GCSE Workbook Part 1 by GoMath Workbooks

MATH GCSE WORKBOOK PART ONE Compiled by Chesley Nell

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Forward: Welcome to “GO MATH WORKBOOKS”. This workbook is designed to be text book and class work book in one. There are sufficient exercises to ensure that learners get the required practice. A detailed memorandum is available at the end of the workbook. The statement “You get out what you put in.” is very apt where math’s is concerned. To succeed in mathematics one must be prepared to invest the time and effort to achieve that success. The partnership that you as a learner and this GOMATH WORKBOOK develop will be profitable if you allow it to be. Chesley Nell : Mathematics Educator  Chesley

Nell 2015

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GCSE MATHEMATICS.

Topic: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21.

Number System Mathematical Notation Algebraic Products Factors Algebraic Fractions Solution of Equations Inequalities Basic Logarithm Simplification Exponents Number Patterns Financial Mathematics Probability Theory Straight Line Graphs Parabola Graphs Hyperbola Graphs Exponential Graphs Trigonometric Graphs Analytical Geometry Transformation Geometry Triangles & Quadrilaterals Circle Geometry

Pages: (4 – 6) (7 - 10) (11 – 15) (16 – 30) (30 – 37) ( 37 – 86) (87 – 98) (99 – 101) (102 - 127) (128 - 156) (157 - 199) (200 – 241) (242 – 253) (254 – 298) (298 – 309) (310 – 336) (337 – 342) (343 – 378) (378 – 405) (406 – 424) (425– 450)

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Rational Numbers and Irrational Numbers

A common fraction can be written in the form: a ; a, b  z; b  0 b A mixed number is made up of two parts a whole number followed by a proper or decimal fraction. A terminating decimal is a decimal fraction that ends after a definite number of digits has been given, e.g. 1.23 To convert a terminating decimal into a common fraction change its denominator into 10, 100 etc. according to the number of decimal digits. E.g.: 1.23 

123 100

A recurring decimal is a decimal fraction or part thereof which goes on repeating itself without end. 

E.g.: 1. 2 3  1,23232323 We use dots to abbreviate recurring fractions e.g. 

 0,34 5  0,34555555 

 0,3 4 5  0,345454545 



 0, 3 4 5  0,345345345

To convert a recurring decimal into a common fraction Method 1: Multiply it by appropriate powers of 10, and then use subtraction to eliminate the repeating part, e.g.

x  1,2323... 100x  123,2323 100x  x  123,2323  1,2323 99x  122 122 99 23 1 99 x

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Method 2: In the denominator of the fraction to be formed write down a “9” for each of the recurring decimal and a “0” for each of the non- recurring decimal. In the numerator write down the decimal value as it is written and subtract the value of 3456  345 3111  the non - recurring decimals from this. e.g. 0,3456  9000 9000 Examples: 1. Convert 0,34 to fraction form: 34  3 31  . 90 90 2. Convert 1,34 5 to fraction form: 345  3 342 171 1,34 5  1 1 1 990 990 495   3. Convert 2,275 to fraction form: 275 2,2 75  2 999 Exercise 1: 1. Convert the following to decimals and whole numbers: 1 2 3 4 5 6 7 8 9 ; ; ; ; ; ; ; ; 2 2 2 2 2 2 2 2 2

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2. Convert the following decimal fractions into common fractions a)

0.45

b) 1.25

c) 6.125

d) 0.824

0.50

f) 2.5

g) 21.25

h) 0.05

3. Use dots to indicate the recurring digits in the following: e. g. 0.4444 = 0,4 a) 0.5555…. b) 0.454 454…. c) 0.714 1414…. d) 123.12333…

e) 123.12323… f) 123. 1231313 …

4. Convert the following recurring decimals into common fractions a)

0,6

e) 2,34 5

b) 0,7 8

c) 0,14

d) 45,345

f) 23,346 5

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2: Mathematical Notations: 1.

Set Builder Notation: Used with brackets and states the range of values including the section of the number system involved. < & > sign means â&#x20AC;&#x153;not includingâ&#x20AC;? ď&#x201A;Ł & ď&#x201A;ł sign means : includingâ&#x20AC;? e.g. ď ťx : ď&#x20AC;5 ď&#x20AC;ź x ď&#x20AC;ź 6; x ď&#x192;&#x17D; ď&#x192;&#x201A;ď ˝ This is read as: the set of all values of x such that x ranges between -5 and +6 where x is a real number.

Interval Notation: Only used with Real numbers: Used with round and square brackets. ( ď&#x192;&#x17E; means â&#x20AC;?not includingâ&#x20AC;? [ ď&#x192;&#x17E; means â&#x20AC;&#x153; including. e.g. x ď&#x192;&#x17D; (ď&#x20AC;5;6) read as all the values of x that range between - 5 and + 6 but not including -5 & +6.

x ď&#x192;&#x17D; [ď&#x20AC;5;6) read as all the values of x ranging between -5 and 6 ; including -5 but not including 6. Exercise 2.1: 1. Express the following in interval notation: 1.1 .{đ?&#x2018;Ľ: đ?&#x2018;Ľ â&#x2030;¤ 2; đ?&#x2018;Ľđ?&#x153;&#x2013;â&#x201E;&#x153;} ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.2

{đ?&#x2018;Ľ: â&#x2C6;&#x2019;3 â&#x2030;¤ đ?&#x2018;Ľ â&#x2030;¤ 0; đ?&#x2018;Ľđ?&#x153;&#x2013;â&#x201E;&#x153;}

{x:đ?&#x2018;Ľ < 2; đ?&#x2018;Ľđ?&#x153;&#x2013;â&#x201E;&#x153;}

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8 1.4

 -1

 3

⃘ -2

 5

2. Express in set builder notation: 2.1 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2.2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2.3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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−4 to 2, including -4 and

2.4 not 2.

3. The set of integers (Z) . NB. Integers must be separately graphed on a number line as it is only the whole numbers involved and no fractions in between.

{ x : 3  x  2; x  }

     -4 -3 -2 -1 0 1

4. Number line graphs:



 means “ including.

O  means ”not including” A set of real numbers can be represented on a line graph as follows: 1. Interval notation: [2;9)

 2

o 9

5. Set Builder notation:

 2

o 9

Exercise 2.2 : 1. Represent the following on a number line graph: 1.1 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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10 1.2

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{𝑥: −2 ≤ 𝑥 ≤ 2; 𝑥 ∈ ℜ}

{𝑥: 𝑥 ≥ 6; 𝑥 ∈ ℜ}

1.1

{𝑥: 𝑥 < −6; 𝑥 ∈ ℜ}

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Algebraic Products : 1. Use the FOIL method for expanding product of binomials. ( x  2)( x  4)  x 2  2 x  4 x  8  x 2  2 x  8 Lasts

Firsts Combination of Outers & Inners 2. If the brackets have co-efficients multiply LASTLY by these.

2( x  3)( x  6)  2( x 2  3x  18)  2 x 2  6 x  36 3. If there is a co-efficient with a binomial squared multiply LASTLY be the co-efficient. 2( x  3) 2

3.1

 2( x  6 x  9) 2

Multiply by 2 lastly

 2 x 2  12x  18

Multiply by -2 lastly

 2( x  3) 2

3.2

 2( x 2  6 x  9)  2 x 2  12x  18

N.B Signs change!

4. The product between a binomial and trinomial must be carried out separately and collect like terms where possible “unless a pattern is reflected.” (a  2)(a 2  2a  3) No pattern .  a 3  2a 2  3a  2a 2  4a  6  a 3  4a 2  7 a  6

4.1 If the trinomial part is formed by the square of each term and the product of the two terms in the binomial then the answer is simply the sum or difference of cubes. ( x  2)( x 2  2 x  4) Trinomial is formed  x3  2x 2  4x  2x 2  4x  8  x3  8

from the squares of each term and product of two terms

In this case leave out the second line and go straight to the answer. (2 x  3 y )(4 x 2  6 xy  9 y 2 )  8 x 3  27 y 3

Simply write down the cube of each term. NB: the sign of the last term takes the sign from the binomial

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Exercise 3.1: 1.

Simplify the following :

1.1

2( x  3)( x  4)

1.2

–3(3 – 5x) + 3(3-5x)

1.3

–2a2(a – b)2

1.4

(x – 3)2 –2(x– 1)2 + 2x

1.5

(a – 2b)( a2 + 2ab + 4b2)

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1.6

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( x  2) 2  3( x  2)( x  2)

1.7

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

1.8

4 x  3 y 2 xy  y 2 

1.9

32 x  y   2x  y x  5 y  2

p





 4 p 2 q  4 p q  16

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1.11

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(ЁЭСе тИТ 2)(ЁЭСе 2 + ЁЭСе + 4)

1.15 (2 ЁЭСе тИТ 3 ЁЭСж) (4 ЁЭСе 2 + 6 ЁЭСеЁЭСж + 9 ЁЭСж 2 ) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.16

(3 ЁЭСе + 4 ЁЭСж) (9 ЁЭСе 2 тИТ 6 ЁЭСеЁЭСж + 16 ЁЭСж 2 )

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1.18 −2(2𝑥 − 5𝑦)(3𝑥 2 − 𝑥 + 2𝑦) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1

1.19 − 5 𝑥(25𝑥 − 10𝑦) + 3 (12𝑥 − 3𝑦) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1

1.20 5 (5 𝑥 − 1) (5𝑥 − 5) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Factors:

4.1. Factors of Algebraic Expressions To factorize is the process of reducing many terms to one term. It is the opposite action of distribution. Procedure: Step 1 : First look for a HCF (if possible): 1.1 Look at the constants and choose the highest possible value, that can divide equally without a remainder, into the constant values in the expression. 1.2 Look at common variables and choose the lowest power (exponential value) of these as HCF. Step 2. Place the HCF outside a bracket and then divide the HCF into each term of the expression placing the answer inside the bracket. NB there must be the same number of terms inside the bracket as there is in the original expression. Step 3. Look at the expression inside the bracket and ascertain whether it can be factorized further. It could be one of the following; 3.1 The difference of two perfect squares. 3.2 A trinomial. 3.3 A quadrinomial ( 4 terms) Examples: HCF plus other expressions. 1.

2 x  4  2( x  2)

2. 3. 4. 5.

2 x  4 x 2  6 x 3  8 x 4  2 x(1  2 x  3 x 2  4 x 3 ) 2ax  4abx  6abcx)  2ax(1  2b  3bc) 2ax 2  8ay 2  2a( x 2  4 y 2 )  2a( x  2 y )( x  2 y ) ab4  ac 4  a(b 4  c 4 )  a(b 2  c 2 )(b 2  c 2 )  a(b 2  c 2 )(b  c )(b  c ) NB: The sum of two squares cannot be factorized at all.

4.2. Difference of two squares: Method: Write down the product ( two brackets) of the summand difference of the roots of each term. NB look at the bracket with the difference of the 2 terms and see if it can factorize further; NB: The sum of two squares cannot be factorized at all. Examples: 1. 2.

a 2  b 2  (a  b)(a  b) (b 4  c 4 )  (b 2  c 2 )(b 2  c 2 )  (b 2  c 2 )(b  c )(b  c )

NB DO NOT FORGET TO BRING DOWN THE SUM OF 2 SQUARES DOWN TO THE NEXT LINE

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Exercise 4.1: 1.

x2  4

x2  9

x 2  16

2 x 2  50

4x 2  9 y 2

x 4  16

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( x  y) 2  ( x  y) 2

4( x  y) 2  9( x  y) 2

____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 9. 4(2𝑥 − 1)2 − 36(𝑥 − 3)2 ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 10. 9𝑚2 (2𝑥 + 3) − 4(2𝑥 + 3) ______________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 11. 3𝑥(𝑥 2 − 𝑦 2 ) − 3𝑥(𝑦 − 𝑥) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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4.3. Sum & Difference of two Cubes: Method: Open two sets of brackets viz. for a binomial and a trinomial. If factorizing the sum of cubes then the binomial will have a plus sign and the middle term of the trinomial will be negative (minus sign) Binomial bracket: In the binomial bracket place the cube root of each term (NB Divide an exponent by 3 to find its cube root.) Trinomial bracket: Square each term in the binomial to get the first and last terms the middle term is the product of the two term in the binomial. Examples:

Open two brackets with the correct signs.

Factorize: a3  b3 = 1.

(

)(

)

(a  b)(a  ab  b ) 2

In the binomial bracket enter the cube roots of the terms

In the trinomial bracket enter the square of each term as Ist & last and then the product of the terms as the middle term.

a3  b3 = (

)(

)

= (a  b)(a 2  ab  b 2 )

8 x 3  27 y 6  ( 

)(





)

 (2 x  3 y 2 )(4 x 2  6 xy 2  9 y 4 )

Exercise 4.2: 1.

x3  y3

x3  y3

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8 x 3  27 y 3

64x 3  125y 3

6. 9𝑥 6 𝑏 − 3 𝑦 3 𝑏 ______________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 7. 4𝑥 2 (8𝑥 6 − 1) − (8𝑥 6 − 1) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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4.4. Trinomials: The method used here is called “trial & error” In products of binomials we use the FOIL method to expand . The middle term of the expression formed is a combination of the products of the inner terms and the outer terms. i.e. the ‘OI’ of FOIL. The reverse procedure is used to factorize trinomials. We find out the correct combination of factors of the First and Last terms of the trinomial. e.g. x 2  2 x  3  ( x  3)( x  1) Method: Draw a table and use the factors of the Ist and Last terms. Cross multiply them and either subtract or add to get the middle term. The sign of the last term informs one whether to add or subtract 1x 1x

x2  2x  3  ( x  3)( x  1)

1 3

x 2  8 x  12  ( x  2)( x  6)

1x 1x

6 12

NB Trinomial + + +

Brackets ( + )( + ) ( - )( - )

+ -

( + )( - ) ( + )( - )

Not necessary to work out

Must work out which bracket is negative and which is positive

Exercise 4.3: 1.

x 2  2x  3

x 2  6x  5

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x 2  6x  8

x 2  2x  8

x 2  7 x  12

x 2  8 x  12

2 x 2  5x  3

2 x 2  5 x  12

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3x 2 яАл 7 x яАл 2

12ЁЭСе 2 + 8ЁЭСе тИТ 15

(ЁЭСе + 3)2 тИТ 3(ЁЭСе + 3) тИТ 4

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4.5. FOUR OR MORE TERMS : Grouping into Brackets This involves a series of HCF’s: Method: 1. Group pairs of terms with common variables into brackets. NB don’t change the operation from addition and subtraction to multiplication. 2. Continue with HCF until fully factorized. Examples: 1.

ab  ac  db  dc  (ab  ac)  (db  dc)  a(b  c )  d (b  c )  (b  c )(a  d ) 3. x 2 (a  b)  y 2 (a  b)

3 px  3 py  x  y  3 p( x  y )  ( x  y )  ( x  y )(3 p  1)

 (a  b)( x 2  y 2 )  (a  b)( x  y )( x  y )

Exercise 4.4: Factorize completely. 1.1 x(a-b) + y(a-b) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2

1.2 p ( x + y) - q ( y + x ) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 1.3 (m  n)  ( pn  pm) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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1.4 a  b  ax  bx ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.5

a2  b2  a  b

4x 2  2x  9 y 2  3 y

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1.8

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(2 x  y ) 2  ( x  2 y ) 2

5 x 2  10xy  5 y 2

1.3

2x2 + 6x – 8x3

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1.4 a2 – 4 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.5

2x2 – 32

(m  n)  ( pn  pm)

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1.9

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a  b  ax  bx

a2  b2  a  b

1.11

(2 x  y ) 2  ( x  2 y ) 2

4( x  y) 2  9( x  y ) 2

1.13

4x 2  2x  9 y 2  3 y

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1.14

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x2  x  6

x 2  7 x  12

2 x 2  24x  70

9 x 2  42x  45

x3  y3

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1.19 8 x 3  27 y 3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

7. Algebraic Fractions: Multiplication & Division: NB. ALL TERMS MUST BE FULLY FACTORISED BEFORE ANY SIMPLIFICATION IS ATTEMPTED. Examples: x2 1. x2  4 ( x  2) = ( x  2)( x  2) 1 = x2

Terms are factorized and simplified

x 2  4 x  1 x 2  2x  4 x x x 2x  4 x3  8

All terms factorized first. And then common values are simplified.

( x  2)( x  2) x  1 ( x 2  2 x  4) x x x 2( x  2) ( x  2)( x 2  2 x  4) x 1  2x 

4 x 2 x 2  8x  4 x 2  16  ( x  4) 4   ( x  4)( x  4 2 x( x  4) 4  2 x ( x  4) 2

Terms all fully factorized first and a –ve sign is used to reverse ( 4 – x ) as well as inverting the fraction after the division sign and changing the operation to multiplication.

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5.1

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Algebraic Fractions Addition & Subtraction

1. NB Before you can find a LCD (lowest common denominator) you must ensure that all denominators are fully factorized. 2. The next step is to get the LCD. Write down the product of the highest power of each type of factor. Remember look at each factor in each separate term to ascertain its power. 1 2 3   Example: 2 ( x  1)( x  2) ( x  2) ( x  1) 2 The highest power of ( x  2) is in the second term and is ( x  2) 2 and the highest power of ( x  1) is in the third term and is ( x  1) 2 thus the LCD = ( x  2) 2 ( x  1) 2 3. Divide each denominator into the LCD and multiply the answer by the numerator of each fraction. 4. Simplify to get the final answer.

Examples.

2 3  x3 x2 2( x  2)  3( x  3)  ( x  3)( x  2) 2 x  4  3x  9  ( x  3)( x  2) 5x  5  ( x  3)( x  2)

LCD

Final Answer

Highest power of (x+3)

1 3 2   2 ( x  3) ( x  3) ( x  3) 3 ( x  3)( x  3)  3( x  3)  2  ( x  3)( x  3)( x  3) x 2  6 x  9  3x  9  2  ( x  3) 3 x 2  3x  2  ( x  3) 3

LCD Final Answer.

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2 3 4  2  x  2 x  1 x  1 ( x  1) 2 2 3 4    ( x  1)( x  1) ( x  1)( x  1) ( x  1)( x  1) 2( x  1)( x  1)  3( x  1)( x  1)  4( x  1)( x  1)  ( x  1)( x  1)( x  1)( x  1)

Denominators fully factorized.

LCD = product of the highest power of each type of factor

2 x 2  4 x  2  3x 2  3  4 x 2  8 x  4  ( x  1) 2 ( x  1) 2 

Expanded form of previous line

3x 2  4 x  5 ( x  1) 2 ( x  1) 2

Final answer.

Examples involving the change of signs.

2 3 4   (b  1) (1  b) (b  1)(1  b) 2 3 4    (b  1) (b  1) (b  1)(b  1) 2(b  1)  3(b  1)  4  (b  1)(b  1) 2b  2  3b  3  4  (b  1)(b  1) b5  (b  1)(b  1)

3 2 1 4    2 2 2 x 1 ( x  1) 1 x (1  x) 3 2 1 4   2   2 2 x 1 ( x  1) x  1 ( x  1) 3 2 1 4     2 2 ( x  1)( x  1) ( x  1) ( x  1) ( x  1) 

3( x  1)  2( x  1)  ( x  1)  4( x  1)( x  1) ( x  1)( x  1)( x  1)



3x  3  2 x  2  x  1  4 x 2  8 x  4 ( x  1)( x  1)( x  1)



 4 x 2  10x  2 ( x  1)( x  1)( x  1)

Use a negative to reverse the order of the denominator (1 – b) So just change the middle sign (1 + b) is the same as (b + 1) so no sign change needed.

Sign change to reverse term (1- x2)

Use a double sign change to change both brackets sign stays a +

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Exercise 5.1: 1

6 x. 12

4x 2 y 3 8x 3 y

2x  4 4

xy  y y

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8x 2  4 x 4x

x2 1 ( x  1) 2

x 2  x  12 x 2  7 x  12

a b ba

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a2 a a2 2

10.

x 2  x  12 4 x

ab  a 2 b 2  ab X b2  a2 a2

x 2  x  6 x3  2x 2 1  2  2 3x  12x x  16 x  4

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2x 2  8 4x  2x 2 4x   2 2 3 x 8 x  2x  4 13. x  4 x  4

x x2  2 x  y y  x2

15. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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 5x 3x 2   2 x 1 x  2x  1 1  x 2

6.EQUATIONS: 6.1

SOLUTION OF LINEAR EQUATIONS:

The method used to solve linear equations is as follows: Remove all variables to the same side of the equal sign and all constants to the opposite side. Solve for the variable concerned

NB: these are equations thus what you do to one side you must do to the other to keep the balance. Examples: 2 x  4  10

2x  6 x3 4 x  5  10  x

4 x  x  10  5 5 x  15 x3 5( x  2)  2( x  1)

5 x  10  2 x  2 3 x  12 x  4

Remove the +4 to the RHS by subtracting 4 from both sides.

Remove the -x to the LHS by adding +x to both sides and the -5 from LHS by adding +5 to both sides4 from both sides. Distribute to remove the brackets. Get the x’s to one side the constants to the other side and solve for x

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Exercise 6.1: 1.

2x  14  5  x

x  5  2x  3

5  3x  6  4x  5  2x

3  2x  6x  1

3( x  2)  2( x  1) ____________________________________________________________________

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1  8( x  5)  2( x  3)  10  4(2 x  3) ____________________________________________________________________

4(2 x  7)  3(2 x  4)  8(5  x)  5( x  7) ____________________________________________________________________

6.2

Linear Equations with Fractions:

Method: Multiply each term by the LCD( lowest common denominator) to convert the question to whole numbers and then solve in the normal way. Examples: 1.

x2 x2 3   LCD= 12 3 4 2 ( x  2) ( x  2) 3   3 4 2 4( x  2)  3( x  2)  18 4 x  8  3 x  6  18 x  20 3 2  3 x   2 1  7 x  LCD = 20 4 5 152  3 x   81  7 x  30  45 x  8  56 x  45 x  56 x  8  30 11x  22 x  2

 ( x  2) 12   ( x  2) 12  3 12       1  3 1 2 1  3

 3 20   2 20    2  3 x     1  7 x  4 1  5 1 

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Exercise 6.2: 1.

y y   1. 2 4

x x  3 5 10

x x 1   4 3 2

x 1 x 3    2 4 4 2

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3x 2 x 1  0 5 5 10

2x  7 x  5  0 6 3

x 1 x 1  40 4 5

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7x  2 9x  2  2 3 5

3x  4 2x  3  1 2 4

10.

3 x  5 2( x  5)  2 3

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11.

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x 1 x  2 x 1 x  2    3 4 2 3

12.

1 3  1 2 x  x  2 10  3  3

1 3  LCD = 2x x 2 2  3x 2 x 3 5 7  LCD = x(x-4) x x4 5( x  4)  7 x 5 x  20  7 x 2 x  20 x  10

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Exercise 6.3:

4 1  0 3 3x ___________________________________________________________________ 1.

3 1 9 2  x 2 2x ____________________________________________________________________ 3.

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2 3  4 x 2x ____________________________________________________________________ 4.

3

1 1 2 x   2 3x 6x

5 2 x3   4 4 3x 12 x ____________________________________________________________________ 6.

x 3 2 x3 x3 ____________________________________________________________________ 7.

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6.3 Quadratic Equations: 6.3.1 Solve by factorizing: Method: 1. Equate to zero, factorize and solve: x2  3x  4 2 e.g. . x  3 x  4  0 ( x  4)( x  1)  0

x  4 or x  -1 2. If already factorized simply solve the equation ( x  4)( x  1)  0 e.g. x  4 or x  -1 3. If not in factorized form do the necessary steps to get the equation into factorized form before solving. x ( x  4)  12 x 2  4 x  12  0

e.g.

( x  6)( x  2)  0 x  6 or

x2 Exercise 6.4: Solve the following equations:

1. ( x  5)( x  2)  0 __________________________________________________________________ __________________________________________________________________

2. (a  6)(a  1)  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

3. x( x  1)  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

4. ( x  2)( x  3)( x  5)  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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5. x(2 x  5)(3x  2)  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 6. y 2  3 y  10  0

7. x 2  5 x  6  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

8. x 2  7 x  6  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

9. x( x  1)  6 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

10. ( x  3)( x  2)  12 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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11. x 2 ď&#x20AC;Ť 2 x ď&#x20AC; 3 ď&#x20AC;˝ 12 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________

12. x( x ď&#x20AC; 16) ď&#x20AC;˝ 3(24 ď&#x20AC; 5 x) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

13.

(2đ?&#x2018;Ľ + 1)2 â&#x2C6;&#x2019; 4 = 0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

14.

đ?&#x2018;Ľ(2đ?&#x2018;Ľ + 1) = 6 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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6.3.2 Solving Quadratic Equations by the method of completing the Square: This is an alternative method that can be utilized to solve equations. NB. Only use this method when instructed to Method: Add to the equation the square of half of the value of the coefficient of x to both sides of the equation. This forms a perfect square trinomial (a binomial squared) on the left hand side and constant values on right hand side. Mathematically zero is being added to the equation thus there is no change at all. Example 1:

Add in the square of half the coefficient of ‘x’ to both sides of the equation Mathematically adding zero as (1)2 –(1)2 = 0

x2  2x  3  0 x2  2x  3 x 2  2 x  (1) 2  3  (1) 2 ( x  1) 2  4 x 1  4

Factorise the left hand side i.e. it forms a binomial squared.

x  1 4 x  1 2 x3 OR x  1 4

Solve for x

x  1 2 x  1

Example 2. 2x2  6x  4  0 x2  3x  2  0 2

 3  3 x  3x     2     2  2 2

 x 

3 17   2 4

3 17  2 2 3  17 x 2 x  3,56 x

or x  0,56

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Example 3. If f ( x )  x 2  2 x , show by completing the square that f ( x  1)  ( x  2) 2  1 f ( x)  x 2  2 x f ( x  1)  ( x  1) 2  2( x  1)  x2  2x  1  2x  2  x2  4x  3 x 2  4 x  ( 2) 2  3  2 

( x  2) 2  1 ( x  2) 2  1  0

 f ( x  1)  ( x  2) 2  1 Exercise 6.5: Use the method of completing the square to solve the following: 1.

x 2  2 x  24  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x 2  9 x  36  0 __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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x 2  8 x  15  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x 2  7 x  12  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

2x 2  7x  6  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

2 x 2  11x  6  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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2 x 2  3x  8  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

6.3.3 SOLVING QUADRATIC EQUATIONS USING A FORMULA: The formula is derived by completing the square with the general quadratic equation: ax 2  bx  c  0 General Quadratic Formula: x 

 b  b 2  4ac : 2a

NB: a; b; & c are constant values WHERE:

a = coefficient of x2 ; b = coefficient

of x ; c is the constant in equations written in the form: ax 2  bx  c  0 NO x –values must be substituted into the formula. I.e. only the constant values are used. Derivation of Quadratic Formula: ax 2  bx  c  0 b c x2  x   0 a a 2

b  b   b  c x  x       a  2a   2a   a  2

b  b2 c   x   2 2a  a 4a  2

b  b 2  4ac  x   2a  4a 2  b   b 2  4ac  x  2a  2a   b  b 2  4ac x 2a

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Examples: Solving quadratic equations using the general quadratic formula. x  x  12  0 2

 b  b  4ac 2a  1  1  4(1)(12 x 2  1  49 x 2 1 7 x 2 x  4 OR 3 x

2x2  7x  6  0 x

 b  b 2  4ac 2a

x

7  49  4( 2)(6) 2( 2)

7 1 4 8 6 x  or 4 4 3 x  2or 2 x

SOLVING QUADRATIC EQUATIONS USING A FORMULA: Extension Work (Optional) The formula is derived by completing the square with the general quadratic equation: ax 2  bx  c  0  b  b 2  4ac : 2a NB: a; b; & c are constant values WHERE: a = coefficient of x2 ; b = coefficient of

General Quadratic Formula: x 

x; c is the constant in equations written in the form: ax 2  bx  c  0 NO x –values must be substituted into the formula. I.e. only the constant values are used. Examples: Solving quadratic equations using the general quadratic formula. Solve: 1. x 2  x  12  0

 b  b 2  4ac x 2a

2x2  7x  6  0

 1  1  4(1)(12 2  1  1  48 x 2  1  49 x 2 1 7 x 2 8 6 x OR 2 2 x  4 OR 3

x

 b  b 2  4ac 2a

x

7  49  4( 2)(6) 2( 2)

x

7 1 4 8 6 x  or 4 4 3 x  2or 2 x

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Exercise 6.6: Solve using x 

 b  b 2  4ac ; 2a

(Answers rounded to 2 decimal places where necessary) NB: First expand if necessary and equate to zero before using the formula. 1.

x2  4x  3  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

2 x 2  x  10 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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3x2  x  2  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x2  6x  4  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

2x2  4  7x __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

2 x ( x  3)  3  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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Exercise 6.6: Solving quadratic equations using a Trial and Improvement process. The method involves substituting the variable with different values until a viable solution is found. Example: Find a positive solution to the equationđ?&#x2018;Ľ 2 + 6đ?&#x2018;Ľ â&#x2C6;&#x2019; 10 = 0. đ?&#x2018;Ľ Has a variable value between 1 and 2. Give the answer correct to 2 decimal places. đ?&#x2018;Ľ 2 + 6đ?&#x2018;Ľ = 10 Substitute the given values into the above equation: (1)2 + 6(1) = 7 (2)2 + 6(2) = 16 The value of 7 is closer to 10 than 16 which indicates that the solution is closer to 1 than 2. Start the trial and improvement using a value between 1 and 2. i.e. 1.5 đ?&#x2018;Ľ 2 + 6đ?&#x2018;Ľ = 10 Try 1.5 (1.5)2 + 6(1.5) = 2.25 + 9 = 11.25 11.25 is greater than 10 so try a value less than 1.5 i.e. (1.5 â&#x2C6;&#x2019; 0.1 = 1.4) Try 1.4 (1.4)2 + 6(1.4) = 1.96 + 8.4 = 10.36 10.36 is greater than 10 so try a value less than 1.4 i.e. (1.4 â&#x2C6;&#x2019; 0.1 = 1.3) Try 1.3 (1.3)2 + 6(1.3) = 1.69 + 7.8 = 9.49 9.49 is smaller than 10 so try a value between 1.3 and 1.4 i.e. 1.35 Try 1.35 (1.35)2 + 6(1.35) = 9.9225 9.9225 is smaller than 10 so try a value between 1.35 and 1.37 i.e. 1.36 Try 1.36 (1.36)2 + 6(1.36) = 10.0096 10.0096 is greater than 10 so try a value between 1.35 and 1.36 i.e. 1.355 Try 1.355 (1.355)2 + 6(1.355) = 9.9225 The best solution is 1.36. Exercise 6.7: Use trial and improvement to find solutions for the following: 1. đ?&#x2018;Ľ 2 + 11đ?&#x2018;Ľ â&#x2C6;&#x2019; 7 = 0 Has a solution between 0 and 1. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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2. đ?&#x2018;Ľ 2 + 15đ?&#x2018;Ľ + 12 = 0 Has a solution between -1 and 0. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 3. đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 6đ?&#x2018;Ľ â&#x2C6;&#x2019; 3 = 0 Has a solution between 6 and 7. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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4. 2đ?&#x2018;Ľ 2 + 4đ?&#x2018;Ľ â&#x2C6;&#x2019; 3 = 0 has a solution between 0 and 1.5 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

Using an iterative formula to solve a quadratic equation. The process of iteration is a numerical method of solving an equation. Iterate means to â&#x20AC;&#x153;repeatâ&#x20AC;? or â&#x20AC;&#x153;perform againâ&#x20AC;?. To find an accurate approximation the process can be repeated at least 5 times. An iterative formula with one approximate solution is usually given. Use this information to find a most accurate approximate solution to the given equation. Example: Find a positive root for the quadratic equation đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 2đ?&#x2018;Ľ â&#x2C6;&#x2019; 1 = 0 to 3 decimal places 1 given the iterative formula: đ?&#x2018;Ľđ?&#x2018;&#x203A;+1 = 2 + đ?&#x2018;Ľ andđ?&#x2018;Ľ0 = 3. đ?&#x2018;&#x203A;

đ?&#x2018;Ľ0 = 3 1 1 đ?&#x2018;Ľ1 = 2 + đ?&#x2018;Ľ = 2 + 3 = 2.3333333 â&#x20AC;Ś 1

đ?&#x2018;Ľ2 = 2 + đ?&#x2018;Ľ = 2 + 2.3333333 = 2.42857 â&#x20AC;Ś 1

đ?&#x2018;Ľ3 = 2 + đ?&#x2018;Ľ = 2 + 2.42857 = 2.411765 â&#x20AC;Ś 2

đ?&#x2018;Ľ4 = 2 +

đ?&#x2018;Ľ3 1

=2+

1 2.411765 1

= 2.414634 â&#x20AC;Ś

đ?&#x2018;Ľ5 = 2 + đ?&#x2018;Ľ = 2 + 2.414634 = 2.414141 â&#x20AC;Ś 4

The solution to 3 decimal places: đ?&#x2019;&#x2122; = đ?&#x;?. đ?&#x;&#x2019;đ?&#x;?đ?&#x;&#x2019;.

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Exercise 6.8: Use the practice of iteration to find a positive solution for the following equations: All answers to 3 decimal places. 4 2. đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 6đ?&#x2018;Ľ â&#x2C6;&#x2019; 4 = 0 using đ?&#x2018;Ľđ?&#x2018;&#x203A;+1 = 6 + đ?&#x2018;Ľ đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ľ0 = 6 đ?&#x2018;&#x203A;

đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 3đ?&#x2018;Ľ â&#x2C6;&#x2019; 2 = 0 using

đ?&#x2018;Ľđ?&#x2018;&#x203A;+1 = 3 + đ?&#x2018;Ľ đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ľ0 = 3 đ?&#x2018;&#x203A;

đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 4đ?&#x2018;Ľ â&#x2C6;&#x2019; 4 = 0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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6.4 Nature of Roots: The discriminant,   b 2  4ac , helps us to determine what type of roots an equation will have. NB We do not solve the equation and we are not concerned about finding the specific roots of a given equation. We want to ascertain what kind of roots an equation will have classified according to the criteria: Real or Non-Real; Equal or Unequal; Rational or Irrational.

 b  b 2  4ac , the value inside the square 2a root sign is important to us as it will enable one to classify the roots. This discriminant is referred to as   b 2  4ac ( delta =) From the general quadratic formula, x 

 is a perfect square then the roots are rational  is not a perfect square then the roots are irrational

  0 then the roots are Real   0 then the roots are non-real.  = 0 then the roots are equal   0 then the roots are unequal. Examples: In the following examples find the value of the discriminant and then make a statement classifying the nature of the roots. Discuss the nature of the roots of the following equation. NB do not solve the equation.

1. x 2  5x  3  0   b 2  4ac   ( 5) 2  4(1)( 3)   37 The roots are real, irrational and unequal. 2. 2 x 2  5x  3  0   b 2  4ac   (5) 2  4(2)(3)   25  24  1 Roots are real ; rational and unequal.

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3. 4x 2  4x  1  0   b 2  4ac   (4) 2  4(4)(1)   16  16 0 The roots are real ; rational and equal. 4. x 2  5x  9  0   b 2  4ac   (5) 2  4(1)(9)   25  36   11 The roots are non-real.

Further examples involving nature of Roots. In the first 2 examples you find delta and then solve for the variable using the clue supplied. In the 3rd and 4th examples you find delta and simply make an observation in the form of statement to answer the question. NB you do not solve for any variable at all. For which values of r will 2 x 2  2 x  r , have equal roots. In this type of question find the discriminant first and then find the value of the variable in question. Example one: 2x 2  2x  r 2x 2  2x  r  0 a  2; b  2; c  r   b 2  4ac   (2) 2  4(2)(r )   4  8r

For equal roots   0 ( must equate the discriminant to zero.) 4  8r  0 8r  4 r

1 2

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Example two: For which values of h will the roots of the equation 3x 2  2hx  3 be non – real?

3 x 2  2hx  3  0 a  3; b  2h; c  3   b 2  4ac   (2h) 2  4(3)(3)   4h 2  36 For non-real roots   0 4h 2  36  0

h2  9  0

-3

(h  3)(h  3)  0

3  h  3 Example 3. Show that the roots of ax 2  (2a  1) x  (a  1)  0 are rational for all rational values of a. NB find delta first and then simply make a statement. ax 2  (2a  1) x  (a  1)  0 a  a; b  (2a  1); c  (a  1)   b  4ac   (2a  1) 2  4(a )(a  1)   4a 2  4a  1  4a 2  4a  1 Delta is a perfect square thus the roots are rational.

Example 4. Prove that the roots of rx 2  x 2  4 x  r  1 are real for all real values of r. rx 2  x 2  4 x  r  1 (rx 2  x 2 )  4 x  ( r  1)  0 (r  1) x 2  4 x  (r  1)  0 a  (r  1); b  4; c  ( r  1)   b 2  4ac   (4) 2  4( r  1)(r  1)   16  4r 2  4   4r 2  12 4r 2  0 For all values of r. 4r 2  12.  12

The roots are real for all values of r.

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Exercise 6.9: 1. Determine the nature of the roots of the following without solving the equations. 1.1

x 2  4x  3  0

1.2

x2  7x  8  0

1.3

4 x 2  8 x  4

4 x 2  12 x  9

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1.5

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Find the value(s) of b if the roots of x 2  5 x  b  0 are equal. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 1.6 If -1 is a root of x 2  4 x  k  0 , calculate k and the other root. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.7

Prove that x 2  ax  a 2  1  0 has non-real roots. __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.8

Show that the roots of ax 2  bx  bx 2  a has rational roots for all rational

values of a and b

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1.9

Determine the value(s) of h for which the graphs of y  3x 2  5h and

y  x  h

will not touch or intersect.

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6.5 Simultaneous Equations: A.

2 linear equations:

Method: 1. 2. 3. 4.

Number the equations as 1 and 2. Rewrite one of them in terms of x or y. ( try not to have fractions involved) Substitute the new x or y value into the other equation and solve. Substitute this value into the changed equation to get the value of the other variable.

Example: Solve for x and y simultaneously if x  2 y  5 and x  y  1 Change x  y  1 Number (1) / x  2 y  5 (2) the the x  y 1 equation equations to the x 1&2 formSubstitute 1 into 2 y 1 2y  5 Change one of the Substitute the value for equations to the x – 3y  6 x in the changed form or y - form y2 equation for y in the other equation x = 1 {from x  2  1} Substitute the value from above into the changed equation and solve for the second variable Exercise 6.10: Solve the following equations simultaneously.

x y 5 1. And x  y  3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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x  3 y  5 And 2 x  3 y  1 2. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

x y 8 3. And 3 x  2 y  21 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3x  2 y  60 And 3 x  3 y  45 4. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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x  y  36 5. And x  2 y  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

x  2y  5 6. And 3x  y  1 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

6 x  y  22 And 4 x  y  8 7. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2y  x  3 8. And 4 x  3 y  10 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3x  y  5  0 And 7 x  3 y  1  0 9. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

10.  m  1 2k And 2k  m  3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2 x  3 y  14 And x  5 y  0 11. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3 x  4 y  24 And 7 x  4 y  16 12. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3x  y  2 13. And 6 x  y  25 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2x  9  y 14. And x  36  4 y ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Level of difficulty increased: Example: Find x and y if y 

x x  11 and 3 y   14 4 5

x  11 4 -----(1) x y  11  4 Substitute 1 into 2: y

3y 

x  14 -----(2) 5

x x 3(11  )   14 4 5 3x x 33    14 4 5 LCD  20 660  15x  4 x  280 19x  380 x  20 y  11 

20 4

y  16

x y x y 1   9 and   0 2 3 3 2 2 LCD = 6 LCD = 6 3 x  2 y  54 3 x  2 y  54 3 x  2 y  54 --------(1) and 2 x  3 y  3  0 -------(2) x

2 y  18 3

Substitute 1 into 2: 2 y  18)  3 y  3  0 3 4y   36  3 y  3  0 3 LCD  3 2( 

 4 y  108  9 y  9  0  13 y  117 y9

x  12

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Exercise continued: x y x y  1  and 4 2 5 3 ____________________________________________________________________

15.

16.

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___________________________________________________________________ 6.6

Problems involving simultaneous equations:

Example: 1. The sum of two numbers is 4 and their difference is 6. 1.1 Let their numbers be x and y . Write down two equations in x and y. 1.2 Solve the equations and find the two numbers. Let one number be x and the other y x  y  4 and x  y  6 x y  4

----(1) and y  4 x Substitute 1 into 2

x  y  6 -------(2)

x  (4  x)  6 x4 x  6 2 x  10 x5 y  1

Exercise 6.11: 1.

The sum of two numbers is 54 and their difference is 6. Find the numbers.

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___________________________________________________________________ 2.

The sum of two numbers is 35 and their difference is 19. Find the numbers

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.

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___________________________________________________________________ 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) .}

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

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____________________________________________________________________ 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?

A boy is 6 years older than his sister. In three yearsâ&#x20AC;&#x2122; time he will be twice her age. What are their present ages?

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____________________________________________________________________ 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?

____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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____________________________________________________________________

Simultaneous Equations Continued: One linear and one quadratic Method: Rewrite the linear equation into either the x – form or y – form and then substitute this value for x or y into the equation of the higher degree. Example: x  y  2 and x 2  y 2  52 ----- ② y  2  x ---- ① Substitute ① into ② x 2  ( 2  x ) 2  52 x 2  4  4 x  x 2  52 2 x 2  4 x  48  0 x 2  2 x  24  0 ( x  6)( x  4)  0 x = 6 or x = -4 y= -4 or y= 6

Exercise 6.12: Solve the following systems of equations simultaneously: 1.

y  x  2 and x 2  2 xy  4  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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x  y  2 and x 2  y 2  20 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x  y  4 and xy  4 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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x  y  3 and xy  4 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x  2 y  1 and x 2  2 xy  2 x  4 y  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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x  y  3 and x 2  y 2  89 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

2 x  y  1 and x 2  2 yx  2 x  y 2  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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Problems involving simultaneous equations: Example: 2. The sum of two numbers is 4 and their difference is 6. 1.3 Let their numbers be x and y. Write down two equations in x and y. 1.4 Solve the equations and find the two numbers. 1.1

Let one number be x and the other y x  y  4 and x  y  6

x y  4

1.2

----(1) and y  4 x Substitute 1 into 2

x  y  6 -------(2)

x  (4  x)  6 x4 x  6 2 x  10 x5 y  1 Exercise 6.13: 9.

The sum of two numbers is 54 and their difference is 6. Find the numbers.

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10.

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The sum of two numbers is 35 and their difference is 19. Find the numbers.

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.

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12.

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

13.

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14.

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Samantha and Warren cycle towards each other along a straight road. They start off 70km apart. Samantha cycles at 15km/h and warren at 20km/h. How far will Samantha have cycled when they meet?

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7.1 LINEAR INEQUALITIES: Linear inequalities are solved in the same manner that linear equalities are. Examples: 1.

Solve the following inequality and illustrate the answer on a number line.

2x  4  6

2 x  10

Solution:

 5 2.

2x  4  6 x5

Solve the following inequality and represent the answer on a number line:

 4  2x  2  8

 4  2x  2  8  2  2 x  10

 -1

1 x  5

 5

Exercise 7.1: Solve the following inequalities and illustrate the answers on a number line: 2x  6  8 1. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ _________________________________________________________________ 2.

3x  6  x  14

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2x  7  5x  14

3x  7  3 x  14 2

 6  2x  14

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 5  2x  3  15

Exercise 7.2: 1. Find the possible solutions of the following linear inequalities and illustrate the answers on a number line. 1.1

3x  2  x  14

5x  3  3x  15

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1.3

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1  2x  x  2

4( x  4)  7( x  2)  1 1.4 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4(2 x  1)  5 x  2 1.5 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2( x  1)  2(2 x  1)  2  x 1.6 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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 2( x  3)  5 x  78 1.7 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 3( x  2) 7( x  3)  3 2 4 ___________________________________________________________________

1.8

1.9

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4  x 2x  1   x2 2 3 ___________________________________________________________________

1.10

1.11

1.12

x

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x  2 1  3x  2 5 3 ___________________________________________________________________

1.13

3

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 1.14  1  x  3  5 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 1.15 1  1  2x  7 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ x3 4 2 ___________________________________________________________________

1.16

3

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2x  3  4 3 ___________________________________________________________________

1.17

2

1.18

2

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7.2 Quadratic Inequalities: Steps: 1. 2. 3. 4. 5.

Equate to zero If fractions involved then combine to make one expression Factorize all expressions fully Test for the solution with whatever method you are comfortable with. Write your solution down.

NB DO NOT CROSS MULTIPLY WITH INEQUALITIES: Examples: 1.

x2  3x  4  0 ( x  4)( x  1)  0

-1

Solution:  1  x  4

2. 4 x 4 x 0 x 2 x 4 0 x ( x  2)( x  2) 0 x  2  x  0 or x  2 x

3 x 0 x8

-2 2

2 2

-8

x  8 or x  3 Exercise 7.3: 1.

x 2  8 x  15  0 . __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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4 x 2  49 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x2 0 x5

2x  1 0 x4 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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 2x  5 0 3 x __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

4 x x3 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

x

9 x

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2 3  x2 x3 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1 2  x5 x7 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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LOGARITHMS:

Calculations using basic logarithms: Calculator work: NB a logarithm is an exponent: 3 2  9 In logarithm form this is written as log3 9  2 And is read: the logarithm of 9 to the base 3 is 2 (I.e. the exponent of the base 3 is 2 to give the number 9)

Example: Solve for x in the following: : x 

log 5 log 2

log 5 log 2 0,6989700043 Answer: x  Answer to 2 decimal digits. 0,3010299957 x  2,32 (Simply enter values as given into your calculator using the fraction and log facility) x

Exercise 8.1: Solve for x giving answers to 2 decimal digits. log1000 . log100 ___________________________________________________________________

x

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100

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log 300 log 200 ___________________________________________________________________

x

log15,5 log 3

x

2 log 25 3 log15

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x

101

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5 log 50 2 log 8  2 log 25 6 log 6

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102

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Exponents:

N.B. The basic laws must always be applied. 1.

Law: a m .a n  a m n

When multiplying like bases you must add the exponents. Do not multiply the bases except in the following case: 2 3.33  8.27  216. e.g. BUT 2 3.33  6 3  216 am = a mn n a When dividing the bases subtract the exponents: The following exception.

Law:

63  6      33  27 3 2 2

3. Law: (a m ) n  a mn When raising a power to a power you must multiply the exponents. 4. 5.

Law: (a m ) 0  1 Any value raised to the power of “zero” will equal 1.

N.B. 1. Never multiply the bases 2. Never divide the bases 3. Never multiply a base by an exponent. 4. If bases are separated by plus or minus signs you MUST FACTORISE before simplifying. Prime Base Factorizing It is important to remember to use prime bases in simplifying with exponents (especially if no calculators are allowed. NB when reducing bases by factorizing, always use the lowest possible bases. E.g. for 16 rewrite as 2 4 and not as 4 2 . Use prime numbers for bases and not composite ones. Once you have factorized using prima bases then the normal laws apply. Example 1: 4

 

Simplify 1253  53 3 (Now use law 3 and raise a power to a power by multiplying the exponents :)

125  5 4 3



4 3 3

5

3 4  1 3

 5 4  625

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103

Example 2: Simplify without the use of a calculator. 5 a ď&#x20AC;Ť 3.5 2 a ď&#x20AC;1 Use law 1 25a ď&#x20AC;1 in the Answer:

5 a ď&#x20AC;Ť 3.5 2 a ď&#x20AC;1 5 3a ď&#x20AC;Ť 2 ď&#x20AC;˝ 2aď&#x20AC; 2 ď&#x20AC;˝ 5 aď&#x20AC;Ť 4 2 a ď&#x20AC;1 (5 ) 5

Prime base factorize and use law 3 to simplify

numerator

Use law 2 to get the answer.

Follow the procedures as set out for any of the following types of simplifications with exponents. Exercise 9.1: 1.

đ?&#x2018;&#x17D;3 Ă&#x2014; đ?&#x2018;&#x17D;2

___________________________________________________________________ ___________________________________________________________________ 2.

đ?&#x2018;?2 Ă&#x2014; đ?&#x2018;?4

___________________________________________________________________ ___________________________________________________________________ 3.

đ?&#x2018;&#x17D;3 đ?&#x2018;? 2 Ă&#x2014; đ?&#x2018;&#x17D;2 đ?&#x2018;? 4

___________________________________________________________________ ___________________________________________________________________ 4.

2đ?&#x2018;&#x17D;đ?&#x2018;? 2 Ă&#x2014; 3đ?&#x2018;&#x17D;4 đ?&#x2018;? 6

____________________________________________________________________ ____________________________________________________________________ 5.

2đ?&#x2018;&#x17D;(3đ?&#x2018;&#x17D; + 2đ?&#x2018;?)

____________________________________________________________________ ____________________________________________________________________ 6.

2đ?&#x2018;&#x17D;2 đ?&#x2018;?(3đ?&#x2018;&#x17D;đ?&#x2018;? 3 â&#x2C6;&#x2019; 2đ?&#x2018;&#x17D;4 đ?&#x2018;?)

____________________________________________________________________ ____________________________________________________________________

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3(2đ?&#x2018;&#x17D;2 đ?&#x2018;? 3 )3

____________________________________________________________________ ____________________________________________________________________ 8.

3đ?&#x2018;&#x17D;0 (2đ?&#x2018;&#x17D;2 đ?&#x2018;? 3 )0

___________________________________________________________________ ___________________________________________________________________ NB. Prime base factorizing first then apply the laws.

2 xď&#x20AC;Ť3 2 x 2 x ď&#x20AC;1

10.

2 x ď&#x20AC;Ť1 2 x ď&#x20AC; 3 2 x ď&#x20AC; 2 .2 x

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4 x 1 8 x  1 32 x 1 ___________________________________________________________________

11.

13.

7 x  2 49 x  2 7 3 x2

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14.

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6 n 1 .12 n1 .2 n 18 n  2 .8 n 1

15.

6 n 212 2 n 1 4 2 n  3 8 3 n1 9 n1 3 n

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If the bases are separated with plus or minus signs one must FACTORISE FIRST... It is easier to split the bases as shown because it is easier to see the HCF. Factories and simplify. NB the base with the variable exponent should always cancel leaving pure numerical values. Exercise 9.2:

2 x3  2 x 2 x 1

2 x 1  2 x  3 2 x2  2 x

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3 x  3 x2 3 x 1  3 x

5 2  x  4 .5  x 5  x  2 .5  x  1

3 n .3 4  6.3 n .3 1 7.3 n .3 2

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109

2 n .2 5  3.2 n .2 2 5.2 n 2 3

Variable in the exponent: Method: 1.1 Equate the bases using prime base factorizing. 1.2 If the bases are now equal then the exponents are also equal, thus Simply equate the exponents and simplify further.

2 2 x  16 Example 1.

22x  24

Rename 16 to the base 2 and equate the bases.

2x  4 x2 2.3 x ( x  3)  54 3 x ( x  3)  27

First divide by the coefficient ‘2’

3 x ( x  2)  33

Example 2:

x 2  2x  3  0 ( x  3)( x  1)  0

Rename 27 to the base 3 and equate the bases.

x3 or x  1

Equate the exponents and solve for x.

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Variable in the base: Method: 2.1 Raise the power of the exponent of the variable base to its multiplicative inverse. This will give a new exponent with the value of 1. As it is an equation you must do the same to both sides of the equal sign. Example. 2 3

x ď&#x20AC;˝ 16 ď&#x192;Ś ď&#x192;§x ď&#x192;§ ď&#x192;¨

2 3

3 2

ď&#x192;ś ď&#x192;ˇ ď&#x20AC;˝ 24 ď&#x192;ˇ ď&#x192;¸

x ď&#x20AC;˝ ď&#x20AC;¨2 4 ď&#x20AC;Š 2 3

x ď&#x20AC;˝ 26 x ď&#x20AC;˝ 64

NB. You must first remove any coefficient values prior to solving for the variable:

More advanced equations: Variables separated by + or â&#x20AC;&#x201C; signs: This involves factorizing. 2 x ď&#x20AC;Ť 3.2 x ď&#x20AC;˝ 16

2 x (1 ď&#x20AC;Ť 3) ď&#x20AC;˝ 16

1.1

2x ď&#x20AC;˝ 4

Use 2 x as the HCF

2 x ď&#x20AC;˝ 22 xď&#x20AC;˝2

Equate the bases thus and then the exponents.

NB Clues to let you know when a simple HCF must be used are: 1. Terms separated by plus and minus signs and 2. The variable exponents have the same value. I.e. the coefficients of the variables exponents are equal in value. 2.

Variables separated by + or â&#x20AC;&#x201C; signs BUT the values of the coefficients of the variables exponents are not equal. I.e. one is double the other. This involves a trinomial and needs to be factorized accordingly. 2.1 22đ?&#x2018;Ľ â&#x2C6;&#x2019; 6. 2đ?&#x2018;Ľ + 8 = 0 The exponent of the 1st (2đ?&#x2018;Ľ â&#x2C6;&#x2019; 4)(2đ?&#x2018;Ľ â&#x2C6;&#x2019; 4) = 0 term is double that of the 2nd . i.e The 2x ď&#x20AC;˝ 4 2x ď&#x20AC; 2 ď&#x20AC;˝ 0 expression is a 2 x ď&#x20AC;˝ 2 2 or 2 x ď&#x20AC;˝ 2 trinomial. The variable base is xď&#x20AC;˝2 x ď&#x20AC;˝1 2x

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2.3 2 x  12.3 x  54  0 (2.3 x  6)(3 x  9)  0

2 .3 x  6  0 3x  3

3x  9

3 x  32 x2

x 1

The exponent of the 1st term is double that of the 2nd . i.e The expression is a trinomial. The variable base is 3x

NB a substitution method can be utilized here: Let 3 x = k [ thus 3 2 x  k 2 ] 3 x 2k 2  12k  54  0 2(k 2  6k  27  0 (k  3)(k  9)  0 k  3 or k  9 3x  9 3x  3 x 1

Substitute 3 x for k at this point and solve for x

or 3 x  3 2 x2

Exercise 9.3: 1. Solve for x without the use of a calculator:

2.1 1.1

2x = 8

3 x  81

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2.3 1.3

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x3 = 27

2.4 1.4

4

= 80

2.5 1.5

2 x  16 x 1

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2.6 1.6

113

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3.2x = 48

2.7 1.7

5.4 2 x  40

2 x  2 x 3  18

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1.9

114

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2ЁЭСе+1 тИТ 2ЁЭСе = 4

1.10

2ЁЭСе + 2ЁЭСетИТ2 = 5

22ЁЭСе+1 тИТ 22ЁЭСе = 2

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1.12

115

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3 2 x ď&#x20AC; 5.3 x ď&#x20AC;Ť 6 ď&#x20AC;˝ 0

22đ?&#x2018;Ľ â&#x2C6;&#x2019; 2. 2đ?&#x2018;Ľ + 1 = 0

4.22đ?&#x2018;Ľ â&#x2C6;&#x2019; 5. 2đ?&#x2018;Ľ + 1 = 0

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9. 32ЁЭСе тИТ 10. 3ЁЭСе + 1 = 0

1.15

1.16

ЁЭСе 3 тИТ 2ЁЭСе 3 + 1 = 0

1.17 ЁЭСе 2 тИТ 5ЁЭСе 4 + 6 = 0 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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117

Equations involving surds: Example: 1. â&#x2C6;&#x161;đ?&#x2018;Ľ â&#x2C6;&#x2019; 2 = 3 đ?&#x2018;Ľâ&#x2C6;&#x2019;2=9 đ?&#x2018;Ľ = 11

Isolate the surd by itself and then square both sides & simplify. Test the answers to get correct solution â&#x2C6;&#x161;2 â&#x2C6;&#x2019; đ?&#x2018;Ľ = đ?&#x2018;Ľ + 4 2 â&#x2C6;&#x2019; đ?&#x2018;Ľ = đ?&#x2018;Ľ 2 + 8đ?&#x2018;Ľ + 16 đ?&#x2018;Ľ 2 + 9đ?&#x2018;Ľ + 14 = 0 (đ?&#x2018;Ľ + 2)(đ?&#x2018;Ľ + 7) = 0

Square both sides & simplify

đ?&#x2018;Ľ = â&#x2C6;&#x2019;2 đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;Ľ â&#x2030; â&#x2C6;&#x2019;7 Test: đ?&#x2018;Ľ == â&#x2C6;&#x2019;2 â&#x2C6;ś đ??żđ??ťđ?&#x2018;&#x2020; = 2 đ?&#x2018;&#x2026;đ??ťđ?&#x2018;&#x2020; = â&#x2C6;&#x2019;2 đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018;Ľ == â&#x2C6;&#x2019;7 â&#x2C6;ś đ??żđ??ťđ?&#x2018;&#x2020; = 3 đ?&#x2018;&#x2026;đ??ťđ?&#x2018;&#x2020; â&#x2C6;&#x2019;

3 đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;Ą đ?&#x2018;&#x17D; đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; Exercise 9.4: 1. â&#x2C6;&#x161;đ?&#x2018;Ľ + 4 = 7 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ ________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2. â&#x2C6;&#x161;3đ?&#x2018;Ľ â&#x2C6;&#x2019; 8 = 15 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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3. 9 â&#x2C6;&#x2019; â&#x2C6;&#x161;3 â&#x2C6;&#x2019; đ?&#x2018;Ľ = 7 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 4. â&#x2C6;&#x161;đ?&#x2018;Ľ + 6 = đ?&#x2018;Ľ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ 5. â&#x2C6;&#x161;2 â&#x2C6;&#x2019; đ?&#x2018;Ľ = đ?&#x2018;Ľ + 4 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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119

SURDS 3

Irrational numbers such as â&#x2C6;&#x161;10 ; â&#x2C6;&#x161;5 ; â&#x2C6;&#x161;17 ; â&#x20AC;Śâ&#x20AC;Śetc, are called surds. Laws of Surds: NB: Surds must always be of the same order to be able t work with them. đ?&#x2018;&#x203A; đ?&#x2018;&#x203A; đ?&#x2018;&#x203A; 1. Multiplying surds of the same order: â&#x2C6;&#x161;đ?&#x2018;&#x17D; Ă&#x2014; â&#x2C6;&#x161;đ?&#x2018;? = â&#x2C6;&#x161;đ?&#x2018;&#x17D;đ?&#x2018;? If you multiply surds of the same order you simply write their product in one root form... đ?&#x2018;&#x2019;. đ?&#x2018;&#x201D;. 1. â&#x2C6;&#x161;4 Ă&#x2014; â&#x2C6;&#x161;9 = â&#x2C6;&#x161;4 Ă&#x2014; 9 = â&#x2C6;&#x161;36 = 6 2. â&#x2C6;&#x161;2 Ă&#x2014; â&#x2C6;&#x161;7 = â&#x2C6;&#x161;2 Ă&#x2014; 7 = â&#x2C6;&#x161;14 đ?&#x2019;?

2. Dividing surds of the same order. 3

2.1

â&#x2C6;&#x161;27 3

â&#x2C6;&#x161;3

2.2

â&#x2C6;&#x161;64 4

â&#x2C6;&#x161;4

â&#x2C6;&#x161;đ?&#x2019;&#x201A; â&#x2C6;&#x161;đ?&#x2019;&#x192;

đ?&#x2019;?

đ?&#x2019;&#x201A;

= â&#x2C6;&#x161; . đ?&#x2019;&#x192;

= â&#x2C6;&#x161; 3 = â&#x2C6;&#x161;9 = â&#x2C6;&#x161; 4 = â&#x2C6;&#x161;16 = 2

3. Mixed Surds & Entire Surds. â&#x2C6;&#x161;12 Is an entire surd . 2â&#x2C6;&#x161;3 Is a mixed surd. To convert from entire to mixed either: 1. Write in factorized form and simplify. â&#x2C6;&#x161;80 = â&#x2C6;&#x161;24 Ă&#x2014; 5 = 22 â&#x2C6;&#x161;5 = 4â&#x2C6;&#x161;5 OR 2. Find the highest possible perfect square that divides into the given surd and Then simplify further. â&#x2C6;&#x161;80 = â&#x2C6;&#x161;16 Ă&#x2014; 5 = 4â&#x2C6;&#x161;5 Reverse the procedure converting from mixed to entire surds. 4. Addition and Subtraction of Surds: In algebra one can only add or subtract LIKE TERMS i.e. 3a + 5a = 8a Surds work in the same way: The surds must be of the same order and have the same value under the root sign. e.g. 1. â&#x2C6;&#x161;7 + â&#x2C6;&#x161;7 = 2â&#x2C6;&#x161;7 2. 5â&#x2C6;&#x161;2 â&#x2C6;&#x2019; 2â&#x2C6;&#x161;2 = 3â&#x2C6;&#x161;2 3 3 3 3. 4â&#x2C6;&#x161;2 + 6â&#x2C6;&#x161;2 = 10 â&#x2C6;&#x161;2 5. Rationalizing the denominators: 1

Irrational denominator: â&#x2C6;&#x161;3 To rationalize the denominator, means one must eliminate any root signs in the denominator.

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120

1. If there is only one term in the denominator then one must multiply the numerator and denominator by the value in the denominator. 1

√3

e.g. →. × = √3 √3 we are

√3 3

[we are not changing the value of the fraction as

Multiplying by 1] →

2√3 √5

2√3

√5

√5 √5

2√15

2. If the denominator is a binomial then the same procedure is used as above HOWEVER use the binomial in the denominator with the opposite sign between the two terms as depicted in the original expression. 3

e.g.2+√3

2(2−√3)

2−√3

× 2−√3 = (2+ 2+√3

= )

√3)(2−√3

4−2√3 4−3

= 4 − 2√3

6. Expressing surds as surds of the same order: First convert the surds to exponential form and then express the indices as equivalent fractions (the same value in the denominators) 1

e.g.

= 56 = √53

√5 = 52 1

= √125

√4 = 43 = 46 = √42 = √16 It can also be seen that √5 is greater than

√4

Exercise 9.5: Surds Worksheet: 1. Simplify: 1.1

√20

1.2

√18

1.3

√245

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1.4

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121 3

√24

√13 5

1.5

1.6

√96

2.2.

√3 + 3√3 − 2√3

2.4

3√8 + 5√50 − 4√32

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2.5

GOMATH WORKBOOKS

122 √245 + 6√5

2.6

√54 − √16

2√8 + 4√32 − 3√50

2.8.

√12−√75+2√3 √3

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3.2

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√3(√6 − √3)

2√5(3√5 − 2√2)

3√2(2√8 − √18)

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3.6 (3√2 + √5)(3√2 − √5) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2

3.7 (√3 + √2) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2

3.8 (√6 + 2√3) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 4. Rationalize the denominators of the following: 3 4.1 √3 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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5√2

4.2 √10 __________________________________________________________________ __________________________________________________________________ ___________________________________________________________ ___________________________________________________________ __________________________________________________________________ ___________________________________________________________ __________________________________________________________________ 4.3

6√18 3√12

4.4 3√12 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2

4.5 2−√3 __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________

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4.6 3+√2 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2−√3

4.7 2+√3 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ √7−3√2

4.8 √7−√2 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2√5+√3

4.9 5√3−3√5 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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127

10. Number Patterns: Number patterns are limited to a first order difference and are basically linear in character. General Formula is: Tn  an  c where d1  a  T2  T1  common difference Tn  an  c General formula for any sequence. To calculate the components of a sequence and its nth formula: Method A: 1. Given a sequence as follows: 4;7;10..... Tn  an  c T1  4

d1  a  3

T2  4  3  7

T3  10  3(3)  c c 1

T3  4  3  3  10

Nth formula for the sequence is Tn  3n  1 General formula for the specific sequence: 4;7;10..... Tn  3n  1 th

The 10 term of the sequence is:

T10  3(10)  1 T10  31

Given the 1st term and the common difference: T1  5 And d1  a  6 T1  5 T2  5  6  11 T3  5  6  6  17 T3  17  6(3)  c c  1 Tn  6n  1 Tn  6n  1

The 10th term of the sequence is: T10  6(10)  1 T10  59 3.

Given the sequence below, find an expression for the general term. 1;6;11;16;21........

d1  a  6  1  5 T4  16  5( 4)  c c  16  20  4 Tn  5n  4

The General term is Tn  5n  4

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In the sequence below, find the common difference and the general (nth) term. 4; 9; 14 ; 19â&#x20AC;Śâ&#x20AC;Ś The common difference is d1 ď&#x20AC;˝ T2 ď&#x20AC; T1 ď&#x20AC;˝ 9 ď&#x20AC; 4 ď&#x20AC;˝ 5 Next 3 terms are 24 ; 29 ; 34. Tn ď&#x20AC;˝ an ď&#x20AC;Ť c Tn ď&#x20AC;˝ an ď&#x20AC;Ť c T1 ď&#x20AC;˝ 4 T3 ď&#x20AC;˝ 5(3) ď&#x20AC;Ť c

d1 ď&#x20AC;˝ a ď&#x20AC;˝ 5

T2 ď&#x20AC;˝ 4 ď&#x20AC;Ť 5 ď&#x20AC;˝ 9 T3 ď&#x20AC;˝ 4 ď&#x20AC;Ť 5 ď&#x20AC;Ť 5 ď&#x20AC;˝ 14

14 ď&#x20AC;˝ 15 ď&#x20AC;Ť c c ď&#x20AC;˝ ď&#x20AC;1

Tn ď&#x20AC;˝ 5n ď&#x20AC; 1

Method B: Use notation: đ?&#x2018;&#x2021;đ?&#x2018;&#x203A; = đ?&#x2018;&#x17D;đ?&#x2018;&#x203A; + đ?&#x2018;? Where: đ?&#x2018;&#x17D; = đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x161;đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018;&#x2018;đ?&#x2018;&#x2013;đ?&#x2018;&#x201C;đ?&#x2018;&#x201C;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;? = đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;Łđ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019; Example: Given sequence 1; 9 ; 14 ; 19 ; 24 ; â&#x20AC;Śâ&#x20AC;Ś. Find the common difference and a formula for the đ?&#x2018;&#x203A;đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;. Write down values for â&#x20AC;&#x153;nâ&#x20AC;? : 1 2 3 4 Write down given sequence: 4 9 14 19 Write down 6n: 5 10 15 20

5 Line 1 24 Line 2 25 Line 3

Subtract the last numbers of Line 3 from Line 2 to get â&#x20AC;&#x153;câ&#x20AC;?: 24 â&#x20AC;&#x201C; 25 = -1 Common difference : a = đ?&#x2018;ťđ?&#x;? â&#x2C6;&#x2019; đ?&#x2018;ťđ?&#x;? đ?&#x2019;?đ?&#x2019;&#x201C; đ?&#x2018;ťđ?&#x;&#x2018; â&#x2C6;&#x2019; đ?&#x2018;ťđ?&#x;? đ?&#x2019;&#x2020;đ?&#x2019;&#x2022;đ?&#x2019;&#x201E; = 5 General (đ?&#x2018;&#x203A;đ?&#x2018;Ąâ&#x201E;&#x17D; ) term for the sequence is đ?&#x2018;ťđ?&#x2019;? = đ?&#x;&#x201C;đ?&#x2019;? â&#x2C6;&#x2019; đ?&#x;? Exercise 10.1: 1. Mina invests R8000,00 in a plan where the growth is as follows: 0 1 2 3 4 5 8000 8440 8880 9320

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

1. Find the rule for this investment plan. 1.1 Use the rule to find the values for the next 3 years. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 1.2

Use the rule to find the value after 23 years.

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1.3

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If the value after n years was R13 280 , what is the value of n?

A class is given the sequence 3; 5; 7;….. to continue for 4 more terms. Explain in a sentence how you would complete it.

Find a rule in the form of Tn  an  c for the given sequence.

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130

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Find the 20th term using the formula in 2.2

2.3

3. 3.1

Find the 50th term

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131

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4. Find a rule or general term for each of the following sequences and then find the 10th term for each one. 4.1 1; 6; 11; 16; 21;…. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 4.2 13; 23; 33; 43;…. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 4.3 6; 9; 12; 15;…. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 4.4 22; 20; 18; 16;…. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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4.5 7; 2; -3 ;-8;….. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 4.6 7; 11; 15; 19;…. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5. Use the general terms below to find the first 5 terms of each sequence and then find the 20th term for each one. 5.1 Tn = n – 5 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5.2

Tn = 3n – 5

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Tn = -2n – 5

Tn = 4n + 2

Tn = 2 – 3n

The following sequences are given: A. For each one write down the next 3 terms B. The nth term ( i.e. Tn = an + c) C. The 50th term. 6.1

3; 9; 15; 21;…..

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2; 9; 16 ; 23; …..

6.3

2; -3; -8; -13;….

3; -1; -5; -9;…

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20; 27; 34; 41;…….

6.5



Find the rule for the nth term Tn



Find the value of the 50th term.



Graph the relationship.

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2. 



 



 



 



     

 

 





       ___________________________________________________________________ 



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4. From the diagram below: 4.1 4.2 4.3

fig1

Determine the next 3 terms. Determine the nth term formula Determine the 20th term.

Fig 2

Fig 3

Fig 4

From the diagram below: 5.1 5.2 5.3

Determine the next 3 terms. Determine the nth term formula Determine the 20th term.

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From the diadram below: 6.1 6.2 6.3

Determine the next 3 terms. Determine the nth term formula Determine the 20th term.

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Number Patterns Continued: 3 basic types of patterns dealt with 1.

Arithmetic Sequences (Linear) : General Term : Tn ď&#x20AC;˝ a ď&#x20AC;Ť (n ď&#x20AC; 1)d These have a common difference: A 1st order sequence : d1 ď&#x20AC;˝ď&#x20AC;˝ a

Geometric Sequence (Exponential) : General Term is: Tn ď&#x20AC;˝ ar nď&#x20AC;1 These have a common ratio.

Quadratic Sequences: General Term is: Tn ď&#x20AC;˝ an2 ď&#x20AC;Ť bn ď&#x20AC;Ť c A 2nd order sequence. d 2 ď&#x20AC;˝ 2a (n ď&#x20AC; 1)(n ď&#x20AC; 2) ď&#x201A;´ d2 Or Tn ď&#x20AC;˝ (n ď&#x20AC; 1)T2 ď&#x20AC; (n ď&#x20AC; 2)T1 ď&#x20AC;Ť 2 NB this sequence has no constant relationship between consecutive terms. However there is a constant relationship at a higher level (2nd level) In quadratic sequences there are two levels of differences: First Difference i.e. subtract T2 ď&#x20AC; T1 Second Difference i.e. subtract the 2nd tier of values. Example: 2

2nd Tier( level) or 2nd difference 1st Tier (level) or 1st difference Sequence

3 5 7 9 1 ; 4 ; 9 ; 16 ; 25

The first difference will be an arithmetic sequence. Example : 1. Given the number pattern 1 ; 4 ; 9 ; 16 ; 25 Find an expression for the nth term. Method A: Formula Method ; đ?&#x2018;&#x2021;đ?&#x2018;&#x203A; = đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;2 + đ?&#x2018;?đ?&#x2018;&#x203A; + đ?&#x2018;? Let 2a ď&#x20AC;˝ d 2 and 3a ď&#x20AC;Ť b ď&#x20AC;˝ T2 ď&#x20AC; T1 and a ď&#x20AC;Ť b ď&#x20AC;Ť c ď&#x20AC;˝ T1 SEQUENCE 1 ; 4 ; 9 ; 16 ; 25 1ST DIFF 3 5 7 9 2ND DIFF 2 2 2 2a ď&#x20AC;˝ d 2

3a ď&#x20AC;Ť b ď&#x20AC;˝ T2 ď&#x20AC; T1

a ď&#x20AC;Ť b ď&#x20AC;Ť c ď&#x20AC;˝ T1

2a ď&#x20AC;˝ 2

3(1) ď&#x20AC;Ť b ď&#x20AC;˝ 3

1ď&#x20AC;Ť 0 ď&#x20AC;Ť c ď&#x20AC;˝ 1

a ď&#x20AC;˝1

bď&#x20AC;˝0

cď&#x20AC;˝0 Tn ď&#x20AC;˝ n 2

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Method B: Table Method: General Formula is đ?&#x2018;&#x2021;đ?&#x2018;&#x203A; (đ?&#x2018;&#x2C6;đ?&#x2018;&#x203A; ) = đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;2 + đ?&#x2018;?đ?&#x2018;&#x203A; + đ?&#x2018;? Table: n 1 2 0 6 17 -1 Tn đ?&#x2018;&#x2018;1

7 4 đ?&#x2018;&#x2018;2 Work back from đ?&#x2018;&#x2021;1 to calculate đ?&#x2018;&#x2021;0 1 đ?&#x2018;&#x17D; = 2 đ?&#x2018;&#x2018;2 = 2 and đ?&#x2018;? = đ?&#x2018;&#x2021;0 = â&#x2C6;&#x2019;1 đ?&#x2018;&#x2021;đ?&#x2018;&#x203A; (đ?&#x2018;&#x2C6;đ?&#x2018;&#x203A; ) = 2đ?&#x2018;&#x203A;2 + đ?&#x2018;?đ?&#x2018;&#x203A; â&#x2C6;&#x2019; 1 đ?&#x2018;&#x2021;1 (đ?&#x2018;&#x2C6;1 ) = 2(1)2 + đ?&#x2018;?(1) â&#x2C6;&#x2019; 1 = 6 đ?&#x2018;?=5 đ?&#x2018;ťđ?&#x2019;? (đ?&#x2018;źđ?&#x2019;? ) = đ?&#x;?đ?&#x2019;?đ?&#x;? + đ?&#x;&#x201C;đ?&#x2019;? â&#x2C6;&#x2019; đ?&#x;?

3 32 15

4 51 19

5 74 23

Refer to Method A: 2a ď&#x20AC;˝ d 2

3a ď&#x20AC;Ť b ď&#x20AC;˝ T2 ď&#x20AC; T1

a ď&#x20AC;Ť b ď&#x20AC;Ť c ď&#x20AC;˝ T1

2a ď&#x20AC;˝ 4

3(2) ď&#x20AC;Ť b ď&#x20AC;˝ 11

2ď&#x20AC;Ť5ď&#x20AC;Ťc ď&#x20AC;˝ 6

aď&#x20AC;˝2

bď&#x20AC;˝5

c ď&#x20AC;˝ ď&#x20AC;1 U n ď&#x20AC;˝ 2n 2 ď&#x20AC;Ť 5n ď&#x20AC; 1

Exercise 10.3: A. B. C.

Find the next two terms in the sequences below. Find an expression for the nth term of each sequence. Find the 10th term of each sequence.

1.1

1 ;3;6.;10;... _______________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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15;30;48;69;...... __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.3

 12;7;1;12;..... __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.4

27;31;37;........ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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12; 17; 24…… __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.6

-13; -4; 8….. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.7 25; 30; 39…….. ____________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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1.8 7; 12; 21…….. ____________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ Exercise 10.4: Write down the next 3 numbers in each of these sequences. Explain the rule in words. 1.1

5; 11; 17; 23 ____________________________________________________

_____________________________________________________________________ 1.2

1; 4; 9; 16; 25___________________________________________________

_____________________________________________________________________ 1.3

10; 5; 0; -5 _______________________________________________________

_____________________________________________________________________

2.1

Look at the pattern below Draw the next pattern.

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2.2 The table below shows the shape number and the sequence of total number of black dots in each shape. Complete the table. Shape number

Number of black dots

3 4 8 10 N Hint: ‘n’ represents a general equation for the nth term of a sequence 2.3

Draw a Graph representing the data in the table in question 2.2 Y

Look at the sequence and the table below: 2; 6; 12; 20; 30; …………. Terms 1st 2nd 3rd 4th 5th 6th 7th 8th

Value of term 2 6 12 20 30 42 56 72

4 6 8 10 12 14 16

2 2 2 2 2 2

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145

How were the answers in the third column obtained? ________________________________________________________________

Value of term 5

1st Difference

2nd Difference

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Question 5: You are given a sequence where the first term is ‘a’ and the constant ratio is ‘r’. The first for terms are written down in general form as : a ; ar ; ar2 ; ar3 5.1

What is the 5th term of this sequence?

______________________________________________________________ 5.2 What is the 20th term of the sequence?

5.2

_______________________________________________________________ Now find an equation that will give the nth term of this sequence: i.e. Tn  .......... _______________________________________________________________ _______________________________________________________________

Exercise 10.5:

Look at each of the following sequences and: A. Find the next 3 terms. B. Find an equation that will give the nth term of the sequence... C. Find the 20th term. 1. 0; 2; 6; 12;……..; __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2. 2; -1; -6; -13;…… __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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3. 5; -1; -7;……. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

4. 8; -4; 2. -----; __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 5. 3; 8; 15;…… __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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Exercise 10.6: 1. The sixth term of an arithmetic sequence is 17 and the tenth term is 33. Determine the first term and the common difference. __________________________________________________________________ ________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2. X; 2x + 1; 11 are three consecutive terms of an arithmetic sequence. Calculate: 2.1 x

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3. The first term of an arithmetic sequence is −3 and the third term is 3. Determine: 3.1 the value of the 25th term of the sequence.

GEOMETRIC PROGRESSIONS: A geometric sequence is a string of values which increase or decrease when multiplied by a constant value. This value is referred to as the “common ratio” (r). The first term of the sequence is referred to as “a”. The last term (nth) in the sequence is referred to as Tn. (N.B. There are an infinite number of terms, however you choose which one you want to use as the last term) The formula for the nth (last) term ( Tn) in a sequence is formulated as follows. If a sequence of numbers is : 7 ; 14 ; 28 ; 56 ; …..(to the nth term). T1 = 7 T2 = 7 .2 = 14 T3 = 7 .2.2 = 28 T4 = 7.2.2.2 = 56 T5 = 7 .24 = 112 T10  7.2 9  3584

second last term Tn  ar n1

last term

first term

common ratio

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Calculations in Geometric Progressions (GP’s) 1.

If a sequence of numbers is : 7 ; 14 ; 28 ; 56 ; …..(to the nth term).

1.1

Find the 20th term in the sequence. a = 7; r = 2; n = 20 ;T20 = ? Tn  ar n1 T20 = 7.219 = 3670016 3670016 is the 20th term in the sequence.

1.2

If 3584 is the nth term in the sequence find out the number of terms (n). Tn = 3584 ; a = 7 ; r = 2 ; n = ? Tn  ar n1 3584 = 7.2n-1 512 = 2n-1 29 = 2n-1 n–1=9 n = 10 3584 is the 10th term in the sequence.

1.3

If 3584 is the 10th term in the sequence find the first term. Tn = 3584 ; ; r = 2 ; n = 10 ; a = ? Tn  ar n1 3584 = a .29 3584 a= 512 a=7

7 is the first term in the sequence. 1.4.

If 3584 is the 10th term in the sequence and 7 is the first term find the common ratio. Tn = 84 ; ; n = 10 ; a = 7 ; r = ? Tn  ar n1 3584 = 7.r9 512 = r9 29 = r9 r=2

the common ratio is 2

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Simultaneous Questions Given two terms calculate the first 3 terms of the sequence Example: Write the term that comes later in the progression first when setting the simultaneous equations up. i.e. T10 on top and T5 on the bottom. If T10  1536 and T5  48 Calculate the first 3 terms of the sequence: Divide to eliminate the “a” value and solve for ‘r’ NB subtract the exponents when dividing.

T10  ar 9  1536 T5  ar 4  48 r 5  32 r2

Sequence : 3 ; 6 ; 12; ….

ar 4  48 48 a 16 a3

Substitute “r” into either one of the equations to solve for “a’

Finding a general formula that satisfies the nth term of a given sequence. I.e. you must be given or can calculate the IST term and the common difference. Example: If the first three terms of a Geometric sequence is 3x  1 ; 2x  3 ; 2x  1 …… NB the constant concept is the common ratio. T T Thus 2 = 3 T1 T2 2x  3 2x  1  3x  1 2 x  3 (2 x  3) 2  (3x  1)(2 x  1)

4 x 2  12x  3  6 x 2  5 x  1 Tn = 23 –4d This is the general term representing the nth term of this specific sequence.

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Exercise 10.7: 1. 2 ; 6 ; 18 ; 54 ; â&#x20AC;Ś. Is a geometric sequence. 1.1 Continue the sequence to the 6th term. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ Find the 20th term. __________________________________________________________________

1.2

__________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2. Determine the 2nd and 3rd terms in the following sequence given that đ?&#x2018;&#x2021;1 = 5 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x2021;4 = 40 . __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 3. In the following geometric sequences: 3.1 đ?&#x2018;&#x17D; = 2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x; = 3, đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019; 5đ?&#x2018;Ąâ&#x201E;&#x17D; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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𝑎 = 1 𝑎𝑛𝑑 𝑟 = 2 , 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 6𝑡ℎ 𝑡𝑒𝑟𝑚. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

3.3

𝑎 = 11 𝑎𝑛𝑑 𝑟 = − 3 , 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 4𝑡ℎ 𝑡𝑒𝑟𝑚.

4.2

𝑡ℎ𝑒 5𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠

7 81

𝑎𝑛𝑑 𝑎 = 7. 𝑓𝑖𝑛𝑑 𝑟.

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𝑡ℎ𝑒 7𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠 192 𝑎𝑛𝑑 𝑡ℎ𝑒 2𝑛𝑑 𝑖𝑠 6, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠.

4.4

𝑡ℎ𝑒 6𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠 − 6 𝑎𝑛𝑑 𝑡ℎ𝑒 9𝑡ℎ 𝑡𝑒𝑟𝑚 𝑖𝑠 32

6 256

, 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠

5.2

Find the first term.

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Find the 10th term. __________________________________________________________________

5.3

Exercise 10.8: 1.

Consider the series of numbers below. You must assume that the number of terms listed will be sufficient to conclude a rule for each sequence. Series A: 1 ; 4 ; 9 ; 16 ; 25 ; 36 ; 49 ; ------------------1 1 1 1 1 Series B: 1 ; ; ; ; ; ;-------------------2 3 4 5 6 1.1

State your observations concerning the behavior of series A:

____________________________________________________ _____________________________________________________________________ 1.2

State your observations concerning the behavior of series B:

_______________________________________________________ _____________________________________________________________________1 .3

What will the tenth term ( 10th Term) in series A be?

_______________________________________________________ _____________________________________________________________________1 .4

What will the 20th term be in series B?

____________________________________________________________________ _____________________________________________________________________

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1.5 If ‘k’ represents the position of a term ( eg. Position 1 will be represents the position of a term ( eg. Position 1 will be k = 1) Write down the general rule, in an equation form, for series A and B, respectively. Series A:_________________________________________________

Series B:_________________________________________________

Find the sum ( addition) of the first 4 terms in series A. ________________________________________________________________.

Find the sum of the first 3 terms in series B. _________________________________________________________________.

Consider the table below Terms 1st 2nd 3rd 4th 5th

Value of term -5 2 11 22 35

2.1

Complete the table.

2.2

What can you conclude about the second difference?

_____________________________________________________________________ 2.3

What can you deduce about all the values that you calculated in the column denoted “first difference”?

_____________________________________________________________________ 2.4 Determine the 10th term in the sequence (the number occupying position ten of the sequence) _____________________________________________________________________ _____________________________________________________________________

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Financial Maths: Calculated Annually: Year 9/10

No business can exist without the information given by figures. Borrowing, Using and making money is the heart of the commercial world thus the Principle of interest and interest rate calculations are extremely important. This leads into an examination of the principals involved in assessing the Value of money over time and how this Information can be utilized in the Evaluation of alternate financial decisions. Remember that the financial decision area is a minefield in the real world, Full of tax implications, depreciation allowances, investment and capital Allowances etc. The basic principles in financial decision making are established through The concept of interest and present value: –

Definition of interest: Interest is the price paid for the use of borrowed money Interest is paid by the user of the money to the supplier of it. It is calculated as a fraction of the amount borrowed or saved over a certain period of time. This fraction is also known as interest rate and is expressed as a percentage per year (per annum).

Simple interest is computed on the principle for the entire term of the loan and is thus due at the end of term. I = Prt I is the interest paid or earned P is the principle or Present value r is the interest rate per annum t is the time or term of loan

A  P(1  in) A  ( Amount) New Price per time period P  Original Price(Cost)(Principle) r i  Interest rate used for growth : i  100 n  the relevanttime period. P

A (1  in)

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Simple Interest ( Growth) Simple Interest:(Growth) NB interest is calculated on the original investment for each time period. i.e. The interest earned is a constant. Formula:

A  P (1  ni) Where:

A  FutureValue (Future amount.) P  present Value (original amount.) r ) 100 n  Number of interest periods.

i  interest level(i 

(When using i use the decimal of the interest rate quoted (or use

r for i) 100

Example 1: A Principle of £300,00 is invested in a Bank at 10% Simple Interest for a number of years; Complete the table below: Principle Interest period(yrs) Simple Interest (%) Interest

300 1

300 2

300 3

300 4

300 5

300 6

300 N

120

150

180

300 100in

Amount

330

360

390

420

450

480

200 

300 100in

Example 2:Invest £300,00 @ 15% per annum( per year) Year 0

£300

Simple Interest p.a

Year 1

£345

Year 2

£390

Year 3

£435

Year 4

£480

Year 5

£525

The value of the investment increases at a constant value of £45 per year. The relevant graph for simple growth is a straight line:

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Exercise 11.1: Question 1: Peter deposits ÂŁ8500, 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. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 1.2

If he had received an interest rate of 20%, how much more would he have earned in interest?

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Question 2: John invests £20 000,00 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? ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ Question 3: 3.1

Mrs. Vezi invests £10 000,00 at 12,5% simple interest per annum.

How much interest will she earn after 5 years?

How much interest will she earn after 10 years?

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What is the final amount in her account after 15 years if she does not have any withdrawals?

Calculate the Principle if James received an amount of £7500, 00 earned in Simple interest at a rate of 10% over 5 years.

Calculate the Principle if James received an amount of £37500, 00 earned in Simple interest at a rate of 15% over 10 years.

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Question 5: A loan of £100 is made at 10% simple interest p.a. 1.21

A line graph is used to display the amount at the end of each year. Draw this graph on the grid below. y

200 190 180 170

A m o u n t

160 150 140 130 120 110 100

Years

5.2

What do you notice about the increase every year? How is this displayed in the graph? ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5.3

If £100 is borrowed at 5% p.a. Simple Interest for 4 years display this on the same graph and show clearly how the two graphs differ from one another. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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5.4

If £100 is borrowed at 15% pea simple interest for 4 years explain without drawing the graph how the graph would differ from the other two (2). ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ COMPOUND INTEREST Compound interest arises when, in a transaction over an Extended period of time, interest due at the end of a payment Period is not paid, but added to the principal. Interest also Earns interest i.e. it is compounded. The amount due at the end of transaction period is referred to As the compounded amount or accrued principal. Interest periods Can vary: daily, monthly, quarterly, half-yearly or yearly. FOR GRADE 10 PURPOSES INTEREST IS ONLY CALCULATED YEARLY

Formula: Compound Growth: A  P (1  i ) n

A = Amount or Future Value P = Principal or Initial value i rate of interest per annum i 

r 100

n = number of years invested Compound Interest (Growth): Interest is calculated on the new amount each time period. I.e. Earn interest on interest, thus interest increases each year.

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Example 1: Year 0

£300

Compound Interest p.a.

Year 1

£345

Year 2

£396,75

51,75

Year 3

£456.26

59,51

Year 4

£524,70

68,44

Year 5

£603,41

78,71

The amount increases at a constant interest rate however it is calculated on the increasing amount each time period. The relevant graph for compound growth is curve shaped.

Example 2: Vusi invests £1000, 00 at 12% compound interest for 5 years: 2.1 Write down a formula for the compound interest to be calculated over n yrs. at r % p.a. n

r   Answer: A  P 1   or  100 

A  P1  i n

2.2 Use the formula to calculate the final amount earned after the 5 years to the nearest cent.

A  P1  i n A  10001  0,125 A  £ 1762,34

Answer: 2.3 How much interest was earned on his investment: i  A P Answer: i  1762,34  1000 Ci  £ 762,34

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2.4 Calculate what he would have earned in simple interest for the same time period: Answer:

Pr t 100 1000  12  5 Si  100 Si  £ 600. It can be deduced that it is more beneficial to invest money at compound Si 

interest: 2.

If the number of interest periods are increased then it can be deduced that the interest earned would increase accordingly:

I.e Calculating interest: 1. annually 2. Half- yearly 3. Quarterly 4. Monthly 5. Daily.

Example: £1000 invested at compound interest for 5 years at 12% p.a. Principle

Interest

A  P (1 

r tm ) 100m

Time Period = 5yrs Yearly

Half-Yearly

Monthly

Daily

m =4

m = 12

m= 365,25

m= no of time periods p.a.

m =1

£1000

£1762,34

£1790,85

£1806,11

£1816,70

R1821,94

£762,34

£790,85

£806,11

£816,70

£821,94

12%

Compound Interest

m =2

Quarterly

Ci = A – P It can be deduced from the table above that the more time periods involved in the calculation of compound interest the better the return on investment: The majority of Banks and investment companies employ daily interest calculations for principles invested with them.

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Exercise 11.2: 1. Charlie wants to invest £7000,00 : There are 2 choices for him: What would the better choice be? Show all your working. 1.1

Invest £7000,00 at 12% simple interest per annum over 5 years.

Invest £7000,00 at 12% compound interest over 5 years.

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3. £20 000,00 is invested for 15 years at 12% p.a. Calculate the value of the investment if the interest is calculated: 3.1

At simple interest.

At compound interest.

Calculate a simple interest rate that would yield the same return.

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

simple interest.

compound interest.

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6. The Ndlovu family uses a loan of £7200, 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?

13% p.a. compound interest?

Compound interest A  P(1  i) n

End of 1st year

End of 2nd year

End of 3rd year

End of 4th year

End of 5th year

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8. Use a table to draw two graphs on the same system of axes showing the difference between simple and compound interest. Use a system of axes below for your graphs.

y 2100 A m o u n t

2000 1900 1800 1700 1600

i n R a n d s

1500 1400 1300 1200 1100 1000 5

no of years

8.1

What kind of graph does the simple interest formula produce?

Find the gradient of the line.

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8.3

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

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

€

R 9.9837

NZ$

USD

0.7747 0.7157

98.1100

1.5253 1.9117

ZAR

0.1002 1

0.0776 0.0717

9.8270

0.1528 0.1915

EUR

1.2909 12.8880 1

GBP

1.3972 13.9492 1.0823 1

137.0793 2.1312 2.6710

JPY

0.0102 0.1018

0.0079 0.0073

0.0155 0.0195

AUD

0.6556 6.5453

0.5079 0.4692

64.3209

NZD

0.5231 5.2225

0.4052 0.3744

51.3213

0.7979 1

0.9239

126.6502 1.9690 2.4678

1.2533

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Examples : R1000 will buy 1.1

$100 (to nearest dollar)

1.2

£72(to nearest pound)

1.3

€78(to nearest Euro)

1.4

¥9827(to nearest Yen)

Exercise 11.3: 1. Copy and complete the following currency table: CURRENCY

USD

ZAR

0.133 1

EUR 2

JPY

7,5

GBP

€

NZD

NZ$

0.0666

300

0.1428

0.933 7

150

O,05

AUD

1 0.333 0.01

0,0105 1

0,3335

Use the table above for questions 2 to 4. 2

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. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3

What will it cost you in South African Rand to purchase the following currencies:

3.1

$800.

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3.2 ¥2000. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3.3 A$500. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 3.4 €2000. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3.5 R5000. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ How many South African Rand can be exchanged for the following? 3.6 $200 ____________________________________________________________________ ___________________________________________________________________ 3.7 ¥1500 ____________________________________________________________________ ___________________________________________________________________

3.8 A$150 __________________________________________________________________ ___________________________________________________________________

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3.9 €350. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

3.10

£750.

Cost of Hire Purchase: A hire purchase agreement is a short - term loan it is calculated using simple interest The lending company also adds insurance to the cost to cover the loan amount. Example: Andy buys a plasma TV set for R12000, 00. He buys the TV using hire purchase agreement involving monthly payments over 3 yrs. The simple interest rate is 14% p.a. calculate Andy’s monthly installment if a monthly insurance premium of R10 is added. Solution:

A  P (1  ni)

A  12000[1  3(0,14)] A  R17040 Monthly installments are = R

17090  10 36

= R473.33 + R1 = R483 .33

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Exercise 11.4: 1.

A couple buy new furniture for their home. The furniture costs ÂŁ120 000,00 They pay for it using a hire purchase loan. The interest rate is 12% p.a. over a 5year period. Calculate the monthly repayment if there is no insurance added.

In South Africa Sihle buys a washing machine for R6500, 00. He pays 15% of the value in cash as a deposit and uses a hire purchase agreement to pay the balance over 36 months. The interest rate is 10% p.a. Calculate his monthly installments if an insurance premium of R6, 50 is added monthly.

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Mr Moosa sees an advert in the newspaper for a surround sound audio system. The advert is worded as follows: “Pay ONLY £299, 99 £7500.

per month for 36 months!” The cash price is

Assume that no insurance premium is added. Calculate: 3.1

How much interest is paid over 3 years?

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

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Ms Bokapane buys a washing machine from an appliance store. She asks 1 for a quote to buy the machine through a hire purchase contract over 1 2 years. 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

Monthly payment

R 273,60

1.1

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

1.2

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)

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Cost of Inflation Inflation is the continuous increase in the cost of goods and services over a period of time. The rate of inflation is given as percentage per annum. It is the average increase (as a percentage) in the cost of goods and services from one year to the next. The compound increase formula is used in inflation problems. Example: A car costs R80 000, 00, and the average rate of inflation is expected to be 8, 05% p.a. over the next 5 years. 1. Calculate the expected price of the same model of car in 5 years’ time, based on the stated inflation rate. 2. How much would the same model car cost 5 years ago, based on the same rate of inflation? Solution:

A  P(1  i ) n A  P (1  i ) n A  80000(1  0,0805)

A  R117818,00

80000  P(1  0,0805) 5 5

P

80000

(1  0,0805) 5 P  R54320,80

Exercise 11.5: 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. 1.1

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

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How much did the 2,5kg packet cost 10yrs ago (the rate of inflation continues at the same level.)

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 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.

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The current annual fees for a Bachelor of Commerce degree at UKZN are R20500, 00. Determine the expected cost of studying the same degree in yearsâ&#x20AC;&#x2122; time if the fees are to increase by 9% p.a. as a result of inflation. Give the answer correct to the nearest rand.

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TIME LINES USED FINANCIAL CALCULATIONS Time lines are useful when dealing with complicated problems, such as changes in the interest rate during an investment period or when several deposits or withdrawals are made from a savings account. It helps to summarize the information and give a visual representation of the data in an ordered manner. Example: R7000 is deposited into a savings account, and 4 years later another R5000 is added to the savings. Calculate the value of the savings at the end of 7 years if the interest rate is 12% p.a. for the first 3 years and then increased to 13,5% for the remaining period. Solution: T0

R7000

R5000 12% p.a.

13,5 % p.a.

T6 Etc. indicates the time period of the investment.

Balance after 3 years:

Balance after 4 years:

Balance after 7 years:

A  7000(1  0,12) 3 A  R9834,496

A  9834,496(1  0,135)  5000 A  R16162,152 A  16162,152(1  0,135) 3 A  R 23631,26(nearest cent)

n (years) Interest rates

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Example 2: In order to save for her sons University fees, Mrs. Gumede deposits £8000 into a savings account at the end of January when her son is 10 years old. The rate of interest is 14% p.a. compounded annually. When her son is 18yrs old he starts a University course which has a duration of 3 years. The first fees are £9000, payable at the end of January. The fees increase by 10% each year. Calculate: 1.

The second and third year’s fees.

The balance in the account after the first years fees have been paid.

The balance in the account after the second years fees are paid.

How much additional cash will be needed to cover the third year’s fees?

T10

n (years) £8000

(£9000) 14% p.a.compounded annually

NB the bracket around £9000 indicates a withdrawal. 2.

Second years fees ( fees at T9): 9000  (0.1)(9000)  £ 9900 Third years fees (fees at T10) 9900  (0.1)(9900)  £ 10890

Balance at T8: (to nearest cent) 8000(1  0. 14) 8  9000  £ 22820 .69 9000  £ 13820.69

Balance at T9: (to nearest cent) 13820,69(1  0,14)  9900  £ 5855.59

Money in savings account at T10: 5855.59(1 + 0.14) = £6675.37 Additional cash required: (to nearest cent) 10890  6675,37  £ 4214,63

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Exercise 11.6: 1. Jack deposits £1000 into a savings account. One year later he adds £2000 to the savings. At the end of the second year he deposits £4000 into the same account, and finally he adds £8000 to the savings account at the end of the 3rd year. Calculate the amount (A) in Jacks account at the end of the 4th year if the interest is calculated at 11, 5% compounded annually. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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2. £6500 is deposited into a savings account, and 3 years later £7400 is added to the savings. At the end of 5 years, £5800 is withdrawn from the account. How much money will be in the account at the end of 10 years if the interest rate is 11% p.a.? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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3. ÂŁ21000 is invested where the interest rate is 7, 5% p.a for the 1st 3 years. The rate is then then increased to 8, 25% p.a. for the next 4 years. Calculate the value of the investment at the end of the 7 years. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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4. Mrs. Smith invests £8000 for in a savings account when her two sons are 7 and 10 years old. She pays each of them £15000 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.

4.2

Is this fair on the sons? Explain your answer.

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5. Mr. Ndlovu places ÂŁ52000 in the bond market as a fixed saving for 12 years. The interest paid during first 5 years is 10, 5% p.a. It is then increased to 12 % p.a. for the next 3 years, and then finally increased to 14% for the last 4 years. In each case the interest compounded annually. Calculate how much Mr. Ndlovu will have in his savings account at the end of the 12 year period. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Compound Interest (Growth) including further interest periods: Year 10/11: Compound interest arises when, in a transaction over an Extended period of time, interest due at the end of a payment Period is not paid, but added to the principal. Thus interest also Earns interest i.e. it is compounded. The amount due at the end of transaction period is referred to As the compounded amount or accrued principal. Interest periods Can vary: daily, monthly, quarterly, half-yearly or yearly.

FORMULAE: NB SOME OF THE FORMULAE USED BELOW DO NOT APPEAR ON THE FORMULAE SHEET SUPPLIED EXTERNALLY: IT WOULD BE TO YOUR ADVANTAGE TO LEARN THEM. Simple Growth A  P (1  ni)

Simple Decay A  P (1  ni)

Compound Growth:

Compound Decay: A  P(1  i) n

A  P(1  i) n

OR J   A  P 1  m  m  

A P tm t m Jm

ONLY These 4 formulae appear on the official formulae sheet. sheet

This formula can be used instead of the compound growth one above. The symbols are explained below.

= Accrued amount / Future value [ S is also used instead of A] = Initial principle / present value = the annual interest rate compounded m times per year = the number of years of investment. = the number of compounded periods per year = the nominal annual interest rate. J m  i

NB: IT IS EASIER TO USE THE DECIMAL VERSION OF % FOR CALCULATIONS: Different Compounding Periods: Annum

 Calculated Once per year

Quartely

 Calculated 4 times per year

Half yearly

 Calculated 2 times per year

Monthly

 Calculated 12 times per year

Weekly

 Calculated 52 times per year

Daily

 Calculated 365 times per year

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Nominal Interest rates: 1.1.

In cases where interest is calculated once a Year, the annual rate quoted is the nominal annual rate Or simply referred to as nominal rate.

Effective Interest rates: 1.2. If the actual interest earned per year is calculated and expressed as a percentage of the relevant principal, then the so-called effective rate is obtained. The effective rate is the actual interest rate per annum taking the number of increased time periods. I.e. monthly; quarterly etc. You effectively earn more than the quoted nominal rate per annum. Converting Nominal Rate to Effective Rate: Method 1: Take the Nominal Rate and divide by the number of time periods involved and apply this to the formula: Eff Rate = [ 100(1  i) n -100] i

nominal rate and n = number of time periods in 1 year. time periods

EG The nominal rate of interest is 22% calculated half yearly. What is the corresponding effective rate of interest? 22  11 % Thus R100(1.11) 2  R123.21 effective Interest rate is 23,21% 2 OR J eff

J eff

m   jm   100 1    1 m   

J m  no min al rate m  number of time periods J eff  the effective rate

2   0.22   100 1    1 = 23,21% 2   

J   OR 1  i   1  m  m   0.22   1  i  1   2  

= 0,2321 = 23,21 %

Converting Nominal Rate to Effective Rate:

J eff

m   jm   100 1    1 m   

J   OR 1  i   1  m  m  

J m  no min al rate m  number of time periods

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Converting Effective to Nominal Rate Jm ď&#x20AC;˝ 100(m)(m 1 ď&#x20AC;Ť eff ď&#x20AC; 1)

Finding the Rate: ď&#x192;Ś a ď&#x192;ś r ď&#x20AC;˝ 100ď&#x192;§ď&#x192;§ n ď&#x20AC; 1ď&#x192;ˇď&#x192;ˇ ď&#x192;¨ p ď&#x192;¸

Further Formulae 1. Finding Principle:

Pv ď&#x20AC;˝ Fv (1 ď&#x20AC;Ť

j m ď&#x20AC;tm ) m

jm and n ď&#x20AC;˝ tm m đ??šđ?&#x2018;Ł = đ??šđ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;˘đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019; đ?&#x2018;&#x153;đ?&#x2018;&#x; đ??´đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;˘đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x192;đ?&#x2018;Ł = đ?&#x2018;&#x192;đ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x2030;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x2019; đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;&#x192;đ?&#x2018;&#x;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2013;đ?&#x2018;?đ?&#x2018;&#x2122;đ?&#x2018;&#x2019; Jm = nominal interest rate. m = no of interest periods involved. n = tm = total no of time periods. t = no of years invested. NB: i ď&#x20AC;˝

Different periods of compound interest. When banks pay interest on an investment, they pay this more frequently than once a year. The effect that different periods of compounding interest can have on an investment can be quite substantial, depending on the size of the investment. Example: ÂŁ100 000, 00 invested at compound interest for 5 years at 12% p.a. Principle

Interest

i Fv ď&#x20AC;˝ Pv (1 ď&#x20AC;Ť ) tm m

m = no of compounding

Time Period = 5years Yearly

Half-Yearly

Quarterly

Monthly

Daily

m=1

m=2

m =4

m = 12

m = 365

periods p.a. ÂŁ100 000

I =12%

Compound InteÂŁest

ÂŁ176234,17 ÂŁ179084,78

ÂŁ180611,12 ÂŁ181669,67 R182193,91

ÂŁ76234,17

ÂŁ80611,12

ÂŁ79084,78

ÂŁ81669,67

Ci = Fv â&#x20AC;&#x201C; Pv It can be deduced from the table above that the more time periods involved in the calculation of compound interest the better the return on investment:

ÂŁ82193,91

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CONVERTING EFFECTIVE TO NOMINAL CAN BE DONE AS FOLLOWS:





J m  m m 1  i   1

e.g. rate:

Convert an effective rate of 23,21% p.a. calculated bi-annually to a nominal





J m  2 2 1  0,2321  1 = 0,22 OR 22% p.a.

Calculating the rate in compound interest: Use the following formula:  A  r  100 m  1 for a compound growth  P  and :  A r  100 1  m  for a compound decay P 

Calculating time one has to use logarithms: A P t i log(1  ) m m log

Example: Thembi deposits £12000, 00 into her savings account which gives an interest rate of 7, 2% p.a. compounded monthly. Her savings grew to £17 181, 47 over a time period. Calculate how long her money was invested for. (Answer in years) A P t i log(1  ) m m 17181,47 log 12000 t log(1  0,072) t = 5 yrs. log

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TIME LINES USED FINANCIAL CALCULATIONS Time lines are useful when dealing with complicated problems, such as changes in the interest rate during an investment period or when several deposits or withdrawals are made from a savings account. It helps to summarize the information and give a visual representation of the data in an ordered manner. Example: ÂŁ7000 is deposited into a savings account, and 4 years later another ÂŁ5000 is added to the savings. Calculate the value of the savings at the end of 7 years if the interest rate is 12% p.a. For the first 3 years and then increased to 13, 5% for the remaining period. Solution: T0

n (years) ÂŁ7000

ÂŁ5000 12% p.a.

13,5 % p.a.

Interest rates

T6 etc indicates the time period of the investment.

Balance after 3 years :

Balance after 4 years :

Balance after 7 years:

A ď&#x20AC;˝ 7000(1 ď&#x20AC;Ť 0,12) 3 A ď&#x20AC;˝ ÂŁ9834,496

A ď&#x20AC;˝ 9834,496(1 ď&#x20AC;Ť 0,135) ď&#x20AC;Ť 5000 A ď&#x20AC;˝ ÂŁ16162,152 A ď&#x20AC;˝ 16162,152(1 ď&#x20AC;Ť 0,135) 3 A ď&#x20AC;˝ ÂŁ23631,26(nearest cent)

Alternative (shorter) Method: đ??´ = 7000(1 + 0.12)3 (1 + 0.135)4 + 5000(1 + 0.135)3 = ÂŁ23631,26

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Example 2: In order to save for her sons University fees, Mrs Gumede deposits £8000 into a savings account at the end of January when her son is 10 years old. The rate of interest is 14% p.a. compounded annually. When her son is 18yrs old he starts a University course which has a duration of 3 years. The first fees are £9000, payable at the end of January. The fees increase by 10% each year. Calculate: 5.

The second and third year’s fees.

The balance in the account after the first years fees have been paid.

The balance in the account after the second years fees are paid.

How much additional cash will be needed to cover the third year’s fees?

T10

n (years) £8000

(£9000) 14% p.a.compounded annually

NB the bracket around £9000 indicates a withdrawal. 6.

Second years fees ( fees at T9):

9000  (0.1)(9000)  £9900 Third years fees (fees at T10) 9900  (0.1)(9900)  R10890 7.

Balance at T8: 8000(1  0.,14) 8  9000  £22820,69  9000  £13820,69 (To nearest cent)

Balance at T9:

13820,69(1  0,14)  9900  R5855,59 (To nearest cent) 9.

Money in savings account at T10:

5855,59( I  0,14)  £6675,37 Additional cash required: 10890  6675,37  £4214,63 (to nearest cent)

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Year 11 ‘time line’ questions: Example 1: Jun invests £50 000, 00 into an interest bearing account for 15 years. For the first 5 years he receives 10% p.a. compounded bi-annually for the next 5 years he gets 12% p.a. compounded quarterly and for the last 5 years he receives 15% compounded monthly. How much does he have in his account at the end of the 15 year period? Solution: NB: If the Principle remains the same BUT the interest rate changes: use multiplication. Pv = £50 000 Tm = 5 x 2 = 10 Tm = 5 x 4= 20 Tm = 5 x 12 = 60

10% (Half-yearly)

12% (Quarterly)

5 10

15% (Monthly)

0,10   0.12   0.15   A  5000001   1   1   2   4   12   A  £309962,68

T1 5

0 60

Example 2: £8000 is invested into a savings account at an interest rate of 12, 5 % p.a. compounded monthly. Three years later £5000 is added to the savings and after a further 3 years R6000 is withdrawn. If a final deposit of £10 000, 00 is made into the account in the beginning of the 8th year how much money will be in the account after 10 years? Solution: If the principle is added to or reduced then addition and subtraction is used. Tm = 10 x 12 = 120 Tm = 7 x 12 = 84 Tm = 4 x 12 = 48

£8000

£5000

Tm = 2 x 12 = 24

(£6000)

15% (Monthly)

£10 000

T 6

0.15 120 0.15 84 0.15 48 0.15 24 )  5000(1  )  6000(1  )  10000(1  ) 12 12 12 12 A  £52298,65 A  8000(1 

T1 0

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Exercise 11.7: 1. Jono wins £3 000 000, 00 on the lottery and invests the money with Barclays Bank. Compounded monthly. At the end of 3 years, Jono withdraws £500 000, 00 to buy a house. Two year later he deposits £120 000, 00 into his account. Calculate how much money he has in his account at the end of 8 years. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2. James invests an amount of £100 000, 00 for 15 years. He receives 10% p.a. compounded quarterly for the first 6 years and 15% compounded monthly for the last 9 years. a.

How much is his investment worth to the nearest rand, at the end of the full term.

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2.2 What rate would he have to receive if his investment had the same final amount after 15 years compounded annually for the duration? (Answer to 1 decimal place). __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 3. Peter invests an amount of ÂŁ200 000, 00 for 20 years. He receives 8% p.a. compounded monthly for the first 14 years and 12% compounded quarterly for the last 6 years. 3.1 How much is his investment worth to the nearest rand, at the end of the full term. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

3.2 What rate would he have to receive if his investment had the same final amount after 20 years compounded annually for the duration? (Answer to 1 decimal place). __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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4. John invests £150 000, 00 into an interest bearing account for 10 years. For the first 2 years he receives 12% p.a. compounded bi-annually for the next 5 years he gets 15% p.a. compounded quarterly and for the last 3 years he receives 20% compounded monthly. How much does he have in his account at the end of the 10 year period? __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 5. A delivery vehicle is purchased for R 1 250 000, 00 in January 2001. It depreciates at 8% per annum. 5.1

Determine its book value in December 2009 if depreciation is calculated according to the straight – line method. (I.e. Simple Decay).

Determine its book value in December 2009 if depreciation is calculated according to A reducing – balance method. (I.e. compound Decay). __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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6. £20 000 is invested into a savings account at an interest rate of 8, 5 % p.a. compounded monthly. 2 years later £15000 is added to the savings and after a further 4 years £16000 is withdrawn. If a final deposit of £5 000, 00 is made into the account in the beginning of the 8th year how much money will be in the account after 10 years? __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 7. Convert the following nominal interest rates to effective annual interest rates: 7.1 15,5% p.a. compounded monthly __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 7.2

32, 4% p.a. compounded half-yearly.

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8. Convert an effective annual rate of 32, 5 % p.a. to a nominal rate per annum compounded monthly. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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12. Probability Theory: The language of probability:       

Probability tells us how likely something is to happen. We often use the word chance for probability. The thing you want to happen is called an event. An experiment etc. that you are carrying out is called a trial. The result of a trial is called an outcome. Some outcomes always happen. We say they are certain to happen. Some outcomes never happen. We say they are impossible. Some outcomes are not certain, but they are not impossible either. They may or may not happen. These probabilities are greater than 0, but less than 1.



A probability of

1 means that there is a 50-50 chance or an even chance 2

of the outcome occurring. We can write probabilities as fractions, decimals or percentages. To compare probabilities we compare the sizes of the fractions, decimals or percentages. Probabilities can be placed on a probability line as shown below: Fractions

1 2

Decimals Percentages

0 0%

0,5 50%

1 100%

Impossible

Even chance

Certain

Example Activity 1: Task: Rank the following activities from lowest to highest probability: Event Probability

Answer: Event Probability %

1 3 5

60%

Answer: 0,25 ; 30%;

2 0,25

3 78%

4 1 3

5 30%

2 0,25

3 78%

5 30%

25%

78%

1 3 ; ; 78% at events 2; 5; 4; 1; 3 3 5

1 3

33 13 %

30%

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Example Activity 2: All the learners in Grade 10 at Northwood were asked what they wanted to do when they left school. Different people worked out the results of the survey and gave the results in different ways. Results: Do nothing Get a job

0,15 60%

Go to University

1 25

Go to Technikon Don’t know

12% 9 100

Arrange the results in order of probability of popularity of options, starting from least popular choice.

The probabilities add up to 1. Explain why this is so.

Answer: Do Nothing

0,15

Get a Job

60%

Go To University

1 25

15 100 60 100 4 100

Go to Technikon

12%

12 100

Don’t Know

9 100

Go to University; don’t know; go to Technikon; do nothing; get a job.

15  60  4  12  9 100  1 100 100

All grade 10 surveyed all gave answers so 100% or value of 1.

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Exercise 12.1: 1.

There is a hot-drink vending machine in an office block. A survey found that the probability that an office worker , buying a drink from this machine, would choose: Soup

Coffee

Tea

Hot Chocolate

1 20 2 5 3 10 1 4

Which drink bought from the machine is: Most likely to be chosen? Least likely to be chosen? Arrange the drinks in order of popularity. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Listing Outcomes: When a coin is tossed into the air there are two possible outcomes: Heads or Tails. When a dice is thrown, there are 6 possible outcomes: 1; 2; 3; 4; 5; 6. For any event, you can usually list all the possible outcomes. Outcomes which give the “event” you are interested in are called favourable outcomes for that event. E.g. Consider “getting an even number” when you throw a dice. 

The activity is to “throw the dice”



All the possible outcomes are : 1, 2, 3, 4, 5, 6



The event you are interested in is “ getting an even number”



The favourable outcomes are 2. 4 and 6.

e.g. A coin is flipped and a dice is thrown at the same time, and you want to get a head and an even number.  

Coin Outcomes  

The activity is “flip” the coin and toss a dice” All the possible outcomes can be worked out in a table as follow.

1 H, 1 T,1

H T

Dice Outcomes 2 3 H, 2 H, 3 T, 2 T, 3

4 H, 4 T, 4

5 H, 5 T, 5

The event you are interested in is “ a head and an even number” The favourable outcomes are (H, 2), (H, 4) and (H, 6).

Example Activity 3: Look at each of the activities in turn:  List all the possible outcomes for the activity.  List all the favourable outcomes for each named event. 3.1

Activity: throw a dice. Event 1:

get a 1

Event 2:

get a multiple of 3.

Event 3: get a prime number. 

The possible outcomes are 1, 2, 3, 4, 5, 6. Event

1. get a 1 2. Get a multiple of 3 3. Get a prime number

Favourable outcomes 1 3;6 2; 3; 5

6 H, 6 T, 6

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Activity: take a coin from a purse containing a 5c coin, a 10c coin, A 20c coin, a 50c coin, a R1 coin, a R2 coin and a R5 coin. Event 4: get a 20c piece.

Event 5: get a silver piece

Event 6: get a coin worth less than 50c. 

The Possible outcomes are : Get a 5c; 10c; 20c; 50c; R1; R2; R5.

Event 4. Get a 20c coin 5. Get a silver coin 6. Get a coin worth less than 50c 3.3

Favourable Outcomes 20c R1; R2; R5 5c; 10c; 50c

Activity: take a letter from the word PROBABILITY Event 7: get a T

Event 8: get a vowel

Event 9: get a consonant. 

The possible outcomes are : P; R ; O ; B ; A ; B ; I ;L; I ; T; Y. Event

7. Get a T 8. Get a vowel 9. Get a consonant

Favourable Outcomes T O; A; I P;R;B;L;T;Y

Exercise 12.2: 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.

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

2.2 I roll dice: 2.2.1

List all the possible outcomes ( that is , possible pairs of numbers)

2.2.2

To get my score I must multiply the value shown on the dice by the value shown on the other dice. List all outcomes that satisfy the condition: the score must be a multiple of 6.

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Sample Spaces: Suppose you have taken the four picture cards of the suit of hearts from a pack. You shuffle them and draw a card at random. You could describe what you have done as a statistical experiment or trial, since there are no laws determining which card you will draw. The fact that you draw an ace, say, is an outcome. We can represent all possible Outcomes in set notation, S  A; K ; Q; J . The possible Outcomes can also be represented In a Venn diagram.

A; K; Q and J are called elements of the sets.

Venn diagram

this set is referred to as the probability Space or sample space

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Example Activity 4: Consider the rolling of an ordinary cubical dice, where the number on the upper face, when it comes to rest, is used. 4.1 Describe the experiment or trial involved. 4.2 Give an example of an outcome in this case. 4.3 Write down the sample space of the experiment. 4.4 Represent the sample space be means of a Venn diagram.

Answers: 4.1 4.2 4.3

Rolling the dice. 4

4.4

S  1;2;3;4;5;6

1 4

2 3 5 6

Exercise 12.3: 3.1

Consider the tossing of a coin: 3.1.1 Describe a trial

Give an example of an outcome.

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Write down the sample space, S of the experiment.

Represent the sample space by means of a Venn diagram.

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.

___________________________________________________________

4 3

___________________________________________________________ 3.2.2

Give an example of an outcome.

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Write down the sample space, S for the experiment.

Give a Venn diagram for S.

Events: A bag contains 3 blue and 5 white counters; the blue counters are labeled b1; b2 and b3. While the white counters are labeled w1; w2; w3…….w5. Suppose the trial consists of drawing out a counter, after the bag has been given a good shake, and then replacing the counter. The sample space here is: S = {b1; b2; b3; w1; w2; w3; w4; w5.} The number of possible outcomes in the sample space, S, is denoted by n(S). Here n(S) = 8 The drawing of a blue counter, no matter what its label is, can be represented by the set B = {b1; b2; b3} A few of the outcomes or elements of S are involved in the situation described by B. We call the drawing of blue counter (no matter the label) an event. The number of outcomes in event B is the number of elements in B. This number is denoted by n (B), which equals 3. Another event related to drawing counters from the bag, would be that of drawing a white counter. We can represent this event by the set: W = {w1; w2; w3; w4; w5} n(W) = 5 The sets B and W are subsets of S. The sample, S, is the universal set. Definition: An event is a set which consists of one or more of the elements of the sample space.

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Union & Intersection: In a shoe box there are cards numbered from1 to 10.

S  1;2;3;4;5;6;7;8;9;10

Consider the events:  Drawing a factor of 6: X = { 1 ;2 ;3 ; 6}  Drawing a factor of 9: Y = {1; 3 ;9} The union of these sets is all the elements in X or Y. X or Y = {2; 6;1;3;9} The intersection of these sets all the elements that are both in X and in Y NB: 1. For an event to occur, it is not necessary for all the outcomes of the event to occur. If any one of the outcomes is the result of a trial, the event is said to have occurred. B thus occurs if any blue counter is drawn. It is not necessary to draw all three blue counters before we can say that B has occurred. 2. The sets W and B are to be disjoint since they have no elements which belong to both sets. Example: 1. A trial for this question is the rolling of a normal dice. Write down the sample space, S. Describe the event, E, where an even number is obtained on rolling the dice, in terms of a set. Use set notation to describe the event, O, of obtaining an odd number. Describe the event, F of obtaining a factor of 6, in set notation. Write down: n(S) n(E) n(O) Represent S and the sets, E, O and F, in a single Venn diagram. Answers: 1.1 1.2 1.3 1.4 1.5.1 1.5.2 1.5.3 1.5.4

S  1;2;3;4;5;6 E  2;4;6 O  1;3;5 F  1;2;3;6 n( S )  6 n( E )  3 n(O )  3 n( F )  4

1.6

5 F

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Exercise 12. 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. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4.1.2 Write down a set for the event, H, of obtaining a head. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4.1.3 Write down a set for the event, t, of obtaining a tail. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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 ) ____________________________

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4.1.5 Represent S, H and T in a single Venn diagram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

4.2

Max has a bag containing three (3) red, four (4) blue and six (6) green marbles. The bag is shaken and a marble is withdrawn. The colour of the marble is noted and the marble is replaced.

4.2.2 Express the event of drawing a green marble, G, in terms of a set. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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4.2.3 Write down a set, B, or R, to represent the event of drawing either a red or a blue marble. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4.2.4 Write down: 4.1.4.1 n(S ) _____________________________ 4.1.4.2 n(G ) _____________________________ 4.1.4.3 n(BorR) _____________________________

Calculating Probabilities:   

Outcomes which have an equal chance of happening are called equally likely outcomes. For example: when you throw a fair dice , each outcome in a sample space S  1;2;3;4;5;6 Is likely to occur. When all the outcomes of an activity are equally likely , you can calculate the probability of an event happening by using the following definition: P (E) =

number of favourable possibilities Total number of possible outcomes

n( E ) n( S )

For example: when you throw a fair dice the possible outcomes are S  1;2;3;4;5;6 I.e. the total number of possible outcomes: n( S )  6 Event 1: Get a 4: The only possible outcome is a 4 i.e. E = {4} I.e. number of favourable outcomes: n( E )  1 Probability of getting a 4 = P (4) =

n( E ) 1  n( S ) 6

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Event 2: Get a number more than 2: Favourable outcomes: E  3;4;5;6 Number of favourable outcomes: n( E )  4 Probability of getting more than 2 = P (more than 2) n( E ) 4 2   = n( S ) 6 3 Event 3: Get a number more than 6: Favourable outcomes written as the “empty set.” E = { } For example: Number of favourable outcomes: n( E )  0 Probability of getting more than 6 = P (more than 6) n( E ) 0  0 = n( S ) 6 Event 4: Get a number less than 10: Favourable outcomes: E  1;2;3;4;5;6 Number of favourable outcomes: n( E )  6 Probability of getting a number less than 10 = P(less than 10) n( E ) 6  1 = n( S ) 6 I.e. it is certain that you will get a number less than 10 when you roll a dice.

Probability can be either found by theory or experiment. The theory method relies on logical thought; the experimental method relies on the results of repetition of many, many trials...

Example: Need to find the probability of getting a head when tossing an unbiased coin. Theory method: Unbiased means that both events (head or tail) are equally likely to occur. There are two (2) possible outcomes: S  H ; T  There is one favourable outcome: E  H  So, the probability of getting a head is

1 or 0, 5. 2

Experimental method: Suppose you toss an unbiased coin 100 times, and count the number of heads you obtain. Suppose you get 49 heads. The relative frequency of getting a head is therefore

49 or 0, 49. Only after many, 100

many trials does the relative frequency give a good value for the probability.

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Mutually exclusive and complimentary events. Mutually Exclusive Events 

The events A and B exclude each other.



If A happens B cannot happen.



If B happens A cannot happen.



Both events cannot happen at any one trial.



The sets of outcomes for the events are disjoint( Have no elements in common)



For two mutually exclusive events A, B: P (A and B) =0

P(A and B) =

n( AorB) n( A)  n( B) = n( S ) n( S ) n( A) n( B)  = = P(A) + P(B) n( S ) n( S )

Complimentary events: For any two(2) complimentary events R and Q. > >

n( R)  n(Q)  n( S ) . n( R) n(Q) P( R)  P(Q)   n( S ) n( S ) n( R)  n(Q) n( S ) 1 = = n( S ) n( S )



It is useful to write this in the form P(Q) = 1- P(R).



Q can also be written R or “not R “

Example: A bag contains three (3) red balls, five (5) white balls, two (2) green balls and four (4) blue balls. 1. Calculate the probability that a red ball will be drawn from the bag. 2.

Calculate the probability of that a ball which is not red will be drawn.

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Solution: Let R be the event that a red ball is drawn: n( R ) 3  1. P(R) = n( S ) 14 2. R and R are complimentary events. 3 11  14 14 P ( R )  P ( B )  P (W )  P (G )

P ( R )  1  P ( R )  1 

Note: Alternatively

4 5 2 11    14 14 14 14 Exercise 12.5: [Use Venn diagrams when necessary:]

5.1



250 tickets were sold for a raffle. Stephanie bought 10 tickets. What is the probability that Stephanie:

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5.2

There are 250 cars in the car park. 165 are Volkswagen. What is the probability that the first car to leave the car park is: 5.2.1 A Volkswagen? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 5.2.2 Not a Volkswagen? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 5.3

Jon has 14 loose socks in a drawer. Four of these socks are black and three are white. Calculate the probability that the first sock taken at random from the drawer is. 5.3.1 black. 5.3.2 not black. 5.3.2. white 5.3.4 not white 5.3.5 black or white 5.3.6 not black nor white.

A packet contains 20 fruit â&#x20AC;&#x201C; flavoured sweets. There are four (4) pineappleflavoured, five (5) melon-flavoured, two (2) lemon-flavoured, three (3) bananaflavoured and six (6) strawberry-flavoured sweets.

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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? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 5.4.2 Albert doesnâ&#x20AC;&#x2122;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? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Probability Trees: Tree diagrams are used to count the various ways in which outcomes happen. Example: A bag has 6 red beads and 4 blue beads in it. First one bead is drawn and then a second without the first been put back. Calculate the probability that : 1. 2. 3.

the first bead drawn is red both beads are blue. one bead is red and the other is blue in any order.

Solution:

5 9

Red 6 10

4 9

6 9 4 10

Blue 3 9

1.1

Blue

4 15

Red

4 15

Blue

2 15

There are 6 red beads and 10 beads in total: P(first red) =

1.2

Red

1 3

6 3  10 5

The probability of getting one blue bead first and blue bead second has to be calculated. The probabilities along the path therefore have to be multiplied. P(BB) =

4 3 2   10 9 15

[NB: there is no replacement so there are 3 blue beads left after the Ist blue bead is drawn and 9 beads in total] 1.3

Either red then blue or blue then red. The probabilities for the separate cases are thus added: P( RBorBR) 

4 4 8   15 15 15

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Exercise 12.6: 1. A coin and a dice are thrown together. 1.1 Draw a tree diagram to show all the possible outcomes. 1.2 What is the probability of obtaining: 1.2.1 A head on a coin and 1 on the dice? 1.2.2 A tail on the coin and an even number on the dice? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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2. A bag has 6 red beads and 4 blue beads in it. First one bead is drawn and then a second without the first being put back. Calculate the probability that: 2.1 the first bead drawn is red. 2.2 both beads are blue. 2.3 one bead is red and the other blue in any order. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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Independent Events: When looking at combining or repeated events, it is said that any event is Independent if the outcome of the event does not affect the outcome of any other event. Probabilities of events that are affected when combining them or when a probability experiment is carried out repetitively. Consider the probabilities of combinations using tables, tree diagrams and by derivation of the product rule where there are replacements. Example activity: Bandile says he is going to toss a coin three times (3) and get heads each time. Sisi says that this is not very likely and she bets him that he will not be able to get three (3) heads in a row. When he hears what his friends say, Bandile is no longer so sure he will be able to toss three (3) heads in a row. He draws a tree diagram to work out his chances of tossing 3 heads. Look at his tree diagram and discuss answers to the questions below. first toss

second toss

1 H 2

1 2

1 2 1

2 1

1 T

(HHH)

(HHT)

(HTH)

(HTT)

(THH)

(THT)

1 2

Outcomes

Third toss

2 1 2 1 2

(TTH)

(TTT)

1. For the first toss the probabilities of getting heads and tails are both

Probability 1 8 1 8 1 8 1 8

1 8 1 8 1 8 1 8

1 . 2

Why is this so? 1 again? 2 3. How many different outcomes are there when a coin is tossed three (3)

2. Why are the probabilities for the second toss all

times? 4. What is the probability of Bandile tossing three (3) heads in a row? 5. What is the probability of Sisi winning the bet? We see from the tree diagram that we can find the probability of each of the Outcomes if we multiply the probabilities P of the outcomes of each toss (t) Together.

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1 . The second toss 2 1 1 1 1 1 1 P(T) = the third toss P(T) = Thus P(TTT) =    . 2 2 2 2 2 8

e.g. to find P(TTT), we look at the first toss: P(T) =

Stated formally, we find the probability of independent events by multiplying the probability of each event together. If A and B are independent events then P (A and B) = P (A) X P (B). We call this the product rule. 6. Use the product rule to find the probability of Bandile tossing three (3) heads in a row. 7. After looking carefully at his tree diagram, Bandile decides, instead, to bet Sisi that he will toss at least two (2) heads out of three (3) tosses. What is the probability of him doing this? Exercise 12.7: [N.B. Use a probability tree to answer the questions] 1

A bag contains 4 yellow counters and 8 white counters. Calculate the Probability that: 1.1

The first counter drawn at random is: 1.1.1 yellow

white.

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For a second draw you get white if first counter was white if: 1.2.1 was replaced

was not replaced before second draw was made.

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1. James cannot decide what to wear to a party. He has three(3) pairs pants â&#x20AC;&#x201C; a grey pair, black pair and blue jeans. He puts his pants on his bed, closes his eyes and chooses a pair of pants to wear. He has four(4) shirts to wear - a white shirt, a white T-shirt, a black shirt and a green T-shirt. He pulls a shirt from his drawer without looking. 1.1

Is James choice of pants independent of his choice of shirt? Explain. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

1.2

Complete the contingency table below to show the different possible combinations of shirt and pants that James has to choose from and answer the questions that follow. White shirt(WS)

Grey pants(G)

White T-shirt(WT)

Black shirt (BS)

Green T-shirt (GT)

G,WS

Black pants(B) Blue jeans(BJ)

2.2.1

BJ, GT

How many outcomes are there in total?

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What is the probability that James will be wearing his green T-shirt?

What is the probability that James will be wearing something

white ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 2.3 Draw a tree diagram to show the probability of each choice that James has. 2.3.1 What is the probability that James chooses his black pants? ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 2.3.2

What is the probability that James chooses his black shirt?

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2.3.3

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What is the probability that James is dressed all in black?

What is the probability that James is not wearing any black at all?

Dependent Events: Events that when combined are effected when they are dependent on one another. Definition: the

Events are dependent if the outcome of the second event is affected by outcome of the first event.

Example Activity: NB P(A then B) = P(A) X P(B given A) is called the product rule for a dependent event. Sandy and her brother, John are doing their homework. Their mother puts out a plate of biscuits on the table next to Sandy. There are three (3) types of biscuits on the plate: 3 plain, 8 chocolate and 4 jam. John watches to see which biscuits Sandy is going to take. After Sandy has eaten her first one, there will be one less biscuit remaining to choose from the second time. So Sandy’s chance of taking a jam one the second time is dependent on her first choice. As the outcome of the first affects the outcome of the next event, the events are dependent.

John draws 2 Venn diagrams representing the choices that Sandy has. He wants to work out the probability that Sandy’s first biscuit will be a chocolate one and her second a jam one.

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Look at John’s Venn diagram and then answer the questions below. Ch

5 Pl

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

What do ‘Ch’; ‘Pl’ and ‘J’ stand for? How many biscuits are there on the plate before Sandy takes any? How many chocolate biscuits are on the plate? Calculate P (Ch), the probability of Sandy choosing a chocolate one first? How many biscuits on the plate after Sandy has taken one? How many jam biscuits would there be on the plate if Sandy took a chocolate one first? Calculate P (J and Ch), the probability of Sandy taking a jam one, given that she took a chocolate one first. Why is the denominator for P(J given Ch) 15 instead of 16? Now use the product rule to find out P (Ch, J), the probability that Sandy’s first biscuit is chocolate and her second one will be a jam one. What is the probability that Sandy chooses a jam one and then a chocolate one? Does the order in which she chooses the biscuits matter?

2. Sandy eats 2 biscuits. Each time she chooses one she reaches out to take One without looking at the plate. She takes a plain one first and a jam one second. a. Draw 2 Venn diagrams to show her options for each biscuit choice. 2.2 What was the probability of her taking this combination? 2.3 Sandy’s mother says she can have another biscuit. Draw a Venn diagram to Show the biscuits for this 3rd choice. 2.4 What is P(Pl , J , J)? 3. How could Sandy’s choices be independent rather than dependent?

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Answers to above questions: 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10

Ch stands for Chocolate biscuit, Pl stands for Plain biscuit and J stands for Jam biscuit. There 16 biscuits on the plate. There are 8 chocolate ones on the plate. 8 1  P(Ch) = 16 2 There are 15 biscuits still on the plate after Sandy has taken one. There would be 5 jam ones still on the plate if Sandy took a chocolate one first. 5 1  P(J given Ch) = 15 3 The denominator for P (J given Ch) is 15 instead of 16 as there are only 15 biscuits on the plate because Sandy ate one. 8 5 1 1 1     Using the product rule P(CH ,then J) = 16 15 2 3 6 5 8 40 1  , the order does not seem to The P (J, then Ch) =   16 15 240 6 matter.

2.1 First Choice Ch

5 Pl

Second Choice Ch

5 Pl

2.2

P(Pl, then J) = P(Pl) x P(J given Pl) =

3 5 15 1    16 15 240 16

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230 Third Choice Ch

4 Pl

2.4 P(Pl, then J, then J) = P(Pl) x P(J given Pl) x P(J given Pl, given J) 3 5 4 60 1      0,018 = 16 15 14 3360 56 3

Sandy was not putting back her biscuits before making her second choice making her next choice so we say that this experiment was without replacement. For the events of choosing biscuits to be independent, there would have to be replacement. That is, each time a biscuit was taken from the plate, one of the same kind would have to be added.

Exercise 12.8: 1.

Are these two events Dependent or Independent? 1.1 rolling a dice ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 1.2

taking 2 names out of a hat

____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 1.3

the first two numbers in a lottery draw

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tossing a coin twice ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Jack has a bag of 5 red, 7 blue, and 8 white and 6 green marbles. He represents this sample space using the Venn diagram below:

2.1

Jack allows some friends to choose 3 marbles each, without looking. Draw a Venn diagram to show the sample space for each choice and then Calculate the probability of each of these events: 2.1.1

Nick chooses first. He chooses red, then white, then green.

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2.1.2

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Vusi chooses second. He chooses green, then green, then green.

2.2

For each question below draw the relevant parts of a tree diagram to Show the marble choices and their probabilities.

2.1.1

Cynthia chooses last. She wants 3 red marbles. What is the probability that she will choose 3 red ones?

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Cynthia chooses red, then white, then white. Was the probability of this Choice greater than the probability of getting 3 red marbles? Explain.

Blake is getting dressed in the dark. His younger brother has been playing with his socks and they are all loose in his drawer. He has 2 grey socks, 2 black and 1 white. 3.1 He reaches into the drawer, he pulls out a sock and puts it on. Draw a Venn diagram to show the sample space for Blakeâ&#x20AC;&#x2122;s First choice. ____________________________________________________________

3.1.1

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3.1.2 How many socks are in the sample space for Blake’s second choice? ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 3.1.3

Are the events dependent or independent?

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3.1.1

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How many different outcomes are there altogether?

3.1.2

How many of these outcomes have two socks the same?

3.1.3

What is the probability then that Blake chooses 2 socks the same

What is the probability that he goes to school with odd socks on?

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Exercise 12.9: [Independent Events] 1. Two coins are flipped together. Using H for heads and T for tails, list all the pairs of combinations from flipping the coins. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 2. Sandwiches are available in chicken(C), beef (B) and ham (H). They can have fillers of Salad(S); chili (CH) and coleslaw (CO) Using Letter codes list all pairs of combinations of sandwiches. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3. Two die are thrown together. The table shows all the combinations of possible outcomes. Find the probability of the following scores happening. a. 12 b. 2 c. 10 d. 7 e. 5

Dice Two

Dice One 1 2 3 4 5 6 1 2 3 4 5 6 7 2 3 4 5 6 7 8 3 4 5 6 7 8 9 4 5 6 7 8 9 10 5 6 7 8 9 10 11 6 7 8 9 10 11 12 ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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4. Two spinners, each with 4 coloured regions (red; blue; green & yellow) are spun one after another. What is the probability of the following outcomes? a) Two colours the same. b) Red followed by red. c) Two different colours. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5. Two coins are tossed up at the same time. What is the probability of the following outcomes? a) One head and one tail b) Two heads c) Both coins land the same way up. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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6. 3 Balls, one yellow, one black and one white are in a cloth bag. One ball is removed from the bag, then another without replacing the first ball. a) What is the probability that a black ball is selected followed by a yellow ball b) On a new selection of two out of three what is the chance that a black ball is left in the bag?

7. A standard pack of cards consists of four suits of 13 cards- diamonds (red), Hearts (red), Clubs (black) and spades (black). If two cards are chosen randomly, a) What are the chances that the second card will be the same suit as the first? b) What is the probability that both cards will be aces? c) What is the probability that both cards will be black? ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Exercise 12.10: [Mutually Exclusive Events] 1. A cloth bag contains a mix of black, white and red balls. Selecting one ball at random, the probability of getting a black ball is 0.2, while the probability of getting a white ball is 0.5. a) Selecting one ball at random, what is the possibility of selecting a red ball? b) If in total there are 5 white balls in the bag, how many black ball are there? c) If there are in total 3 red balls in the bag, how many balls are there all together? ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2. A paper bag contains 5 red sweets. 7 green sweets and 8 blue sweets. If a sweet is selected at random from the bag, find the probability that the sweet is: a) Red or blue b) Not green c) Green or red d) Not blue ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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3. A tall metal box contains beads of many colours . The probability of selecting a bead of a particular colour is as follows: White 0.2 Red 0.3 Black 0.4. What is the probability that a bead taken from the box will be: a) Black or Red b) Not White or Black c) A different colour to white, Black or Red. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 4. A pack of cards consists of different numbers of white, black or red cards. If the probability of choosing a white or red card is 0.6 and the probability of choosing a white or black card is 0.7, what is the probability of choosing each of the coloured cards individually? ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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5. In a herd of 30 cattle there are 8 coloured black, 12 coloured white and 7 without horns. Find the probability that: a) A cow is coloured white or black b) A cow has horns. c) A cow is neither white nor black. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 6. Scientists examining climate classify winters as mild, normal, hard or severe. The probability that a winter will be mild or normal is 0.4 The probability that a winter will be severe is 0.1 a) What is the probability of having a hard winter? b) If the probability that a winter will be mild is three times the probability that it is normal, what is the probability that it will be mild or severe? ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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STRAIGHT LINE GRAPHS:

General Equation : y  mx  c Written in this form : y  mx  c

y - intercept

Gradient The Table Method: Draw a table using x-values of -2 ; 0;& +2. Change the equation to the y-form and by substitution calculate the corresponding y-values. Example: Sketch the graph of

X Y

-2 -1

0 3

2 5

Plot the above points on a Cartesian plane. Lable and name the graph. 4

fx = 2x+3

-5

-2

Exercise 13.1: Change the following equations to the y-form. 1.1

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1.3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2.

Sketch the following linear functions using the table method:

y  x3 2.1 ___________________________________________________________________

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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22. y  2x  4 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2.3 2 y  4x  8 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2.4 3y  6x  3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2.5 2y  x  6 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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4 y  8x  8  0 2.6 ___________________________________________________________________

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Method 2 : Dual Intercept Method. Sketching linear graphs (straight line) the best method to us is the Dual Intercept method. This method finds the values where the line cuts (intercepts) the x- axis and y-axis. NB The only time that this method is not used is sketching any line passing through the origin. Use substitution to find another point or use the gradient. METHOD: To find the x – intercept substitute zero for x – intercept substitute zero for y and To find the y – intercept substitute zero for x. Example: Sketch the graph of 2 y  3x  6 using the dual intercept method. 2 y  3(0)  6

Let x = 0  2 y  6 y3

2(0)  3x  6 and Let y = 0 3x  6

x2 y

2y + 3x = 6

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Exercise 13.2: Use the” dual intercept method” to sketch the following: 1. y  x3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2. y  2x  4 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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3 2 y  4x  8 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 4. 3y  6x  3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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5. 2y  x  6 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 6.

4 y  8x  8  0

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Gradients of straight lines: 1.

In the y- form of the equation simply use the m- value (co-efficient of x) as this represents the gradient. Given at least 2 points that lie on a given line, then calculate the gradient y  y1 y y1  y 2  or 2 using: m  x x1  x 2 x 2  x1 I.e the difference between the y – values divided by the difference between the x – values.

Parallel lines have the same gradients i.e. their m – values are the same

i.e y  2 x  6 . Is parallel to y  2 x  10 4.

Perpendicular lines have inverse gradients. i.e.

m1  m2  1

y  2 x  c is perpendicular to y  

1 xc 2

m1  m2  1 2 1    1 1 2 Domain & Range Domain represents all the x – values Range represents all the y – values. The domain is also referred to as the independent value and is made up of all the numbers in the number system. The range is referred to as the dependent value because you can only find why if you know x. The x and y values together form an ordered pair and represent a point on a Cartesian plane. X is always written first.

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Functions and Relations: A function occurs when the x-value is not repeated i.e there is one x-value for one yvalue. e.g. 1. (3 ; 4) ; (4 ;5) ; (5;6) 2. (3;4) ; (4;4) ; 5;4) [the y-value can be repeated in a function. A non- function occurs when the x-value is repeated. I.e more than one x-value for a yvalue. e.g. 1. (2;3) ; (2;4); (3;5) 2. (3;4) ; (3;5); (3;6) Graph representations of functions and non- functions. Graphs of Functions:

Graphs of Non- Functions:

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A pencil test can be used to see if a particular graph is function or a non- function. I.e. a vertical line test: 1. If the line cuts the graph at only once then the graph represents a function. 2. If it cuts more than once then the graph represents a non-function. The reason however is due to the fact that the x-value is not repeated for a function and is repeated for a non-function .

Domain and Range of graphs: It is easy to read the domain and range from graph representations.

(-2;4)

Domain: x  R Range: y  R

Domain: { x  2 ; x  R } Range: { y  4 ; y  R }

(4 ; 6)

Domain: { x  4 ; x  R }

{Domain: x  R }

Range: { y  6 ; y  R }

Range: { y  8 ; y  R }

-6

 6  x  6 ; x  R .} Range: {  6  y  6 ; y  R }

-6

Domain:{

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Parabola Graphs:

General Equation: y  ax  c OR y  ax  q 2

Two methods can be utilized to sketch a parabola graph: 1. Table method. 2. A ‘5 point’ method. Method: Table: A. Sketch the graph of y  x  4 2

X Y

-3 5

-2 0

-1 -3

0 -4

1 -3

2 0

3 5

y 4

f x  = x 2-4

-10

-5

-2

-1

-2

-3 -4

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Exercise 14.1: Sketch the following graphs using a table method: Use x –values of -3 to 3 for each table. (As in the example above)

yx 1. ___________________________________________________________________ 2

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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y  x 1 2. ___________________________________________________________________ 2

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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y x 2 3. ___________________________________________________________________ 2

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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y  x 1 4. ___________________________________________________________________ 2

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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y  x2  2

___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Exercise 14.2: A. Sketch the following graphs on the same set of axes: 1.

y  x2

y  x2  1

y  x 1 3. What deduction can be made? ___________________________________________________________________ 2

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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B. Sketch the following graphs on the same set of axes: 1.

y  x2

y  2x 2

y

1 2 x 2

What deduction can be made? ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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C. Sketch the graphs of the following equations: 1.

y  x2

y  x2

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D. Sketch the graphs of the following on the same set of axes: 1. y   x  1 and y  x  2 2. Write down the coordinates of the point(s) of intersection for the two graphs. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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

y   x and y  x  2 1. 2. Write down the coordinates of the point(s) of intersection for the two ___________________________________________________________________ 2

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Sketch of a parabola using a 5 point method. 1.

x –intercepts at y = 0 :

y- intercept at x = 0

Turning point: for y  ax  c the y- intercept and the turning point are the same.

Get two further points using substitution:

Example: A. Sketch the graph of y  x  4

Method: 1.

x – intercepts at y = 0: x2  4  0 ( x  2)( x  2)  0 x = 2 or x = -2

y – intercept at x = 0: y in = -4

Turning point: (0; -4)

Substitution:

qx  = x 2-4

-5

-2

-4

At x = 1 thus y = (1)2 – 4 = -3 plot point ( 1; -3) At x = -1 thus y = (-1)2 –4 = -3 plot point ( -1; -3) 8

B. Sketch the graph of y  2 x  18 2

x – intercepts at y = 0: 2 x 2  18  0

2( x 2  9)  0 2( x  3)( x  3)  0 x = 3 or x = -3

-4

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y – intercept at x = 0. y-int = -18

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-18

Turning point : ( 0; -18)

By substitution: at x = 2 y = 2(2)2 – 18 =8 – 18 =- 10. plot point (2; -10) at x = -2 y = 2(-2)2 –18 = 8 – 18 = -10 plot point ( -2 ; -10)

-20

qx  = 2x 2-18

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Exercise 14.3: Use a point by point method to sketch:

y  x 9 9. ___________________________________________________________________ 2

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y  x 2  9

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y  2x  8 1.3 ___________________________________________________________________ 2

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2 -1

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y  3x 2  12

1.4

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2 -1

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Finding Equations of Parabolas:

Given the turning point and any other point Method: Use the general form of the equation: y  ax2  q Example :

-2

2 -8

y  ax2  q y  ax 2  8

sub Minimum value of - 8 for q

0  a (2) 2  8

sub point (2;0) for x and y

0  4a  8

solve for a and sub back into equation.

4a  8 a2 y  2x 2  8

Given the two x – intercepts and any other point. Method: Use the general form of the equation: NB

and

y  a( x  x1 )(x  x2 )

refer to the roots or x – intercepts

Example (1;3) -2

y  a( x  x1 )(x  x2 ) y  a ( x  2)( x  2) 3  a (1  2)(1  2)

Sub the x – intercepts changing their signs when you do so. Sub the other point in for the x and y values

3  a (1)(3) 3  3a

Solve for a and sub back into equation

a  1 y  1( x  2)( x  2) y  x2  4

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Quadratic Function Continued : Sketching a Parabola using 5 points. A parabola can be neatly sketched using the following 5 points: 2. 3. 3.1 3.2

The two roots i.e. x – intercepts. The y – intercept at x = 0

(2 points) (1 point)  sum of roots  The axis of symmetry(a.o.s) at either  b or   2 2a   The corresponding y-value by substituting (a. o. s) into the original expression. These two values are the turning point: TP(x ; y) (1 point) Axis of symmetry (ado’s) minimum value

4. The mirror image of the y- intercept.

(1 point)

Example: 1.

Sketch the graph of y = x2 –2x –3

x – intercepts at y = 0

x 2  2x  3  0 ( x  1)( x  3)  0

y B.

y – intercept = -3

a.o.s =  b   (2)  1 2a

X = -1 or x = 3

y= x2-2x-3

Y – value @ f(1)= (1)2 –2(1) – 3= -4 TP ( 1 ; -4)

D. mirror image of the y – intercept; (2; -3) -

2; -3 1;-4

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Exercise 14.4: Sketch the following functions using a “5 – point method”. 1.

y  x 2  3x  4 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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y  x 2  4x  5

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2 -1

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y  x2  x  6

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2 -1

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y  x 2  3x  10

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y  x 2  2x  8

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-10

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y  x 2  4 x  12

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-8

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-14

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y  2x 2  7 x  6

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2 -1

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y  2 x 2  5x  3 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 6

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2 -1

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Parabola Graphs: Sketching Using Completing the Square: General Equation: y  a( x  p)  q 2

q is the maximum or minimum value.

p is the axis of symmetry ( a.o.s) Do not use the same setting out method for solving equations: Ensure that the coefficient of x2 is one (1). If it is larger than one place the value outside a bracket and divide each term in the equation by the value. Now complete the square inside the bracket. Finally distribute the value outside the bracket with the two terms inside. At this point the co-ordinates of the turning point can be written down. Example:

To solve equation:

2x 2  4x  6  0

[(x  1)2  4]  0

2[ x  2 x  3]  0 2

( x  1) 2  4

2[( x  1) 2  (1) 2  3]  0

x  1  2

2[(x  1) 2  4]  0

x  1 2

2( x  1) 2  8  0

x  1 or x  3

Only Use the value inside the square brackets to solve

for x pq(1 ; -8) (Turning point)

To sketch the graph x-intercepts: (-1;0) and (3;0) y-intercept: -6 Turning point(pq): (1;-8) Mirror image of y-intercept for 5th point.

-1

-6 Y = 2x2 –4x-6

(1; -8)

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Exercise 14.5: a) Write the following equations in the form: y  a( x  p)  q i.e in the completed square form of the equation. b) Write down the coordinates of the turning point: c) Solve the equation and write down the x and y intercepts. d) Sketch the graphs of the equations. 2

1. y   (2 x  1)( x  1) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 6

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2. y  x  2 x  3 __________________________________________________________________ 2

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2 -1

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3. y  x  6 x  8 __________________________________________________________________ 2

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2 -1

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4. y  x  5 x  6 __________________________________________________________________ 2

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2 -1

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5. y  2 x  4 x  6 __________________________________________________________________ 2

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2 -1

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6. y   x  2 x  3 __________________________________________________________________ 2

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2 -1

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7. y  3x  2 x  1 __________________________________________________________________ 2

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2 -1

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Shifting parabolas: Horizontal ( Left or Right) In the completed square form of the equation simply change the “p” value and multiply the equation out for the ax 2  bx  c form ( if required) Example: 2 x 2  12 x  10  0

2[ x 2  6 x  5]  0 2[( x  3) 2  4]  0 2( x  3) 2  8  0 pq(3;8)

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 5 units to the left.

p3

p1  3  5

New equation

p1  2

y  2( x  2) 2  8 y  2 x 2  8x

NB. Don’t forget to change the sign when substituting back into y  a( x  p)  q 2

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 2 units to the right

p3 p1  3  2 p1  5

New equation

y  2( x  5) 2  8 y  2 x 2  20x  42

Exercise 14.6: 2. Shift questions 1 & 2 in Exercise 11.5 by 4 moves to the left and write down the new equation in the form of y  ax  bx  c __________________________________________________________________ 2

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Shift questions 3 & 6 in Exercise 11.5 by 5 moves to the right and write

down the new equation in the form of y  ax  bx  c __________________________________________________________________ 2

2[ x 2  6 x  5]  0 2[( x  3) 2  4]  0 2( x  3) 2  8  0 pq(3;8)

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 5 units upwards.

q  8

q1  8  5 q1  3 B:

y  2( x  3) 2  3 y  2 x 2  12x  15

New equation

Instruction: Shift the graph of, 2 x 2  12x  10  0 , 5 units downwards.

q  8

q1  8  5 q1  13

y  2( x  3) 2  13 y  2 x 2  12x  5

New equation

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Exercise 14.7: 2.

Shift questions 2 & 3 in Exercise 11.5 by 3 moves upwards. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

Shift questions 5 & 6 in Exercise 11.5 by 3 moves downwards. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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Finding Equations of Parabolas: A. Given the x- intercepts and one other point. Method: Use the general form of the equation : y  a( x  r1 )(x  r2 ) and substitute the roots(xvalues) and the x & y values of the given point into this formula and solve for a. e.g. Find the equation of the parabola that has x-intercepts –3 and 4 which passes through point(1 ; -24) y  a ( x  r1 )( x  r2 )  24  a (1  3)(1  4)  24  12a a2 y  2( x  3)( x  4) y  2 x 2  2 x  24 B. Given the turning point and one other point.

Method: use the general form : y  a( x  p)  q i.e the completed square form of the general equation. Substitute the TP and the other point into this form to solve for a. e.g. Find the equation of a parabola that has a turning point (2 ; 3) And passes through point (1 ; 2) 2

y  a( x  p) 2  q y  a ( x  2) 2  3 2  a (1  2) 2  3 2  a3 a  1 y  1( x  2) 2  3 y  x 2  4x  1

C: Given a sketch : Use the information supplied on the sketch to find the equation of the parabola and straight line:

y=x + 1

C(-3 ; 0)

A (0 ; y=ax2 +bx +c

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Method: A:

Straight line : y  mx  c y0

x 1 0 x  1 B ( 1;0) B:

Parabola: y  a( x  r1 )(x  r2 ) [Use this form as roots are known] y  a ( x  r1 )( x  r2 ) Substitute the roots into the y  a ( x  1)( x  3) equation and one other point:  3  a (0  1)(0  3) Then solve for ‘a’  3  3a a  1 y  1( x  1)( x  3) y   x 2  4x  3

Exercise 14.8: Find the equations of the following given: 1. Turning Point (2;10) passing through (0 ;2) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 2. Turning point ( -1;5) passing through (1;13) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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Turning point ( -4;-1) passing through (-3;2)

x – intercepts (1;0) and (-3;0) passing through (-1;-4

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x – intercepts (2;0) and (-4;0) passing through (3;-14)

x – intercepts (1;0) and 5;0) and y –intercept (0;-5)

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Exercise 14.9: Quadratic Function: Parabolas. Sketch the graphs of y   x  x  12 and y  3 x  12 on the same 2

1.1

System of axes. ___________________________________________________________________ __________________________________________________________________ ________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 12

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-2

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1.2

Write down the co-ordinates of the points of intersection.

3.1

Write

y  x 2  2 x  8 in the form of y  a( x  p) 2  q .

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Write down the co-ordinates of the turning point.

3.3

Write down the roots (x-intercepts) of the graph.

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3.5 Sketch the graph of y  x  2 x  8 2

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Find the new equation if y  x  2 x  8 is moved 5 units to the left 2

2.6

Write y   x  4 x  5 in the form y  a( x  p)  q . 2

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Sketch the graph of

y  x 2  4x  5 .

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-6

Sketch the graph of y  x  5 on the same system of axes.

3.3 3.3.1

Write down the co-ordinates of the points of intersection of the two graphs.

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3.4 Find the equation if y   x  4 x  5 is moved 4 units downwards. 2

Hyperbola Graphs:

Sketching: Use a table method (on the calculator or on paper) Example: Sketch the graph of y 

y

1. X Y

6 x -3 -2

-2 -3

6 x

-1 -6

1 6

2 3

3 2

f x  =

6 x

-10

-5

-2

-4

-6

-8

  

Asymptotes are : x  0 and y  0 Domain: x  (;  ) x  0 . y  (;  ) y  0 . Range:

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Exercise 15.1: Complete the tables and use the values to sketch the graphs of: 1.

y

X Y

-3 2

6 x

-2 3

-1

3 -2

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

y

2. X Y

8 x

-8 -1

-4

-2

-1

8 1

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

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X Y

-8 1

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8 x

-4

-2

-1

8 -1

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

y

X Y

-4 -1

4 x

-2

-1

1 4

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

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X Y

-4

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4 x

-2 2

-1

4 -1

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-8

-6

-4

-2

-1

-2

-3

-4

-5

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Shifting the hyperbola graph: 1.

If a constant is added to the equation after k then this will cause a vertical

This shift will change the horizontal asymptote BUT not the vertical. Asymptotes are: x  0 (y- axis). And y  3 (New horizontal asymptote)

x 4 4 shift: e.g. y   3 : the graph of y  is shifted upwards by 3 units. x x Asymptotes of y  4 are y  0 (x – axis) and x  0 (y- axis). x

(1;7) 6

(2;5) 4

h x  =

(4;4)

(1;4)

horizontal asymptote is y = 3 (-4;2) 2

(2;2)

(-2;-1)

(4;1)

f x  =

4 x

horizontal asymptote is y = 0 -10

-5

(-4;-1)

(-1;-1) (-2;-2)

(-1;-4)

-2

-4

-6

Horizontal shifts are caused when a constant value is added to the x- value in the denominator of the hyperbola equation.

The sift is in the opposite direction of the integer in the equation. I.e if the integer is +ve then the shift is to the left and if the integer is – ve then the shift is to the right.

4 x

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Shift the graph of y  4 as follows: y 

4 x3

This shift is 3 units to the right.

The horizontal asymptote stays the same BUT the vertical asymptote changes to the line x  3

vertical asymptote x = 0 vertical asymptote x = -3

(-2;4)

(-1;2)

(1;4)

(2;2) (4;1)

f x  =

(1;1)

4 x

horizontal asymptote is y = 0 -10

-5

(-1;-1)

(-4;-1)

h x  =

(-7;-1) (-5;-2)

(-4;-4)

(-2;-2)

(-1;-4)

-2

-4

-6

Exercise 15.2: Write down the: Asymptotes, Domain & Range of each of the following: 8 3 x ___________________________________________________________________

1. y 

4 x+3

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10 6 x ___________________________________________________________________

2. y 

3. y  

___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ Exercise 15.3 Copy and complete the following table and use it to sketch the graph of y  6

x Y

-6 -1

-3

-2 -3

-1

2 3

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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Write down the equations of the asymptotes. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

1.2

Sketch the graph of y 

4 ( lateral move to the right) x3 6

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.3

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306

Copy and complete the following table and use it to sketch the graph of

6 y x

x y

-6 1

-3

-2 3

-1

2 -3

2.1 Write down the equations of the asymptotes. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ 6

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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307

2.2 Sketch the graph of y   6 ( lateral move to the left) x3

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

2.3 Write down the equations of the asymptotes. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

Gcse Mathematics

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308

Complete the following table and use it to sketch the graph of y  8

x y

-8

-4 -2

-2

-1 -8

2 4

-12

-10

-8

-6

-4

-2

-4

-6

-8

-10

3.1 Write down the equations of the asymptotes. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

Gcse Mathematics

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309

3.2 Sketch the graph of y 

8 1 x2

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

3.3 Write down the equations of the new asymptotes. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

Gcse Mathematics

16.

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310

Exponential Graphs:

General formula :

a>0,a1,x  y>0

y = ax

A. To sketch the graph y = ax (a) x  and y > 0 y = ax lies above the x – axis in quadrants 1 and 2. b) (i) If a > 1 , then as x increases , y increases. (ii) If 0 < a < 1 , then as x increases , y decreases. (i)

y = 2x [Where ‘a’ is a whole number]

X Y

-2

-1

1 4

1 2

(ii)

y = ( 12 )x. [Where ‘a’ is a fraction]

X Y

-2 4

-1 2

0 1

1 2

e.g. y = 2 x e.g. y = ( 12 )x.

2 4

1 2

1 4

x 1 y    or y  0,5  2 These 2 graphs are mirror images of each other, the axis of symmetry being the y- axes (x = 0 ). The x- axes is a vertical asymptote, as y will never equal zero.

y = 0,5x

y = 2x y

1 x

Gcse Mathematics

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311

Shifting the exponential graph: x x 1. Vertical Shifts: y  a or y  a

y  2 x Original graph y  2 x  2 Graph shifted 2 units upwards.

1. 2.

y  2  4 Graph shifted 4 units downwards. 3. NB: the values are added after the base (2x) for vertical movement. x

g x  = 2 x +2

f x  =

new horizontal asymptote for g(x) =2x + 2

2. -5

h x  = 2 x -4

-2

3. new horizontal asymptote for h(x) =2x - 4

-4

Exercise 16.1: Complete the tables and use the values to sketch the graphs of:

y  2x

1. X Y

-2

-1

0 1

1 4

2 4

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

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312

y  2x  1

2. X Y

-2

-1

5 4

0 2

2 5

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

3. X Y

y  2x  1 -2

-1

3  4

0 0

2 3

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

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313

y  2.2 x

4. X Y

-2

-1

0 2

1 2

2 8

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

5. X Y

y  2.2 x -2

-1

1  2

0 -2

2 8

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

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314

Horizontal Shifts: y  a

y  ax

y  2 x Original graph y  2 x  4 Graph shifted 4 units to the right. y  2 x  4 Graph shifted 4 units to the left.

2.1

2.3 NB: the movement is added or subtracted in the exponent for lateral shifts. 6

s x  =

r x  = 2 x-4

2 x+4

f x  = 2 x 4

3. New vertical Asymptote x = -1

New vertical Asymptote x= 7

2. -5

-2

3. -4

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315

Exercise 16.2: 2. Sketch the graph of y  2 on a Cartesian plane. x

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.1 Shift y  2 by 4 units to the left and sketch this graph. x

1.2 Shift y  2 by 4 units to the right and sketch this graph. x

2 Write down the equations of the asymptotes after the shifts in question 1. __________________________________________________________________ __________________________________________________________________

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316

3 Sketch the graph of y  2

x

on a Cartesian plane,

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

3.1 Shift y  4 3.2 Shift y  4

x

by 3 units to the left and sketch this graph. by 4 units to the right and sketch this graph.

4 Write down the equations of the asymptotes after the shifts in 3. __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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317

5 Sketch the graph of y  3 __________________________________________________________________ x

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

5.1

Shift the graph 4 to the right and 3 upwards and sketch the new position. i.e. the graph of y  2

x 4

3

Gcse Mathematics

318

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Functional Notation Graph Interpretation: Exercise 16.3: 1. 1.1

If f ( x)  4 x  3 find: f ( 4)

f (7)

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319

2.2 g (3) ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 2.3 g ( a  b) ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 3. Sketch the graph of 2 y  4 x  8 using the dual intercept method. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Gcse Mathematics

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320

Sketch the graphs of y  x  3 and 2 y  3x  6 on the same system of axes and:

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.1

write down the co-ordinates of the point of intersection. _______________________________________________________________

___________________________________________________________________ ____________________________________________________________________

Gcse Mathematics

GOMATH WORKBOOKS

321 Sketch the graph of y 

4.2

8 8 and y  on the same system of axes. x x

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.2.1 Write down the domain and range for the two graphs above. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4.2.2

Write down the asymptotes of the two graphs above.

Gcse Mathematics

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322

Sketch the graphs of y  x  4 and y   x  4 on the same system of axes. ____________________________________________________________________ 2

4.3

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.3.1

Write down the domain and range of the graphs above:

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323

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.4.1 Write down the asymptotes of the two graphs ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Write down the gradients; x and y intercepts of the following graphs: 5.1

5.2 8 6 -2

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324

5.3

5.4

y 8

-3

-2 -9

6.2

(2 ; -9)

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325

6.3

6.4

-12

-4

6.5

6.6

-7

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326

6.7

6.8 O (3; 9)

• 0

-4

•(-4 ; -4)

• 4

-9

7.2 y (4 ; 2)

-4

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327 y (2 ; 9)

7.3 (0;1 0

(2 ; 5)

Gcse Mathematics

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328

7.5 y

0 -2 (2 ; -4)

-2

h f -6

-12

8.1 Find the equation of â&#x201E;&#x17D;. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Gcse Mathematics

329

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8.2 Write down the domain and range of đ?&#x2018;&#x201C;. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 8.3 For which value(s) of x is đ?&#x2018;&#x201C; decreasing. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Use algebraic methods to find the coordinates of đ?&#x2018; , the point of intersection of đ?&#x2018;&#x201C; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x201E;&#x17D;. ____________________________________________________________________ 8.4

Gcse Mathematics

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330

8.5 Give the new equation of đ?&#x2018;&#x201D; if đ?&#x2018;&#x201C; is reflected on the x â&#x20AC;&#x201C; axis. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Exercise 16.4 Further Graph Interpretation: Find a and c in the following: 1. đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?

đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?

Gcse Mathematics

𝑦 = 𝑎𝑥 2 + 𝑐

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331 4.

𝑦 = 𝑎𝑥 2 + 𝑐

𝑦 = 𝑎𝑥 + 𝑐

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332 𝑎

7. 𝑦 = 𝑥

GOMATH WORKBOOKS 𝑎

8. 𝑦 = 𝑥 + 𝑐

9. 𝑦 = 𝑥 + 𝑐

10. 𝑦 = 𝑎𝑐 𝑥

Gcse Mathematics

333

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11. The graphs of đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) = â&#x2C6;&#x2019;đ?&#x2018;Ľ 2 + 4 and đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) = 3đ?&#x2018;Ľ + 4 are sketched alongside. 11.1 Calculate the coordinates of A, B, C and D. _______________________________________________ _______________________________________________ _______________________________________________ _______________________________________________ _______________________________________________ _______________________________________________ _______________________________________________ _______________________________________________ _______________________________________________

11.2 Determine the coordinates of E. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

11.3 Calculate the length of FB. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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334

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11.4 List an expression for đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) â&#x2C6;&#x2019; đ?&#x2018;&#x201C;(đ?&#x2018;Ľ). ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 11.5 For which values of x is đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) â&#x2030;Ľ đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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335

12. D

B x

A H g

The graphs of đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) = đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 9 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) = 2đ?&#x2018;Ľ + 6 are sketched above. 12.1

Calculate the coordinates of A, B , C and D.

State the lengths of OC, AB and OD

Calculate the coordinates of E.

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12.4

336

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IF đ?&#x2018;&#x201A;đ?&#x2018;&#x2020; = 1 Find FT.

If đ?&#x2018;&#x201A;đ?&#x2018;&#x192; = â&#x2C6;&#x2019;4 Find GH.

Calculate the length of AD.

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337

17.Trigonometric Functions: Method: Sketch the graphs using the values of the special angles. Viz using 0º ; 90º ; 180º ; 270º and 360º (You do not have to set up a table) Example 1: Sketch the graph of f ( x)  sin x where x  [0 ;360 ] 

X y = sinx

0º 0

90º 1

180º 0

270º -1



360º 0

0

90 

180

270

360

f x  = sin x 

Example 2 Sketch the graph of f ( x)  sin x where x  [360 ;360 ]

f x  = sin x 

-360

-270

-180

-90

-1

90

180

270

360

Gcse Mathematics

Example 3:

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338

Sketch the graph of f ( x)  cos x where x  [360 ;360 ]

g x  = cos x  -360

-270

-180

-90

90

180

270

360

-1

Example 4:

Sketch the graph of f ( x)  tan x where x  [360 ;360 ]

h x  = tan x  1

-360

-270

-180

-90

-1

90

180

270

360

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339

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Exercise 17.1: 1. Sketch the graphs of y  sin x and y  cos x on the same set of axes for the interval x  [0 ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  



From the sketch find the following: 1.1

the period of y  sin x

1.2

the range of y  cos x

1.3

the amplitude of y  sin x

1.4

the value for x for sin x  cos x

y 2

x 0 90

-1

-2

180

270

360

Gcse Mathematics

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340

2. Sketch the graphs of y ď&#x20AC;˝ 2 sin x and y ď&#x20AC;˝ cos x on the same set of axes for the interval x ď&#x192;&#x17D; [0 ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x ď&#x20AC; axis : 10mm represents 30 ď Ż ď Ż

ď Ż

From the sketch find the following: 2.1

the period of y ď&#x20AC;˝ 2 sin x

2.2

the range of đ?&#x2018;Ś = đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;Ľ

2.3

the amplitude of y ď&#x20AC;˝ 2 sin x

2.4

the value for x for 2đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;Ľ = đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;Ľ

y 2

x 0 90ď&#x201A;°

-1

-2

180ď&#x201A;°

270ď&#x201A;°

360ď&#x201A;°

Gcse Mathematics

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341

3. Sketch the graphs of y  sin 2 x and y  cos x on the same set of axes for the interval x  [0 ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  



From the sketch find the following: 3.1

the period of y  sin 2 x

3.2

the range of y  cos x

3.3

the amplitude of y  sin 2 x

3.4

the value for x for sin 2x  cos x

y 2

x 0 90

-1

-2

180

270

360

Gcse Mathematics

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342

4. Sketch the graphs of y   sin x and y  cos 2 x on the same set of axes for the interval x  [0 ;360 ] . Use the scale : y-axis: 20mm represents 1 unit And x  axis : 10mm represents 30  



From the sketch find the following: 4.1

the period of y   sin x

4.2

the range of y  cos 2 x

4.3

the amplitude of y   sin x

4.4

the value for x for  sin x  cos 2x

y 2

x 0 90

-1

-2

180

270

360

Gcse Mathematics

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343

ANALYTICAL GEOMETRY

Analytical geometry - Studies the properties of geometric figures algebraically. This is pursued by the means of examining significant points (co-ordinates) of these figures in a Cartesian plane. Hence also referred to as Co-ordinate Geometry.

Formulae: 1.

Length of a line:

A(2 ; 5)

B(-4 ; -3)

Length of AB = (x1  x 2 ) 2  (y1  y 2 ) 2 = (6) 2  (8) 2 = 100 = 10 2.

Mid – Point of a line

 (x  x 2 ) (y 1  y 2 )  ; Mid – point =  1  2 2    A(2 ; 5) C (x ; y )

B(-4 ; -3)

y Mid – Point AB = C (-1; 1) 3.

Gradient of Straight Line: Gradient is represented using the symbol ‘m’ [from y= mx+c] M=

y [i.e the difference in y divided by the difference in x] x

A( 2;4)

m

B (3:6)

y y1  y 2 6  (4) 10 2     x x1  x2 3  (2) 5 1

Parallel Lines have the same gradients: m1  m2 Perpendicular lines have inverse gradients: m1  m2  1

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344

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Exercise 18.1: Distance between points: 1.

Find the distance between the given pairs of points: 1.1 (3;7 ) and (5 ; -2)

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345

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1.4 (-7 ; 1) and (0 ; 0) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.5 (-3 ; 2) and (-6 ; -7) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.6 (-5 ;-4) and (3 ; -8) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Gcse Mathematics

346

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Given the coordinates of the vertices of ď &#x201E;ABC , in each case ( 2.1 to 2.5) Determine: A. the perimeter of the triangle. B. Whether the triangle is equilateral, isosceles or scalene. C. Whether or not the triangle has a right angle. 2.1

A(2 ; -5) ; B(5 ; 5); C(-2 ; 4)

A(5 ;1) ; B(1 ; 3) ; C(1 ; -2)

Gcse Mathematics

347

GOï&#x201A;·MATH WORKBOOKS

2.3 A(2 ; 1) ; B(2 ; -2) ; C(7 ; -2) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2.4

A(0 ; 0) B( 3 ; 1) ; C( 3 ; -1)

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2.5

348

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A(-4 ; 1) B(-1 ;-3) ;C(0; 4)

Show that: 3.1

A(-3 ; 2) , B(3 ;6), C(9 ;-2) and D(3 ; -6) are vertices of a parallelogram.

Gcse Mathematics

349

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3.2 (6 ;-4) , (5 ;3) (-2 ; 2) and (-1 ; -5) are vertices of a square. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ Mid-points of lines: 4.

Calculate the coordinates of the midpoints of the line joining the following points: 4.1 (-3 ;1) and (1 ; 5)

Gcse Mathematics

350

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4.2 (-2 ; 3) and (6 ; 3) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.3 (4 ; -1) and (-1 ; 3) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.4 (0 ;0 ) and (3 ; -8) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.5 ( 3;1) and (3 3;1) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Gcse Mathematics

351

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

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

Gcse Mathematics

5.3

352

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(x ; y) is the centre of a circle on diameter AB where A(-2 ; -1) and B(-1 ; 9).

____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 5.4 (x ; 3) is the centre of a circle with diameter MN. M (5 ; -2) and N(-7 ; y) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Gcse Mathematics

353

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Exercise 18.2: Formulae:

AB  ( x1  x2 ) 2  ( y1  y2 ) 2 Gradient = m =

 x  x2 y1  y2  ; Mid-point =  1  2   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:

1.1

the length of AB in units.

1.2

The co-ordinates of the mid – point (C) of line AB.

The gradient of line AB.

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1.4

354

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show that points A; B and C are collinear.

____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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355

2. y

A C 0

The Points A(-4 ;3) ; B(-4 ; -4) ; C(6 ; 1) and D(6 ; 8) lie on a Cartesian plane. Determine: 2.1

the length of AD.

2.2

the mid-point of DC

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2.3

356

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the gradient of BC

2.4

show that ABCD is a parallelogram.

2.5

the co-ordinates of the point of intersection of the diagonals AC & BD

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357

A(-3;3) C(8;2)

-5

-2

B(3;-3) -4

The figure above represents a triangle on a Cartesian plane. A( -3 ;3) ; B(3 ; -3) and C(8 ; 2) 3.1 Calculate the perimeter (distance around) of ď &#x201E;ABC .

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358

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3.2 Prove that triangle ABC is right angled at B.

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

Gcse Mathematics

359

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

4.1

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

P (4;1) , Q (3;0) and R (1;3) 4.2 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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360

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U (5;2) , V (1;1) and W (13;1) 4.3 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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361

The gradients and Inclinations of straight lines:

y y1  y 2  x x1  x2 y y1  y 2  Inclination = Tan  = x x1  x2 Gradient = slope =

NB: The inclination of a line is measured from the positive x –axes to the line in question.

Angle of Inclination 

Exercise 18.3: Calculate the gradients of the lines joining the following points: 1.1 (-3 ; 2) and (1 ; 1) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 1.2

(4 ; 3 ) and (-1 ; 8) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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1.3

362

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(-3 ; -5) and (1 : 3) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Write down the gradients of the lines perpendicular to the lines in 7. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3 3.1

Calculate the inclinations of the line AB in each of the following cases. A(-3 ; 2) and B(-5 ; 0) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3.2

A(-2 ; 1 ) and B(1 ; -2) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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3.3

363

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A( 3 ; 1) and B((2 3 ; -2) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3.4

A(-1 ;2) and B(1 ; -1) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3.5

A(-5 ; 2) and B(3 ; -1) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

4 4.1

Calculate the gradients of lines with inclinations of: 45º ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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4.2

364

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60º ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

4.3

150º ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

4.4

110º ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Calculate the gradients of the following lines and state whether they are A. Parallel B. Perpendicular C. Neither.

5.1

A(0 ; -1) , B(-4 ; -2) , C( -3 ; 1) and D ( 1 ; 2) ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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5.2

365

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A(6 ; -10) , B(0 ; 4) , C( 6 ; 0) and D ( -4 ; -3) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

5.3

A(-3 ; 5) , B(5 ; -1) , C( -2 ; -1) and D ( 1 ; 3) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

5.4

A(-2 ; -4) , B(3 ; 1) , C( 5 ; -1) and D ( -2 ; -8) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

6 6.1

Show that the following points are collinear: ( lie on the same line) A(-2 ; -6) , B(2 ; -4) , C( 4 ; -3) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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6.2

366

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A(-5 ; 5) , B(1 ; 1) , C( 4 ; -1) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ Equations of straight lines:

7 7.1

Determine the equation of a line where the gradients and a point on the line are given as follows: 1 ; ( 2;3) 2 ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

7.2

3 ; (3;1) 2

 2; (1;3) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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8 8.1

367

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Lines passing through the following points: (-2 ; 4) and (2 ; 2) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

8.2

(-1 ; 1) and (1 ; 5) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

8.3

(-3 ; -2) and (-1 ; -1) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

8.4

(3 ; -3) and (3 ; -6) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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368

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A line with the slope of 3 and intersecting the y – axes at 2 ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

parallel to y  3x  2 and passing through (3 ; 1) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Through (-2 -1) and perpendicular to 3 y  2 x  6 ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Through (-1 ; 3 ) and an inclination of 120º ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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13.

369

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A(-2 ; 1), B(3 ; 3) and C(6 ; -3) are the vertices of a triangle . Determine: 13.1 The coordinates of M, the mid-point of AC. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 13.2 the gradient of AC. ____________________________________________________________ ___________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ 13.3 the equation of the perpendicular bisector of AC. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 13.4 the equation of the median BM ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 13.5 the equation of the altitude from B to AC. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Gcse Mathematics

370

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Circle centre the origin: 14.

Determine the equation of a circle with centre origin and: 14.1 radius = 3 cm ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 14.2 radius = 3 2 cm ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 14.3 Passing through point (-2 ; 3) ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 14.4

Passing through point ( -4 ; -2)

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371

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15. A(-3 ; 4) is a point on a circle with centre at the origin: 15.1 Determine the equation of the circle. ____________________________________________________________ ____________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 15.2

Determine the coordinates of B if AB is a diameter.

Show that the point C(0 ; 5) lies on the circle. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

15.4 Prove that đ??´đ??śĚ&#x201A; đ??ľ is a right angle. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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372

Tangents to circle centre origin: y

B( 3;4 )

O (0;0)

x C

In the figure O is the centre of the circle and the origin of the set of axes. ABC is a tangent to the circle at point B. OBď &#x17E; AB Determine: 16.1 The gradient of OB. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

16.2

The equation of AC, the tangent. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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16.3

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373

The equation of the circle centre O. ____________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ ___________________________________________________________

Exercise 18.4: QUESTION 1: The Points A(-8 ;3) ; B(-1 ; 1) ; C(1 ; -4) and D(-6 ; -2) lie on a cartesian plane. 4

A(-8 ; 3)

B(-1 ;1)

-10

-5

-2

D(-6 ; -2)

-4

C(1 ; -4)

-6

Determine: 1.1

the length of AD.

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1.2

374

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the mid-point of DC

1.3

The gradient of BC

1.4

The length of BC ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

1.5

the inclination of BC ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ___________________________________________________________

1.6

the equation of BC ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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1.7

375

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The perimeter of ABCD

1.8

State what shape is represented in the diagram ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

QUESTION 2: 2.1

Show that A(-5 ; -3); B(-1 ; 0) and C(3 ; 3) lie on the same straight line.

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2.2

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376

P(13 ; t) , Q(7 ; 2) and R(4 ; 1) are points in a Cartesian plane. If P , Q and R are collinear, then determine the value of t. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

QUESTION 3: B(2 ; 3) A(-5 ;2)

D( t ; t-3 ) C(-3 ;-2)

Three points A 5;2 ; B(2 ; 3)and C  3;2 in a Cartesian plane are given. 3.1 Calculate the distance AB. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ 3.2

Calculate the gradient of AC. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

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3.3

377

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Calculate the value of t if the point D(t ; t- 3) is such that AC // BD. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3.4

Calculate the mid-point of BC. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3.5

Determine, by calculation , whether the quadrilateral is a parallelogram. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

3.6

Give the equation of BC. ____________________________________________________________ ____________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________

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3.7

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378

What is the size of the angle of inclination of BC with the positive x – axes. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ___________________________________________________________ ____________________________________________________________

3.8

Calculate the size of ACˆ B . ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

19.1 Transformation: The process of transformation occurs when you move all the points of a geometric shape using certain rules. Transformation results in an image of the original shape. If we transform a point (P) the transformation is referred to as (P´) (pronounced P prime). Horizontal and vertical translations: A translation changes the position of a point or shape by “sliding” it to another position under a given Rule. Translating a shape (region) changes its position BUT not its size or shape. 2.1.1 The point P is translated by moving it 3 units to the right. The new position is P´(5;1)

P(2;1) -5

P'(5;1) 5

-2

-4

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2.1.2

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379

The point P(2;1) is now translated by moving it 4 units upward. The new position is P´(2;5)

P'(2;5) 4

P(2;1) -5

-2

-4

2.1.3

The point P(2;1) is translated by moving 3 units upward and 3 units to the left. The new position is P´(-1;4)

P'(-1;4) 2

P(2;1) -5

-2

-4

The rules that govern the 3 moves above can be written as follows. 1. ( x  3; y ) or ( x; y )  ( x  3; y ) 2. ( x; y  4) or ( x; y )  ( x; y  4) 3. ( x  3; y  3) or ( x; y )  ( x  3; y  3)

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380

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Exercise 19.1: 1. Write down the coordinates of P(5;7) if it is translated as follows and write down the rule governing each translation.: 1.1 3 units to the right and 3 units upwards. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 1.2

3 units to the left and 3 units upward.

3 units to the right and 3 units downwards.

( x; y )  ( x  3; y  3) 2.1 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

( x; y )  ( x  3; y  3) 2.2 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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381

( x; y )  ( x  3; y ) 2.3 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

( x; y )  ( x; y  3) 2.4 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2.2

Translation of shapes(regions):

NB: Translating a shape (region) changes its position BUT not its size or shape. Examples: 2.2.1. Translate the ABC in quadrant 1 using the rule ( x; y )  ( x  6; y  2) Position 1: ABC . Coordinates are A(4;3) ; B( 3;1) & C(5;1) Position 2:Translated position: A' B' C' . coordinates are A´(-2;5) ; B´( -4;3) & C´(-1;3) A'(-2;5) 4

A(4;3) B'(-4;3)

C'(-1;3) 2

C(5;1)

B(3;1) -5

-2

-4

2.2.2.

Translation of a parallelogram PQRS: Coordinates: P(-5;-3) ; Q(-2;-3)

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382

R(-1;-1) & S(-5;-1) Translate using Rule: ( x; y )  ( x  8; y  5) Translation: P´Q´R´S´ : P´ (3;2) ; Q´ (6;2) ; R´ (7;4) & S´ (4;4)

S'(4;4)

R'(7;4)

P'(3;2)

-5

S(-5;-1)

R(-1;-1) -2

P(-5;-3)

Q(-2;-3) -4

Q'(6;2)

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383

Exercise 19.2: 1. The shape is drawn on the Cartesian plane for you and is called region A. Task: Draw the image of region A after each of the following translations: Indicate each translation clearly using an arrow. i.e. ( x; y )  ( x  7; y  2) 1.1

( x; y )  ( x  2; y  2)

1.2

( x; y )  ( x  2; y  2)

1.3

( x; y )  ( x  2; y  2)

1.4

( x; y )  ( x  2; y  2)

1.5

( x; y )  ( x  3; y  1)

Example 4

(-2 ;3)

(0;3)

A 2

-10

-5

(0 ;2)

-2

-4

-6

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384

2. Determine the translations on the grid below:

P'' 4

P' 2

-5

P''' -2

-4

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2.3

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385

Reflections of points and shapes.

The line of reflection or the mirror line or the axis of symmetry : Commonly the xaxis; the y-axis and the lines y = x and y = -x. 2.3.1. Point P(4;2) is reflected in the y-axis and the new position is P´(-4;2) What do we notice about the x – coordinate of the refection.

P(4;2)

P'(-4;2)

-5

-2

-4

The rule for above is : ( xy)  ( x; y ) 2.3.2. Point P(4;2) is reflected in the x-axis and the new position is P´(4;-2) What do we notice about the x – coordinate of the refection?

P(4;2)

-5

-2

-4

The rule for above is: ( x; y )  ( x; y )

P'(4;-2)

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386

2.3.3. Point P (4; 2) is reflected in the line y = x and the new position is P´ (2; 4) What do we notice about the x – coordinate of the refection?

line y = x 4

P'(2;4)

P(4;2)

-5

-2

The rule for above is: (x ; y)  (y ; x) 2.3.4. Point P (4; 2) is reflected in the line y = -x and the new position is P´ (2; 4) What do we notice about the x – coordinate of the refection?

P(4;2)

-5

-2

y = -x -4

P'(-2;-4)

The rule for above is :

( x; y )  ( y; x)

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387

Exercise 19.3: 1. Reflect the points in the x; y – plane as instructed and write down the new coordinates and the rule for the reflection. Use the grid below to illustrate your answers. 1.1

Reflect P(5.2) in the x – axis.

____________________________________________________________________ ___________________________________________________________________ 1.2 1.3

Reflect P(5.2) in the y – axis.

____________________________________________________________________ ___________________________________________________________________ 1.4

Reflect P(5.2) in the line y = x

____________________________________________________________________ ___________________________________________________________________ 1.4 Reflect P(5.2) in the line y = -x ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

P(5;2)

-5

-2

-4

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388

2. Reflect the shape A as follows and in each case give the new coordinates of the reflected shape and the rule for its refection. 2.1

Reflect the shape A in the y- axis to give shape B.

-5

-2

-4

2.2

Reflect the shape A in the x – axis to give shape C.

-5

-2

-4

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2.3

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389

Reflect the shape A in the line y = x to give shape D.

y=x 2

-5

-2

-4

2.4

Reflect the shape A in the line y = -x to give shape E.

y = -x

-5

-2

-4

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390

3. In each of the diagrams below draw the image according to the specific rule and identify the type of transformation used. Examples: _6

_A(-2;4)

_A(2;4)

B(-5;3) _ 2. _

_1.

_B(5;3)

_C(-1;1)

_C(1;1)

_- 5

C(-5;-1) _

_C(1;-1)

_- 2

_B(-1;-3) _4.

A(-4;-4) _

_B(5;-3)

_- 4

_A(2;-4)

Diagram 1 is reflected about the y- axis using the rule: ( x; y )  ( y; x) To give diagram 2. Diagram 2 is reflected about the x- axis using the rule ( x; y )  ( x; y ) To give diagram 3. Diagram 3 is translated using the rule ( x; y )  ( x  6; y ) To give diagram 4.

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391

In each of the diagrams below draw the image according to the specific rule and identify the type of transformation used. 3.1

( x; y )  ( x  3; y )

A(4;4)

A(2;2) B(5;1)

-5

-2

-4

3.2

( x; y )  ( x  3; y  3)

A(-1;2)2

B(-3;1)

D(2;1)

-5

C(-2;-1) -2

-4

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3.3

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392

( x; y )  ( x; y ) 4

-5

A(-4;-2)

-2

B(-2;-3) -4

C(-5;-5) -6

3.4

( x; y )  ( y; x) 4

-5

A(2;-1)

-2

-4

-6

B(5;-2)

C(1;-4)

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393

2.4 A Glide Reflection is a combined transformation of a reflection followed by a translation parallel to the line of reflection of a region. Example 1. Express a glide reflection from: 1. A to B 2. B to A Solution: 1. The translation from A to C is ( x; y )  ( x  6; y ) The line of reflection is y = 0 (x – axis) given by rule ( x; y )  ( x; y ) The glide reflection from A to B is written as ( x; y )  ( x  6; y ) 2. The translation from B to A is ( x; y )  ( x  6; y ) The line of reflection is y = 0 (x – axis) given by rule ( x; y )  ( x; y ) The glide reflection from B to A is written as ( x; y )  ( x  6; y )

(6;3)

(0;3) 2

-5

D -2

B (0;-3)

-4

(6;-3)

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394

Example 2. Express a glide reflection from: 3. A to B 4. B to A In a single statement. Solution: 1. The line of reflection is y = x given by rule ( x; y )  ( y; x) The translation from A to C parallel to y = x is ( x; y )  ( x  3; y  3) The glide reflection from A to B is written as ( x; y )  ( y  3; x  3)

(3;6)

C 4

(6;3)

(0;3) B 2

(3;0)

-5

-2

-4

2. The translation from B to A parallel to y = x is ( x; y )  ( x  3; y  3) The line of reflection is y = x given by rule ( x; y )  ( y; x) The glide reflection from B to A is written as ( x; y )  ( y  3; x  3)

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395

Example 3: Interpret the transformation ( x; y )  ( x; y  3) as two possible glide reflections. Solution: ( x; y )  ( x; y  3) can be seen as ( x; y )  ( x  0; y  3) so there is a translation of (3;0). The sign has changed so there is a reflection about the y – axis: ( x; y )  ( x; y ) The two glide reflections are: a) a reflection about the y – axis followed by a translation of (0;3) OR: b) a translation of (0;3) followed by a reflection about the y – axis.

NB: Look at the grid below and read what the transformations actually take place:

(5;4)

(0;3) 2

-5

-2

(0;-3) -4

B (5;-4)

The transformation from A to B or from B to A is not a glide reflection, because the translation is not parallel to the line of reflection y = 0 It is a transformation made up of a reflection and a translation and visa versa. The transformation form A to B can be seen as:

( x; y )  ( x; y ) a reflection about y = 0 followed by ( x; y )  ( x  5; y  1) a translation of (5;-1) which gives a single rule: ( x; y )  ( x  5; y  1)

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396

Rotations of shapes:

1. Rotations about the origin through ± 90º : + 90º indicates a rotation in an anti-clockwise direction: - 90º indicates a rotation in a clockwise direction. 2. Rotations about the origin through ± 180º: It is immaterial whether rotation is through ± the final resting position will be the same.

Rotations about the origin and its effects on co-ordinates To rotate a point through +90  about the origin: The co-ordinate s will reve rse place s and the sign of the y-value changes: x y and y  x e.g. P(3;4)  P'(-4;3) To rotate a point through -90  about the origin: The co-ordinate s will reve rse place s and the sign of the x-value changes: x y and y  x e.g. P(3;4)  P'(4;-3) To rotate a point through  180 about the origin: The co-ordinate s remain in the same positions however both the ir s igns change. P(3;4)  P'(-3;-4) 10

NB: To rotate a point 90 about the origin: the co-ordinates w ill rev erse places and the sign of the y v alue change s. x  y and y  x 8 the sign of the y co ordinate changes.

90 rotations about origin: P(2;1)  P'(-1;2) Q(-5;2)  Q'(-2;-5) R(-5;-3) R'(3;-5)

To rotate 180  about the origin: the co-ordinates remain the same however both their signs cha nge.

180 about origin: 6

P(2;1)  P''(-2'-1) Q(-5;2) Q''(5;-2) R(-5;-3) R''(5;3)

R''(5;3) Q(-5;2)

P'(-1;2)

P(2;1)

-10

-5

P(-2;-1)

Q''(5;-2)

-2

R(-5;-3)

-4

R'(3;-5)

Q'(-2;-5) -6

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397 8

A(5;6)

B'(-4;6)

A'(-6;5) A(0;4)

Rotation of shape thru 90

C(3;4)

D'(-3;3)

B(6;4)

D(3;3)

C'(-4;3)

Q(5;2)

-10

-5 B(-4;0)

A'(-4;0) C(-1;-1)

Q(5;2)  Q'(2;-5) the x and y co-ordinates interchange\ and the sign of the y c o-ordinate changes.

-2

B'(-6;-4) C'(-3;-4)

ROTATING CLOCKWISE ie thru -90 

C'(1;1)

-4

B'(0;-4)

ROTATING Sha pes thru 180 Q'(2;-5)

Rotation of ABC thru 180

the x and y co-ordinates stay the s ame BUT change signs

-6

A'(-5;-6) -8

Exercise 19.4: 1.

Use the grid below and show the following rotations: 1.1 Rotate the point P(3;4) about the origin through 90º 1.2 Rotate the point P(3;4) about the origin through -90º 1.3 Rotate the point P(3;4) about the origin through 180º

P(3;4) 4

-5

-2

-4

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398

2. Rotate the shape on the grid below as indicated. 2.1 Rotate the shape about the origin through -90º 2.2 Rotate the shape about the origin through -180º

A(3;3) D(1;2) 2

C(2;1)

B(4;1)

-5

-2

-4

3. Transform the shape as indicated: Use the grid below: 3.1 Translate the shape ABCD using the rule: ( x; y )  ( x  6; y  2) 3.2 Rotate the shape about the origin through -90º and then reflect this in The y – axis.

A(3;3) D(1;2) 2

C(2;1) -5

B(4;1) 5

-2

-4

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4. Example of a shape ABC below rotated through 90º (anticlockwise) about the point ‘A’. NB: The shape swivels about the point A. i.e. A remains fixed and the other points rotate around it.

_B _2

_C

_- 5

_- 2

_- 4

5. Example of a shape ABC below rotated through -90º (clockwise) about the point ‘A’. B

C' -5

-2

-4

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400

2.6 Enlarging or Reducing Shapes: Shapes are enlarged by a factor ‘k’: Method : 

Multiply each of the vertices by the factor k.



And using these new co-ordinates redraw the shape.



the area of the enlarged shape will increase/decrease by the square of the factor ‘k’:

Examples:

L=8

(-4;4)

(4;4)

k = 4 and Are a increase d by 4 2 =16 times

L=6

-3;3

3;3

k = 3 and Are a increase d by 32 = 9 times 2

L=2

-1;1

-10

L=4 1;1

k = 2 and Are a increase d by 22 = 4 times

-5

-1;-1

1;-1 -2

3;-3

-3;-3 -4

(-4;-4)

(4;-4)

In Scaling a figure : The side s are incr ease d or de cre ased -6 in dimension by a factor re ferre d to as 'k'. This increase s or de cre ase s the are a of the figure as follows: Incre ase by the squar e of the re le v ant factor. -8 i.e . increase/ decreases by multiplying original area by k2 -10

Triangle enlarged by a factor of 3 Ratio of areas is 1: 9 5

(4;-1)

(1;-1) SHAPE -2

(3;-3)

(12;-3)

(2;-3) -4

ENLARGEMENT -6

-8

(6;-9) -10

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Example: Task: Complete the following transformations as indicated below on the grid supplied. 1.4 The shape ABC is rotated about the origin through 90º and then scaled 1 through the origin by a factor of . State the ratio of the area of ABC 2

To ABC 1.5 Rotate ABC about the origin through 180º and then scaled through the origin by a factor of 2. State the ratio of the area of ABC to

ABC Answers: A(4;4)

A'(-2;2) B'(-1;2) 2 1

C(2;2)

B(2;4)

C'(-1;1)_ -5

-2

B''(-8;-4)

C''(-4;-4)

-4

2 -6

-8

A''(-8;-8) -10

Answers: 1. 2.

1 or 4 : 1 4 1 Ratio of areas is 1 : 4 or : 1 4

Ratio of areas is 1 :

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402

Exercise 19.5: 1. A triangle with vertices (2;-2); (3;-2) and (2;-4) is scaled through the origin by a factor of 2. 1.1 What will the vertices of the new triangle be? 1.2 By what factor will each side of the triangle increase? 1.3 By what factor will the area of the triangle increase? 1.5 Sketch the enlargement on the grid below. 8

-10

-5

-2

-4

-6

(2;-2)

(3;-2)

(2;-4)

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403

2.7 Putting Shadow sizes on paper: A square piece placed between a flashlight and a wall will have a shadow projected on the wall The shadow will be double the size of the square if the distance from the flashlight to the wall is equal to the distance from the square paper to the wall. Shadow size will have sides of 20cm if the paper size is 10cm If the distance from the paper to the wall is double the distance from the flashlight to the paper then the shadow will be triple the size. i.e the shadow dimensions would be 30cm in this case.

W S is on side of the s quare S

1 metre

WL is the shadow on the wall

F F is the flashlight U L

Summary: In the above diagram: 1. Using a 10cm square piece of paper: If FA  2 FQ then WL will be 20cm. 2. Using a 10cm piece of paper: If FA  3FQ ; then WL will be 30cm This theory can be applied to project any required size shadow on to a wall. The facts above are based on the theorem of similarity in polygon.

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404

Rotations using trigonometry:

14 8

12 6

P'[ rcos(90 + );rsin(90 + )] P'(-4;4) 10 4

P(4;4)

r=4 2

(90 +)

6 -10

P'[rcos( + );rsin( + )] P'(a 1;b1)

cos(+) =

r a1=r cos(+)



-5

sin(+)=

P(a;b)

-24

r b = r sin(+)

cos =

r -42

r a=r cos 

sin =

r b = r sin

  -6 -10

-5

Given the point P (4; 4) with an angle at the origin of. TASK is to rotate the point through 90º about the origin and find the co-ordinates of the new point P(x; y) Solution: P [r cos(  90 ); r sin(  90 )] P[4 2 cos120 ;4 2 sin120 ]  1 3 P4 2  ;4 2   2 2   P   2 2 ;2 6





P (2,83;4,90)

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405

Rotation of an angle other than 90º on a Cartesian plane. Example 1. Rotate a point P (2; 4) about the origin through an angle of 30º.

P ' ( x cos  y sin  ; y cos  x sin  ) P ' ( x cos 30  y sin 30 ; y cos 30  x sin 30 ) 2 3 4 1 4 3 2 1 P '     ;     1 2 1 2 1 2 1 2  P'





3  2;2 3  1

Example 2: Rotate a point P (2; 4) about the origin through -135º. P ' ( x cos  y sin  ; y cos  x sin  )

P '[ x cos(135 )  y sin(135 ); y cos(135 )  x sin(135 )] 2  2 4 2 4  2 2 2  P '     ;     1 2 1 2 1 2 1 2  



P'  2  2 2 ;  2 2  2



P ' ( 2 ; 3 2 ) Example 3: Rotate a point P (4; 2) about the origin through 90º in an anticlockwise direction and then rotate P ( x ; y ) a further 120º anticlockwise and find the coordinates of P ( x ; y ) Rotation of point P (4; 2) through 90  about the origin. New co-ordinates are: P (2;4) Now rotate P (-2; 4) a further 120  anticlockwise: P  x cos  y sin  ; y cos  x sin   P ( x cos120  y sin120 ; y cos120  x sin120 )  1 3 1 3  P  (2   )  (4  ); (4   )  (2  )  2 2 2 2   P  1  2 3 ; 2  3





New co-ordinates are:

P (1 - 2 3 ; -2 -

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406

20.Triangles: Theorems: The sum of the angles of a triangle equals 180ď&#x201A;° A

2 1

Given: â&#x2C6;&#x2020;ABC. RTP: đ??´Ě&#x201A; + đ??ľĚ&#x201A; + đ??śĚ&#x201A; = 180ď&#x201A;° Proof: Produce BC to D and draw CE parallel to AB. đ??śĚ&#x201A;2 = đ??´Ě&#x201A; (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;Ą ď&#x192;?, đ?&#x2018; đ??´đ??ľ ď źď ź đ??śđ??¸) đ??śĚ&#x201A;3 = đ??ľĚ&#x201A; (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x;đ?&#x2018;&#x; ď&#x192;?, đ?&#x2018; đ??´đ??ľď źď ź đ??śđ??¸) But đ??śĚ&#x201A;1 + đ??śĚ&#x201A;2 + đ??śĚ&#x201A;3 = 180ď&#x201A;° ( ď&#x192;?, đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018;&#x17D; đ?&#x2018; đ?&#x2018;Ą đ?&#x2018;&#x2122;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2019;) ď &#x153;đ??´Ě&#x201A; + đ??ľĚ&#x201A; + đ??śĚ&#x201A; = 180ď&#x201A;° The exterior angle of a triangle is equal to the sum of the interior opposite angles A E

2 1

Given: â&#x2C6;&#x2020;ABC with BC produced to D. RTP: đ??´đ??śĚ&#x201A; đ??ˇ = đ??´Ě&#x201A; + đ??ľĚ&#x201A; Proof: Draw CE parallel to AB. đ??śĚ&#x201A;1 = đ??´Ě&#x201A; (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;Ą ď&#x192;?, đ?&#x2018; đ??´đ??ľ đ??źđ??ź đ??śđ??¸ đ??śĚ&#x201A;2 = đ??ľĚ&#x201A; (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x;đ?&#x2018;&#x; ď&#x192;?, đ?&#x2018; đ??´đ??ľ đ??źđ??ź đ??śđ??¸

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407

In an isosceles triangle , the sides opposite the equal angles are equal. A 1 2

ď&#x201A;ˇ

Given: â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;Ąâ&#x201E;&#x17D; đ??ľĚ&#x201A; = đ??śĚ&#x201A; RTP: AB = AC. Proof: Construct AD to bisect đ??´Ě&#x201A; , cutting BC at D. In â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??´đ??śđ??ˇ Ě&#x201A;1 = đ??´ Ě&#x201A;2 (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;˘đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;) đ??´ đ??ľĚ&#x201A; = đ??śĚ&#x201A; ( đ?&#x2018;&#x201D;đ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) AD is common â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??śđ??ˇ (đ??´đ??´đ?&#x2018;&#x2020;) â&#x2C6;´ đ??´đ??ľ = đ??´đ??ś Conversely: The angles opposite equal sides of a triangle are equal. The Mid â&#x20AC;&#x201C; Point Theorem. The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half its length. 1

RTP: DE â&#x20AC;&#x2013; đ??ľđ??ś and đ??ˇđ??¸ = 2 đ??ľđ??ś A

Proof: Construct EF = DE and join FC. đ??źđ?&#x2018;&#x203A; â&#x2C6;&#x2020;đ??´đ??ˇđ??¸ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??śđ??šđ??¸ AE = CE ( Given) DE = EF ( by construction) đ??¸Ě&#x201A;1 = đ??¸Ě&#x201A;2 (đ?&#x2018;&#x2030;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;Ś đ?&#x2018;&#x153;đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; ) â&#x2C6;&#x2020;đ??´đ??ˇđ??¸ â&#x2030;Ą â&#x2C6;&#x2020;đ??śđ??šđ??¸ ( đ?&#x2018;&#x2020;đ??´đ?&#x2018;&#x2020;) Ě&#x201A; đ??¸ = đ??śđ??šĚ&#x201A; đ??¸ â&#x2C6;´ đ??´đ??ˇ â&#x2C6;´ đ??´đ??ľâ&#x20AC;&#x2013;đ??šđ??ś ( đ??´đ?&#x2018;&#x2122;đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x2019;đ?&#x2018;&#x17E;đ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;) BD = DA (given) BD= FC ( proved above) â&#x2C6;´ đ??ˇđ??śđ??šđ??ľ đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x17D; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x161; ( đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;&#x2019; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x2013;đ?&#x2018;&#x; đ?&#x2018;&#x153;đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;&#x17E;đ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;) â&#x2C6;´ đ??ˇđ??š = đ??ľđ??ś ( đ?&#x2018;&#x153;đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x17D; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x161;) 1 đ??ˇđ??¸ = 2 đ??ľđ??ś and đ??ˇđ??¸â&#x20AC;&#x2013;đ??ľđ??ś

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Congruence: Definition: Two triangles are said to be congruent if they are equal in all respects. There are four cases of congruence: i.e. methods of proving triangles are congruent. â&#x;šTwo angles and one corresponding side are equal. ( AAS). â&#x;šTwo sides and the angle between them are equal. (SAS). â&#x;šThree sides are equal. (SSS) â&#x;šRight angle, hypotenuse and one further side are equal. (RHS) NB: If two parallel lines are cut by a transversal the angles formed are: Alternate angles, which are equal. ď&#x201A;ˇ ď&#x201A;ˇ

Corresponding angles, which are equal. ď&#x201A;ˇ

ď&#x201A;ˇ

Co-interior angles, which are supplementary. x x + y = 180ď&#x201A;° y

1 4

2 3

If two lines intersect then two types of angles are formed. ( đ?&#x2018;łđ?&#x;? & đ?&#x2018;łđ?&#x;? ) 1. Adjacent supplementary angles ( angles on a straight line) i.e. 1 + 2 = 180Â° & 3 + 4 = 180Â°. 2. Vertically opposite angles. đ?&#x2018;&#x2013;. đ?&#x2018;&#x2019; 1 = 3 & 2 = 4

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Exercise 20.1: In the sketch below, AB = CD andđ??´đ??ľĚ&#x201A; đ??ś = đ??ˇđ??śĚ&#x201A; đ??ľ . A

D E

Prove that: 1.1 â&#x2C6;&#x2020;đ??´đ??ľđ??ś â&#x2030;Ą â&#x2C6;&#x2020;đ??ˇđ??śđ??ľ _________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 1.2 â&#x2C6;&#x2020;đ??´đ??ľđ??¸ â&#x2030;Ą â&#x2C6;&#x2020;đ??ˇđ??śđ??¸ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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2.In the sketch below đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; = đ?&#x2018;&#x2020;đ?&#x2018;&#x2021; ; đ?&#x2018;&#x192;đ?&#x2018;&#x2026; = đ?&#x2018;&#x192;đ?&#x2018;&#x2020; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x192;đ?&#x2018;&#x2026;Ě&#x201A; đ?&#x2018;&#x2020; = đ?&#x2018;&#x192;đ?&#x2018;&#x2020;Ě&#x201A;đ?&#x2018;&#x2026;. P

2.1

Prove â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2026;đ?&#x2018;&#x201E; â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2020;đ?&#x2018;&#x2021;.

Deduce that PQ = PT

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Prove that â&#x2C6;&#x2020;đ??´đ??¸đ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??ľđ??śđ??ˇ in the diagram below: Given: AD = BD ; AE = BC; AE ď &#x17E; BD and BC ď &#x17E; AD. A C

F D E

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412

Prove that â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??ľđ??ś in the diagram below: Given: AD = BC ; đ??ľđ??´Ě&#x201A;đ??ˇ = đ??´đ??ľĚ&#x201A; đ??ś D

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413

Similarity: Two triangles are said to be similar if they are equiangular or conversely if the ratios of their corresponding sides are in proportion. Examples: A

A r

D x

E 6cm

8cm

3cm

2cm

E 4cm

y x

Юљ╝ЮЉЏ РѕєРђ▓ ЮЉаЮљ┤ЮљхЮљХ ЮЉјЮЉЏЮЉЉ ЮљИЮљиЮљ╣: Юљ┤╠ѓ = ЮљИ╠ѓ (ЮЉћЮЉќЮЉБЮЉњЮЉЏ) ╠ѓ (ЮЉћЮЉќЮЉБЮЉњЮЉЏ) Юљх╠ѓ = Юљи ЮљХ╠ѓ = Юљ╣╠ѓ (ЮЉћЮЉќЮЉБЮЉњЮЉЏ) Рѕ┤ РѕєЮљ┤ЮљхЮљХ ле РѕєЮљИЮљиЮљ╣ ( Юљ┤Юљ┤Юљ┤) Юљ┤Юљх ЮљхЮљХ Юљ┤ЮљХ Рѕ┤ ЮљИЮљи = ЮљиЮљ╣ = ЮљИЮљ╣

4cm

Юљ╝ЮЉЏ РѕєРђ▓ ЮЉаЮљ┤ЮљхЮљХ ЮЉјЮЉЏЮЉЉ ЮљИЮљиЮљ╣: Юљ┤Юљх 6 =3=2 ЮљИЮљи ЮљхЮљХ

ЮљиЮљ╣ Юљ┤ЮљХ

=2=2 8

=4=2 Рѕ┤ РѕєЮљ┤ЮљхЮљХ ле РѕєЮљИЮљиЮљ╣ (ЮЉаЮЉќЮЉЉЮЉњЮЉа ЮЉќЮЉЏ ЮЉЮЮЉЪЮЉюЮЉЮЮЉюЮЉЪЮЉАЮЉќЮЉюЮЉЏ) ЮљИЮљ╣

Triangles: Properties: 1. The sum of the interior angles is 180№ѓ░. 2. The exterior angle of a triangle is equal to sum of the interior opposite angles. 3. Isosceles triangles have two sides equal and the angles opposite these sides are also equal. 4. Equilateral triangles have all three sides and all three angles equal. 5. Scalene triangles have no sides or angles equal.

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414

Quadrilaterals:

Number None

opposites sides parallel

Both pairs

One pair Both pairs

 

One pair

Square

Rectangle

Rhombus

  



 



 



 







 



 







 



 



  

Both pairs None





All four sides None

Opposite angles equal



One pair

None

Adjacent sides equal



One pair

None opposites sides equal

Parallelogram

Property

Trapezium

Kite

Properties of Quadrilaterals

 



  



One pair Consecutive angles equal

Both pairs All four angles none only one

Line of symmetry



 

  



  

only two



 

four

Diagonals are equal

Diagonals bisect each other



Diagonals are perpendicular







 



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415

Relationship between sides and angles:

Number of sides of a polygon 3 4 5 6 7 8 9 10 n Sum of interior angles 180º 360º 540º 720º 900º 1080º 1260º 1440º (n-2)180º Size of one interior angle of regular polygons 60º

90º

108º 120º 128,570º

135º

140º

144º

(n-2)180º n

The pattern is (number of sides-2)180º Sum of interior Angles = (n-2)180º Number of sides of a polygon 3 4 5 6 7 8 9 Sum of exterior angles 360º 360º 360º 360º 360º 360º 360º

Number of sides of a polygon 4 5 6 7 8 9 Number of Triangles in a Regular Polygon 2 3 4 5 6 7

3 1

10 360º

10 8

Exercise 20.2: 1

Find the values of the variables in the following (i.e. x ; y & z) 1.1

AB//DC and AB=DC A



62 D



y C

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1.2

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416 PQRS is a rectangle: P



36

y x



ABCD is a rhombus: A

35

x D

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417

1.4LMNO is a trapezium with LM // PO and LM = LP ; đ??żĚ&#x201A; = 105Â° Ě&#x201A; đ?&#x2018;&#x201A; = 65Â° and đ?&#x2018;&#x192;đ?&#x2018;&#x20AC; L 105Â°

ď&#x201A;Ž

65Â°

y z

ď&#x201A;Ž

1.5

ABCD is a kite. AB = AC & BD = CD. đ??´đ??śĚ&#x201A; đ??ľ = 60ď&#x201A;°, đ??ˇđ??ľĚ&#x201A; đ??ś = 70Â° A x

60ď&#x201A;°

y 70ď&#x201A;°

Gcse Mathematics

1.6

418

GOï&#x201A;·MATH WORKBOOKS

ABCDEF is a regular hexagon.

Gcse Mathematics

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419

Paralellograms: Theorem A Given: Parallelogram ABCD Ě&#x201A; RTP: AB = CD ; AD = BC; đ??´Ě&#x201A; = đ??śĚ&#x201A; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ľĚ&#x201A; = đ??ˇ Proof: Construct diagonal BD Ě&#x201A;2 đ??ľĚ&#x201A;1 = đ??ˇ (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; đ??´đ??ľâ&#x20AC;&#x2013;đ??śđ??ˇ) Ě&#x201A; Ě&#x201A; đ??ˇ1 = đ??ľ2 (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; đ??´đ??ˇâ&#x20AC;&#x2013;đ??ľđ??ś) đ??ľđ??ˇ = đ??ľđ??ˇ (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x161;đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;) â&#x2C6;´ â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??śđ??ˇđ??ľ (đ??´đ??´đ?&#x2018;&#x2020;) â&#x2C6;´ đ??´đ??ˇ = đ??ľđ??ś â&#x2C6;´ đ??´đ??ľ = đ??śđ??ˇ Ě&#x201A;2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ˇ Ě&#x201A;1 = đ??ľĚ&#x201A;2 đ??ľĚ&#x201A;1 = đ??ˇ Ě&#x201A; đ??ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??´đ??ˇ Ě&#x201A; đ??ľ = đ??śđ??ľĚ&#x201A; đ??ˇ đ??´đ??ľĚ&#x201A; đ??ˇ = đ??śđ??ˇ Theorem B: Converse to A: States that if the opposite sides of a quadrilateral are equal then the quadrilateral is a parallelogram. Theorem C: Converse to A: States that if the opposite angles of a quadrilateral are equal then the quadrilateral is a parallelogram. To prove that a quadrilateral is a parallelogram one must be able to prove at least two of the properties of this particular shape, Properties concerned are: 1.

Both pairs of opposite sides are parallel.

Both pairs of opposite sides are equal.

Opposite angles are equal.

The diagonals bisect each other.

Hint: a good method is to prove that one pair of opposite sides are equal and parallel. Examples: 1. Prove that ABCD is a parallelogram.

B (0 ; 3) 2

A (-4 ; 0) -10

C (1 ; 0)

-5

-2

D (-3 ; -3) -4

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420

Proof: đ??´đ??ľ = â&#x2C6;&#x161;(0 + 4)2 + (3 â&#x2C6;&#x2019; 0)2 = 5 đ??ˇđ??ś = â&#x2C6;&#x161;(1 + 3)2 + (0 + 3)2 = 5

đ?&#x2018;&#x161;đ??´đ??ľ = 4 3

đ?&#x2018;&#x161;đ??ˇđ??ś = 4

ABCD is a parallelogram (one pair of opposite sides are equal and parallel) 2. ABCD is a quadrilateral with AB = CD & AD = BC A 2

B 1

2.1

2.2

2 1 C

RTP: â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Ťđ?&#x2018;Ş â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Šđ?&#x2018;Ş Proof: đ??źđ?&#x2018;&#x203A; â&#x2C6;&#x2020;đ??´đ??ˇđ??ś đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ??´đ??ˇ = đ??ľđ??ś ( đ?&#x2018;&#x201D;đ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) đ??śđ??ˇ = đ??´đ??ľ ( đ?&#x2018;&#x201D;đ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) đ??´đ??ś = đ??´đ??ś (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x161;đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;) â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Ťđ?&#x2018;Ş â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Šđ?&#x2018;Ş (đ?&#x2018;şđ?&#x2018;şđ?&#x2018;ş) RTP: đ?&#x2018;¨đ?&#x2018;Šâ&#x20AC;&#x2013;đ?&#x2018;Ťđ?&#x2018;Ş đ?&#x2019;&#x201A;đ?&#x2019;?đ?&#x2019;&#x2026; đ?&#x2018;¨đ?&#x2018;Ťâ&#x20AC;&#x2013;đ?&#x2018;Šđ?&#x2018;Ş Proof: đ??´Ě&#x201A;2 = đ??śĚ&#x201A;2 (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; 2.1 â&#x2C6;&#x2020;đ??´đ??ˇđ??ś â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ??´đ??ˇâ&#x20AC;&#x2013;đ??ľđ??ś (Alternate angles equal) đ??´Ě&#x201A;1 = đ??śĚ&#x201A;1 (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; 2.1 â&#x2C6;&#x2020;đ??´đ??ˇđ??ś â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ??´đ??ľâ&#x20AC;&#x2013;đ??ˇđ??ś (Alternate angles equal)

2.2

)

RTP: ABCD is a parallelogram Proof: đ??´đ??ľ = đ??śđ??ˇ & đ??´đ??ľâ&#x20AC;&#x2013;đ??śđ??ˇ (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;) đ??´đ??ˇ = đ??ľđ??ś & đ??´đ??ˇâ&#x20AC;&#x2013;đ??ľđ??ś (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;) ABCD is a parallelogram (both pairs opposite sides equal and parallel)

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421

Exercise 20.3: 1.

A (0 ; 4)

B( 5 ; 4)

C ( 2 ; -3)

D ( -3 ; -3)

Prove with reasons that the shape above is a parallelogram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2

PRSQ is a parallelogram. RS = ST & TR is a straight line. Prove: 2.1 QA = AT. 2.2 PT // QS. T



 R

2.1 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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422

2.2 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 3. FBCD is a parallelogram. AF = FB. Prove that FE = ED

1 2

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423

A 2

B 2

In the figure above, ABCD is a parallelogram. đ??´đ??ˇ = đ??´đ??¸ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??šđ??ś = đ??ľđ??ś. Prove that đ??´đ??¸đ??śđ??š is a parallelogram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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424

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P 2

Q 1

In the figure above â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2020;đ?&#x2018;&#x2026; are isosceles triangles with đ?&#x2018;&#x192;đ?&#x2018;&#x2020; = đ?&#x2018;&#x192;đ?&#x2018;&#x2026; = đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x192;đ?&#x2018;&#x201E;â&#x20AC;&#x2013;đ?&#x2018;&#x2020;đ?&#x2018;&#x2026;. Prove that PQRS is a parallelogram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Gcse Mathematics

425

21.

GOď&#x201A;ˇMATH WORKBOOKS

CIRCLE GEOMETRY

Introduction: Circle Geometry is the study of shapes and angles formed within circles.

Shapes involved: Quadrilaterals Parallelograms Rectangle Square Rhombus Kite Trapezium

Triangles: Equilateral Isosceles Right Angled Scalene

Parallel lines and angles formed Corresponding Angles Alternate Angles Co-interior Angles

Gcse Mathematics

426

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Geometry Theorems:

Chord Theorems: Theorem 1: (proof required) The line segment joining the centre of a circle to the mid-point of a chord is perpendicular to the chord. (reason: Mid-pt chord theorem)

Given: Circle centre O and C, the midpoint of AB. RTP: OC  AB Proof: Join AO and OB In ∆AOC & BOC AC = CB ( C is midpt) OC is common AO = OB (radii) ∆AOC ≡∆BOC ( SSS) C1 = C2 But C1 + C2 = 180 ( L’s on a str line) C1 = C2 = 90 OC  AB 2.

Converse of 1. (proof required) The perpendicular from the centre of a circle to a chord bisects the chord. ( Reason: Mid pt chord theorem)

Given:

Circle centre O and OC  AB

R.T.P.

AC = CB

Proof:

Join AO and OB In ∆AOC & BOC AC = CB ( C is midpt) OC is common C1 = C2 (both 90) ∆AOC ≡∆BOC ( RHS) AC = CB

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427

Theorem 3: (proof required) The angle which an arc of a circle subtends at the centre of the circle is twice the angle it subtends at any point on the circle. ( Reason: Angle at centre) C C

A O O 1

O 1

1 2

B A A

Fig 1 fig 2 fig 3 Ě&#x201A; Given: Circle centre O , đ??´đ?&#x2018;&#x201A;đ??ľ the angle at the centre subtended by AB and đ??´đ??śĚ&#x201A; đ??ľ the at the circumference. RTP: đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ľ = 2đ??´đ??śĚ&#x201A; đ??ľ Proof: Join CO and produce to D AO = OC (Radii) đ??´Ě&#x201A; = đ??´đ??śĚ&#x201A; đ?&#x2018;&#x201A; (Lâ&#x20AC;&#x2122;s opp = sides) đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = đ??´Ě&#x201A; + đ??´đ??śĚ&#x201A; đ?&#x2018;&#x201A; (ext L triangle) ď &#x153; đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ľ = 2đ??´đ??śĚ&#x201A; đ??ľ Similarly đ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??ľđ??śĚ&#x201A; đ?&#x2018;&#x201A; Thus in figure 1 & 2 đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ + đ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??´đ??śĚ&#x201A; đ?&#x2018;&#x201A; + 2đ??ľđ??śĚ&#x201A; đ?&#x2018;&#x201A; = 2(đ??´đ??śĚ&#x201A; đ?&#x2018;&#x201A; + 2đ??ľđ??śĚ&#x201A; đ?&#x2018;&#x201A;) = 2đ??´đ??śĚ&#x201A; đ??ľ Thus in fig3. đ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ â&#x2C6;&#x2019; đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??ľđ??śĚ&#x201A; đ?&#x2018;&#x201A; â&#x2C6;&#x2019; 2đ??´đ??śĚ&#x201A; đ?&#x2018;&#x201A; = 2(2đ??ľđ??śĚ&#x201A; đ?&#x2018;&#x201A; â&#x2C6;&#x2019; đ??´đ??śĚ&#x201A; đ?&#x2018;&#x201A;) = 2đ??´đ??śĚ&#x201A; đ??ľ

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428

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Theorem 4: (proof not required) The angle subtended at the circle by a diameter is a right angle. (Reason: Angle in a semi-circle) C

Given: Circle centre O and AB the diameter. đ??´đ??śĚ&#x201A; đ??ľ is an angle in the semi-circle. RTP: đ??´đ??śĚ&#x201A; đ??ľ = 90Â° Proof: đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ľ = 2đ??´đ??śĚ&#x201A; đ??ľ ( L at centre = 2 L at circumference) đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ľ = 180Â° ( St line) 2đ??´đ??śĚ&#x201A; đ??ľ = 180Â° đ??´đ??śĚ&#x201A; đ??ľ = 90Â° Theorem 5: (proof required) Angles in the same segment are equal: (Reason: ď&#x192;? in same segt) B

D A

Given: Circle centre O and đ??´đ??ľĚ&#x201A; đ??ˇ and đ??´đ??śĚ&#x201A; đ??ˇ angles in the same segment. RTP: đ??ľĚ&#x201A; = đ??śĚ&#x201A; Proof: Join AO and OD đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??ľĚ&#x201A; ( L at centre = 2 L at circle) đ??´đ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??śĚ&#x201A; ( L at centre = 2 L at circle) 2đ??ľĚ&#x201A; = 2đ??śĚ&#x201A; đ??ľĚ&#x201A; = đ??śĚ&#x201A;

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429

Theorem 6: (proof not required)_ Equal angles at the centre stand on equal chords: A B A

O O

O C D B

Theorem 7: (proof not required) Equal angles are subtended by equal chords: (Reason: equal ,s; equal chord) F

E A D

Theorem 8: (proof not required) If two chords of a circle are equal , then they are equidistant from the centre: (Reason :equal chords; equidistant) D

A E

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430

Theorem 9: (proof not required) Two chords of a circle are equal (or of two different circles) if they subtend supplementary angles at the circumference. i.e. if đ?&#x2018;Ľ + đ?&#x2018;Ś = 180Â° then AB = DE C y

A D E x B

Cyclic Quadrilaterals: (quads within a circle) Theorem 10: (proof required) The opposite angles of a cyclic quad are supplementary: (Reason: opp ď&#x192;?,s cyclic quad) A

Given: Cyclic quadrilateral ABCD with circle centre O. Ě&#x201A; = 180Â° RTP: đ??´Ě&#x201A; + đ??śĚ&#x201A; = 180Â° & đ??ľĚ&#x201A; + đ??ˇ Proof: Join BO and DO đ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??śĚ&#x201A; ( L at centre = 2 L at circle) đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;Ľđ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 2đ??´Ě&#x201A; ( L at centre = 2 L at circle) đ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ + đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;Ľđ??ľđ?&#x2018;&#x201A;Ě&#x201A;đ??ˇ = 360Â° ( Lâ&#x20AC;&#x2122;s at a point) 2đ??śĚ&#x201A; + 2đ??´Ě&#x201A; =360Â° đ??śĚ&#x201A; + đ??´Ě&#x201A; = 180Â° Ě&#x201A; = 180Â° Similarly it can be proved that đ??ľĚ&#x201A; + đ??ˇ

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431

Theorem 11: (proof required) The exterior angle of a cyclic quad is equal to the interior opposite angle. (Reason: ext ď&#x192;? cyclic quad) A

C E

Given:

Cyclic quad ABCD with BC produced to E.

RTP: đ??ˇđ??śĚ&#x201A; đ??¸ = đ??´Ě&#x201A; Proof: đ??ˇđ??śĚ&#x201A; đ??¸ + đ??ˇđ??śĚ&#x201A; đ??ľ = 180Â° (Lâ&#x20AC;&#x2122;s on a st line) đ??´Ě&#x201A; + đ??ˇđ??śĚ&#x201A; đ??ľ = 180Â° (opp Lâ&#x20AC;&#x2122;s of a cyclic quad) đ??ˇđ??śĚ&#x201A; đ??¸ = đ??´Ě&#x201A;

Tangents to Circles: Theorem 12. (proof not required) A radius( or Diameter) is always perpendicular to a tangent at the point of tangency. (Reason: rad ď &#x17E; Tan)

Gcse Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

432

Theorem 13: (proof not required) Tangents drawn from the same point to a circle are equal in length: (Reason: tan from same point)E

Theorem 14. (proof required) The angle formed between a tangent and a chord is equal to the angle in the alternate segment (Reason: Alt Segt Thm)

D E O

Given: Tangent AB touching circle centre O at C. Chord CE Ě&#x201A; & đ??¸đ??śĚ&#x201A; đ??´ = đ??šĚ&#x201A; RTP: đ??¸đ??śĚ&#x201A; đ??ľ = đ??ˇ Proof: Draw diameter COG and join DG. Ě&#x201A;2 = đ?&#x2018;Ś (i) Let đ??şĚ&#x201A; = đ?&#x2018;Ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ś đ??şđ??¸Ě&#x201A; đ??ś = 90Â° ( L in a semi-circle) đ?&#x2018;Ľ + đ?&#x2018;Ś = 90Â° ( L sum â&#x2C6;&#x2020;) đ?&#x2018;Ś + đ??¸đ??śĚ&#x201A; đ??ľ = 90Â° ( Rad ď &#x17E; Tan) đ?&#x2018;Ľ = đ??¸đ??śĚ&#x201A; đ??ľ Ě&#x201A; ( Lâ&#x20AC;&#x2122;s in same segt) BUT đ?&#x2018;Ľ = đ??ˇ Ě&#x201A; đ??¸đ??śĚ&#x201A; đ??ľ = đ??ˇ Ě&#x201A; + đ??šĚ&#x201A; = 180Â° ( opp Lâ&#x20AC;&#x2122;s cyclic quad) (iii) đ??ˇ đ??¸đ??śĚ&#x201A; đ??ľ + đ??¸đ??śĚ&#x201A; đ??´ = 180Â° (Lâ&#x20AC;&#x2122;s on st line) Ě&#x201A; (proved in i) But đ??¸đ??śĚ&#x201A; đ??ľ = đ??ˇ đ??¸đ??śĚ&#x201A; đ??´ = đ??šĚ&#x201A;

Gcse Mathematics

433

GOMATH WORKBOOKS

GEOMETRIC RYDERS (PROBLEMS)

O is the centre of the circle. AB = 60mm ; OM = 40mm Calculate the radius of the circle and the Length of CD if ON = 30mm. ON CD and OM  AB

N C O

M A

CD = 80 mm; AB = 60mm and AB // CD If the radius is 50mm find the distance between the chords

B A

Gcse Mathematics

434 E

D A

GOď&#x201A;ˇMATH WORKBOOKS

CD // AB Prove CD = 2AB NB: A and B are centres of the circles.

M is the mid-point of AB, O is the centre. Prove â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;´đ?&#x2018;Ş â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;Šđ?&#x2018;´đ?&#x2018;Ş

_M _C

Gcse Mathematics

GOď&#x201A;ˇMATH WORKBOOKS

435 O is the centre of the circles 5 to 10. Find the sizes of x and y in each case.

6. A

120ď&#x201A;° C

8. O O

70ď&#x201A;°

110ď&#x201A;°

x A C

đ?&#x2018;¨đ?&#x2018;śâ&#x20AC;&#x2013; đ?&#x2018;Šđ?&#x2018;Ş ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Gcse Mathematics

436

GOď&#x201A;ˇMATH WORKBOOKS

9. C

đ?&#x2018;¨đ?&#x2018;Şâ&#x20AC;&#x2013; đ?&#x2018;Šđ?&#x2018;ś

70ď&#x201A;° x 10ď&#x201A;°

Prove that BË&#x2020;1 ď&#x20AC;Ť AË&#x2020; ď&#x20AC;˝ 90ď Ż O

1 B

1 C

Gcse Mathematics

GOMATH WORKBOOKS

437 Find the values x ; y and z in 11 to 16.

11.

12. A

D D

x z

60

2 1

20 B

x C

BOˆ C  130 and DOˆ C  60 ____________________________________________________________

Gcse Mathematics

GOMATH WORKBOOKS

438

13. F x A 2

1 20

4 B

85 D

____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Gcse Mathematics

GOMATH WORKBOOKS

439

14. 100

15. O is the centre of circle ABCD D

A z

O x

B E

120 D

y C

Gcse Mathematics

440

GOMATH WORKBOOKS

16. DE is a tangent to circle ABC. A z

O y

x 62 D

Gcse Mathematics

441

\ 17.

GOMATH WORKBOOKS

EF is tangent to circle ABCD and BC = CD . Find 5 angles equal to Cˆ1 , (giving reasons.)

2 1

4 E

3 2 C

1 F

Gcse Mathematics

442

18.

GOMATH WORKBOOKS

D A E

Chord AB is parallel to chord CD. Cˆ1  Cˆ 2

Prove that BC = DE

1 2 B

Gcse Mathematics

443

19.

GOMATH WORKBOOKS

D 1

Chord AD equals chord AC and Sˆ1  Sˆ3 Prove: 19.1 Eˆ  Bˆ 19.2 AE  AB

1 1

2 A

____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Gcse Mathematics

444

GOMATH WORKBOOKS

20. AB and CD are two chords which intersect at T. AK  CD and DL  AB .

A T L

Prove: 20.1 AKLD is a cyclic quad 20.2 KL // CB. 20.3 If AK and Dl produced cut CB at M and N respectively, prove AMND is a cyclic quad.

____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________

Gcse Mathematics

445

GOMATH WORKBOOKS

21.

P 1 2

F 1 2

PA and PC are tangents to the circle at A and C. AD //PC and PD cuts the circle at B. CB is produced to meet AP at F. AB, AC and DC are joined.

A 4 3

4 3 2 B 1

2 3

1 2 D

Prove: 21.1 AC is the bisector of PAˆ D . 21.2 Bˆ1  Bˆ 3 . 21.3 AP = AC. 21.4 APˆ C  ABˆ D . 21.5 Aˆ 4  Pˆ2

Gcse Mathematics

446

GOMATH WORKBOOKS

22. FEˆ C  ADˆ C (i.e.Eˆ 3  Dˆ 1 2 )

B 3 1 2

2 A

2 E1 3 6 4 5

Prove: 22.1 CEBF is a cyclic quad. 22.2 CGAF ARE concyclic. 22.3 AC bisects BCˆ G

1 2 D

Gcse Mathematics

23.

447

AB is a diameter. ADP and BCP are straight lines. PQTR is a straight line.

1 2

D 1 2

GOMATH WORKBOOKS

4 1 T 3 2

2 6 Q 3 4 5

1 2

Prove: 23.1 DQCP is a cyclic quadrilateral. 23.2 If Qˆ 5  TBˆ C , then PT  AB 23.3 DATQ is a cyclic quad.

1 2 C

Gcse Mathematics

448

GOMATH WORKBOOKS

24. AOD and EOB are diameters. AF  EB Prove : 24.1 EFHD is cyclic. 24.2 BAˆ D  DHˆ C . 24.3 Cˆ 3  Aˆ1 .

A 1 E

2 3

3 1

2 O F 1

24.4 2

2 2 1 H 1 2 D

EB bisects ABˆ C (i.e Bˆ 3  Bˆ1 2 )

1 1

3 C

Gcse Mathematics

GOMATH WORKBOOKS

449

25. T

In the figure BOD is a diameter of the circle with centre O. BA and BC are chords of the circle. BA produced and CD produced meet in T and AD produced and BC produced meet in S. Prove:

25.1 25.2 25.3

D O

ATSC is a cyclic quad

ADˆ B  ATˆS OA is a tangent to circle ATSC.

C S

Gcse Mathematics

GOMATH WORKBOOKS

450

26. D 3 2

1 T 2 1

3 2 4

1 B W

In the figure above, TD is a tangent to circle ABCD at D. AD // BC, AB and DC produced meet at W. TBS is a straight line. If WBˆ T  CBˆ D , Prove that: 26.1 BWTD is a cyclic quadrilateral. 26.2 TBS is a tangent to the circle at ABCD. 26.3 TW // BC. ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ___________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________ ____________________________________________________________