------------- 3

5 3+----------------+


Maths terms 6

Maths terms 6base• The term which is operated on by the

index.eg for xn, x is the base

for 53, 5 is the base.

conjugate• The binomials that multiply to give the

difference of two squares are the ‘conjugate’ of each other.eg (a − b) and (a + b)

and These are conjugate pairs.

exponent• Another term for a power or index.• Equations which involve a power are

called exponential equations.eg 3x = 27

fractional indices• Another way of writing the ‘root’ of a

number or term.

• , ,

index• A number indicating how many of a base

term need to be multiplied together.eg for xn, n is the index

xn = x x x x x x . . . . . . x x

n factors• The plural of index is indices.

irrational numbers• Numbers that cannot be expressed in the

form where a and b are integers.

• They cannot be given an exact decimal value.eg π, ,

negative indices• Indicate the reciprocal of a term.

eg ,

ie ,

power• Another term for an index or exponent.

rational numbers• Numbers that can be written in the form

where a and b are integers (b ≠ 0).

• They can be expressed as a terminating or repeating decimal.eg integers, fractions, percentages

real numbers• The combination of rational and irrational

numbers.

scientific (standard) notation• A useful way to write very big or very small

numbers.• Numbers are written as the product of a

number between 1 and 10 and a powerof 10.eg 76 000 000 = 7·6 × 107

0·000 0054 = 5·4 × 10−6

surds• Numerical expressions that involve

irrational roots.

eg , ,

zero index• A term or number with a zero index is

equal to 1.eg x0 = 1, 40 = 1

maths terms

6

3 5+( ) 3 5–( )

x12---

x= x13---

x3= x1n---

xn=

⎧ ⎨ ⎩

ab--

3 3 2 1+

x 1– 1x---= x n– 1

xn-----=

5 1– 15---= 2 3– 1

23----- 1

8---= =

ab--

3 53 2 7 5+


Diagnostic Test 6: | Indices and Surds• These questions reflect the important skills introduced in this chapter.• Errors made will indicate areas of weakness.• Each weakness should be treated by going back to the section listed.

These questions can be used to assess outcomes PAS5·1·1, PAS5·2·1, NS5·1·1, NS5·3·1.

tsetcitsongai d

6

1 Express in index form:

a 3 × 3 × 3 × 3 b 5 × 5 c m × m × m

2 Evaluate:a 32 b 24 c 103

3 Simplify:a 32 × 35 b x3 × x2 c 6m2n × mn4

4 Simplify:a x7 ÷ x2 b 15a5 ÷ 3a2 c 20a3b2 ÷ 10ab

5 Simplify:a (a4)2 b (x3)4 c (2a4)3

6 Simplify:a 70 b 5p0 c 18x3 ÷ 6x3

7 Simplify:a 3−2 b 5−1 c ( )−3

8 Simplify, writing answers without negative indices:a x7 × x−3 b 6x2 ÷ 3x4 c (3x−1)2

9 Simplify:

a b c

10 If x > 0 and m > 0, simplify:

a b c

11 Express in scientific notation:a 243 b 67 000 c 93 800 000

12 Write as a basic numeral:a 1⋅3 × 102 b 2⋅431 × 102 c 4⋅63 × 107

13 Express in scientific notation:a 0⋅043 b 0⋅000 059 7 c 0⋅004

14 Write the basic numeral for:a 2⋅9 × 10−2 b 9⋅38 × 10−5 c 1⋅004 × 10−3

15 Simplify, giving answers in scientific notation:a (3⋅1 × 108)2 b (8⋅4 × 106) + (3⋅8 × 107)

c d

Section6:01

6:01

6:01

6:01

6:01

6:01

6:02

6:02

6:03

6:03

6:04

6:04

6:04

6:04

6:05

23---

2512---

2713---

813---

3x12---

4x12---

× 49m6( )12---

8x3( )13---

1·96 1024× 1·44 10 6–×


16 For each, write rational or irrational.

a b c d

17 Evaluate correct to 3 decimal places:

a b c d

18 Simplify each surd.

a b c d

19 Express each of the following as an entire square root.

a b c d

20 Simplify completely:

a b c d

21 Simplify:

a b c d

22 Simplify:

a b c d

23 Expand and simplify:

a b

c d

24 Rationalise the denominator of:

a b c d

6:06

6:06

6:07

6:07

6:08

6:09

6:09

6:10

6:11

50 0·4̇ 2 16

5 13 21 47

20 27 3 8 2 75

2 5 3 2 5 7 4 5

2 3 4 3+ 6 5 5– 8 2– 27 2 3+

5 6× 3 12× 2 3 5× 3 8 2 2×

12 2÷ 32 8÷ 5 3 3÷ 10 10 2 5÷

2 5+( ) 3 5+( ) 2 3 2+( ) 3 3 2–( )7 3+( )2 5 3–( ) 5 3+( )

3

2------- 5

5------- 3 1+

2 3---------------- 5 2–

5 2----------------

• Can you use your calculator to find the value of 2500? What is the largest power of 2 that can be calculated using your calculator?


Chapter 6 | Revision Assignment1 Simplify, writing the answers in index

form:

a a2 × a3 b 3a2 × 4a3

c a2b × ab d 3a2b × 4ab2

e 32 × 33 f a6 ÷ a3

g 7m2 ÷ m h 12y6 ÷ 3y2

i 20a4b3 ÷ 10a2b2 j 47 ÷ 43

k (32)4 l (x2)3

m (a3)2 × a5 n m7 ÷ (m2)3

o p

2 Express in simplest form:a (2x2)0 b 6x0 c (5x3)3

d (10a2)3 e (4x2)3 ÷ 8x5

3 Write each of the following in standard form (scientific notation).a 21 600 b 125c 0·000 07 d 0·000 156

4 Write each of the following as a basic numeral.a 8·1 × 105 b 1·267 × 103

c 3·5 × 10−2 d 1·06 × 10−4

5 Use your calculator to evaluate:a 210 b 312

c 55 × 66 d 73 × 45

6 Find the value of n, if:a 2n = 128 b 3n = 243c 10n = 100 000 000

7 Simplify and evaluate:a 32 × 35 b 107 ÷ 104 c (24)2

8 Simplify:

a b (2a3)3 × (3a4)2

c

9 Noting that , evaluate without using a calculator:

a b c d

10 Simplify:

a b 10x2 ÷ 5x4

c

11 Simplify each expression.

a b

c d

e f

g h

i

j k

l

12 Rationalise the denominator of each expression:

a b

c d

Extension

13 Rationalise the denominator of each expression.

a b

c d

14 Simplify each expression, writing your answer with a rational denominator.

a b

c

tnemngissa

6A

12x2

6x----------- 4a3

8a--------

m7 m6×m10

--------------------

8x7 9x4×6x6 6x5×-----------------------

x32---

x3( )12---

=

432---

823---

932---

100053---

5x12---

4x12---

×

36m4n6( )12---

5 20 28 3 7+

3 8 2 18– 2 7 3 2×

2 5 20× 15 12 5 3÷

3 3 15+( ) 7 2+( )2

5 6+( ) 2 5+( )

m n× m n+( )2

m n+( ) m n–( )

5

2 5---------- 2 3

3 2----------

1

2------- 1

3-------+ 3

2 5---------- 2

3 2----------–

2

5 1–---------------- 1

7 2+---------------------

5 2–

5 2+---------------- 2 3+

2 3–---------------------

5

3------- 2

3 1–----------------– 3

5 2+---------------- 2

5 1+----------------×

7 2+

3 1+---------------- 2 3 1+

2 7 1–--------------------÷


Chapter 6 | Working Mathematically1 Use ID Card 7 on page xxii to identify:

a 5 b 8 c 17 d 18 e 19f 20 g 21 h 22 i 23 j 24

2 Use ID Card 6 on page xxi to identify numbers 1 to 12.

3 a How many diagonalscan be drawn from onevertex of a regularhexagon? How manyvertices has a hexagon?

b Each diagonal joins two vertices and adiagonal cannot be drawn from a vertex to the two adjacent vertices or to itself.The number of diagonals of a hexagon

is . How many diagonals has:

i a regular octagon?ii a regular decagon?iii a regular polygon that has 30 sides?

4 Tom was given acheque for an amountbetween $31 and $32.The bank teller madea mistake and exchangeddollars and cents on thecheque. Tom took themoney without examiningit and gave 5 cents tohis son. He now foundthat he had twice thevalue of the originalcheque. If he had nomoney before entering the bank, what was the amount of the cheque?

5 In the decibel scale, for measuring noise, 10 decibels is a noise that is barely audible. A noise 10 times as intense is 20 decibels, and so on up to 140 decibels, which is the threshold of pain. Study the table and answer the questions below.

a If ordinary conversation has a relative intensity of 106, what is its loudness in decibels?b If a lawn mower has a relative intensity of 1012, what is its loudness in decibels?c By how many times is the relative intensity of the mower greater than that of conversation?d By how many times is the relative intensity of heavy traffic (loudness 80 dB) greater than

that of bush quiet?e From the above it would appear that heavy traffic (80 dB) is four times as noisy as a

whisper at 1 metre (20 dB). However, a rise of 10 dB corresponds to a doubling in the subjective loudness to the human ear. How much louder to the human ear is:i the average office (50 dB) than bush quiet (30 dB)?ii heavy traffic (80 dB) than a whisper at 1 metre (20 dB)?iii a rock group (110 dB) than a business office (60 dB)?

assignment

6B

6 6 3–( )2

--------------------

$31.62 ...... $62.31?

Noise Relative intensity Decibels

Minimum of audible sound 1 0

Soft wind on leaves 10 10

Whisper at 1 metre 102 20

Bush quiet 103 30

1 Index laws

2 Negative indices

3 Fractional indices

4 Simplifying surds

5 Operations with surds

6 Indices and Surds - NafNothgielnafnothgiel.weebly.com/uploads/5/9/3/7/59373103/maths... ·...

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