GOMATHGR10W1 by admin Bookbuzz

Compiled by Chez Nell

Grade 10 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 10 Core Mathematics

GO MATH WORKBOOKS

GRADE 10 CORE MATHEMATICS. PAPER ONE.

Topic:

Pages:

Number System

(4 – 6)

Mathematical Notation

(7 - 10)

Algebraic Products

(11 – 13)

Factors

(14 – 26)

Algebraic Fractions

(27 – 32)

Solution of Equations

( 33 – 68)

Basic Logarithm Simplification

( 68 – 71)

Exponents

( 71 – 86)

Number Patterns

(87 – 98)

10. Straight Line Graphs

(99 – 110)

11. Parabola Graphs

(111 – 126)

12. Hyperbola Graphs

(127 – 131)

13. Exponential Graphs

(132 – 153)

14. Financial Mathematics

(154 – 167)

15. Probability Theory

(185- 206)

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Rational Numbers and Irrational Numbers

A common fraction can be written in the form: a ; a, b  z ; b  0 b A mixed number is made up of two parts a whole number followed by a proper or decimal fraction. A terminating decimal is a decimal fraction that ends after a definite number of digits has been given, e.g. 1,23 To convert a terminating decimal into a common fraction change its denominator into 10, 100 etc. according to the number of decimal digits. Type equation here. 123 E.g.: 1.23  100 A recurring decimal is a decimal fraction or part thereof which goes on repeating itself without end. 

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

 0,34 5  0,34555555 

 0,3 4 5  0,345454545 



 0, 3 4 5  0,345345345 To convert a recurring decimal into a common fraction Method 1: Multiply it by appropriate powers of 10, and then use subtraction to eliminate the repeating part, e.g.

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

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

1 2 3 4 5 6 7 8 9 ; ; ; ; ; ; ; ; 2 2 2 2 2 2 2 2 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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

0,45

b) 1,25

c) 6,125

d) 0,824

0,50

f) 2,5

g) 21,25

h) 0,05

3. Use dots to indicate the recurring digits in the following: e. g. 0,4444 = 0,4 a) 0,5555…. b) 0,454 454…. c) 0,714 1414…. d) 123,12333…

e) 123,12323… f) 123, 1231313 …

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

0,6

e) 2,34 5

b) 0,7 8

c) 0,14

d) 45,345

f) 23,346 5

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2: Mathematical Notations: 1.

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

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

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

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

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

Grade 10 Core Mathematics

1.4

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8  -1

 3

⃘ -2

 5

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

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−4 to 2, including -4 and

2.4 not 2.

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

{ x : 3  x  2; x  }

     -4 -3 -2 -1 0 1

4. Number line graphs:



 means “ including.

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

 2

o 9

5. Set Builder notation:

 2

o 9

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

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1.2

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{ЁЭСе: тИТ2 тЙд ЁЭСе тЙд 2; ЁЭСе тИИ тДЬ}

{ЁЭСе: ЁЭСе тЙе 6; ЁЭСе тИИ тДЬ}

1.1

{ЁЭСе: ЁЭСе < тИТ6; ЁЭСе тИИ тДЬ}

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

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

2( x  3)( x  6)  2( x 2  3x  18)  2 x 2  6 x  36 3. If there is a co-efficient with a binomial squared multiply LASTLY be the co-efficient. 2( x  3) 2  2( x  6 x  9) 2

3.1

Multiply by 2 lastly

 2 x 2  12 x  18

Multiply by -2 lastly

 2( x  3) 2  2( x 2  6 x  9)

3.2

 2 x 2  12 x  18

N.B Signs change!

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

4.1 If the trinomial part is formed by the square of each term and the product of the two terms in the binomial then the answer is simply the sum or difference of cubes. ( x  2)( x 2  2 x  4) Trinomial is formed  x3  2x 2  4x  2x 2  4x  8  x 8 3

from the squares of each term and product of two terms

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

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

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Exercise 3.1: 1.

Simplify the following :

1.1

2( x  3)( x  4)

1.2

–3(3 – 5x) + 3(3-5x)

1.3

–2a2(a – b)2

1.4

(x – 3)2 –2(x– 1)2 + 2x

1.5

(a – 2b)( a2 + 2ab + 4b2)

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1.6

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( x  2) 2  3( x  2)( x  2)

1.7

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

1.8

4 x  3 y 2 xy  y 2 

1.9

32 x  y   2x  y x  5 y  2

p



 4 p 2q  4 p q  16



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Factors :

4.1. 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 ( 4 terms) 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.

4.2. Difference of two squares: Method: Write down the product ( two brackets) of the summand difference of the roots of each term. NB look at the bracket with the difference of the 2 terms and see if it can 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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Exercise 4.1: 1.

x2  4

x2  9

x 2  16

2 x 2  50

4x 2  9 y 2

x 4  16

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( x  y) 2  ( x  y) 2

4( x  y) 2  9( x  y) 2

Open two brackets with the correct signs.

Factorise: 1. a 3  b3 =

(

)(

)

(a  b)(a  ab  b ) 2

In the binomial bracket enter the cube roots of the terms

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

a 3  b3 = (

)(

)

= (a  b)(a 2  ab  b 2 )

8 x 3  27 y 6  ( 

)(





)

 (2 x  3 y )(4 x  6 xy  9 y ) 2

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Exercise 4.2: 1.

x3  y3

x3  y3

8 x 3  27 y 3

64 x 3  125 y 3

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4.4. Trinomials: The method used here is called “trial & error” In products of binomials we use the FOIL method to expand . The middle term of the expression formed is a combination of the products of the inner terms and the outer terms. i.e. the „OI‟ of FOIL. The reverse procedure is used to 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

Exercise 4.3: 1.

x 2  2x  3

x 2  6x  5

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x 2  6x  8

x 2  2x  8

x 2  7 x  12

x 2  8x  12

2 x 2  5x  3

2 x 2  5x  12

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3x 2  7 x  2

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

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

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

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

1.2 p ( x + y) - q ( y + x ) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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1.3 (m  n)  ( pn  pm) ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 1.4 a  b  ax  bx ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.5

a2  b2  a  b

4x 2  2x  9 y 2  3 y

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1.7 4( x  y) 2  9( x  y) 2 ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.8

(2 x  y ) 2  ( x  2 y ) 2

5x 2  10 xy  5 y 2

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1.3

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2x2 + 6x – 8x3

2x2 – 32

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1.8

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(m  n)  ( pn  pm)

a  b  ax  bx

a2  b2  a  b

1.11

(2 x  y ) 2  ( x  2 y ) 2

4( x  y) 2  9( x  y) 2

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1.13

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4x 2  2x  9 y 2  3 y

1.14

x2  x  6

x 2  7 x  12

2 x 2  24 x  70

9 x 2  42 x  45

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1.18

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x3  y3

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

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5. 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

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

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

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

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

5.1

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.

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.

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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)

LCD

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

x 2  3x  2 ( x  3) 3

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) 2

Final Answer.



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



3x 2  4 x  5 ( x  1) 2 ( x  1) 2

Denominators fully factorised.

LCD = product of the highest power of each type of factor Expanded form of previous line Final answer.

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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 +

Exercise 5.1: 1

6 x. 12 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

4x 2 y 3 8x 3 y __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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2x  4 4 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

xy  y y

8x 2  4 x 4x

x2 1 ( x  1) 2 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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x 2  x  12 x 2  7 x  12

a b ba __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

a2 a a2 2

10.

x 2  x  12 4 x

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11.

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ab  a 2 b 2  ab X b2  a2 a2

12.

x x2  2 x  y y  x2 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

14.

7x 3x  2 x  2 y 5 y  5x

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6. EQUATIONS: 6.1

SOLUTION OF LINEAR EQUATIONS:

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

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

2x  6 x3

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

5 x  10  2 x  2 3x  12 x  4

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

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

Exercise 6.1: 2 x  14  5  x ____________________________________________________________________

x  5  2x  3 ____________________________________________________________________

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

3  2x  6x  1 ____________________________________________________________________

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

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

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

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6.2

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Linear Equations with Fractions:

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

x2 x2 3   LCD= 12 3 4 2 ( x  2) ( x  2) 3   3 4 2 4( x  2)  3( x  2)  18 4 x  8  3 x  6  18 x  20

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

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

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

Exercise 6.2: 1.

y y   1. 2 4

x x  3 5 10

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x x 1   4 3 2

x 1 x 3    2 4 4 2

3x 2 x 1  0 5 5 10

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2x  7 x  5  0 6 3

x 1 x 1  40 4 5

7x  2 9x  2  2 3 5

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3x  4 2x  3  1 2 4

10.

3 x  5 2( x  5)  2 3

11.

x 1 x  2 x 1 x  2    3 4 2 3

Grade 10 Core Mathematics

12.

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

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

Exercise 6.3:

4 1  0 3 3x ___________________________________________________________________ 1.

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

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

2 3  4 x 2x ____________________________________________________________________ 4.

3

1 1 2 x   2 3x 6x

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

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

6.3 Quadratic Equations: 6.3.1 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

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Exercise 6.4: Solve the following equations:

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

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

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

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

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

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

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

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

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

11. x 2  2 x  3  12 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________

12. x( x  16)  3(24  5x) __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ _________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

Grade 10 Core Mathematics

6.3.2

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SOLVING QUADRATIC EQUATIONS USING A FORMULA: Extension Work (Optional)

The formula is derived by completing the square with the general quadratic equation: ax 2  bx  c  0 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. Examples: Solving quadratic equations using the general quadratic formula. Solve: 1. x 2  x  12  0

 b  b 2  4ac x 2a

2x2  7x  6  0

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

x

 b  b 2  4ac 2a

x

7  49  4( 2)(6) 2( 2)

x

7 1 4 8 6 x  or 4 4 3 x  2or 2 x

Exercise 6.5:

 b  b 2  4ac Solve using x  ; 2a (Answers rounded to 2 decimal places where necessary) NB: First expand if necessary and equate to zero before using the formula.

Grade 10 Core Mathematics

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x 2  4x  3  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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

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

Grade 10 Core Mathematics

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x 2  6x  4  0 __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________ __________________________________________________________________

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

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

Grade 10 Core Mathematics

6.4 A.

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Simultaneous Equations: 2 linear equations:

Method: 1. 2. 3. 4.

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

Example: Solve for x and y simultaneously if x  2 y  5 and x  y  1 Change x  y   1 Number (1) / x  2 y  5 (2) the the x  y 1 equation equations to the x 1&2 formSubstitute 1 into 2 y 1 2y  5 Change one of the Substitute the value for equations to the x – 3y  6 x in the changed form or y - form y2 equation for y in the other equation x = 1 {from x  2  1} Substitute the value from above into the changed equation and solve for the second variable Exercise 6.6: Solve the following equations simultaneously.

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

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

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

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

Grade 10 Core Mathematics

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

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

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

Grade 10 Core Mathematics

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

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

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

Grade 10 Core Mathematics

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

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

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

Grade 10 Core Mathematics

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

Grade 10 Core Mathematics

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

x x  11 and 3 y   14 4 5

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

3y 

x  14 -----(2) 5

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

20 4

y  16

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

2 y  18 3

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

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

y9 x  12

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Grade 10 Core Mathematics

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Exercise continued:

x y x y and  1  4 2 5 3 ____________________________________________________________________ 15.

x8 x y x y  2 y  8 and  2 3 2 3 ____________________________________________________________________ 16.

Grade 10 Core Mathematics

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1 1 1 1  9   1 and x y x y ____________________________________________________________________

17.

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

Grade 10 Core Mathematics

6.5

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

x y  4

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

x  y  6 -------(2)

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

Exercise 6.7: 1.

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

Grade 10 Core Mathematics

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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.

Grade 10 Core Mathematics

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The length of a rectangle is twice the breadth, while the perimeter is 6m. Find the length and breadth of the rectangle. { Hint: P ď&#x20AC;˝ 2(l ď&#x20AC;Ť b) .}

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

Grade 10 Core Mathematics

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

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

Grade 10 Core Mathematics

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Tumi is twice as old as John. Two years ago she was three times as old as John was then. What are their present ages?

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

Grade 10 Core Mathematics

6.6

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Linear Inequalities :

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

Solve the following inequality and illustrate the answer on a number line. 2x  4  6 2x  4  6

2 x  10

Solution:

 5 2.

x5

Solve the following inequality and represent the answer on a number line:  4  2x  2  8  4  2x  2  8

 2  2 x  10

 -1

1 x  5

 5

Exercise 6.8: Solve the following inequalities and illustrate the answers on a number line: 2x  6  8 1. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3x  6  x  14

Grade 10 Core Mathematics

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2 x  7  5x  14

3x  7  3x  14 2

 6  2 x  14

Grade 10 Core Mathematics

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 5  2 x  3  15

Exercise 6.9: 1. Find the possible solutions of the following linear inequalities and illustrate the answers on a number line . 1.1

3x  2  x  14

5x  3  3x  15

Grade 10 Core Mathematics

1.3

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1  2x  x  2

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

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

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

Grade 10 Core Mathematics

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

3( x  2) 7( x  3)  3 2 4 ___________________________________________________________________ 1.8

x2 7 3 x  3 8 2 ___________________________________________________________________ 1.9

Grade 10 Core Mathematics

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4  x 2x  1   x2 2 3 ___________________________________________________________________ 1.10

3x  1 3x  3 19   4 8 8 ___________________________________________________________________ 1.11

5  3x  3( x  5) 2 ___________________________________________________________________ 1.12

x

Grade 10 Core Mathematics

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x  2 1  3x  2 5 3 ___________________________________________________________________ 1.13

3

x3 4 2 ___________________________________________________________________ 1.16

3

Grade 10 Core Mathematics

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2x  3  4 3 ___________________________________________________________________ 1.17

2

2  2x 3 4 ___________________________________________________________________ 1.18

2

LOGARITHMS:

Calculations using basic logarithms : Calculator work: NB a logarithm is an exponent: 32  9 In logarithm form this is written as log 3 9  2 and is read : the logarithm of 9 to the base 3 is 2 ( i.e. the exponent of the base 3 is 2 to give the number 9)

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

log 5 log 2

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

Grade 10 Core Mathematics

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Exercise 7.1: Solve for x giving answers to 2 decimal digits. log 1000 . log 100 ___________________________________________________________________

x

Grade 10 Core Mathematics

x

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log 15,5 log 3

x

2 log 25 3 log 15

x

5 log 50 2 log 8  2 log 25 6 log 6

Grade 10 Core Mathematics

x

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10 log 450 log 500  12 log 60 6 log 65

Exponents:

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

Law: a m .a n  a m n

When multiplying like bases you must add the exponents. Do not multiply the bases except in the following case: 23.33  8.27  216. e.g. BUT 23.33  63  216 am = a mn n a When dividing the bases subtract the exponents: the following exception.

Law:

63  6      33  27 3 2 2

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

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

N.B. 1. Never multiply the bases 2. Never divide the bases 3. Never multiply a base by an exponent. 4. If bases are separated by plus or minus signs you MUST FACTORISE before simplifying.

Grade 10 Core Mathematics

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Prime Base Factorising 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 factorised using prima bases then the normal laws apply. Example 1:

ď&#x20AC;¨ ď&#x20AC;Š

4 3

4 3 3

Simplify 125 ď&#x20AC;˝ 5 (Now use law 3 and raise a power to a power by multiplying the exponents:)

125 ď&#x20AC;˝ ď&#x20AC;¨5 4 3

ď&#x20AC;Š

4 3 3

ď&#x20AC;˝5

3 4 ď&#x201A;´ 1 3

ď&#x20AC;˝ 5 4 ď&#x20AC;˝ 625

Example 2: Simplify without the use of a calculator. 5 a ď&#x20AC;Ť 3.5 2 a ď&#x20AC;1 Use law 1 25 a ď&#x20AC;1 in the numerator 5 a ď&#x20AC;Ť 3.5 2 a ď&#x20AC;1 53a ď&#x20AC;Ť 2 aď&#x20AC;Ť 4 ď&#x20AC;˝ 2aď&#x20AC;2 ď&#x20AC;˝ 5 Answer: (5 2 ) a ď&#x20AC;1 5

Prime base factorize 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. Exercise 8.1: 1.

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

___________________________________________________________________ ___________________________________________________________________ 2.

đ?&#x2018;?2 Ă&#x2014; đ?&#x2018;?4

___________________________________________________________________ ___________________________________________________________________ 3.

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

___________________________________________________________________ ___________________________________________________________________

Grade 10 Core Mathematics

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

____________________________________________________________________ ____________________________________________________________________ 5.

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

____________________________________________________________________ ____________________________________________________________________ 6.

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

____________________________________________________________________ ____________________________________________________________________ 7.

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

____________________________________________________________________ ____________________________________________________________________ 8.

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

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

2 xď&#x20AC;Ť3 2 x 2 x ď&#x20AC;1

Grade 10 Core Mathematics

10.

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2 x 1 2 x  3 2 x  2 .2 x

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

11.

12.

Grade 10 Core Mathematics

14.

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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

16.

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

Grade 10 Core Mathematics

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If the bases are separated with plus or minus signs one must FACTORISE FIRST.. It is easier to split the bases as shown because it is easier to see the HCF. Factorise and simplify. NB the base with the variable exponent should always cancel leaving pure numerical values. Exercise 8.2:

2 x3  2 x 2 x 1

2 x 1  2 x  3 2 x2  2 x

Grade 10 Core Mathematics

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3 x  3 x2 3 x 1  3 x

5 2 x  4.5  x 5  x  2.5  x 1

3 n .3 4  6.3 n .31 7.3 n .3 2

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2 n .2 5  3.2 n .2 2 5.2 n 2 3

Variable in the exponent: Method: 1.1 Equate the bases using prime base factorizing. 1.2 If the bases are now equal then the exponents are also equal, thus simply equate the exponents and simplify further. 2 2 x  16

Example 1.

22x  24 2x  4

Rename 16 to the base 2 and equate the bases.

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

First divide by the coefficient „2‟

3 x ( x  2 )  33

Example 2:

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

Rename 27 to the base 3 and equate the bases.

Equate the exponents and solve for x.

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

x  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:

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

2 x (1  3)  16

1.1

2x  4 2 x  22 x2

Use 2 x as the HCF

Equate the bases thus and then the exponents.

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

Variables separated by + or – signs BUT the values of the coefficients of the variables exponents are not equal. i.e. one is double the other. This involves a trinomial and needs to be factorised accordingly. 22x  2 x  8  0 The exponent of the 1st 2.1 (2 x  4)(2 x  2)  0 term is double that of 2x  4  0 the 2nd . i.e The 2x  4 2x  2  0 expression is a trinomial. 2 x  2 2 or 2 x  2 The variable base is 2 x x2 x 1

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

2.3 x  6  0 3x  3

3x  9

3 x  32 x2

x 1

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

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

or 3 x  3 2 x2

Exercise 8.3: 1. Solve for x without the use of a calculator:

1.1

2x = 8

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1.2

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3 x  81

1.3

x3 = 27

1.4

4

= 80

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1.5

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2 x  16 x 1

1.6

3.2x = 48

1.7

5.42 x  40

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1.1

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2 x яАл 2 xяАл3 яА╜ 18

2ЁЭСе+1 тИТ 2ЁЭСе = 4

1.3

2ЁЭСе + 2ЁЭСетИТ2 = 5

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1.4

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22đ?&#x2018;Ľâ&#x2C6;&#x2019;1 â&#x2C6;&#x2019; 22đ?&#x2018;Ľ = 2

1.5

32 x ď&#x20AC; 5.3 x ď&#x20AC;Ť 6 ď&#x20AC;˝ 0

1.6

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

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1.7

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

1.8

9. 32đ?&#x2018;Ľ â&#x2C6;&#x2019; 10. 3đ?&#x2018;Ľ + 1 = 0

1.9

2 3

1 3

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

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1.10 đ?&#x2018;Ľ 2 + 5đ?&#x2018;Ľ 4 + 6 = 0 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

1.11 đ?&#x2018;Ľ 5 + 4đ?&#x2018;Ľ 5 â&#x2C6;&#x2019; 12 = 0 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Number Patterns:

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

Given a sequence as follows: 4;7;10.....

Tn  an  c

T1  4

d1  a  3

T2  4  3  7

T3  10  3(3)  c c 1

T3  4  3  3  10

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

The 10th term of the sequence is:

T10  3(10)  1 T10  31

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

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

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

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

The General term is Tn  5n  4

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In the sequence below , Find the common difference and the general (nth) term. 4; 9; 14 ; 19…… The common difference is d1  T2  T1  9  4  5 Next 3 terms are 24 ; 29 ; 34. Tn  an  c Tn  an  c T1  4 T3  5(3)  c

d1  a  5

T2  4  5  9 T3  4  5  5  14

14  15  c c  1

Tn  5n  1

Exercise 9.1: 1. Mina invests R8000,00 in a plan where the growth is as follows: 0 1 2 3 4 5 8000 8440 8880 9320

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

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

Use the rule to find the value after 23 years.

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

A class is given the sequence 3; 5; 7;….. to continue for 4 more terms. 2.1 Explain in a sentence how you would complete it.

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

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Find the 20th term using the formula in 2.2

Given the sequence 7; 10; 13;â&#x20AC;Ś.. 3.1 Find a general term or rule for this sequence.

Find the 50th term

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

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

Tn = 3n – 5

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Tn = -2n – 5

Tn = 4n + 2

Tn = 2 – 3n

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

3; 9; 15; 21;…..

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6.2

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2; 9; 16 ; 23; …..

6.3

2; -3; -8; -13;….

3; -1; -5; -9;…

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20; 27; 34; 41;…….

6.5



Find the rule for the nth term Tn



Find the value of the 50th term.



Graph the relationship.

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2. 



 



 



 



     

 

 





       ___________________________________________________________________ 



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4. From the diagram below: 4.1 4.2 4.3

fig1

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

Fig 2

Fig 3

Fig 4

From the diagram below: 5.1 5.2 5.3

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

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From the diadram below: 6.1 6.2 6.3

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

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STRAIGHT LINE GRAPHS:

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

y - intercept

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

X Y

-2 -1

0 3

2 5

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

fx = 2x+3

-5

-2

Exercise 10.1: Change the following equations to the y-form. 1.1

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1.3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2.

Sketch the following linear functions using the table method:

y  x3 2.1 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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y  2x  4 22. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2 y  4x  8 2.3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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3 y  6x  3 2.4 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

2y  x  6 2.5 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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4 y  8x  8  0 2.6 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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

2(0)  3x  6 2 y  3(0)  6 Let x = 0  2 y  6 and Let y = 0 3x  6 x y 2 y3

2y + 3x = 6

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Exercise 10.2: Use the” dual intercept method” to sketch the following:

y  x3 1. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

y  2x  4 2. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2 y  4x  8 3 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3 y  6x  3 4. ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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

4 y  8x  8  0

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Gradients of straight lines: 1. 2.

In the y- form of the equation simply use the m- value( co-efficient of x) as this represents the gradient . Given at least 2 points that lie on a given line, then calculate the gradient y  y1 y y1  y 2  or 2 using: m  x x1  x2 x2  x1 i.e the difference between the y – values divided by the difference between the x – values.

Parallel lines have the same gradients i.e. their m – values are the same i.e y  2 x  6 . Is parallel to y  2 x  10

Perpendicular lines have inverse gradients. i.e. m1  m2  1

1 y  2 x  c is perpendicular to y   x  c 2

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

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

Graphs of Non- Functions:

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

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

(-2;4)

Domain: x  R Range: y  R

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

(4 ; 6)

Domain: { x  4 ; x  R }

{Domain: x  R }

Range: { y  6 ; y  R }

Range: { y  8 ; y  R }

-6

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

-6

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111

Parabola Graphs:

General Equation: y  ax 2  c OR y  ax 2  q Two methods can be utilized to sketch a parabola graph: 1. Table method. 2. A „5 point‟ method. Method: Table: A. Sketch the graph of y  x 2  4 X Y

-3 5

-2 0

-1 -3

0 -4

1 -3

2 0

3 5

y 4

f x  = x 2-4

-10

-5

-2

-1

-2

-3 -4

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Exercise 11.1: Sketch the following graphs using a table method: Use x –values of -3 to 3 for each table.( as in the example above) 1. y  x2 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2. y  x2  1 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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3. y  x2  2 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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4. y  x2 1 ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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y  x2  2

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

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Exercise 11.2: A. Sketch the following graphs on the same set of axes: 1. y  x2 2. y  x2 1 3. y  x2  1 What deduction can be made? ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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

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C. Sketch the graphs of the following equations: 1. y  x2 2. y  x2 What deduction can be made? ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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

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

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122

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

x –intercepts at y = 0 :

y- intercept at x = 0

Turning point: for the same.

Get two further points using substitution:

y  ax 2  c the y- intercept and the turning point are

Example: A. Sketch the graph of y  x 2  4

Method: 2

x – intercepts at y = 0: x2  4  0

qx  = x 2-4

-5

( x  2)( x  2)  0 x = 2 or x = -2 2.

y – intercept at x = 0: y in = -4

Turning point: (0; -4)

Substitution:

-2

-4

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

B. Sketch the graph of y  2 x 2  18

x – intercepts at y = 0: 2 x 2  18  0 2( x 2  9)  0 2( x  3)( x  3)  0 x = 3 or x = -3

-4

-2

-4

-6

-8

-10

y – intercept at x = 0. y-int = -18

-12

-14

-16

-18

Turning point : ( 0; -18) -20

-2

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

qx  = 2x 2-18

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123

Exercise 10.3: Use a point by point method to sketch: y  x2  9

-14

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

y  x 2  9

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124

-12

-10

-8

-6

-4

-2

-1

-2

-3

-4

y  2x  8 1.3 ___________________________________________________________________ 2

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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125

y  3x 2  12

1.4

-12

-10

-8

-6

-4

-2

2 -1

-2

-3

-4

-5

-6

-7

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Finding Equations of Parabolas:

Given the turning point and any other point Method: Use the general form of the equation : y  ax 2  q Example :

-2

2 -8

y  ax 2  q

y  ax 2  8

sub Minimum value of - 8 for q

0  a (2)  8

sub point (2;0) for x and y

0  4a  8

solve for a and sub back into equation.

4a  8 a2 y  2x 2  8 B.

Given the two x – intercepts and any other point. Method: Use the general form of the equation: y  a( x  x1 )( x  x2 ) NB x1 and x 2 refer to the roots or x – intercepts Example (1;3) -2

y  a( x  x1 )( x  x2 ) y  a( x  2)( x  2) sub the x – intercepts changing their signs when you do so. 3  a(1  2)(1  2) sub the other point in for the x and y values 3  a(1)(3) 3  3a solve for a and sub back into equation a  1 y  1( x  2)( x  2) y  x 2  4

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12.

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127

Hyperbola Graphs:

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

y

1. X Y

6 x -3 -2

-2 -3

6 x

-1 -6

1 6

2 3

3 2

f x  =

6 x

-10

-5

-2

-4

-6

-8

  

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

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128

Exercise 12.1: Complete the tables and use the values to sketch the graphs of: 1.

y

X Y

-3 2

6 x -2 3

-1

3 -2

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

y 2. X Y

8 x

-8 -1

-4

-2

-1

8 1

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Grade 10 Core Mathematics

y

X Y

-8 1

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8 x -4

-2

-1

8 -1

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

y

X Y

-4 -1

4 x -2

-1

1 4

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Grade 10 Core Mathematics

y

X Y

-4

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4 x -2 2

-1

4 -1

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Shifts with hyperbola graphs:

k q x The q- value gives the vertical shift of the graph. It also gives the new horizontal asymptote. y

Example:

y   

4 2 x Asymptotes are : x  0 and y  2 Domain: x  (; ) x  0 . y  (; ) y  2 . Range:

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131

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Exercise 12.2: Write down the : Asymptotes, Domain & Range of each of the following:

8 3 x ___________________________________________________________________ 1.

y

10 6 x ___________________________________________________________________ 2.

y

9 6 x ___________________________________________________________________ 3. y  

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13.

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132

Exponential Graphs:

General formula :

a>0,a1,x  y>0

y = ax

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

y = 2x [Where „a‟ is a whole number]

X Y

-2

-1

1 4

1 2

(ii)

y = ( 12 )x. [Where „a‟ is a fraction]

X Y

-2 4

-1 2

0 1

1 2

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

2 4

1 2

1 4

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

y = 0,5x

y = 2x y

1 x

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Exercise 13.1: Complete the tables and use the values to sketch the graphs of: y  2x -2 1 4

1. X Y

-1

0 1

2 4

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

y  2x  1 -2 -1 5 4

2. X Y

0 2

2 5

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Grade 10 Core Mathematics

y  2x  1 -2 -1 3  4

3. X Y

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134 0 0

2 3

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

y  2.2 x -2 -1 1 2

4. X Y

0 2

2 8

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Grade 10 Core Mathematics

5. X Y

y  2.2 x -2 -1 1  2

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135 0 -2

2 8

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Functional Notation Graph Interpretation: Exercise 13.2: 1. 1.1

If f ( x)  4 x  3 find: f (4)

f (7)

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136

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f (a  b) 1.3 ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 2. If g ( x)  3x 2  x  6 . Find: g (2) 2.1 ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

g (3) 2.2 ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

g (a  b) 2.3 ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 3. Sketch the graph of 2 y  4 x  8 using the dual intercept method. ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

Grade 10 Core Mathematics

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137 6

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Sketch the graphs of y  x  3 and 2 y  3x  6 on the same system of axes and:

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

Grade 10 Core Mathematics

4.1

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138

write down the co-ordinates of the point of intersection. _______________________________________________________________

___________________________________________________________________ ____________________________________________________________________

8 8 and y  on the same system of axes. x x ____________________________________________________________________ Sketch the graph of y 

4.2

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.2.1 Write down the domain and range for the two graphs above. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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4.2.2

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139

Write down the asymptotes of the two graphs above.

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.3.1

Write down the domain and range of the graphs above:

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140

Sketch the graphs of y  2 x and y  2 x  2 on the same system of axes. ____________________________________________________________________ 4.4

-10

-8

-6

-4

-2

-1

-2

-3

-4

-5

4.4.1 Write down the asymptotes of the two graphs ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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141

Write down the gradients; x and y intercepts of the following graphs: 5.1

5.2 8 6 4

-2

5.4

y 8

-3

-2 -9

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142

6. Write down the domain and range; the mapping and whether the graph represents a function or a no-function for each of the following: 6.1

6.2

(2 ; -9)

6.3

6.4

-12

-4

Grade 10 Core Mathematics

6.5

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143

6.6

-7

6.7 O (3; 9)

6.8

• 0

•(-4 ; -4)

-4

• 4

-9

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144

7. Find the equations of the following graphs: 7.1

7.2 y (4 ; 2)

-4

7.3 (0;1 0

Grade 10 Core Mathematics

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145

7.4 y (0;9)

(2 ; 5)

0 -2 (2 ; -4)

Grade 10 Core Mathematics

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146

8. đ?&#x2018;&#x201C; đ?&#x2018;Ľ = 3đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 12 and đ?&#x2018;&#x2022; đ?&#x2018;Ľ = đ?&#x2018;&#x161;đ?&#x2018;Ľ + đ?&#x2018;? are depicted below.

-2

h f -6

-12

8.1 Find the equation of đ?&#x2018;&#x2022;. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 8.2 Write down the domain and range of đ?&#x2018;&#x201C;. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 8.3 For which value(s) of x is đ?&#x2018;&#x201C; decreasing. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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147

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Use algebraic methods to find the coordinates of đ?&#x2018; , the point of intersection of đ?&#x2018;&#x201C; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x2022;. ____________________________________________________________________ 8.4

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Further Graph Interpretation: Find a and c in the following: 1. đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?

đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?

đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;Ľ 2 + đ?&#x2018;?

Grade 10 Core Mathematics

5. đ?&#x2018;Ś = đ?&#x2018;&#x17D; đ?&#x2018;Ľ

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149 6.

đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;Ľ + đ?&#x2018;?

đ?&#x2018;&#x17D; đ?&#x2018;Ľ

đ?&#x2018;&#x17D;

8. đ?&#x2018;Ś = + đ?&#x2018;? đ?&#x2018;Ľ

Grade 10 Core Mathematics đ?&#x2018;&#x17D;

9. đ?&#x2018;Ś = đ?&#x2018;Ľ + đ?&#x2018;?

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150

10. đ?&#x2018;Ś = đ?&#x2018;&#x17D;đ?&#x2018;? đ?&#x2018;Ľ

11. The graphs of đ?&#x2018;&#x201C; đ?&#x2018;Ľ = â&#x2C6;&#x2019;đ?&#x2018;Ľ 2 + 4 and đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) = 3đ?&#x2018;Ľ + 4 are sketched alongside. 11.1 Calculate the coordinates of A, B, C and D.

y g

_______________________________________________ _______________________________________________

_______________________________________________ _______________________________________________ _______________________________________________

_______________________________________________

B x

_______________________________________________ f

_______________________________________________ _______________________________________________ E

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11.2 Determine the coordinates of E. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

11.3 Calculate the length of FB. ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 11.4 List an expression for đ?&#x2018;&#x201D; đ?&#x2018;Ľ â&#x2C6;&#x2019; đ?&#x2018;&#x201C; đ?&#x2018;Ľ . ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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11.5 For which values of x is đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) â&#x2030;Ľ đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ E

12. D

B x

A H g

The graphs of đ?&#x2018;&#x201C; đ?&#x2018;Ľ = đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 9 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x201D; đ?&#x2018;Ľ = 2đ?&#x2018;Ľ + 6 are sketched above. 12.1

Calculate the coordinates of A, B , C and D.

Grade 10 Core Mathematics

12.2

153

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State the lengths of OC, AB and OD

Calculate the coordinates of E.

IF đ?&#x2018;&#x201A;đ?&#x2018;&#x2020; = 1 Find FT.

If đ?&#x2018;&#x201A;đ?&#x2018;&#x192; = â&#x2C6;&#x2019;4 Find GH.

Calculate the length of AD.

Grade 10 Core Mathematics

14.

154

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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 thus 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 etc. The basic principles in financial decision making are established through the concept of interest and present value: –

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

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

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

A (1  in )

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

A  P(1  ni) Where:

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

i  interest level(i 

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

r for i) 100

Example 1: A Principle of R300,00 is invested in a Bank at 10% Simple Interest for a number of years; Complete the table below: Principle Interest period(yrs) Simple Interest (%) Interest

300 1

300 2

300 3

300 4

300 5

300 6

300 N

120

150

180

300 100in

Amount

330

360

390

420

450

480

200 

300 100in

Example 2:Invest R300,00 @ 15% per annum( per year) Year 0

R300

Simple Interest p.a

Year 1

R345

Year 2

R390

Year 3

R435

Year 4

R480

Year 5

R525

The value of the investment increases at a constant value of R45 per year. The relevant graph for simple growth is a straight line:

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Exercise 14.1: Question 1: Peter deposits R8500,00 for 8 years into a savings account. 1.1 If the money earns a simple interest rate of 12% per annum. How much will he have at the end of the investment period. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 1.2

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

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157

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Question 2: John invests R20 000,00 into a capital venture that will earn him a simple interest of 10% for the first 5 years and then 15% for the last 5 years. What is his investment worth at the end of the 10 years? ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ Question 3: 3.1

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

How much interest will she earn after 5 years?

How much interest will she earn after 10 years?

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3.3

158

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

Calculate the Principle if James received an amount of R7500,00 earned in simple interest at a rate of 10% over 5 years.

Calculate the Principle if James received an amount of R37500,00 earned in simple interest at a rate of 15% over 10 years.

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Question 5: A loan of R100 is made at 10% simple interest p.a. 5.1

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

200 190 180 170

A m o u n t

160 150 140 130 120 110 100

Years

5.2

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

If R100 is borrowed at 5% p.a. Simple Interest for 4 years display this on the same graph and show clearly how the two graphs differ from one another. ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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5.4

If R100 is borrowed at 15% p.a simple interest for 4 years explain without drawing the graph how the graph would differ from the other two (2). ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ COMPOUND INTEREST Compound interest arises when, in a transaction over an extended period of time, interest due at the end of a payment period is not paid, but added to the principal. Interest also earns interest i.e. it is compounded. The amount due at the end of transaction period is referred to as the compounded amount or accrued principal. Interest periods can vary : daily, monthly, quarterly, half-yearly or yearly. FOR GRADE 10 PURPOSES INTEREST IS ONLY CALCULATED YEARLY

Formula: Compound Growth:

A  P(1  i ) n A = Amount or Future Value P = Principal or Initial value i rate of interest per annum i 

r 100

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

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161

Example 1: Year 0

R300

Compound Interest p.a.

Year 1

R345

Year 2

R396,75

51,75

Year 3

R456.26

59,51

Year 4

R524,70

68,44

Year 5

R603,41

78,71

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

Example 2: Vusi invests R1000,00 at 12% compound interest for 5 years: 2.1 Write down a formula for the compound interest to be calculated over n yrs at r % p.a. n

r   n Answer: A  P 1   or A  P1  i   100 

2.2 Use the formula to calculate the final amount earned after the 5 years to the nearest cent. A  P1  i n

Answer:

A  10001  0,125 A  R1762,34

2.3 How much interest was earned on his investment:

i  A P Answer: i  1762,34  1000 Ci  R762,34

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162

2.4 Calculate what he would have earned in simple interest for the same time period: Pr t 100 1000  12  5 Si  100 Si  R600. Si 

Answer:

It can be deduced that it is more beneficial to invest money at compound interest:

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

i.e Calculating interest: 1. annually 2. half- yearly 3. quarterly 4. monthly 5. daily.

Example: R1000 invested at compound interest for 5 years at 12% p.a. Principle

Interest

A  P(1 

r tm ) 100m

Time Period = 5yrs Yearly

Half-Yearly

Monthly

Daily

m =4

m = 12

m= 365,25

m= no of time periods p.a.

m =1

R1000

R1762,34

R1790,85

R1806,11

R1816,70 R1821,94

R762,34

R790,85

R806,11

R816,70

12%

Compound Interest

m =2

Quartely

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

R821,94

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163

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Exercise 14.2: 1. Charlie wants to invest R7000,00 : There are 2 choices for him: What would the better choice be? Show all your working. 1.1

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

Invest R7000,00 at 12% compound interest over 5 years.

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3. R20 000,00 is invested for 15 years at 12% p.a. Calculate the value of the investment if the interest is calculated: 3.1

At simple interest.

At compound interest.

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

Grade 10 Core Mathematics

4.2

165

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

simple interest.

compound interest.

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166

6. The Ndlovu family uses a loan of R7200,00 to buy furniture They repay the loan at the end of 3 years. How much would they have to repay if the interest is calculated as: 6.1

16% p.a. simple interest?

13% p.a. compound interest?

Compound interest A  P(1  i) n

End of 1st year

End of 2nd year

End of 3rd year

End of 4th year

End of 5th year

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

y 2100 A m o u n t

2000 1900 1800 1700 1600

i n R a n d s

1500 1400 1300 1200 1100 1000 5

no of years

8.1

What kind of graph does the simple interest formula produce?

Find the gradient of the line.

Grade 10 Core Mathematics

8.3

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168

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

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

€

R 9.9837

NZ$

USD

0.7747 0.7157

98.1100

1.5253 1.9117

ZAR

0.1002 1

0.0776 0.0717

9.8270

0.1528 0.1915

EUR

1.2909 12.8880 1

GBP

1.3972 13.9492 1.0823 1

137.0793 2.1312 2.6710

JPY

0.0102 0.1018

0.0079 0.0073

0.0155 0.0195

AUD

0.6556 6.5453

0.5079 0.4692

64.3209

NZD

0.5231 5.2225

0.4052 0.3744

51.3213

0.7979 1

0.9239

126.6502 1.9690 2.4678

1.2533

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Examples : R1000 will buy 1.1

$100 (to nearest dollar)

1.2

£72(to nearest pound)

1.3

€78(to nearest Euro)

1.4

¥9827(to nearest Yen)

Exercise 14.3: 1.Copy and complete the following currency table:

CURRENCY

USD

ZAR

0.133 1

EUR 2

JPY

7,5

GBP

€

NZD

NZ$

0.0666

300

0.1428

0.933 7

150

O,05

AUD

1 0.333 0.01

0,0105 1

0,3335

Use the table above for questions 2 to 4. 2

You have R1000 to spend in each of the above countries. How much of the local money can be purchased in each of the countries. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3

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

3.1

$800.

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3.2 ¥2000. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3.3 A$500. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 3.4 €2000. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ 3.5 £5000. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ How many rands can be exchanged for the following: 3.6 $200 ____________________________________________________________________ ___________________________________________________________________ 3.7 ¥1500 ____________________________________________________________________ ___________________________________________________________________

3.8 A$150 __________________________________________________________________ ___________________________________________________________________

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3.9 €350. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

3.10

£750.

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

A  P(1  ni)

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

17090  10 36

= R473.33 + R1 = R483 .33

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Exercise 14.4: 1.

A couple buy new furniture for their home. The furniture costs R120 000,00 They pay for it using a hire purchase loan. The interest rate is 12% p.a. over a 5year period. Calculate the monthly repayment if there is no insurance added.

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

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Mr Moosa sees an advert in the newspaper for a surround sound audio system. The advert is worded as follows: â&#x20AC;&#x153;Pay ONLY R299,99 R7500.

per month for 36 months!â&#x20AC;? The cash price is

Assume that no insurance premium is added. Calculate: 3.1

how much interest is paid over 3 years?

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

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Ms Bokapane buys a washing machine from a appliance store . She asks 1 for a quote to buy the machine through a hire purchase contract over 1 2 years. The following quote was received by her:

QUOTATION: Purchase price

R3800,00

10% cash deposit

R 380,00

Balance owing

R3420,00

Finance charges(interest)

R 820,00

Insurance over 18 months

R 684,00

Total amount to be paid

R4924,80

Monthly payment

R 273,60

1.1

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

1.2

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

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

A  P(1  i ) n

80000  P(1  0,0805) 5

A  80000(1  0,0805) 5 A  R117818,00

P

80000

(1  0,0805) 5 P  R54320,80

Exercise 14.5: 1.

The average rate of inflation over the last 10 years was 7,3% p.a. The current price of a 2,5kg packet of white sugar is R10,25. 1.1

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

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

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

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

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

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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.

T6 etc indicates the time period of the investment.

Balance after 3 years :

Balance after 4 years :

Balance after 7 years:

A  7000(1  0,12) 3 A  R9834,496

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

n (years) Interest rates

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Example 2: In order to save for her sons University fees , Mrs Gumede deposits 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. 2.

Second years fees ( fees at T9):

9000  (0.1)(9000)  R9900 Third years fees ( fees at T10) 9900  (0.1)(9900)  R10890 3.

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) 5.

Money in savings account at T10:

5855,59( I  0,14)  R6675,37 Additional cash required: 10890  6675,37  R4214,63 (to nearest cent)

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Exercise 14.6: 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. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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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. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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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.

4.2

Is this fair on the sons? Explain your answer.

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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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15 Probability Theory: The language of probability:       

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



A probability of

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

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

1 2

Decimals Percentages

0 0%

0,5 50%

1 100%

Impossible

Even chance

Certain

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

Answer: Event Probability %

1 3 5

60%

Answer: 0,25 ; 30%;

2 0,25

3 78%

4 1 3

5 30%

2 0,25

3 78%

5 30%

25%

78%

1 3

33 13 %

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

30%

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

0,15 60%

Go to University

1 25

Go to Technikon Don‟t know

12% 9 100

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

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

Answer: Do Nothing

0,15

Get a Job

60%

Go To University

1 25

15 100 60 100 4 100

Go to Technikon

12%

12 100

Don‟t Know

9 100

Go to University; don‟t know; go to Technikon; do nothing; get a job.

15  60  4  12  9 100  1 100 100

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

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Exercise 15.1: 1.

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

Coffee

Tea

Hot Chocolate

1 20 2 5 3 10 1 4

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

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

The activity is to “throw the dice”



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



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



The favourable outcomes are 2. 4 and 6.

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

Coin Outcomes  

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

1 H, 1 T,1

H T

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

4 H, 4 T, 4

5 H, 5 T, 5

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

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

Activity: throw a dice. Event 1:

get a 1

Event 2:

get a multiple of 3.

Event 3: get a prime number. 

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

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

Favourable outcomes 1 3;6 2; 3; 5

6 H, 6 T, 6

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

Event 5: get a silver piece

Event 6: get a coin worth less than 50c. ď&#x192;&#x2DC;

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

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

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

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

Event 8: get a vowel

Event 9: get a consonant. ď&#x192;&#x2DC;

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

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

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

Exercise 15.2: 2.1 In my left-hand pocket I have a 50c coin, a 20c coin and a 10c coin. In my right-hand pocket I have R1 coin, a R2 coin and a R5 coin. 2.1.1

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

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

2.2 I roll dice: 2.2.1

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

2.2.2

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

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

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

Venn Diagram

This set is referred to as the probability space or sample space

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

Answers: 4.1 4.2 4.3

Rolling the dice. 4

4.4

S  1;2;3;4;5;6

1 4

2 3 5 6

Exercise 15.3: 3.1

Consider the tossing of a coin: 3.1.1 Describe a trial

Give an example of an outcome.

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Write down the sample space, S of the experiment.

Represent the sample space by means of a venn diagram.

The diagram below shows a four sided spinner with sides labeled 1, 2, 3and 4 respectively. 3.2.1 Describe a trial using such a spinner.

___________________________________________________________

4 3

___________________________________________________________ 3.2.2

Give an example of an outcome.

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Write down the sample space, S for the experiment.

Give a venn diagram for S.

Events: A bag contains 3 blue and 5 white counters; The blue counters are labled b1; b2 and b3. While the white counters are labled w1; w2; w3â&#x20AC;Śâ&#x20AC;Ś.w5. Suppose the trial consists of drawing out a counter, after the bag has been given a good shake, and then replacing the counter. The sample space here is: S = {b1; b2; b3; w1; w2; w3;w4 ;w5.} The number of possible outcomes in the sample space, S , is denoted by n(S). Here n(S) = 8 The drawing of a blue counter , no matter what its lable is, can be represented by the set B ={ b1; b2; b3} A few of the outcomes or elements of S are involved in the situation described by B. We call the drawing of blue counter ( no matter the lable) an event. The number of outcomes in event B is the number of elements in B. This number is denoted by n(B), which equals 3. Another event related to drawing counters from the bag, would be that of drawing a white counter. We can represent this event by the set : W = { w1; w2; w3;w4 ;w5} n(W) = 5 The sets B and W are subsets of S. The sample , S, is the universal set. Definition: An event is a set which consists of one or more of the elements of the sample space.

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

S  1;2;3;4;5;6;7;8;9;10 Consider the events:  Drawing a factor of 6: X = { 1 ;2 ;3 ; 6}  Drawing a factor of 9: Y = {1; 3 ;9}

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

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

1.6

5 F

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Exercise15. 4: 4.1

The trial to be considered in this instance is the spinning of a coin.

4.1.1 Represent the sample space, S, in set notation. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4.1.2 Write down a set for the event, H, of obtaining a head. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 4.1.3 Write down a set for the event, t , of obtaining a tail. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

4.1.4 What is the value of each of the following: 4.1.4.1 n(S )

____________________________

4.1.4.2 n(H ) ____________________________

4.1.4.3 n(T ) ____________________________

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

4.2

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

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

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

Calculating Probabilities:   

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

number of favourable possibilities Total number of possible outcomes

n( E ) n( S )

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

n( E ) 1  n( S ) 6

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

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

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

1 or 0,5. 2

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

49 or 0,49. Only after 100

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

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Mutually exclusive and complimentary events. Mutually Exclusive Events 

The events A and B exclude each other.



If A happens B cannot happen.



If B happens A cannot happen.



Both events cannot happen at any one trial.



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



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

P(A and B) =

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

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

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



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



Q can also be written R or “not R “

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

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

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

P( R )  1  P( R)  1 

Note: Alternatively

4 5 2 11    14 14 14 14 Exercise 15.5: [Use venn diagrams when necessary:]

5.1



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

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5.2

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

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

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

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5.4.1 Derek picks a sweet from the packet without looking.What is the probability that he picks either a melon-flavoured or lemon-flavoured sweet? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 5.4.2 Albert doesnâ&#x20AC;&#x;t like banana-flavoured or melon-flavoured sweets. He likes all other flavours. What is the probability that he picks a sweet that he likes? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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204

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

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

Solution:

5 9

Red 6 10

Blue 4 9

6 9 4 10

Blue 3 9

1.1

1 3 4 15

Red

4 15

Blue

2 15

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

1.2

Red

6 3  10 5

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

4 3 2   10 9 15

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

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

4 4 8   15 15 15

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Exercise 15.6: 1. A coin and a dice are thrown together. 1.1 Draw a tree diagram to show all the possible outcomes. 1.2 What is the probability of obtaining: 1.2.1 A head on a coin and 1 on the dice? 1.2.2 A tail on the coin and an even number on the dice? ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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

Compiled by Chez Nell

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

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GRADE 10 CORE MATHEMATICS. PAPER TWO:

Topic:

Pages:

Analytical Geometry

(4 – 21)

Triangles & Quadrilaterals

(22 – 40)

Trigonometry

(41 - 67)

Data Handling

(68 – 100)

Volumes & Surface Area

(101 - 115)

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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)

y Mid – Point AB = C (-1; 1) 3.

Gradient of Straight Line: Gradient is represented using the symbol „m‟ [from y= mx+c] M=

y [ i.e the difference in y divided by the difference in x] x

A(2;4)

m

B(3:6)

y y1  y 2 6  (4) 10 2     x x1  x2 3  (2) 5 1

Parallel Lines have the same gradients: m1  m2 Perpendicular lines have inverse gradients: m1  m2  1

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Exercise 1.1: Distance between points: 1.

Find the distance between the given pairs of points: 1.1 (2 ; 3) and (4 ; 5)

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1.4 (-4 ; 3) and (0 ; 0) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.5 (-2 ; 1) and -4 ; -1) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 1.6 (-3 ;-1) and (4 ; -6) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Given the coordinates of the vertices of ď &#x201E;ABC , in each case ( 2.1 to 2.5) Determine: A. the perimeter of the triangle. B. Whether the triangle is equilateral, isosceles or scalene. C. Whether or not the triangle has a right angle. 2.1

A(1 ; -3) ; B(7 ; 3); C(4 ; 6)

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

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2.3 A(-2 ; -3) ; B(-4 ; 1) ; C(4 ; 5) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2.4

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

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2.5

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

Show that: 3.1

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

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3.2 (6 ;-4) , (5 ;3) (-2 ; 2) and (-1 ; -5) are vertices of a square. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ Mid-points of lines: 4.

Calculate the coordinates of the midpoints of the line joining the following points: 4.1 (-3 ;1) and (1 ; 5)

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4.2 (-2 ; 3) and (6 ; 3) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.3 (4 ; -1) and (-1 ; 3) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.4 (0 ;0 ) and (3 ; -8) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.5 ( 3;1) and (3 3;ď&#x20AC;1) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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

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

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5.3

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

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

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Exercise 1.2: Formulae: AB  ( x1  x2 )2  ( y1  y2 )2

Gradient = m =

 x  x y  y2  Mid-point =  1 2 ; 1  2   2

y1  y2 x1  x2

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

1.1

the length of AB in units.

1.2

the co-ordinates of the mid – point ( C ) of line AB.

the gradient of line AB.

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1.4

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show that points A;B and C are collinear.

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

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2. y

A C 0

The Points A(-4 ;3) ; B(-4 ; -4) ; C(6 ; 1) and D(6 ; 8) lie on a cartesian plane. Determine: 2.1

the length of AD.

2.2

the mid-point of DC

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2.3

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the gradient of BC

2.4

show that ABCD is a parallelogram.

2.5

the co-ordinates of the point of intersection of the diagonals AC & BD

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A(-3;3) C(8;2)

-5

-2

B(3;-3) -4

The figure above represents a triangle on a Cartesian plane. A( -3 ;3) ; B(3 ; -3) and C(8 ; 2) 3.1 Calculate the perimeter (distance around) of ď &#x201E;ABC .

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3.2 Prove that triangle ABC is right angled at B.

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

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

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

P(ď&#x20AC;4;1) , Q(3;0) and R(1;ď&#x20AC;3) 4.2 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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4.3

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U (5;2) , V (1;1) and W (13;1)

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

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1. Triangles: Theorems: 1.

The sum of the angles of a triangle equals 180ď&#x201A;° A

2 1

Given: â&#x2C6;&#x2020;ABC. RTP: đ??´ + đ??ľ + đ??ś = 180ď&#x201A;° Proof: Produce BC to D and draw CE parallel to AB. đ??ś2 = đ??´ (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;Ą ď&#x192;?, đ?&#x2018; đ??´đ??ľ ď źď ź đ??śđ??¸) đ??ś3 = đ??ľ (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x;đ?&#x2018;&#x; ď&#x192;?, đ?&#x2018; đ??´đ??ľď źď ź đ??śđ??¸) But đ??ś1 + đ??ś2 + đ??ś3 = 180ď&#x201A;° ( ď&#x192;?, đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018;&#x17D; đ?&#x2018; đ?&#x2018;Ą đ?&#x2018;&#x2122;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x2019;) ď &#x153;đ??´ + đ??ľ + đ??ś = 180ď&#x201A;° 2.

The exterior angle of a triangle is equal to the sum of the interior opposite angles A E

2 1

Given: â&#x2C6;&#x2020;ABC with BC produced to D. RTP: đ??´đ??ś đ??ˇ = đ??´ + đ??ľ Proof: Draw CE parallel to AB. đ??ś1 = đ??´ (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;Ą ď&#x192;?, đ?&#x2018; đ??´đ??ľ đ??źđ??ź đ??śđ??¸ đ??ś2 = đ??ľ (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x;đ?&#x2018;&#x; ď&#x192;?, đ?&#x2018; đ??´đ??ľ đ??źđ??ź đ??śđ??¸

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In an isosceles triangle , the sides opposite the equal angles are equal. A 1 2

ď&#x201A;ˇ

Given: â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2022; đ??ľ = đ??ś RTP: AB = AC. Proof: Construct AD to bisect đ??´ , cutting BC at D. In â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??´đ??śđ??ˇ đ??´1 = đ??´2 (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018; đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;˘đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;) đ??ľ=đ??ś ( đ?&#x2018;&#x201D;đ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) AD is common â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??śđ??ˇ (đ??´đ??´đ?&#x2018;&#x2020;) â&#x2C6;´ đ??´đ??ľ = đ??´đ??ś Conversely: The angles opposite equal sides of a triangle are equal. 4.

The Mid â&#x20AC;&#x201C; Point Theorem. The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to half its length. 1

RTP: DE â&#x20AC;&#x2013; đ??ľđ??ś and đ??ˇđ??¸ = 2 đ??ľđ??ś A

Proof: Construct EF = DE and join FC. đ??źđ?&#x2018;&#x203A; â&#x2C6;&#x2020;đ??´đ??ˇđ??¸ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??śđ??šđ??¸ AE = CE ( Given) DE = EF ( by construction) đ??¸1 = đ??¸2 (đ?&#x2018;&#x2030;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;Ś đ?&#x2018;&#x153;đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; ) â&#x2C6;&#x2020;đ??´đ??ˇđ??¸ â&#x2030;Ą â&#x2C6;&#x2020;đ??śđ??šđ??¸ ( đ?&#x2018;&#x2020;đ??´đ?&#x2018;&#x2020;) â&#x2C6;´ đ??´đ??ˇ đ??¸ = đ??śđ??š đ??¸ â&#x2C6;´ đ??´đ??ľâ&#x20AC;&#x2013;đ??šđ??ś ( đ??´đ?&#x2018;&#x2122;đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x2019;đ?&#x2018;&#x17E;đ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;) BD = DA (given) BD= FC ( proved above) â&#x2C6;´ đ??ˇđ??śđ??šđ??ľ đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x17D; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x161; ( đ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;&#x2019; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x2013;đ?&#x2018;&#x; đ?&#x2018;&#x153;đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;&#x17E;đ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;&#x2122; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;) â&#x2C6;´ đ??ˇđ??š = đ??ľđ??ś ( đ?&#x2018;&#x153;đ?&#x2018;?đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x17D; đ?&#x2018;?đ?&#x2018;&#x17D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;&#x201D;đ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;&#x161;) 1 đ??ˇđ??¸ = 2 đ??ľđ??ś and đ??ˇđ??¸â&#x20AC;&#x2013;đ??ľđ??ś

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Congruence: Definition: Two triangles are said to be congruent if they are equal in all respects. There are four cases of congruence: i.e. methods of proving triangles are congruent. â&#x;šTwo angles and one corresponding side are equal. ( AAS). â&#x;šTwo sides and the angle between them are equal. (SAS). â&#x;šThree sides are equal. (SSS) â&#x;šRight angle, hypotenuse and one further side are equal. (RHS) NB: If two parallel lines are cut by a transversal the angles formed are: Alternate angles, which are equal. ď&#x201A;ˇ ď&#x201A;ˇ

Corresponding angles, which are equal. ď&#x201A;ˇ

ď&#x201A;ˇ

Co-interior angles, which are supplementary. x x + y = 180ď&#x201A;° y

1 4

2 3

If two lines intersect then two types of angles are formed. ( đ?&#x2018;łđ?&#x;? & đ?&#x2018;łđ?&#x;? ) 1. Adjacent supplementary angles ( angles on a straight line) i.e. 1 + 2 = 180Â° & 3 + 4 = 180Â°. 2. Vertically opposite angles. đ?&#x2018;&#x2013;. đ?&#x2018;&#x2019; 1 = 3 & 2 = 4

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Exercise 3.1: 1. In the sketch below, AB = CD and đ??´đ??ľ đ??ś = đ??ˇđ??ś đ??ľ . A

D E

Prove that: 1.1 â&#x2C6;&#x2020;đ??´đ??ľđ??ś â&#x2030;Ą â&#x2C6;&#x2020;đ??ˇđ??śđ??ľ _________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 1.2 â&#x2C6;&#x2020;đ??´đ??ľđ??¸ â&#x2030;Ą â&#x2C6;&#x2020;đ??ˇđ??śđ??¸ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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In the sketch below đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; = đ?&#x2018;&#x2020;đ?&#x2018;&#x2021; ; đ?&#x2018;&#x192;đ?&#x2018;&#x2026; = đ?&#x2018;&#x192;đ?&#x2018;&#x2020; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x192;đ?&#x2018;&#x2026; đ?&#x2018;&#x201E; = đ?&#x2018;&#x192;đ?&#x2018;&#x2020;đ?&#x2018;&#x201E;. P

2.1

Prove â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2026;đ?&#x2018;&#x201E; â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2020;đ?&#x2018;&#x2021;.

Deduce that PQ = PT

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Prove that â&#x2C6;&#x2020;đ??´đ??¸đ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??ľđ??śđ??ˇ in the diagram below: Given: AD = BD ; AE = BC; AE ď &#x17E; BD and BC ď &#x17E; AD. A C

F D E

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Prove that â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??ľđ??ś in the diagram below: Given: AD = BC ; đ??ľđ??´đ??ˇ = đ??´đ??ľ đ??ś D

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Similarity: Two triangles are said to be similar if they are equiangular or conversely if the ratios of their corresponding sides are in proportion. Examples: A

A r

D x

E 6cm

8cm

3cm

2cm

E 4cm

y x

Юљ╝ЮЉЏ РѕєРђ▓ ЮЉаЮљ┤ЮљхЮљХ ЮЉјЮЉЏЮЉЉ ЮљИЮљиЮљ╣: Юљ┤ = ЮљИ (ЮЉћЮЉќЮЉБЮЉњЮЉЏ)

Юљ╝ЮЉЏ РѕєРђ▓ ЮЉаЮљ┤ЮљхЮљХ ЮЉјЮЉЏЮЉЉ ЮљИЮљиЮљ╣: Юљ┤Юљх 6 = =2 ЮљИЮљи 3 ЮљхЮљХ

Юљх = Юљи (ЮЉћЮЉќЮЉБЮЉњЮЉЏ) ЮљХ = Юљ╣ (ЮЉћЮЉќЮЉБЮЉњЮЉЏ) Рѕ┤ РѕєЮљ┤ЮљхЮљХ ле РѕєЮљИЮљиЮљ╣ ( Юљ┤Юљ┤Юљ┤) Юљ┤Юљх ЮљхЮљХ Юљ┤ЮљХ Рѕ┤ ЮљИЮљи = ЮљиЮљ╣ = ЮљИЮљ╣

4cm

ЮљиЮљ╣ Юљ┤ЮљХ

=2=2 8

=4=2 Рѕ┤ РѕєЮљ┤ЮљхЮљХ ле РѕєЮљИЮљиЮљ╣ (ЮЉаЮЉќЮЉЉЮЉњЮЉа ЮЉќЮЉЏ ЮЉЮЮЉЪЮЉюЮЉЮЮЉюЮЉЪЮЉАЮЉќЮЉюЮЉЏ) ЮљИЮљ╣

Triangles: Properties: 1. The sum of the interior angles is 180№ѓ░. 2. The exterior angle of a triangle is equal to sum of the interior opposite angles. 3. Isosceles triangles have two sides equal and the angles opposite these sides are also equal. 4. Equilateral triangles have all three sides and all three angles equal. 5. Scalene triangles have no sides or angles equal.

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Quadrilaterals:

Number None

opposites sides parallel

Both pairs

One pair Both pairs

 

One pair

Square

Rectangle

Rhombus

  



 



 



 







 



 







 



 



  

Both pairs None





All four sides None

Opposite angles equal



One pair

None

Adjacent sides equal



One pair

None opposites sides equal

Parallelogram

Property

Trapezium

Kite

Properties of Quadrilaterals

 



  



One pair Consecutive angles equal

Both pairs All four angles none only one

Line of symmetry



 

  



  

only two



 

four

Diagonals are equal

Diagonals bisect each other



Diagonals are perpendicular







 



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Relationship between sides and angles:

Number of sides of a polygon 3 4 5 6 7 8 9 10 n Sum of interior angles 180º 360º 540º 720º 900º 1080º 1260º 1440º (n-2)180º Size of one interior angle of regular polygons 60º

90º

108º 120º 128,570º

135º

140º

144º

(n-2)180º n

The pattern is (number of sides-2)180º Sum of interior Angles = (n-2)180º Number of sides of a polygon 3 4 5 6 7 8 9 Sum of exterior angles 360º 360º 360º 360º 360º 360º 360º

Number of sides of a polygon 4 5 6 7 8 9 Number of Triangles in a Regular Polygon 2 3 4 5 6 7

3 1

10 360º

10 8

Exercise 3.2: 1

Find the values of the variables in the following (i.e. x ; y & z) 1.1

AB//DC and AB=DC A



62 D



y C

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PQRS is a rectangle: P



36

y x



ABCD is a rhombus: A

35

x D

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1.3

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LMNO is a trapezium with LM // PO and LM = LP ; đ??ż = 105Â° and đ?&#x2018;&#x192;đ?&#x2018;&#x20AC;đ?&#x2018;&#x201A; =L 65Â° 105Â°

ď&#x201A;Ž

65Â°

y z

ď&#x201A;Ž

1.5

ABCD is a kite. AB = AC & BD = CD. đ??´đ??ś đ??ľ = 60ď&#x201A;°, đ??ˇđ??ľ đ??ś = 70Â° A x

60ď&#x201A;°

y 70ď&#x201A;°

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1.6

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ABCDEF is a regular hexagon.

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Paralellograms: Theorem A Given: Parallelogram ABCD RTP: AB = CD ; AD = BC; đ??´ = đ??ś đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ľ = đ??ˇ Proof: Construct diagonal BD đ??ľ1 = đ??ˇ2 (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;Ąđ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; đ??´đ??ľâ&#x20AC;&#x2013;đ??śđ??ˇ) đ??ˇ1 = đ??ľ2 (đ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; đ??´đ??ˇâ&#x20AC;&#x2013;đ??ľđ??ś) đ??ľđ??ˇ = đ??ľđ??ˇ (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x161;đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;) â&#x2C6;´ â&#x2C6;&#x2020;đ??´đ??ľđ??ˇ â&#x2030;Ą â&#x2C6;&#x2020;đ??śđ??ˇđ??ľ (đ??´đ??´đ?&#x2018;&#x2020;) â&#x2C6;´ đ??´đ??ˇ = đ??ľđ??ś â&#x2C6;´ đ??´đ??ľ = đ??śđ??ˇ đ??ľ1 = đ??ˇ2 đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??ˇ1 = đ??ľ2 đ??´đ??ľ đ??ˇ = đ??śđ??ˇđ??ľ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??´đ??ˇđ??ľ = đ??śđ??ľ đ??ˇ Theorem B: Converse to A: States that if the opposite sides of a quadrilateral are equal then the quadrilateral is a parallelogram. Theorem C: Converse to A: States that if the opposite angles of a quadrilateral are equal then the quadrilateral ios a parallelogram. To prove that a quadrilateral is a parallelogram one must be able to prove at least two of the properties of this particular shape, Properties concerned are: 1.

both pairs of opposite sides are parallel.

both pairs of opposite sides are equal.

opposite angles are equal.

the diagonals bisect each other.

Hint: a good method is to prove that one pair of opposite sides are equal and parallel. Examples: 1. Prove that ABCD is a parallelogram.

B (0 ; 3) 2

A (-4 ; 0) -10

C (1 ; 0)

-5

-2

D (-3 ; -3) -4

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Proof: đ??´đ??ľ =

(0 + 4)2 + (3 â&#x2C6;&#x2019; 0)2 = 5

đ??ˇđ??ś =

(1 + 3)2 + (0 + 3)2 = 5

đ?&#x2018;&#x161;đ??´đ??ľ = 4 3

đ?&#x2018;&#x161;đ??ˇđ??ś = 4

ABCD is a parallelogram ( one pair of opposite sides are equal and parallel) 2. ABCD is a quadrilateral with AB = CD & AD = BC A 2

B 1

2.1

2.2

2 1 C

RTP: â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Ťđ?&#x2018;Ş â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Šđ?&#x2018;Ş Proof: đ??źđ?&#x2018;&#x203A; â&#x2C6;&#x2020;đ??´đ??ˇđ??ś đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ??´đ??ˇ = đ??ľđ??ś ( đ?&#x2018;&#x201D;đ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) đ??śđ??ˇ = đ??´đ??ľ ( đ?&#x2018;&#x201D;đ?&#x2018;&#x2013;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;) đ??´đ??ś = đ??´đ??ś (đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;&#x161;đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x2018;đ?&#x2018;&#x2019;) â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Ťđ?&#x2018;Ş â&#x2030;Ą â&#x2C6;&#x2020;đ?&#x2018;¨đ?&#x2018;Šđ?&#x2018;Ş (đ?&#x2018;şđ?&#x2018;şđ?&#x2018;ş) RTP: đ?&#x2018;¨đ?&#x2018;Šâ&#x20AC;&#x2013;đ?&#x2018;Ťđ?&#x2018;Ş đ?&#x2019;&#x201A;đ?&#x2019;?đ?&#x2019;&#x2026; đ?&#x2018;¨đ?&#x2018;Ťâ&#x20AC;&#x2013;đ?&#x2018;Šđ?&#x2018;Ş Proof: đ??´2 = đ??ś2 (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; 2.1 â&#x2C6;&#x2020;đ??´đ??ˇđ??ś â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ??´đ??ˇâ&#x20AC;&#x2013;đ??ľđ??ś ( alternate angles equal) đ??´1 = đ??ś1 (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; 2.1 â&#x2C6;&#x2020;đ??´đ??ˇđ??ś â&#x2030;Ą â&#x2C6;&#x2020;đ??´đ??ľđ??ś đ??´đ??ľâ&#x20AC;&#x2013;đ??ˇđ??ś ( alternate angles equal)

2.2

)

RTP: ABCD is a parallelogram Proof: đ??´đ??ľ = đ??śđ??ˇ & đ??´đ??ľâ&#x20AC;&#x2013;đ??śđ??ˇ (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;) đ??´đ??ˇ = đ??ľđ??ś & đ??´đ??ˇâ&#x20AC;&#x2013;đ??ľđ??ś (đ?&#x2018;?đ?&#x2018;&#x;đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x2018; đ?&#x2018;&#x17D;đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;Łđ?&#x2018;&#x2019;) ABCD is a parallelogram ( both pairs opposite sides equal and parallel)

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Exercise 3.3: 1.

A (0 ; 4)

B( 5 ; 4)

C ( 2 ; -3)

D ( -3 ; -3)

Prove with reasons that the shape above is a parallelogram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 2

PRSQ is a parallelogram. RS = ST & TR is a straight line. Prove: 2.1 QA = AT. 2.2 PT // QS. T



 R

2.1 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2.2 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 3. FBCD is a parallelogram. AF = FB. Prove that FE = ED

1 2

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A 2

B 2

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In the figure above, ABCD is a parallelogram. đ??´đ??ˇ = đ??´đ??¸ đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ??šđ??ś = đ??ľđ??ś. Prove that đ??´đ??¸đ??śđ??š is a parallelogram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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P 2

Q 1

In the figure above â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; â&#x2C6;&#x2020;đ?&#x2018;&#x192;đ?&#x2018;&#x2020;đ?&#x2018;&#x2026; are isosceles triangles with đ?&#x2018;&#x192;đ?&#x2018;&#x2020; = đ?&#x2018;&#x192;đ?&#x2018;&#x2026; = đ?&#x2018;&#x201E;đ?&#x2018;&#x2026; đ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x2018;&#x2018; đ?&#x2018;&#x192;đ?&#x2018;&#x201E;â&#x20AC;&#x2013;đ?&#x2018;&#x2020;đ?&#x2018;&#x2026;. Prove that PQRS is a parallelogram. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Trigonometry:

Use the sketches that follow and complete the tables. The angle sizes at A are 30º ; 45º and 60º. Measure the sides accurately and write the ratios down in decimal form. The task is to compare the ratios of the different sides of the 3 triangles formed. Decimal answers to 3 places. WHAT CONCLUSIONS CAN BE MADE? Task 1: 30º E

F G

30  A

Measure the sides as indicated in the table below and represent the answers first as a fraction and then as a decimal. Compare your answers and come to a conclusion. NB: it is important to be very accurate.

AGB BG @30º= _____= AB

AFC FC @30º=______= AC

AED ED @30º=_____= AD

AB @30º=______= AG

AC @30º=______= 0, AF

AD @30º=_____= AE

BG @30º=______= AG

FC @30º=______= AF

ED @30º=_____= AE

AGB

AFC

AB BG @30º= _____=

AC FC @30º=______=

AED AD ED @30º=_____=

AG AB @30º=______=

AF AC @30º=______= 0,

AE AD @30º=_____=

AG BG @30º=______=

AF FC @30º=______=

AE ED @30º=_____=

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Task 2: 45º

45  A

AGB BG @45º= _____= AB

AFC FC @45º=______= AC

AED ED @45º=_____= AD

AB @45º=______= AG

AC @45º=______= AF

AD @45º=_____= AE

BG @45º=______= AG

FC @45º=______= AF

ED @45º=_____= AE

AGB

AFC

AB BG @45º= _____=

AC FC @45º=______=

AED AD ED @45º=_____=

AG AB @45º=______=

AF AC @45º=______=

AE AD @45º=_____=

AG BG @45º=______=

AF FC @45º=______=

AE ED @45º=_____=

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Task 3: 60º E

60 A

AGB BG @60º= _____= AB AB @60º=______= AG BG @60º=______= AG

AFC FC @60º=______= AC AC @60º=______= AF FC @60º=______= AF

AED ED @60º=_____= AD AD @60º=_____= AE ED @60º=_____= AE

AGB

AFC

AB BG @60º= _____=

AC FC @60º=______=

AED AD ED @60º=_____=

AG AB @60º=______=

AF AC @60º=______=

AE AD @60º=_____=

AG BG @60º=______=

AF FC @60º=______=

AE ED @60º=_____=

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Summary:

C h

r y

30Âş

x ď&#x20AC;˝ a ď&#x20AC;˝ adjacent y ď&#x20AC;˝ o ď&#x20AC;˝ opposite r ď&#x20AC;˝ h ď&#x20AC;˝ hypotenuse In general format ď &#x201E;ACB y BC @30Âş = tan30Âş = x AB AB x @30Âş = cos30Âş = AC r y BC @30Âş = sin30Âş= r AC

Correct Trigonometric format ď &#x201E;ACB y o tan30Âş = OR x a x a cos30Âş = OR r h y o sin30Âş = OR r h

ď &#x201E;ACB đ??´đ??ľ đ??ľđ??ś đ??´đ??ś đ??ľđ??ś đ??´đ??ś đ??ľđ??ś

@30Âş = cot30Âş = @30Âş = sec30Âş =

đ?&#x2018;Ľ

cot30Âş =

đ?&#x2018;Ś đ?&#x2018;&#x;

@30Âş = cosec30Âş=

đ?&#x2018;Ľ

sec30Âş = đ?&#x2018;&#x;

đ?&#x2018;Ś

đ?&#x2018;Ľ đ?&#x2018;Ś đ?&#x2018;&#x;

ď &#x201E;ACB OR

đ?&#x2018;&#x17D;

đ?&#x2018;&#x153; đ?&#x2018;&#x2022;

OR đ?&#x2018;Ľ đ?&#x2018;Ľ đ?&#x2018;&#x; đ?&#x2018;&#x2022; cosec30Âş = OR đ?&#x2018;Ś đ?&#x2018;&#x153;

N B: In the above table , the right hand side is the correct way to represent the different trigonometric ratios. It important to remember that a trig ration must be followed by a specific angle size or a variable representing an angle. After the equal sign a ratio is written: e.g. đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x203A;60Â° = 1,732 â&#x20AC;Ś ..A trig function cannot stand on its own and must be written with an angle or a variable (denoting an angle).

Grade 10 Core Mathematics

sin x 

y r

cos x 

x r

tan x 

y x

 cos ecx   sec x 

r x

 cot x 

x y

r y

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1 cos ecx 1 cos x  sec x

sin x 

tan x 

1 cot x

tan x 

sin x cos x

cot x 

cos x sin x

cos 2 x  sin 2 x  1

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4.2 Pythagoras in trigonometry: Solving basic equations & solution of triangles. Example: Using Pythagoras:

 (3 ;4)



Complete the triangle by joining the point (3;4) to the x – axis to form a right –angled triangle. Use Pythagoras theorem to calculate r (hypotenuse). You are now ready to find the values of the different ratios for . As follows below:

 (3 ;4) 4

r 2  x 2  y 2 (pythagoras theorem) 4 sin   5 r 2  32  4 2 3 cos   r 2  25 5 r5 4 Tan  3



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Exercise 4.1: In each of the cases below calculate the length of the radius and the find all the trig ratios for  y 15 e.g tan    x 8 2.

 (6 ;8)

 (8 ;15)



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 (7 ;24)

 (9 ;12)



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y  (44 ;33)

 (40;9)

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In each of the cases below calculate the length of the radius and use the values from the sketch to answer the questions that follow: 7.

ď&#x201A;ˇ (11;60) ;40)

ď&#x201A;ˇ (20;21) ;16)

ď ą

0 7.1 đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x153;&#x192; =

0 8.1 đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x203A;đ?&#x153;&#x192; =

___________________________________________________________________ ___________________________________________________________________ 7.2

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x153;&#x192; đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x153;&#x192;

8.2 đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;2 đ?&#x153;&#x192; =

8.3 1 â&#x2C6;&#x2019; đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; 2 đ?&#x153;&#x192; =

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4.3 Equations in trigonometry: Solving for  given a ratio: i.e Solving basic equations: [NB:  is a variable and represents an angle.] Example: Solve for  where   [0  ;90  ] 1.

sin   0,5

  30 

Method: use you calculator as follows: enter {shift} {sin-1 } followed by the ratio 0,5 then equals (=) 2. Tan x = 3,456 x = 73,86º Method: use you calculator as follows: enter {shift} {tan-1 } followed by the ratio 0,5 then equals (=) NB: The functions sin;cos;tan on the calculator convert trig functions to decimal ratios. The functions sin-1;cos-1;tan-1 convert trig ratios back to angles. Exercise 4.2: Solve for x where x  [0  ;90  ] sin x  0,336 . 1. 7. sin( x  10 )  0,800 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

cos x  0,786 2. 8. cos( x  15 )  0,642 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

tan x  1,732 3. 9. tan(x  25 )  5,482 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2 sin x ď&#x20AC;˝ 0,412 4. đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ľ = 1,1223 10. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

3 cos x ď&#x20AC;˝ 1,236 5. đ?&#x2018; đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ľ = 1,743 11. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 6. đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018;Ąđ?&#x2018;Ľ = 2,1445 12. 2 tan(x ď&#x20AC; 30ď Ż ) ď&#x20AC;˝ 11,3426 ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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4.4 Solving Triangles:

B h y  A y r x cos   r y tan   x

sin  

BC AB AC cos   AB BC tan   AC

sin   or

Exercise 4.3 : 1.

C D

BC AC BD sin A  AB sin A 

Task complete all ratios for above triangle: ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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54 _S

_B _T 1.

2. _C

_N _0 3.

In diagram 1 & 2: Find the ratios for: 2.1

sin A

2.7

sinP

___________________________________________________________________ ____________________________________________________________________ 2.2

cosA

2.8

cosP

___________________________________________________________________ ____________________________________________________________________ 2.3

tanA

2.9

tanP

___________________________________________________________________ ____________________________________________________________________ 2.4

sinB

2.10

sinT

___________________________________________________________________ ____________________________________________________________________ 2.5

cosB

2.11

sinS

___________________________________________________________________ ____________________________________________________________________ 2.6

tanB

2.12

cosS

___________________________________________________________________ ____________________________________________________________________

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In diagram 3: Write down two possible values for: 3.1 sinL __________________________________________________________ ____________________________________________________________________ 3.2 cosL ___________________________________________________________________ ____________________________________________________________________ 3.3 tanL ___________________________________________________________________ ____________________________________________________________________ 3.4 sinN ___________________________________________________________________ ____________________________________________________________________

3.5 cosN ___________________________________________________________________ ____________________________________________________________________

3.6 tanN ___________________________________________________________________ ____________________________________________________________________

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Exercise 4.4 1.

2. C

30cm 15 cm 35ď&#x201A;°

56ď&#x201A;°

Calculate the lengths of BC and AB

Calculate the lengths of AC & AB

C 24 cm

14 cm

12 cm

26ď&#x201A;° 32ď&#x201A;°

8 cm

In the diagram above: PQ = 12 cm; In the diagram above: AB = 24 cm BAË&#x2020; C ď&#x20AC;˝ 26 ď Ż and CAË&#x2020; D ď&#x20AC;˝ 32 ď Ż QR = 8 cm and SR = 14 cm. đ?&#x2018;&#x2026; = 90Â° đ??ľ & đ??ˇ = 90Â° Calculate: Calculate : Ë&#x2020; Ë&#x2020; 3.1. the size of Q1 and Q2 4.1. AC ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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3.2. the length of PS. 4.2. the length of CD ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ _____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 3.5

Solution of Triangles ( problem/reality questions)

A ladder rests 4 metres up a wall and has an angle at the foot of the ladder of 45. Calculate the length of the ladder.

Ladder 4m 45º Answer: 4m 4  Ladder L  L sin 45  4 4 L sin 45  Ladder  5,7 metres NB: sin 45  

L sin 45  4   sin 45 sin 45  4 L sin 45 

A x

Angle of depression

y B

Angle of elevation

Line 1 Lines 1 and 2 are parallel. The angle of elevation from B to A is yº The angle of depression from A to B is xº

Line 2

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Exercise 4.5: 1.

A swimmer crosses a swiftly flowing river and is washed downstream by the current. He reaches the opposite bank and by swimming a distance of 100m.

100m

River Current flow

35ยบ Task: If the angle he swam was 35ยบ with the near side bank , calculate the actual width of the river. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________

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Calculate the height of the taller building if the distance between the buildings is 10m. The angle between a line from the top of the tall building (excluding the roof) to the bottom of the short one is 35ยบ. Task: Calculate the height of the taller building.

35ยบ

Height of building 10m

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Cliff Face

Height of cliff

Surface of Sea

50ยบ

1200m 3.

In the diagram above , a yacht is anchored 1200m from the base of a cliff. The angle of elevation from the yacht to the top of the cliff is 50ยบ. Calculate the height of the cliff. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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4.6 Graphs in Trigonometry : Sketching graphs of trigonometric functions by table method where x  [0  ;360  ] Method: Draw up a table using x –values of [ 0º ; 30º ; 60º ; 90º ;120º ; 150º ; 180º ;210º ; 240º ; 270º ;300º ; 330º ; 360º] Use your calculator to get the corresponding y – values Plot these points on a Cartesian plane and make an accurate sketch joining the points in a smooth curve shape. Example 1: Sketch the graph of f ( x)  sin x where x  [0  ;360  ] X y = sinx

0º 0

90º 1

180º 0

270º -1

360º 0

0

90 

180

270

360

f x  = sin x 

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Exercise 4.4: Copy and complete the following table and sketch the graph of y  sin x 0º 90º 180º 270º 360º 0 -1

1. X y = sinx y 2

x 0 90

180

360

270

-1

-2

Copy and complete the following table and sketch the graph of y  cos x Decimals to nearest 1 decimal place

0º 1

y  cos x

90º 0

180º

270º

360º 1

y 2

x 0 90

-1

-2

180

270

360

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3. X y = sinx +1 y  sin x  1

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Copy and complete the following table and sketch the graph of y  sin x  1 0º 90º 180º 270º 360º 0 1 0 -1 0 +1 +1 +1 +1 +1

y 2

x 0 90

180

360

270

-1

-2

Copy and complete the following table and sketch the graph of y  cos x  1

X y = cos x +1

0º 1 +1

90º 0 +1

180º +1

270º 0 +1

360º +1

y  cos x  1 y 2

x 0 90

-1

-2

180

270

360

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Copy and complete the following table and sketch the graph of y  sin x  1

X y = sinx -1 y  sin x  1

0º 0 -1

90º -1

180º 0 -1

270º -1

360º 0 -1

y 2

x 0 90

180

270

360

-1

-2

Copy and complete the following table and sketch the graph of y  cos x  1 0º 90º 180º 270º 360º

x y = cosx -1

y  cos x  1 y 2

x 0 90

-1

-2

180

270

360

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Copy and complete the following table and sketch the graph of y   cos x

X y = -cosx

0º -1

90º

180º 1

270º

360º -1

y 2

x 0 90

180

270

360

-1

-2

8. X y = -sinx

Copy and complete the following table and sketch the graph of y   sin x 0º 90º 180º 270º 360º 0 -1 1 0

y 2

x 0 90

-1

-2

180

270

360

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Copy and complete the following table and sketch the graph of y   cos x  1

x y = -cosx +1 y   cos x  1

0º

90º

180º

270º

360º

y 2

x 0 90

180

360

270

-1

-2

10.

Copy and complete the following table and sketch the graph of y   sin x  1

x y = -sinx +1 y   sin x  1

0º

90º

180º

270º

360º

y 2

x 0 90

-1

-2

180

270

360

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11. X y = tanx

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Copy and complete the following table and sketch the graph of y  tan x 0º 0

90º ∞

180º

270º ∞

360º 0

y 2

x 0 90

-1

-2

180

270

360

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DATA HANDLING:

Statistics : A branch of mathematics concerned with collection, interpretation and analyses of data. “The science of making decisions in face of uncertainty” Data: Name given to the collection of information usually expressed in numerical form. Data Handling involves: 

Collecting data for a particular purpose.



Sorting data.



Representing data in tables, charts or graphs.



Analyzing the results.



Coming to conclusions.

Once collected data must be “interpreted” or analyzed . Only then can conclusions be made . Interpreting is Pictoral or Arithmetic. 

Pictoral Methods: Graphs: Bar; Histograms; Frequency Polygons; Pie Charts; Line and Broken Line graphs.



Arithmetic:



Measures of Central Tendency: Mean; Median Mode.



Measures of Dispersion: Range; deciles; percentiles; quartiles and interquartile range.

5.1

Displaying Data: 

A Bar Graph is a diagram consisting of a series of columns(bars) that are parallel(vertical or horizontal) that the length show frequency.

Example: 80 households surveyed & number of children in each household. Number of children Per family 0 1 2 3 4 5

Frequency 8 14 20 17 10 11

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Vertical Bar Graph depicts the data:

Number of children per family 80 households f r e q u e n c y

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

How many learners in the class?

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

Draw a bar graph to show the results.

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

Compound Bar Graphs:

Dual bar graphs : Used when two different sets of information given for connected topics: Example: 20 people recorded which TV station they watched at 8:15 p.m. on two consecutive nights in July 2004. First Night Second Night TV1 6 7 TV2 4 1 TV3 6 6 M-NET 3 4 E-TV 1 2

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Dual Bar Graph :

f r e q u e n c y

TV 1

TV 2

TV 3 TV Stations

MNET

E - TV

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Sectional Bar Graphs:

Airport survey: 100 passengers traveling overseas - “which countries were traveled to” Passengers USA Male 12 Female 6 Total 18

UK 15 17 32

GERMANY HOLLAND INDIA 8 5 4 3 11 5 11 16 9

AUSTRIA 9 5 14

f r e q u e n c y

USA

GER HOLL COUNTRIES

IND

AUST

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Exercise 5.2:

Ave Temp in ยบC

Jan July

The Average day-time temperature for 9 provincial capital cities were recorded. Eastern Free State Gauteng KwaZulu- Mpuma- Northern Limpopo North west Cape BloemNatal langa Cape fontein JHB PMB KimberPoloMafikeng Bisho Nelspruit ley kwane 22 14

23 8

20 10

23 13

24 15

25 11

23 12

24 12

Task: 1.

Draw a dual bar graph to illustrate the above information.

Which province has the greatest difference in temperature between January and July?

Western Cape Cape Town 21 12

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Why would somebody need to know the average temp in ºC of the various cities in Jan and July?

Water Land-Habit Land – cold Land – Mountainous Land – Dry

Calculation of % of 360º

Angle written correct to whole no

70 11 6 4 9 100

252º 39.6º 21.6º 14.4º 32.4º 360º

252º 40º 22º 14º 32º 360º

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

Copy and complete the table:

Go to University

No of learners 80

Go to Technikon

Get a job

Don‟t Know

Total

240

Calculation 80  360  240

Angle 120º

64  360  240 15º ……

Illustrate this information on a pie chart.

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Illustrate the same information as a bar graph.

Compare the two displays and identify:

One feature that the pie chart shows better. One feature that the bar graph shows better. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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Broken Line Graphs.

Can Help:  Find Patterns or Trends  Spot relationships between data  Find & predict values not given in the data Example: Norma is in hospital. Every 3 hrs her temperature is taken and the points are plotted on a graph. When data is collected in time intervals it is usual to plot them in this way.

39.5

39 38.5

38 37.5

37 0:00

3:00

6:00

9:00

12:00

15:00

18:00

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Misleading Statistics:

Statistics are notorious for being a way that people use to make false and misleading arguments. 1. 2.

Statistics can be made up. Charts and graphs are drawn so that they also mislead.

CHECK: 1. Does it have a title. 2. Are both axes labeled. 3. Are the units included. 4. Are the scale units equal distances apart. 5.7

Measures of Central Tendency(M.O.C.T.)

Measures of central tendency of ungrouped data If one number can be compared with another ---- then this is called a measure of central tendency. 3 different Measures Of Central Tendency are: 1. Mean The average of the scores. 2.

Mode The most repeated score.

Median ( Midlemost) in a ordered set of data.

5.8

Frequency Tables:

NB Must use all the data when finding one of the M.O.C.T‟s. Example: Value Frequency

3 1

4 4

5 3

6 2

Mode = 4 4,5 Median = 4,5 [3 4 4 4 4 5 5 5 6 6] Mean = 4,6 5.9

Discrete and continuous data. 1.

Discrete Data: Information collected by “counting”. Number of children cannot include fractions. But scores can include fractions. Continuous Data: Collected by “measurement” Suitable degree of accuracy( not always exact.)

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Stem & Leaf Diagrams

Example: Heights in cms. 147; 145; 157; 159; 154; 164; 161; 164; 162; 166; 163; 162; 171; 172; 171 Stem 14 15 16 17

Leaf 57 579 1223446 112

Tally Table Hts in cms 140 – 149 150 – 159 160 – 179 180 – 189

// /// ///// // ///

Measures of central tendency (MOCT) from stem and leaf diagrams. Median & Mode. Example: 33; 57; 46; 28; 68; 32; 60; 65; 54; 54; 40; 46; 45 ;26. Stem 2 3 4 5 6

Leaf 68 23 0566 447 0589

There are 2 modes (46) and (54) thus Bimodal Number: 15 data items. Median is 46 Mean = 48,2

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Grouped Data:

If dealing with many values it is easier to group data into sub sets. Groups of „5‟ as an example. 1–5 6 – 10 11 – 15 16 – 20

Measures of Central Tendency in Grouped Data: Example: Width of leaf (mm) No of leaves

11 – 20

21 - 30

31 – 40

41 - 50

51 – 60

Mode: Modal class is the class (group) with the highest frequency e.g. 31 – 40 above. Mean: To find the mean we must find a value to represent each class thus take the average of each class by adding 1st and last terms. 11  20 e.g  15,5 2 Width of leaf (mm) Mid point No of leaves

11 – 20

21 - 30

31 – 40

41 - 50

51 – 60

15.5 2

25.5 6

35.5 8

45.5 5

55.5 4

sum of midpo int s no of leaves 917.5 mean  25 mean  36.7 mean 

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HISTOGRAMS:

Bar graph used for Discrete Data. Histogram used for Continuous Data.

F r e q u e n c y 1-5

6-10 11-15 16-20 21-25 26-30 31-35

Number of items bought

5.13

FREQUENCY POLYGONS:

Histograms and Frequency Polygons are frequency graphs. Histograms-----Bar graph. Frequency polygon-----Line graph. A frequency Polygon is a 2 dimensional shape made of line segments joined up. NB: We can use the midpoints of the bars of a histogram as the points for the frequency polygon. Donâ&#x20AC;&#x;t start the histogram at zero BUT leave a space then start the bars. The Frequency polygon will then start and end on the bottom axis.

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MEASURES OF DISPERSION.

RANGE: This is a simple measure of the spread of data. Range = Maximum Value minus Minimum Value. Range cannot be used for grouped data as it ignores the distribution of values lying between the maximum and minimum values. Example: 2 classes scores in a maths test are as follows: Class 1:

22; 48; 52 ; 54; 64; 66; 76; 86; 88; 92; 100. Range = 100 â&#x20AC;&#x201C; 22 = 78

Class 2:

56; 58; 609; 62; 66; 66; 78; 78; 78; 80; 82. Range = 82 â&#x20AC;&#x201C; 56 = 26

Class 1 :

Scattered between 22 and 100

8 4 2 6 4 6 8 6 2

1 2 3 4 5 6 7 8 9 10

Class 2: Bunched between 56 and 82.

6 0 8 0

8 2 6 6 8 8 2

Measures of Dispersion: A Median is sometimes a better measurement of Central Tendency than the mean. Median divides the ordered data into 2 halves. Further subdivisions are : Quartiles: Points that divide into quarters. Deciles: points that divide into tenths Percentiles: points that divide into hundredths

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There are 3 quartiles. Q1 – Lower quartile Q2 – Median (M) Q3 – Upper Quartile Percentiles are first and second.

75 for a test. To compare her marks with the rest of her class we can 90 calculate the percentile in which her mark falls. Arrange all marks in descending order. Meg gets

100 learners 80% got better than 75  thus Megs mark is in the 80th percentile. If a mark is in the 99th percentile then 99% of the marks are below that



value. If a mark is in the 60th percentile then 60% of the learners scored lower



than this mark. The lower quartile is 25 percentile. The upper quartile is 75 percentile. Study of the values of quartiles, deciles and percentiles give us an idea of the spread of data.

5.15

CALCULATING QUARTILES:

22; 48; 52; 54; 64; 66; 76; 88; 92; 100 82. Q1

Q3 Dividing data into 4 quartiles: ± 25% data below Q1 ±

50% data below M.

± 75% data below Q3 Class 1: ± 25% marks less than 52 ± 50% marks less than 66

56; 58; 60; 62; 66; 66; 78; 78; 78; 80;

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± 75% marks below 78. Class 2: ± 25% marks less than 60 ± 50% marks less than 66 ± 75% marks below 78 INTER-QUARTILE RANGE: Lie between the two quartiles (lower and upper) The Inter- quartile range = Q3 – Q1 It measures a better dispersion than the range as it is not affected by extreme values. It is based on middle half of the data. i.e 50% of the data lies here. Example; Class 1: IQR : 88 -52 = 36 Class 2: IQR : 78 – 60 = 18 Class 2 has a smaller IQR which tells us that marks are not as spread out around the median as for class 1. Semi Inter-quartile range: It is half the IQR. Q  Q1 SIQR = 3 2 Values for Quartiles are not necessarily part of the data given. e.g. Class 3:

20; 39; 40; 43; 43; 46; 53; 58; 63; 70; 75; 91; . Q1

46  53  49,5 2 40  43 Q1   41,5 2 63  70 Q3   66,5 2 Range = 91 – 20 = 70 M

Range shows marks widely spread 25% lie below Q1 = 41,5 50% lie below Q2 = M = 49,5 75% lie below Q3 = 66,5 IQR = 66,5 – 41,5 = 25 25  12,5 SIQR = 2

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Exercise 5.4: 1.

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

Trained: 121 137 120 118 Untrained: 135 142 134 139 1.1

131 125

135 134

130

128

130

126

132

127

129

126 140

147 142

145

156

152

153

149

145

144

Draw a back â&#x20AC;&#x201C; to â&#x20AC;&#x201C; back stem & leaf diagram to show the two sets of data.

Find the medians and quartiles for both sets of data.

Find the Inter-quartile Range for both sets of data.

____________________________________________________________________ ___________________________________________________________________

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Comment on the results.

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

Frequency 2 5 10 17 19 15 4 2 1

Draw a histogram to illustrate this data: ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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2.2 The heights, measured to nearest cm, of 75 boys picked at random at Glory High School, Are shown on the following frequency table: Height (h) in cm Frequency 0 135  h < 140 1 140  h < 145 7 145  h < 150 11 150  h < 155 15 155  h < 160 18 160  h < 165 15 165  h < 170 6 170  h < 175 2 175  h < 180 On the same set of axes as the histogram draw a frequency polygon to illustrate this data. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

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Use the frequency tables and the two graphs to help answer the following questions. For each group of learners state: 2.3.1 The modal class.

The median height.

The lower quartile

The upper quartile

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Exercise 5.5: 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.

The following marks were recorded for a maths class:

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45 75 84

54 53 75 63

46 58 75 92

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91 44 81 78 67

22 60 60 68

28 54 37 68

37 71 56 69

56 71 25 76

45 44 90 98

2.1 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.

2.4 Calculate:

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the interquartile range.

2.4.2

the semi-interquartile range.

2.4.3

the range for the class.

2.4 Write down the maximum and minimum scores.

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Do a box and whisker diagram using the five-number summary (L;Q1;M;

Q3;H) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ 3.

34 28 34 3.1

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

15 12 42 65

34 45 23 28

22 65 50 19

56 33 54 39

23 24 25 32

22 9 8 40

20 18 20 31

Do a stem and leaf diagram for the data.

3.2

Find the median, mode and mean for the data.

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Find the lower and upper quartile.

Calculate: 3.4.1

the interquartile range.

3.4.2

the semi-interquartile range.

3.4.3

the range for the class.

3.4 Write down the maximum and minimum scores.

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

(L;Q1;M; Q3;H) ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4. Girls 34 Boys 75 4.1

The following marks for a class of Girls and Boys were recorded : 72 85 77 72

65 92 42 65

44 90 85 68

72 65 50 79

66 63 74 89

80 54 65 62

58 55 85 70

70 72 80 71

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

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Find the median, mode and mean for both sets of data

4.3

Find the lower and upper quartile of each set of data

4.4 Calculate: 4.4.1

The inter-quartile ranges for:

4.4.1.1

girls

4.4.1.2

boys

class

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The semi-interquartile ranges for:

4.4.2.1 girls ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4.4.2.2 boys ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________ ____________________________________________________________________

4.4.2.3

class

4.4.3

The ranges for:

4.4.3.1 girls ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 4.4.3.2

boys

class

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4.5

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Write down the maximum and minimum scores of each set of data

4.6

Do separate box and whisker diagrams for the girls and the boys

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

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5. 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

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

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

fX 67.5

5.1 Complete the last column of the table i.e (fX) 5.2 Find the modal class ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 5.3 Find the median class ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 5.4 Find the interval where Q1 and Q3 lie. ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ 5.5 Calculate the estimated mean. ď&#x192;Ľ fX NB estimated mean = n ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________

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100

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5.6 Use the grouped data to display the data on a histogram ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ________________________________________________________________ ___________________________________________________________________ 5.7 Draw the relevant frequency polygon on the histogram.

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101

Volume and Surface Area of 3-D Shapes Formulae: Volumes of Rectangular Prisms:

V  (area of base)  height 1. Square Base: V  (side of base)2  height (length) V  length  breadth  height (length 2. Rectangular Base : 1 3. Trapezium Base : V  sum of parallel sides   height (length) 2 1 4. Triangular Base : 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: 4 V   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

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102

Net : Square base prism:

s sxh

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s sx s

s sxh

sxh

s sx s Surface area = 4(h x s) + 2( s x s) Net of a Cylinder

h 2r

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

Sh  h 2  r 2

Cone:

Surface Area = r h 2  r 2 Circumference = 2  r OR Surface area =

1 circumference x slant height 2

NB Slant height =

h2  r 2

Slant height Net Diagram of a cone:

Arc length

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103

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

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

S  bh  3bl

 If triangle is isosceles.  If equilateral.

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) OTHER 3- D SHAPES Cone:

CurvedSurface   r h 2  r 2 or 1 CurvedSurface  circumference  slant height 2 Sphere: S  4 r 2 Tetrahedron: 1 Vol = (Area of base x height) 3 4 3 1  (base) 2 Surface Area = 2 2

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104

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Pictures of Different 3-Dimensional Shapes:

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.

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105

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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 Exercise 6.1: 1. Calculate the volume and surface area of the following closed prisms:

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

T 36 14 58

VolumeR VolumeS 2.4 VolumeS VolumeT ________________________________________________________________ 2.3

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106

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SurfaceAre aP SurfaceAre aQ 3.2 SurfaceAre aQ SurfaceAre aR ________________________________________________________________

3.1

3.3

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. ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________

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107

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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 : 6.1 to reduce the dimensions of the prisms.

6.2

To enlarge the dimensions of the prisms.

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108

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

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1 , then calculate 2

the volume and surface area of the new prism X. ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 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.

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

Grade 10 Core Mathematics

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109

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) 24 48 48 48 96 96 96 192

Vxk

V Vx2 Vx2 Vx2 Vx4 Vx4 Vx4 Vx8

Factor No of Sides Doubled 0 k 1 k 1 k 1 K2 2 2 K 2 k2 2 3 k 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 Prism L (cm) b (cm) h (cm) Volume Vxk 3 (cm ) A 4 3 2 24 V B 4 3 6 72 Vx3 C 4 9 2 72 Vx3 D 12 3 2 72 Vx3 E 12 9 2 216 Vx9 F 8 3 4 216 Vx9 G 4 9 6 216 Vx9 H 12 9 6 648 V x 27

Factor k k k k2 k2 k2 k3

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10.

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110

Copy and complete the following tables:

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 A B C D E F G H

L (cm) 6 12 6 6 12 12 4 12

b (cm) 4 4 8 4 8 4 8 8

h (cm) 3 3 3 6 3 6 6 6

Volume (cm3) 72

Vxk V Vx2

Factor k k k2

Volume of prism = l  b  h Factor k =______ Prism A B C D E F G H

L (cm) 4 4 4 12 12 4 12 12

b (cm) 3 3 9 3 9 9 9 9

h (cm) 2 6 2 2 2 6 6 6

Volume (cm3) 24 72 216 216

Vxk Factor V Vx3 Vx Vx Vx 9 Vx Vx V x 27

K k2 K3

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111

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Key Ideas: The circumference of a circle is proportional to its area: arc length sector area sector  circumfere nce of circle area circle

To find the area of a sector of a circle one must first find the ratio:

arc . circumfere nce

Multiply this ratio by  r2 to get the area of the sector.

Area of a base that is a hexagon shape: It takes 6 equilateral triangle shapes to form 3 3.a a OR Area  the hexagon base : Area  ( where „a‟ is the base of the 2 2 equilateral triangle) height  3.a    height Volume of Prism with Hexagon Base = 6   2 

a ( base of ) NB: The perpendicular height of any equilateral triangle with side „a‟ is

3  a by 2

Pythagoras.

Exercise continued: 11.

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). The writing on the can says that it contains 200 ml of liquid. How much air space is there in the can? ( 1ml  1cm 3 ) ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ________________________________________________________________

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112

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11.2 What is the height of the liquid in the can? ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ 12.1

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

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12.2

113

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

he manufacturer of Lemon Twist wants to double the volume of the can , but keep the radius as it is. By which factor must the height be increased?

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114

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14.1 If the radius is increased by a scale factor of 2, but the height is kept the same by which factor will the volume increase?

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115

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14.3 By which factor will the area of the lateral surface ( the area of the curved side) increase? ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ ___________________________________________________________________ ____________________________________________________________________ ___________________________________________________________________ __________________________________________________________________