GOMATHMEMOGR10

Page 1

Compiled by Chez Nell

2 Topic:

Pages:

1.

Number System

3

2.

Mathematical Notation

3

3.

Algebraic Products

4

4.

Factors

5

5.

Algebraic Fractions

6.

Solution of Equations

10

7.

Basic Logarithm Simplification

22

8.

Exponents

9.

Number Patterns

24

10. Straight Line Graphs

30

8

24

11. Parabola Graphs

35

12. Hyperbola Graphs

41

13. Exponential Graphs

44

14. Financial Mathematics

53

15. Probability Theory

66

16. Analytical Geometry

70

17. Transformation Geometry

72

18. Triangles & Quadrilaterals

82

19. Trigonometry

84

20. Data Handling

93

21.

Volumes & Surface Area

ďƒ“ Norma Nell 2011

104

3 Page 4: Exercise 1 1.1

0,5 ; 1 ; 1,5 ; 2 ; 2,5 ; 3 ; 3,5 ; 4 ; 4,5.

1.2 45

9

1.2.1 100 = 20 824

1.2.4

103

1000

= 125

25

5

1.2.5 1

21 100 = 21 4

1.2.7

25

1

125

1100 = 1 4

1.2.2.

10

1.2.8

1

=2 5 100

1

1.2.3 6 1000 = 6 8 5

1

1.2.6 2 10 = 2 2 1

= 20

2. 2.1 0,5555 = 0, 5 2.4 2.6

0,454454 = 0. 454

2.2.

123,12333 = 123,123

2.5

0,7141414 = 0,714

2.3

123,1232323 = 123,123

123,1231313 = 123,1231

3. 6

2

3.1

0, 6 = 9 = 3

3.4

45,345 = 45

3.6

23,3465 = 23

3.2 345−34 900

𝑥𝜖(−∞; 2] 𝑥𝜖[−3 ; 0] 𝑥𝜖(−∞ ; 2) 𝑥𝜖[1− ; 3] 𝑥𝜖(−2 ; 5]

2.1

𝑥: 3 ≤ 𝑥 ≤ 7 , 𝑥𝜖𝑍

2.2

𝑥: −8 ≤ 𝑥 ≤ −4 , 𝑥𝜖𝑍

2.3

𝑥: 10 < 𝑥 < 16, 𝑥𝜖𝑁

2.4

𝑥: −4 ≤ 𝑥 < 2 , 𝑥𝜖𝑍

9900

311

= 300

3465 −34

Page 6: Exercise 2.1: 1.1 1.2 1.3 1.4 1.5

0, 78 =

78 99

3.3

0,14 =

3.5 2,345 = 2 3431

= 23 9900

14−1

345−3 990

90

13

= 90 19

= 2 990

4 Page 8: Exercise 2.2:

⃘    ⃘

1.1 1 1.2

2

3

4

5

-2

2

6

1.3

6

1.4

-6 Page 12: Exercise 3.1 1.1

2( x  3)( x  4)  2 x 2  2 x  24

1.2

–3(3 – 5x) + 3(3-5x)

 9  15 x  9  15 x 0 1.3

–2a2(a – b)2  2a 4  4a 3b  2a 2 b 2

1.4

(x – 3)2 –2(x– 1)2 + 2x

 x 2  6x  9  2x 2  4x  2  2x  x 2  7 1.5

(a – 2b)( a2 + 2ab + 4b2)  a 3  8b 3

1.6

( x  2) 2  3( x  2)( x  2)

 x 2  4 x  4  3x 2  12  4x 2  4x  8

1.7

3( x  2)( x  3)  2( x  4) 2  6  x

 3x 2  3x  18  2 x 2  16 x  32  6  x  x 2  20 x  44

5

4 x  3 y 2 xy  y 2 

1.8

 8 x 2 y  4 xy 2  6 xy 2  3 y 3  8 x 2 y  2 xy 2  3 y 3 32 x  y   2x  y x  5 y  2

1.9

3(4 x 2  4 xy  y 2 )  2( x 2  4 xy  5 y 2 )  12 x 2  12 xy  3 y 2  2 x 2  8 xy  10 y 2  10 x 2  4 xy  13 y 2

1.10

p

q



 4 p 2q  4 p q  16

 p 3q  4 p 2 q  16 p q  4 p 2 q  16 p q  64  p 3q  64 Page 14 :Exercise 4.1 1. 𝑥 + 2 (𝑥 − 2) 2. 𝑥 + 3 (𝑥 − 3) 3. 𝑥 + 4 (𝑥 − 4) 4. 2 𝑥 2 − 25 = 2 𝑥 + 5 (𝑥 − 5) 5. 2𝑥 + 3𝑦 (2𝑥 − 3𝑦) 6. 𝑥 2 + 4 𝑥 2 − 4 = 𝑥 2 + 4 𝑥 + 2 (𝑥 − 2) 7. 𝑥 − 𝑦 + 𝑥 + 𝑦 𝑥 − 𝑦 − 𝑥 − 𝑦 = 2𝑥 −2𝑦 = −4𝑥𝑦 8. 2 𝑥 − 𝑦 + 3 𝑥 + 𝑦 2 𝑥 − 𝑦 − 3 𝑥 + 𝑦 = 2𝑥 − 2𝑦 + 3𝑥 + 3𝑦 2𝑥 − 2𝑦 − 3𝑥 − 3𝑦 = 5𝑥 + 𝑦 (−𝑥 − 5𝑦) Page 15: Exercise 4.2 1. 2. 3. 4.

𝑥 − 𝑦 (𝑥 2 + 𝑥𝑦 + 𝑦 2 ) 𝑥 + 𝑦 (𝑥 2 − 𝑥𝑦 + 𝑦 2 ) 2𝑥 − 3𝑦 (4𝑥 2 + 6𝑥𝑦 + 9𝑦 2 ) 4𝑥 + 5𝑦 (16𝑥 2 − 20𝑥𝑦 + 25𝑦 2 )

Page 17.Exercise 4.3 1. 𝑥 − 3 (𝑥 + 1) 2.

𝑥 − 1 (𝑥 − 5)

3.

𝑥 + 2 (𝑥 + 4)

4.

𝑥 ∓ 4 (𝑥 − 2)

5.

𝑥 − 3 (𝑥 − 4)

6.

𝑥 + 2 (𝑥 + 6)

6 7.

2𝑥 + 1 (𝑥 − 3)

8.

2𝑥 + 3 (𝑥 − 4)

9.

3𝑥 + 1 (𝑥 + 2)

Page 19:Exercise 4.4 1.1 𝑎 − 𝑏 (𝑥 + 𝑦) 1.2 𝑥 + 𝑦 (𝑝2 − 𝑞 2 ) = 𝑥 + 𝑦 𝑝 + 𝑞 (𝑝 − 𝑞) 1.3 (m  n)  p(m  n) = 𝑚 − 𝑛 (1 + 𝑝) 1.4

(a  b)  x(a  b) = 𝑎 − 𝑏 (1 − 𝑥)

1.5

(a  b)(a  b)  (a  b) = 𝑎 − 𝑏 (𝑎 + 𝑏 + 1)

1.6

= (4 x 2  9 y 2 )  (2 x  3 y ) = 2𝑥 − 3𝑦 2𝑥 + 3𝑦 + (2𝑥 + 3𝑦) = 2𝑥 + 3𝑦 (2𝑥 − 3𝑦 + 1)

1.7

4( x  y) 2  9( x  y) 2

= 2 𝑥−𝑦 +3 𝑥+𝑦

2 𝑥−𝑦 −3 𝑥+𝑦

= 2𝑥 − 2𝑦 + 3𝑥 + 3𝑦 2𝑥 − 2𝑦 − 3𝑥 − 3𝑦 = 5𝑥 + 𝑦 (−𝑥 − 5𝑦) 1.8

(2 x  y ) 2  ( x  2 y ) 2

= 2𝑥 + 𝑦 + 𝑥 − 2𝑦 2𝑥 + 𝑦 − 𝑥 + 2𝑦 = 3𝑥 − 𝑦 (𝑥 + 3𝑦)

Page 21: Exercise 4.5 1.11

3a +6 = 3(a  2)

1.12

5x 2  10 xy  5 y 2 = 5( x 2  2 xy  y 2 ) = 5( x  y)( x  y)

1.13

2x2 + 6x – 8x3 = 2 x( x  3  4 x 2 )

7

1.14

a2 – 4 = (a  2)(a  2)

1.15

2x2 – 32 = 2( x 2  16) = 2( x  4)( x  4)

1.16

x(a-b) + y(a-b) = (a  b)( x  y)

1.17

p2( x + y) - q2( y + x ) = ( x  y)( p 2  q 2 ) = ( x  y)( p  q)( p  q)

1.18

(m  n)  ( pn  pm) =

(m  n)  p(n  m) ( m  n)  p ( m  n)

= (m  n)(1  p)

(a  b)  (ax  bx) 1.19

a  b  ax  bx = (a  b)  x(a  b) (a  b)(1  x)

1.20

a2  b2  a  b =

1.21

[(2 x  y )  ( x  2 y )][(2 x  y )  ( x  2 y )] (2 x  y)  ( x  2 y) = (2 x  y  x  2 y )(2 x  y  x  2 y ) (3x  y )( x  3 y )

1.22

[2( x  y )  3( x  y )][2( x  y )  3( x  y )] 4( x  y)  9( x  y) = (2 x  2 y  3x  3 y )(2 x  2 y  3x  3 y ) (5 x  y )( x  5 y )

2

2

(a  b)(a  b)  (a  b) (a  b)(a  b  1)

2

2

(4 x 2  9 y 2 )  (2 x  3 y )

1.23

4 x 2  2 x  9 y 2  3 y = (2 x  3 y )(2 x  3 y )  (2 x  3 y )

(2 x  3 y )(2 x  3 y  1)

1.24

x 2  x  6 = ( x  3)( x  2)

8 1.25

x 2  6 x  8 = ( x  2)( x  4)

1.26

x 2  7 x  12 = ( x  3)( x  4)

1.27

2 x 2  24 x  70 =

2( x 2  12 x  35) 2( x  5)( x  7)

1.28

9 x 2  42 x  45 =

3(3x 2  14 x  15) 3(3x  5)( x  3)

1.29

x 3  y 3 = ( x  y)( x 2  xy  y 2 )

1.30

x 3  y 3 = ( x  y)( x 2  xy  y 2 )

1.31

27 x 3  64 y 3 = (3x  4 y)(9 x 2  12 xy  14 y 2 )

1.32

1 3 8 6 1 2 1 1 4 x  y = ( x  y 2 )( x 2  xy  y 2 ) 8 27 2 3 4 3 9

Page 29: Exercise 5.1.

6x x  12 2 4x 2 y 3 y 2 2. 3  2x 8x y 2 x  4 2( x  2) x  2 3.   4 4 2 xy  y y ( x  1) 4.   x 1 y y

1.

8 x 2  4 x 4 x(2 x  1)   2x  1 4x 4x x 2  1 ( x  1)( x  1) x  1 6.   ( x  1) 2 ( x  1)( x  1) x  1

5.

x 2  x  12 ( x  4)( x  3) x  4 7. 2   x  x  12 ( x  4)( x  3) x  4 a b ( a  b) 8.   1 b  a  ( a  b) a2 (a  2) 1 9. 2   a  a  2 (a  2)(a  1) a  1 10.

x 2  x  12 ( x  4)( x  3)   ( x  3) or  x  3 4 x  ( x  4)

11.

ab  a 2 b 2  ab a (b  a ) b(b  a ) b(b  a ) X  X  2 2 2 (b  a )(b  a ) a (b  a ) b a a a2

9

3 2 3b  2a 12.   a b ab 2 1 1  2  2 x  3x  2 x  x  2 x  1 2 1 1   ( x  1)( x  2) ( x  2)( x  1) ( x  1)( x  1) 2( x  1)  1( x  1)  1( x  2) ( x  1)( x  1)( x  2) 2x  2  x  1  x  2 ( x  1)( x  1)( x  2) 5 ( x  1)( x  1)( x  2)

14.    

2

15.

x x2  2 x  y y  x2

.

x x2  ( x  y ) ( y  x)( y  x)

x( y  x)  x 2 ( x  y )( y  x)

xy  x 2  x 2 ( x  y )( y  x) xy  ( x  y )( y  x) 7x 3x 16.  2x  2 y 5 y  5x 7x 3x   2( x  y )  5( x  y )  5(7 x)  2(3 x)   10( x  y )  35 x  6 x   10( x  y )  41x   10( x  y ) 

10 1. 3ЁЭСе = тИТ9 ЁЭСе = тИТ3 2. тИТЁЭСе = 8

ЁЭСе = тИТ8

3. 4. 5. 6.

7.

ЁЭСе = тИТ4 тИТ8ЁЭСе = тИТ4 1 ЁЭСе=2 3ЁЭСе + 6 = 2ЁЭСе тИТ 2 ЁЭСе = тИТ8 1 тИТ 8ЁЭСе + 40 тИТ 2ЁЭСе + 6 = 10 тИТ 8ЁЭСе + 12 тИТ2ЁЭСе = тИТ25 ЁЭСе = 12,5 8ЁЭСе тИТ 28 тИТ 6ЁЭСе тИТ 12 = 40 тИТ 8ЁЭСе тИТ 5ЁЭСе тИТ 35 15ЁЭСе = 45 ЁЭСе=3

Page 28: Exercise 5.2 1.

2.

3.

y y яАн яА╜ 1 . X LCD = 4 2 4 2ЁЭСж тИТ ЁЭСж = 4 ЁЭСж=4 x x яАл яА╜3 X LCD = 10 5 10 2ЁЭСе + ЁЭСе = 30 ЁЭСе = 10 x x 1 яА╜ яАн 4 3 2

X LCD = 12

3ЁЭСе = 4ЁЭСе тИТ 6 ЁЭСе=6 4.

x 1 x 3 яАл яА╜ яАл 2 4 4 2

X LCD = 4

2ЁЭСе + ЁЭСе = ЁЭСе + 6 ЁЭСе=3 8.

7 x яАл 2 (9 x яАн 2) яАн яА╜ 2 X LCD = 15 3 5

11 35𝑥 + 10 − 27𝑥 + 6 = 30 8𝑥 = 14 7

𝑥=4 9.

3x  4 2x  3   1 X LCD = 4 2 4 6𝑥 − 8 = 2𝑥 − 3 + 4 4𝑥 = 9 9

𝑥=4

10.

3 x  5 2( x  5)  X LCD = 6 2 3 9𝑥 − 15 = 4𝑥 − 20 5𝑥 = −5 𝑥 = −1

11.

x  1 ( x  2) x  1 x  2 X LCD = 12    3 4 2 3

4𝑥 − 4 − 3𝑥 + 6 = 6𝑥 + 6 + 4𝑥 + 8 −9𝑥 = 12 4

𝑥 = −3 12.

1 3  1 2  x     x   X LCD = 6 2 10  3  3 9

4

3𝑥 − 10 = 2𝑥 − 3 90𝑥 − 27 = 60𝑥 − 40 30𝑥 = −13 13

𝑥 = − 30

X LCD = 30

12

Page 32: Exercise 5.3 1.

4 1   0 X LCD = 3x 3 3x 𝟒𝒙 − 𝟏 = 𝟎 𝟏

𝒙=𝟒 2.

2 4 2   x 3 3x

X LCD = 3x

𝟔 − 𝟒𝒙 = 𝟐 𝒙=𝟏 3.

3 1 9 2  x 2 2x

X LCD = 2x

𝟔 − 𝟒𝒙 = 𝒙 − 𝟗 −5𝑥 = −15 𝑥=3 4.

3

2 3  4 x 2x

X LCD = 2x

6𝑥 − 4 + 3 = 8𝑥 −2𝑥 = 1 1

𝑥 = −2 5.

1 1 2 x   X LCD = 6x 2 3x 6x 3𝑥 + 2 = 2 − 𝑥 4𝑥 = 0 𝑥=0

6.

5 2 x3   4 4 3x 12 x

X LCD = 12x

13 15𝑥 + 10 = 48𝑥 − 𝑥 + 3 −32𝑥 = −7 7

𝑥 = 32 7.

x 3 2 x3 x3

X LCD = 𝑥 − 3

𝑥 + 2𝑥 − 6 = 3 3𝑥 = 9 𝑥=3

Page 35: Exercise 5.4

1. ( x  5)( x  2)  0 𝑥 = 5 𝑜𝑟 𝑥 = 2

2. (a  6)(a  1)  0 𝑎 = −6 𝑜𝑟 1

3. x( x  1)  0 𝑥 = 0 𝑜𝑟 𝑥 = 1

5. x(2 x  5)(3x  2)  0 5

4. ( x  2)( x  3)( x  5)  0 𝑥 = 2 𝑜𝑟 𝑥 = −3 𝑜𝑟 𝑥 = 5 6. y 2  3 y  10  0

𝑥 = 0 𝑜𝑟 𝑥 = 2

𝑦+2 𝑦−5 =0

𝑜𝑟 𝑥 =

𝑦 = −2 𝑜𝑟 5

2 3

7. x 2  5x  6  0 𝑥+2 𝑥+3 =0 𝑥 = −2 𝑜𝑟 − 3

9. x( x  1)  6 𝑥2 − 𝑥 − 6 = 0 𝑥−3 𝑥+2 =0 𝑥 = 3 𝑜𝑟 𝑥 = −2

8. x 2  7 x  6  0 𝑥−1 𝑥−6 =0 𝑥 = 1 𝑜𝑟 𝑥 = 6.

10. ( x  3)( x  2)  12 𝑥 2 + 5𝑥 − 6 = 0 𝑥+6 𝑥−1 =0 𝑥 = −6 𝑜𝑟 𝑥 = 1

11. x 2  2 x  15  0 𝑥−3 𝑥+5 =0 𝑥 = 3 𝑜𝑟 − 5 12. x( x  16)  3(24  5x) 𝑥 2 − 16𝑥 = 72 − 15𝑥 𝑥 2 − 𝑥 − 72 = 0

13. (2 x  5)(3x  2)  2(3x 11) 6𝑥 2 − 11𝑥 − 10 = 6𝑥 − 22 6𝑥 2 − 17𝑥 + 12 = 0

14 𝑥−9 𝑥+8 =0 𝑥 = 9 𝑜𝑟 − 8

2𝑥 − 3 3𝑥 − 4 = 0 3 4 𝑥 = 2 𝑜𝑟 3

Page 38: Exercise 5.5

x 2  4x  3  0

1.

=

−𝑏± 𝑏 2 −4𝑎𝑐

=

2𝑎 4± 16−4 1 (3)

= 2 𝑥 = 3 𝑜𝑟 𝑥 = 1

x 2  6x  4  0

4.

−𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 −6± 36−4 1 (4)

= 2 𝑥 = −0,76 𝑜𝑟 𝑥 = −5,24

2 x 2  x  10  0

2.

=

𝑏± 𝑏 2 −4𝑎𝑐

=

2𝑎 1± 1−4 2 (−10)

= 4 𝑥 = 1,4 𝑜𝑟 𝑥 = −0,90

3x 2  x  2  0

3.

= =

−𝑏±

𝑏 2 −4𝑎𝑐

2𝑎 1± 1−4 1 (−2) 2

6

1

𝑥 = 3 𝑜𝑟 𝑥 = − 3

2.

x  3y  5

2𝑎 7± 49−4 2 (4)

2𝑥 2 + 6𝑥 + 3 = 0

6. =

−𝑏± 𝑏 2 −4𝑎𝑐

=

2𝑎 −6± 36−4 2 (3) 4

𝑥 = −0,63 𝑜𝑟 𝑥 = −2,37

x y 5 and x  y  3 y  5 x x  (5  x)  3 x5 x  3 2x  8 x4 y 1

x  3 y  5

−𝑏± 𝑏 2 −4𝑎𝑐

= 4 𝑥 = 2,7 𝑜𝑟 𝑥 = 0,72

Page 41: Exercise 5.6.

1.

2𝑥 2 − 7𝑥 + 4 = 0

5.

and 2 x  3 y  1

15

2(3 y  5)  3 y  1 6 y  10  3 y  1 9y  9 y 1 x  2 3.

x y 8 and 3x  2 y  21 x  8 y 3(8  y )  2 y  21 24  3 y  2 y  21  y  3 y3 x5

3x  3 y  45 4.

5.

6.

3x  2 y  60 and 3x  3 y  454 x  y  15 3 x  2 y  60 3( y  15)  2 y  60 3 y  45  2 y  60 5 y  15 y3 x  18 x  y  36

and x  2 y  12 y  36  x x  2(36  x)  12 x  72  2 x  12 3x  60 x  20 y  16

x  2y  5 x  5  2y

and 3x  y  1

16

3(5  2 y )  y  1 15  6 y  y  1  7 y  14 y2 x 1 7.

6 x  y  22 y  22  6 x

and 4 x  y  8

4 x  (22  6 x)  8 4 x  22  6 x  8 10 x  30 x3 y4

8.

2y  x  3  x  3  2 y and 4 x  3 y  10 x  2y  3 4(2 y  3)  3 y  10

8 y  12  3 y  10 11 y  22 y2 x 1

9.

10.

3x  y  5  0  y  3x  5 and 7 x  3 y  1  0 y  3x  5 7 x  3(3x  5)  1 7 x  9 x  15  1 16 x  16 x 1 y  2

 m  1 2k and

2k  m  3 m  3  2k

17

 (3  2k )  1  2k  3  2k  1  2k 4k  4 k 1 m 1

x  5y  0

11.

2 x  3 y  14 and

12.

3 x  4 y  24 3 x  4 y  24 and 7 x  4 y  16

x  5y 2(5 y )  3 y  14 10 y  3 y  14 7 y  14 y2 x  10

x

4 y8 3 4 7( y  8)  4 y  16 3 28  y  56  4 y  16 3  28 y  168  12 y  48

 40 y  120 y  3 x  4

13.

3x  y  2

and 6 x  y  25 y  2  3x 6 x  (2  3x)  25 6 x  2  3x  25 9 x  27 x3 y  4

18

2 x  9  y and x  36  4 y 2(36  4 y )  9  y 72  8 y  9  y  9 y  63 y7 x8

14.

Page 49: Exercise 5.7

1.

The sum of two numbers is 54 and their difference is 6. Find the numbers. Let one number be x and the other y

x  y  54

x y  6 x y6

x  y  54 y  6  y  54 2 y  48

The numbers are 24 and 30

y  24 x  30

2.

The sum of two numbers is 35 and their difference is 19. Find the numbers. Let one number be x and the other y

x  y  19

x  y  35

x  y  19

x  y  35 y  19  y  35 2 y  16

The numbers are 8 and 27

y8 x  27

3.

In a two digit number, the sum of the digits is 12 and their difference is 4. Find the number if the tens digit is larger than the units digit. Let one digit be x the other is y

x  y  12

x y  4 x y4

19

y  4  y  12 2y  8 The number is 84. y4 x8

4.

The length of a rectangle is twice the breadth, while the perimeter is 6m. Find the length and breadth of the rectangle. { Hint: P  2(l  b) .} Let the length be x and the breadth y

P  2x  2 y 6  2( 2 y )  2 y 6  6y

x  2y

y 1 x2 The length is 2 and breadth 1

5.

The perimeter of a rectangular flower bed is 26m. If the length exceeds the breadth by 3m, find its dimensions.

x y 3 x  3 y

P  2x  2 y 26  2(3  y )  2 y 26  6  2 y  2 y 20  4 y

Length is 8 and breadth is 5

y5 x8 6.

A number consisting of two digits has the following properties. When the number is added to twice the tens digit the answer is 33. If the digits are reversed, the number obtained exceeds the original number by 63. What is the original number? Let x represent the tens digit thus tens digit is 10x Let y represent the units digit thus units digit is

10 x  y  2 x  33 y  33  12 x

10 y  x  10 x  y  63 9 y  9 x  63 y  x7

y

20

33  12 x  x  7  13 x  26 x2 y9 The original number is 29

7.

A boy is 6 years older than his sister. In three years time he will be twice her age. What are their present ages? Let the boys age be x The sisters age is x - 6 In three years time : Boy is x+3 Sister is x -3

x  3  2( x  3) x  3  2x  6 x9 Boy is 9 and sister is 3

8.

Tumi is twice as old as John. Two years ago she was three times as old as John was then. What are their present ages? Let John’s age be x Tumi is 2x Two years ago: John is x - 2 Tumi is 2x – 2

3( x  2)  2 x  2 3x  6  2 x  2 x4 Page 54: Exercise 5.8 1.

2x  6  8 2� > 14 �>7

2.

3x  6  x  14 2đ?‘Ľ < −8

John is 4 and Tumi is

21 𝑥 < −4

3.

2 x  7  5x  14 −21 ≥ 3𝑥 −7 ≥ 𝑥

4.

3x  7  3x  14 2

3𝑥 − 7 ≤ 6𝑥 + 28 −35 ≤ 3𝑥 −

5.

35 3

≤𝑥

 6  2 x  14 = −3 < 𝑥 < 7

6.

 5  2 x  3  15

= −8 ≤ 2𝑥 ≤ 12 = −4 ≤ 𝑥 ≤ 6 Page 56: Exercise 5.9 .

1.1

3x  2  x  14 2𝑥 ≥ −12 𝑥 ≥ −6

1.2

5x  3  3x  15 2𝑥 ≤ −18 𝑥 ≤ −9

1.3

1  2x  x  2 3 > 3𝑥 1>𝑥

1.4

4( x  4)  7( x  2)  1 4𝑥 − 16 ≤ 7𝑥 − 14 + 1 −3 ≤ 3𝑥 −1 ≤ 𝑥

1.5

4(2 x  1)  5x  2 8𝑥 − 4 < 5𝑥 + 2 3𝑥 < 6 𝑥<2

22 1.6

2( x ๏ ซ 1) ๏ ณ 2(2 x ๏ ญ 1) ๏ ญ 2 ๏ ซ x 2๐ ฅ + 2 โ ฅ 4๐ ฅ โ 2 โ 2 + ๐ ฅ 6 โ ฅ 3๐ ฅ 2โ ฅ๐ ฅ

1.7

๏ ญ 2( x ๏ ญ 3) ๏ พ 5x ๏ ญ 78 โ 2๐ ฅ + 6 > 5๐ ฅ โ 78 84 > 7๐ ฅ 12 > ๐ ฅ

1.8

1.9

3( x ๏ ญ 2) 7( x ๏ ญ 3) ๏ ญ ๏ ฃ 3 X LCD = 4 2 4 6๐ ฅ โ 12 โ 7๐ ฅ + 21 โ ค 12 โ ๐ ฅ โ ค 3 ๐ ฅ โ ฅ โ 3 x๏ ญ2 7 3 X LCD = 24 ๏ ญx๏ ณ ๏ ญ 3 8 2 8๐ ฅ โ 16 โ 24๐ ฅ โ ฅ 21 โ 36 โ 16๐ ฅ โ ฅ 1 1 ๐ ฅโ คโ 16

1.10

1.11

4 ๏ ญ x 2x ๏ ญ 1 ๏ ญ ๏ ผ x ๏ ญ 2 X LCD = 6 2 3 12 โ 3๐ ฅ โ 4๐ ฅ + 2 < 6๐ ฅ โ 12 26 < 13๐ ฅ 2<๐ ฅ 3x ๏ ญ 1 3x ๏ ญ 3 19 X LCD = 8 ๏ ซ ๏ ฃ 4 8 8 6๐ ฅ โ 2 + 3๐ ฅ โ 3 โ ค 19 9๐ ฅ โ ค 24 8 ๐ ฅโ ค 3

1.12

5 ๏ ฃ 3x ๏ ซ 3( x ๏ ซ 5) X LCD = 2 2 2๐ ฅ โ 5 โ ค 6๐ ฅ + 6๐ ฅ + 30 โ 35 โ ค 10๐ ฅ 7 โ โ ค๐ ฅ x๏ ญ

2

1.13

x ๏ ซ 2 1 ๏ ญ 3x ๏ พ ๏ ซ 2 X LCD = 15 5 3 45 โ 3๐ ฅ โ 6 > 5 โ 15๐ ฅ + 30 12๐ ฅ > โ 4 3๏ ญ

23 1

๐ ฅ > โ 3 1.14

๏ ญ1 ๏ ฃ x ๏ ญ 3 ๏ ผ 5 2โ ค๐ ฅ<8

1.15

1.16

1.17

1 ๏ ฃ 1 ๏ ญ 2x ๏ ผ 7 0 โ ค โ 2๐ ฅ < 6 0 โ ฅ ๐ ฅ > โ 3 x๏ ญ3 ๏ ญ3๏ ฃ ๏ ผ4 2 2x ๏ ญ 3 ๏ ฃ ๏ ญ4 X LCD = 3 3 6 < 2๐ ฅ โ 3 โ ค โ 12 9 < 2๐ ฅ โ ค โ 9 9 9 <๐ ฅโ คโ 2๏ ผ

2

1.18

2

2 ๏ ญ 2x ๏ ฃ 3 X LCD = 4 4 โ 8 โ ค 2 โ 2๐ ฅ โ ค 12 โ 10 โ ค โ 2๐ ฅ โ ค 10 5 โ ฅ ๐ ฅ โ ฅ โ 5 ๏ ญ2๏ ฃ

Page 62: Exercise 6.1 1. 2. 3. 4.

log 1000 ๐ .=๐ log 100 log 12 = 1,79 x๏ ฝ log 4 log 300 = 1,076 x๏ ฝ log 200 log 15,5 = 2,49 x๏ ฝ log 3 x๏ ฝ

5.

x๏ ฝ

2 log 25 = 0,79 3 log 15

6.

x๏ ฝ

5 log 50 2 log 8 ๏ ซ 2 log 25 6 log 6

7.

x๏ ฝ

10 log 450 log 500 = 1,49 ๏ ซ 12 log 60 6 log 65

= 3,43

24 Page 65: Exercise 7.1 1.

2 x яАл3 2 x яА╜ 2 x яАл 4. x яАн1 2

2.

2 x яАл1 2 x яАн3 2 2 x яАн2 яА╜ 2 x яАн2 яА╜ 1 2 x яАн2.2 x 2

3.

4 x яАн18 x яАл1 (2 2 ) x яАн1 (2 3 ) x яАл1 2 2 x яАн2 2 3 x яАл3 2 5 x яАл1 1 1 яА╜ яА╜ яА╜ 5 x яАл5 яА╜ 4 яА╜ x яАл1 5 x яАл1 5 x яАл5 16 32 (2 ) 2 2 2

4.

5 x 25 x яАн1 (5 x.(2 2 ) x яАн1 5 x 5 2 x яАн2 53 x яАн2 1 1 яА╜ яА╜ яА╜ 3 x яАл1 яА╜ 3 яА╜ x 3 x 3x 125 5.125 5.(5 ) 5.2 5 5 xяАл4

5. 6.

7 x яАн2 49 x яАл 2 7 x яАн2.(7 2 ) x яАл 2 7 x яАн2..7 2 7 3xяАл2 яА╜ яА╜ яА╜ 3xяАл2 яА╜ 7 0 яА╜ 1 3 xяАл2 3xяАл2 3xяАл2 7 7 7 7 n яАл1 n яАн1 n n яАл1 2 n яАн1 n n яАл1 n яАл1 2 n яАн 2 n яАн1 n 6 .12 .2 (2.3) .(2 .3) .2 2 .3 .2 .3 .2 2 4 nяАн1.32 n 1 1 яА╜ яА╜ яА╜ яА╜ 4 яА╜ n яАл 2 n яАн1 2 nяАл 2 3 n яАн1 n яАл 2 2 n яАл 4 3 n яАн3 4 n яАн1 2 n яАл 4 81 18 .8 (2.3 ) .(2 ) 2 .3 .2 2 .3 3

7. 6 nяАл 212 2 nяАл14 2 nяАн3 (2.3) nяАл 2 (2 2.3) 2 nяАл1 (2 2 ) 2 nяАн3 2 nяАл 2 3nяАл 2 2 4 nяАл 2 32 nяАл12 4 nяАн6 2 9 nяАняАн2 33nяАл3 3 3 яА╜ яА╜ яА╜ 9 n яАл3 3n яАл 2 яА╜ 5 яА╜ 3n яАл1 n яАл1 n 3 3n яАл1 2 n яАл1 n 9 n яАл3 2 n яАл 2 n 32 8 9 3 (2 ) (3 ) 3 2 3 3 2 3 2

Page 68: Exercise 7.2 2 x яАл 3 яАл 2 x 2ЁЭСе (8+1) 1. = ЁЭСе 1 = 18 2 2 x яАн1 2 2.

2 x яАл1 яАл 2 x яАн 3 2ЁЭСе (2+8) = ЁЭСе 1 = 2 +1 2 xяАн2 яАл 2 x 4

3.

3 x яАл 3 xяАл2 3ЁЭСе (1+9) = = 1 3ЁЭСе +1) 3 x яАн1 яАл 3 x 3

4.

5 2яАн x яАн 4.5 яАн x 5тИТЁЭСе 25тИТ4 21 = = тИТЁЭСе яАнx яАн x яАл1 5 1+10) 11 5 яАл 2.5

5.

3 n .3 4 яАн 6.3 n .31 3ЁЭСе 81тИТ18 = 3ЁЭСе .63 = 1 7.3 n .3 2

6

2 n .2 5 яАн 3.2 n .2 яАн2 = 5.2 n 2 3

1

Page 72: Exercise 7.3 1. 2x = 8

17 8 5 4

10

17

= 10 =

4 3

3 4

2ЁЭСе 32тИТ 2ЁЭСе .40

15 2

=

125 4

40

25

= 32

25 2ЁЭСе = 23 ЁЭСе=3 2. 3 x яА╜ 81 3ЁЭСе = 34 ЁЭСе=4 3. x3 = 27 ЁЭСе 3 = 33 ЁЭСе=3 4. 5 x

яАн4

3

= 80

4

ЁЭСе тИТ3 = 24 ЁЭСе = 24

3 4

тИТ

1

ЁЭСе = 2тИТ3 = 8 5. 2 x яА╜ 16 x яАн1 2ЁЭСе = 24ЁЭСетИТ4 ЁЭСе = 4ЁЭСе тИТ 4 тИТ3ЁЭСе = тИТ4 ЁЭСе=

4 3

6. 3.2x = 48 2ЁЭСе = 24 ЁЭСе=4 7.

5.42 x яА╜ 40

24ЁЭСе = 23 3

ЁЭСе=4

26

Page 76: Exercise 8.1 1. Mina invests R8000,00 in a plan where the growth is as follows: 0 1 2 3 4 5 6 8000 8440 8880 9320 9760 10200 10640

No of yrs passed :n Value of investment (in R) :A

Find the rule for this investment plan. Tn  8000  8000  0,055  n

Tn  8000  440n Use the rule to find the values for the next 3 years. R9760; R10200; R10640 Use the rule to find the value after 23 years. Tn  8000  440n

T23  8000  440(23)  R18120 If the value after n years was R13 280 , what is the value of n? Tn  8000  440n

13280  8000  440n 440n  5280 n  12 yrs

Tn

2. 3; 5; 7; 9; 11; 13; 15;….. Add 2 to each term to get the next term. Tn  an  c a  d1  2 9  2(4)  c

an 

c

T4  9

c 1 Tn  2n  1 T20  2(20)  1 T20  41

3. Given the sequence 7; 10; 13;….. Tn  an  c a  d1  3 10  3(2)  c c4 Tn  3n  4

T2  10

T50  3(50)  4  154 4.

Find a rule or general term for each of the following sequences and then find the 10 term for each one.

27 4.1

1; 6; 11; 16; 21;…. Tn  an  c 6  5(2)  c c  4 Tn  5n  4

4.2

13; 23; 33; 43;… Tn  an  c

23  10(2)  c 23  20  c . c  13 Tn  10n  13 4.3

6; 9; 12; 15;…. Tn  an  c 12  3(3)  c c3 Tn  3n  3

4.4

22; 20; 18; 16;…. Tn  an  c 20  2(2)  c c  24 Tn  2n  24

4.5

7; 2; -3 ;-8;….. Tn  an  c 2  5(2)  c c  12 Tn  5n  12

4.6

7; 11; 15; 19;…. Tn  an  c 15  4(3)  c c3 Tn  4n  3

5.1

Tn = n – 5

28

T1  1  5

T2  2  5

d 1 T1  4 T2  3 Sequence: -4 ; -3; -2; -1; 0;…. 5.2

Tn = 3n – 5 T1  3(1)  5

T2  3(2)  5

T1  2 T2  1 Sequence: -2 ; 1; 4 ; 7; 10;…. 5.3

Tn = -2n – 5 T1  2(1)  5

d=3

T2  2(2)  5

d = -2 T1  7 T2  9 Sequence: -7 ; -9; -11; -13; 15;…. 5.4

Tn = 4n + 2 T1  4(1)  2

T2  4(2)  2

Tn = 2 – 3n T1  2  3(1)

T2  2  3(2)

d=4 T1  6 T2  10 Sequence : 6; 10; 14; 18 22;…… 5.5

d = -3 T1  1 T2  4 Sequence = -1; -4; -7; -10; -13;…..

6. a) 3; 9; 15; 21; 27; 33; 39;….. Tn  an  c b)

c)

9  6( 2)  c c  3 Tn  6n  3 T50  6(50)  3

T50  297

2; 9; 16 ; 23; ….. Tn  an  c 16  7(3)  c c  5 Tn  7 n  5

2; -3; -8; -13;….

29 Tn  an  c  8  5(3)  c c7 Tn  5n  7

3; -1; -5; -9;… Tn  an  c  5  4(3)  c c7 Tn  4n  7

20; 27; 34; 41;……. Tn  an  c 34  7(3)  c c  13 Tn  7 n  13

Page 83: Exercise 8.2 Questions 1 to 3

1. the sequence is given by the number of shaded squares

Tn  2n  6 T50  106

2. The sequence is given by the number of matchsticks

 

 

 

 

 

     

 

 

 

 

  

   

30

Tn  4n  1 T50  201 3. The sequence is given by the perimeter of each rectangle.

Tn  4n  6 T50  206

4

fig1 4.1 4.2

d = 2 and T5  11

Tn  2n  1 Determine the 20th term. T20  2(20)  1  41

5.

Fig 3

Determine the next 3 terms. …..9; 11; 13;….. Determine the nth term formula 11  2(5)  c

c 1 4.3

Fig 2

Fig 4

31

5.1

Determine the next 3 terms. ‌21; 25; 29;‌..

5.2

Determine the nth term formula 9  4(2)  c

c 1 Tn  4n  1 5.3

Determine the 20th term. T20  4(20)  1  81

6.

6.1 6.2

Determine the next 3 terms. ‌9; 11 ;13;‌‌ Determine the nth term formula 5  2(3)  c

c  1 Tn  2n  1 6.3 Determine the 20th term. T20  2(20)  1  39 Page 88: Exercise 9.1. 1.

Change the following equations to the y-form. 1.1 đ?‘Ś = 3đ?‘Ľ − 6

32 1.2

1.3

2.

8đ?‘Ś = 4đ?‘Ľ − 16 1 đ?‘Ś = 2đ?‘Ľ −2 3đ?‘Ś = 5đ?‘Ľ + 9 5 đ?‘Ś = 3đ?‘Ľ +3

Sketch the following linear functions using the table method: x y

-1 2

0 3

1 4

y  x3

2.1

4

2

-10

f x  = x+3

-5

5

10

-2

-4

x y

22.

-1 -6

0 -4

1 -2

y  2x  4

4

2

-10

-5

f x  = 2ďƒ—x-4

5

-2

-4

10

33 x y

2.3

-1 2

0 4

1 6

2 y  4x  8 y  2x  4 8

6

f x  = 2x+4 4

2

-10

-5

5

10

-2

X Y

2.4

-1 -1

0 1

1 3

3y  6x  3 3y  6x  3 y  2x  1 8

6

f x  = 2x+1 4

2

-10

-5

5

10

-2

X Y

-1 3,5

0 3

1 -2.5

34

2.5

2y  x  6 2 y  x  6 1 y   x3 2 8

6

4

f x  = -0.5x+3 2

-10

-5

5

10

-2

X Y

2.6

-1 4

0 2

1 0

4 y  8x  8  0 4 y  8 x  8 y  2 x  2 8

6

4

f x  = -2x+2 2

-10

-5

5

-2

Page 93: Exercise 9.2. 1.

y  x3

10

35 At x = 0 ∴ y - intercept = 3 At y = 0 ∴ x – intercept = -3

2

-5

5

-2

2.

y  2x  4 At x = 0 ∴ y – intercept = -4 At y = 0 ∴ x – intercept = 2

-5

5

-2

-4

3

2 y  4x  8 At y = 0 ∴ x – intercept = -2 At x = 0 ∴ y – intercept = 4

4

36

4.

3 y  6x  3 đ?&#x;?

At y = 0 ∴ x – intercept = − đ?&#x;? At x = 0 ∴ y – intercept = 1

4

2

5.

2y  x  6 At y = 0-5∴ x – intercept = 6 At x = 0 ∴ y – intercept = 3

5

6

4

2

-5

5

-2

6.

4 y  8x  8  0

10

37 At x = 0 ∴ y – intercept = 2 At y = 0 ∴ x – intercept = 1 4

2

-5

5

-2

Page 100:Exercise 10.1 4

1. x y

y  x2 -2 4

0 0

2 4

fx = x2

2

-5

y  x2  1 x -2 đ?&#x;? 5 y=đ?’™ +đ?&#x;?

5

2.

0 1

2 5

-2

4

gx = x2+1 2

-5

5

6

y  x2  2 x -2 6 y = đ?’™đ?&#x;? + đ?&#x;?

3.

0 2

2 6

4

gx = x2+2

2

y  x2 1 x -2 đ?&#x;? 3 y=đ?’™ −đ?&#x;?

4

4.

0 -1

2 3

5 2

10

gx = x2-1

5

38

4

y  x2  2 x -2 đ?&#x;? 2 y=đ?’™ −đ?&#x;?

5.

0 -2

gx = x2-2

2

2 2

5

10

-2

Page 105: Exercise 10.2 A. Sketch the following graphs on the same set of axes: 1. y  x2 2. y  x2 1 3. y  x2  1 What deduction can be made? 8

6

4

hx  = x 2+1 gx  = x 2

-10

2

-5

qx  = x 2-1

5

10

-2

Deduction: The ‘c’ value is the only difference in each equation and this change causes a vertical shift of the graph.

39

B. Sketch the following graphs on the same set of axes: 1. y  x2 2. y  2x 2 1 3. y  x2 2 What deduction can be made? 8

qx  = 2ďƒ—x 2 6

hx  = 0.5ďƒ—x 2 4

gx  = x 2 2

-10

-5

5

10

-2

Deduction: The coefficient of đ?’™đ?&#x;? changes and this causes the slope of the arms to change C. Sketch the graphs of the following equations: 1. y  x2 2. y  x2 What deduction can be made? 4

gx  = x 2 2

-10

-5

5

10

qx  = -1ďƒ—x 2 -2

-4

-6

đ?&#x;?

Derduction: The coefficient of � changes sign and this inverts the graph; i.e. a +ve coefficient produces a minimum graph and a –ve produces a maximum graph.

D. Sketch the graphs of the following on the same set of axes:

40

y   x  1 and y  x  2

1.

4

2

-10

gx  = x-2

-5

5

10

qx  = -x+1 -2

-4

-6

2.

Write down the coordinates of the point(s) of intersection for the two graphs. đ?&#x;? đ?&#x;? đ?’‘đ?’?đ?’Šđ?’?đ?’• đ?’?đ?’‡ đ?’Šđ?’?đ?’•đ?’†đ?’“đ?’”đ?’†đ?’„đ?’•đ?’Šđ?’?đ?’? = (đ?&#x;? đ?&#x;? ; − đ?&#x;?)

E. Sketch the graphs of the following on the same set of axes: 1. đ?’š = −đ?’™đ?&#x;? and y  x  2

4

2

-10

gx  = x-2

-5

5

10

qx  = -x 2 -2

-4

-6

2.

Write down the coordinates of the point(s) of intersection for the two đ?‘?đ?‘œđ?‘–đ?‘›đ?‘Ąđ?‘ đ?‘œđ?‘“ đ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘&#x;đ?‘ đ?‘’đ?‘?đ?‘Ąđ?‘–đ?‘œđ?‘› đ?‘Žđ?‘&#x;đ?‘’: −2 ; −4 đ?‘Žđ?‘›đ?‘‘ (1 ; −1)

Page 111: Exercise 10.3 1. y  x2  9

41

𝐴𝑡 𝑦 = 0 ∶ 𝑥 2 − 9 = 0 𝑥−3 𝑥+3 =0 𝑥 = 3 𝑜𝑟 𝑥 = −3

𝐴𝑡 𝑥 = 0 ∶ 𝑦𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −9 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = (0 ; −9)

4

2

-4

-2

2

4

6

8

10

12

14

16

18

-2

qx  = x 2-9

-4

-6

-8

-10

y  x 2  9

2.

𝐴𝑡 𝑦 = 0 ∶ 𝑥 2 − 9 = 0 𝑥−3 𝑥+3 =0 𝑥 = 3 𝑜𝑟 𝑥 = −3

𝐴𝑡 𝑥 = 0 ∶ 𝑦𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 9 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = (0 ; 9)

8

6

4

2

-6

-4

-2

2

4

qx  = -x 2+9 -2

-4

-6

6

8

10

12

14

16

18

20

42 y  2x 2  8 1.3 𝐴𝑡 𝑦 = 0 ∶ 2(𝑥 2 − 4) = 0 𝑥−2 𝑥+2 =0 𝑥 = 2 𝑜𝑟 𝑥 = −2

𝐴𝑡 𝑥 = 0 ∶ 𝑦𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −8 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = (0 ; −8)

4

2

-5

5

10

15

20

qx  = 2x 2-8

-2

-4

-6

-8

-10

1.4

y  3x 2  12

𝐴𝑡 𝑦 = 0 ∶ −3(𝑥 2 − 4) = 0 𝑥−2 𝑥+2 =0 𝑥 = 2 𝑜𝑟 𝑥 = −2

𝐴𝑡 𝑥 = 0 ∶ 𝑦𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 12 𝑇𝑢𝑟𝑛𝑖𝑛𝑔 𝑝𝑜𝑖𝑛𝑡 = (0 ; −12

12

10

8

rx  = -3x 2+12 6

4

2

-5

5

-2

10

15

20

43 Page 115: Exercise 11.1 1.

y

X Y

-3 2

6 x

-2 3

-1 6

1 -6

2 -3

3 -2

8

6

4

rx  =

-6 x 2

-10

-5

5

10

15

10

15

-2

-4

-6

2. y

X Y

-8 -1

8 x

-4 -2

-2 -4

-1 -8

1 8

2 4

4 2

8 1

8

6

4

rx  =

8 x

2

-10

-5

5

-2

-4

-6

44

3.

y

X Y

-8 1

8 x

-4 2

-2 4

-1 8

1 -8

2 -4

4 -2

8 -1

8

6

4

rx  =

-10

-8 x

2

-5

5

10

15

10

15

-2

-4

-6

4.

y

X Y

-5 -2

10 x

-2 -5

-1 -10

1 10

2 5

5 2

8

6

4

2

rx  =

-10

-5

10 x

5

-2

-4

-6

45

5.

y

X Y

-5 2

 10 x

-2 5

-1 10

1 -10

2 -5

5 -2

8

6

4

rx  =

-10

-10

2

x

-5

5

-2

-4

-6

Page 11.2: Exercise 11.2 1. 𝑕𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 ∶ 𝑦 = −3 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑥 = 0 𝐷𝑜𝑚𝑎𝑖𝑛: 𝑥𝜖 −∞ ; ∞ ; 𝑥 ≠ 0 𝑅𝑎𝑛𝑔𝑒: 𝑦𝜖 −∞ ; ∞ ; 𝑦 ≠ −3 2. 𝑕𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 ∶ 𝑦 = 6 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑥 = 0 𝐷𝑜𝑚𝑎𝑖𝑛: 𝑥𝜖 −∞ ; ∞ ; 𝑥 ≠ 0 𝑅𝑎𝑛𝑔𝑒: 𝑦𝜖 −∞ ; ∞ ; 𝑦 ≠ 6 3. 𝑕𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒 ∶ 𝑦 = 6 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑥 = 0 𝐷𝑜𝑚𝑎𝑖𝑛: 𝑥𝜖 −∞ ; ∞ ; 𝑥 ≠ 0

10

15

46 𝑅𝑎𝑛𝑔𝑒: 𝑦𝜖 −∞ ; ∞ ; 𝑦 ≠ 6

Page 122: Exercise 12.1 1. X Y

y  2x -2 1 4

-1 1 2

0 1

1 2

2 4

8

6

4

2

-10

rx  = 2 x

-5

5

10

15

-2

-4

-6

2. X Y

y  2x  1 -2 -1 5 3 4 2

0 2

1 3

2 5

8

6

4

2

-10

-5

rx  = 2x+1

5

-2

-4

-6

10

15

47

y  2x  1 -2 -1 3  4

3. X Y

0 -1

1

2 3

8

6

4

rx  = 2x-1

2

-10

-5

5

10

15

-2

-4

-6

y  2.2 x -2 -1 1 1 2

4. X Y

0 2

1 4

2 8

8

6

4

f x  = 22x

2

-10

-5

5

-2

5.

y  2.2 x

10

48 X Y

-2 

-1 -1

1 2

0 -2

1 -4

2 8

2

-10

-5

5

f x  = -22x -2

-4

-6

-8

Page 128: Exercise 12.2 1.1 𝑓 𝑥 = 4𝑥 − 3 𝑓 4 = 4 4 −3 𝑓 4 = 13

1.2

𝑓 𝑥 = 4𝑥 − 3 𝑓 −7 = 4 −7 − 3 𝑓 −7 = −31

1.3

𝑓 𝑥 = 4𝑥 − 3 𝑓 𝑎−𝑏 =4 𝑎−𝑏 −3 𝑓 −7 = 4𝑎 − 4𝑏 − 3

2,

𝑔 𝑥 = 3𝑥 2 − 𝑥 − 6

2.1

𝑔 2 = 3(2)2 − (2) − 6 𝑔 2 =4

10

49

2.2

𝑔 𝑥 = 3𝑥 2 − 𝑥 − 6 𝑔 −3 = 3(−3)2 − (−3) − 6 𝑔 2 = 27 + 3 − 6 𝑔 2 = 24

2.3

𝑔 𝑥 = 3𝑥 2 − 𝑥 − 6 𝑔 𝑎 + 𝑏 = 3(𝑎 + 𝑏)2 − (𝑎 + 𝑏) − 6 𝑔 2 = 3𝑎2 + 6𝑎𝑏 + 3𝑏 2 − 𝑎 − 𝑏 − 6 𝑔 2 = 24

3. 𝑙𝑒𝑡 𝑥 = 0 𝑦 = 4 𝑙𝑒𝑡 𝑦 = 0 𝑥 = 2 6

4

f x  = -2x+4

2

-10

-5

5

-2

4. 𝑙𝑒𝑡 𝑥 = 0 ∴ 𝑦 = −3

𝑙𝑒𝑡 𝑥 = 0 ∴ 𝑦 = 3

𝑙𝑒𝑡 𝑦 = 0 ∴ 𝑥 = 3

𝑙𝑒𝑡 𝑦 = 0 ∴ 𝑥 = 2

10

50

6

5

gx  = -1.5x+3 4

3

2

f x  = x-3

1

-10

-8

-6

-4

-2

2

4

6

8

10

12

4

6

8

10

12

-1

-2

-3

-4

-5

(2,5 ; 0.5)

4.1

4.2

6

5

h x  =

-8 x

4

q x  =

8 x

3

2

1

-10

-8

-6

-4

-2

2

-1

-2

-3

-4

-5

4.2.1 𝑥 ∈ 𝑅; 𝑥 ≠ 0 𝑦 ∈ 𝑅; 𝑦 ≠ 0

51 𝑥 = 0 𝑎𝑛𝑑 𝑦 = 0

4.2.2

4.3 6

5

rx  = x 2-4

4

3

2

1

-10

-8

-6

-4

-2

2

4

6

8

10

12

4

6

8

10

12

-1

-2

-3

-4

s x  = -x 2+4 -5

𝑥 ∈ (−∞; ∞)

and

𝑦 ∈ [−4; ∞]

𝑥 ∈ (−∞; ∞) 𝑦 ∈ (−∞; 4]

4.4 6

t x  = 2 x 5

4

3

ux  = 2x+2 2

1

-10

-8

-6

-4

-2

2

-1

-2

-3

-4

-5

4.4.1 𝑦 = 0

𝑎𝑛𝑑

𝑦=2

52 5.1 6

𝑚=2=3 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −2 𝑎𝑛𝑑𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 6

5.2 8

𝑚 = − 4 = −2 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 4 𝑎𝑛𝑑𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 8

5.3 9

𝑚 = − 3 = −3 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −3 𝑎𝑛𝑑𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −9

5.4 8

𝑚=2=4 𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −2 𝑎𝑛𝑑𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = 8

6.1

6.2

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ (−∞; ∞)

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ (−∞; 2]

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ (−∞; ∞)

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ [−9; ∞)

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 1 − 1

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 1 − 1

𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

6.3

6.4

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ [−12; 12]

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ [−4; ∞)

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ [−12; 12]

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ (−∞; ∞)

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 𝑚 − 𝑚

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 1 − 𝑚

𝑛𝑜𝑛 − 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝑛𝑜𝑛 − 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

6.5

6.6

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ (−∞; ∞)

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ (−∞; ∞)

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ (−∞; 6]

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ (−∞; ∞)

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 𝑚 − 1

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 𝑚 − 𝑚

𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝑛𝑜𝑛 − 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

53

6.7

6.8

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ [−4; 3)

𝑑𝑜𝑚𝑎𝑖𝑛: 𝑥 ∈ [−4; 4]

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ [−4; 9)

𝑟𝑎𝑛𝑔𝑒: 𝑦 ∈ [−9; 8]

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 1 − 1

𝑚𝑎𝑝𝑝𝑖𝑛𝑔: 𝑚 − 1

𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛

7.1 𝑦 = 𝑎𝑥 2 + 𝑐 0 = 𝑎(2)2 − 4 4𝑎 = 4 𝑎=1 𝑦 = 𝑥2 − 4

7.2 𝑥𝑦 = 𝑘 2 (4) = 8 𝑥𝑦 = 8

7.3 𝑦 = 𝑎𝑥 9 = 𝑎2 𝑎=3 𝑦 = 3𝑥 7..4 𝑦 = 𝑎𝑥 2 + 𝑐 5 = 𝑎(2)2 + 9 4𝑎 = −4 𝑎 = −1 𝑦 = −𝑥 2 + 9

54 7.5 𝑘

𝑦=𝑥

𝑘

−4 = 2 − 2 𝑘 2

=2

𝑘=4 4

𝑦 =𝑥−2 8.1 6

𝑚=2=3 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = −6 𝑦 = 3𝑥 − 6 8.2 𝑥 ∈ (−∞; ∞) 𝑦 ∈ [−12; ∞)

8.3 𝑥<0

8.4 3𝑥 2 − 12 = 3𝑥 − 6 3𝑥 2 − 3𝑥 − 6 = 0 𝑥2 − 𝑥 − 2 = 0 𝑥−2 𝑥+1 =0 𝑥 = 2 𝑜𝑟 − 1 𝑆(−1 ; −9)

8.5 𝑦 = −3𝑥 2 + 12

55 Further Graph Interpretation: 4

1. 𝑚 = 2 = 2 2

2. 𝑚 = − 3

∴𝑎=2 2

∴ 𝑎 = −3

𝑎𝑛𝑑 𝑐 = 4. 𝑎𝑛𝑑 𝑐 = −2.

3. 𝑦 = 𝑎𝑥 2 + 𝑐 𝑐 = −4 𝑦 = 𝑎𝑥 2 − 4 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑝𝑜𝑖𝑛𝑡 2; 0

∴ 0 = 𝑎(2)2 − 4 4𝑎 = 4 𝑎 = 1 𝑎𝑛𝑑 𝑐 = −4

4. 𝑦 = 𝑎𝑥 2 + 𝑐 𝑐 = 9 𝑦 = 𝑎𝑥 2 + 9 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑝𝑜𝑖𝑛𝑡 3; 0

∴ 0 = 𝑎(3)2 + 9 9𝑎 = −9 𝑎 = −1 𝑎𝑛𝑑 𝑐 = 9

5. 𝑦 = 𝑎 𝑥 9 = 𝑎2 𝑎 = 3 𝑎𝑛𝑑 𝑐 = 0 6. 𝑦 = 𝑎 𝑥 + 𝑐 𝑐 = −2 𝑦 = 𝑎𝑥 − 2 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑝𝑜𝑖𝑛𝑡 (1; 2) 2 = 𝑎1 − 2 𝑎=4 𝑎

7. 𝑦 = 𝑥

𝑥𝑦 = 𝑘 𝑘 = 12

∴

3 4 =𝑘 𝑥𝑦 = 12 𝑜𝑟 𝑦 =

12 𝑥

56 𝑎

8. 𝑦 = 𝑥 + 𝑐 𝑐 = −2 𝑎

𝑦 = 𝑥 −2 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑝𝑜𝑖𝑛𝑡 (3; 0) 𝑎

0= 3−2

∴𝑎=6

𝑎

9. 𝑦 = 𝑥 + 𝑐 𝑐 = −3 𝑎

𝑦 = 𝑥 −3 𝑠𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑝𝑜𝑖𝑛𝑡 (3; 0) 𝑎

0 = −3 − 3

∴ 𝑎 = −9

10. 𝑦 = 𝑎𝑐 𝑥 −2 = 𝑎𝑐 0 ∴ 𝑎 = −2 1 2

= −2𝑐 2 1

𝑐2 = 4 1

𝑐=2 11. 1 −𝑥 2 + 4 = 0 𝑥2 − 4 = 0 𝑥−2 𝑥+2 =0 𝑥 = 2 𝑜𝑟 𝑥 = −2 𝐴 −2; 0 𝑎𝑛𝑑 𝐵(2; 0) 𝑦 = 3𝑥 + 4 𝑎𝑡 𝑥 = 0 ∴ 𝑦 = 4 4

𝑎𝑡 𝑦 = 0 ∴ 𝑥 = − 3 4

𝐶 0; 4 𝑎𝑛𝑑 𝐷 − 3 ; 0

57 11.2 3𝑥 + 4 = −𝑥 2 + 4 𝑥 2 + 3𝑥 = 0 𝑥 𝑥+3 =0 𝑥 = −3 𝑜𝑟 𝑥 = 0 𝑦 = −5 𝑜𝑟 𝑦 = 4 𝐸(−3; −5)

11.3 𝑔 2 = 3 2 + 4 = 10 𝐹𝐵 = 10 𝑢𝑛𝑖𝑡𝑠

11.4 𝑔 𝑥 − 𝑓 𝑥 = 3𝑥 + 4 − (−𝑥 2 + 4) 𝑔 𝑥 − 𝑓 𝑥 = 𝑥 2 + 3𝑥

11.5 𝑔 𝑥 ≥ 𝑓 𝑥 : 𝑥 ≤ −5 𝑜𝑟 𝑥 ≥ 0

12.1 𝑥2 − 9 = 0

2𝑥 + 4 = 0

𝑥−3 𝑥+3 =0

𝑎𝑡 𝑥 = 0 ∴ 𝑦 = 4

𝑥 = 3 𝑜𝑟 𝑥 = −3

𝑎𝑡 𝑦 = 0 ∴ 𝑥 = −2

𝐴 −3; 0 𝑎𝑛𝑑 𝐵(3; 0)

𝐷(0; 4)

𝐶(0; −9) 12.2 𝑂𝐶 = 9 𝑢𝑛𝑖𝑡𝑠 ; 𝐴𝐵 = 6 𝑢𝑛𝑖𝑡𝑠 & 𝑂𝐷 = 4 𝑢𝑛𝑖𝑡𝑠. 12.3 𝑥 2 − 9 = 2𝑥 + 6 𝑥 2 + 2𝑥 − 15 = 0 𝑥+5 𝑥−3 =0 𝑥 = −5 𝑜𝑟 𝑥 = 3 𝑦 = −4 𝑜𝑟 𝑦 = 12

𝐸(3; 12)

58 12.4 đ?‘‚đ?‘† = 1 đ?‘“ 1 = 1 − 9 = −8 đ?‘Žđ?‘›đ?‘‘ đ?‘” 1 = 2 + 6 = 8 đ??šđ?‘‡ = 16 đ?‘˘đ?‘›đ?‘–đ?‘Ąđ?‘

12.5 đ?‘‚đ?‘ƒ = −3 đ?‘“ −4 = 16 − 9 = 7 đ?‘Žđ?‘›đ?‘‘ đ?‘” −4 = −8 + 6 = −2 đ??šđ?‘‡ = 9 đ?‘˘đ?‘›đ?‘–đ?‘Ąđ?‘ .

12.6 đ??´đ??ˇ2 = đ?‘‚đ??ˇ2 + đ?‘‚đ??´2 (đ?‘ƒđ?‘Śđ?‘Ąđ?‘•đ?‘Žđ?‘”đ?‘œđ?‘&#x;đ?‘Žđ?‘ ) đ??´đ??ˇ2 = 62 + 32 = 45 đ??´đ??ˇ = 45

OR 3 5 OR 6.71‌‌.

Page 156: Exercise 14.1 Question 1: Peter deposits R8500,00 for 8 years into a savings account. 1.1 If the money earns a simple interest rate of 12% per annum. How much will he have at the end of the investment period. A  P(1  ni )

A  8500[1  8(0.12)] A  R16660,00 1.2

If he had received an interest rate of 20% , how much more would he have earned in interest? A  P(1  ni )

A  8500[1  8(0.2)] A  R 22100,00 Received R5440 more.

59 Question 2: John invests R20 000,00 into a capital venture that will earn him a simple interest of 10% for the first 5 years and then 15% for the last 5 years. What is his investment worth at the end of the 10 years? A  P(1  ni )

A1  20000[1  5(0.1)] Method 1:

A  P(1  ni ) A1  R30000,00 Method 2: A1  20000(1  5  0.10)(1  5  0.15) A2  30000[(1  5(0.15)] A2  R52500,00 A2  R52500,00

Question 3: Mrs Vezi invests R10 000,00 at 12,5% simple interest per annum. 3.1

How much interest will she earn after 5 years? A  P(1  ni )

A  10000(1  5  0.125) A  R16250 Interest  A  P  R6250

3.2

How much interest will she earn after 10 years? A  P(1  ni )

A  10000(1  10  0.125) A  R 22500 Interest  A  P  R12500,00

3.3

What is the final amount in her account after 15 years if she does not have any withdrawals? A  P(1  ni )

A  10000(1  15  0.125) A  R 28750,00 Question 4: 4.1

Calculate the Principle if James received an amount of R7500,00 earned in simple interest at a rate of 10% over 5 years. A  P (1  ni )

A (1  ni ) 7500 P (1  5  0.1) P  R5000 p

60 4.2

Calculate the Principle if James received an amount of R37500,00 earned in simple interest at a rate of 15% over 10 years. A  P (1  ni )

A (1  ni ) 37500 P (1  10  0.15) P  R15000 p

Question 5: 200 190 180 I 170 N T 160 E R E 150 S T 140

5.1

5.3

130 120 110 100

2

4

6

8

10

12

14

16

YEARS

5.2 The interest is a fixed amount and is displayed by a straight line. 5.3

On the above graph.

5.4

The graph would have a steeper gradient.

18

20

22

24

61 COMPOUND INTEREST Page 163: Exercise 14.2: 1. Charlie wants to invest R7000,00 : There are 2 choices for him: 1.1 Invest R 7000,00 at 12% simple interest pennum over 5 years.

A  P(1  ni ) A  7000(1  5  0.12) A  R11200 1.2 Invest R7000,00 at 12% compound interest over 5 years.

A  P(1  i ) n A  7000(1  0.12) 5 A  R12336,39 Choose 1.2 2. R12000,00 is invested in a savings account in FNB. The interest is compounded annually at 15%. How much money will be in the savings account after 10 years?

A  P(1  i ) n A  12000(1  0.15)10 A  R 48546,69 3. R20 000,00 is invested for 15 years at 12% p.a. Calculate the value of the investment if the interest is calculated:

A  P(1  ni ) At simple interest. A  20000(1  15  0.12) A  R56000

A  P(1  i ) n At compound interest. A  20000(1  0.12)15

A  R109471,32

62 4. R5000 was invested in order to fund a small business. After 5 years R7000 was paid out from the profits. The amount was profit only 4.1 Calculate a simple interest rate that would yield the same return. A  P(1  ni ) A P Pn i  28% i

4.2 Calculate a compound interest rate that would provide the same return.

A  P (1  i ) n  A  i  100 n  1  P   12000  i  100 5  1  5000  i  19.14% 5. An investment doubles over a period of 8 years. Determine a rate of interest correct to 2 decimal places which would make this possible if the interest was calculated as follows: 5.1 simple interest i 

2 1  12,5% 1 8

 2000  i  100 8  1 5.2 compound interest.  1000  i  9.05%

6. The Ndlovu family uses a loan of R7200,00 to buy furniture They repay the loan at the end of 3 years. How much would they have to repay if the interest is calculated as:

6.1

16% p.a. simple interest?

6.2

13% p.a. compound interest?

A  7200(1  3  0.16) A  R10656 A  7200(1  0.13) 3 A  R10388,86

63 7. Use the simple and compound interest formulae for this question: R1000 for 5years at 15% p.a Simple interest A  P(1  ni)

Compound interest A  P(1  i) n

End of 1st year

A =R1150

End of 1st year

End of 2nd year

A =R1300

End of 2nd year

A =R1323

End of 3rd year

A =R1450

End of 3rd year

A =R1521

End of 4th year

A =R1600

End of 4th year

A =R1749

End of 5th year

A =R1750

End of 5th year

A =R2011

A =R1150

8. Use a table to draw two graphs on the same system of axes showing the difference between simple and compound interest . Use a the system of axes below for your graphs.( Add in two more rows to 2100) 2100 Compound Interest 2000 1900 Simple Interest I N T E R E S T

1800 1700 1600 1500 1400 1300 1200 1100 1000

2

4

6

YEARS

8

10

12

14

16

18

20

22

24

64 9.1

What kind of graph does the simple interest formula produce? Straight Line

9.2

Find the gradient of the line. 150

9.3

What is the difference in the value of A for simple and compound interest after 5 years? R261

9.4

Use the formula to calculate the difference in A, ( the accumulated value of the investment), after 10 years. R1545,56

Exchange Rates in the Money Market: Page 169:Exercise 14.3: 1. Copy and complete the following currency table: CURRENCY

$

R

€

£

¥

A$

NZ$

USD

1

7,5

0.5769

0.5

150

1.0718

1.499

ZAR

0.133

1

0.077

0.0666

20

0.143

0.2

EUR

1.7333

13

1

0,866

260

1.858

2.599

GBP

2

15

1.154

1

300

2.14

2.999

JPY

O,007

O,05

0.013

0.0035

1

0.008

0.0105

AUD

0.933

7

0.333

0.4665

140

1

1.399

NZD

0.667

5

0.01

0.3335

100

0.715

1

65 Use the table above for questions 2 to 4. 1

You have R1000 to spend in each of the above countries. How much of the local money can be purchased in each of the countries. USD = 133.33 EUR = 76.92 GBP = 66.67 JPY = 20000 AUD = 142.86 NZD = 200

2

What will it cost you in Rands to purchase the following currencies:

2.1

$800  R 6000

2.2

¥2000  R100

2.3

A$500  3500

2.4

€2000  26000

2.5

£5000  75000

3

How many rands can be exchanged for the following:

3.1

$200 = R 1500

3.2

¥1500 = R75

3.3

A$150 = R1050

3.4

€350 = R4550

3.5

£750 = R11250

66 Cost of Hire Purchase: Page 172: Exercise 14.4: 1.

A  120000[1  (0.12)(5)]  192000 / 60  R3200 p.m 2. Dep = R975 Balance owing = R5525

A  5525[1  (0.10)(3)]  R7182.5 / 36  R199.51 p.m Insurance = R6.50 Total installment = R206.01 3.

Price + interest = 299.99 x 36 = R10799.64 Interest = R10799.64 - R7500 Interest = R3299.64

The interest rate per annum if the advertisement refers to a hire purchase agreement.

Si (100) 7500(3) r  14,67% r

4.

Ms Bokapane buys a washing machine from a appliance store . She asks for a quote to buy the machine through a hire purchase contract over 1

The following quote was received by her:

QUOTATION: Purchase price

R3800,00

10% cash deposit

R 380,00

Balance owing

R3420,00

Finance charges(interest)

R 820,00

Insurance over 18 months

R 684,00

Total amount to be paid

R4924,80

1 years. 2

67 Monthly payment

R 273,60

Why do you think Ms Bokopane has to pay insurance costs?

Calculate the interest rate per year. (Note: the interest charged is calculated on the balance owing i.e. R3420,00 .The insurance is not included in this calculation)

Page 175 : Exercise 14.5 Cost of Inflation 1.

The average rate of inflation over the last 10 years was 7,3% p.a. The current price of a 2,5kg packet of white sugar is R10,25.

Calculate the expected price of sugar in 10 years time if the rate of inflation continues at the same level.

A  P(1  i ) n A  10.25(1  0.073)10 A  R 20.74 How much did the 2,5kg packet cost 10yrs ago (the rate of inflation continues at the same level.)

A  P (1  i ) n A P (1  i ) n 10.25 P (1  0.073)10 P  R5.07 2.

A kettle costs R180,00. Determine the expected cost of a similar kettle in 5 years time , based on an inflation rate of 18% p.a.

A  P(1  i ) n A  180(1  0.18) 5 A  R 411.80

68 3.

A block of 500grams margarine costs R16,19 . Determine the cost in 10 years time if the price is expected to rise by 9% p.a. as a result of inflation.

A  P(1  i ) n A  16.19(1  0.09)10 A  R38.33

4.

The current annual fees for a Bachelor of Commerse degree at UKZN are R20500,00. Determine the expected cost of studying the same degree in 5 years time if the fees are to increase by 9% p.a. as a result of inflation. Give the answer correct to the nearest rand.

A  P(1  i ) n A  20500(1  0.09) 5 A  R31541.79 Page 180: Exercise 14.6 TIME LINES USED FINANCIAL CALCULATIONS 1.

đ??´ = 1000(1,115)4 + 2000(1,115)3 + 4000(1,115)2 + 8000(1,115)1 đ??´ = đ?‘…18210.90

69

2.

𝐴 = 6500(1,11)10 + 7400(1,11)7 − 5800(1,11)5 𝐴 = 𝑅24046.48

3.

𝐴 = 21000 1.075 𝐴 = 𝑅35822.53

3

1.0825

4

70 4 4. Mrs Smith invests R8000 for in a savings account when her two sons are 7 and 10 years old. She pays each of them R15000 in the year they turn 21. 4.1

Calculate how much money is in the savings account after she has paid her younger son. The interest rate is 14% p.a. compounded annually.

đ??´= đ??´ = 8000 1.14

14

− 15000 1.14

3

− 15000

đ??´ = đ?‘…12867.63 4.2

Yes it is fair. She fulfilled her promise to each son.

5.

đ??´ = 52000(1.105)5 (1.12)3 (1.14)4 đ??´ = đ?‘…203276.98

71 PROBABILITY THEORY Page 187: Exercise 15.1: Soup

1 20 2 5 3 10 1 4

Coffee Tea Hot Chocolate

5% 40% 30% 25%

Answers: a) i) Coffee most likely to be chosen ii) Soup is least likely to be chosen b) Ascending order: Soup; hot chocolate; tea; coffee. Page 189: Exercise 15.2: 2.1 In my left-hand pocket I have a 50c coin, a 20c coin and a 10c coin. In my right-hand pocket I have R1 coin, a R2 coin and a R5 coin. 2.1.1

I choose one coin at random from either pocket. List all the possible outcomes.

2.1.2

I choose all outcomes that satisfy the condition : the total must be greater than R2,50.

a) R5; R2; R1; 50c; 20c; 10c b) i)

R5,50; R5,20; R5,10 R2,50; R2,20; R2,10 R1,50; R1,20; R1,10

ii)

R5,50; R5,20; R5,1

72

2.2 a)

Dice 1

Dice 2 1

2

3

4

5

6

1

1;1

1;2

1;3

1;4

1;5

1;6

2

2;1

2;2

2;3

2;4

2;5

2;

3

3;1

3;2

3;3

3;4

3;5

3;6

4

4;1

4;2

4;3

4;4

4;5

4;6

5

5;1

5;2

5;3

5;4

5;5

5;6

6

6;1

6;2

6;3

6;4

6;5

6;6

b)

2;3, 3;2, 3;4, 4,3; 1;6, 6,1, 2;6, 6;2, 3;6, 6;3, 4;6, 6;4, 5;6, 6;5, 6;6

Page 192: Exercise 15.3: 3.1

Consider the tossing of a coin: 3.1.1 Describe a trial Answer: a) Tossing the coin 3.1.2 Give an example of an outcome. Answer: H 3.1.3 Write down the sample space, S of the experiment. Answer: S = {4;7} 3.1.4 Represent the sample space by means of a venn diagram.

S H T Answer: 3.2

The diagram below shows a four sided spinner with sides labeled 1, 2, 3and 4 respectively. 3.2.1 Describe a trial using such a spinner. Answer: Spinning the spinner 3.2.2 Give an example of an outcome. Answer: 3 3.2.3 Write down the sample space, S for the experiment. Answer: S = {1;2;3;4}

73 3.2.4

Give a venn diagram for S.

S 12 43 Answer:

Events: Page 196: Exercise 15.4: 4.1 The trial to be considered in this instance is the spinning of a coin. 4.1.1 Represent the sample space, S, in set notation. Answer: S = {h; t} 4.1.2 Write down a set for the event, H, of obtaining a head. Answer: H = {h} 4.1.3 Write down a set for the event, t , of obtaining a tail. Answer: T = {t} 4.1.4 What is the value of each of the following: 4.1.4.1 n(S ) 4.1.4.2 n(H ) 4.1.4.3 n(T ) Answer: 4.1.4.1 n(S )  2 4.1.4.2 n( H )  1 4.1.4.3 n(T )  1 4.1.5 Represent S, H and T in a single venn diagram.

S Hh Tt Answer:

4.2 4.2.1 Write down a sample space , S, for this experiment. Answer: S  {r1 ; r2 ; r3 ; b1 ; b2 ; b3 ; b4 ; g1 ; g 2 ; g 3 ; g 4 g 5 g 6 } 4.2.2 Express the event of drawing a green marble, G, in terms of a set. Answer: G  {g1 ; g 2 ; g 3 ; g 4 g 5 g 6 } 4.2.3 Write down a set, B, or R, to represent the event of drawing either a red or a blue marble. Answer: BorR  {b1 ; b2 ; b3 ; b4 ; r1 ; r2 ; r3 } 4.2.4 Write down: Answer: 4.1.4.1 4.1.4.2 n(G)  6 4.1.4.3 n( BorR)  7 n(S )  13

74 Page 201: Exercise 15. 5 5.1 5.1.1 wins the prize? 10 1  250 25 Does not win the prize?

Answer: P( wins )  5.1.2

Answer:

P(does not win) = 1 

1 24  25 25

5.2 5.2.1 a Volkswagen? Answer: P(Volkswagen) 

165 33  220 50

5.2.2 not a Volkswagen? Answer: P(not a Volkswagen) = 1 

33 17  50 50

5.3 2 7

5.3.1

P(Black) =

5.3.4

P(not white) =

5.3.6

P(not black nor white) =

11 14

5 7

5.3.2

P(not black) =

5.3.5

P(black or white) =

5.3.2. P(white) =

3 14

1 2

1 2

5.4 5.4.1 Derek picks a sweet from the packet without looking.What is the probability that he picks either a melon-flavoured or lemon-flavoured sweet? 7 Answer: 20 5.4.2 Albert doesn’t like banana-flavoured or melon-flavoured sweets. He likes all other flavours. What is the probability that he picks a sweet that he likes? 3 Answer: 5

75 WORKBOOK TWO: Page 5 : Exercise 1.1: Distance between points: 1. Find the distance between the given pairs of points: (2 ; 3) and (4 ; 5)

Dis tan ce  ( x1  x2 ) 2  ( y1  y 2 ) 2 Dis tan ce  (2  4) 2  (3  5) 2 Dis tan ce  8 Dis tan ce  2,82 (6 ; 1) and ( -6 ;6)\ Dis tan ce  (6  6) 2  (1  6) 2 Dis tan ce  144  25 Dis tan ce  13

(3 ; -7) and (-1 ; 3)

Dis tan ce  (3  1) 2  (7  3) 2 Dis tan ce  16  100 Dis tan ce  10,77 (-4 ; 3) and (0 ; 0) Dis tan ce  (4  0) 2  (3  0) 2 Dis tan ce  16  9 Dis tan ce  5

(-2 ; 1) and -4 ; -1)

Dis tan ce  (2  4) 2  (1  1) 2 Dis tan ce  8 Dis tan ce  2,82 (-3 ;-1) and (4 ; -6)

Dis tan ce  (3  4) 2  (1  6) 2 Dis tan ce  49  25 Dis tan ce  8,6

76

Given the coordinates of the vertices of ABC , in each case ( 2.2 to 2.5)

2.

A(1 ; -3) ; B(7 ; 3); C(4 ; 6) AB  36  36  72

BC  9  9  18

Perimeter = 22,2

AC  9  81  90 AB 2  BC 2  AC 2 Right angled Scalene triangle A(5 ;1) ; B(1 ; 3) ; C(1 ; -2)

AB  4 2  (2) 2  20 BC  0  25  5

Perimeter = 14,5

AC  16  9  5 Isosceles Triangle Not Right angled A(-2 ; -3) ; B(-4 ; 1) ; C(4 ; 5) AB  4  16  20 \

BC  64  16  80

Perimeter = 23,4

AC  36  64  100 Scalene Triangle AB 2  BC 2  AC 2 Right angled A(0 ; 0) B( 3 ; 1) ; C( 3 ; -1)

AB  3  1  2 BC  0  4  2

Perimeter = 6

AC  3  1  2 Equilateral Triangle Not Right angled A(2 ; -1) B(-3 ;4) ;C(4; 5) AB  25  25  50

BC  49  1  50 AC  4  36  40 Isosceles Triangle Not Right angled

Perimeter = 20,5

77 3.

Show that:

A(-3 ; 2) , B(3 ;6), C(9 ;-2) and D(3 ; -6) are vertices of a parallelogram. AB  36  16  52 AB = DC DC  36  16  52 AD  36  64  10 AD = BC BC  36  64  10 ABCD is a parallelogram ( Both prs opp sides equal) (6 ;-4) , (5 ;3) (-2 ; 2) and (-1 ; -5) are vertices of a square. AB  1  49  50

DC  1  49  50 AD  49  1  50 BC  49  1  50 AB = DC = AD = AB ABCD is a square ( All sides are equal) Mid-points of lines: 4. Calculate the coordinates of the midpoints of the line joining the following points: (-3 ;1) and (1 ; 5)  31 x  1 2 Mid point ( -1 ; 3) 1 5 y 3 2 (-2 ; 3) and (6 ; 3) 26 x 2 2 Mid – point (2 ; 3) 3 3 y 3 2 (4 ; -1) and (-1 ; 3)

4 1 3  2 2 1 3 y 1 2

x

Mid point ( 1,5 ; 1)

(0 ;0 ) and (3 ; -8)

03 3  2 2 08 y  4 2

x

Mid Point ( 1,5 ; -4)

78 ( 3;1) and (3 3;1)

33 3 2 3 2 Mid point ( 2 3 ; 0) 11 y 0 2

x

5.1 Determine the values of x and y if: (-3 ; 2) is the mid-point of the line joining (-1 ; 5) and (x ; y). 1 x 5 y  3 2 2 2  1  x  6 5 y  4 x  5 y  1

5.2

(-1 ; y) is the mid-point of the line joining (0 ; -2) and x ; 8) 0 x  1 28 2 y x  2 2 y3

5.3

(x ; y) is the centre of a circle on diameter AB where A(-2 ; -1) and B(-1 ; 9).  2 1 1 9 x y 2 2 2 x  3 2y  8 3 y4 x 2

5.4

(x ; 3) is the centre of a circle with diameter MN. M (5 ; -2) and N(-7 ; y) 2 y 3 2 57 x 2 y  6 2 x  1 y8

mAC tan

79 Page 14: Exercise 1.2

ďƒŚ x  x2 y1  y2 ďƒś Mid-point = ďƒ§ 1 ; ďƒˇ 2 ďƒ¸ ďƒ¨ 2

AB  ( x1  x2 )2  ( y1  y2 )2 Gradient = m = 1.

1.1

1.2

y1  y2 x1  x2

AB is a straight line on a Cartesian plane where A(-3; -4) and B( 2 ; 6) Calculate the following:

AB  ( x1  x2 )2  ( y1  y2 )2 the length of AB in units. đ??´đ??ľ = (−3 − 2)2 + (−4 − 6)2 đ??´đ??ľ = 5 5 the co-ordinates of the mid – point ( C ) of line AB. ďƒŚ x1  x2 y1  y2 ďƒś ; ďƒ§ ďƒˇ 2 ďƒ¸ m= ďƒ¨ 2 C=

1.3

the gradient of line AB. y  y2 m= 1 x1  x2 MAB =

1.4

−1

( 2 ; 1)

đ?&#x;?đ?&#x;Ž đ?&#x;“

=đ?&#x;?

show that points A;B and C are collinear. đ?‘€đ??´đ??ľ = 2 đ?‘€đ??ľđ??ś = 2 Points are collinear

2.1

the length of AD. = 100  25  11.18

2.2

the mid-point of DC = (6;4 12 )

2.3

The gradient of BC

2.4

Prove that ABCD is a parallelogram.

=

ď „y 5 1   ď „x 10 2

AD  125 BC  125 đ??´đ??ˇ = đ??ľđ??ś ď „y 1 ď „y 1 mAD   mBC   ď „x 2 ď „x 2 đ?‘¨đ?‘Ť đ?’Šđ?’” đ?’†đ?’’đ?’–đ?’‚đ?’? đ?’‚đ?’?đ?’… đ?’‘đ?’‚đ?’“đ?’‚đ?’?đ?’?đ?’†đ?’? đ?’•đ?’? đ?‘Šđ?‘Ş âˆ´ đ?‘¨đ?‘Šđ?‘Şđ?‘Ť đ?’Šđ?’” đ?’‚ đ?’‘đ?’‚đ?’“đ?’Ž.

80 2.5

đ?‘€đ?‘–đ?‘‘ − đ?‘?đ?‘œđ?‘–đ?‘›đ?‘Ą đ??´đ??ś = 1; 2

đ?‘€đ?‘–đ?‘‘ −point BD = 1; 2

Co-ordinates = (1 ; 2)

3.1 Calculate the perimeter (distance around) of ď „ABC .

AB  36  36  8,49 BC  16  1  4,12 AC  121  1  11,05 Perimeter  23,66

3.2 Prove that triangle ABC is right angled at B.

6  1 6 5 M BC   1 5 M AB  mBC  1 M AB 

AB ď ž BC Right Angled at B

3.3 Give the coordinates of the mid -point of AC, AB & BC

5 5 Mid  Pt AC  ( ; ) 2 2

Mid  Pt AB  (0;0)

11 1 Mid  Pt BC  ( ; ) 2 2

Question 4. Determine whether the following triangles are Isosceles, Equilateral or Scalene. 4.1

A(1;2) , B(6;3) and C (6;1)

81

AB  25  1  26 AC  25  1  26

ISOSCELES TRIANGLE

BC  4  2

4.2

P(4;1) , Q(3;0) and R(1;3) PQ  49  1  50 QR  4  9  13

SCALENE TRIANGLE

PR  25  16  52

4.3

U (5;2) , V (1;1) and W (13;1) UV  16  1  17 UW  64  9  73

SCALENE TRIANGLE

VW  144  4  148

Triangles & Quadrilaterals Page 25: Exercise 3.1 1. 𝐼𝑛 ∆𝐴𝐸𝐷 𝑎𝑛𝑑 ∆𝐵𝐶𝐷 𝐴𝐷 = 𝐵𝐷 (𝑔𝑖𝑣𝑒𝑛) 𝐴𝐸 = 𝐵𝐶 (𝑔𝑖𝑣𝑒𝑛) 𝐴𝐸 𝐷 = 𝐵𝐶 𝐷 ( 𝑔𝑖𝑣𝑒𝑛) ∆𝐴𝐸𝐷 ≡ ∆𝐵𝐶𝐷 (𝑅𝐻𝑆) 2. 𝐼𝑛 ∆𝐴𝐵𝐷 𝑎𝑛𝑑 ∆𝐴𝐵𝐶 𝐴𝐷 = 𝐵𝐶 (𝑔𝑖𝑣𝑒𝑛) 𝐴𝐵 = 𝐴𝐵 ( 𝑐𝑜𝑚𝑚𝑜𝑛 𝑠𝑖𝑑𝑒) 𝐵𝐴𝐷 = 𝐴𝐵 𝐶 (𝑔𝑖𝑣𝑒𝑛) ∆𝐴𝐵𝐷 ≡ ∆𝐴𝐵𝐶 (𝑆𝐴𝑆) Page 31: Exercise 3.2 1.1

𝑥 = 62° ( opp L’s of parm). 𝑧 = 180 − 62 (𝐶𝑜𝑟𝑟 𝐿′ 𝑠 𝐴𝐵  𝐶𝐷) = 118° 𝑦 = 118° ( 𝑜𝑝𝑝 𝐿′ 𝑠 𝑜𝑓 𝑎 𝑝𝑎𝑟𝑚) 𝑥 = 36° 𝐴𝑙𝑡 𝐿′ 𝑠 𝑃𝑄 𝑆𝑅) 𝑦 = 90° − 36° ( 𝐿 𝑆𝑢𝑚 ∆)

1.2

1.3

𝑦 = 55° (𝐿 𝑆𝑢𝑚 ∆) 𝑥 = 35° ( 𝐷𝑖𝑎𝑔 𝑏𝑖𝑠𝑒𝑐𝑡 𝑜𝑝𝑝 𝐿′ 𝑠)

82

1.4

1.5

𝑥 = 𝑦 ( 𝐿′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠) 𝑥 + 𝑦 = 180° − 105° ( 𝐿′ 𝑆𝑢𝑚 ∆) = 75° 𝑥 = 𝑦 = 37,5° 𝑀 + 𝑧 = 180° ( 𝐶𝑜𝑟𝑟 𝐿′ 𝑠 𝐿𝑀 𝑃𝑂) 𝑧 = 180° − (65° + 37,5°) 𝑧 = 77,5 𝑦 = 60° ( 𝐿′ 𝑠 𝑜𝑝𝑝 = 𝑠𝑖𝑑𝑒𝑠) 𝑥 = 30° (L Sum ∆)

1.6 𝐼𝑛𝑡𝑒𝑟𝑖𝑜𝑟 𝐴𝑛𝑔𝑙𝑒 = 180° 6−2 6

= 120 𝒙 = 𝟏𝟐𝟎°

180°(𝑛 − 2) 𝑛

𝑥 + 𝑦 = 180° ( 𝐿′ 𝑠 𝑜𝑛 𝑎 𝑠𝑡 𝑙𝑖𝑛𝑒) 𝑦 = 180° − 120° = 𝟔𝟎 Page 37: Exercise 3.3 1. 𝑀𝐴𝐷 =

7 7 𝑀𝐵𝐶 = 3 3 𝐴𝐵 𝐵𝐶

𝐴𝐷 = 32 + 72 = 58 𝐵𝐶 = 32 + 72 = 58 𝐴𝐷 = 𝐵𝐶 𝐴𝐵𝐶𝐷 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙𝑜𝑔𝑟𝑎𝑚 ( 1𝑝𝑟 𝑜𝑝𝑝 𝑠𝑖𝑑𝑒𝑠 = 𝑎𝑛𝑑 )

2. 2.1 𝑅𝑇𝑃: 𝑄𝐴 = 𝐴𝑇 𝑆𝑅 = 𝑃𝑄 ( 𝑂𝑝𝑝 𝑠𝑖𝑑𝑒𝑠 𝑜𝑓 𝑎 𝑝𝑎𝑟𝑚) 𝑇𝑆 = 𝑆𝑅 (𝐺𝑖𝑣𝑒𝑛) ∴ 𝑇𝑆 = 𝑃𝑄 𝑇𝑆 𝑃𝑄 (𝐺𝑖𝑣𝑒𝑛) 𝑃𝑇𝑆𝑄 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑚 ( 1 𝑝𝑟 𝑜𝑝𝑝 𝑠𝑖𝑑𝑒𝑠 = & ) 𝑄𝐴 = 𝑇𝐴 ( 𝐷𝑖𝑎𝑔 𝑏𝑖𝑠𝑒𝑐𝑡) 2.2

𝑅𝑇𝑃: 𝑃𝑇 𝑄𝑆 𝑃𝑇𝑆𝑄 𝑖𝑠 𝑎 𝑝𝑎𝑟𝑚 ( 𝑃𝑟𝑜𝑣𝑒𝑑 𝑎𝑏𝑜𝑣𝑒) 𝑃𝑇 𝑄𝑆 (𝑜𝑝𝑝 𝑠𝑖𝑑𝑒𝑠 𝑎𝑟𝑒 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙)

83 3. đ?‘…đ?‘‡đ?‘ƒ: đ??¸đ??š = đ??¸đ??ˇ đ??šđ??ľ = đ??ˇđ??ś ( đ?‘œđ?‘?đ?‘? đ?‘ đ?‘–đ?‘‘đ?‘’đ?‘ đ?‘œđ?‘“ đ?‘Ž đ?‘?đ?‘Žđ?‘&#x;đ?‘š) đ??´đ??š = đ??šđ??ľ ( đ??şđ?‘–đ?‘Łđ?‘’đ?‘›) đ??źđ?‘› ∆đ??´đ??šđ??¸ & ∆đ??¸đ??ˇđ??ś đ??´đ??š = đ??śđ??ˇ (đ?‘ƒđ?‘&#x;đ?‘œđ?‘Łđ?‘’đ?‘‘ đ?‘Žđ?‘?đ?‘œđ?‘Łđ?‘’) đ??¸1 = đ??¸2 (đ?‘‰đ?‘’đ?‘&#x;đ?‘Ąđ?‘–đ?‘?đ?‘Žđ?‘™đ?‘™đ?‘Ś đ?‘œđ?‘?đ?‘? đ??żâ€˛ đ?‘ ) đ??š1 = đ??ˇ đ??´đ?‘™đ?‘Ą đ??żâ€˛ đ?‘ đ??´đ??ľ đ??ˇđ??ś) ∆đ??´đ??šđ??¸ ≥ ∆đ??¸đ??ˇđ??ś ( đ??´đ??´đ?‘†) đ??´đ??š = đ??šđ??ľ

Page 47:Exercise 4.1 1. r  x 2  y 2 (pythagoras theorem) đ?‘&#x; 2 = 82 + 152 đ?‘&#x; = 17 2

15

2. r 2  x 2  y 2 (pythagoras theorem) đ?‘&#x; 2 = 6 + 82 đ?‘&#x; = 10 4

sin đ?œƒ = 17

sin đ?œƒ = 5

cos đ?œƒ = 17

cos đ?œƒ = 5

tan đ?œƒ =

tan đ?œƒ = 3

8

3

15

4

8

3. r 2  x 2  y 2 (pythagoras theorem) đ?‘&#x; 2 = 7 + 242 đ?‘&#x; = 25 24

4. r 2  x 2  y 2 (pythagoras theorem) đ?‘&#x; 2 = 9 + 122 đ?‘&#x; = 15 12

sin đ?œƒ = 25

sin đ?œƒ = 15

cos đ?œƒ = 25

cos đ?œƒ = 25

tan đ?œƒ =

tan đ?œƒ = 5

7

9

24

4

7

1. r  x 2  y 2 (pythagoras theorem) đ?‘&#x; 2 = 442 + 332 đ?‘&#x; = 55 2

3

6. r 2  x 2  y 2 (pythagoras theorem) đ?‘&#x; 2 = 40 + 92 đ?‘&#x; = 41 9

sin đ?œƒ = 5

sin đ?œƒ = 41

cos đ?œƒ = 5

cos đ?œƒ = 41

tan đ?œƒ = 4

tan đ?œƒ = 40

4

40

3

9

11

7.1 đ?‘?đ?‘œđ?‘ đ?œƒ = 61 7.2 đ?‘ đ?‘–đ?‘›2 đ?œƒ + đ?‘?đ?‘œđ?‘ 2 đ?œƒ = 3600

60 2 61

+

121

= 3721 + 3721

11 2 61

21

8.1 đ?‘Ąđ?‘Žđ?‘›đ?œƒ = 20 8.2 1 − đ?‘?đ?‘œđ?‘ 2 đ?œƒ =1−

21 2 29 441

= 1 − 841

84 3721

400

= 3721 =1

= 841

Page 51:Exercise 4.2 1. 1.1

1.3

1.5

Solve for x where x  [0  ;90  ] sin x  0,333 . 1.2 x  19,5

cos x  0,666

tan x  4,900

sin x  0,417

x  78,5

cos x  0,222 x  77,2

1.4

x  48,2 

x  24,6  tan 2 x  8,544

1.6

2 x  83,3 x  41,7 

2 sin( x  30  )  1,666

1.7

3 cos( x  25  )  1,233

sin( x  30  )  0,833

1.8

x  30   56,4  x  86,4 

tan(2 x  25 )  3,0964 

2 x  25   72,1

1.10

2 sin 3 x  0,412 3 sin 3 x  0,618

2 x  97,1

3 x  38,2 

x  48,6 

x  12,7 

1.11

x  25   65,7 x  90,7 

5 tan(2 x  25  )  15,482 1.9

cos( x  25  )  0,411

3 cos x  1,236 cos x  0,412 x  65,7 

4 tan(2 x  50  )  3,3426 3 tan(2 x  50  )  2,50695

1.12

2 x  50   68,3 2 x  118,3 x  59,1

85 Solving Triangles: Page 53: Exercise 4.3 1. 𝑰𝒏 ∆′ 𝒔𝑨𝑩𝑪 ; 𝑨𝑩𝑫 & 𝐵𝐷𝐶 𝐴𝐵 𝐴𝐷 𝑐𝑜𝑠𝐴 = 𝐴𝐶 𝑂𝑅 cos 𝐴 = 𝐴𝐵 𝐵𝐶

𝐵𝐷

𝑡𝑎𝑛𝐴 = 𝐴𝐵 𝑂𝑅 𝑡𝑎𝑛𝐴 = 𝐴𝐷 𝑠𝑖𝑛𝐶 = 𝑐𝑜𝑠𝐶 =

𝐴𝐵 𝐴𝐶

𝐵𝐷

𝑂𝑅 𝑠𝑖𝑛𝑐 =

𝐵𝐶 𝐴𝐶

𝐶𝐷

𝐵𝐶

𝑠𝑖𝑛𝐴 = 𝐴𝐵

2.3

𝑡𝑎𝑛𝐴 = 𝐴𝐶

2.5

𝑐𝑜𝑠𝐵 = 𝐴𝐵

2.7

𝑠𝑖𝑛𝑃 = 𝑃𝑆 𝑜𝑟

2.9

𝑡𝑎𝑛𝑃 = 𝑃𝑅 𝑜𝑟

2.11

𝑠𝑖𝑛𝑆 =

𝐵𝐶

𝑆𝑅

𝑇𝑄 𝑃𝑇

𝑆𝑅

𝑇𝑄 𝑃𝑄

𝑃𝑅

𝑀𝑁

𝑠𝑖𝑛𝐿 =

3.3

𝑡𝑎𝑛𝐿 =

3.5

𝑐𝑜𝑠𝑁 =

𝐿𝑁 𝑀𝑁 𝐿𝑀 𝑀𝑁 𝐿𝑁

𝑂𝑅 𝑂𝑅 𝑂𝑅

𝑂𝑀 𝐿𝑀 𝑀𝑂 𝐿𝑂 𝑂𝑁 𝑀𝑁

𝐵𝐷

𝐵𝐶 𝐶𝐷 𝐵𝐶

𝑂𝑅 cos 𝐵 =

𝐴𝐵 𝐴𝐷

𝐵𝐷 𝐵𝐶

𝐶𝐷

𝐴𝐶

𝑐𝑜𝑠𝐴 = 𝐴𝐵 𝐴𝐶

2.4

𝑠𝑖𝑛𝐵 = 𝐴𝐵

2.6

𝑡𝑎𝑛𝐵 = 𝐵𝐶

2.8

𝑐𝑜𝑠𝑃 =

𝐴𝐶

2.10 𝑠𝑖𝑛𝑇 = 2.12

𝑃𝑆

𝐶𝐷

𝑡𝑎𝑛𝐵 = 𝐵𝐷 𝑂𝑅 𝑡𝑎𝑛𝐵 = 𝐵𝐷

2.2

𝐵𝐶

3.1

1𝐷

𝑠𝑖𝑛𝐵 = 𝐴𝐵 𝑂𝑅 𝑠𝑖𝑛𝐵 = 𝑐𝑜𝑠𝐵 =

𝐵𝐶

𝑂𝑅 cos 𝐶 = 𝐵𝐶

2.1

1𝐷

𝑠𝑖𝑛𝐵 = 𝐴𝐵 𝑂𝑅 𝑠𝑖𝑛𝐵 =

3.2

𝑃𝑅

𝑜𝑟

𝑃𝑆

𝑃𝑄 𝑃𝑇

𝑃𝑄 𝑃𝑇 𝑆𝑅

𝑐𝑜𝑠𝑆 = 𝑃𝑆 𝑐𝑜𝑠𝐿 =

𝑀𝐿 𝐿𝑁 𝐿𝑀

𝑂𝑅

3.4

𝑠𝑖𝑛𝑁 =

3.6

𝑡𝑎𝑛𝑁 = 𝑀𝑁 𝑂𝑅

𝐿𝑁 𝐿𝑀

𝑂𝑅

𝐿𝑂 𝐿𝑀 𝑀𝑂 𝑀𝑁 𝑀𝑂 𝑂𝑁

86 Page 56: Exercise 4.4 1. Calculate the lengths of BC and AB đ??ľđ??ś sin 35° = 30

2. Calculate the lengths of AC & AB 15 tan 56° = đ??´đ??ľ

đ??ľđ??ś = 30đ?‘ đ?‘–đ?‘›35° đ??ľđ??ś = 17.2 đ?‘?đ?‘š

đ??´đ??ľ = đ?‘Ąđ?‘Žđ?‘› 56° đ??´đ??ľ = 10.12 đ?‘?đ?‘š

cos 35° =

đ??´đ??ľ

15

15

sin 56° = đ??´đ??ś

30

15

đ??´đ??ľ = 30đ?‘?đ?‘œđ?‘ 35° đ??´đ??ľ = 24.6 đ?‘?đ?‘š

đ??´đ??ľ = đ?‘ đ?‘–đ?‘› 56° đ??´đ??ľ = 18.1 đ?‘?đ?‘š

3.

4.

Calculate the lengths of BC and AB 14 1. tan đ?‘„1 = 8

Calculate the lengths of AC & AB 24 1. cos 26° = đ??´đ??ś 24

đ?‘„1 = 60.3°

đ??´đ??ś = đ?‘?đ?‘œđ?‘ 26°

đ?‘„2 = 119.7°

đ??´đ??ś = 26.7 đ?‘?đ?‘š

2. đ?‘ƒđ?‘† 2 = 142 + 202 đ?‘ƒđ?‘† = 24.4 đ?‘?đ?‘š

đ??ˇđ??ś

2. sin 32° = 26.7 đ??ˇđ??ś = 26.7đ?‘ đ?‘–đ?‘›32° đ??ˇđ??ś = 14.2 đ?‘?đ?‘š

Page58:Exercise 4.5

1.

2.

3.

���� �

đ?‘ đ?‘–đ?‘›35° = 100 đ?‘¤đ?‘–đ?‘‘đ?‘Ąđ?‘• = 100đ?‘ đ?‘–đ?‘›35° đ?‘¤đ?‘–đ?‘‘đ?‘Ąđ?‘• = 57.4

Calculate the height of the taller building if the distance between the buildings is 10m. The angle between a line from the top of the tall building (excluding the roof) to the bottom of the short one is 35Âş. Task: Calculate the height of the taller building. 10 đ?‘Ąđ?‘Žđ?‘›35° = đ?‘•đ?‘Ą 10

đ?‘•đ?‘’đ?‘–đ?‘”đ?‘•đ?‘Ą = đ?‘Ąđ?‘Žđ?‘› 35° đ?‘•đ?‘’đ?‘–đ?‘”đ?‘•đ?‘Ą = 14.3 4. đ?‘•đ?‘Ą

đ?‘Ąđ?‘Žđ?‘›50° = 1200 đ?‘•đ?‘’đ?‘–đ?‘”đ?‘•đ?‘Ą = 1200đ?‘Ąđ?‘Žđ?‘›50° đ?‘•đ?‘’đ?‘–đ?‘”đ?‘•đ?‘Ą = 1430.1 đ?‘š

87

4.6 TRIGONOMETRIC GRAPHS: Page 62: Exercise 4.4: X y = sinx

0º 0

90º 1

180º -1

270º -1

360º 0

2 y

1 f x  = sin x 

0

90

180

270

360

x

-1

X y  cos x

0º 1

90º 0

180º -1

270º 0

360º 1

2 y

1 f x  = cos x 

0

-1

90

180

270

360

x

88 X 0º y = sinx 0 +1 +1 y  sin x  1 1

90º 1 +1 2

180º 0 +1 1

270º -1 +1 0

360º 0 +1 1

2 y

1 f x  = sin x +1

0

90

180

270

360

x

-1

X y = cos x +1 y  cos x  1

0º 1 +1 2

90º 0 +1 1

180º -1 +1 0

270º 360º 0 1 +1 +1 1 2

2 y f x  = cos x +1

1

0

-1

90

180

270

360

x

89

X y = sinx -1 y  sin x  1

0º 0 -1 -1

90º 1 -1 0

180º 0 -1 -1

270º -1 -1 -2

360º 0 -1 -1

2 y

1

0

90

180

270

-1

x y = cosx -1 y  cos x  1

0º 1 -1 0

90º 0 -1 -1

180º -1 -1 -2

270º 0 -1 -1

360

x

f x  = sin x -1

360º 1 -1 0

2 y

1

0

90

180

270

360 f x  = cos x -1

-1

x

90

X y = -cosx

0º -1

90º 0

180º 1

270º 0

360º -1

2 y

1

0

90

180

270

360

x

f x  = -cos x 

-1

X y = -sinx

0º 0

90º -1

180º 0

270º 1

360º 0

2 y

1 f x  = -sin x 

0

90

-1

x y = -cosx +1

0º -1 1

90º 0 1

180º 1 1

270º 0 1

360º -1 1

180

270

360

x

91 y   cos x  1

0

1

2

1

0

2 y

1 f x  = -cos x +1

0

90

180

270

360

x

-1

x y = -sinx +1 y   sin x  1

0º 0 1 1

90º -1 1 0

180º 0 1 1

270º 1 1 2

360º 0 1 1

2 y

1 f x  = -sin x +1

0

-1

90

180

270

360

x

92

X y = tanx

0º 0

90º ∞

180º 0

270º ∞

360º 0

2 y

1 f x  = tan x 

0

-1

90

180

270

360

x

93 Page 69 :Exercise 5.1: An educator was trying to ascertain why certain learners were not doing set homework . She asked the learners to calculate the number of hours spent watching TV the previous night. The time was rounded up to the nearest hour. The following data was collected by an educator. 1 2 2 1 0 1 3 1 0 0 2 1 0 1 1 1 3 1 1 1 0 1 2 3 3 1 0 0 0 1 1 1 1 1 0 0 0 1 0 1 3 3 1 2 0 1 3 0 1 3 1.

How many learners in the class?

Answer = 50

2.

Draw up table to tally the numbers of learners who watched TV for 0 hrs, 1 hr, 2hrs and for 3 hrs.

3.

No of hours

Frequency

0

14

1

23

2

5

3

8

Draw a bar graph to show the results. 26 24 22 20 F 18 r e 16 q u 14 e n c 12 y 10 8 6 4 2

0

1 Number of hours

2

3

94 4.

Was the educator correct in thinking that the learners were not doing the set homework because they were watching too much TV. Validate your answers.

The majority of the learners watched TV for 1 hour or less. This is probably not a reason for them not doing their homework. Page 74: Exercise 5.2: The Average day-time temperature for 9 provincial capital cities were recorded. Average Temp in ยบC

Eastern Cape

Jan July

Gauteng

Bisho

Free State Bloemfon tein

22 14

23 8

20 10

JHB

KwaZuluNatal PMB

Mpumalanga

Limpopo

North west

Polokwane

Mafikeng

Nelspruit

Northern Cape Kimberley

24 15

25 11

23 12

24 12

23 13

Western Cape Cape Town 21 12

Task: 1.

Draw a dual bar graph to illustrate the above information. January Temperature

July Temperature

26 24 22

T e m p e r a t u r e

20 18 16 14 12 10 8 6 4 2

Bisho

2. 3.

Bloem m

Jhb

Pmb

Nels

Kim

Polo

Maf

Capital Cities

Which province has the greatest difference in temperature between January and July? Answer = Bloemfontein Why would somebody need to know the average temp in ยบC of the various cities in Jan and July? Answer = Tourists would want to know what the weather was. Farmers might want to know what the temperature was etc, etc.

CT

95

Page 76: Exercise 5.3: 240 learners were asked what they intended doing on leaving school. The results were: 80 wanted to attend University 86 wanted to attend Technikon 64 wanted to get a job 10 did not know 1.

Copy and complete the table:

Go to University

No of learners 80

Go to Technikon

86

Get a job

64

Don’t Know

10

Total

240

Calculation 80  360  240 86  360  240 64  360  240 10  360  240

Angle 120º 129 96 15º

360 2.

Illustrate this information on a pie chart.

University Technikon Job Don't Know

96 3.

Illustrate the same information as a bar graph.

L e a r n e r s U n i v e r s i t y

T e c h n i k o n

J o b

Don't Know

Activity

4.

Compare the two displays and identify:

One feature that the pie chart shows better. Answer: Segments stand out One feature that the bar graph shows better. Answer: Easier to read of exact values.

97 Page 86: Exercise 5.4: 1.

A company wanted to evaluate the training programme in its factory. They gave the same task to trained and untrained employees and timed each one in seconds.

Trained: 121 137 11 125 Untrained: 135 142 139 140 1.1

131 134

135

130

128

130

126

132

127

129

120

126 142

147

145

156

153

152

149

145

144

134

Draw a back – to – back stem & leaf diagram to show the two sets of data. Traine d

Untraine d 10 8

11

6 5 1 0

12

7 5 4 2 1 0 0

13

4 5

9

14

0

2

15

2

9 8 7

1.2

1.3

1.4

6

2 3

4

5

5 7

9

6

Find the medians and quartiles for both sets of data. Answer :Trained: Median = 129 Q1 = 125 and Q3 = 132 Untrained: Median = 144 Q1 = 139 and Q3 = 149 Find the Inter-quartile Range for both sets of data. Trained : IQR = 7 Untrained : IQR = 10 Comment on the results. Answer: The IQR measures a better dispersion than the range as it is not affected by extreme values

98 2..1

The heights, measured to nearest cm, of 75 girls picked at random at Glory High School, Are shown on the following frequency table:

Girls Height (h) in cm 135  h < 140 140  h < 145 145  h < 150 150  h < 155 155  h < 160 160  h < 165 165  h < 170 170  h < 175 175  h < 180

Frequency 2 5 10 17 19 15 4 2 1

Cumulative Frequency 2 7 17 34 53 68 72 74 7

Boys Height (h) in cm 135  h < 140 140  h < 145 145  h < 150 150  h < 155 155  h < 160 160  h < 165 165  h < 170 170  h < 175 175  h < 180

Frequency 0 1 7 11 15 18 15 6 2

Cumulative Frequency 0 1 8 19 34 52 67 73 75

Histogram : Girls

Frequency Polygon : Boys

18 16 F 14 r e q 12 u e n 10 c y 8

6 4

2

0

135-140 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180

Heights in centimeters

99 2.2 The heights, measured to nearest cm, of 75 boys picked at random at Glory High School, Are shown on the following frequency table: On the same set of axes as the histogram draw a frequency polygon to illustrate this data. 2.3

Use the frequency tables and the two graphs to help answer the following questions. For each group of learners state: 2.3.1 The modal class. Answer: Boys: 160 to 165 Girls : 155 to 160 2.3.2 The median height. Answer: Boys : 162,5 Girls : 157,5 2.3.3 The lower quartile Answer : Boys: 150-155 Girls : 150-155 2.3.4 The upper quartile Answer : Boys : 165-170 Girls : 160-165 2.4 Are the girls in the sample taller than the boys or are the boys taller? Answer : Boys are taller Use the statistical measures in 2.3 to back up your conclusions. Answer : Modal class Boys = 160 to 165 Modal class Girls = 155 to 165 Page 90: Exercise 5.5 2. The following marks were recorded for a maths class: 54 53 75 63

45 75 84

46 58 75 92

44 81 78 67

22 60 60 68

2.1 Do a stem and leaf diagram for the data 2

258

3

77

4

44556

5

344668

6

0037889

7

1155568

8

14

9

0

28

28 54 37 68

37 71 56 69

56 71 25 76

45 44 90 98

100 2.2

Find the median, mode and mean for the data

Median = 60 mode = 75 mean = 60,7

2.3

Find the lower and upper quartile Q1 = 45

Q3 = 75

2.4 Calculate: 2.4.1

the interquartile range.

IQR = 30 2.4.2

the semi-interquartile range.

SIQR = 15 2.4.3

the range for the class.

Range = 76 2.5 Write down the maximum and minimum scores. LOWEST = 22

HIGHEST = 98

2.6

98

22 45 0

10

20

30

40

60 50

60

75

70

80

90

101 3

The following marks were recorded for a maths class: 12 15 37 80

34 28 34

15 12 42 65

34 45 23 28

22 65 50 19

56 33 54 39

23 24 25 32

1.1 Do a stem and leaf diagram for the data 0

8 9

1

225589

2

0022334588

3

12344479

4

025

5

046

6

55

7 8

0

9

3.2 Find the median, mode and mean for the data Median = 28 Mean = 31,8 Mode = 34 3.3 Find the lower and upper quartile Q1 = 20

Q3 = 40

3.4 Calculate:

3.5

3.4.1

the interquartile range. IQR = 20

3.4.2

the semi-interquartile range. SIQR = 10

3.4.3

the range for the class. Range = 72

Write down the maximum and minimum scores. L = 8 H = 80

3.6 Box and Whisker diagram

8

80 20

0

10

20

28

30

40

40

50

60

70

80

90

22 9 8 40

20 18 20 31

102 4

The following marks for a class of Girls and Boys were recorded :

Girls 34 Boys 75

72 85 77 72

65 92 42 65

44 90 85 68

72 65 50 79

66 63 74 89

80 54 65 62

4.1 Do a back to back stem and leaf diagram for the data GIRLS

BOYS 4

3

4

4

2

8 54

5

0

6553

6

25

2220

7

0124579

50

8

0459

20

9

58

4.2 Find the median, mode and mean for both sets of data Boys: Mean = 70.1 Girls: Mean = 67.5

Median = 72 Median = 66

Mode = 65 Mode = 72

4.3 Find the lower and upper quartile of each set of data Boys : Q1 = 65 Q3 = 79.5 Girls :

Q1 = 56.5 Q3 = 76

4.4 Calculate: 4.4.1

the interquartile ranges for:

4.4.1.1

girls: 19.5

4.4.1.2

boys 14.5

4.4.1.3

class

4.4.2

the semi-interquartile ranges for:

4.4.2.1 girls 9.75 4.4.2.2 boys 7.25 4.4.2.3

class

58 55 85 70

70 72 80 71

103 4.4.3

the ranges for:

4.4.3.1 girls 58 4.4.3.2

boys 47

4.4.3.3

class

4.5 Write down the maximum and minimum scores of each set of data Girls Max = 92 Min = 34 Boys Max = 89

Min = 42

4.6

92

34 56.5

66

76

89

42 65 0

10

20

30

40

50

60

72 79.5

70

80

90

5. The following table represents the maths scores for the entire grade 10 maths group at Northwood School. The data is grouped due to the size of group. Class 0 to 9 10 to 19 20 to 29 30 to 39 40 to 49 50 to 59 60 to 69 70 to 79 80 to 89 90 to 99 100 to 109 Totals 5.1 5.2 5.3 5.4

Frequency(f) 15 10 17 40 35 22 20 20 15 5 1 200

Cf 15 25 42 82 117 139 159 179 194 199 200

Mid-points(X) 4.5 14.5 24.5 34.5 44.5 54.5 64.5 74.5 84.5 94.5 104.5

fX 67.5 145 416.5 1380 1557.5 1199 1290 1490 1267.5 472.5 104.5 9390

Complete the last column of the table i.e (fX) Find the modal class 30 to 39 Find the median class 40 to 49 Find the interval where Q1 and Q3 lie. Q1 lies in 30 to 39 and Q3 lies in 60 to 69

104 5.5

1.6

Calculate the estimated mean. ďƒĽ fX = 46,95 NB estimated mean = n Use the grouped data to display the data on a histogram

5.7

Draw the relevant frequency polygon on the histogram.

0

10

20

30

40

50

60

70

80

90

100

110

105 Page 105: Exercise 6.1 2. Calculate the volume and surface area of the following closed prisms:

Prism P Q R S 52 47 43 39 Length (mm) 20 18 17 15 Breadth (mm) 85 77 70 64 Height (mm) 88400 65142 51170 37440 Volume 14320 11702 9862 8082 Surface Area Determine the following ratios correct to 2 decimals. VolumeP VolumeQ 2.1 2.2  1,36  1,27 VolumeR VolumeQ 2.3

VolumeR  1,37 VolumeS

2.4

VolumeS  1,28 VolumeT

3.

SurfaceAre aP  1,22 SurfaceAre aQ

3.2

SurfaceAre aQ  1,19 SurfaceAre aR

3.3

SurfaceAre aR  1,22 SurfaceAre aS

3.4

SurfaceAre aS  1,19 SurfaceAre aT

4.

T 36 14 58 29232 6808

Are the volumes of the prisms approximately in proportion? Give reasons for your answers. Answer: Yes , approximately, the ratios around 1,3

5.1

5.2

How much smaller in volume is prism T than prism P? Give the scale factor (not the 1 change in volume).Answer: About smaller. 3 Are the surface areas of the prisms in proportion? Give reasons for pour answers. Answer: Yes, approximately, ratios are around 1,2. How much smaller in surface area is prism T than prism P? Give the scale factor (not 1 the change in area). Answer: About smaller. 2

5.3

6.

Determine the scale factor used : to reduce the dimensions of the prisms. Answer: 0,9 To enlarge the dimensions of the prisms. Answer 1,1 7. What is “The Golden Ratio” ? Answer: the ratio is 1,6 1   (1  5 )  1,6180 2 Determine which ratio of the faces comes closest to this ratio. NB: You must choose a ratio greater than 1. Answer: The height to length ratio is 1,6 in each case.

106

8.1 Reduce each of the dimensions of prism P by a factor of

1 , then calculate the 2

volume and surface area of the new prism X. Answers: V  26  10  42,5  11050mm3 Answers: SA  3580mm2 8.2 How much smaller in volume and in surface area is this new prism X? Give the scale factor in each case. 1 Answers: Vol Pr ismX  volume of prism P. 8 1 SA of X = SA of prism P. 4 9.1 Using the answers to question 8 , estimate the volume and surface area of prism Y, where each dimension of prism P has been enlarged by a factor of 2. Answers: Volume of Y = 8 x volume of prism P. = 8 x 88400 = 707200mm3 Answers: SA of prism X = 4 x SA of prism P = 4 x 14320 = 57 280 mm2 9.2 Calculate the volume and surface area of prism Y, and compare your answers to your answers to question 9.1 Vol Y = 104 x 40 x 170 = 707200 SA of Y = 2(104 x 40) + (40 x 170) + (170 x 104)= 57 280mm2

10.

Copy and complete the following table:

Dimensions of the prism is length; breadth and height. i.e. l ; b ; h Volume of prism = l  b  h Factor k =_2_____ Prism l (cm) b (cm) h (cm) A B C D E F G H

6 12 6 6 12 6 12 12

4 4 8 4 8 8 8 8

3 3 3 6 3 6 3 6

Volume (cm3) 72 144 144 144 288 288 288 576

Vxk V Vx2 Vx2 Vx2 Vx4 Vx4 Vx4 Vx8

Factor k k k k k2 k2 k2 k3

107 Volume of prism = l  b  h Factor k =_3_____ Prism l (cm) b (cm) h (cm) A B C D E F G H

4 4 4 12 12 12 4 12

3 3 9 3 9 3 9 9

2 6 2 2 2 6 6 6

Volume (cm3) 24 72 72 72 216 216 216 648

Vxk Factor V Vx3 Vx3 Vx3 Vx9 Vx9 Vx9 V x 27

k k k k2 k2 k2 K3

11. A cold - drink can measures approximately 65 mm in diameter and 75mm in height. Calculate the volume of the can ( in mm2 and cm2). Volume of can = r 2 h  32,5 2  75  248873mm3 V  248,873cm 3  249cm 3 The writing on the can says that it contains 200 ml of liquid. How much air

space is there in the can? ( 1ml  1cm 3 ) Air space is approximately 49cm3 What is the height of the liquid in the can? Height  60mm 12.1 Calculate the total surface area of the can (in cm3 ) , assuming that the can is a closed cylinder. SA  2( r 2 )  D h Surface area  220m2 12.2 If the metal to make the can costs 0,25 cents per square centimeter, calculate the cost of making each can. Cost of a can = 55 cents 13 he manufacturer of Lemon Twist wants to double the volume of the can , but keep the radius as it is. By which factor must the height be increased? Increase height by a factor of 2 14.1 If the radius is increased by a scale factor of 2, but the height is kept the same by which factor will the volume increase? vol of new can  ( 2r ) 2 h  4 vol of original can  (r ) 2 h 14.2 By which scale factor will the area of the top of the can increase? surface area of new lid  ( 2r ) 2  4 surface area of original lid  (r ) 2 14.3 By which factor will the area of the lateral surface ( the area of the curved side) increase? surface area of new side 2 (2r )h  2 surface area of original side 2 (r )h

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