GOMATHGR11W1 by admin Bookbuzz

Compiled by Chez Nell

Grade 11 Core Mathematics

GO MATH WORKBOOKS

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 booklet is available for each workbook. The statement “You get out what you put in.” is very apt where maths 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. Chez Nell: Mathematics Educator: Northwood School  Norma Nell 2011

Grade 11 Core Mathematics

GO MATH WORKBOOKS

GRADE 11 CORE MATHEMATICS CONTENTS: Paper One: Topic

Pages

Exponents

(4 - 20)

Algebraic Factors

(21 –24 )

Quadratic Equations

(25 – 33)

Simultaneous Equations

(34 – 41)

Inequalities

(41 – 45)

Algebraic Fractions

(45 - 52)

Number Patterns

(53 – 72)

Financial Math

(73 – 88)

Functions & Graphs

(89 – 141)

10.

Probability Theory

(142 –155)

Grade 11 Core Mathematics

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Exponetial Laws and Examples:

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

23.33  8.27  216.

e.g.

BUT 23.33  63  216 2.

Law:

am = a mn n a

When dividing the bases subtract the exponents: the following exception. 3

63  6      33  27 3 2 2

Law: (a m ) n  a mn

When raising a power to a power you must multiply the exponents. 4.

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 prime bases then the normal laws apply.

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Grade 11 Core Mathematics

Example 1:

 

Simplify 125 3  53

4 3

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

125  5 4 3

Example 2:



4 3 3

5

3 4  1 3

 5 4  625

Simplify without the use of a calculator. 5 a  3.5 2 a 1 25 a 1

Answer:

Use law 1 in the numerator

5 a  3.5 2 a 1 53a  2  2a2  5a 4 2 a 1 (5 ) 5

Prime base factorise and use law 3 to simplify

Use law 2 to get the answer.

Follow the procedures as set out for any of the following types of simplifications with exponents. NB. Prime base factorizing first then apply the laws. Exercise 1.1: 2 x3 2 x 1. 2 x 1 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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2 x 1 2 x  3 2 x  2 .2 x

5 x 25 x 1 5.125 x

Grade 11 Core Mathematics

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7 x  2 49 x  2 7 3 x2 ____________________________________________________________________

6 n1 .12 n1 .2 n 18 n 2 .8 n1

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Grade 11 Core Mathematics

Exercise 1.2: If bases are separated using plus or minus signs one must FACTORISE FIRST.. It is easier to separate the bases as shown because it is easier to see the HCF. Factorise and simplify. NB the base with the variable exponent should always cancel leaving pure numerical values. Example: 2 x3  2 x = 2 x 1

2 x.2 3  2 x = 2 x.2 1

2 x (2 3  1) = 2 x.2 1

9 1 2

 18

2 x 1  2 x  3 1. 2 x2  2 x ____________________________________________________________________

3 x  3 x2 3 x 1  3 x

Grade 11 Core Mathematics

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5 2 x  4.5  x 5  x  2.5  x 1 ____________________________________________________________________

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Grade 11 Core Mathematics

EXPONENTIAL EQUATIONS There are two types of exponential equations: 1. The unknown(variable) in the exponent: 2. The unknown(variable) in the base: 1.

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. Example 1.

2 2 x  16

Rename 16 to the base 2 and equate the bases.

22x  24 2x  4 x2 Example 2: 2.3 x ( x  3)  54

First divide by the coefficient „2‟

3 x ( x  3)  27 3 x ( x  2 )  33 x  2x  3  0 ( x  3)( x  1)  0 x3 or 2

x  1

Rename 27 to the base 3 and equate the bases.

Equate the exponents and solve for x.

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. In an equation you must do the same to both sides of the equal sign. Example . 2

x 3  16 3

 23  2  x   24     x  2 4  2 3

x  26 x  64

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

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Grade 11 Core Mathematics

More advanced equations: Variables separated by + or – signs: This involves factorizing. 2 x  3.2 x  16

2 x (1  3)  16

Use 2 x as the HCF

2x  4

1.1

2 x  22 x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 are equal in value.

Variables separated by + or – signs, BUT the values of the coefficients of the exponents are not equal. i.e. one is double the other. This involves a trinomial and needs to be factorised accordingly. 22x  2 x  8  0 (2 x  4)(2 x  2)  0

2.1

2x  4  0 2x  4 2x  2  0

2 x  2 2 or 2 x  2 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 2 x

2.32 x  12.3 x  54  0

2.2

(2.3 x  6)(3 x  9)  0

2.3 x  6  0 3x  3 x 1

3x  9

3 x  32 x2

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

NB a substitution method can be utilized here: Let 3 x = k [ thus 32 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 x 3 3 or 3 x  3 2 x 1 x2

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

Grade 11 Core Mathematics

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Exercise 1.3: 1. Solve for x without the use of a calculator: 1.1

3 x  81

1.2

x3 = 27

1.3

4

= 80

Grade 11 Core Mathematics

1.4

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2 x  16 x 1

1.5

5.42 x  40

1.6

1 1  x2 8 (2 ) x

Grade 11 Core Mathematics

1.7

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3.4 2 x  48 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.8. 2 x1  2 x  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.9 32 x1  2.32 x  45 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Grade 11 Core Mathematics

SURDS 3

Irrational numbers such as 10 ; 5 ; 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: đ?&#x2018;&#x17D; Ă&#x2014; đ?&#x2018;? = 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. 4 Ă&#x2014; 9 = 4 Ă&#x2014; 9 = 36 = 6 2. 2 Ă&#x2014; 7 = 2 Ă&#x2014; 7 = 14

2. Dividing surds of the same order. 3

2.1

2.2

3 4 4

27 3 64 4

= =

27 3

64 4

đ?&#x2019;?

đ?&#x2019;&#x201A;

đ?&#x2019;?

đ?&#x2019;&#x192;

16 = 2

đ?&#x2019;?

đ?&#x2019;&#x201A; đ?&#x2019;&#x192;

3. Mixed Surds & Entire Surds. 12 is an entire surd . 2 3 is a mixed surd. To convert from entire to mixed either: 1. Write in factorized form and simplify. OR and

80 = 24 Ă&#x2014; 5 = 22 5 = 4 5 2. Find the highest possible perfect square that divides into the given surd

then simplify further. 80 = 16 Ă&#x2014; 5 = 4 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. 7+ 7 = 2 7 2. 5 2â&#x2C6;&#x2019;2 2 = 3 2 3 3 3 3. 4 2 + 6 2 = 10 2 5. Rationalizing the denominators: 1

Irrational denominator : 3 To rationalize the denominator , means one must eleiminate any root signs in the denominator.

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Grade 11 Core Mathematics

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

e.g. →. we are

3 3

[we are not changing the value of the fraction as

multiplying by 1] →

2 3 5

2 3

2 15 5

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 inn the original expression. 3

e.g.2+

3 3

2+ 3

2− 3

× 2−

2 2− 3 2+ 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.

5 = 52

= 56 = 1 3

2 6

125

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

Grade 11 Surds Worksheet: 1. Simplify: 1.1 20 1.2 18 1.3 245 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.4

1.5

13 5

1.6

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Grade 11 Core Mathematics

2. Simplify. 2.1. 2+3 2

3+3 3−2 3

2.2.

2 + 18

2.4

3 8 + 5 50 − 4 32

245 + 6 5

2.6

54 − 16

2.7

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Grade 11 Core Mathematics

2 8 + 4 32 − 3 50

2.8.

12− 75+2 3 3

6− 3

2 5 3 5−2 2

3 2 2 8 − 18

Grade 11 Core Mathematics

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3.6 3 2+ 5 3 2â&#x2C6;&#x2019; 5 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 2

3.7 3+ 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 2

3.8 6+2 3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4. Rationalize the denominators of the following: 3 4.1 3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 5 2

4.2 10 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 6 18

4.3 3 12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4 3

4.4 3 12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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4.5 2− 3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 7

4.6 3+ 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2− 3

4.7 2+ 3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.8

7−3 2 7− 2

2 5+ 3

4.9 5 3−3 5 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

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Factors of Algebraic Expressions

To factorise 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 factorised further. It could be one of the following; 3.1

The difference of two perfect squares.

3.2

A trinomial.

3.3

A quadrinomial

3.4

The sum & or difference of two cubes.

Examples: HCF plus other expressions. 1. 2. 3. 4. 5.

2 x  4  2( x  2) 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 ) ab 4  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 factorised at all.

Difference of two squares: 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 factorise further; NB: The sum of two squares cannot be factorised 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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Grade 11 Core Mathematics

Trinomials: The method used here is referred to as “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 factorise 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

Revision Exercise 2.1: Factorise the following completely: 1.1

2x2 – 32 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

Grade 11 Core Mathematics

1.2

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p2( x + y) - q2( y + x ) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.3

a2  b2  a  b ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.4

(2 x  y ) 2  ( x  2 y ) 2

1.5

4( x  y) 2  9( x  y) 2

Grade 11 Core Mathematics

1.6

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x 2  7 x  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.7

2 x 2  24 x  70 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

1.8

9 x 2  42 x  45 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Quadratic Equations:

1. Solve by factorizing: Method: 1. Equate to zero, factorise and solve: x2  3x  4 e.g. .

x2  3x  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

e.g.

x 2  4 x  12  0 ( x  6)( x  2)  0 x  6 or x2

Exercise 3.1 : 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  5x  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  2 x  3  12 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

12. x( x  16)  3(24  5x) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

13. (2 x  5)(3x  2)  2(3x  11) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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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

3 17  x   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 3.2 : 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  7 x  6  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

2 x 2  11x  6  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2 x 2  3x  8  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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.

2x2  7x  6  0

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

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

Exercise 3.3: 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.

x 2  4x  3  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

2 x 2  x  10 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

Grade 11 Core Mathematics

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3x 2  x  2  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

x 2  6x  4  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

2x 2  4  7x ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

2 x( x  3)  3  0 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Simultaneous Equations:

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 4.1: 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: 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. 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 4.2: 1.

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

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

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?.

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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?

Quadratic Inequalities:

Steps: 1. 2. 3. 4. 5.

Equate to zero If fractions involved then combine to make one expression Factorise 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

Solution:  1  x  4

-1

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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 5.1: 1.

x 2  8 x  15  0 . ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

4 x 2  49 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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x2 0 x5 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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

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

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4 x x3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

9 x ____________________________________________________________________ x

2 3  x2 x3 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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Grade 11 Core Mathematics

1 2  x5 x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 factorised and simplified

x2  4 x 1 x 1 x x 2 x 2x  4 x  3x  2

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

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



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 factorised 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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Grade 11 Core Mathematics

Algebraic Fractions Addition & Subtraction 1. NB Before you can find a LCD (lowest common denominator) you must ensure that all denominators are fully factorised. 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 factorised.

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  10 x  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 6.1:

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

11.

ab  a 2 b 2  ab X b2  a2 a2

Grade 11 Core Mathematics

12.

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

13.

x x2  2 x  y y  x2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Grade 11 Core Mathematics

14.

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7x 3x  2 x  2 y 5 y  5x

____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________

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Grade 11 Core Mathematics

Number Patterns:

3 basic types of patterns dealt with 1.

Arithmetic Sequences: General Term : Tn  a  (n  1)d These have a common difference: A 1st order sequence : d1  a

Geometric Sequence : General Term is: Tn  ar n1 These have a common ratio.

Quadratic Sequences: General Term is: Tn  an 2  bn  c A 2nd order sequence. d 2  2a (n  1)(n  2) Or Tn  (n  1)T2  (n  2)T1   d2 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  T1 Second Difference ie 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. Let 2a  d 2 and 3a  b  T2  T1 and

a  b  c  T1

SEQUENCE 1 ; 4 ; 9 ; 16 ; 25 1ST DIFF 3 5 7 9 ND 2 DIFF 2 2 2 2a  d 2

3a  b  T2  T1

a  b  c  T1

2a  2 a 1

3(1)  b  3

1 0  c  1 c0

b0

Tn  n 2

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Exercise 7.1: 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;... _________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.2

15;30;48;69;...... ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

1.3

 12;7;1;12;..... ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________

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1.4

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27;31;37;........ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________

1.5

12; 17; 24…… ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.6

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

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1.7

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

1.8 7; 12; 21…….. ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ Exercise 7.2: 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

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Grade 11 Core Mathematics

Look at the pattern below Draw the next pattern.

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: â&#x20AC;&#x17E;nâ&#x20AC;&#x2122; 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

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Grade 11 Core Mathematics

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

3.1

Value of term 2 6 12 20 30 42 56 72

4 6 8 10 12 14 16

2 2 2 2 2 2

How were the answers in the third column obtained? __________________________________________________________________

________________________________________________________________________ ________________________________________________________________________

3.2 What type of pattern of pattern do the answers in 3.1 form: Give reasons: ________________________________________________________________________ ________________________________________________________________________ ________________________________________________________________________

3.3 What do you notice about the fourth column? ________________________________________________________________________ ________________________________________________________________________ 4. Use question 3. as a reference to complete the table below for the sequence: 5; 11; 20; 32 ; 47; 65; 86; …………… Terms 1st

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 7.3:

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. -----;

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Grade 11 Core Mathematics

ARITHMETIC PROGRESSIONS. 1. Sequences: An arithmetic sequence is a string of values which increase or decrease by a constant value. This value is referred to as the “common difference” ( d). 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 ; 10 ; 13 ; 16 ; …..(to the nth term). T1 = 7 T2 = 7 + 3 = 10 T3 = 7 + 3 + 3 = 13 T4 = 7 + 3 + 3 + 3 = 16 T5 = 7 + 4(3) = 19 T10 = 7 + 9(3) = 34

Tn  a  (n  1)d last term

first term

common difference

second last term

Calculations in Arithmetic Progressions (AP‟s). It is a good idea to develop the formula for the nth term of a progression prior to any calculation. It is an easier formula to work with and it is relevant to the progression in question. Example: A progression is given and you want to develop the nth term formual for it. 5;9;13;17...... Simply use the general nth term formula for an arithmetic progression and substitute the „a‟ and „d‟ values into it and simplify. Tn  a  (n  1)d

Tn  5  (n  1)4 Tn  5  4n  4 Tn  4n  1 NOW if I need to find the 20th term simply substitute 20 for n in the formula above: i.e. Tn  4n  1 T20  4(20)  1 T20  81

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Further Examples: 1.

If a sequence of numbers is : 7 ; 10 ; 13 ; 16 ; â&#x20AC;Ś..(to the nth term).

1.1

Find the 20th term in the sequence. a = 7; d = 3; n = 20 ;T20 = ? Tn = a + (n-1)d T20 = 7 + 19(3) = 64 64 is the 20th term in the sequence.

1.2

If 94 is the nth term in the sequence find out the number of terms (n). Tn = 94 ; a = 7 ; d = 3 ; n = ? Tn = a + (n-1)d 94 = 7 + (n-1)3 94 = 7 +3n â&#x20AC;&#x201C; 3 90 = 3n n = 30

1.3

94 is the 30th term in the sequence. If 94 is the 30th term in the sequence Find the first term. Tn = 94 ; ; d = 3 ; n = 30 ; a = ? Tn = a + (n-1)d 94 = a + 29(3) a=7 7 is the first term in the sequence.

1.4.

If 94 is the 30th term in the sequence and 7 is the first term find the common difference. Tn = 94 ; ; n = 30 ; a = 7 ; d = ? Tn = a + (n-1)d 94 = 7 + 29d 29d = 87 d=3 the common difference is 3

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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 an arithmetic sequence is 3x  1 ; 2 x  3 ; 2 x  1 …… NB the constant concept is the common difference. Thus T2 – T1 = T3 - T2

2x  3  3x  1  2x  1  2x  3 x= 8 T1 = 23; T2 = 19 and T3 = 15

Tn = a + (n-1)d Tn = 23 +(n-1)d Tn = 23 –4d This is the general term representing the nth term of this specific sequence. 3.

Simultaneous Equations in AP‟s Given that Tn = 25 and T11 = 81 find the arithmetic progression. i. e. find the first three terms of the sequence. Start with the nth term formula and write down a specific formula for the terms supplied. Tn = a + (n-1)d T11 = a + 10d = 81 T4 = a + 3d = 25 ( subtract the 2 equations to solve for „d‟) 7d = 56 d=8 a = 1 ( by substitution) AP = 1 ; 9 ; 17 ;….

Exercise 7.4: 1. The sixth term of an arithmetic sequence is 17 and the tenth term is 33. Determine the first term and the common difference. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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2. x; 2x + 1; 11 are three consecutive terms of an arithmetic sequence. Calculate: 2.1 x

3.1 the value of the 25

term of the sequence.

3.2 which term of the sequence will be equal to 57?

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Grade 11 Core Mathematics

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  7.29  3584

second last term Tn  ar n1

last term

common ratio

first term

Calculations in Geometric Progressions (GP‟s) 2.

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

2.1

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.

2.2

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.

Grade 11 Core Mathematics

2.3

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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. 3.4.

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 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:

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

48 16 a3 a

Divide to eliminate the “a” value and solve for „r‟ NB subtract the exponents when dividing. Sequence : 3 ; 6 ; 12; …. Substitute “r” into either one of the equations to solve for “a‟

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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 ; 2 x  3 ; 2 x  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  12 x  3  6 x 2  5x  1 Tn = 23 –4d This is the general term representing the nth term of this specific sequence. Exercise 7.5: 1. 2 ; 6 ; 18 ; 54 ; …. is a geometric sequence. 1.1 Continue the sequence to the 6th term. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

1.2

Find the 20th term.

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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;Ąđ?&#x2018;&#x2022;đ?&#x2018;&#x2019; 5đ?&#x2018;Ąđ?&#x2018;&#x2022; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3.2

đ?&#x2018;&#x17D; = 1 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x; = 2 , đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x2022;đ?&#x2018;&#x2019; 6đ?&#x2018;Ąđ?&#x2018;&#x2022; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;.

3.3

đ?&#x2018;&#x17D; = 11 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x; = â&#x2C6;&#x2019; 3 , đ?&#x2018;&#x201C;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x2022;đ?&#x2018;&#x2019; 4đ?&#x2018;Ąđ?&#x2018;&#x2022; đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x161;.

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____________________________________________________________________ 4. In a geometric progression of which: 4.1 𝑡𝑕𝑒 6𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 96 𝑎𝑛𝑑 𝑎 = 3, 𝑓𝑖𝑛𝑑 𝑟. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4.2

𝑡𝑕𝑒 5𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠

7 81

𝑎𝑛𝑑 𝑎 = 7. 𝑓𝑖𝑛𝑑 𝑟.

4.3

𝑡𝑕𝑒 7𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 192 𝑎𝑛𝑑 𝑡𝑕𝑒 2𝑛𝑑 𝑖𝑠 6, 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠.

4.4

𝑡𝑕𝑒 6𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠 − 32 𝑎𝑛𝑑 𝑡𝑕𝑒 9𝑡𝑕 𝑡𝑒𝑟𝑚 𝑖𝑠

6 256

, 𝑓𝑖𝑛𝑑 𝑡𝑕𝑒 𝑓𝑖𝑟𝑠𝑡 3 𝑡𝑒𝑟𝑚𝑠

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____________________________________________________________________ 5. In a geometric progression with first 3 terms: đ?&#x2018;&#x2DC; â&#x2C6;&#x2019; 4; đ?&#x2018;&#x2DC; + 2; 3đ?&#x2018;&#x2DC; + 1 5.1 find the value(s) of k. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

5.2

find the first term.

5.3

Find the 10th term.

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Exercise 7.6: 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 behaviour of series A:

____________________________________________________ _____________________________________________________________________ 1.2

State your observations concerning the behaviour 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? ____________________________________________________________________ _____________________________________________________________________ 1.5 If â&#x20AC;&#x17E;kâ&#x20AC;&#x; 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. _________________________________________________________________.

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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 â&#x20AC;&#x153;first differenceâ&#x20AC;??

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

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Financial Maths:

No business can exist without the information given by figures. Borrowing, using and making money is the heart of the commercial world. The principle of interest and interest rate calculations are extremely important. This leads into an examination of the principles 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. 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 (Growth) is computed on the principle for the entire term of the loan and is thus due at the end of term. Si = Prt Si = 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

Other Formulae: Fv  Pv(1  ni) Pv 

Fv 1  ni

Pv  Fv(1  dt )

Key to Symbols Fv = Future Value = A Pv = Present Value = P i = Interest Rate n = Time period d = discounted rate

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Compound Interest (Growth) 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)

A  P(1  i) n

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 Quartely Half yearly Monthly

 Calculated Once per year  Calculated 4 times per year  Calculated 2 times per year  Calculated 12 times per year

Weekly Daily

 Calculated 52 times per year  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: m   jm  J eff  100 1    1 m  

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

J ď&#x192;ś ď&#x192;Ś OR 1 ď&#x20AC;Ť i ď&#x20AC;˝ ď&#x192;§ 1 ď&#x20AC;Ť m ď&#x192;ˇ m ď&#x192;¸ ď&#x192;¨

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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: R100 000,00 invested at compound interest for 5 years at 12% p.a. Principle

Interest

Fv ď&#x20AC;˝ Pv (1 ď&#x20AC;Ť

i tm ) m

Time Period = 5years Yearly

m = no of compounding m = 1

Half-Yearly

Quarterly

Monthly

Daily

m=2

m =4

m = 12

m = 365

periods p.a. R100 000

I =12%

Compound Interest Ci = Fv â&#x20AC;&#x201C; Pv

R176234,17 R179084,78

R180611,12 R181669,67 R182193,91

R76234,17

R80611,12

R79084,78

R81669,67

R82193,91

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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: CONVERTING EFFECTIVE TO NOMINAL CAN BE DONE AS FOLLOWS:





J m  m m 1  i   1

e.g.

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





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 R12000,00 into her savings account which gives an interest rate of 7,2% p.a. compounded monthly. Her savings grew to R17 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 summarise 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.

n (years) 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;˝ R9834,496

A ď&#x20AC;˝ 9834,496(1 ď&#x20AC;Ť 0,135) ď&#x20AC;Ť 5000 A ď&#x20AC;˝ R16162,152 A ď&#x20AC;˝ 16162,152(1 ď&#x20AC;Ť 0,135) 3 A ď&#x20AC;˝ R 23631,26(nearest cent )

Alternative (shorter) Method: đ??´ = 7000 1 + 0.12 3 (1 + 0.135)4 + 5000(1 + 0.135)3 = đ?&#x2018;&#x2026;23631,26

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Example 2: In order to save for her sons University fees , Mrs Gumede deposits R8000 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 R9000 , payable at the end of January . The fees increase by 10% each year. Calculate: 1.

the second and third years 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 years fees?

R8000

T10

(R9000)

n (years)

14% p.a.compounded annually NB the bracket around R9000 indicates a withdrawal. 1.

Second years fees ( fees at T9):

9000  (0.1)(9000)  R9900 Third years fees ( fees at T10) 9900  (0.1)(9900)  R10890 2.

Balance at T8: 8000(1  0.,14)8  9000  R22820,69  9000  R13820,69 (to nearest cent)

Balance at T9:

13820,69(1  0,14)  9900  R5855,59 (to nearest cent) 4.

Money in savings account at T10:

5855,59( I  0,14)  R6675,37

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Additional cash required: 10890 ď&#x20AC; 6675,37 ď&#x20AC;˝ R4214,63 (to nearest cent)

Exercise 8.1: Revision of Grade 10 time line problems 1. Jack deposits R1000 into a savings account . One year later he adds R2000 to the savings. At the end of the second year he deposits R4000 into the same account, and finally he adds R8000 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. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ _________ ___________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2. R6500 is deposited into a savings account , and 3 years later R7400 is added to the savings. At the end of 5 years , R5800 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. R21000 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. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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

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

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Is this fair on the sons? Explain your answer.

___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 5. Mr Ndlovu places R52000 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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____________________________________________________________________ ____________________________________________________________________

Further „time line‟ questions: Example 1: Jun invests R50 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 = R50 000 Tm = 5 x 2 = 10 Tm = 5 x 4= 20 Tm = 5 x 12 = 60

10% (Half-yearly)

12% (Quarterly)

15% (Monthly)

0,10   0.12   0.15   A  500000 1   1   1   2   4   12   A  R309962,68

T1 5

0 60

Example 2: R8000 is invested into a savings account at an interest rate of 12,5 % p.a. compounded monthly. Three years later R5000 is added to the savings and after a further 3 years R6000 is withdrawn. If a final deposit of R10 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

R8000

R5000

Tm = 2 x 12 = 24

(R6000)

R10 000

15% (Monthly)

T1 0

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0.15 120 0.15 84 0.15 48 0.15 24 )  5000(1  )  6000(1  )  10000(1  ) 12 12 12 12 A  R52298,65 A  8000(1 

Exercise 8.2: 1. Jono wins R3 000 000,00 on the lottery and invests the money with Standard Bank. The bank pays 6 % p.a. compounded monthly. At the end of 3 years , Jono withdraws R500 000,00 to buy a car. Two year later he deposits R120 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 R100 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. 2.2 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 R200 000,00 for 20 years. He receives 8% p.a. compounded monthly for the first 14 years and 12% compounded quartery 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. Sipho invests R150 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 â&#x20AC;&#x201C; line method. (i.e.Simple Decay).

Determine its book value in December 2009 if depreciation is calculated according to A reducing â&#x20AC;&#x201C; balance method . ( i.e compound Decay). ___________________________________________________________________

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__________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 6. R20 000 is invested into a savings account at an interest rate of 8,5 % p.a. compounded monthly. 2 years later R15000 is added to the savings and after a further 4 years R16000 is withdrawn. If a final deposit of R5 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. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 9. Sam wins R1 500 000,00 on the lottery and invests the money with Standard Bank. The bank pays 9 % p.a. compounded monthly. After 2 years , Sam withdraws R500 000,00 to buy a car. 4 year later he deposits R550 000,00 into his account. Calculate how much money he has in his account at the end of 10 years. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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FUNCTIONS & GRAPHS

General Formulae for Functions and Graphs & the effects of movement to functions: 1. Parabola: y  a( x  x1 )( x  x2 ) . Use this form of the equation to calculate the 1.1 equation of a given parabola sketch if you are supplied with the x – intercepts and one other point that the graph passes through. 1.2 y  a( x  p) 2  q : Use this form of the equation to calculate the equation of a given parabola sketch if you are supplied with the Turning point and one other point that the graph passes through. 2.Shifting the parabola: Lateral shifts affect the axis of symmetry i.e. the p- value in the equation y  a( x  p) 2  q . Vertical shifts affect the q – value in the equation y  a( x  p) 2  q i.e the maximum or minimum value of the graph.

f ( x)  g ( x  3) is an instruction to shift the graph 3 moves to the RIGHT ( positive direction) f ( x)  g ( x  3) is an instruction to shift the graph 3 moves to the LEFT ( negative direction).

f ( x)  g ( x)  3 is an instruction to shift the graph vertically upwards (positive direction) f ( x)  g ( x)  3 is an instruction to shift the graph vertically downwards (negative direction) 3.The exponential graph:

ya

x

1 / y  a

To get the equation of a given sketch a point must be supplied . Simply substitute into the general equation and calculate the a – value.

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4.Shifting the exponential graph: y  a x  p is a horizontal shift to the LEFT ( negative direction) y  a x  p is a horizontal shift to the RIGHT ( positive direction) y  a x  q is a shift vertically upwards (positive direction ) y  a x  q is a shift vertically downwards (negative direction)

5.The hyperbola graph:

y

k x

or y 

k x

To get the equation from a given sketch a point must be supplied. Simply substitute into general equation and calculate k –value. 6.Shifting the hyperbola graph: y

k is a horizontal shift to the LEFT ( negative direction) x p

y

k is a horizontal shift to the RIGHT ( positive direction) x p

y

k  q is a vertical shift upwards (positive direction) x

y

k  q is a vertical shift downwards (negative direction) x

7. Trigonometric Graphs:

y  asinb(x  c)  d ; ETC. a  affects the amplitude of the graph and also inverts the graph when negative.

b  affects the period or frequency of the graph c  shifts the graph horizontally

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d  shifts the graph vertically

Quadratic Function: Sketching a Parabola using 5 points. A parabola can be neatly sketched using the following 5 points: 1. 2. 3.1 3.2

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

(2 points) (1 point) b  sum of roots  The axis of symmetry(a.o.s) at either  or   2a 2   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

y – intercept = -3

 b  (2)  1 a.o.s = 2a 2 Y – value @ f(1)= (1)2 –2(1) – 3= -4 TP ( 1 ; -4)

X = -1 or x = 3

y= x2-2x-3

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

2; -3 1;-4

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Exercise 9.1: 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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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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y  x 2  4 x  12 ___________________________________________________________________

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y  2x 2  7 x  6 ___________________________________________________________________

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y  2 x 2  5x  3 ___________________________________________________________________

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

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 in side 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  2 x  3]  0

( x  1) 2  4 x  1  2 x  1 2

2[( x  1) 2  (1) 2  3]  0 2[(x  1) 2  4]  0 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 9.2: a) Write the following equations in the form: y  a( x  p) 2  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. 1. y  (2 x  1)( x  1) ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ 6

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

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

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

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

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

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7. y  3x 2  2 x  1 ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 6

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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  12 x  10  0 , 5 units to the left. p3 y  2( x  2) 2  8 p1  3  5 New equation y  2 x 2  8x p1  2

NB. Don‟t forget to change the sign when substituting back into y  a( x  p) 2  q B:

Instruction: Shift the graph of, 2 x 2  12 x  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  20 x  42

Exercise 9.3: 1. Shift questions 1 & 2 in Exercise 9.2 by 4 moves to the left and write down the new equation in the form of y  ax 2  bx  c ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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Shift questions 3 & 6 in Exercise 9.2 by 5 moves to the right and write down the new equation in the form of y  ax 2  bx  c

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  12 x  10  0 , 5 units upwards. q  8 y  2( x  3) 2  3 q1  8  5 New equation y  2 x 2  12 x  15 q1  3

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Instruction: Shift the graph of, 2 x 2  12 x  10  0 , 5 units downwards. q  8 y  2( x  3) 2  13 q1  8  5 New equation y  2 x 2  12 x  5 q1  13

Exercise 9.4: 1.

Shift questions 2 & 3 in Exercise 9.2 by 3 moves upwards. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

Shift questions 5 & 6 in Exercise 9.2 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) 2  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)

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)

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Method: Straight line : y  mx  c y0

x 1 0 x  1 B (1;0) 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 9.5:

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 â&#x20AC;&#x201C; 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 9.6: Quadratic Function: Parabolas. 1.1

Sketch the graphs of y   x 2  x  12 and y  3x  12 on the same system of axes.

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1.2

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

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2.1

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

2.2

Write down the co-ordinates of the turning point.

2.3

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

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___________________________________________________________________ __________________________________________________________________ 2 2.5 Sketch the graph of y  x  2 x  8 4

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

2.6

3.1

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

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Sketch the graph of

y  x 2  4x  5 .

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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 2  4 x  5 is moved 4 units downwards. ___________________________________________________________________ __________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ Exponential Graphs: y = ax

General formula :

a>0,a1,x  y>0

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. e.g. y = 2 x (ii) If 0 < a < 1 , then as x increases , y decreases. e.g. y = ( 12 )x. (i)

y = 2x

x y

-2

-1

1 4

1 2

(ii)

y = ( 12 )x.

x y

-2 4 x

1 y    or  2

-1 2

0 1

1 2

2 4

0 1

1 2

1 4

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y = 0,5x y = 2x y

1 x

These 2 graphs are mirror images of each other, the axis of symmetry being the y- axes (x = 0 ). The x- axes is a horizontal asymptote, as y will never equal zero. Log Graphs: Extension work ( Grade 12 Sylabus) General formula : y  log a x To sketch the graph of

y  log a x

a) x>0 and y   thus y  log a x lies on the right of the y- axes in quadrants 1 and 4. b) If a > 1, then as x increases so y- increases c) If o < a < 1 , then as x increases so y decreases. (1)

y  log 2 x ( x  2 y ) X Y

(2)

1 4

1 2

-2

-1

y  log 1 x ( x   12  ) y

1 4

1 2

-1

-2

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y y = log 2x

1 1

y = logo,5x The graph y   log 2 x is exactly the same as y  log 1 x 2

These two graphs are mirror images of each other, the axix of symmetry being the x – axes. The y- axes is a vertical asymptote, as x will never equal zero.

Shifting the exponential graph: 1. Vertical Shifts: y  a x or y  a  x 1. y  2 x Original graph 2. y  2 x  2 Graph shifted 2 units upwards. 3. y  2 x  4 Graph shifted 4 units downwards. NB: the values are added after the base (2x) for vertical movement.

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g x  = 2 x +2

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

f x  = 2 x 2. -5

h x  = 2 x -4

-2

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

-4

Exercise 9.7: Sketch the graph of y  3 x on a Cartesian plane.

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1.1 Shift y  3 x by 2 units upwards and sketch this graph on the same system of axes. 1.2 Shift y  3 x by 4 units downwards and sketch this graph on the same system of axes.

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Write down the equations of the asymptotes after the shifts in question 1.

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3.1 Shift y  2  x by 3 units upwards and sketch this graph on the same system of axes, 3.2 Shift y  2  x by 4 units downwards and sketch this graph on the same system of axes. Write down the equations of the asymptotes after the shifts in 3.

___________________________________________________________________

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Horizontal Shifts: y  a x or y  2 x Original graph

y  ax

2.1 y  2 x  4 Graph shifted 4 units to the right. 2.3 y  2 x  4 Graph shifted 4 units to the left. 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

Exercise 9.8: 1. Sketch the graph of y  2 x on a Cartesian plane. ___________________________________________________________________ __________________________________________________________________

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1.1 Shift y  2 x by 4 units to the left and sketch this graph. 1.2 Shift y  2 x by 4 units to the right and sketch this graph. 2 Write down the equations of the asymptotes after the shifts in question 1. ___________________________________________________________________ __________________________________________________________________

3 Sketch the graph of y  2  x on a Cartesian plane,

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3.1 Shift y  4  x by 3 units to the left and sketch this graph. 3.2 Shift y  4  x 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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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

The Hyperbola Graph. General equation: y 

k k or y  x x

Sketching the graphs:

Method 1. Table Method

y

1. X Y

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4 x -4 -1

-2 -2

-1 -4

1 4

f x  =

2 2

4 1

2 -2

4 -1

4 x

-5

-2

-4

Sketch of y 

2. X Y

4 x

-4 1

-2 2

-1 4

1 -4

f x  =

-4 x

-5

-2

-4

Shifting the hyperbola graph: 1.

If a constant is added to the equation after shift: e.g. y 

k then this will cause a vertical x

4 4  3 : the graph of y  is shifted upwards by 3 units. x x

Asymptotes of 2.

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y

4 are y  0 ( x – axis) and x  0 ( y- axis). x

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

(1;7) 6

(2;5) 4

h x  =

(4;4)

(1;4)

4 x

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 gthe shift is to the right.

Shift the graph of y 

4 4 as follows: y  x3 x

This shift is 3 units to the right.

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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;4)

(-1;2)

(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 9.9 Copy and complete the following table and use it to sketch the graph of y 

1. x Y

-6 -1

-3

-2 -3

-1

2 3

6 x

4 x+3

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-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

1.1

Write down the equations of the asymptotes. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

1.2

Sketch the graph of y 

4 ( lateral move to the right) x3 6

1.3

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Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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

2. x y

-6 1

-3

-2 3

-1

2.1 Write down the equations of the asymptotes.

2 -3

6 x 3

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-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

2.2 Sketch the graph of y 

6 ( lateral move to the left) x3 6

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2.3 Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Complete the following table and use it to sketch the graph of y ď&#x20AC;˝

3. x y

-8

-4 -2

-2

-1 -8 10

2 4

8 x 4

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3.1 Write down the equations of the asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

3.2 Sketch the graph of y 

8 1 x2

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-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

3.3 Write down the equations of the new asymptotes. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Trigonometric Functions: Method: Sketch the graps 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

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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 

-270

-360

-180

-90

90

180

270

360

-1

Example 3:

Sketch the graph of f ( x)  cos x where x  [360 ;360 ]

g x  = cos x  -360

-270

-180

-90

90

180

270

360

Example 4:

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Sketch the graph of f ( x)  tan x where x  [360 ;360 ]

h x  = tan x  1

-360

-270

-180

-90

90

180

270

360

-1

Exercise 9.10: 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:

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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

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y 2

x 0 90

180

270

360

-1

-2

2. Sketch the graphs of y  2 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: 2.1

the period of y  2 sin x

2.2

the range of y  sin x

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2.3

the amplitude of y  2 sin x

2.4

the value for x for sin x  cos x

y 2

x 0 90

180

270

360

-1

-2

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

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the value for x for sin 2 x  cos x

y 2

x 0 90

180

270

360

-1

-2

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

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the value for x for  sin x  cos 2 x

y 2

x 0 90

180

270

360

-1

-2

11.

Probability Theory: 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 effect 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.

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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 . Why is 2

this so?

1 again? 2 3. How many different outcomes are there when a coin is tossed three(3) times? 2. Why are the probabilities for the second toss all

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.

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.

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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 11.1: [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.

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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______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.3 On drawing two(2) counters, the first is yellow and the second is white, if the first counter: 1.3.1 was replaced. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.3.2 Was not replaced before the second draw. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 2. 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. 2.6

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

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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?

2.2.2

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

What is the probability that James will be wearing something white

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What is the probability that James chooses his black shirt?

2.3.3

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

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

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

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

Look at John‟s venn diagram and then answer the questions below. Ch

5 Pl

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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.

Answers to above questions: 1.1 1.2 1.3 1.4 1.5 1.6

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.

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5 1  15 3

1.7

P(J given Ch) =

1.8

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 P(J, then Ch) =    , The order does not seem to 16 15 240 6 matter.

1.9 1.10

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) =

2.3

Third Choice

3 5 15 1    16 15 240 16

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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 11. 2: 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

tossing a coin twice

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Jack has a bag of 5 red, 7 blue, 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.

2.1.2

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

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2.2

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

2.2.1

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

2.2.2

Cynthia chooses red, then white , then white. Was the probability of this

Grade 11 Core Mathematics

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choice greater than the probability of getting 3 red marbles? Explain. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ _____________________________________________________________ 3.

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. 3.1.1

Draw a venn diagram to show the sample space for Blakesâ&#x20AC;&#x;s first choice.

How many socks are in the sample space for Blakeâ&#x20AC;&#x;s second

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choice? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.1.3

Are the events dependent or independent?

3.2 Draw a tree diagram showing Blakeâ&#x20AC;&#x;s sock choices, then answer the questions that follow: ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.2.1

How many different outcomes are there altogether?

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3.2.2

How many of these outcomes have two socks the same?

3.2.3

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?

Compiled by Chez Nell

Grade 11 Core Mathematics

GO MATH WORKBOOKS

Forward: Welcome to “GO MATH WORKBOOKS”. This workbook is designed to be a text book and class work book in one. There are sufficient exercises to ensure that learners get the required practice. A detailed memorandum booklet is available for each workbook. The statement “You get out what you put in.” is very apt where maths 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. Chez Nell: Mathematics Educator : Northwood School  Norma

Nell 2011

Grade 11 Core Mathematics

GO MATH WORKBOOKS

GRADE 11 CORE MATHEMATICS CONTENTS:

Paper Two: Topic

Pages

Analytical Geometry

(4 – 29)

Trigonometry

(30 – 73)

Data Handling

(74 – 117)

Grade 12 Statistical Data

(118 – 129)

Volumes & Surface Area

(130 – 150)

Circle Geometry

(151 – 176)

GO MATH WORKBOOKS

Grade 11 Core Mathematics

PAPER TWO 1.

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  (y 1  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)

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

Equation of a circle centre origin on a cartesian plane. Given by x2 + y2 = r2

Grade 11 Core Mathematics

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Exercises 1.1: Distance between points: 1.

Find the distance between the given pairs of points: 1.1 (2 ; 3) and (4 ; 5) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.2 (6 ; 1) and ( -6 ;6) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.3 (3 ; -7) and (-1 ; 3) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 1.4 (-4 ; 3) and (0 ; 0) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

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1.5 (-2 ; 1) and -4 ; -1) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.6 (-3 ;-1) and (4 ; -6) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 2.

Given the coordinates of the vertices of ď &#x201E;ABC , in each case ( 2.2 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(1 ; -3) ; B(7 ; 3); C(4 ; 6)

Grade 11 Core Mathematics

2.2

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A(5 ;1) ; B(1 ; 3) ; C(1 ; -2)

A(-2 ; -3) ; B(-4 ; 1) ; C(4 ; 5)

A(0 ; 0) B( 3 ; 1) ; C( 3 ; -1)

Grade 11 Core Mathematics

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Show that: 3.1

A(-3 ; 2) , B(3 ;6), C(9 ;-2) and D(3 ; -6) are vertices of a parallelogram.

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 3.2 (6 ;-4) , (5 ;3) (-2 ; 2) and (-1 ; -5) are vertices of a square. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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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) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.2 (-2 ; 3) and (6 ; 3) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.3 (4 ; -1) and (-1 ; 3) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.4 (0 ;0 ) and (3 ; -8) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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4.5 ( 3;1) and (3 3;ď&#x20AC;1) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.

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). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.2

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

(x ; y) is the centre of a circle on diameter AB where A(-2 ; -1) and B(-1 ; 9).

Grade 11 Core Mathematics

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5.4 (x ; 3) is the centre of a circle with diameter MN. M (5 ; -2) and N(-7 ; y) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Calculate the lengths of the medians of ď &#x201E;ABC in which the coordinates of the vertices are as follows: A(-3 ;1), B(-5 ; -3) and C(1 ; -5). (NB: a median is the line from a vertex drawn to the mid-point of the side opposite the vertex) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

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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 

Calculate the gradients of the lines joining the following points: 7.1 (-3 ; 2) and (1 ; 1) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 7.2 (4 ; 3 ) and (-1 ; 8) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

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7.3 (-3 ; -5) and (1 : 3) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 8.

Write down the gradients of the lines perpendicular to the lines in 7. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Calculate the inclinations of the line AB in each of the following cases. 9.1 A(-3 ; 2) and B(-5 ; 0) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 9.2

A(-2 ; 1 ) and B(1 ; -2)

Grade 11 Core Mathematics

9.3

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A( 3 ; 1) and B((2 3 ; -2)

A(-1 ;2) and B(1 ; -1)

A(-5 ; 2) and B(3 ; -1)

Calculate the gradients of lines with inclinations of: 10.1 45ยบ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

10.2

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60ยบ

150ยบ

110ยบ

Calculate the gradients of the following lines and state whether they are A. Parallel B. Perpendicular C. Neither. 11.1

A(0 ; -1) , B(-4 ; -2) , C( -3 ; 1) and D ( 1 ; 2)

Grade 11 Core Mathematics

11.2

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A(6 ; -10) , B(0 ; 4) , C( 6 ; 0) and D ( -4 ; -3)

A(-3 ; 5) , B(5 ; -1) , C( -2 ; -1) and D ( 1 ; 3)

A(-2 ; -4) , B(3 ; 1) , C( 5 ; -1) and D ( -2 ; -8)

Show that the following points are collinear: ( lie on the same line) 12.1 A(-2 ; -6) , B(2 ; -4) , C( 4 ; -3) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

12.2

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A(-5 ; 5) , B(1 ; 1) , C( 4 ; -1)

Determine the equaition of a line where the gradients and a point on the line are given as follows: 1 13.1 ; (2;3) 2 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

13.2

3 ; (3;1) 2

 2; (1;3)

Grade 11 Core Mathematics

14.

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Lines passing through the following points: 14.1 (-2 ; 4) and (2 ; 2) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 14.2

(-1 ; 1) and (1 ; 5)

(-3 ; -2) and (-1 ; -1)

(3 ; -3) and (3 ; -6)

Grade 11 Core Mathematics

15.

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A line with the slope of 3 and intersecting the y – axes at 2 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

16.

parallel to y  3x  2 and passing through (3 ; 1) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

17.

Through (-2 -1) and perpendicular to 3 y  2 x  6 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

18.

Through (-1 ; 3 ) and an inclination of 120º ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

19.

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A(-2 ; 1), B(3 ; 3) and C(6 ; -3) are the vertices of a triangle . Determine: 19.1 The coordinates of M, the mid-point of AC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 19.2 the gradient of AC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 19.3 the equation of the perpendicular bisector of AC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 19.4 the equation of the median BM ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 19.5 the equation of the altitude from B to AC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

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Circle centre the origin: 20.

Determine the equation of a circle with centre origin and: 20.1 radius = 3 cm ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 20.2 radius = 3 2 cm ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 20.3 Passing through point (-2 ; 3) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 20.4

Passing through point ( -4 ; -2)

Grade 11 Core Mathematics

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A(-3 ; 4) is a point on a circle with centre at the origin: 21.1 Determine the equation of the circle. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 21.2

Determine the coordinates of B if AB is a diameter.

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Grade 11 Core Mathematics

Tangents to circle centre origin: 22.

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: 10.1 The gradient of OB.

10.2

The equation of AC, the tangent. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

10.3

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Grade 11 Core Mathematics

The equation of the circle centre O. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Exercise 1.2: 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.

Grade 11 Core Mathematics

1.2

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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 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

1.7

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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.

2.2

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Grade 11 Core Mathematics

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. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

3.3 3.4

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Calculate the value of t if the point D(t ; t- 3) is such that AC // BD. Calculate the mid-point of BC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.5

Determine, by calculation , whether the quadrilateral is a parallelogram. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.6

Give the equation of BC. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.7

What is the size of the angle of inclination of BC with the positive x â&#x20AC;&#x201C; axes. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

3.8

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Calculate the size of ACË&#x2020; B .

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Grade 11 Core Mathematics

TRIGONOMETRY

In trigonometry there are 3 different approaches to finding answers without the use of a calculator.

2.1

Using Pythagoras:

A ratio will be supplied and a clue in the form of a restriction stating in which quadrants the angle can be situated. Combining the two clues locate the precise quadrant for the sketch. 3 sin x Example: If sin x  and x  [90 ;360 ] find the value of 5 cos x

y 2  r 2  x 2 ( pythagoras ) y  5 3 2

y 2  16

5 3

-4

y  4

X X 3 sin x 3 5 3 Answer:  5    cos x  4 5  4 4 5 Exercise 2.1: 8 1.1 If sin x   and 90 < x < 270 17 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ Find without the use of a calculator: 1.1.1 cosx ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

1.1.2

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sin2x + cos2 x

1.2 If cos A 

3 , and A  [180 ;360 ] Determine then value of: 5

sin A  cos A

Grade 11 Core Mathematics

1.2.4

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sin A cos A

Grade 11 Core Mathematics

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Negative angles: Clockwise rotation of the terminal ray gives rise to negative angles.

Co – terminal Angles: These are angles which share the same terminal ray. Thus all ratios of co-terminal angles must be equal.

210

Eg.

-150º Co-terminal angles. 210º -150

cos(-150º) = cos 210º = cos(180º + 30º) = - cos 30º tan ( -150º) = tan 210º = tan(180º + 30º) = tan 30º When finding a ratio of a negative angle, convert it to the ratio of the positive , co-terminal angle eg. sin ( -261º) = sin 99º

Grade 11 Core Mathematics

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Angles Greater than 360º The terminal ray of these angles is rotated through one or more rotation(s). Again coterminal angles arise. 

Eg.

Are Co-terminal angles. All ratios of both are equal.  + 360º Thus when finding a ratio of an angle greater than 360º convert it to the ratio of  Where   [0º ; 360º] Likewise, if the terminal arm is rotated clockwise through 360º , we get:  Are Co-terminal angles. All ratios of both are equal.  - 360º If the terminal ray is rotated through more than one positive or negative revolution , then the angles formed ; [eg  + 2.360º ;  - 2. 360º ;  + 3. 360º] have equal ratios to those of  [0º ; 360º] e.g sin( + 720º) = sin( -720º) = sin ( + k.360º) = sin  cos( + 720º) = cos( -720º) = cos ( + k.360º) = sin  tan( + 720º) = tan( -720º) = tan ( + k.360º) = sin  cos 730º = cos(2.360º + 10º) = cos (720 +10º) = cos 10º sin 380º = sin(360º + 20º) = sin 20º tan 1030º = tan(3.360º - 50º) = -tan 50º

Grade 11 Core Mathematics

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2.2. Using reduction function: In these types of questions specific angle sizes are not given. The angles are given as variables and the different reduction functions are utilized. i.e

sin(180   x)   sin x sin(90   x)  cos x

Example:

tan(180   x) sin(360   x) cos(90   x) ( tan x)( sin x)(sin x) sin x     tan x    ( sin x)(cos x)(tan x) cos x sin(180  x) sin(90  x) tan(180  x) Exercise 2.2: 2.1

cos(180   x). sin(90   x). tan(180   x) cos(360   x). sin 270

2.2

sin(180   x) tan(360   x) sin(90   x) sin(180   x) cos(90   x) tan(360   x)

Grade 11 Core Mathematics

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sin(360   x) sin(90   x) cos(90   x)

2.3

tan(180   x)( sin x) cos(180   x)

2.4

Using Special Angles: Specific angle sizes given :

A. Special angles and their multiples: 1 sin 30   2 e.g. 1 sin 330    sin 30    2 B. Complimentary angles. e.g. sin 40  cos 50 sin 25 sin 25 cos 65  or 1 cos 65 sin 25 cos 65

Grade 11 Core Mathematics

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Exercise 2.3: 3.1 3sin30º tan45º cos30º ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.2

sin30º cos30º tan60º ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.3

4sin60º + tan45º + 2cos30º ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.4

4sin 2 45º - 3 sin 2 30º

3.5

cos 2 0º +cos 2 30º + sin 2 45º ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

3.6

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cos 30º tan 2 45º + tan 2 30º + tan 0

3.7

1 2 sin 60   cos 30  tan 2 30   sin 45 tan 2 60  2 3 3

3.8

   cos 60 . cos 40 . tan 330    sin 210 sin 50 . sin 270

3.9

   sin 315 . cos 135  cos 210   tan 135 . sin 270

Grade 11 Core Mathematics

3.10

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   sin 315 . cos 20 . sin 240    tan 135 . sin 80 cos 180

Grade 11 Core Mathematics

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2.4. Reduction Function: NB: All angles greater than 90º can be written as ratios of acute angles using then relevant reduction function. Example: represent sin 240 as a ratio of an acute angle. 3/sin 240   sin(180   60  )

 - sin60  Exercise 2.4: Represent the following as ratios of acute angles: 4.1 sin 230 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.2 cos150 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.3 tan125 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.4 cos230 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.5 tan320 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 4.6 sin145 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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2.5. Calculator Work: Exercise:2.5 Use a calculator to Find: 5.1 3 sin 120  ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.2

2 tan135

cos 240  ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.3

2 tan125 cos 150 

sin 139  ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.4

If x  25,7  and y  137,4  calculate the value of the following: sin 2 x cos 2 y 3 tan y ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

2.6

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Solution of trig equations:

It is important to note the restrictions given ,to arrive at the correct solution. Find a Key L ( reference angle) and use the relevant reduction function to supply the correct answer in the correct quadrant. [NB: The Key L must not be manipulated or changed ] Example:1 Solve for x where x  [0 ;360 ] if :

Ex 1.

sin x  0,5

2 sin( x  30  )  1

KeyL  30 

sin( x  30  )  0,5

x  180   30 

Key ( x  30  )  30 

x  210  OR x  360   30  x  330



Ex 2:

x  30   180   30  x  240  OR x  30   360   30  x  360 

Exercise 2.6: Use a calculator to determine the values of x for x  [0  ;360  ] [ 1 decimal place.] 6.1

2 sin x  0,545

6.2

2 cos x  0,147 3

Grade 11 Core Mathematics

6.3

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3 tan x  6,605

6.4

1 cos x  0,245  0 2

6.5

tan x  8,213  0

Grade 11 Core Mathematics

6.6

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3 sin x  0,369 2

6.7

 2 sin x  0,546  0

6.8

2 tan x  8,442  0 3

Grade 11 Core Mathematics

6.9

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4 sin x  3,208  0

6.10

2 sin( x  20 )  1,636

6.11

2 cos( x  30  )  0,262 3

Grade 11 Core Mathematics

6.12

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2 sin( x  25)  0,345

6.13

3 tan(x  75 )  6,147 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.14

 2 cos( x  15)  1,605

Grade 11 Core Mathematics

6.15

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2 sin( x  30)  0,445  0 3 ______________________________________________________________

6.16

3 tan( x  54)  21,213 4 ______________________________________________________________

2.7.

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Grade 11 Core Mathematics

General Solution in trig equations

The general solution of an equation are not in specific form. i.e they do not give the final vale of the given variable. They are solutions that satisfy the given equation in a general form and when the parameters for the variable are supplied then the specific values can be found. [They involve k.360º which simply refers to the number of extra revolutions required to obtain a specific answer in a given restriction] Find the general solution for x in the following: Example: +ve 1/2

2 sin 2 x  1,630 sin 2 x  0,815

KeyL (2 x)  54,6  In Q1: 2x = 54,6 + k .360º x = 27,3º + k .180º In Q2 2x = (180º - 54,6º) + k.360º 2x = 125,4º + k.360º x = 62,7º + k.180º

General solutions:

If the specific solution is required then the period (Domain)must be stated: Example: Now find the value(s) for x when x  [0 ;360 ] If: k = 0 then x = 27,3º or 62,7º 

k = 1 then x = 207,3º or 242,7º Exercise 2.7: Find the general solution of the following: 7.1

k=  2



k=  1 k= 0

k- represents the number of revolutions required to obtain the answers that fall within the stated restriction

tan 2 x  2,6 ______________________________________________________________

Grade 11 Core Mathematics

7.2.1

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2 cos x  0,66  0 ______________________________________________________________

7.3.1

sin

1 x  0,825 2 ______________________________________________________________

Grade 11 Core Mathematics

7.3.2

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Find the specific solutions for 7.3.1 if x  [270 ;270 ] ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

7.4

Find the general solution for the following: tan 3x  4,302 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

7.6.1

Find the general solution for the following:

2 cos(2 x  60 )  0,684 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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7.6.2 Find the specific solutions for 7.6.1 if x  [360 ;360 ] ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 7.7.1 Find the general solution for the following:

4 sin(3x  120 )  2,812 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 7.7.2 Find the specific solutions for 7.7.1 if x  [180 ;360 ] ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Exercise 2.8. Use the fundamental identities to simplify the following: 8.1

sin x tan x

1-sin2x

______________________________________________________________ ______________________________________________________________ ______________________________________________________________

8.3

1-cos2x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

8.4

sin 2 x ď&#x20AC; 1 ď&#x20AC;Ť cos 2 x

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cos x ) tan x ______________________________________________________________

sin x(sin x 

8.5

8.6

Proving Fundamental Identities:

Example: Prove the following: sin 4 x  cos 4 x  sin 2 x  cos 2 x

LHS  sin 4 x  cos 4 x  (sin 2 x  cos 2 x)(sin 2 x  cos 2 x)  sin x  cos x  RHS 2

NB sin 2 x  cos 2 x  1

Exercise 2.9: 9.1

cos x 1 1   2 tan x cos x sin x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

9.2

(1  cos 2 x) 

1 2

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 cos 2 x

tan x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 9.3

1  1  cos x   sin x  sin x  tan x 

9.4

sin 3 x  cos 3 x  (sin x  cos x)(1  sin x cos x ) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

9.5

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(sin x  cos x) 2  1  2 sin x cos x ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

9.6

. (1  sin 2 x)(1  tan 2 x)  1 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

9.7

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Grade 11 Core Mathematics

cos x(1  tan 2 x) 

1 cos x

Finding values trigonometric functions in terms a given of a variable : Example: 10

If cos 40 = p , then represent the following in terms of p

10.1

sin 50 = cos 40º = p

10.2

cos140 = -cos40º = -p

NB: MAKE A SKETCH

sin 2 40   1  cos 2 40 

10.3

sin 40 

sin 2 40   1  p 2 

sin 40  1  p

10.4

tan 40 =

sin 2 40  cos 2 40 



1 p2 p2

1 - p2

40 p

If the answer cannot be found using the diagram above THEN make use of the Reduction formulae to get back to the given clue i.e, 40 in this case. 180 ±  ; 360 -  ; 90 ± 

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Exercise 2.10: 10.5 If sin 20 = k , then represent the following in terms of k: 10.1

sin 160 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

10.2

cos70 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

10.3

cos 20 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

10.4

tan 20 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

10.5

If cos 40  a determine the following in terms of a: 10.5.1 tan 40 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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10.5.2 cos 220 ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 10.5.3 sin 40  ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

2.11 11.1

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Solution of Triangles: Area Rule:

Theorem: The area of a triangle is half the product of the lengths of two sides of the triangle and the sine of the included angle. Example: Calculate the area of ABC

A 8cm 40º B 5cm C 1 ( AB )( BC ) sin Bˆ 2 1 Area of ABC  (8)(5) sin 40  2 ABC 

ABC  12,86cm 2

Exercise 2.11: Calculate the area of: 11.1.1

ABC in which BC = 7cm; AC = 6cm and Cˆ  27,6 .

11.1.2

EFG in which EG = 29cm ; EF = 54cm and Eˆ  61,4

Grade 11 Core Mathematics

11.1.3

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FGH in which Hˆ  61,4 ; GH = 9,5cm and FH = 2,3cm

11.2

Calculate the area of a parallelogram in which two adjacent sides measure 100mm and 120mm and the angle between them is 65º. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

11.3

If the area of XYZ is 3000m2 and x  80m and y  150m , calculate two possible sizes of Zˆ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.12

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Grade 11 Core Mathematics

Sine Rule: a b c ď&#x20AC;˝ ď&#x20AC;˝ sin A sin B sin C

In any ď &#x201E;ABC :

Theorem:

Example:

15cm 40Âş B

8cm

Find AË&#x2020; and AC

8 AC ď&#x20AC;˝ sin 20 sin 120 8 sin 120 AC ď&#x20AC;˝ sin 20 AC ď&#x20AC;˝ 20,3cm

8 15 ď&#x20AC;˝ sin A sin 40 8 sin 40 sin A ď&#x20AC;˝ 15 ď Ż Ë&#x2020; A ď&#x20AC;˝ 20

The Ambiguous Case: This comes into effect when the side opposite the given angle is smaller than the side adjacent to the given angle. One angle and two sides are given.

30ď&#x201A;°

30ď&#x201A;° 5

Ambiguous Case đ?&#x;&#x2019;

đ?&#x;&#x201C;

= đ?&#x2019;&#x201D;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2018;¨ đ?&#x2019;&#x201D;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x;&#x2018;đ?&#x;&#x17D;Â° đ?&#x2019;&#x201D;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x2018;¨ =

đ?&#x;&#x201C;đ?&#x2019;&#x201D;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x;&#x2018;đ?&#x;&#x17D;Â° đ?&#x;&#x2019;

Normal Case đ?&#x;&#x201C;

đ?&#x;&#x2019;

đ?&#x;&#x2019;đ?&#x2019;&#x201D;đ?&#x2019;&#x160;đ?&#x2019;?đ?&#x;&#x2018;đ?&#x;&#x17D;Â° đ?&#x;&#x201C;

đ?&#x2018;¨ = đ?&#x;&#x2018;đ?&#x;&#x2013;, đ?&#x;&#x2022;Â° đ?&#x2019;?đ?&#x2019;&#x201C; đ?&#x;?đ?&#x;&#x2019;đ?&#x;?, đ?&#x;&#x2018;Â°

đ?&#x2018;¨ = đ?&#x;?đ?&#x;&#x2018;, đ?&#x;&#x201D;Â° đ?&#x2019;?đ?&#x2019;&#x201C; đ?&#x;?đ?&#x;&#x201C;đ?&#x;&#x201D;, đ?&#x;&#x2019;Â°(reject)

đ?&#x2018;Ş = đ?&#x;?đ?&#x;?đ?&#x;?, đ?&#x;&#x2018;Â° đ?&#x2019;?đ?&#x2019;&#x201C; đ?&#x;&#x2013;, đ?&#x;&#x2022;Â°

đ?&#x2018;Ş = đ?&#x;?đ?&#x;?đ?&#x;&#x201D;, đ?&#x;&#x2019;Â°

Two possible answers For đ?&#x2018;¨ đ?&#x2019;&#x201A;đ?&#x2019;?đ?&#x2019;&#x2026; đ?&#x2018;Ş

Only one possible answer For đ?&#x2018;¨ đ?&#x2019;&#x201A;đ?&#x2019;?đ?&#x2019;&#x2026; đ?&#x2018;Ş

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Exercise 2.12: Solve the following triangles: 12.1.1

ABC in which Aˆ  30 ; Bˆ  45 and a  2

PQR in which Pˆ  115 ; Qˆ  20  and q  15,3

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12.1.3 XYZ in which Yˆ  64  ; Zˆ  21 and x  30

12.1.4

ABC in which ; Bˆ  50,1 ; Cˆ  72,3

and AC  5,34

Grade 11 Core Mathematics

12.2

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In ď &#x201E;PQR , đ?&#x2018;&#x192; = 38,2; đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; = 5,2đ?&#x2018;?đ?&#x2018;&#x161; and PR = 7,4 cm. Solve the triangle. i.e. find all missing values. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

12.3 In ď &#x201E;KMS , KË&#x2020; ď&#x20AC;˝ 30,7 ď Ż ; SË&#x2020; ď&#x20AC;˝ 19,1ď Ż and KM = 4,2m. Calculate the length of KS. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.13

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Grade 11 Core Mathematics

Cosine Rule: In any ABC :

Theorem:

a 2  b 2  c 2  2bc cos A b 2  a 2  c 2  2ac cos B c 2  a 2  b 2  2ab cos C

Example:

Calculate the length of AB:

AB 2  BC 2  AC 2  2( BC )( AC ) cos 40  AB 2  12 2  15 2  2(12)(15) cos 40  AB 2  93,22400048

15cm

AB  9,7cm

40º B

12cm

Exercise 2.13: Solve the following triangles: 13.1

ABC in which Aˆ  60 ; AB = 5cm and AC = 8cm ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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13.2

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PQR in which Qˆ  135 ;QR = 3 2 and PQ = 1

13.3

GHK in which GH = 8cm ; HK = 9cm and GK = 10cm ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 11 Core Mathematics

LMN in which LM = 7cm ; MN = 13cm and NL = 8cm

13.4

13.5

20 m

120

ABCD is a trapezium in which: AD = 20 m. BC = 30 m, DAˆ B  120 and ABˆ D  30 . AD // BC.

30 B

Show that : 13.5.1

BD = 20 3m

30 m

[Hint: use special angles)

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Grade 11 Core Mathematics

13.5.2 DC = 10 3m ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 13.6

In the following sketch AC  BC  x and Cˆ  y  C y x

13.6.1 Show that the distance AB  x 2(1  cos y) and hence,

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Grade 11 Core Mathematics

13.6.2 Calculate AB if x  150m and y  112

In the diagram QP = 10,28 cm PR = 5,73 cm and Qˆ  32  P

10,28cm 5,73cm 32 Q

Calculate Pˆ

Grade 11 Core Mathematics

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14.2 In the diagram AB = 5cm , AC = 4cm and BC = 6cm A

4cm 5cm

C 6cm B

Calculate all the angles of ď &#x201E;ABC ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 11 Core Mathematics

14.3 In the figure PQR is right agled at Q: RS = 10cm, Rˆ  40  and PSˆQ  68 P 1

68

40 10cm

Calculate: 14.3.1 PS

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Grade 11 Core Mathematics

______________________________________________________________ 14.4 In the diagram A B & C lie on the same horizontal plane . HC is a vertical height equal to 100m.

100m  ACB=104,5

26,5

A 21,8 P B

Calculate: 14.4.1 AB

14.4.2 CP

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Grade 11 Core Mathematics

______________________________________________________________ 14.5 In the figure B and D are points in the same horizontal plane as C, the foot of a A

vertical tower AC. AC = BD = x and BDˆ C  90  

C 

Show that: BC  x

1 sin 2 

 2 cos 

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DATA HANDLING: Data Collection: 1. Understand data collection as a potential source of bias. 2. Choose a sample that is free from bias. 3. Understand the need for random sampling methods. 4. Develop appropriate and relevant survey questions. Choosing your source. 

must be appropriate



whole group of things called :



the reference group



Target population.



Researchers that need to find information to help them understand an area will do research under specific conditions, defined Carefully ahead of time.



Data involving peoples opinions collected by:



1. A population census is collected from every member of the target population.



2. Survey – a sample or selected part of the target population.

A census requires a lot of planning ,is time consuming and costly. A survey is a more convenient way of collecting data from a ample of the community. Chosen people must be representative of the community. Too many of a certain section of the community will produce a biased or unrepresentative outcome. Biased data - an imbalance towards a particular result in the data. Not a true reflection of target population. The group chosen for a survey is called the sample. NB: Sample must be large enough to be representative but small enough to be convenient tom work with. NB:

Free from bias: Need to generalize findings Representative sample are more accurate.

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Random Numbers: A number chosen by chance – each number must have an equal chance of being the result. Random number tables: Lists from 0 to 9 that have random selection. Simple random samples: Chosen from target population where every person in population has had an equal chance of being chosen. Stratified Random Samples: Where target population separated into groups according to relevant information can help to make sample selection more representative—less biased. Univariate Numerical Data: Univariate data involves working with single variables or frequencies of single variables. Know: mean ; median; mode; quartiles; range inter-quartile range and semi-quartile range; Notation used: Mean of a set of data : x 

Sum Total of values x  number of the values n

SKILLS NEEDED: 1.

Summarise sets of data by calculating The 5 –number summary.

Calculate the variance and standard deviation of sets of data manually and using technology(calculator)

Representing data – box & whisker diagrams – Ogives – histograms – and frequency tables.

Compare sets of data.

Effectively communicate conclusions from analysed data.

Differentiate between symmetrical and skewed data.

Critically analyze data misrepresentations.

Represent data with 2 variables as a Scatter plot and suggest types of graphs that will fit the data.

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Exercise 3.1: 1.

Class results for a test out of 30 are recorded in the table below:

10A 16 12 16 11 14 15 22 16 17 15 26 23 16 22 16 17 24 19 16 10B 20 19 14 10 14 9 8 13 14 30 27 23 24 28 17 29 20 16 14 18 10C 5 20 14 12 7 2 12 21 14 26 14 14 12 14 21 24 14 14 1.1

Calculate the mean for each class.

1.2

Calculate the mode for each class.

1.3

Calculate the median for each class. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.4

Calculate the range for each class. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

1.5

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Calculate the lower quartile (Q1) for each class. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.6

Calculate the upper quartile (Q3) for each class. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.7

Calculate the inter-quartile range for each class.

1.8

Calculate the semi-quartile range for each class ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 11 Core Mathematics

The table below shows the masses ( in kg) of people who go to gym on a regular basis. Complete the table and hence calculate the estimated mean.

Class interval: kg‘s

Frequency

10 – 19 20 – 19 30 – 39 40 – 49 50 – 59 60 – 69 70 – 79 80 – 89 90 – 99 100 – 109

0 1 3 7 18 31 38 49 13 0

Midpoint class interval 14,5 24,5 34,5 44,5 54,5 64,5 74,5 84,5 94,5 104,5

Frequency x midpoint 0 24,5 103,5 311,5 981 1999,5 2831 4140,5 1228,5 0

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Example: Using energy as an issue. Energy consumed per person in same sub-Saharan African countries. Country Angola Benin Cameroon Congo Congo-Dem Rep Cote dâ&#x20AC;&#x2DC;voire Eritrea Ethiopia Gabon Ghana Kenya Mozambique Namibia Nigeria Senegal South Africa Sudan Tanzania Toga Zambia Zimbabwe 1.

Energy use per person 323 164 280 141 204 151 108 6 659 206 243 278 98 585 138 272 205 312 57 309 430

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

(a) Which has least residential energy per person? Why so low. Answer: Ethiopia has smallest residential sector energy per person. A poor country that uses a lot of wood as fuel source. (b) Which has highest per person? Is it extreme? Why? Answer: Gabon has the highest residential sector energy use per person. The mean per person in Gabon is not an extreme value as there is only a moderate gap between it and its closest data value ( Nigeria: 585)

What is the range for the data set? Answer: Range is 659 -6 = 653.

What is the size of the data set? Answer: n = 21

(a) Ranked data set â&#x20AC;&#x201C; get the median. What does this mean? Answer: Median = 206 When the values are ordered from smallest to largest, the median is the middlemost value. (b) How many countries above the median and how many below? Answer: 21 data values. Middle value is 11th = 206. 10 countries lie above and 10 below.

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(c) What percentage of the data lies below the median? Answer: Of data values 50% lie above and 50% lie below the median. 5.

(a) Find Q1 : What % below? Answer: Q1 = 141. 25% of the data lies below 1st quartile. (b) Find Q3: What % above? Answer: Q3 = 309. 25% of data lies above 3rd quartile. 6. Calculate the inter-quartile range. What does this mean? Answer: Inter-quartile range : 309 – 141 = 168. It is the middle 50% of the data. 7.

5 number summary Answer: Minimum Value 6 Q1 = 141 Median = 206 Q3 = 309 Maximum Value =

659

Key Ideas: 1. Extreme Data value – lies far away from rest of data (maybe errors or a genuine observation) 2. Range: Max – Min. 3. Median: Middlemost score: divides data set into 2 halves. 4. Median of Ist half called First quartile (Q1): lower quartile. Median of 2nd half called Q3 or third quartile: upper quartile. 5. Inter-quartile Range : Q3 – Q1 Measures the spread of middle 50% of data. If large then there is a high level of variation in the data. If Small then data is widely spread. 6. 5 number summary – Min; Q1; Median; Q3; Max. 25% of data lies below Q1 thus 75% above 75 % of data lies below Q3 thus 25% above.

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Grade 11 Core Mathematics

Representing 5-number summary: Median

Box and whisker diagram:

Highest Value

Lowest Value

659

Q1 141 100

309

206 200

600

300

Ogive curve drawn from the Box & Whisker Diagram:

100

0 100

200

300

700

600

659

141 100

206 200

309 300

600

Key Ideas: 1.

Quartiles in 5 - Number Summary gives an indication of frequency of data.

Ogive or Cummulative Frequency Graph of 5- Number Summary gives a rough idea of how the data values in a set are distributed.

Slope will be steeper where quartile range closest together indicating a higher concentration of data values. Slope thus a measure of the spread of data. Steep slope closer with small range. A gradual slope shows spread out data with a greater range.

Grade 11 Core Mathematics

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MEAN, VARIANCE AND STANDARD DEVIATION USED TO SUMMARIZE DATA SET. Learners in Portia‘s class record the number of hours out of 24 that they switch lights on at home . Portia wants to compare her use of electricity to others. She wants to calculate the average or mean electricity use of lights among the learners taking part in the exercise. First Column Second Column Third Column Fourth Column Data Value (Data value – mean) (Data value – mean)² Portia 7 1 1 Nomfundiso 7 1 1 Unathi 8 2 4 Busiswe 4 -2 4 Pearl 5 -1 1 Sonwabo 6 0 0 Zuzeka 2 -4 16 Bulelani 9 3 9 Total 48 0 36 mean 

48 6 6

36  4,5 8 sum of squared Variance  n-1 36 Variance  7 Variance  5,14

Squared deviation =

Std Deviation  5,14 Std Deviation  2,27

Key Ideas: 

If one adds up all the differences between the data values and the means the answer will be zero (0) because the negative values will ‗:cancel out‖ the positive ones. For this reason, the square of the differences is used for the average.



The variance is defined as the sum of the squared differences between each data value and the sample mean, this sum then divided by one less than the number of data values in the sum.



The standard deviation is defined as the square root of the variance.



The square root of 5,44 is 2,27 Because the square root has been taken the variance has been reduced to a measurement that has the same units as the data. If the data is symmetrical, a SD of 2,27 suggests that roughly two-thirds of the data values should fall between 2,27 units below the mean and 2,27 units above the mean.

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Second Column Third Column Household energy (Data value – mean) use per person(kgoe)

Fourth Column (Data value – mean)²

Grade 11 Core Mathematics

Measures of Central Tendency:

First Column Northern and Western European Countries Poland Ireland Netherlands United Kingdom Germany Denmark Norway Sweden Belgium Iceland Total

503 647 654 716 774 822 854 903 937 2114 8924

-389,4 -245,4 -238,4 -176,4 -118,4 -70,4 -38,4 10,6 44,6 1221,6 0

151632,36 60221,16 56834,54 31116,96 14018,56 4956,16 1474,56 112,36 1989,16 1492306,56 1814662,4

mean  892,4 1814662,4 Variance   201629,1556 9 StdDev  449,031 5 number summary;

503;654;798;903;2114

100

200

400

600

800

1000

1200

1400

1600

1800

2000

2200

2114

503 654

798

903

Grade 11 Core Mathematics

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Key Ideas:     



M.O.C.T.describe the middle or typical value in a data set. M.O.D. measure the spread of data. Outlier values: data values that are very different from others- i.e. they lie far from the median. Outlier values have a large effect on calculations of mean and std deviation. Causing statistics to change as outliers are included or omitted form data sets. Generally speaking if data is symmetrically distributed the mean will be a 2 good measure of location and ± of the data will often lie between 1 (one) 3 std deviation above and below the mean. If skewed the median becomes a better measure of central tendency and the inter-quartile range a better measure of spread as they are not affected by outliers as mean and std dev are.

Grade 11 Core Mathematics

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Grouped Data: Frequency tables and histograms. Examine represented data after grouping data values into convenient intervals. Representations will include : 1.

Histograms

Frequency polygons

Ogive curves.

For the above we first need a frequency table. Choose a class interval to group the data values. Example: Class Interval

Class width

0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900

100 100 100 100 100 100 100 100 100

Class Mark (midpoint of interval) 0 50 150 250 350 450 550 650 750 850

1300 – 1400

100

1350

2500 – 1600

100

2550

Total NB: Class intervals 0 – 100 include up to 99. 100 is in the next set 100 – 200.

Frequency 0 0 1 3 6 3 2 2 1 1 0 1 0 1 0 21

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Grade 11 Core Mathematics



Histograms and frequency tables give a quick picture of the spread of data.



Skewed data – median a better measure of middle data.



For symmetrical data use the mean as a measure. HISTOGRAM 8

Median (419)

F r e q

Mean (587,62)

u e n c y

200

400

600

800

1000

1200

1400

Per capita energy use (kgoe) for Sub-Saharan Africab coun tries (n = 21)

   

Data is skewed to the right Median is 419 Mean is 587,62 The median is a better measure of middle or centre of the data. FREQUENCY POLYGON 8

F r e q

u e n c y

200

400

600

800

1000

1200

Per capita energy use (kgoe) for Sub-Saharan Africab coun tries (n = 21)

1400

Grade 11 Core Mathematics

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CUMULATIVE FREQUENCY TABLE: Class Interval 0 – 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900 1300 – 1400 2500 – 1600 Total

Frequency 0 0 1 3 6 3 2 2 1 1 0 1 0 1 0 21

Cumulative Frequency

CUMULATIVE FREQUENCY GRAPH - OGIVE CURVE:

200

400

600

800

2600 1200 1000 1400 Per capita energy use (kgoe) for Sub-Saharan Africab coun tries (n = 21)

0 1 4 10 13 15 17 18 19 19 20 20 21 21 198

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The Normal distribution curve: In general for a large sample the graph will be bell shaped. The reason for this is that a few scores will be high and a few low. The majority of scores will be in the central region. The results are thus said to be normally distributed. If we have a normal distribution then:  The mean , median and mode will be the same.  It is symmetric and has an equal number of scores on either side of the mean.  The majority of scores lie within 3 std dev‘s from the mean.  Two thirds (67%) of sample lie within one std dev of mean.  95% lies within two std dev‘s of mean.  99% of sample lies within three std dev‘s of mean. Symetrical and Skewed Data A curve is symmetrical when one half is a mirror image of the other. NB: Not all corves are symmetrical. If extremely high or low scores are added to distribution the the mean tends to shift towards these scores and the curve becomes skewed. 

If the greater number of scores are massed on the right and a few are much lower scores than most others then the distribution is negatively skewed ( Skewed to the left)



If the greater number of scores are massed on the left and a few are much higher scores than most others then the distribution is positively skewed ( Skewed to the right) The relative position of the mean and median gives an idea whether the distribution is symmetrical, negative of positively skewed. (mean – median)= 0 Symmetric (mean – median) > 0 Positively skewed (skewed to right) (mean – median) < 0 negatively skewed. ( skewed to left)

   

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Grade 11 Core Mathematics

Example: 1.

The following marks were recorded for a maths class: 28 53 75 63

45 75 63

36 58 75 63

36 60 78 67

36 60 81 68

38 60 83 68

45 71 84 69

42 71 84 76

45 75 90 79

1.1 1.2 1.3 1.4

Do a stem and leaf diagram for the data Find the median, mode and mean for the data Find the lower and upper quartile Calculate: 1.4.1 the interquartile range 1.4.2 the semi-interquartile range 1.4.3 the range for the class 1.5 Write down the maximum and minimum scores. 1.6 Do a box and whisker diagram using the five-number summary

Answer: Stem 2 3 4 5 6 7 8 9

Leaf 8 6668 2555 38 0003337889 115555689 1344 0

Mode = 75 ; Mean = 62.9 ; Number = 35 Interquartile range = 30 ; Semi- interquartile ; range = 15 Range = 62 Standard Deviation = 16.6 Lowest = 28 ; Q1 = 45 ; Median = 67 ; Q3 = 75 ; Highest = 90

28 45 0

67 50

75 80

100

x  Q2  62.9  67  4.1  0 Data is negatively skewed i.e. skewed to the left. The marks are concentrated to the right of the median and spread out to the left of median.

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Grade 11 Core Mathematics

Exercise 3.2: 1. The number of points scored by four (4) Formula One racing drivers over a number of races is in the table below A B C D

1 1 1 2

1 2 1 2

1 6 2 2

2 8 2 4

6 8 4 4

6 8 4 6

8 8 6 6

8 8 6 8

1.1

Calculate the mean for each of the drivers.

8 8 8 8

8 10 8 10

10 10 10 10

10 10 10

10 -

List the Five Number Summary for each driver.

1.3

Calculate the difference between the mean and median for each driver.

Grade 11 Core Mathematics

1.4

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Draw a Box and Whisker plot for each driver.

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

1.5

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Discus each drivers distribution of scores in terms of the spread about the median and mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

1.6

Compare the performance results for each driver by using the information obtained above. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Grade 11 Core Mathematics

The following set of data records the number of chocolates sold by a convenience store over a period of 44 days.

2.1

Draw a Stem and Leaf Plot to organize the data.

List the Five Number Summary. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.3

Draw a Box and Whisker Plot. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

2.4

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Draw the cumulative frequency graph (Ogive Curve) using the Box and Whisker plot as a starting point. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

% Frequency

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Grade 11 Core Mathematics

The following table (grouped frequency distribution) shows the mark obtained by 220 learners in a Science exam. 1-10 2 3.1

11-20 6

21-30 11

31-40 22

41-50 39

51-60 59

61-70 45

71-80 20

81-90 91-100 11 5

Complete the cumulative frequency table below for this data:

Marks

Class midpoint

Frequency

1-10 11-20 21-30 31-40 41-50 51-60 61-70 71-80 81-90 91-100 Total

5.5 14.5 25.5 35.5 45.5 55.5 65.5 75.5 85.5 95.5

2 6 11 22

Cumulative Frequency 2 8 19 41

3.2 On a set of axes draw the cumulative frequency graph ( Ogive Curve) for the data.

3.3

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Determine the lower quartile (Q1). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.4

Determine the median.

3.5

Determine the upper quartile (Q3). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

The following is list of heights of learners in a class. Heights are in centimeters(cm) 152 149 151 4.1

153 155 142

147 153 183

151 167 168

138 180 150

181 132 145

159 157 145

Determine: 4.1.1 median ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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4.1.2

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arithmetic mean

4.1.3

standard deviation

4.1.4

first and third quartiles.

Draw a box-and-whisker diagram for the data set.

Complete the table below:

Interval 0  x  10 10  x  20 20  x  30 30  x  40 40  x  50 50  x  60 5.1

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Class midpoint 4,5 24,5 44,5 54,5

Frequency 7 11 22 25 10 0

Cumulative frequency 7 18 65 75

Draw a histogram of the information.

5.2

Determine the mean and standard deviation.

Grade 11 Core Mathematics

5.3

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Draw a graph to determine whether the median is likely to be closer to 30 or to 40.

Grade 11 Core Mathematics

6.1

100

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The ages to nearest year of 27 members of a Cricket Club are : 17 21 23 19 27 18 20 21 28 18 21 24 30 25 19 22 27 35 27 22 20 30 27 21 23

31 18

Organize the ages using a stem & leaf diagram. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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6.2

101

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Using 5 classes, and starting at 16, construct a frequency table and histogram for the data. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.3

Use the histogram to construct a frequency polygon on the same set of axes.

6.4

Describe the shape of the frequency polygon. Use the shape to predict the relation ship between the median and mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

102

Grade 11 Core Mathematics

6.5 6.5.1

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Find: Median ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.5.2

An estimate of the mean using grouped data. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.5.3

Mean â&#x20AC;&#x201C; median ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

103

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The following table shows the prices , correct to nearest rand, of second-hand VW Golf cars for sale in Car Finder . a magazine listing second-hand cars for sale. Prices in Rands 0 - 19 999 20 000 – 39 999 40 000 – 59 999 60 000 – 79 999 80 000 – 99 999 100 000 – 119 999 120 000 – 139 999 1400 – 159 999

7.1

Frequency 8 19 45 25 23 9 11 1

Find estimates of: 7.1.1 The mean and standard deviation of the prices. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

7.1.2

The median price.

Calculate mean – median.

Grade 11 Core Mathematics

1.2.2

104

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Use the answer to predict whether the distribution is symmetrical, positively skewed or negatively skewed.

7.3.1

Draw a histogram to illustrate the data.

Describe the shape of the histogram. Does the answer confirm the prediction in 7.2.2.

Grade 11 Core Mathematics

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105

Bivariate Data and Scatter Plots: Relationships between two variables can cause problems in investigation and interpretation. 

Scatter plots can be utilized to show the relationship between 2 sets of information.



If the cloud of data produced is gently sloping before rising sharply the relationship may represent an exponential function.



If the cloud has a steep positive slope and then decreases rapidly and becomes steeply negative it could be represented by a quadratic function. For linear relationships:



.If correlation is positive the cloud slopes to the right.



If correlation is negative the cloud slopes to the left.



No correlation points scattered all over.

The following representations numbered A to H are all scatter plots. A.

Grade 11 Core Mathematics

A. B. C. D. E. F. G. H.

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106

Parabola Function (with a < 0 ; steeply positive then steeply negative) Exponential Function. No Relationship or correlation Linear Function (positive correlation) Linear Function (negative correlation) Linear Function (No correlation) Quadratic Function. (with a > 0; Steeply negative going to steeply positive) Linear Function (No correlation)

Exercise 3.3: 1.

Botanists compared the use of growth hormones and the length of growth in plants:

Look at the table below and answer the questions that follow: Amount of growth hormones in ml 0,5 1 2 1,5 2 2,5 3 2,5 3 1,5 1.1

Plant growth in cm 1,5 2 2,5 3,5 4 5 8 7 6 3

Draw a scatter plot of the data.

Grade 11 Core Mathematics

1.2

107

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Why do we represent the amount of growth hormone on the x â&#x20AC;&#x201C; axis?

Discuss the correlation between the variables.

1.4

Suggest whether a linear, quadratic or exponential function would best fit the data? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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108

Grade 11 Core Mathematics

The following marks were recorded for a maths class:

64 88 75 2.1

80 75 67 80

75 62 72 65

74 65 74 68

72 55 50 79

66 73 64 89

53 84 75 72

82 90 80

80 78 80

Do a stem and leaf diagram for the data

2.2

Find the median, mode and mean for the data

2.3

Find the lower and upper quartile ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

2.4

109

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Calculate:

2.4.1 the interquartile range.

2.4.2

the semi-interquartile range.

2.4.3

the range for the class.

2.4.4 Write down the maximum and minimum scores.

Grade 11 Core Mathematics

2.5

110

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Do a box and whisker diagram using the five-number summary (L;Q1;M; Q3;H). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.6

Standard Deviation. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.7

What % of scores lie within 1 standard deviation from the mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.8

What % of scores lie within 2 standard deviations of the mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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111

Grade 11 Core Mathematics

The following marks were recorded for a maths class: 53 68 70 76

3.1

60 75 55 65

45 70 67 74

50 67 64 67

62 50 60 75

75 73 70 54

85 90 82 67

76 80 85 68

69 67 64

Do a stem and leaf diagram for the data

3.2

Find the median, mode and mean for the data ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

3.3

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3.4

112

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Calculate:

3.4.1 the interquartile range.

3.4.2

the semi-interquartile range.

3.4.3

the range for the class.

3.5

Write down the maximum and minimum scores.

Grade 11 Core Mathematics

3.6

113

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3.7

3.8

3.9

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114

Grade 11 Core Mathematics

The following marks were recorded for a maths class: 23 48 30 46

4.1

30 45 25 45

31 50 35 44

52 47 24 53

42 20 30 45

15 43 40 54

45 60 52 35

36 40 75 28

29 37 34

Do a stem and leaf diagram for the data

4.2

4.3

Find the lower and upper quartile

54 38 45

Grade 11 Core Mathematics

4.4

115

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Calculate:

4.4.1 the interquartile range.

4.4.2

the semi-interquartile range.

4.4.3

the range for the class.

4.5

Write down the maximum and minimum scores.

Grade 11 Core Mathematics

4.6

116

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4.7

4.8

4.9

Grade 11 Core Mathematics

117

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Misrepresenting Data: 

Misrepresenting data can happen both intentionally and unintentionally. It is always important to find out the source of the data to be able to critically evaluate its graphs and tables.



In reports, the actual data is not usually given in detail. It is up to us to read the graphs critically and, if we feel the graph is a misrepresentation, to question why.



Changing the information on the axes of the graphs ( or simply leaving any axes labeling out altogether) is a common used technique for distorting information.



The kinds of changes that can be made to information given on sets of axes:



Adjusting the scales on one or both of the axes.



Varying the scale on an axis



Changing the scale around on one of the axes.

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118

Grade 11 Core Mathematics

OPTIONAL TO COMPLETE IN GRADE 11 IF TIME AVAILABLE:

GRADE 12 STATISTICAL DATA.

MODIFYING BOX AND WHISKER DIAGRAMS. Box and Whisker diagrams: The inter-quartile range gives the spread of the middle 50% of data values and is not affected by the extremes. Outliers are values that stand apart from the rest of the values. An outlier is a value that is more than 1,5 times the interquartile range from the nearest quartile. E.G. If Q1 = 46 and Q3 = 60 then the IQR = 14. IQR x 1,5 = 14 x 1,5 = 21 Q1 – 21 =46 – 21 = 25. : Any value less than 25 will be an outlier Q3 + 21 = 60 + 21 = 81 : Any value greater than 81 will be an outlier. Outliers greatly affect the mean but have no more affect on the median or mode than any other value. THE BOX AND WHISKER DIAGRAM CAN BE MODIFIED TO MAKE IT MORE DESCRIPTIVE BY EXCLUDING OUTLIERS. If Q1 = 20 and Q3 = 34 then the IQR = 14. IQR x 1,5 = 14 x 1,5 = 21 Q1 – 21 =20 – 21 = -1. : Any value less than -1 will be an outlier Q3 + 21 = 34+ 21 = 55 : Any value greater than 55 will be an outlier.

Original Box & Whisker

80 20

Modified Box & Whisker 12

55 20

34 30

100

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119

Grade 11 Core Mathematics

STANDARD DEVIATION: From a given table of data: 67; 70; 71; 71; 73; 74; 75; 75; 75; 77; 78; 78; 78; 78; 79; 80; 81; 82; 82; 83; 86; 86; 87; 91

67 70 71 73 74 75 77 78 79 80 81 82 83 86 87 91 x  78,2

1 1 2 1 1 3 1 4 1 1 1 2 1 2 1 1

n  24

F x1 67 70 142 73 74 225 77 312 79 80 81 164 83 172 87 91

xx

( x  x )2

11,2 8,2 7,2 5,2 4,2 3,2 1,2 O,2 -0,8 -1,8 -2,8 -3,8 -4,8 -7,8 -8,8 -12,8

125,44 67,24 51,84 27,04 17,64 10,24 1,44 0,04 0,64 3,24 7,84 14.44 23,04 60,84 77,44 163,84

 (x  x

STD DEV

s

 ( x  x1 ) 2 n 1



 (x  x ) 1

= 652,24

 5,8

s  5.9

Standard Deviation is a Measure of Dispersion about the mean: It measures how far each data item is from the mean and takes into account all data items. If the differences of the scores above the mean are added to the differences below the mean the answer eill be zero. The differences are recorded as positives whether the score is above or below the mean. Variance is defined as OR

 (x  x ) 1

n 1

, when working with a sample of a population.

when working with a population n Variance is called the standard deviation and is considered the best measure of dispersion. The symbol “  ” is used to denote Standard Deviation when referring to a population AND “s” when referring to a sample of a population. A small standard deviation indicates that the data items are clustered around the mean. While a large standard deviation indicates that the items are more spread out.

120

Grade 11 Core Mathematics

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STATISTICAL DATA & THE CALCULATOR Mean & Std Deviation using a calculator The Casio f(x) 82ES and STD Dev Key MODE  2: STAT THEN 1 –VAR To enter Data into table:

key (data list) into each row.

To calculate the MEAN : key SHIFT -1 THEN No 5 : VAR : 2: x To calculate Std Dev: key SHIFT : 1 then 5 : VAR then 3 xn [OR 4 : xn  1] Mean & Standard Deviation on a Frequency Table using a Casio: Key MODE 2 : STAT then 1: 1 – VAR. Set a table up: Key SHIFT SETUP Scroll down to new screen: then key 3:STAT ; 1 : ON. Enter Data into column 1 and frequency into column 2 When complete key [AC] For the MEAN: key Shift 1 then 5:VAR then 2 : x . For Std Dev: key Shift 1 then 5 : VAR 3: xn [OR 4 : xn  1] Example: Speed in Kph 5060708090100110120-

Midpoint of Interval X 55 65 75 85 95 105 115 125

No of cars „f‟ 20 27 25 54 21 15 8 5

mean  82,48 StdDEv  17,5

n  175 175  87,5 2

Median lies in interval 80 – 89 thus Median = 84,5

Total No fX 1100 1755 1875 4590 1995 1575 920 625

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Grade 11 Core Mathematics

SYMMETRIC & SKEWED DATA:

a) Equal spread either side of the median in a box & Whisker diagram portrays a symmetrical spread. b) If data values are spread out more on one side than the other of the median then the data is said to be skewed.

xM 0

SYMMETRICAL

NORMAL DISTRIBURION

Q2 DIAGRAM SKEWED TO LEFT

X X M 0 NEGATIVELY SKEWED DISTRIBUTION

DIAGRAM SKEWED TO RIGHT

 X  X M 0

POSITIVELY SKEWED DAT

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122

Grade 11 Core Mathematics

Using Standard Deviation to reach conclusions: Provided that a sample is reasonably large and the data is not too skewed ( that is , it does not have some very large or very small values), it is possible to make the following approximate statements. 

About 66% of the individual observations will lie within one standard

deviation of the mean. 

For most sets of data, about 95% of the individual observations will lie within

2 standard deviations of the mean. 

Almost all of the data will lie within 3 standard deviations of the mean.

Exercise: 4.1 5. The following marks were recorded for a maths class:

45 75 84

54 53 75 63

46 58 75 92

44 81 78 67

22 60 60 68

28 54 37 68

37 71 56 69

56 71 25 76

45 44 90 98

5.1 Do a stem and leaf diagram for the data.

Grade 11 Core Mathematics

123

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5.3 Find the lower and upper quartile.

5.4 Calculate: 5.4.1

the interquartile range.

5.4.2

the semi-interquartile range.

the range for the class.

______________________________________________________________ ______________________________________________________________ 5.4.4

Write down the maximum and minimum scores.

Grade 11 Core Mathematics

5.4.5

124

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Do a box and whisker diagram using the five-number summary

(L;Q1;M; Q3;H) ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.4.6

Standard Deviation ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.4.7 What % of scores lie within 1 standard deviation from the mean. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.4.8

5.4.9

State whether the data is negatively or positively skewed and give a reason for your decision. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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125

Grade 11 Core Mathematics

6. The following table represents the maths scores for the entire grade 11 maths group at Northwood School. The data is grouped due to the size of group.

Class 0 to 9 10 to 19 20 to 29 30 to 39 40 to 49 50 to 59 60 to 69 70 to 79 80 to 89 90 to 99 100 to 109 Totals 6.1 6.2

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

Mid-points(X) 4.5 9.5 14.5 19.5 24.5 29.5 34.5 39.5 44.5 49.5 54.5

fX 67.5

Complete the last column of the table i.e (fX) Find the modal class

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 6.3

Find the median class

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 6.4

Find the interval where Q1 and Q3 lie.

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 6.5

Calculate the estimated mean. ď&#x192;Ľ fX NB estimated mean = n ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Grade 11 Core Mathematics

6.6

126

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Use the grouped data to display the data on a histogram

Draw the relevant frequency polygon on the histogram.

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127

Grade 11 Core Mathematics

Complete the table and calculate the variance and the standard deviation.

40; 50; 65; 65; 70; 75; 75; 75; 75; 78; 78; 78; 78; 78; 79; 80; 81; 81; 82; 82; 82; 86; 88; 90

40 50 65 70 75 78 79 80 81 82 86 88 90 x  75,5

1 1 2 1 4 5 1 1 2 3 1 1 1

n  24

F x1 40 50 130 70 300 390 79 80 162 246 86 88 90

xx

( x  x )2

35,5 25,5

1260.25 650.25

-12,5 -14,5

156.25 210.25

 (x  x

STD DEV

s s

 ( x  x1 ) 2 n 1

 

 ( x  x1 ) 2 n

 

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128

Grade 11 Core Mathematics

Complete the table

Height (h) in cm 135  h < 140 140  h < 145 145  h < 150 150  h < 155 155  h < 160 160  h < 165 165  h < 170 170  h < 175 175  h < 180 8.1

Mid points

Frequency

137,5

Cumulative Frequency 2

142,5

147,5

Coordinates (140 ; 2)

Calculate the estimated mean.

8.2

Draw a histogram of the data.

Draw a frequency polygon on the histogram.

Grade 11 Core Mathematics

8.4

129

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State the modal group, median height ,upper and lower quartiles for the data.

8.5

Sketch the Ogive Curve for the data.

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130

Grade 11 Core Mathematics

Volume and Surface Area of 3-D Shapes Formulae: Volumes of Rectangular Prisms: V  (area of base)  height

1. 2.

Square Base: Rectangular Base :

Trapezium Base :

Triangular Base :

V  (side of base)2  height (length) V  length  breadth  height (length 1 V  sum of parallel sides   height (length) 2 1 V  (base  height)  length 2

Volume of cylinders: V  (area of base)  height

V   r 2h Volume of a Cone: 1 V   r 2h 3 Volume of Pyramid: 1 V  (area of base)  height 3 Volume of Sphere: V 

4  r3 3

Surface Areas of Shapes: Hint Draw a net diagram of the shape: Net Diagrams of 3 D shapes: Rectangular Prisms:

Volume  side1 side2  side3 Volume = area of base X height l Net: Rectangular Prisms: b

hxl

b h

hxl

hxb

b hxl

hxl Surface area = 2(h x l) + 2( h x b) + 2(l x b)

hxb

131

Net : Square base prism:

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s sx s

sxh

s sx s Surface area = 4(h x s) + 2( s x s)

Net of a Cylinder

2r

Surface Area = area of circles plus area of reactangle Surface Area = r 2  r 2  2rh Surface Area = 2r 2  2rh

Cone: Surface Area = r h 2  r 2

Sh  h 2  r 2

Or Surface area =

1 circumference x slant height 2

NB Slant height =

h2  r 2 Circumference = 2  r

Net Diagram of a cone:

Slant height

Arc length

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Grade 11 Core Mathematics

Formulae for Surface Area of Shapes. PRISMS: Rectangular bases:

S  2lb  4hb Triangular Bases: 1. S  bh  bl  2l b 2  h 2

perpendicular height

 If triangle is isosceles.

Slant height Length

Base

1.2 S  bh  3bl  If equilateral.

perpendicular height

Base Length

Base

Cylinders:

S  2 r 2  2 rh S  2( r 2   rh) Square base pyramid:

1 SurfaceAre a  ( side) 2  4( base  sh) 1 slant height  ( base) 2  (height ) 2 2 2 SurfaceAre a  ( side) 2  2(base  sh) (Pythagoras theorem)

Grade 11 Core Mathematics

133

OTHER 3- D SHAPES Cone:

CurvedSurface   r h 2  r 2 or 1 CurvedSurface  circumference  slant height 2 Sphere: S  4 r 2 Pictures of Different 3-Dimensional Shapes:

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134

Grade 11 Core Mathematics

Surface area and volume of cylinder with section removed: Calculate the surface area of the shape: Calculate the ratio of arc to circle 

Arc Leave  as a symbol Circumference

Area of Arc(sector) = Use the ratio above and multiply  r2 ( area of full circle) Area of flat sections = l  b (length x breadth) they are rectangles. Area of curved surface = Arc  height Total Surface area = 2( area of arc) + 2( area of rectangular flat sections + area of curved surface Find the volume of the shape. Volume of shape = Area of arc x height of cylinder. Surface area and volume of cone with section removed: Calculate the ratio of arc to circle

Arc Leave  as a symbol Circumference

Area of Arc(sector) = Use the ratio above and multiply  r2 ( area of full circle) Calculate the Slant Height as follows ( Pythagoras) Slant height = h 2  r 2 1 Area of curved surface = ( area of curved sector) x slant height 2 1 Area of straight side =  base  height 2 Total area = area of sector( arc) + area ofn curved side + 2( area of straight side) Volume of Cone =

1 (Area of Arc) x height 3

Volumes of prisms & the effect of the factor--k

Grade 11 Core Mathematics

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Exercise 5.1: 1. Calculate the volume and surface area of the following closed prisms:

Prism P Q R S T 52 47 43 39 36 Length (mm) 20 18 17 15 14 Breadth (mm) 85 77 70 64 58 Height (mm) Determine the following ratios correct to 2 decimals. VolumeP VolumeQ 2.1 2.2 VolumeQ VolumeR ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

2.3

VolumeR VolumeS

2.4

VolumeS VolumeT

SurfaceAre aP SurfaceAre aQ 3.2 SurfaceAreaQ SurfaceAre aR ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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3.3

136

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SurfaceAre aR SurfaceAre aS 3.4 SurfaceAre aS SurfaceAre aT ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Are the volumes of the prisms approximately in proportion? Give reasons for your answers. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

5.1 How much smaller in volume is prism T than prism P? Give the scale factor (not the change in volume). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.2 Are the surface areas of the prisms in proportion? Give reasons for pour answers. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 5.3 How much smaller in surface area is prism T than prism P? Give the scale factor (not the change in area). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

Determine the scale factor used :

Grade 11 Core Mathematics

6.1

137

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to reduce the dimensions of the prisms. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

6.2

To enlarge the dimensions of the prisms. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

7. What is â&#x20AC;&#x2022;The Golden Ratioâ&#x20AC;? ? ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 7.1

Determine which ratio of the faces comes closest to this ratio. NB: You must choose a ratio greater than 1. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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

1 , then calculate the 2

volume and surface area of the new prism X. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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8.2 How much smaller in volume and in surface area is this new prism X? Give the scale factor in each case. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 9.1

Using the answers to question 8 , estimate the volume and surface area of prism Y, where each dimension of prism P has been enlarged by a factor of 2. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

9.2

Calculate the volume and surface area of prism Y, and compare your answers to your answers to question 9.1

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Grade 11 Core Mathematics

Examples: h b l Dimensions of the prism is length; breadth and height. i.e. l ; b ; h Volume of prism = l  b  h [ k is a factor of 2] Prism l (cm) b (cm) H (cm)

A B C D E F G H

4 8 4 4 8 8 4 8

3 3 6 3 6 3 6 6

2 2 2 4 2 4 4 4

Volume (cm3)

Vxk

24 48 48 48 96 96 96 192

V Vx2 Vx2 Vx2 Vx4 Vx4 Vx4 Vx8

No of Factor Sides Doubled 0 k 1 k 1 k 1 K2 2 2 K 2 2 k 2 k3 3

It is noticed : When 1 dimension is doubled then the volume is doubled as well When 2 dimensions were doubled then the volume is 4 times the original. When all 3 dimensions are doubled the volume is 8 times the original. This holds for any factor value. i.e. k = 3 the volumes increase accordingly : 1 trebled thus volume trebled 2 trebled thus volume is 9 times original 3 trebled thus volume is 27 times the original. Factors affecting the volumes are: k k2 k3 Volume of prism = l  b  h [ k is a factor of 3] Prism l (cm) b (cm) h (cm) A B C D E F G H

4 4 4 12 12 12 4 12

3 3 9 3 9 3 9 9

2 6 2 2 2 6 6 6

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

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

k k k k2 k2 k2 k3

10.

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Complete the following table:

h b l Dimensions of the prism is length; breadth and height. i.e. l ; b ; h Volume of prism = l  b  h Factor k =______. Prism

l (cm)

b (cm)

h (cm)

Volume (cm3) 72

Vxk V

Factor k

Vx2

Volume of prism = l  b  h Factor k =______. Prism

l (cm)

b (cm)

h (cm)

Volume (cm3) 24

Vxk Factor V Vx3

Vx Vx

216

Vx 9

Vx 216

Vx Vx

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11.

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A cold - drink can measures approximately 65 mm in diameter and 75mm in height. 11.1 Calculate the volume of the can ( in mm2 and cm2). ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ 11.2

The writing on the can says that it contains 200 ml of liquid. How much air space is there in the can? ( 1ml ď&#x20AC;˝ 1cm 3 )

What is the height of the liquid in the can?

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12.1

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Calculate the total surface area of the can (in cm3 ) , assuming that the can is a closed cylinder. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

12.2

If the metal to make the can costs 0,25 cents per square centimeter, calculate the cost of making each can. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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Exercise 5.2: Use the formulae provided at the beginning of this section to do this exercise. Find the volumes and Surface areas of the following solids: 5.2.1

đ??śđ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;&#x2019; đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2022; â&#x2C6;ś đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2018;đ?&#x2018;&#x2013;đ?&#x2018;˘đ?&#x2018; = 4đ?&#x2018;?đ?&#x2018;&#x161; & đ?&#x2018;&#x2022;đ?&#x2018;&#x2019;đ?&#x2018;&#x2013;đ?&#x2018;&#x201D;đ?&#x2018;&#x2022;đ?&#x2018;Ą = 11đ?&#x2018;?đ?&#x2018;&#x161;

5.2.2

đ?&#x2018;&#x2020;đ?&#x2018;?đ?&#x2018;&#x2022;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2022; â&#x2C6;ś đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2018;đ?&#x2018;&#x2013;đ?&#x2018;˘đ?&#x2018; = 8đ?&#x2018;?đ?&#x2018;&#x161;

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5.2.3

144

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𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑝𝑟𝑖𝑠𝑚: 𝑕𝑒𝑖𝑔𝑕𝑡 = 10𝑐𝑚 ; 𝑙𝑒𝑛𝑔𝑡𝑕 = 25𝑐𝑚; 𝑕𝑒𝑖𝑔𝑕𝑡 = 5𝑐𝑚

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Apothem From Wikipedia, the free encyclopedia

Apothem of a hexagon The apothem of a regular polygon is a line segment from the center to the midpoint of one of its sides. Equivalently, it is the line drawn from the center of the polygon that is perpendicular to one of its sides. The word "apothem" can also refer to the length of that line segment. Regular polygons are the only polygons that have apothems. Because of this, all the apothems in a polygon will be congruent and have the same length. For a regular pyramid, which is a pyramid whose base is a regular polygon, the apothem is the slant height of a lateral face; that is, the shortest distance from apex to base on a given face. For a truncated regular pyramid (a regular pyramid with some of its peak removed by a plane parallel to the base), the apothem is the height of a trapezoidal lateral face. [1] For a triangle (necessarily equilateral), the apothem is equivalent to the line segment from the midpoint of a side to any of the triangle's centers, since an equilateral triangle's centers coincide as a consequence of the definition.

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5.2.4

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𝑆𝑞𝑢𝑎𝑟𝑒 𝑏𝑎𝑠𝑒 𝑝𝑦𝑟𝑎𝑚𝑖𝑑: 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑕𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑝𝑦𝑟𝑎𝑚𝑖𝑑 = 71,2 𝑚𝑒𝑡𝑟𝑒𝑠 𝐿𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑏𝑎𝑠𝑒 = 233,5 𝑚𝑒𝑡𝑟𝑒𝑠 𝑇𝑕𝑒 𝒂𝒑𝒐𝒕𝒉𝒆𝒎 𝑖𝑠 𝑡𝑕𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡𝑕𝑒 𝑐𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑡𝑕𝑒 𝑏𝑎𝑠𝑒 𝑙𝑖𝑛𝑒 𝑜𝑓 𝑡𝑕𝑒 𝑝𝑦𝑟𝑎𝑚𝑖𝑑 𝑡𝑜 𝑖𝑡𝑠 𝑐𝑒𝑛𝑡𝑟𝑒 .

Grade 11 Core Mathematics

5.2.5

147

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𝑆𝑕𝑎𝑝𝑒 𝑖𝑠 𝑎 𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑎𝑛𝑑 𝑎 𝑐𝑜𝑛𝑒: 𝑟𝑎𝑑𝑖𝑢𝑠 = 5𝑐𝑚 ; 𝑠𝑙𝑎𝑛𝑡 𝑕𝑒𝑖𝑔𝑕𝑡 = 20𝑐𝑚

89 𝑐𝑚 𝑎𝑛𝑑 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟 𝑕𝑒𝑖𝑔𝑕𝑡 =

Grade 11 Core Mathematics

5.2.6

148

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𝑆𝑕𝑎𝑝𝑒 𝑖𝑠 𝑎𝑐𝑜𝑛𝑒 𝑜𝑛 𝑎 𝑕𝑒𝑚𝑖𝑠𝑚𝑖𝑠𝑝𝑕𝑒𝑟𝑒: 𝑟𝑎𝑑𝑖𝑢𝑠 = 3,5𝑐𝑚 𝑎𝑛𝑑 𝑕𝑒𝑖𝑔𝑕𝑡 𝑜𝑓 𝑐𝑜𝑛𝑒 = 10𝑐𝑚

Grade 11 Core Mathematics

5.2.7

149

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𝑆𝑕𝑎𝑝𝑒 𝑖𝑠 𝑕𝑎𝑙𝑓 𝑎 𝑐𝑦𝑙𝑖𝑛𝑑𝑒𝑟: 𝐿𝑒𝑛𝑔𝑡𝑕 𝑖𝑠 20 𝑐𝑚 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑏𝑎𝑠𝑒 = 6 𝑐𝑚

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5.2.8

Grade 11 Core Mathematics

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8. CIRCLE GEOMETRY

Introduction: Circle Geometry is the study of shapes and angles formed within circles.

Shapes involved: Quadrilaterals Parallogram Rectangle Square Rombus Kite Trapezium

Triangles: Equilateral Isosceles Right Angled Scalene

Parallel lines and angles formed Corresponding Angles Alternate Angles Co-interior Angles

152

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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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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 Given: Circle centre O , đ??´đ?&#x2018;&#x201A;đ??ľ the angle at the centre subtended by AB and đ??´đ??ś đ??ľ the at the circumference. RTP: đ??´đ?&#x2018;&#x201A;đ??ľ = 2đ??´đ??ś đ??ľ Proof: Join CO and produce to D AO = OC (Radii) đ??´ = đ??´đ??ś đ?&#x2018;&#x201A; (Lâ&#x20AC;&#x2DC;s opp = sides) đ??´đ?&#x2018;&#x201A;đ??ˇ = đ??´ + đ??´đ??ś đ?&#x2018;&#x201A; (ext L triangle) ď &#x153; đ??´đ?&#x2018;&#x201A;đ??ľ = 2đ??´đ??ś đ??ľ Similarly đ??ľđ?&#x2018;&#x201A;đ??ˇ = 2đ??ľđ??ś đ?&#x2018;&#x201A; Thus in figure 1 & 2 đ??´đ?&#x2018;&#x201A;đ??ˇ + đ??ľđ?&#x2018;&#x201A;đ??ˇ = 2đ??´đ??ś đ?&#x2018;&#x201A; + 2đ??ľđ??ś đ?&#x2018;&#x201A; = 2(đ??´đ??ś đ?&#x2018;&#x201A; + 2đ??ľđ??ś đ?&#x2018;&#x201A;) = 2đ??´đ??ś đ??ľ Thus in fig3. đ??ľđ?&#x2018;&#x201A;đ??ˇ â&#x2C6;&#x2019; đ??´đ?&#x2018;&#x201A;đ??ˇ = 2đ??ľđ??ś đ?&#x2018;&#x201A; â&#x2C6;&#x2019; 2đ??´đ??ś đ?&#x2018;&#x201A; = 2(2đ??ľđ??ś đ?&#x2018;&#x201A; â&#x2C6;&#x2019; đ??´đ??ś đ?&#x2018;&#x201A;) = 2đ??´đ??ś đ??ľ

154

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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??ś đ??ľ is an angle in the semi-circle. RTP: đ??´đ??ś đ??ľ = 90Â° Proof: đ??´đ?&#x2018;&#x201A;đ??ľ = 2đ??´đ??ś đ??ľ ( L at centre = 2 L at circumference) đ??´đ?&#x2018;&#x201A;đ??ľ = 180Â° ( St line) 2đ??´đ??ś đ??ľ = 180Â° đ??´đ??ś 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??ľ đ??ˇ and đ??´đ??ś đ??ˇ angles in the same segment. RTP: đ??ľ = đ??ś Proof: Join AO and OD đ??´đ?&#x2018;&#x201A;đ??ˇ = 2đ??ľ ( L at centre = 2 L at circle) đ??´đ?&#x2018;&#x201A;đ??ˇ = 2đ??ś ( L at centre = 2 L at circle) 2đ??ľ = 2đ??ś đ??ľ=đ??ś

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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 ď&#x192;?,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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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. RTP: đ??´ + đ??ś = 180Â° & đ??ľ + đ??ˇ = 180Â° Proof: Join BO and DO đ??ľđ?&#x2018;&#x201A;đ??ˇ = 2đ??ś ( 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;đ??ˇ = 2đ??´ ( L at centre = 2 L at circle) đ??ľđ?&#x2018;&#x201A;đ??ˇ + đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;&#x201C;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;Ľđ??ľđ?&#x2018;&#x201A;đ??ˇ = 360Â° ( Lâ&#x20AC;&#x2DC;s at a point) 2đ??ś + 2đ??´ =360Â° đ??ś + đ??´ = 180Â° Similarly it can be proved that đ??ľ + đ??ˇ = 180Â°

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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??ś đ??¸ = đ??´ Proof: đ??ˇđ??ś đ??¸ + đ??ˇđ??ś đ??ľ = 180Â° (Lâ&#x20AC;&#x2DC;s on a st line) đ??´ + đ??ˇđ??ś đ??ľ = 180Â° (opp Lâ&#x20AC;&#x2DC;s of a cyclic quad) đ??ˇđ??ś đ??¸ = đ??´

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)

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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 RTP: đ??¸đ??ś đ??ľ = đ??ˇ & đ??¸đ??ś đ??´ = đ??š Proof: Draw diameter COG and join DG. (i) Let đ??ş = đ?&#x2018;Ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ś2 = đ?&#x2018;Ś đ??şđ??¸ 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??ś đ??ľ = 90Â° ( Rad ď &#x17E; Tan) đ?&#x2018;Ľ = đ??¸đ??ś đ??ľ BUT đ?&#x2018;Ľ = đ??ˇ ( Lâ&#x20AC;&#x2DC;s in same segt) đ??¸đ??ś đ??ľ = đ??ˇ (i) đ??ˇ + đ??š = 180Â° ( opp Lâ&#x20AC;&#x2DC;s cyclic quad) đ??¸đ??ś đ??ľ + đ??¸đ??ś đ??´ = 180Â° (Lâ&#x20AC;&#x2DC;s on st line) But đ??¸đ??ś đ??ľ = đ??ˇ (proved in i) đ??¸đ??ś đ??´ = đ??š

159

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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

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

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3. E

C A

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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 AMC  BMC

_M _C

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O is the centre of the circles 5 to 10. Find the sizes of x and y in each case. 5.

6. A

A x O

110

70 C

x B

x A B

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O 70 x 10

Prove that Bˆ1  Aˆ  90 O

1 B

1 C

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163

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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 ______________________________________________________________

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164

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13. F x A 2

1 20

4 B

85 2 D

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

14. 100

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165

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15. O is the centre of circle ABCD D

A z

O x

B E

120 D

y C

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16. DE is a tangent to circle ABC. A z

O y

x 62ď&#x201A;° D

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17.

EF is tangent to circle ABCD and BC = CD . Find 5 angles equal to CË&#x2020;1 , (giving reasons.)

2 1

4 E

3 2 C

1 F

168

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18.

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D A E

Chord AB is parallel to chord CD. Cˆ1  Cˆ 2

Prove that BC = DE

1 2 B

169

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19.

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D 1

Chord AD equals chord AC and Sˆ1  Sˆ3 Prove: 19.1 Eˆ  Bˆ 19.2 AE  AB

1 1

2 A

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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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.

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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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 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

______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________

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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

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23.

AB is a diameter. ADP and BCP are straight lines. PQTR is a straight line.

1 2

D 1 2

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4 1 T 3 2

1 Q 2 6 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

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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

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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

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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. ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________ ______________________________________________________________