Gr10 Via Afrika Mathematics Gr10 Study Guide LR

Via Afrika Mathematics Grade 10 Teacher’s Guide To me, the most important thing about teaching is that you lead the development of the learner in so many different ways. This is why my lessons open Grade 10 Study Guide up to so much more. M. Pillay, L.J. Schalekamp, G. du Toit, C.R. Smith, L. Bruce, M.L. Botsane, — Popsy Naidoo, Teacher J. Bouman, N.S. Muthige, S.F. Carletti Via Afrika understands, values and supports your role as a teacher. You have the most important job in education, and we Via Afrika Mathematics realise that your responsibilities involve far more than just teaching. We have done our utmost to save you time and make your life easier, and we are very proud to be able to help you teach this subject successfully. Here are just some of the things we have done to assist you in this brand-new course: 1. The series was written to be aligned with CAPS. See page XX to see how CAPS requirements are met. 2. A possible work schedule has been included. See page XX to see how much time this could save you. 3. Each topic starts with an overview of what is taught, and the resources you need. See page XX to ﬁnd out how this will help with your planning. 4. There is advice on pace-setting to assist you in completing all the work for the year on time. Page XX shows you how this is done. 5. Advice on how to introduce concepts and scaffold learning is given for every topic. See page XX for an example. 6. All the answers have been given to save you time doing the exercises yourself. See page XX for an example. 7. Also included are a full-colour poster and a CD ﬁlled with resources to assist you in your teaching and assessment. See the inside front cover. 8. A question bank with tests you may photocopy will help you assess your learners effectively. See the Question Bank on page XX. The accompanying Learner’s Book is written in accessible language and contains all the content your learners need to master. The exciting design and layout will keep their interest and make teaching a pleasure for you. We would love to hear your feedback. Why not tell us how it’s going by emailing us at maths@viaafrika.com? Alternatively, visit our teacher forum at www.viaafrika.com. Language: English www.viaafrika.com

Authors • Authors • Authors • Authors S tu d y G u i d e Mathematics Grade 10 • Via Afrika Maths Study Guide Gr10 1 16/01/2012 12:43

Contents Introduction to Mathematics.......................................................................................5 Chapter 1 Algebraic expressions�� 6 Overview��6 Unit 1 The number system��7 Unit 2 Multiplying algebraic expressions��8 Unit 3 Factorisation��10 Unit 4 Algebraic fractions��13 Questions��15 Chapter 2 Exponents�� 16 Overview��16 Unit 1 Laws of exponents��17 Unit 2 Simplifying exponents��18 Unit 3 Solving equations that contain exponents��20 Questions��23 Chapter 3 Number patterns�� 24 Overview��24 Unit 1 Linear number patterns��25 Unit 2 More complicated linear number patterns��27 Questions��29 Chapter 4 Equations and inequalities�� 30 Overview��30 Unit 1 Linear equations��31 Unit 2 Quadratic equations��34 Unit 3 Literal equations��35 Unit 4 Simultaneous equations��37 Unit 5 Linear inequalities��38 Questions��40 Chapter 5 Trigonometry�� 41 Overview��41 Unit 1 Right-angled triangles��42 Unit 2 Definitions of the trigonometric ratios��43 • Via Afrika Maths Study Guide Gr10 2 16/01/2012 12:43

Unit 3 Special triangles��44 Unit 4 Using your calculator��45 Unit 5 Solving trigonometric equations��46 Unit 6 Extending the ratios to 0° ≤ θ ≤ 360°��48 Unit 7 Graphs of the trigonometric functions��53 Questions��55 Chapter 6 Functions�� 57 Overview��57 Unit 1 What is a function?��58 Unit 2 Graphs of functions��60 Unit 3 The graph of y = ax2 + q ��63 Unit 4 The graph of y =  __ax + q��64 Unit 5 The graph of y = abx + q, b > 0 and b ≠ 1��66 Unit 6 Sketching graphs��68 Unit 7 Finding the equations of graphs��71 Unit 8 The effect of a and q on trigonometric graphs��72 Questions��74 Chapter 7 Polygons�� 75 Overview��75 Unit 1 Similar triangles��76 Unit 2 Congruent triangles��78 Unit 3 Quadrilaterals��80 Questions��83 Chapter 8 Analytical geometry�� 84 Overview��84 Unit 1 The distance formula��85 Unit 2 The gradient formula��86 Unit 3 The midpoint formula��88 Questions��89 Chapter 9 Finance, growth and decay�� 91 Overview��91 Unit 1 Simple interest��92 Unit 2 Compound growth��94 Questions��97 • Via Afrika Maths Study Guide Gr10 3 16/01/2012 12:43

Chapter 10 Statistics�� 98 Overview��98 Unit 1 Measures of central tendency: ungrouped data��99 Unit 2 Measures of dispersion��101 Unit 3 Box-and-whisker plot��103 Unit 4 Grouped data��104 Questions��107 Chapter 11 Measurement�� 110 Overview��110 Units 1 and 2 Right prisms and cylinders��111 Unit 3 The volume and surface area of complex-shaped solids��113 Questions��116 Chapter 12 Probability�� 117 Overview��117 Unit 1 Probability��118 Unit 2 Combination of events��120 Questions��122 Exam Papers�� 123 Answers to questions�� 133 Answers to exam papers�� 152 Glossary�� 162 • Via Afrika Maths Study Guide Gr10 4 16/01/2012 12:43

Introduction to Mathematics There once was a magnificent mathematical horse. You could teach it arithmetic, which it learned with no difficulty, and algebra was a breeze. It could even prove theorems in Euclidean geometry. But when it tried to learn analytic geometry, it would rear back on its hind legs, and make violent head motions in resistance. The moral of this story is that you can’t put Descartes before the horse. If you have a study routine that you are happy with and you are getting the grade you want from your mathematics class you might benefit from comparing your study habits to the tips presented here. Mathematics is Not a Spectator Sport In order to learn mathematics you must be actively involved in the doing and feeling of mathematics. Work to Understand the Principles ●● LISTEN During Class. No the study guide is not enough. In order to get something out of the class you need to listen while in class. Sometimes important ideas will not be written down, but instead be spoken by the teacher. ●● Learn the (proper) Notation. Bad notation can jeopardise your results. Pay attention to them in the worked examples in this study guide, ●● Practise, Practise, Practise. Practise as much as possible. The only way to really learn how to do problems is work through lots of them. The more you work, the better prepared you will be come exam time. There are extra practise opportunities in this study guide. ●● Persevere. You might not instantly understand every topic covered in a mathematics class. There might be some topics that you will have to work at before you completely understand them. Think about these topics and work through problems from this study guide. You will often find that, after a little work, a topic that initially baffled you will suddenly make sense. ●● Have the Proper Attitude. Always do the best that you can. The AMA of Mathematics ●● ABILITY is what you’re capable of doing. ●● MOTIVATION determines what you do. ●● ATTITUDE determines how well you do it. It is not pure intellectual power that counts, it’s commitment. – Dana Scott © Via Afrika – Mathematics 5 • Via Afrika Maths Study Guide Gr10 5 16/01/2012 12:43

Chapter 1 Algebraic expressions Overview In this unit, we discuss real numbers, which are divided into rational and irrational numbers. Here, you will also learn about surds, and how to round off real numbers. You will also learn how to multiply integers, monomials and binomials by a polynomial. Finally, we discuss factorisation and how to work with algebraic fractions. UNIT 1 Page 7 • Real numbers The number system • Surds • Rounding off UNIT 2 Page 8 • Multiply integers and monomials by Multiplying algebraic polynomials expressions • The product of two binomials • Multiplying a binomial by a trinomial • The sum and difference of two cubes CHAPTER 1 Page 6 Algebraic expressions UNIT 3 Page 10 • Common factors Factorisation • Difference between two squares • Perfect squares • Trinomials of the form x2 + bx + c • Trinomials of the form ax2 + bx + c • Factorising by grouping • Factorising the sum and difference of cubes UNIT 4 Page 13 • Simplifying fractions Algebraic fractions • Products of algebraic fractions • Adding and subtracting algebraic fractions 6 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 6 16/01/2012 12:43

Unit 1 The number system 1.1 Real numbers ●● Real numbers are divided into rational and irrational numbers. Natural Whole Rational Integers (Z) numbers (N) numbers (N0) numbers (Q) Real numbers (R) Negative Fractions of the form Irrational Zero   ba__, a, b ∊ Z, b ≠ 0 numbers numbers (Qʹ) ●● We can write a rational number as a fraction,  ba__, where a and b ∈ Z, and b ≠ 0. ●● We cannot express an irrational number as a fraction. 1.2 Surds ●● A surd is a root of an integer that we cannot express as a fraction. ●● Surds are irrational numbers. __ __ ●● Examples of surds are √3 and √2 . 1.3 Rounding off ●● If the number after the cut-off point is 4 or less, then we leave the number before it as it is. ●● If the number is equal to 5 or more, then increase the value of the number before it by 1. © Via Afrika – Mathematics 7 • Via Afrika Maths Study Guide Gr10 7 16/01/2012 12:43

Unit 2 Multiplying algebraic expressions 2.1 Multiply integers and monomials by polynomials ●● Each term inside the bracket is multiplied by the term in front of the bracket. –  1__6 a2b(6abc – 12ac + 18b) = (– 1__6 a2b ¿ 6abc) – (– 1__6 a2b ¿ 12ac) + (–  61__a2b ¿ 18b) = –a3b2c + 2a3bc – 3a2b2 2.2 The product of two binomials ●● Each term inside the first set of brackets is multiplied by each term inside the second set of brackets. (a – 3b)(a + 7b) = (a × a) + (a × 7b) – (3b × a) – (3b × 7b) = a2 + 7ab – 3ab – 21b2 = a2 + 4ab – 21b2 ●● When squaring a binomial: (m + n)2 = (m + n)(m + n) A common error is to think that = (m × m) + (m × n) + (n × m) + (n × n) (m + n)2 = m2 + n2 = m2 + mn + nm + n2 = m2 + 2mn + n2 The square of any binomial produces the following three terms: 1 The square of the first term of the binomial: m2 2 Twice the product of the two terms: 2mn 3 The square of the second term: n2 2.3 Multiplying a binomial by a trinomial Multiply each term in the first set of brackets by each term in the second set of brackets. Then we simplify by collecting the like terms. (8 – 3y)(12 – 2y + 8y2 – 4y3) = (8 × 12) + (8 × –2y) + (8 × 8y2) + (8 × –4y3) + (–3y × 12) + (–3y × –2y) + (–3y × 8y2) + (–3y × –4y3) = 96 – 16y + 64y2 – 32y3 – 36y + 6y2 – 24y3 + 12y4 = 96 – 52y + 70y2 – 24y3 + 12y4 8 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 8 16/01/2012 12:43

Unit 2 2.4 The sum and difference of two cubes ●● The expression in the following form gives the difference of two cubes: (a – b)(a2 + ab + b2) = a3 + a2b + × ab × 2 – a2b – × ab × 2 – b3 = a3 – b3 ●● The expression in this form gives the sum of two cubes: (a + b)(a2 – ab + b2) = a3 – a2b + ab2 + a2b –ab2 + b3 = a3 + b3 © Via Afrika – Mathematics 9 • Via Afrika Maths Study Guide Gr10 9 16/01/2012 12:43

Unit 3 Factorisation Factorisation is the opposite process to the one you learnt in the previous unit when multiplying out algebraic expressions. expand y(x + 5) xy + 5y factorise 3.1 Common factors When finding a common factor, we find and take out a factor that can divide into each term in the expression. For example, in the expression 6h3 +18h2, 6h2 can divide into each term, so it becomes the common factor: 6h3 ÷ 6h2 = h 18h2 ÷ 6h2 = +3 Therefore: 6h3 +18h2 = 6h2(h + 3) 3.2 Difference between two squares A difference of two squares occurs when we have two perfect squares separated by a minus sign. An expression in the form a2 – b2 has two factors, (a – b)(a + b). For example: ____ 2 4 2 4a – 81b = (2a + 9b )(2a – 9b ) 2 √_____ 4a2 = 2a √81b4 = 9b2 3.3 Perfect squares In Unit 2 you learnt that (x + y)2 = x2 + 2xy + y2. Therefore, a trinomial will factorise into a perfect square if: ●● the first and last terms are perfect squares (e.g. x × x = x2) ●● the middle term is equal to 2 × first term × last term. For example: x2 + 10x + 25 = (x + 5)(x + 5) = (x + 5)2 10 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 10 16/01/2012 12:43

Unit 3 3.4 Trinomials of the form x2 + bx + c When factorising a trinomial: ●● If the third term of the trinomial is positive, we use the sum of the factors of the third term to give us the middle term. ●● If the third term of the trinomial is negative, we use the difference between the factors of the third term to give us the middle term. Use the following rules to decide on the signs in the binomials: ●● If the third term of the trinomial is positive, then the signs between the terms in the binomials are the same (both positive or negative). ●● If the third term of the trinomial is negative, then the signs between the terms of the binomials are different. For example: x2 + 3x + 2 = (x + 2)(x + 1) 3.5 Trinomials of the form ax2 + bx + c In this case, the rules are the same as in the previous section, except that now we need to consider the factors of the coefficient of x2 and the last term. For example: 6x2 + x – 15 ●● Multiply the coefficient of the squared term (6) by the last term (–15) = –90. ●● Look for the factors of –90 that will give +1 (the coefficient of the middle term) when added: –9 × 10 = –90 and –9 + 10 = 1 Remember to change ●● Rewrite 6x2 + x – 15 as 6x2 +10x – 9x – 15. the sign inside the 2 brackets when you ●● Group the terms: (6x +10x) – (9x + 15) divide by –1. ●● Take out common factors from each set of brackets: 2x(3x + 5) – 3(3x + 5) = (3x + 5)(2x – 3) Therefore: 6x2 + x – 15 = 6x2 +10x – 9x – 15 = 2x(3x + 5) – 3(3x + 5) = (3x + 5)(2x – 3) 3.6 Factorising by grouping If an expression has four or more terms, but has no factor common to all of them, we can often group the terms, factorise each group, and then remove a common factor. © Via Afrika – Mathematics 11 • Via Afrika Maths Study Guide Gr10 11 16/01/2012 12:43

Unit 3 x3 – 3x2 + 2x – 6 = (x3 – 3x2) + (2x – 6) (Group terms together using brackets) = x2(x – 3) + 2(x – 3) (Factor out the common factors from each group) = (x – 3)(x2 + 2) 3.7 Factorising the sum and difference of cubes ●● The difference between_________ two cubes factorises as: a3 – b3 = (a – b)(a2 + ab + b2) ___________ 3 3 ●● The first factor is: (√ first term ) – (√ second    term ) ●● The second factor is: (Square of the first term) (Opposite sign) (Product of the two terms) + (Square of the last term) For example: y3 – 64 = (y – 4)(y2 + 4y + 16) ●● The sum of two cubes factorises as: a3 + b3 = (a + b)(a2 – ab + b2) For example: y3 + 64 = (y + 4)(y2 – 4y + 16) 12 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 12 16/01/2012 12:43

Unit 4 Algebraic fractions 4.1 Simplifying fractions Special fractions: Negative fractions: __   1a = a  –a – _ba = __b = __ a  –b –a   1__a =  1__a __=  a   –b _ b –a   a_a = 1 But: – _ba ≠ __  –b   0__a = 0 But: __   0a is undefined To simplify a fraction: 1 Factorise the expressions in the numerator and denominator 2 Cancel the terms common to both. (x2 – 4) (x + 2)(x – 2) For example: _____  (x – 2) = _________   (x – 2) = (x + 2) 4.2 Products of algebraic fractions Multiplying and dividing algebraic fractions works in the same way as ordinary fractions that contain numbers: ●● Factorise each expression. ●● Cancel any like terms. ●● Multiply what is left. 2 2 (4n – 9) _________ ______ (2n – n – 3) ___________ (2n + 3)(2n – 3) __________ (2n – 3)(n + 1)   (2n + 3) ×   =   (n + 1) ×   (2n + 3) (n + 1) (2n – 3) (2n – 3) = ______   1 × ______   1 = (2n – 3)2 Multiply each term in the first set of brackets by each term in the second set of brackets. Then we simplify by collecting the like terms. © Via Afrika – Mathematics 13 • Via Afrika Maths Study Guide Gr10 13 16/01/2012 12:43

Unit 4 4.3 Adding and subtracting algebraic fractions ●● As always, factorise and simplify where possible. ●● Find a common denominator. ●● Express each fraction in terms of that denominator. ●● Add and subtract the like terms. 3p – q _____ 2p – 3q 2x + 5 _____   2p +   4q   5 –   x – _____ ____ 2 3 3(2x + 5) – 5(x – 2) 2q(3p – q) + p(2p – 3q) =  _______________    =  _____________    15 4pq 6pq – 2q2 + 2p2 – 3pq 6x + 15 – 5x + 10 ____________ = ______________    =    15 4pq x + 25 2p2 + 3pq – 2q2 =  __________ = _____   15 4pq (2p – q)(p + 2q) =  __________ 4pq 14 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 14 16/01/2012 12:43

Questions 1 Multiplying expressions: a Square the binomial (3x − 4). b Expand: (2x − 5)(2x + 1) c Expand: (2a + b)(4a2 – 2ab + b2) d Find the products and simplify: (2a – 5b)(4a – 3b) – (a + 3b)(5a – 12b)  34 )(k – __ e Find the product: (k2 + __  12 ) 2 Factorise the following expressions: a 25x2 + 30x + 9 b 5x2 − 7x – 6 c x3 –  1__4 x d (2t – 5)3 – (2t – 5)2 e 3x3 + x2 − 3x – 1 f m3 – m2 – mn2 + n2 g (y + 3)(2y – 3) + (y – 1)(3 – 2y) h x2(2x – 1) – 2x(2x – 1) – 3(2x – 1) 3 Algebraic fractions: 2 a Simplify  xx2 –– ________ x–6 4x + 3 3 2  4x – 9x b Simplify to the lowest terms: _______ 4x3 + 6x2 2 c Simplify as far as possible:  x –47x ________ –x + 12 a(a + 1) + (a + 1) 2 2 d Simplify as far as possible: ____________    ¿ ____   a2 – 2a + 1   a a–2 1 ÷   aa____ 2 +a –a e ____ +   –4 Calculate and simplify:  x –1 _____ –   1 ____ 2 (x + 2)2 x+2 p–3 ____–   3 f Calculate and simplify:  ________ +   2 p2 – p – 12 3 + p _____ 8 – 2p © Via Afrika – Mathematics 15 • Via Afrika Maths Study Guide Gr10 15 16/01/2012 12:43

Chapter 2 Exponents Overview As you know, we can describe numbers in terms of their factors. For example, 125 = 5 × 5 × 5. We can also write this another way, in exponential form: 125 = 53. Here, 5 is the base, and 3 is the exponent. An exponent is also sometimes called an index or a power. In this chapter, you learn about the laws of exponents, how to simplify exponents and how to solve equations that contain exponents. UNIT 1 Page 17 • What is an exponent? Laws of exponents • The laws of exponents UNIT 2 Page 18 • The product rule: bm ¿ bn = b(m + n) Simplifying exponents • The quotient rule: bm ÷ bn = b(m + n) • Rules for 0 and 1 CHAPTER 2 Page 16 • Power rule: (bm)n = bmn • All inside brackets raised to a power: an Exponents (ab)n=anbn and (  __ba )n =  __ bn • Negative power rule: a–n = __   1 , provided an that a ≠ 0 UNIT 3 Page 20 • Solving exponential equations using Solving equations that factorisation contain exponents • Exponents containing fractions • Three common mistakes to avoid when working with exponents 16 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 16 16/01/2012 12:43

Unit 1 Laws of exponents 1.1 What is an exponent? In general, we define exponents as follows: xn = x × x × x × x × … × x, n ∈ N This definition states that x is multiplied by itself n times. Here, x is the base, and n is the exponent. 1.2 The laws of exponents ●● bm × bn = b(m + n) (The product rule) m n (m – n) Note that, for these ●● b ÷b =b where m ≥ n and b ≠ 0 (The quotient rule) rules, the base stays 0 ●● a = 1, when a ≠ 0 the same. ●● (bm)n = bmn (The power rule) n n n ●● (ab) = a b ●● a–1 = __  1a , when a ≠ 0 © Via Afrika – Mathematics 17 • Via Afrika Maths Study Guide Gr10 17 16/01/2012 12:43

Unit 2 Simplifying exponents In this unit, we revise how to use the laws of exponents to simplify expressions. 2.1 The product rule: bm × bn = b(m + n) The product rule tells you that, when multiplying powers that have the same base, you can add the exponents. Example 1 x5 × x3 = x5 + 3 = x8 2 23 × 2–2 = 23 + (–2) = 21 = 2 3 2 × 23x × 22 = 21 + 3x + 2 = 23x + 3 2.2 The quotient rule: bm ÷ bn = b(m – n) The quotient rule tells you that you can divide powers with the same base by subtracting the exponents. Example 5 1   x__ x = x 5–1 = x4 3 An exponent 2   2__ 25 = 23 – 5 = 2–2 is also called a 12a _______b 6 6 power. 3   –4a ×–2b3 = –3a6 – (–2)b6 – 3 = –3a8b3 2.3 Rules for 0 and 1 Rules Anything to the power zero x1 = x 1n = 1 10 = 1 is equal to 1, 121 = 12 14 = 1 × 1 × 1 × 1 = 1 x0 = 1, x ≠ 0 except 00, which is undefined. Example 1 16a0, a ≠ 0 Everything 2 16 × 1 = 16 inside a bracket to the power of 3 (2 + x)0 = 1 zero is always 4 (–3)0 – 30 = 1 – 1 = 0 [(–3)0 = 1 and 30 = 1] one. ( __ 3a ) ___0 +   b 0 5   3a 2   =   3__2 + 1 =  5__2 18 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 18 16/01/2012 12:43

Unit 2 2.4 Power rule: (bm)n = bmn The power rule tells us that to raise a power to a power, just multiply the exponents. Example 1 5(a2)3 = 52 × 3 = 56 Always express __2 __ 2 2 (27) 3= (33) 3= 32 = 9 answers 3 with positive  bc 3 (a–3.b 3.c)3 = a–3 × 3.b 3× 3.c1 × 3 = a–9.b.c3 = ___ __ 1 __1 a9 exponents. 2.5 All inside brackets raised to a power: (ab)n = anbn and ( _b )n = __ a an  bn Example 1 (5a2)3 = 53a2 × 3 = 125a6 Switch the 2 (2x2y)5 = 25x5 × 2y1 × 5 = 32x10y5 denominator and numerator __  x =  3 ( ) x x 3   3a b2 ___ a b2x and multiply. 4 2x.3x = (2.3)x = 6x 5 (2a6b)3(3a2b3)2 = (23a6(3)b3)(32a2(2)b3(2)) = (8a18b3)(9a4b6) = 72a18+4b3+6= 72a22b9 2.6 Negative power rule: a–n = ___ 1 a  –n , provided that a ≠ 0 Any non-zero number raised to a negative power equals its reciprocal raised to the opposite positive power. Example 2 1   2__ 3 _____ =   1 =   22 1 __ – 3 12 5 2   ab __ =   b 5 __ = b ×  a __ 1 __ = a5b 1   a5 x2 a2 2 x  __ 1 2 x  __ x2 x3 ___ ___ a2 __ 3   a3x1 =   3 1__ =  __ a2 __ __ x __ 3 =   2 +   3 =   5 a   x   a a a a x 3 2 3 4 3x 3 ____ 4   (3x) _____ =   1 2 =   ___ 9x2.x3 3x5  4a ____ b 3ab1 =   4b ___ 3a3 © Via Afrika – Mathematics 19 • Via Afrika Maths Study Guide Gr10 19 16/01/2012 12:43

Unit 3 Solving equations that contain exponents The key to solving equations that contain exponents is to ensure that the bases on both sides of the equation are the same. This enables us to compare the exponents, and so solve for the variable. The rule is: If ax = an, then x = n Example 1 3x = 81= 34 [Rewrite 81 in exponential form: 81 = 34] Therefore, x = 4 2 7x – 3 = 49 [Rewrite 49 in exponential form: 49 = 72] 7x – 3 = 72 x – 3 = 2 [Exponents are equal if bases are equal] x = 5 3 3.9x – 1 = 27–x [Make the bases the same by applying the laws of exponents] 3.32(x – 1) = 33(–x) 31 + 2x – 2 = 3–3x 32x – 1 = 3–3x 2x – 1 = –3x 2x + 3x = 1 5x = 1 x =  1__5 4 2x + 1 = 9 [Move terms without x to the right side of the equation] x 2 = 9 – 1 =8 2 = 23 x [8 = 23] Therefore, x = 3 according to the principle of same bases. 3.1 Solving exponential equations using factorisation Before we remove a common factor from an expression, we sometimes need to apply the laws of exponents in reverse. Example 1 2x + 1 = 2x = 24 [Use the reverse of the product rule: 2x + 1 = 2x × 21] 2x.2 + 2x = 24 [Take out a common factor of 2x] 2x(2 + 1) = 24 3.2x = 24 2x =  24 __ 3 20 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 20 16/01/2012 12:43

Unit 3 2x = 8 2x = 23 x=3 2 3x + 1 = 3x – 1 = 90 3x(31 + 3–1) = 90 [Take out a common factor of 3x] 3x(3 +  31__) = 90 3x (   10 3 ) __  = 90 3 __ 3x = 90  10 3x = 27 [The bases are the same, so exponents are equal] 3x = 33 x=3 3 2 x+1 = 2x + 4 [Move terms containing x to the left side of the equation] 2x + 1 – 2x = 4 2x(2 – 1) = 4 [Take out a common factor of 2x] 2x = 4(4 ÷ 1 = 4) 2x = 22 Therefore, x = 2 3.2 Exponents containing fractions Fractions often make expressions look more complicated, but we use all the same rules and laws of exponents. Example 5 __ 5 __ 1 4 2= (22) 2= 25 = 32 2  x98 (33x–12) = 32x–8 = __ [Remember to express exponents as positive numbers] __  1__3 1 3 __   13 3   27 = (3 ) = 31 = 3  x98 4 (27x12) 3= (33)2.(x12)2 = 32x8 = __ __ 2 y  2 __  2y  2 __  4 2a y 2y 2 .a ___ 2a 4a 2 4 5  ____ ______ ___ y y y =   y 2 =   1 =    __  b2  __  –  __  b b b2 y 3.3 Three common mistakes to avoid when working with exponents 3.3.1 Brackets The exponent next to a number only applies to that number. However, if the numbers are inside brackets, and the exponent lies outside the brackets, then the exponent applies to everything inside the brackets. For example: –32 ≠ (–3)2 –3 ¿ 3 ≠ (–3)(–3) © Via Afrika – Mathematics 21 • Via Afrika Maths Study Guide Gr10 21 16/01/2012 12:43

Unit 3 3.3.2 The product rule 1 The product rule only applies when the bases are the same. For example: 23 ¿ 32 ≠ 65 2 The product rule only applies when multiplying, not when adding. For example: 23 + 25 ≠ 28 3.3 Working backwards to go forwards Remember, we need the bases to be the same before we can apply the laws of exponents. Therefore, first write all bases as a product of their prime factors, and then apply the laws of the exponents to those bases that are the same. 27 = 33 81 = 34 32 = 25 36 = 22 × 33 40 = 23 × 51 Example 9 = 32 12 = 3 × 4 = 3.22 6=2×3 4 = 22 Therefore: n n+1 (32)n ¿ (3.22)n + 1   9 4¿ 12 _______ ¿ 6n =  ___________ 22 ¿ (2.3)n 2n n+1 2n + 2 3 ¿3 ¿2 =  ____________    22 ¿ 2n ¿ 3n = 32n + n + 1 – n ¿ 22n + 2 – 2 – n = 32n + 1 ¿ 2n 22 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 22 16/01/2012 12:43

Questions 1 Simplify the following expressions by applying the laws of exponents. All indices must be positive, and all variables represent positive rational numbers. 5x0 + 8 3 – ( __   21   ) .1x __ –2 –2 a (2an – 2)2 b –2 a 12x2.x4 – (–3x3)2 c   2a _____ 6a 3 d  ___________ 3x3 x+1 x–1 e   327 .9 –3 ________ ____ f  32 x x–1 –1 –2 .81x + 1 x a–1 a+1 4 .8 g  ________ ___– 3a0 + 27 3 – 1 3 1 __ 1 __ 2 2a – 2.6a + 1 h   –2   ___ 83 2a – b 4a i   x____ x __ x b – 2a ÷   2b x n–3 5 2 a Calculate b if b =  ____ 5n + 1 b If 5x = f, express 2.25x + 1 in terms of f. c If 3m = 5, find the value of i 3m + 2 ii 9–2m d Remove the brackets and simplify: (32x – 3)(32x + 3) 3 Solve the following equations: a 2x + 1 = 2 b 9x – 2 = 271 – 2x c 52x + 1 = 0,04 d 5x + 1 + 5x – 150 = 0 4 An astronomer studying a region of space needs to determine the volume of a cubic region. The edges of the region measure 3 × 108 miles long. Find the volume. © Via Afrika – Mathematics 23 • Via Afrika Maths Study Guide Gr10 23 16/01/2012 12:43

Chapter 3 Number patterns Overview A number pattern is a list of numbers. The numbers are in a definite order, and follow a fixed rule, or pattern. Examples of number patterns are: ●● Even numbers 2; 4; 6; 8; 10; 12; … ●● Odd numbers 1; 3; 5; 7; 9; 11; … ●● Square numbers 1; 4; 9; 16; 25; 36; 49; … ●● Cube numbers 1; 8; 27; 64; 125; 216; … ●● Fibonacci numbers 1; 1; 2; 3; 5; 8; 13; 21; 34; … In a number pattern: ●● the numbers in the sequence are called terms ●● each term has a position in the sequence. A linear number pattern is a special kind of number pattern, in which the difference between the terms is always a fixed value. UNIT 1 Page 25 • The general term of linear number Linear number patterns patterns CHAPTER 3 Page 24 Number patterns UNIT 2 Page 27 More complicated • Non-numeric sequences linear number patterns 24 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 24 16/01/2012 12:43

Unit 1 Linear number patterns 1.1 The general term of linear number patterns A linear number pattern has a constant difference between consecutive terms. For example, consider the pattern: 2; 5; 8; 11; … We obtain each term of the pattern by adding 3 to the previous term. Therefore, the difference between the second and first term is 3. The difference between the third and the second term is also 3, and so on. We represent the first term in the pattern by the letter a. We represent the constant difference between terms as d. Therefore, in our previous example: a=2 (first term) d=3 (constant difference) We define the first term (a) as T1, the second term as T2, the third term as T3, and so on. In our example, It is relatively easy to find the sixth term in the pattern (17), but it is not as easy to find, for example the 1 000th term. To do so, we need to develop a rule, or formula. The general formula to calculate the value of a term in a linear number pattern is: Tn = a + (n – 1) × d Therefore, the 1 000th term is: T1 000 = 2 + (999) × 3 = 2 999 Example The general term (Tn) of a sequence is T1 = 6(1) + 2 = 8 Replace n with “1” for Term 1 given by Tn = 6n + 2. Find the first four T2 = 6(2) + 2 = 14 Replace n with “2” for Term 2 terms. T3 = 6(3) + 2 = 20 T4 = 6(4) + 2 = 26 ∴ sequence is 8; 14; 20; 26; … Given the sequence 3; 7; 11; 15 … 3 ; 7 ; 11 ; 15 … Determine the formula for Tn. +4 +4 +4 Constant difference Tn = an + d a = constant difference Therefore: Tn = 4n + d Substitute n = 1: T1 = 4(1) + d = 3 ∴ b = –1 Therefore, Tn = 4n – 1 © Via Afrika – Mathematics 25 • Via Afrika Maths Study Guide Gr10 25 16/01/2012 12:43

Unit 1 Given the sequence 10; 2; –6; –14; … 10 ; 2 ; –6 ; –14 Find Tn. –8 –8 –8 Constant difference Tn = an + d a = constant difference Therefore: Tn = –8n + d Substitute n = 1: T1 = –8(1) + d = 10 b = 18 Therefore, Tn = –8n + 18 26 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 26 16/01/2012 12:43

Unit 2 More complicated linear number patterns 2.1 Non-numeric sequences Matches are used to build squares around a central area (in black), as shown below. Complete the table that shows how wide the squares are, and how many matches are used in each square. Number of matches wide 3 4 5 7 156 n Number of matches used 24 36 48 ? ? ? There is a constant difference of 12 between matches used. Therefore: 24 = 12(1) + d d =12 Tn = 12n + 12 = 12(n + 1) Term 7 = 12(7) + 12 = 96 matches (if the square is 27 matches wide) Term 156 = 12 (157) = 1 884 matches (if the square is 156 matches wide) Rapunzel created a chain by forming hexagons with matches. She predicts that the number of matchsticks needed to form a chain of 20 hexagons will be 121. To test this prediction, we need a rule: H1 = 6 matches H2 = 6 + 5 = 11 matches H3 = 6 + 5 + 5 = 16 matches © Via Afrika – Mathematics 27 • Via Afrika Maths Study Guide Gr10 27 16/01/2012 12:43

Unit 12 Note that 6 is the starting value and then we 5 each time, one less than the number of hexagons. Therefore: Hn = 6 + (n – 1) × 5 = 5n + 1 Apply the rule for 20 hexagons: 5(20) + 1 = 101 Therefore, the prediction is false. 28 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 28 16/01/2012 12:43

Questions 1 Complete the following patterns by writing down the next two terms. a 12 ; 19 ; 26 ; … b  2__3 ;  3__5 ;  74__; … 2 Determine the nth term and 16th term of the following patterns: a 25; 22; 19; 16;... b 1; 2; 4; 8; 16;... 3 Chains of squares can be built with matchsticks as shown below. a How many matches are used to create a chain of 4 squares? b How many matches are used to create a chain of 5 squares? What is the general formula for calculating how many matches there will be in a c chain of squares? d Now determine how many matches will be needed to build a chain of 100 squares. 4 You are stacking polystyrene cups, which fit into each other as in the diagram below. The first cup is 7 cm high. Two cups stacked together are 9 cm high, and three cups stacked together are 11 cm high. 11 cm 9 cm 7 cm a How high would a stack of four cups be? b Write down the sequence of numbers in stacks with 1, 2, 3, 4, 5 and 6 cups. c Find a general formula for the height of a stack of n cups. d How many cups will stack on a shelf if the height between shelves is 22 cm? e Can you build a stack of cups that has a height of 81 cm? If so, how many cups do you need? © Via Afrika – Mathematics 29 • Via Afrika Maths Study Guide Gr10 29 16/01/2012 12:43

Chapter 4 Equations and inequalities Overview In this chapter, we investigate equations and inequalities. We begin with linear equations before moving on to more compicated equations. An inequality is similar to an equation, except that it contains an inequality sign rather than an equals sign. The rules are very similar, but there are also a few important differences. So here you will learn: ●● how to solve linear equations ●● how to solve quadratic equations ●● how to solve simultaneous equations ●● how to work with linear inequalities, including how to represent them on a number line. UNIT 1 Page 31 Linear equations • Simplifying linear equations UNIT 2 Page 34 • Linear equations with fractions Quadratic equations • Exponential equations CHAPTER 4 Page 30 Equations and UNIT 3 Page 35 inequalities Quadratic equations UNIT 34 Page UNIT Page XX 37 • The elimination method Simultaneous Literal equations • The substitution method equations UNIT 5 Page 38 • Illustrating inequalities Linear inequalities • Solving inequalities 30 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 30 16/01/2012 12:43

Unit 1 Linear equations 1.1 Simplifying linear equations Use the following steps to solve a linear equation: Steps to follow Example Simplify by removing brackets and gathering 3(5x – 2) = 2(4x + 11) like terms. 15x – 6 = 8x + 22 Add and subtract terms on both sides until 15x – 8x = 22 + 6 all variables are on the left of the equals sign 7x = 28 and the constant terms are on the right. 7x Divide both sides by the coefficient of the __   28   = __ 7 7 variable (x). ∴x=4 Check the solution by substituting the value LHS: 3( 5 ¿ 4 – 2) = 3(20 – 2) = 3(18) = 54 back into the original equation. RHS: 2(4 ¿ 4 + 11) = 2(16 + 11) = 2(27) = 54 Since LHS = RHS, the solution is correct. Special cases: Identities and false statements The following statement is true for all values of x, and so we call it an identity. 2(3x + 4) = 6x + 8 ∴ 6x + 8 = 6x + 8 ∴ 6x – 6x = 8 – 8 ∴0=0 ∴x∈R The following statement can never be true, so we call it a false statement. 2(3x + 4) = 6x – 8 ∴ 6x + 8 = 6x – 8 ∴ 6x – 6x = –8 – 8 ∴ 0 = –16 ∴x∉R 1.2 Linear equations with fractions When working with fractions, always remember that you cannot divide by 0. Steps to follow Example 3 Write down any restrictions (that is, values for  ____ ____ =   1 +   2 x2 – 9 x + 3 ____ x–3 x that will make the denominator 0): x ≠ ±3 3 Multiply both sides of the equation by the  ____ ____ =   1 +   2 x2 – 9 x + 3 ____ x–3 LCM of all the denominators: (x + 3)(x – 3) 3  _________ ____ =   1 +   2 (x + 3)(x – 3) x + 3 ____ x–3 ∴ 3 + 2(x – 3) = x + 3 © Via Afrika – Mathematics 31 • Via Afrika Maths Study Guide Gr10 31 16/01/2012 12:43

Unit 1 Steps to follow Example Simplify the equation (remove the brackets ∴ 3 + 2x – 6 = x + 3 and collect the like terms together) ∴ 2x – x = 3 – 3 + 6 ∴x=6 Check that the answer is allowed (by referring Yes, x ≠ ±3 back to any restrictions) Check the solution by substituting it into the LHS: RHS: original equation. 3 =  ____ + ____   2 ____ =   6 –1 x2 – 9 x + 3 3 =  ____ 3 ____ +   2 = __   13 62 – 9 6 + 3 3 = _____   36 – 9 + __   29 3 =  __ __ 2 27 +   9 =   1__3 LHS = RHS, therefore the solution is correct. Example 1  _____ x+2 = ____   1 x2 – 2x x – 2 [Restriction: x ≠ 0; x ≠ 2] 2 (x – 2)(x + 2) = 1(x – 2x) x2 – 4 = x2 – 2x 2x = 4 x=2 However, x ≠ 2, therefore there is no solution. 2   a –3 ____ –   2 1 ____= 0 a+1 [Restriction: a ≠ 1; a ≠ –1] 3(a + 1) – 2(a – 1) = 0 3a + 3 – 2a + 2 = 0 a = –5 Check solution:  –53 _____ –   2 –1 _____ = – __ –5 + 1 + __ 1 1 2 2   = 0 3   x 2x – ____ 3 x– + ____   3x2 = 1 [Restriction: x ≠ 0] 3(x – 3) + 2(x – 2) = 6x 3x – 9 + 2x – 4 = 6x –x = 13 x = –13 Check solution:  –13 ______– 3 –13 __ +   15   –39– 2 =   16 + ______ __ +   5 __ =   8 __= 1 –26 26 39 13 13 1.3 Exponential equations You learnt how to solve exponential equations in Chapter 2. In this section, we will show you how to solve an equation when you are not able to make the bases the same on both sides of the equation. In this case, you need to use the trial-and-error method, as shown in the following example. 32 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 32 16/01/2012 12:43

Unit 1 Example Help table 2x – 1 = 13 x x–1 Accuracy so, x – 1 = 3,7 x = 2,7 5 16 Too big 4 8 Too small 3,5 11,313 … Too small 3,6 12,125 … Still too small 3,7 12,996 … Very close 3,71 18,086 Too big © Via Afrika – Mathematics 33 • Via Afrika Maths Study Guide Gr10 33 16/01/2012 12:43

Unit 2 Quadratic equations A quadratic equation has square term in it, for example, x2 + x – 12 = 0. We say that the equation is of degree 2, or is a second-degree polynomial. When you solve a quadratic equation, you may have 0, 1 or 2 solutions. Steps to follow Example Remove any brackets and take all terms to the left of 2(2a2 – 1) = 7a the equals sign so that there is only a 0 on the right. 4a2 – 2 = 7a The equation will now be in the form: 4a2 – 7a – 2 = 0 ax2 + bx + c = 0 Factorise the equation. (4a + 1)(a – 2) = 0 We find the solution to the equation by letting each (4a + 1) = 0 or (a – 2) = 0 factor equal 0. Why? ∴ a = – 1__4 or a=2 If ab = 0, then this statement will be true if a = 0 or if b = 0. Note: Do not fall in the trap of trying to solve an equation without first removing the brackets and making the right-hand side of the equation equal to 0. Example Solve for x: x(x + 3) = 10 Here, the solution is not x = 10 and x + 3 = 10. The correct solution is: x2 + 3x – 10 = 0 (x – 2)(x + 5) = 0 x = 2 or x = –5 34 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 34 16/01/2012 12:43

Unit 3 Literal equations A literal equation is usually a formula, such as E = mc2. Since there is more than one variable, we cannot solve a literal equation in the same way as an equation containing only one variable and numbers. What you are expected to be able to do is to change the subject of the formula. All this means is that you need to change the formula so that one variable appears on the left-hand side of the equation, and all the other variables and numbers are on the right-hand side of the equation. Example Steps to follow Example If needed, remove fractions by multiplying by the LCM (abc) of the Solve for x: denominators.  _ax +  1__c = _  bx bcx + ab = acx Write all the terms with the new subject on one side of the equals bcx – acx = ab sign and all the other terms on the other side. Factorise with the new subject (x) as a common factor. x(bc – ac) = ab Divide both sides by the coefficient (bc – ac) of the new subject. _____ x =  bcab – ac Let’s work through a few more examples. Steps to follow Example Make b______ the subject of the formula: 3x = √2b – y  ______ If there is a square root (or cube root), square (or cube) both sides. (3x)2 = (√2b – y )2 9x2 = 2b – y Write all the terms with the new subject on one side of the equals 2b = 9x2 + y sign, and all the other terms on the other side. 9x2 + y Divide both sides by the coefficient of the new subject. b =  _____ 2 Steps to follow Example Make r the subject of the formula: 4 V = __  3 π(R3 – r3) If needed remove fractions by multiplying by the LCM of the 3V = 4π(R3 – r3) denominators. Multiply out the brackets to remove them. 3V = 4πR3 – 4πr3 Write all the terms with the new subject on one side of the equals 4πr3 = 4πR3 – 3V sign, and all the other terms on the other side. 4πR3 – 3V Divide both sides by the coefficient of the new subject. r3 = _______   4π _______ √ 3 4πR3 – 3V Find the required root (cubed root) again, if necessary. r = _______   4π   © Via Afrika – Mathematics 35 • Via Afrika Maths Study Guide Gr10 35 16/01/2012 12:43

Unit 13 Steps to follow Example Make h __the subject of this formula: T = 2π   h_g   √__ If there is a square root (or cube root), square (or cube) both sides after placing the root sign on one side by itself. T = 2π   h_g   __ √ __ T 2π  = √ h _    g __    ) = ( √_      ) ( __ T 2 2π h 2 g 2 ___   T = _  h 4π2 g Remove any fractions by multiplying both sides of the equation 4π h = T2g 2 by the LCM of the denominators. T2g Write all the terms with the new subject on one side of the h = ___  4π2 equal sign and all the other terms on the other side. Divide both sides by the coefficient of the new subject. 36 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 36 16/01/2012 12:43

Unit 4 Simultaneous equations Solving two equations simultaneously means finding one set of values for the variables that satisfy both equations. There are two ways to solve equations simultaneously (at the same time), namely the elimination method and the substitution method. 4.1 The elimination method Steps to follow Example Rearrange the equations so that the coefficients Solve the pair of equations simultaneously: of one of the variables are the same in both 5x – 2y = 6 ................... (1) equations. 3x + 4y = 14 ................. (2) Tip: Number the equations Multiply equation 1 by 2: (The coefficient of y in both cases is now 4) 10x – 4y = 12 ............... (3) [(1) × 2] Add or subtract the equations to eliminate one of Add (3) and (2): 13x = 26 the variables. Solve for the unknown variable (x) x=2 Substitute this solution (x) back into either of the Use 5x – 2y = 6 ............. (1) original equations to solve for the other variable 5(2) – 2y = 6 (y). –2y = 6 – 10 y=2 Therefore, x = 2 and y = 2. 4.2 The substitution method Steps to follow Example Solve the pair of equations simultaneously: 3x – y = 10 .................. (1) 3x – 4y = –8 ................. (2) Rearrange at least one of the equations in the form Use equation (1): y=… 3x – y = 10 Tip: Number the equations 3x – 10 = y ................... (3) Substitute the value of y into the other equation. Substitute (3) into (2): 3x – 4(3x – 10) = –8 Solve for the unknown variable (x). 3x – 12x + 40 = –8 3x – 12x = –8 – 40 9x = –48 48 x = __   9 x = 5 __13 Substitute this solution (x) back into either of the Use 3x – y = 10: original equations to solve for the other variable (y). ( 3 3 )   16 __  – y = 10 16 – 10 = y y=6 Therefore, x = 5 1__3 and y = 6. © Via Afrika – Mathematics 37 • Via Afrika Maths Study Guide Gr10 37 16/01/2012 12:43

Unit 5 Linear inequalities 5.1 Illustrating inequalities Linear inequalities have a range of solutions. As we will see later, we can represent the solution to an inequality graphically on a number line. Example Solve for x: x + 3 < 2 Even though we have an inequality rather than an equals sign, we use the same principles to solve the equation. x+3–3<2–3 x < –1 5.2 Solving inequalities Note: When multiplying or dividing an inequality by a negative number, you must reverse the direction of the inequality sign. Example 1 Solve for x:  x – ____ 2 4 –   5x _____ 3 +1 ≥ 1 3(x – 4) – 2(5x + 1) ≥ 6 3x – 12 – 10x – 2 ≥ 6 –7x ≥ 6 + 12 + 2 –7x ≥ 20 [Divide by –7: reverse the inequality] 20 __ x ≤   7 2 Solve for x: 3(x + 4) < 5x + 9 3x + 12 < 5x + 9 –2x < –3 [Divide by –2: reverse the inequality] 3 __ x >  2 As we mentioned earlier, we can show the solution of a linear inequality on a number line. 38 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 38 16/01/2012 12:43

Unit 5 Example Solve for n and indicate the solution on a number line: –1 < 2 – n < 2 –1 – 2 < –n < 2 – 2 +3 > n > 0 0 1 2 3 0<n<3 When the inequality is < or >, we show this as an open circle on the number line. When the inequality is ≥ or ≤, then we show this as a closed circle on the number line. © Via Afrika – Mathematics 39 • Via Afrika Maths Study Guide Gr10 39 16/01/2012 12:43

Questions 1 Solve the following equations and inequalities: a   5a __– 2 =  a 3 __+ 15 4 __+ 15 <  5m b  m ___– 2 4 3 c (2b + 1)(b + 8) = 27 d ____ =   3 2 –  d –1 ____ 1 d+1 e b² – 7b + 15 = 3 5y – 2 3y – 1 y+1 f  _____ 3 + _____   2 – ____   2 < –  1__3 5 g   9h2 3 ______ –   – 16 _________ =      ______________ 2 3h2 + 5h – 12 h(3h + 4) + 3(4 + 3h) 2 Solve simultaneously: 2a = 24 + 7b and 3a + 5b = 5 3 The formula for finding the volume, V, of material used to make a hollow sphere is given by V =  34__π(R3 – r3), where R is the outer radius and r is the inner radius. Make R the subject of the formula, giving the expression in its simplest form. 4 Solve for in the following equations: a x2 + 3(2x – 5) = x2 – 2(7 + 3x) y – 3x b  _____ x + z = 2 c   x + ____ 6 5 + 1 =  x – 35 ______ –x d  x2 3 ____ +   2 ____ =   1 –9 x+3 x–3 ____ e x –  ____   7x ≥  32 + __ x – 2 __ 3 8 5. Solve for a and b: 7(a + 2) + 3(b – 5) = 34 and 3(a + 2) – 2(b – 5) = 8 40 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 40 16/01/2012 12:43

Chapter 5 Trigonometry Overview Here, we will introduce you to trigonometry. Trigonometry is the field of mathematics in which we study the relationship between the sides and angles of triangles. The word trigonometry is derived from the Greek words trigonon, which means triangle, and metron, which means measurement. In ancient times, mathematicians, astronomers and surveyors in Egypt, Babylon, India and China used trigonometry for navigation, surveying and astronomy. For example, they were able to calculate: ●● the height of mountains ●● the distance and direction across the sea (navigation) ●● the dimensions of large areas of land for construction ●● astronomical distances, for example, between the Earth, the moon and the sun. You will also learn how to sketch the graphs of each of the trigonometric ratios. UNIT 1 Page 42 • Trigonometric ratios in right-angled Right-angled triangles triangle UNIT 2 Page 43 • Definitions Definitions of the • Reciprocals of the three trigonometric trigonometric ratios ratios UNIT 3 Page 44 Special triangles UNIT 4 Page 45 CHAPTER 5 Page 41 Using your calculator Trigonometry UNIT 5 Page 46 • Solve diagrams in two dimensions Solving trigonometric equations UNIT 6 Page 48 Extending the ratios to 0° ≤ θ ≤ 360° UNIT 7 Page 53 • The sine and cosine graphs Graphs of the • The tan graph trigonometric functions © Via Afrika – Mathematics 41 • Via Afrika Maths Study Guide Gr10 41 16/01/2012 12:43

Unit 1 Right-angled triangles 1.1 Trigonometric ratios in right-angled triangles Trigonometry is the measurement of triangles, specifically right-angled triangles. When we use the Cartesian plane in trigonometry, we call the positive x-axis the 0° line. If we move anti-clockwise by 360°, we have completed a full circle. The line that forms any angle (θ) in this way is known as the terminal side and the angle (θ) is known as the inclination angle. y(90°) Terminal side III Inclination angle θ 180° x (0° or 360°) 0 IIIIV 270° y In a right-angled triangle, each side has a P specific name: β ●● The hypotenuse (OP) is always opposite the 90° angle. ●● The opposite side is opposite the angle. In this example, PR is opposite θ and OR is θ opposite β. 0 R x ●● The adjacent side forms the angle together with the hypotenuse. Therefore OR is adjacent to θ, and PR is adjacent to β. Adjacent is always hypotenuse θ next to the angle. se θ nu adjacent te po op hy po jac Opposite is opposite sit en e t the angle. ad opposite 42 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 42 16/01/2012 12:43

Unit 2 Definitions of the trigonometric ratios 2.1 Definitions In Trigonometry, the ratios between the sides are given certain names. In ΔABC with ∠BAC = α, we ca define the ratios as: opposite A sin α = _________  hypotenuse α adjacent cos α = _________  hypotenuse opposite tan α = _______  adjacent β B C 2.2 Reciprocals of the three trigonometric ratios The reciprocal ratios are the inverse versions of each of the main ratios. The inverse of sin θ is cosec θ, the inverse of cos θ is sec θ, and the inverse of tan θ is cot θ. The reciprocal ratios are defined as follows: Trigonometric ratio Reciprocal ratio Conclusion sin θ =  ho_ cosec θ =  h_o ____ cosec θ =  sin1 θ cos θ =  a__h sec θ =  ah__ ____ 1 sec θ =  cos θ tan θ =  ao__ cot θ =  a__o cot θ =  ____ 1 tan θ © Via Afrika – Mathematics 43 • Via Afrika Maths Study Guide Gr10 43 16/01/2012 12:43

Unit 3 Special triangles The ratios of certain angles are well known, and we can use them to solve problems. These angles are 0°, 30°, 45°, 60° and 90°. We can show these angles (except for 0°) in two triangles, as shown below. 45° 30° __ 1 √ 2  __ 2 √3  45° 1 60° 1 The ratios formed by each of these special angles is shown below. 0° 30° 45° 60° 90° __ __   √   √ __ __ 2  __ 3  sin θ 0   1 2 2 2 1 __ __ √   √ 3  2  cos θ 1  __ 2 __ 2  __12 0 __ __ √ 3  tan θ 0  __ 3 1 √3  undefined 44 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 44 16/01/2012 12:43

Unit 4 Using your calculator When all you know is the size of an angle, you can use a calculator to work out the value of a ratio. Your scientific calculator has keys for the trigonometric ratios. Make sure your calculator is in DEG mode when working with degrees. Note: sin–1 on your calculator is not the same as ___ 1 sin . In trigonometry, ___ 1 sin refers to the –1 reciprocal ratio for sin, namely cosec. The sin key on the calculator is the reciprocal function of sin, which is what you use to calculate the value of an angle when you know the value of the ratio. The following example shows how to use your calculator to work out the value of a ratio for a particular angle. Example Answer Calculator p In ΔPQR; ∠Q is 90°, q = 12 cm sin P = __q sin ( 1 ÷ 3 ) = and p = 4 cm� Find the values 4 sin P = __ of ∠P and ∠R� 12 ∠P = sin–1 1__3 P ∠P = 19,47° ∠R = 90° – ∠P q = 12 cm = 70,53° Q p = 4 cm R Note: We name the sides of a right-angled triangle according to the angle names� So for example, the side opposite angle Q is q, and the side opposite angle P is p� © Via Afrika – Mathematics 45 • Via Afrika Maths Study Guide Gr10 45 16/01/2012 12:43

Unit 5 Solving trigonometric equations 5.1 Solve diagrams in two dimensions Use the following tools to solve right-angled triangles: A ●● Theorem of Pythagoras: a2 + b2 = c2 α ●● Sine: sin A =  a__c , sin B =  b__c c b ●● Cosine: cos A = __ bc , cos B = __ ac ●● Tangent: tan A =  ba__, tan B =  b__a β C a B Remember: In triangle ABC, a is opposite ∠A, b is opposite ∠B, and c opposite ∠C. You can calculate an unknown side or angle when you know the sizes of: ●● an angle and a side ●● two sides and we need the angle. Example 1 Using the diagram alongside, calculate the length of A BC if AB is adjacent to ∠A and BC is the opposite side. 50° BC tan 50° =  ___ AB 4 cm BC = AB.tan 50° BC = 4.tan 50° B C BC = 4,77 cm (to 2 decimal numbers) 2 Calculate the length of AC, given that AB is adjacent to ∠A and AC is the hypotenuse. cos 50° = ___  AB AC AC.cos 50° = AB AC = cos 50° = A cos 50° AC = 6,22 cm (to 2 decimal numbers) We can also use calculate an angle if we know the value of a ratio. 46 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 46 16/01/2012 12:43

Unit 5 Example Using the diagram alongside, calculate the size of ∠B if ∠C = 90°. A 3,4 cm AC is the opposite side to ∠B and AB is the hypotenuse: C AC sin B = ___  AB 3,4 sin B = ___  7,2 7,2 cm ∠B = sin–1 ___ ( 3,4 )   7,2   Therefore, ∠B = 28,18° (to two decimal places) B Let’s work through another example. Example 5 __. In the triangle alongside, tan θ =  12 P Calculate the following using the triangle: 1 ∠θ 2 cos θ 3 The length of the hypotenuse θ Q R 1 θ = tan–1 = 22,62° 2 cos θ = cos 22,62° = 0,92 opposite 3 The tan ratio is  ______ adjacent . Therefore, we know that the side opposite θ (PR) is 5 units long, and the adjacent side is 12 units long (QR). Now we can use the theorem of Pythagoras to calculate the length of the hypotenuse, PR: PR2 = 5____ 2 + 122 = 25 + 144 = 169 PR = √169  PR = 13 © Via Afrika – Mathematics 47 • Via Afrika Maths Study Guide Gr10 47 16/01/2012 12:43

Unit 6 Extending the ratios to 0° ≤ θ ≤ 360° 6.1 Trigonometry ratios in four quadrants By convention we label the four quadrants of a Cartesian plane as follows: y 5 4 Quadrant 2 3 Quadrant 1 2 1 x –5 –4 –3 –2 –1 0 1 2 3 4 5 –1 –2 Quadrant 3 –3 Quadrant 4 –4 –5 In trigonometry, we tend to think of an angle as created by a rotating radius (arm). The beginning position of the “arm” is called the initial side (usually, the initial side coincides with the positive x-axis; angles with such an initial side are said to be in standard position). When the “arm” rotates and ends up in one of the quadrants, the side (“arm”) is now called the terminal side. The measure of the angle is a number which describes the amount of rotation. If the rotation of the “arm” was clockwise, the angle measure is a negative number; if the rotation was anti-clockwise, the angle measure is positive. Angles greater than 90° We define the trigonometric functions for angles greater than 90° in the following way: y-axis (x; y) Terminal side y r θ x-axis x Initial side 48 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 48 16/01/2012 12:43

Unit 6 The trigonometric definitions do not rely on the lengths of the sides of the triangle, but only on the angle. When negative or obtuse angles are used in trigonometric functions, they will sometimes produce negative values. The CAST diagram will help you to remember the signs of trigonometric functions for different angles. The functions will be negative in all quadrants except those that indicate that the function is positive. For example, when the angle is between 0° and 90°, the side r is in the first quadrant. All functions will be positive in this quadrant. When the angle is between 90° and 180°, the terminal side is in the second quadrant. This means that only the sine function is positive. All other functions will be negative. When the angle is between 180° and 270°, the terminal side is in the third quadrant. This means that only the tan function is positive. All other functions will be negative. When the angle is between 270° and 360°, the terminal side is in the fourth quadrant. This means that only the cosine function is positive. All other functions will be negative. y Quadrant 1 4 y Quadrant 2 Quadrant 1 sine of an angle =  _r 3 cosine of an angle =  _xr y 2 tan of an angle =  _x –y r y All positive, because x, y and r > 0 1 x –4 –2 –x 0 x 2 4 –1 –2 –3 Quadrant 3 Quadrant 4 –4 y Quadrant 2 4 y Quadrant 2 Quadrant 1 sine of an angle = _ r 3 cosine of an angle =   –x __ r y 2 tan of an angle = __   –x –y r y Because x is negative in the second quadrant, all the 1 functions containing x will be negative, i.e. cos and x tan. –4 –2 –x 0 x 2 4 –1 Only the sine function is positive, because y and r > 0. –2 –3 Quadrant 3 Quadrant 4 –4 © Via Afrika – Mathematics 49 • Via Afrika Maths Study Guide Gr10 49 16/01/2012 12:43

Unit 6 y Quadrant 3 4 –y Quadrant 2 Quadrant 1 sine of an angle = __   r 3   –x cosine of an angle = __ r –y _ y 2 tan of an angle =  __ –x =  x –y r y Only the tan function is positive in the third quadrant 1 (neg ÷ neg = pos). –x x –4 –2 0 x 2 4 –1 –y –2 –3 Quadrant 3 Quadrant 4 –4 y Quadrant 4 4 –y Quadrant 2 Quadrant 1 sine of an angle = __   r 3 cosine of an angle =   x_r –y 2 tan of an angle = __   x –y r y In quadrant 4 only the cos function is positive, 1 because x and r > 0. –x x –4 –2 0 x 2 4 –1 –y –2 –3 Quadrant 3 Quadrant 4 –4 The CAST rule There is a simple rule by which you can remember all of these results. Notice in the first quadrant, All the functions are positive; in the second quadrant, only the Sine is positive; in the third, only the Tangent is positive; and in the fourth quadrant, only the Cosine is positive. This is called the CAST Rule and tells you which function is positive S A in each quadrant. And the C, A, S and T stand for cosine, all, sine T C and tangent. y (90°) Quadrant 2 Quadrant 1 S A sin all 180° x (0° or 360°) 0 T C tan cos Quadrant 3 Quadrant 4 270° 50 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 50 16/01/2012 12:43

Unit 6 Examples using the CAST rule 1. Without using a calculator, use the figure to 1.1 Use Pythagoras’ theorem to calculate the calculate the following: length of OP: 1.1 OP y OP2 = 33 + (–2)2 1.2 tan A _____ ___ P(–2; 3) ∴ OP = √9 + 4 = √13  1.3 cos A y 3 3 1.4 sin2A – cos2A A 1.2 tan A = _  x = __   –2 = – __2 (tan is negative in the 2nd quadrant!) x 0 1.3 cos A =   x_r =   –2___ ___ √13  (cos is also negative in the 2nd quadrant!) y 3 1.4 sin A = _  r = ___   ___ Notice that the point P is in the SECOND √13  S A therefore, sin2A – cos2A quadrant, only sine is positive in the second quadrant. T C 4 =  __ __ 9 13 +   13 13 = __   13 =1 __ and A 2. If sin A = – 12 ^ ∈ [90°; 270°] determine, 13 It is useful to make a drawing: without the use of a calculator, the values of y 2.1 cos A 2.2 tan A A x Notice that the angle A could be in the S A –12 13 second or third quadrants, but sine is negative in the third quadrant. So, angle T C A has to be in the THIRD quadrant. The x-value = –5 (Pythagoras) –5 2.1 cos A =   x_r = __   13 (cos is negative in the 3rd quadrant!) y 2.2 tan A = _  x = ___   –12 __12 –5 =   5 (tan is positive in the 3nd quadrant!) 5 3. If tan x = – __ 12 , 0° < x < 180°, use a sketch The terminal side: 13 (Pythagoras) (no calculator) to determine the value of y 3 sin x – 2 cos x. 5 x x –12 Sine is positive in the 2nd quadrant and cosine is negative. 3 sin x – 2 cos x = 3 ( __   13   )– 2 ( ___13   ) 5   –12 15 24 Notice that the tan function is negative S A =  __ __ 13 +   13 and the given domain covers the first two 39 T C = __   13 quadrants. Tan is negative in the second quadrant. =3 © Via Afrika – Mathematics 51 • Via Afrika Maths Study Guide Gr10 51 16/01/2012 12:43

Unit 6 4. If 4 sin θ + 3 = 0 and 0 > 270° determine, without 4 sin θ + 3 = 0 y using a calculator: 3 ∴ sin θ = – __4 4.1 tan θ θ x 4.2 cos2θ _________ x = √______ 42 – (–3)2  3 4 Notice that angle is greater than 270°.   = √16 – 9  (Pythagoras) __ S A   = √7  So, the operating quadrant must be the fourth quadrant and the sine function is T C also negative in the fourth quadrant. 3 4.1 tan θ = – __ __ (tan is negative in the 4th √7  quadrant) __ ( 7  ) 2 4.2 cos20 =   √__ 7 __ 4   =   16 (cos is positive in the th 4 quadrant) 52 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 52 16/01/2012 12:43

Unit 7 Graphs of the trigonometric functions 7.1 The sine and cosine graphs 1. y = sin θ y 2 1 x –360 –270 –180 –90 0 90 180 270 360 –1 –2 The graph repeats itself every 360° Domain: All real numbers Range: [–1, 1] Maximum value: 1 when θ = –270°; 90° Minimum value: –1 when θ = –90°; 270° Zero: when the graph cuts the θ-axis, i.e. when θ = –360°; –180°; 0°; 180°; 360° 2. y = cos θ y 2 1 x –360 –270 –180 –90 0 90 180 270 360 –1 –2 © Via Afrika – Mathematics 53 • Via Afrika Maths Study Guide Gr10 53 16/01/2012 12:43

Unit 7 The graph repeats itself every 360° Domain: All real numbers Range: [–1, 1] Maximum value: 1 when θ = –360°; 0°; 360° Minimum value: –1 when θ = –180°; 180° Zero: when the graph cuts the θ-axis, i.e. when θ = –270°; –90°; 90°; 270° 7.2 The tan graph 1. y = tan θ y 2 1 x –360 –270 –180 –90 0 90 180 270 360 –1 –2 The graph repeats itself every 180° No maximum or minimum turning points Zero: when the graph cuts the θ-axis, i.e. when θ = –360°; –180°; 0°; 180°; 360° Asymptotes: at θ = –270°; –90°; 90°; 270° 54 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 54 16/01/2012 12:43

Questions 1 In ΔABC, DC is perpendicular to AB, ∠A = 55°, A AD = BD and DC = 15 cm. (Use your calculator to answer following questions.) a Find the length of AD and then AB. D b Find the length of AC. c Find the length of BC. (What does you answer imply about ΔABC?) B d Find ∠DBC. C 2 A rocket is launched vertically and changes direction through an angle of 55° to the vertical line after 2 minutes. If the average speed of the rocket is 6 000 km/h, calculate: a The height of the rocket after 2 minutes. b The height of the rocket after 3 minutes. c The horizontal distance from A, the launching tower, to B, a point directly under the rocket after 3 minutes. A E 3 From the top of a perpendicular cliff, AB, 60 m 1 20° 3 above sea level, the angles of depression to two 25° 2 boats at C and D, in the same vertical plane as B, are 20° and 25°, respectively. Calculate the distance between the two boats at 60 m C and D (correct to one decimal place). 1 B C D 4 In the figure, M and N represent service vehicles for a P cellular telephone company and are parked on opposite sides of a cellular phone mast, PT. MTN is a straight line. The angles of elevation of mast PT from the vehicles M and N are 52° and 38° respectively. If MTN = 160 m, calculate: 52° 38° a the magnitude (size) of ∠MPN M T N b the length of MP, to the nearest metre c the height of the mast PT, to the nearest metre © Via Afrika – Mathematics 55 • Via Afrika Maths Study Guide Gr10 55 16/01/2012 12:43

5 Use your calculator to solve the following equations, correct to two decimal numbers (0° ≤ x ≤ 90°). a sin x = 0,57 b 3cos x = 0,96 c  __23 tan x = 4,2 d sin (x + 25°) = 0,813 e tan x = sin 42° 6 Using the special__ angles, solve the following equations for 0° ≤ x ≤ 90°. a 2sin x = √3  b cos 3x =  __12 c 4sin (2x – 10°) – 1 = 1 d x.tan 60° = cos 30° 56 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 56 16/01/2012 12:43

Chapter 6 Functions Overview In this chapter, you will learn how to: ●● represent functions using function notation: f (x) ●● graph the basic functions f (x) = ax2 + b, f (x) = _  kx , f (x) = bx ●● graph transformations of the basic functions, including vertical shifts, stretches, and shrinks, as well as reflections across the x- and y-axes ●● investigate and explain the characteristics of a function: domain, range, zeros, intercepts, intervals of increase and decrease, maximum and minimum values and end behaviour ●● relate to a given context the characteristics of a function, and use graphs and tables to investigate its behaviour. UNIT 1 Page 58 • Dependent and independent variables What is a function? • Relationship between variables UNIT 2 Page 60 • The graph of the parabola, Graphs of functions • The graph of y = x __1 • The graph of y = bx, b > 0 and b ≠ 1 UNIT 3 Page 63 • The effect of a The graph of • The effect of q y = ax2 + q UNIT 4 Page 64 • The effect of a CHAPTER 6 Page 57 The graph of y =  a__x + q • The effect of q Functions UNIT 5 Page 66 • Possible values of b The graph of y = abx + q, • The effect of a b > 0 and b ≠ 1 • The effect of q UNIT 6 Page 68 • The parabola Sketching graphs • The hyperbola • The exponential function UNIT 7 Page 71 Finding the equations of graphs UNIT 8 Page 72 The effect of a and q on trigonometric graphs © Via Afrika – Mathematics 57 • Via Afrika Maths Study Guide Gr10 57 16/01/2012 12:43

Unit 1 What is a function? 1.1 Dependent and independent variables A function is like a machine: ●● it has an input and an output ●● there is a relationship between input and output values ●● the output depends on the input, so we say that the output is a dependent variable and the input is the independent variable. ●● “f (x) = …” is the classic way of writing a function. Rule or function • Input • Output • Independent variable • Dependent variable 1.2 Relationship between variables ●● A function relates inputs to outputs ●● A function is a special type of relation in which: ●● the output value is dependent on the input value ●● any input produces only one output (not this or that) ●● an input and its matching output are together called an ordered pair ●● a function can also be seen as a set of ordered pairs For example: y = 2x + 3 is a function. Input Relationship Output 0 ×2 + 3 3 First, it is useful to give a function a name. 1 ×2 + 3 5 The most common name is “f ”, but you can 2 ×2 + 3 7 have other names, such as “g”. But f is used 3 ×2 + 3 9 most often in mathematics. … … … f (x) = 2x + 3 function name input what to output We read the function statement as: “f of x equals two x plus three”. 58 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 58 16/01/2012 12:43

Unit 1 What goes into the function is put inside brackets (  ) after the name of the function. So, f (x) shows you the function is called “f ”, and “2x + 3” tells you that function “f ” takes “x”, multiplies it by two and adds three. The answer is the output. The output value uniquely depends on the input value. An input of 3 gives an output of 9: f (3) = 9. The input and output values can be plotted as coordinates on a graph, in the form (input; output) or (x; f (x)): (0; 3), (1; 5), (2; 7), (3; 9), etc. y 10 9 8 7 6 5 4 3 2 1 x –2 –1 0 1 2 3 4 5 6 7 8 9 10 –1 –2 So, when plotted in the Cartesian plane, the function y = 2x + 3 forms a straight line of the kind y = ax + q, also written as f (x) = ax + q. Example A function f is defined by f (x) = 2 + x – x2. What is the value of f (–3)? f (–3) = 2 + (–3) –(–3)2 = 2 – 3 – 9 = –10 © Via Afrika – Mathematics 59 • Via Afrika Maths Study Guide Gr10 59 16/01/2012 12:43

Unit 2 Graphs of functions At first, it may be helpful to generate a table of values using numbers from the domain (x-values). Your x-values should include examples of negative numbers and fractions. 2.1 The graph of the parabola, y = x2 ●● The graph forms a parabola ●● Axis of symmetry: y = 0 (y‑axis) ●● Turning point at (0; 0) ●● Intercept with x‑axis at x = 0 and intercept with y‑axis at y = 0 x (domain) –2 –1 0 1 2 y (range) 4 1 0 1 4 y 10 5 x –3 –2 –1 0 1 2 3 Equation 1: y = x2 60 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 60 16/01/2012 12:43

Unit 12 2.2 The graph of y = _x 1 ●● The graph forms a hyperbola ●● Axis of symmetry: y = x ●● Intercept with y‑axis at y = 1 x (domain) –2 –1 0 1 2 y (range) –0,5 –1 undefined 1 0,5 y 1 0.5 x –10 –8 –6 –4 –2 0 2 4 6 8 10 –0.5 –1 Equation 1: y =   1__x 2.3 The graph of y = bx, b > 0 and b ≠ 1 ●● The graph forms an exponential graph ●● Axis of symmetry: y = x ●● Intercept with y‑axis at y = 1 x (domain) –2 –1 0 1 2 y (range) 0,25 0,5 1 2 4 © Via Afrika – Mathematics 61 • Via Afrika Maths Study Guide Gr10 61 16/01/2012 12:43

Unit 12 y 5 4 3 2 b=2 1 x –6 –4 –2 0 2 4 6 –1 Equation 1: y = 2 x 62 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 62 16/01/2012 12:43

Unit 3 The graph of y = ax2 + q 3.1 The effect of a ●● The effect of a is a vertical stretch and/or a reflection about the x‑axis. ●● The value of a gives an indication of the steepness (wide or narrow) of the two “arms”. ●● For a > 0: ●● The parabola will have a minimum value. ●● The “arms” point upwards. ●● Range is y ≥ q, y ∈ R ●● For a < 0: ●● The parabola will have a maximum value. ●● The “arms” point downwards. ●● Range is y ≤ q, y ∈ R 3.2 The effect of q ●● The y-intercept (where the graph cuts the y‑axis) is equal to the value of q or y = q (this is also the maximum or minimum value). ●● The effect of q is a vertical shift. ●● The turning point of the graph is at (0; q). ●● The axis of symmetry is the y‑axis (x = 0). ●● Domain is x ∈ R. y 10 5 x –3 –2 –1 0 1 2 3 © Via Afrika – Mathematics 63 • Via Afrika Maths Study Guide Gr10 63 16/01/2012 12:43

Unit 4 The graph of y = _ ax + x 4.1 The effect of a ●● Graphs that represent the equation y =  a__x + q form a hyperbola. ●● We call the x- and y-axes y 1 the asymptotes. ●● The line x = 0 (y‑axis) is called a vertical asymptote of the 0.5 function if y approaches infinity (positive or negative) as x approaches 0. x –10 –8 –6 –4 –2 0 2 4 6 8 10 ●● The line y = 0 (parallel to x‑axis) is called a horizontal asymptote –0.5 of the function. ●● For a > 0, the graph lies in the 1st and 4th quadrants –1 ●● For a < 0, the graph lies in the 2nd and 3rd quadrants. 4.2 The effect of q ●● For q = 0, the graph will never intercept the x- or y‑axis because division by 0 is undefined. ●● For q ≠ 0, the line y = q is the horizontal asymptote and x = 0 the vertical asymptote. ●● Domain: all real x-values, except 0 ●● Range: all real y-values, except y = q Example Sketch the graph of y =  6__x – 1 Domain: x ∈ R, x ≠ 0 Range: y ∈ R, y ≠ 1 x-intercept (let y = 0): x = 6 y-intercept: none Asymptotes: ●● The graph is only decreasing, for x ∈ R, x ≠ 0 ●● The graph shifts down one unit: new asymptote, y = –1 ●● Therefore, the asymptotes are: y‑axis and y = –1 64 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 64 16/01/2012 12:43

Unit 4 y 8 Asymptote: x = 0 6 4 xy = 6 2 x –8 –6 –4 –2 0 2 4 6 8 –2 Asymptote: y = –1  6x – 1 y = __ –4 –6 –8 © Via Afrika – Mathematics 65 • Via Afrika Maths Study Guide Gr10 65 16/01/2012 12:43

Unit 5 The graph of y = abx + q, b > 0 and b ≠ 1 5.1 Possible values of b ●● Graphs that represent the equation y 5 y = abx + q form an exponential graph in two quadrants of the Cartesian plane. 4 ●● When b > 1 the y-values increase as x 3 increases and the function is called an increasing function. 2 ●● The graph will incline to the right. 1 ●● When 0 < b < 1 the y values will decrease and the function is a decreasing function. x –3 –2 –1 0 1 2 ●● The graph will decline to the right. –1 ●● The y-intercept (where the graph cuts the y‑axis) is equal to the value of a for q = 0 and is therefore a + q for any other value. ●● There are no x-intercepts. ●● The line y = q is the asymptote for the graph. ●● Domain: x ∈ R ●● Range: y > 0, x ∈ R 5.2 The effect of a ●● a affects the steepness of the graph ●● y-intercept: y = a ●● If a < 0, then the graph reflects about the y‑axis y 6 5 4 y = 3(2x) 3 2 y = 2x 1 y = 0.5(2x) x –3 –2 –1 0 1 2 3 4 –1 66 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 66 16/01/2012 12:43

Unit 5 5.3 The effect of q ●● The value of q shifts the graph up or down (vertical shift) ●● y-intercept: 1 + q y 6 5 4 3 y = 2x + 1 2 y = 2x 1 y = 2x – 1 x –3 –2 –1 0 1 2 3 –1 © Via Afrika – Mathematics 67 • Via Afrika Maths Study Guide Gr10 67 16/01/2012 12:43

Unit 6 Sketching graphs 6.1 The parabola To plot graph of any quadratic function (parabola), we need answers to these questions: ●● What is the sign of a (the coefficient of x² in quadratic function)? ●● Does the graph of quadratic function intersect the x‑axis? And if it does, at what point does it intersect? ●● Does the graph of quadratic function intersect the y‑axis? ●● What is the maximum or minimum value of function? For a > 0: For a < 0: The parabola will have a minimum The parabola will have a maximum value. The “arms” point upwards. value. The “arms” point downwards. x-intercept(s): y-intercept(s): Let y = 0 and solve for x Let x = 0 and solve for y Because y = ax2 + q has no term in x, the axis of The turning point of the parabola is (0; q) symmetry is the y‑axis. 6.2 The hyperbola a > 0 a < 0 x-intercepts: y-intercepts: None, except when q ≠ 0. Then put y = 0 and None, the y‑axis is an asymptote solve for x Asymptotes: The y‑axis (x = 0) The 2nd asymptote is the line y = q The__ closest __ points to__the axes: __ Domain: x ∈ R; x ≠ 0 (√a ; √a  + q) and (–√a ; –√a  + q) Range: y ∈ R; y ≠ q Example Given y =  4__x , sketch the graph of the function. ●● Show the coordinates where x = 1, x = –1, x = 4 and x = –4 clearly on your graph. To do so, work out the y-values and plot the points: (1; 4), (–1; –4), (4; 1), (–4; –1) 68 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 68 16/01/2012 12:43

Unit 6 ●● Draw two axes of symmetry on your sketch and give their equations. (y = x and y = –x) ●● Find the point(s) of intersection of the hyperbola and the axes of symmetry. __ __ __ __ (√a ; √a  + q) and (–√a ; –√a  + q) = (–2; –2) and (2; 2) ●● The graph will be in quadrants 1 and 3, because a > 0 ●● The domain is x ∈ R, x ≠ 0 ●● The range is y ∈ R, y ≠ q y 6 y = –x y=x (1, 4) 4 2 (2, 2) (4, 1) x –12 –10 –8 –6 –4 –2 0 2 4 6 8 10 12 (–4, –1) (–2, –2) –2 –4 (–1, –4) –6 6.3 The exponential function Use the characteristics of the graph to determine the general shape. a b Characteristic a<0 0 < b < 1 (a positive fraction) decreasing a>0 0 < b < 1 (a positive fraction) Increasing a<0 b>1 Decreasing a>0 b>1 Increasing ●● If a < 0, the graph is below the asymptote and if a > 0, the graph is above the asymptote. ●● asymptote: y = q ●● y-intercept: y = a + q ●● Domain: x ∈ R ●● Range if a < 0: y < q, y ∈ R ●● Range if a > 0: y > q, y ∈ R © Via Afrika – Mathematics 69 • Via Afrika Maths Study Guide Gr10 69 16/01/2012 12:43

Unit 6 Example Given: f (x) = 3, sketch the graph of f. Show any intercepts with the axes. General form: y = abx + q a > 0 (=1) b > 1 (=3) Increasing Therefore: ●● Asymptote: y = 0 (q = 0) ●● y-intercept: y = 1 (a = 1) ●● Domain: x ∈ R ●● Range if a > 0: y > q, y ∈ R On the same system of axes, sketch the graph of g(x) = 2.3x. Clearly show any intercepts with the axes. y 6 5 4 3 2 y = 2(3x) 1 y = 3x – 1 y = 3x x –3 –2 –1 0 1 2 3 –1 70 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 70 16/01/2012 12:43

Unit 7 Finding the equations of graphs To find the equation of a graph, use the general form of the equation and the characteristics of the graph to write down what you know. Then, substitute point(s) from the graph into the equation to find the values of any unknown variables. Example The graph alongside is the graph of y =  a__x + q. y 1 Find the equation of the graph. 2 Give a reason why y = 2 is an asymptote. 2 1 The asymptote is at y = 2, so we know q = 2. We can then use the point (2; 0) in the 2  1__2 x general equation to find a: 0 =  a__5__ + 2   2 __ a   __5 = –2   2 a = –2 ×  __52 a = –5 –5 Therefore, y =  __x + 2 The range of the graph is y ∈ R; y ≠ 2 2 If we make x the subject of the equation, we have: –5 y =  __ x + 2 xy = –5 + 2x x(y – 2) = –5 x =  y–5 ____ –2 Therefore, we cannot have y = 2, because division by 0 is undefined. © Via Afrika – Mathematics 71 • Via Afrika Maths Study Guide Gr10 71 16/01/2012 12:43

Unit 8 The effect of a and q on trigonometric graphs ●● The trigonometric functions form periodic graphs ●● a.sin x + q and a.cos x + q have the form of a repeating wave with period 360° ●● For q > 0, the graph shifts q units up, and for q < 0, it shifts q units down. ●● The variable a stretches the graphs of the trigonometric functions. ●● Multiplying sin x or cos x by a positive constant, a, increases the amplitude to the value of a y 4 y = sin x + 2 3 2 1 y = a sin x + q x –180° –90° 0° 90° 180° 270° 360° –1 y = sin x –2 y = 2 sin x –3 –4 y 4 y = cos x + 2 3 2 y = a cos x + q 1 x –180° –90° 0° 90° 180° 270° 360° –1 y = cos 2x y = cos x –2 –3 –4 72 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 72 16/01/2012 12:43

Unit 8 y 4 3 y = tan x + 2 y = tan x 2 1 x –180° –90° 0° 90° 180° 270° 360° –1 –2 y = 2tan x y = tan x – 2 –3 –4 © Via Afrika – Mathematics 73 • Via Afrika Maths Study Guide Gr10 73 16/01/2012 12:43

Questions 1 Given the equation y = x2 – 4: a Describe the form of the graph. b Does the graph have a minimum or maximum value? Give a reason for your answer. c What is the y-value for the minimum or maximum? d For which values of x will the graph intercept the x‑axis? e Sketch the graph of the function. f Write down the equation if the graph moves up 4 units. g What will the x-intercept(s) be of the new graph? h Write down the domain and range of the function y = x2 – 4 in interval notation. i Write down the equation of the axis of symmetry. 2 Given a parabola with turning point (0; 3) and x-intercept (–3; 0): a Write down the equation of the graph. b Write down the domain and range of the graph. c Write down the equation of the axis of symmetry of the graph. 74 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 74 16/01/2012 12:43

Chapter 7 Polygons Overview In this chapter, we will first revise what you have learnt previously about polygons and the concepts of similarity and congruence in triangles. Then, we extend this knowledge to quadrilaterals, including the kite, parallelogram, rectangle, rhombus, square and trapezium. Here, you will investigate and learn about some of the important geometric properties of each of these shapes. Here, you will learn: ●● how two triangles are similar if they are the same shape, but different sizes ●● about corresponding angles and sides ●● how two triangles are congruent if their corresponding sides and angles are the same size ●● about different types of quadrilaterals (trapezium, parallelogram, rhombus, rectangle, square, kite) UNIT 1 Page 76 • Parallelogram Similar triangles • Rectangle • Rhombus • Square • Kite • Trapezium CHAPTER 7 Page 75 Polygons UNIT 2 Page 78 Congruent triangles UNIT 3 Page 80 Quadrilaterals © Via Afrika – Mathematics 75 • Via Afrika Maths Study Guide Gr10 75 16/01/2012 12:43

Unit 1 Similar triangles Similar figures have the same shape, but are different sizes. In the case of triangles: ●● Corresponding angles are equal. ●● Corresponding lengths are in the same ratio. B 70° 5 8 60° 50° A 10 C E 10 16 70° D 60° 50° 20 F There are three ways to determine if two triangles are similar. 1. AAA similarity: If in two triangles, the corresponding angles are equal, the triangles are similar. (The 3rd set of angles will be equal, because of the sum of the interior angles of a triangle being 180°.) 2. SSS similarity: If the corresponding sides of two triangles are proportional the triangles are similar. 3. SAS similarity: If one angle of a triangle is equal to one angle of the other triangle and the sides containing these angles are proportional, the triangles are similar. Finding the scale factor We use the scale factor to find missing lengths of similar figures. Here, we can: ●● Find a corresponding side in each shape when we know the length of 15 cm both. 10 cm 12 cm 8 cm ●● Divide the length in the shape we are going to by the length in the shape we are coming from. 6 cm 9 cm ¿ 1,5 ¿  __23 76 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 76 16/01/2012 12:43

Unit 1 Example To find the scale factor from A to B, divide the length in B by the length in A: 9 ÷ 6 = 1,5 The scale factor from A to B is 1,5. Similarly, to find the scale factor from B to A, divide the length in A by the length in B: 6 ÷ 9 =  2__3 The scale factor from B to A is  2__3 . To prove that two triangles are similar, we have to show that one (not all) of the following statements is true: ●● The three sides are in the same proportion. ●● Two sides are in the same proportion, and their included angle is equal. ●● The three angles of the first triangle are equal to the three angles of the second triangle. Notes: ●● The symbol for similarity is ||| ●● The order of the points in the names of the triangles is important. Equal angles of the two triangles must coincide. © Via Afrika – Mathematics 77 • Via Afrika Maths Study Guide Gr10 77 16/01/2012 12:43

Unit 2 Congruent triangles Congruent triangles have the same size and the same shape. The corresponding sides and the corresponding angles of congruent triangles are equal. There are four methods to prove if two triangles are congruent. 1. The Side-Side-Side principle (SSS): A D The three corresponding sides of two triangles must be the same. B C E F 2. The Side-Angle-Side principle (SAS): A D The two corresponding sides and the included angle of two triangles must be equal. B C E F 3. The Angle-Side-Angle principle (ASA): A D The two corresponding angles and the included side of two triangles must be equal. B C E F 4. The Right-Angle-Hypotenuse-Side principle A D (RHS): The hypotenuse and one corresponding side of the two right-angled triangles must be equal. B C E F Example ∆ABC ≡ ∆DFE because: B E 53° 53° 5 cm 3 cm 5 cm 3 cm 37° 37° C 4 cm A F 4 cm D ∠A = ∠D = 90° ∠B = ∠E = 53° ∠C = ∠F = 37° 78 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 78 16/01/2012 12:43

Unit 12 Notes: ●● The symbol for congruency is ≡ ●● The order of the points in the names of the triangles is important. Equal angles of the two triangles must coincide. © Via Afrika – Mathematics 79 • Via Afrika Maths Study Guide Gr10 79 16/01/2012 12:43

Unit 3 Quadrilaterals ●● A polygon with four sides is called a quadrilateral. ●● The special types of quadrilaterals include the parallelogram, rectangle, rhombus, square, trapezoid and kite. ●● It is important to understand the relationship of these figures and their properties in order to properly classify or identify a figure. ●● A conjecture is an assertion that is likely to be true but has not been formally proven. ●● To prove conjectures you need to complete a proof that will always be true. ●● To show that a conjecture is false, show one counter example. 3.1 Parallelogram Properties: ●● opposite sides are parallel ●● opposite angles are congruent ●● opposite sides are congruent ●● diagonals bisect each other ●● consecutive angles are supplementary Theorems: 1. If one pair of opposite sides of a quadrilateral is equal and parallel, then the quadrilateral is a parallelogram. 2. If both pairs of opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram. 3. If both pairs of opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram. 4. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. 3.2 Rectangles Properties: ●● has all the properties of a parallelogram ●● diagonals are congruent ●● contains four right angles Theorem: 1. If the diagonals of a parallelogram are equal, then the parallelogram is a rectangle. 80 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 80 16/01/2012 12:43

Unit 13 3.3 Rhombus Properties: ●● has all the properties of a parallelogram ●● four sides are equal in length ●● diagonals are perpendicular ●● diagonals bisect each pair of opposite angles Theorems: 1. If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. 2. If each diagonal of a parallelogram bisects a pair of opposite angles, then the parallelogram is a rhombus. 3.4 Square Properties: ●● has all the properties of a parallelogram ●● diagonals are congruent and perpendicular ●● is a rectangle with all sides congruent ●● is a rhombus with four right angles 3.5 Kite Properties: ●● is a quadrilateral that has exactly two distinct pairs of adjacent congruent sides 3.6 Trapezium Properties: ●● one pair of opposite sides that are parallel ●● two parallel sides are called bases and the nonparallel sides are the legs ●● isosceles trapezoid has one pair of congruent sides and congruent diagonals © Via Afrika – Mathematics 81 • Via Afrika Maths Study Guide Gr10 81 16/01/2012 12:43

Unit 13 Example 1 PQRS is a parallelogram with SY = QX. P Prove that PXRY is a parallelogram. Y Q Steps to follow: ●● Identify what is given and what to prove. S X ●● Redraw the diagram, filling in the information given. ●● Plan your proof using the drawing. R ●● Write up the proof explaining your reasoning using symbols and valid reasons in brackets. QR = PS (opp sides of parallelogram) QX = YS = PY = XR And PY||XR (given) Therefore, PYRX is a parallelogram (one pair of opp. sides || and =) 2 ABCD is a rhombus and AE = CE. B Prove that ΔADE = ΔCDE. Use congruency to prove the required angles are equal. Remember, a rhombus is a parallelogram with all sides A C equal. AE = CE (given) D Then AD = DC (all 4 sides equal) DE = DE (common side) ∴ ΔADE ≡ ΔCDE (SSS) E ∴ ΔADE = ΔCDE 82 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 82 16/01/2012 12:43

Questions 1 In the diagram, WX||ZY and ∆WXZ = ∆WYZ. W X Prove that WXYZ is a parallelogram. Z Y 2 In the diagram alongside, ABCD is a rhombus with A B BG = AE and CG = DE. Prove that EBGC is a rectangle. E G D C 3 Use the diagram alongside to prove that the D E diagonals of a parallelogram bisect each other. O G F 4 In the diagram alongside, AD = DB and AE = EC. Prove that: A a ∆ADE|||∆ABC b DE||BC c DE = BC D E B C © Via Afrika – Mathematics 83 • Via Afrika Maths Study Guide Gr10 83 16/01/2012 12:43

Chapter 8 Analytical geometry Overview When answering questions on analytical geometry: · ALWAYS make a sketch. It can be a rough sketch, but at least have the points in the correct quadrants of the Cartesian plane. · Read carefully and put all the information in the sketch. · Always ask yourself: Is this answer possible if I look at my drawing? · When using a calculator, always round off at the final answer only. UNIT 1 Page 85 • Using the distance formula The distance formula CHAPTER 7 Page 84 UNIT 2 Page 86 • Parallel and perpendicular lines Analytical geometry The gradient formula • Collinear points UNIT 3 Page 88 The midpoint formula 84 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 84 16/01/2012 12:43

Unit 1 The distance formula The distance between the two points is given by the formula: __________________ d=√ (x1 – x2)2 +    (y1 – y2)2  (Remember, the answer is a length) 1.2 Using the distance formula Example 1 Determine the distance between P(2; 5) and Q(–4; 1) on the Cartesian coordinate system. PQ2 = (xp – xq)2 + (yp – yq)2 = (2 – (–4))2 + (5 – 1)2 (Use the given values) 2 2 = (6) + (4) = 36 + 16 = 52 ___ ∴ PQ = √52 = 7,21 units (Usually rounded off to two decimal places) 2 The distance between A(–5; k) and B(7; –3) is 13 units. Determine the value(s) of k. Here, point A can be anywhere along the y vertical dashed line in the graph, as long as B(7; –3) the distance between A and B is 8 units. AB2 = (xa – xb)2 + (ya – yb)2 Substitute in the values given: A(–5; k) 132 = (–5 – 7)2 + (k – (–3))2 x 0 169 = 144 + (k + 3)2 = 144 + k2 + 6k + 9 k2 + 6k – 16 = 0 (k + 8)(k – 2) = 0 ∴ k = –8 or 2 © Via Afrika – Mathematics 85 • Via Afrika Maths Study Guide Gr10 85 16/01/2012 12:43

Unit 2 The gradient formula The gradient of the line joining the points is given by the following formula: y –y _____2 m =  x1 –x 1 2 2.2 Parallel and perpendicular lines ●● When two lines are parallel, their gradients y are the same. m 1 m 1 x 0 ●● When two lines are perpendicular then the m1m2 = –1 y product of their gradients equals –1. y = m1x + b1 0 x y = m2x + c1 ●● A vertical line has an undefined gradient. y Undefined gradient 0 x y ●● A horizontal line has a gradient of 0. Zero gradient 0 x 86 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 86 16/01/2012 12:43

Unit 12 2.3 Collinear points Points A, B and C are collinear if AB + BC = AC or if mAB = mBC AND point B is a common point. C B A Example 1 Show that ABC is a straight line (in other words, show that A(4; –2), B(2; 3) and C(–6; 7) are collinear). y 8 We need to calculate the C gradients between AB and BC. 6 If the gradients are the same, 4 the points are collinear. 2 B y – yB A _____ 2– ____ 3 __ 1 mAB =  xA – xB =   4 – 2 = –  2 A x –6 –4 –2 0 2 4 6 y – yB _____ 7– _____ 3 __ 1 mBC =  x C – xB =   –6 – 2 = –  2 –2 C Therefore, ABC is a straight line. 2 Determine the value(s) of k if P(8; –4); Q(k; 1) and R(–1; 3) are collinear. If the points are collinear, then mPQ must equal mPQ. Therefore: yp – yq yr – yq =  _____ _____ x – x =  x – x p q r q 3–1   –4 = _____ – 1 _____ 8–k =  –1 –k (Replace with coordinate values) –3(–1 – k) = 2(8 – k) (solve for k) 3 + k = 16 – 2k 3k = 13 ∴ k = 4 © Via Afrika – Mathematics 87 • Via Afrika Maths Study Guide Gr10 87 16/01/2012 12:43

Unit 3 The midpoint formula The coordinates of the midpoint of the line joining the points is given by: x +x y +y ( _____2 ;   1 M   1 2 _____2   2 ) (Remember, the answer is a set of coordinates) Example 1 Determine the midpoint, M, of CD, if the points are C(–4: 5) and D(2; –3). ( x +x y +y ) _____d   = M   –4 _____d ;   c M   c 2 2 ( + 2 ______ _____ 2 5 + (–3) ;   2   ) = M(–1; 1) 2 Determine the coordinates of P if Q is (–1; 5) and point R(3; 7) is the midpoint of PQ. Suppose the coordinates for P are (xp; yp). Use the midpoint formula to solve two equations, one for the x-value of P and one for the y-value for P. xp + xq yp + yq ( (3; 7) = _____   2 ; _____   2   ) x-coordinate of the midpoint: y-coordinate of the midpoint: xp + xq yp + yq xR = _____   2 yR = _____   2 xp + (–1) yp + 5 3 = ______   2 7 = ____   2 6 = xp – 1 14 = yp + 5 ∴ xp = 7 ∴ yp = 9 ∴P is the point (7; 9) 88 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 88 16/01/2012 12:43

Questions 1 Given: P(–4; –1); Q(6; 3); R(6; b) and S(–4; –3): a Determine the gradient of PQ. b If PQ is parallel to SR, determine the value of b. c Show that PQ = SR. d Is quadrilateral PQRS a parallelogram? Give a reason for your answer. e Calculate the midpoint of: i PR ii SQ f What can you subsequently deduce regarding the diagonals of a parallelogram? g A rhombus is a parallelogram with two consecutive sides that are equal in length. Show that PQRS is not a rhombus. 2 Using the sketch alongside (not drawn to scale), A(–4; 5) y calculate: a AB b the midpoint, K, of AB c the gradient of A 0 x 3 BD and AC are the diagonals of a parallelogram. If B(2; –3) B = (2; 3), D = (6; 0), C = (7; 5) and A = (x; y), find the values of x and y. 4 Given points A(3; 7), B(5; 11) and C(6; 3), find: a the length of AB, leaving your answer as a surd. b the length of BC, correct to two decimal places c the midpoint of AC d the gradient of AB and BC 5 Triangle ABC (alongside) is made up of the points y B(2; 6) A(–2; 3), B(2; 6) and C (8; k). Find the value of k, given that ΔABC = 90°. A(–2; 3) 6 G(3; 7), H(–5; 1), K(1; –3) and D(x; y) are the 0 x vertices of a parallelogram. C(8; k) a Calculate the length of the line HK. b Calculate the length of KD. c Find the coordinates of D. d Find the coordinates of the midpoint of HD. © Via Afrika – Mathematics 89 • Via Afrika Maths Study Guide Gr10 89 16/01/2012 12:43

7 The points M(–4 0), N(3; –7) and P(7; 4) are shown y P(7; 4) in the diagram alongside. Calculate the following: a Calculate the midpoint, Q, of MN. b Calculate the gradients of MN and PQ M(–4; 0) c Calculate the product of the gradients of MN 0 x and PQ. d What can you say about MN and PQ? N(3; –7) e Calculate the lengths of PM and PN. f Draw a conclusion about ΔPMN. 90 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 90 16/01/2012 12:43

Chapter 9 Finance, growth and decay Overview Interest is a fee paid to borrow money. It is usually charged or paid as a percentage of the total amount borrowed or invested. An interest rate is the cost stated as a percentage of the amount borrowed/invested per period of time, usually one year. Simple interest and compound interest are examples of how interest can be calculated. The compound interest formula can also be used in other areas in which compound growth occurs, such as inflation, exchange rates and population growth. UNIT 1 Page 92 • Calculating simple interest Simple interest • Hire purchase CHAPTER 9 Page 91 Finance, growth and decay UNIT 2 Page 94 • Compound interest Compound growth • Applications of compound growth © Via Afrika – Mathematics 91 • Via Afrika Maths Study Guide Gr10 91 16/01/2012 12:43

Unit 1 Simple interest Simple interest is calculated as a constant percentage based only on the amount of money invested/borrowed. Here, we do not receive or pay interest on any interest accrued during the period of the loan or investment. 1.1 Calculating simple interest ●● Simple interest is calculated on the original principal amount only. ●● The formula is: simple interest = P × i × n, where ●● P = principal (original amount borrowed or loaned) ●● i = interest rate for one period ●● n = number of periods Example 1 You borrow R10 000 for 3 years at 5% simple interest, paid annually. 5 ___= 0,05) P = R10 000; n = 3; i = 0,05 (5% =   100 interest = P × i × n = 10 000 × 0,05 × 3 = R1 500 2 a You borrow R10 000 for 60 days at 5% simple interest per year (assume a 365 day year). P = R10 000; n = (60 ÷ 365); i = 0,05 interest = P × i × n = 10 000 × 0,05 × (60 ÷ 365) = R1 500 Note: interest is calculated per year, but the loan is only for 60 days. This is a fraction part of the year. Therefore, n = (60 ÷ 365). b After the 60 days, you have to pay the money back, with interest. A = P + simple interest = R 10 000 + R82,19 = R10 082,19 A = P + P × i × n (take out a common factor) A = P(1 + in) Example An amount, b, is invested at 7,5% p.a. for 3 years to yield simple interest amounting to R500. Find the value of b. 92 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 92 16/01/2012 12:43

Unit 1 P × i × n = Simple interest 7,5 P = b; i = ___  100 = 0,075; n = 3 b ¿ 7,5 ¿ 3 _______   100 = 500 b = R2 222,22 1.2 Hire purchase Hire purchase is a system in which a buyer pays for something in regular instalments while enjoying the use of it. During the repayment period, the buyer does not own whatever was bought. The buyer only takes ownership once the loan has been paid in full. Example Jean wants to buy a motorbike for R10 500. He pays a deposit of R2 000. He wishes to pay the balance using a hire purchase agreement over 3 years. The interest charged on the loan is 18% per annum. Included in the agreement, is an insurance cost of 2% per annum on the purchase price of the motor bike. Calculate his monthly instalment. Balance owing after deposit = R8 500 Insurance = (2% of R10 500) × 3 years = R630 Interest on R8 500 = P ¿ i ¿ n = R8 500 × 18 × 3 = R4 590 8 500 + 630 + 4 590 Monthly Payment = ______________     36 = R381,11 © Via Afrika – Mathematics 93 • Via Afrika Maths Study Guide Gr10 93 16/01/2012 12:43

Unit 2 Compound growth 2.1 Compound interest ●● Compound interest is calculated each period on the original principal and all interest accumulated during past periods. ●● Although the interest may be stated as a yearly rate, the compounding periods can be yearly, semi-annually, quarterly, or even continuously. ●● You can think of compound interest as a series of back-to-back simple interest contracts. The interest earned in each period is added to the principal of the previous period to become the principal for the next period. For example, you borrow R10 000 for three years at 5% annual interest compounded annually: ●● interest year 1 = P × i × n = 10 000 × 0,05 × 1 = 500 ●● interest year 2 = (P2 = P1 + i1) × i × n = (10 000 + 500) × 0,05 × 1 = 525 ●● interest year 3 = (P3 = P2 + i2) × i × n = (10 500 + 525) × 0,05 × 1 = 551,25 ●● Total interest earned over the 3 years = 500 + 525 + 551,25 = 1 576,25 Compare this to 1 500 paid over the same period using simple interest. The formula for compound interest is A = P(1 + i)n, where ●● A = the amount at the end of a loan/investment period ●● P = the principle (initial amount borrowed/invested) ●● i = the interest rate (expressed as a decimal number) ●● n = the period of the loan/investment Example An amount of R1 500,00 is deposited in a bank paying an annual interest rate of 4,3%, compounded quarterly. What is the balance after 6 years? Using the compound interest formula, we have: P = 1 500, i = 0,043, n = 6 However, because the interest is calculated quarterly, we divide i by 4 and multiply n by 4. Therefore: ( A = 1 500 1 + _____ ) 0,043 4(6)   4   ≈ R1 938,84 So the balance after 6 years is approximately R1 938,84 94 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 94 16/01/2012 12:43

Unit 12 2.2 Applications of compound growth 2.2.1 Inflation Inflation refers to an ongoing general increase in prices. Use the compound growth formula if you need to work out how much something will cost in future: A = P(1 + i)n Inflation means that we can buy less in future, so if you need to work out what is something worth in future, just an adjusted formula: A = P(1 – i)n Example If the average rate of inflation for the past few years was 7,3%, and your family’s water and electricity bill is on average R1 425, what can you expect to pay in 6 years’ time? A = P(1 + i)n 2.2.2 Exchange rates ●● An exchange rate gives the relationship between two countries’ currencies. Example 1 Ruff wants to import a TV from England at a price of £507, plus a delivery cost of 15% and import duty of 20%. The equivalent TV locally costs R13 000. The exchange rate is R14,08 to the pound. Show, with necessary calculations, whether Ruff should import or buy locally. £507 × 14,08 = R7 138,56 Delivery cost = R7 138,56 × 15% = R1 070,78 Import duty = R7 138,56 × 20% = R1 427,71 Total cost = R 9 637,05 Conclusion: It costs less to import the TV 2 If the exchange rate is 1 euro = R8,1671 and 1 pound sterling = R12,1668, determine the exchange rate between the euro and the pound. 1 euro = R8,1671 1 rand = 1 ÷ 8,1671 = 0,1224 euro 1 pound sterling = R12,1668 1 rand = £(1 ÷ 12,1668) = £0,0822 0,1224 euro = £0,0822 1 euro = £0,6713 © Via Afrika – Mathematics 95 • Via Afrika Maths Study Guide Gr10 95 16/01/2012 12:43

Unit 12 2.2.3 Population growth ●● Since we are referring to growth, we use the formula A = P(1 + i)n. ●● Here, i = rate of population growth and P = initial size of the population. Example There are 12 500 people in a small town. The population of the town increases every year by 5,5%. What will the population of the town be after 5 years? A = 12 500(1 + 0,055)5 A ≈ 16 337 96 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 96 16/01/2012 12:43

Questions 1 Mr du Toit decided to invest an amount of money at an interest rate of 9% p.a. compounded annually. He would like R70 000 after six years. How much money must he invest now? (4) 2 A person invests R30 000 in a savings account for eight years at a simple interest rate of 6% p.a. Calculate the accumulated amount. (4) 3 A motor vehicle is purchased for R105 000. Of this, 20% is paid in cash and the balance is paid using a hire purchase agreement for seven years by means of equal monthly payments. The bank decides to charge an interest rate of 14,5% p.a. for the seven-year period. Calculate: a The loan amount (4) b The full amount including interest (2) c The monthly repayments (3) d The total amount paid if the insurance on the car per month is R780 (4) 4 A mother decides to buy a baby’s pram from Kuwait. The price of the praxm is 80 Kuwaiti dinars. Calculate how much she will pay if the exchange rate is 1 Kuwaiti dinar to R26,09. (4) 5 A tourist from the United Kingdom buys computers from South Africa at a cost of R530 000. He wants to pay for these computers in British pounds. If the exchange rate is 1£ = R11,72, how much will he pay in pounds? (To the nearest pound) (4) 6 Mr Dube wants to buy a car. So he borrows R420 000 and agrees to settle this amount in eight years together with simple interest charged at 12,5% p.a. How much will he have to pay in eight years from now? (4) 7 Ronnie wants to receive an amount of R180 000 after 10 years. How much must he invest now at an interest rate of 7,25% p.a. compounded annually (to the nearest rand)? (4) 8 The population of Phoenix is growing at a rate of 4% p.a. compounded annually. If the population is now 50 000, what will be the percentage growth in five years time? (Give your answer correct to two decimal places) (4) 9 Oil costs $76 a barrel. The exchange rate is 1$ is equal to R7,23. If South Africa were to import 40 000 barrels , how much will the oil cost (in rand)? (4) 10 Matthew invests R30 000. The institution offers Matthew two investment options: Option 1: 8 years at 12,5% p.a. simple interest. Option 2: 8 years at 12,5% p.a. compound interest. Which option should Matthew choose? Motivate your answer. (9) © Via Afrika – Mathematics 97 • Via Afrika Maths Study Guide Gr10 97 16/01/2012 12:43

Chapter 10 Statistics Overview Statistics is the branch of Mathematics that is used daily over a broad spectrum of common everyday life to gather and interpret information. This information or data enables us to predict the future, to make choices and to improve existing frameworks. UNIT 1 Page 99 Measures of central • The mean, median and mode tendency: ungrouped data UNIT 2 Page 101 • The range Measures of dispersion • Quartiles • Interquartile range CHAPTER 10 Page 98 • Semi-interquartile range • Percentiles Statistics • The five number summary UNIT 3 Page 103 Box-and-whisker plot UNIT 4 Page 104 • Measures of central tendency Grouped data • More about organising data 98 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 98 16/01/2012 12:43

Unit 1 Measures of central tendency: ungrouped data There are mainly two different types of numerical which could be collected, namely discrete and continuous data. ●● Discrete data is data with specific values normally gathered by means of counting, e.g. we are three children in our family. ●● Continuous data is data that can occupy any value between two points and is usually obtained through measurement, e.g. I am 1,76 m tall. 1.1 The mean, median and mode Mean The arithmetic mean is commonly known as the average. It is calculated by the sum of the values in the sample group (∑x), divided by the number of values in the sample group (n). ∑x Mean = __   n Median The median is the middle value in the sample group, after they have been arranged in ascending order. In other words, it is the value in the middle of the ordered data. If there is an even number of values, the median is the average of the two middle values. Mode The mode is the value that appears most often. In other words, it is the value with the highest frequency. How do we decide which measure to use? It is important to use the most appropriate measure of central tendency in order to make a sound conclusion. (You sometimes have to answer such a question in a test or exam.) Measure Advantages Disadvantages Average/Mean • Most common measure • If an extreme (very large or very • Used widely in media small) value is added, the average • Easy to calculate changes drastically • All values in the sample group are • Cannot be used when the data value used is simply a category Median • Easy to identify if the sample group • Time consuming if the sample if very is small big and not in ascending order Mode • No calculation needed • A small sample usually does not • Easy to deduce in grouped data have a mode • Is used to describe any kind of data • More than one mode is possible © Via Afrika – Mathematics 99 • Via Afrika Maths Study Guide Gr10 99 16/01/2012 12:43

Unit 1 Example How long is the average name? The number of letters in the first names of 26 learners was recorded as follows: 4 6 3 4 7 6 5 6 8 4 3 9 4 5 4 6 7 8 8 4 4 4 6 6 5 9 Arithmetic mean/average: sum of all the letters 145 ______________    = ___ Mean =     number of names   26 = 5, 57… ≈ 6 letters Median: First arrange in ascending order: 3 3 4 4 4 4 4 4 4 4 5 5 5 6 6 6 6 6 6 7 7 8 8 8 9 9 There is an even number of values, therefore: middle terms 5 + sum of the ________________ Median =     2   2 6 = 5,5 ≈ 6 letters = ____ Mode: 4 occurs most often and in thus the mode. 100 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 100 16/01/2012 12:43

Unit 2 Measures of dispersion 2.1 The range The range = largest value – smallest value. It gives us an indication of how big the span of the data is. For example, does the data lie between 1 and 10 or 1 and 100? 2.2 Quartiles The quartiles divide the data (in ascending order) into four quarters. Every quarter has the same number of items. There are three quartiles: ●● The lower quartile (Q1) ●● The middle quartile or median (Q2) ●● The upper quartile (Q3) 2.3 Interquartile range The interquartile range is the difference between the third quartile and the first quartile. It measures the range of the middle 50% of the data (in ascending order). A large interquartile range indicates the data values are widely dispersed. ●● Interquartile range = Third quartile – First quartile ●● IQR = Q3 – Q1 2.4 Semi-interquartile range The semi-interquartile range is half of the range of the middle half of the data. It is seldom influenced by extreme values in a set of data, and so is a good measure of dispersion. Third quartile – First quartile Semi-interquartile range = ____________________     2 2.5 Percentiles ●● The percentiles divide the data into 100 equal parts. ●● Every part has the same number of items. ●● Percentiles give an indication of how many values in the data set are smaller than a certain value. ●● For example, the 56th percentile of 70 means that 56% of the other values are less than 70. © Via Afrika – Mathematics 101 • Via Afrika Maths Study Guide Gr10 101 16/01/2012 12:43

Unit 12 2.6 The five-number summary The five-number summary is five-point scale that we can use to summarise the information about a data set. The five-number summary contains: ●● Minimum value ●● First or lower quartile: 25% of the data lies below the first quartile ●● Median: 50% of the data lies below and above the median ●● Third or upper quartile: 25% of the data lies above the third quartile ●● Maximum value Example Consider the following values 12 6 4 9 8 4 9 8 5 9 8 10 7 9 12 14 5 8 4 6 6 4 8 2 6 5 10 The first thing to do when confronted with a new set of data is to arrange it in ascending order: 2 4 4 4 4 5 5 5 6 6 6 6 7 8 8 8 8 8 9 9 9 9 10 10 12 12 14 ●● Range = 14 – 2 = 12 ●● Minimum value = 2 ●● First quartile (Q1) = T7 = 5 (For discrete data: 0,25 ¿ 27 = 6,75 thus 7th term) ●● Median = 8 ●● Third quartile(Q3) = T21 = 9 (For discrete data: 0,75 ¿ 26 = 20,25 thus 21st term) ●● Maximum value = 14 ●● Inter-quartile range = 9 – 5 = 4 Third quartile – First quartile ●● Semi-inter-quartile range = ____________________     2 =2 102 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 102 16/01/2012 12:43

Unit 3 Box-and-whisker plot The box-and-whisker plot is a graphical representation of the five-number summary. ●● The left side of the box is the lower quartile ●● The vertical line inside the box is the median ●● The right side of the box is the upper quartile. ●● The two lines on either side of the box extend out to the minimum and maximum values. Example Below is the box-and-whisker plot for our earlier earlier example: ●● Minimum value = 2 ●● First quartile (Q1) = T7 = 5 (For discrete data: 0,25 ¿ 27 = 6,75 thus 7th term) ●● Median = 8 ●● Third quartile (Q3) = T21 = 9 (For discrete data: 0,75 ¿ 26 = 20,25 thus 21st term) ●● Maximum value = 14 0 5 10 15 © Via Afrika – Mathematics 103 • Via Afrika Maths Study Guide Gr10 103 16/01/2012 12:43

Unit 4 Grouped data 4.1 Measures of central tendency A frequency table is used to sort and count information. Earlier, we collected data about the first names of 26 learners: 4 6 3 4 7 6 5 6 8 4 3 9 4 5 4 6 7 8 8 4 4 4 6 6 5 9 We can count how many times each length of name occurs and summarise this information in a frequency table. In this way, we group the data into categories, or class intervals. Total letters Tally Frequency 3 || 2 4 |||| ||| 8 5 ||| 3 6 |||| | 6 7 || 2 8 ||| 3 9 || 2 The maths marks of 220 grade 10 learners at a school can be summarised as follows: Percentage (Class interval) Number of learners (frequency) 4 10 37 43 36 26 24 20 ∑f = 200 How to calculate the approximate mean of grouped data: ●● Step 1: Determine the midpoint for each interval. ●● Step 2: Multiply the class midpoint by the frequency. ●● Step 3: Add up the results from Step 2. ●● Step 4: Divide the total from Step 3 by the frequency. 104 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 104 16/01/2012 12:43

Unit 4 Class intervals Frequency (f ) (Step 1) Midpoints (x) (Step 2) f × x 4 9,5 38 10 24,5 245 37 34,5 1 276,5 43 44,5 1 913,5 36 54,5 1 962 26 64,5 1 677 24 74,5 1 788 20 84,5 1 690 0 94,5 0 ∑f = 200 ∑fx =10 590 (Step 3) Step 4: Approximate mean = ∑fx ÷ ∑f = 10 590 ÷ 200 = 52,95 Central tendency In which class it falls Median 50 ≤ x < 60 (The middle (100th) score is in this class) Lower quartile 30 ≤ x < 40 (25% or less of the marks are in this class) Upper quartile 60 ≤ x < 70 (More than 75% of the marks are in this class or lower) 13th percentile 30 ≤ x < 40 (13% is about 26 marks and lower – the frequency up to this class is 51) Modal class 40 ≤ x < 50 (The class with the highest frequency) highest value of the interval – lowest value ●● The midpoint of a class: _____________________________         2 ●● ∑f: Sum of the frequencies ●● ____ The position of median:  n + 1 2 More about organising data The following table shows the costs (in rand) of 50 new members of a cellphone company. The company would like touse the data to help with its planning. 42,19 5,64 83,26 99,56 3,43 38,45 91,1 27,56 99,5 106,84 95,73 16,44 3,69 2,42 115,78 104,8 99,03 5,1 23,41 21 22,57 14,34 18,49 21,13 72,02 92,97 20,55 15,3 6,48 15,42 88,62 117,69 29,23 10,88 64,78 115,5 33,69 103,15 109,08 19,34 119,63 92,17 74,01 29,24 3,03 13,26 11,27 63,7 79,52 85,67 When large amounts of data need to be studied and interpreted, it is sometimes necessary to group the data, and then study the each group. © Via Afrika – Mathematics 105 • Via Afrika Maths Study Guide Gr10 105 16/01/2012 12:43

Unit 4 It is important to decide how many groups you would like to use before you start organising the data. In this case, we will use six groups for the data. The size of each group is: largest value – smallest value divided by the number of groups. Largest number – smallest number 119,63 – 2,42 Group size: _______________________ number of groups = __________ 6 = 19,5 ≈ 20 The next step is to draw up a frequency table. Telephone costs (R) Frequency (f) Midpoint (x) f×x 0 – 19 16 9,5 152 20 – 39 10 29,5 295 40 – 59 1 4,5 49,5 60 – 79 5 69,5 347,5 80 – 99 10 89,5 895 100 – 120 8 109,5 876 n = 50 2 615 ∑f ¿ x Estimated average = ____ 50 = 2____ 615 50 = 52,3 ∑x 2 647,7 Arithmetic mean = __ 50 = _____ 50 = 53 We can represent this data graphically in a histogram or a frequency polygon. 20 15 Frequency polygon 10 5 histogram 0 10 20 30 40 50 60 70 80 90 100 110 120 We can also represent the data in a pie chart, as follows. Telephone costs in rand 0–19 20–39 40–59 60–79 80–99 100–120 106 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 106 16/01/2012 12:43

Questions 1 Consider the pie chart below showing the number of deaths on South African roads over the Easter weekend of 2009. Road fatalities per Province (Easter Weekend, 2009) Northern Cape, 2 Gauteng, 26 Limpopo, 35 KwaZulu-Natal, 37 North West, 23 Western Cape, 9 Mpumalanga, 22 Source: Road Traffic Management Corporation, Easter 2009 Road Eastern Cape, 21 Free State, 22 Traffic Report; Wed, 15 April 2009 a How many people lost their lives due to road accidents in South Africa during the Easter weekend of 2009? b Which province had the highest number of road fatalities? c Why does Northern Cape such a small number of road fatalities? d What percentage of road deaths occurred in KwaZulu-Natal? 2 The following table shows the heights (in cm) of the learners in a class. 163 137 146 166 163 134 146 141 134 157 147 141 134 157 136 158 157 188 158 144 157 145 144 158 144 146 137 134 137 188 a Determine the range, mean, median and mode of the heights in the class. b Based on the range, do you think that the heights of the learners vary considerably, or do the learners have fairly similar heights? Explain your answer. 3 The following table shows the percentage test marks for the learners in a class. The marks have been organised separately for the female and male learners. Females 70% 52% 48% 64% 54% 66% 50% 68% 62% 52% 70% 60% 66% 56% 58% 56% 70% Males 52% 74% 60% 22% 60% 24% 90% 58% 62% 70% 46% 74% 40% 34% © Via Afrika – Mathematics 107 • Via Afrika Maths Study Guide Gr10 107 16/01/2012 12:43

a Determine the range for the marks of the female learners. b Determine the range for the marks of the male learners. c Compare the range values for the female and male learners. What does the difference between these values tell you about the spread of the marks for each of these groups. d i Complete the following frequency table to organise the marks for the female and male learners. Females Males Mark category No. of learners who achieved marks No. of learners who achieved marks in this category in this category 0 – 29% 30 – 39% 40 – 49% 50 – 59% 60 – 69% 70 – 79% 80 – 89% 90 – 100% ii Draw two separate histograms to represent the marks scored by the females and the males. You will need to construct your own axes for the graphs. iii Compare the two graphs and describe what you notice about the difference between the spread of the marks for the females and the males. iv Do you think the female learners or the male learners performed better in the test? Explain your answer. 4. Three learners are applying for a bursary and have to submit marks for all their subjects in their application. The marks are given below: Learner A: 55% ; 33% ; 22% ; 69% ; 58% ; 29% Learner B: 44% ; 11% ; 33% ; 22% ; 55% ; 49% Learner C: 53% ; 54% ; 64% ; 49% ; 45% ; 64% a Determine the average (mean) of the marks for each learner. b Determine the range of the marks for each learner. c If you were responsible for deciding which student should get the bursary, who would you choose? Give reasons for your answer using the mean and the range values. 108 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 108 16/01/2012 12:43

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Chapter 11 Measurement Overview In this chapter, we focus on three-dimensional (3D) solids. The surface area of a 3D solid refers to the area of the outside surface of the solid. The volume refers to the amount of space inside the 3D solid. UNIT 1 and 2 Page 111 Right prisms and CHAPTER 11 Page 110 cylinders Measurement UNIT 3 Page 113 • A Sphere The volume and • A right circular cone surface area of • A pyramid complex-shaped solids 110 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 110 16/01/2012 12:43

Unit 1 and 2 Right prisms and cylinders ●● A prism is a polyhedron (solid) with two congruent base faces, called bases, that lie in parallel planes. lateral edges ●● The other faces, called lateral faces, are parallelograms formed by connecting the lateral height faces corresponding vertices of the bases. ●● The segments connecting these vertices are lateral edges. base ●● A prism can be cut into slices, which are all the Right rectangular prism same shape. Prism Volume Surface area Rectangular prism length × breadth × height 2lw + 2lh + 2wh height len gth width Triangular prism base ¿ height __________ b.h + (S1+ S2 + S3) × h   2 ¿ height of prism Area length Cylinder V = π r2 ¿ h 2π rh + 2π r2 base radius r π r2 base areas lateral height h area 2πrh π r2 base Cube V = s3 Area of six cubes: s = side length of cube Surface area = 6s2 © Via Afrika – Mathematics 111 • Via Afrika Maths Study Guide Gr10 111 16/01/2012 12:43

Unit 1 and 2 Example Find the height of a cylinder which has a radius of 6,5 cm and a surface area of 592,19 cm2. h 1 Make h the subject of the formula. 2 Substitute values into the formula. 3 Do not leave answer in terms of π , unless specifically asked to do so. 6,5 cm 4 Do not forget the units. Surface area = 2π rh + 2π r2 2π rh = SA – 2π r2 2 SA – 2π r h =  _______ 2π r h = 592,19 – 2π(6,5)2 h ≈ 8 cm Example A right circular prism (cylinder) has a volume of 50 units3, a radius r and height h. If the radius is tripled and the height halved, what is the new volume of the cylinder? Vnew = π(3r)2 (  __2h   ) = 9r2π (   2h__  ) = ( __   92   )πr2h (πr2h = 50) = (   9__2   )¿ 50 = 225 units3 112 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 112 16/01/2012 12:43

Unit 3 The volume and surface area of complex-shaped solids 3.1 A sphere ●● Surface area = 4πr2 ●● V =  4__3 πr3 center 3.2 A right circular cone A cone is simply a pyramid with a V =  1__3 π r2h πrs + πrs2 circular base. • Area of the cone section + area of circle V • Curved surface area (without the base) = πrs • s = slant height slant height s r Circumference of the base = 2πr 3.3 A pyramid A pyramid is made by connecting a base to an apex. igh t height nt h e sla perimeter © Via Afrika – Mathematics 113 • Via Afrika Maths Study Guide Gr10 113 16/01/2012 12:43

Unit 3 Rectangular base pyramid V =  1__3 Ah Area of the base +  1__2 A = Area of the base perimeter of base ¿ slant height h l s Triangular base pyramid (also called a V =  1__3 Ah 4 ¿ Area triangle: tetrahedron) A = Area of the base Area __ of regular triangle √3  = __   4 ¿ side__2 V = __  1 Ah 3 √3  SA = 4 ¿ __   4 ¿ side2 V =  1__3 ¿   1__2 ¿ b ¿ hbase ¿ Hpyramid __ ∴ SA = √3 ¿ side2 V =  1__6 bh ¿ H Example 1 Consider two types of containers, each 15 cm deep. One is a rectangle pyramid, with a base of 4 cm by 7 cm, and the other is a cone with radius 3 cm. Determine which container holds more water when full, and by how much? Volume of pyramid =  __13 × 4 × 7 × 15 = 140 cm3 Volume of cone =  __13 × 3 × 3 × 15 × π = 45π cm3 Difference = 45π cm3 – 140 cm3 = 1,37 cm3 Therefore, the cone holds more water. 18 2 The sketch alongside shows the cross-section of a swimming pool. Determine the surface area and the 2,6 m 1,3 m volume of the pool. (Remember: the top is open!). 6m 4m 7m ______ Surface area = (2,6 × 4) + (1,3 × 4) + (6 × 4) + (7 × 4) + (4 × √ 26,69 ) + 2(1,3 × 18) 1,3 + 2 ¿ ___   2 (12 + 7) = 159,76 m2 1,3 Volume = [(1,3 ¿ 18) +  ___ 2 (12 + 7)] ¿ 4 3 = 143 m 114 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 114 16/01/2012 12:43

Unit 3 Note: If you have a cone with no base, don’t add the base area. There are different ways to calculate the surface area of a cone. Remember the basic concept: you add the cone’s slanted area to the cone’s base area. © Via Afrika – Mathematics 115 • Via Afrika Maths Study Guide Gr10 115 16/01/2012 12:43

Questions 1 A cube with a side length of 12 cm is filled to the top with water. The water is carefully poured into a rectangular prism with a length of 18 cm and a width of 8 cm. Calculate the height of the water in the rectangular prism. 2 A cylindrical hole has been drilled through the centre of a 10 cm solid cube (see figure alongside). The diameter of the cylindrical hole is 3 cm and its height is perpendicular to the two opposite faces of the cube. What is the total surface area of the cube (correct to two decimal places)? 3 A hollow sphere (e.g. a tennis ball) has an interior radius of 15 mm and an exterior radius of 20 mm. Calculate the volume of the material forming the sphere in cubic centimetres. 4 A metal top in the shape of a cone has a perpendicular height of 70 mm. If it displaces 4 224 cm3 of water when fully immersed, calculate the total surface area. 116 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 116 16/01/2012 12:43

Chapter 12 Probability Overview Probability refers to the likelihood or chance of an event taking place. Probability is used on a daily basis by businesses, insurance companies, engineers, climatologists and even doctors to predict future events. UNIT 1 Page 118 • The language of probability Probability • Theoretical probability • Relative frequency CHAPTER 12 Page 117 Probability UNIT 2 Page 120 • Venn Diagrams Combinations of • Using Venn diagrams to solve probability events problems © Via Afrika – Mathematics 117 • Via Afrika Maths Study Guide Gr10 117 16/01/2012 12:43

Unit 1 Probability ●● Probability is the study of chance. number of favourable outcomes ________________________ ●● Probability is calculated as follows: P(A) =         total number of outcomes 1.1 The language of probability To illustrate each of the following terms, consider rolling a normal six-sided die. Suppose we want to know the probability of rolling an odd number: ●● Outcome: the result of an experiment (roll an odd number) ●● Sample space: the set of all possible outcomes (set {1; 2; 3; 4; 5; 6}) ●● Event: a subset of the sample space (set {1; 3; 5}) ●● An event that will definitely take place has a probability of 1. ●● An event that can never take place has a probability of 0. ●● All other events fall somewhere between these two extremes. ●● Probability can be expressed as a fraction, decimal fraction or percentage. 1.2 Theoretical probability The outcome of rolling a die will be a number from 1 to 6. We want to know the probability of a successful outcome, which is one that matches an event. If we let n(S) be the number of items in the sample space, n(A) be the number of items in event A, and P(A) be the probability that A will occur: number of successful outcomes n(A) P(A) = ____________________________    = ____      total number of all possible outcomes   n(S) In our example, A is the possibility of rolling an odd number. Since there are three odd numbers between 1 and 6, we have: n(A) P(rolling an odd number) = ____   36 = __   = __   12 n(S) 1.3 Relative frequency The relative frequency is the actual frequency of an event during an experiment. In practice, many instances occur where the probability of an event is not easy to calculate. The only way to determine the probability of such an event is to determine its relative frequency through many experiments. ●● Relative frequency = __(where: x = total favourable outcomes and N = total possible x N outcome in the sample space.) 118 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 118 16/01/2012 12:43

Unit 1 Example Cat and his friend Ratatouille learnt in their Maths class that the probability of a coin falling on heads is 50%, but struggle to believe it. Cat is the captain of the U16 soccer team and has lost many a coin toss on a Saturday morning. The two friends decide to test the theory experimentally. Cat flipped the coin and Ratatouille recorded the results. The results after 80 flips were as follows: H H T H T T H H H T H T T T H T H T H T T H T T H H T T T H H T T T H H H H T T T T H H T H T H T H T T T H T H T T T H T H T H T H H T H T H H T T H H H H H H According to the data, what is the relative frequency of heads (H) after: 1 10 throws? 2 20 throws? 3 30 throws? 4 80 throws? 6 __= 0,6 1 After 10 throws: RF =  10 2 After 20 throws: RF =  10 __= 0,5 20 3 After 30 throws: RF =  14 __= 0,47 30 4 After 80 throws: RF =  40 __= 0,5 80 The relative frequency of an event is predicted with the aid of an experiment or an investigation. © Via Afrika – Mathematics 119 • Via Afrika Maths Study Guide Gr10 119 16/01/2012 12:43

Unit 2 Combination of events 2.1 Venn diagrams Venn diagrams are used to visually represent the sample space of an experiment and the relationship between the events in the sample space. Remember: The sample space contains all the possible outcomes of an event. For example, suppose a container contains 7 red marbles and 6 blue marbles. The collection of red marbles is a subset of all the marbles. The same is true for the collection of blue marbles. ● Number of marbles in the sample space S Blue marbles = n(S) = 13 ● Number of marbles in the red 7 subset= n(R) = 7 6 ● Number of marbles in the blue subset = n(B) = 6 Red marbles ● thus: n(S) = n(R) + n(B) 2.2 Using Venn diagrams to solve probability problems ● The union of two collections, A and B, contains all elements that occur in A or B: Union of A or B = A ∪ B ● The intersection of two collections, A and B, S A B contains all elements that occurs in both A and B: Intersection of A and B = A ∩ B ● Two collections are disjoint if there are no elements belonging to both collections. In other words: n(A ∩ B) = 0. Mutually exclusive events In this case, if A happens, B cannot happen, and S A B vice versa. ● P(A OR B) = P(A) + P(B) ● The events cannot both happen at the same time. ● The sets of outcomes for the events are disjoint (have no elements in common). Therefore P(A B) = 0 n(A) + n(B) ● P(A ∪ B) = ________ n(S) = P(A) + P(B) ● n(A ∪ B) = n(A) + n(B) 120 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 120 16/01/2012 12:43

Unit 2 Complementary events Events A and B are complementary if, together, A and B form the S A complete sample space. ●● n(A) + n(B) = n(S) ●● P(A) + P(B) = 1 P(A) = 1 – P(B) ●● A = B′ = complement of A ●● If the events A and B are not mutually exclusive, then: P(A or B) = P(A) + P(B) – P(A and B) Independent events The outcomes of independent events do not influence each other. When A and B are independent events: P(A AND B) = P(A) × P(B). © Via Afrika – Mathematics 121 • Via Afrika Maths Study Guide Gr10 121 16/01/2012 12:43

Questions 1 Forty-four learners plan to go camping over the long weekend. Two of the learners say that they have too many tests to write the next week and are going to stay at home. The rest then decide to camp in the Cederberg mountains. They have to decide if they want to camp in tents (T) or hire bungalows (R). Of the learners, 28 voted for camping in tents and 17 for bungalows (Note: some learners voted for both). a Draw a Venn diagram to represent the information. b Use the Venn diagram to determine the following: i n(T ∩ R) ii P(T) iii P(T ∪ R) iv n(R′) v n(S) vi n(T or R) 2 Suppose you flip a coin and flick the pointer in the diagram alongside. 1 a How many outcomes are possible? List these. b Determine P(heads and the value of 2). c Determine P(tails and an uneven number). 2 3 122 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 122 16/01/2012 12:43

Questions Mid-Year Exam Paper Question 1 a   5x Give a value for x that will make ______+4 ______ √x – 4x  i imaginary ii undefined (2) b Consider the following three rows containing true statements: Row 1: 32 – 2 = 3 + 22 Row 2: 42 – 3 = 4 + 32 Row 3: 52 – 4 = 5 + 42 i Find two other rows of numbers for which similar statements are true. ii Describe a conjecture about these rows of numbers. iii Use a variable to write an algebraic statement for your conjecture to generalise the relationship between the left-hand side and the right-hand side. iv Hence prove or disprove your conjecture. v Write down the nth row of numbers. (11) [13] Question 2 a Simplify: n–2 n+3 2 .3 i  _______ 6n (4) ii   4x + ______ 3 12 –   3x + ______ 16 4 (4) b Multiply and simplify: (2x + 3)(3x2 + 2x – 1) (4) c Factorise: i 2x2 + 5x – 12 (2) ii 12an – 8a + 3mn – 2m (3) x2(x + 2) – 4(x + 2) iii (4) [21] Question 3 a 10 – 2a ≤ 84 (3) b 5.27 – 2a = 40 (3) [6] Question 4 Solve for x and y: 3x + 2y = 5 and 2x – 3y = 12 [6] Question 5 Solve for x: a (x – 2)(x + 3) = 6 (4) © Via Afrika – Mathematics 123 • Via Afrika Maths Study Guide Gr10 123 16/01/2012 12:43

Questions 2 x – 4x + 3 ____ b  ________ 3–x =  x +2 1 (6) c x(x + 2) = 5x(x + 2) (2) [12] Question 6 A conical tank is shown alongside. The height of water at any point is h = 6 –  3__4 r, where r is the radius of the water surface in metres. a Find the depth of the full tank. b i How does the height (h) change with every 1 m decrease in r? ii Find the radius of the bottom of the tank. iii What is the circumference of the bottom of the tank? Give your answer correct to two decimal places. iv How much water does the tank hold when full? [10] Question 7 In the following sketch, the lines AB and y CD intersect on the x-axis, with AB⊥CD. A(–2; 3) D Calculate: a the length of the line AB leaving your answer in surd form. x (3) b the midpoint of AB (2) c the equation of AB B(1; –3) (4) C d the equation of CD (4) [13] Question 8 Consider the following raw data: 2; 4; 7; 11; 11; 15; 18; 19; 22; 31 Calculate and write down: a the range b the mode c the median d the upper quartile e the mean [8] 124 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 124 16/01/2012 12:43

Questions Question 9 Two Grade 7 classes recorded their mass as listed below: Mass of Grade 7 boys Frequency 36–40 3 41–45 8 46–50 12 51− 55 20 56–60 10 61–65 6 66–70 1 a Write down the class boundaries of the modal class. b On the graph paper provided draw a cumulative frequency curve. c Use your curve to estimate i the median ii the 70th percentile [11] [100] © Via Afrika – Mathematics 125 • Via Afrika Maths Study Guide Gr10 125 16/01/2012 12:43

Questions End-of-year sample exam paper: Paper 1 Question 1 . a 7as a fraction  ba__where a, b ∈ Z. Showing all work, express 0.5     (4) b Simplify each of the following algebraic expressions: i (5a2)3 (2) ii 5(a2)3 (2) 6 6 12a b iii  ______ –4a–2b3 (3) c Without using a calculator, ___ explain fully how you would determine between 3 which two integers √90 lies. (3) [14] Question 2 Given: 5; 8; 11; 14; … a What are the next three terms? (1) b Determine the rule for the general term of the row. (3) c Determine the 49th term. (2) [6] Question 3 a Shrek decides to invest his hard-earned bonus of R1 005 000 from playing in the Rugby World Cup immediately. The recent interest rate at The People’s Bank is 6,5% simple interest per year. For how many years must he invest his money in order to have at least R2,2 million to buy Fiona the engagement ring she would like to have? (3) b Ben inherited R8 000 from his grandparent’s estate. In 3 years, Ben needs R10 000 to pay for college expenses. What rate of compound interest is necessary for the inheritance to grow to the required amount? (4) c On January 1, 2007 a Kia Sorento had a value of R320 000. Each year after that, the car’s value will decrease 20% of the previous year’s value. What is the value of the car on January 1, 2011? (4) d In South Africa, the cost of a new Honda Civic is R161 300. In England the same vehicle costs £11 200 and in the USA $21 300. In which country is the car the cheapest if you compare it to the South African rand? Note the following exchange rates: £1 @ R14,70 and $1 @ R7,30 (3) [14] Question 4 a Remove the brackets and simplify: (2d + 3)(4d2 – 8d + 9) (3) 126 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 126 16/01/2012 12:43

Questions b Factorise the following expressions: i 14n2 – 5n – 1 (2) ii x2 – 2x + 1 – 9a2 (3) c Simplify as far as possible: 2   4n –9 _____ 2n + 3 ________ ÷   2n2n–+ 1 n–3 (3) [11] Question 5 a Solve the following equations: i x(x + 3) = 10 (3) ii 23x – 1 = 4x + 5 (3) b Solve for x: ____ i   x – 1 4 ≥ 5 (2) ii 2 ≤ 3x – 5 < 9 (3) iii 3T = ax6 (3) c Solve for x and y in the following simultaneous equations: 4x + 3y = 7 and 6x – 2y = –9 (5) [19] Question 6 a Given the functions f (x) = –x2 + 4 and g(x) = –2x + 4. i Draw these functions on the same set of axes. (4) ii Use the graph to determine for which values of x is f ≥ g. (2) iii Find the equation of the reflection of f in the x-axis. (2) iv Find the equation of the line parallel to g and passing through x = –2. (2) b The curve of the exponential function f in the accompanying diagram cuts the y-axis at the point A(0; 1), and B(2; 4) is on f. y B(2; 4) A(0; 1) x Determine: i The equation of the function f (2) ii The equation of h, the reflection of f in the x-axis. (2) iii The range of h. (1) [15] © Via Afrika – Mathematics 127 • Via Afrika Maths Study Guide Gr10 127 16/01/2012 12:43

Questions Question 7 a i Make a neat sketch of the graph f: y = 3.2x + 2, clearly labelling intercepts with axes and any asymptotes. (3) ii Write down the equation of the asymptote of f. (1) b i Sketch the graph of h(x) =  4__x ; x > 0 (2) ii Give the domain of h. (1) iii Give the equation of k, the reflection of h about y = –x. (2) c The following graph shows the path of the a golf ball. y 150 140 130 120 110 100 90 Height (feet) 80 70 60 50 40 30 20 10 0 1 2 3 4 5 6 7 x Time (seconds) i What is the range of this function? (2) ii After how many seconds does the ball reach its maximum height ? (1) iii What is the maximum height that the golf ball travelled in metres given that 1 foot = 30,48 cm? (2) iv For how long was the golfer’s ball airborne? (1) d John had 15 coins in R5 and R2 pieces. He had 3 more R2 coins than R5 coins. He wrote a system of equations to represent this situation, letting x represent the number of R5 coins and y represent the number of R2 coins. Then he solved the system by graphing the equations. i Write down the system of equations. (2) ii Draw the graphs on the same set of axes. (2) iii What is the solution? (2) [21] [100] 128 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 128 16/01/2012 12:43

Questions End-of-year sample exam paper: Paper 2 Question 1 a This is the sketch of the base of the Webb Ellis trophy (World Cup Rugby Trophy). Front View 12 cm Bottom part of the base 8 cm Top part of the base 12 cm x cm x cm 25 cm 30 cm 30 cm i Calculate the volume of the cylindrical piece. (2) ii Calculate the surface area of the cylindrical piece. (3) iii What is the effect on the volume of the entire base if the radius of the cylindrical piece is doubled? (2) b A rectangular prism has its length, width and height doubled in length. What effect does this have on the volume of the prism? (2) c A cylindrical hole is drilled through the centre of a rectangular metal prism, as illustrated. The cross-section of the prism is a square. The radius, x mm, of the hole is a quarter of the length of the cross-section and the height of the prism is 50 mm. 4x x 4x 50 mm i Show that the formula for the volume, V, of the remaining metal is: V = 50x2(16 – π) mm2 (3) ii Calculate V if x = 20 mm. (2) [14] [5] © Via Afrika – Mathematics 129 • Via Afrika Maths Study Guide Gr10 129 16/01/2012 12:43

Questions Question 2 Given A(1; 1), B(–1; 0), C(2; –2) and D(4; –1). y A(1; 1) B(–1; 0) 0 x D(4; –1) C(2; –2) a Calculate: i Gradient of AD (2) ii Gradient of BC (2) iii What can you deduce from your calculations in ii and iii? (1) b If E is the mid-point of AD, calculate the coordinates of E. (2) c Find the equation of EF if F is on BC and EF is perpendicular on line AD. [3] d Calculate the length of line CD. Leave your answer in surd form. (3) e Is it possible for ABCD to be a rhombus? Justify your answer with suitable calculations. (3) [13] Question 3 ^ In the diagram alongside, AF = FD, AE = EB and A E   B = 60°. ABCD is a parallelogram. F D A E 60° C B Prove that: ^ a F D    C = EBC (5) b ΔEBC ≡ ΔCDF (8) [13] 130 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 130 16/01/2012 12:43

Questions Question 4 5 __. Find, using a suitable diagram: a You are given that tan θ =  12 i cos θ (3) ii tan θ.sin θ (2) iii θ (2) b In ΔABC, ∠A = 43°; ∠B = 90° and AC = 11 cm. Find the area of the triangle. (5) c From the external lift at the top of the Sun Hotel A 72° at A, the angle of depression to the entrance to an underground shopping complex across x the street, at B, is 72°. After the lift has descended x metres to D (40 m above C), the angle of depression D 44° to B, is 44°. i Calculate the width of the street, BC. (2) ii Calculate the height of the hotel, AC. 40 m (3) d Using the yellow diagram sheet provided, draw a sketch of the graph of: y = 2 sin x; B C x ∈ [0°; 360°]. Show all your critical points. (3) [20] Question 5 a You have entered a Blackjack competition at a casino. The dealer deals you one card, you look at it. Then he deals you another card. (Using a normal deck of cards, each card (2 to 10) represents its numerical value, with the face cards (Jack, Queen, King) value = 10, and Ace = 11. Calculate the probability of: i The first card being a diamond (1) ii Getting two aces (3) iii Getting Blackjack (the value of the two cards = 21) (5) b When you go on holiday, the probability that it will be a warm day is  2__3 . If it is a warm day, the probability that you will eat ice-cream is  5__6 . If it is not a warm day, the probability that you will not have ice-cream is __  12 . The owner of the ice-cream shop wants to know the probability that you eat ice-cream when you go on holiday to the town, regardless of the temperature. Draw a tree diagram to help the owner answer the question. (6) [15] © Via Afrika – Mathematics 131 • Via Afrika Maths Study Guide Gr10 131 16/01/2012 12:43

Questions Question 6 A survey was conducted at Newlands Cricket Stadium, where 22 spectators were asked how much money they spent on food and beverages during a one-day international. The results were as follows (amount shown in rand). 18 19 24 24 30 32 32 4 41 41 41 50 51 57 58 75 78 78 80 87 94 100 a Using any method, calculate (show all your workings): i The range (2) ii The mode (1) iii The median (2) iv The mean (3) v The semi-interquartile range (4) b The heights of 80 Grade 10 Boys were recorded as follows: Height (cm) Number of Boys 131–140 1 141–150 4 151–160 10 161–170 20 171–180 34 181–190 7 191–200 3 201–210 1 i Draw a cumulative frequency curve representing the above data. (5) ii Use your graph to estimate the median and the upper quartile (3) iii Only 15% of the boys were tall enough to play lock for the 1st Rugby team. (2) How tall do you have to be to qualify to play lock? [22] [100] 132 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 132 16/01/2012 12:43

Answers Answers to questions Chapter 1 1 a (3x − 4)2 = 9x2 − 24x + 16 9x2 is the square of 3x −24x is twice the product of 3x and −4 16 is the square of −4 b 4x2 − 8x – 5 c 8a3 + b3 d (2a – 5b)(4a – 3b) – (a + 3b)(5a – 12b) = (8a2 – 26ab + 15b2) – (5a2 +3ab – 36b2) = 3a2 – 29ab + 51b2 ( k +  3__4   )( k –  1__2   ) = k2 –  1__2 k +  3__4 k –  3__8 = k2 +  1__4 k –  3__8 e 2 a (5x + 3)2 b (5x + 3)(x – 2) c  14 x = x ( x2 – __ x3 – __  14   ) = x ( x + __  12   )( x – __  12   ) d (2t – 5)3 – (2t – 5)2 = (2t – 5)2 (2t – 5 – 1) = (2t – 5)2 (2t – 6) = 2(2t – 5)2 (t – 3) e 3x3 + x2 − 3x – 1 = x2(3x + 1) − (3x + 1) = (x2 − 1)(3x + 1) = (x + 1)(x − 1)(3x + 1) f m – m – mn + n = (m3 – m2) – (mn2 – n2) 3 2 2 2 = m2(m – 1) – n2(m – 1) = (m – 1)(m2 – n2) = (m – 1)(m – n)(m + n) g (y + 3)(2y – 3) + (y – 1)(3 – 2y) = (y + 3)(2y – 3) – (y – 1)(2y – 3) (Note! (3 – 2y) = –(2y – 3)) = (2y – 3)(y + 3 – y +1) = (2y – 3)(4) = 4(2y – 3) h x2(2x – 1) – 2x(2x – 1) – 3(2x – 1) Reminder: How to invert a fraction = (2x – 1)(x2 – 2x – 3)   5__8 →   1__5__ = 1 ÷ __  58 = 1 ¿  8__5 =   8__5   8 = (2x – 1)(x – 3)(x + 1) Switch the numerator and the denominator © Via Afrika – Mathematics 133 • Via Afrika Maths Study Guide Gr10 133 16/01/2012 12:43

Answers 2 (x – 3)(x + 2) 3 a   xx2 –– ________x–6 = _________   ____ =   x + 4x + 3 (x – 3)(x – 1) x – 1 2 (x – 3) _____   = 1 and the factors ‘cancel (x – 3) b  4x 3 – x2(4x – 9) 9x2 ________ _______ =   =   4x –9 _______ out’. 4x3 + 6x2 2x2(2x + 3) 2(2x + 3) 2 (x – 4)(x – 3) c   x –47x ________ –x + 12 _________ =   –(x – 4) = –1(x – 3) = 3 – x a(a + 1) + (a + 1) 2 2 (a + 1)(a + 1) (a + 1)(a – 1) a(a – 1) (a + 1)2 d  ____________    a2 – 2a + 1 ¿   a_____ –1 a2 ÷   aa____ 2 + a _________ _________ ______ _____ =   – a (a – 1)(a – 1) ¿   a2 ¿   a(a + 1) =   a2 (x + 2)2 – 4(x – 2) – (x + 2)(x – 2) e  ____ 1   –4 – ____ + _____ x – 2 (x + 2)2   x +1 2 = ______________________        (x – 2)(x + 20)2 2 2 x + 4x + 4 – 4x + 8 – x + 4 __________________ =         (x – 2)(x + 2)2 16 __________ =   (x – 2)(x + 2)2 p–3____–   3 ____ +   3 p–3 f  ________ +   2 p2 – p – 12 3 + p _____= _________   +   2 ______ 8 – 2p (p – 4)(p + 3) p + 3 2(p – 4) 2(p – 3) + 4(p – 4) + 3(p + 3) ___________________ =         2(p + 3)(p – 4) 2p – 6 + 4p – 16 + 3p + 9 _________________ =         2(p + 3)(p – 4) __________ 9p – 13 =   2(p + 3)(p – 4) Chapter 2 1 a 22a2(n – 2) = 4a2n – 4 __ 2 b 5.1 + (23)–  3 – (2–1)–2.1 = 5 + 2–2 – 22 = 4 +  __14 – 4 = __54   –2 a a 2a 3 c  _____ 6a  a22(2¿¿33)a = _______ 1 a 2 ¿3 ________ = a2   ¿ 22 ¿ 3a 1–2 a–a 2 ¿3 ________ =   a2 –1 0 2 ¿3 ______ =  a2 ___ 1 =   2a2 12x2 + 4 – (–3)2(x3)2 6 6 =   12x – 9x 6 d  _____________    3x3 _______ 3x3 =   3x ___ = x6 – 3 = x3 3x3 x+1 x–1 27 .9 3(x + 1) 2(x – 1) e  ________ 3x – 1.81x + 1   3 x – 1 .34(x + 1) = _________ 3 .3 = 33x + 3 + 2x – 2 – x + 1 – 4x – 4 [Be careful with negatives!] = 3–2 = 1__9   134 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 134 16/01/2012 12:43

Answers 2 3   3x.x   3x8 –3 f  ____ 2 x 3–1x–2 = ____23 = ___ a–1 a+1 2(a – 1) 3(a + 1) g   24 .8 ________ a–2 .16a + 1 =   2_________ .2 a – 2 4(a + 1) 2 .2 = 22a – 2 + 3a + 3 – a + 2 – 4a – 4 = 2–1 =  __12 ___– 3a0 + 27 3 – 1 3 =   1 h   1 ____– 3 ¿ 1 + (33)  3 – 1 __ 1 __ 2 __ 1 2  – __ 2  – __ 3 8 3 (2 ) 3 =  ___ 1 2–2 – 3 + 3 – 1 = 22 – 1 =4–1 =3 2a – b 4a 2a – b 2b i   x____ xb – 2a ÷  xx___ ____ x___ x 2b =   b – 2a ¿  x4a x = x2a – b + 2b – b + 2a – 4a = x0 =1 n–3–n–1 2 a b = 5 = 5–4 =  __ 1 54 b 2.25x + 1 = 2. 52(x + 1) [25 = 52] = 2.52x + 2 = 2.52x.52 = 2.52.52x = 2.25.52x = 50.(5x)2 = 50f 2 c i 3m + 2 = 3m.32 = 5 × 9 = 45 __ =   1 9–2m = 32(–2m) = 3m(–4) = 5(–4) =  51 ii ___ 4 625 d (32x – 3)(32x + 3) = (32x)2 – 32 = 34x – 9 If we let 32x = a and 3 = b, then we have (a – b)(a + b), which, when multiplied out, gives us the difference between two squares (a2 – b2). 3 a 2x + 1 = 2 2x = 2 – 1 2x = 1 = 20 Therefore x = 0 b 9x – 2 = 271 – 2x 32(x – 2) = 33(1 – 2x) © Via Afrika – Mathematics 135 • Via Afrika Maths Study Guide Gr10 135 16/01/2012 12:43

Answers The bases are the same, therefore: 2(x – 2) = 3( 1 – 2x) 2x – 4 = 3 – 4x 2x + 4x = 3 + 4 6x = 7 x =  7__6 4 ___ =   1 __= 5–2] c 52x + 1 = 0,04 [Note: 0,04 =  100 25 52x + 1 = 5–2 Therefore: 2x + 1 = –2 2x = –3 x = – 3__2 d 5x + 1 + 5x – 150 = 0 5x(5 + 1) = 150 5x = 25[25 = 52] Therefore, x = 2 4 (3 ¿ 108)3 = 33 ¿ (108)3 = 27 ¿ 108 ¿ 3 = 27 ¿ 1024 = 27 ¿ 10 ¿ 1024 = 27 ¿ 1024 + 1 = 27 ¿ 1025 Therefore, the cubic region of space has a volume of 2,7 × 1025 cubic miles. Chapter 3 1 a 33; 40 (pattern is +7) b  __59 ; __6  11 (pattern in numerator + 1; denominator consecutive odd numbers) 2 a Common difference = –3; a = 25 General term: Tn = –3n + 28 and T16 = –3(16) + 28 = –20 b Tn = (2)n – 1 T16 = (2)16 – 1 = 32 768 3 a 13 matches b 16 matches c 4; 7; 10; 13; … difference is +3 Therefore: Tn = 3n + 1 d T100 = 3(100) + 1 = 301 4 A table of values will help to see a pattern more easily: # Cups 1 2 3 4 5 6 n Height in cm 7 9 11 13 15 17 22 136 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 136 16/01/2012 12:43

Answers a 13 cm b 7; 9; 11; 13; 15; 17; ... c Tn = 2n + 5 d 22 > 2(n) + 5 (The maximum height of shelf is 22 cm) n < 17 ÷ 2 n = 8 cups e 81 = 2(n) + 5 n = 38 Theoretically you can stack 38 cups to reach a height of 81 cm. Chapter 4 1 a  5a __– 2 =  a 3 __+ 15 4 ∴ 4(5a – 6) = 3(a + 60) ∴ 20a – 3a = 24 + 180 ∴ 17a = 204 ∴ a = 12 __+ 15 <  5m b  m ___– 2 4 3 ∴ 3m + 180 < 20m – 24 Note: ∴ –17m < –204 the inequality sign will swap over, because ∴ m > 12 we are dividing by a negative number. c (2b + 1)(b + 8) = 27 Step 1: remove brackets 2b2 + 17b + 8 = 27 2b2 + 17b – 19 = 0 Factorise when a quadratic equation is = 0. (2b + 19)(b – 1) = 0 b =1 or b = – 19 __ 2 d  2dd– 2–1 _______ –1 =   d +3 ____ 1 (2d – 3)(d + 1) = 3(d – 1) ∴ 2d2 – d – 3 = 3d – 3 ∴ 2d2 – 4d = 0 ∴ d(d – 2) = 0 ∴ d = 0 or d = 2 e b² – 7b + 12 = 0 (b – 3)(b – 4) = 0 b = 3 or b = 4 © Via Afrika – Mathematics 137 • Via Afrika Maths Study Guide Gr10 137 16/01/2012 12:43

Answers 5y – 2 3y – 1 y+1 f  _____ 3 + _____   2 – ____   2 < – __13 10y – 4 + 9y – 3 – 3y – 3 _________________    < – __ 1   6 3 16y – 10 ______ < – __ 1   6 3 8y – 5 _____ < –  1__   3 3 ∴ 24y – 15 < –3 ∴ 24y < 12 ∴ y <  1__2 g Restrictions: h ≠ ± __43 ; –3 3 ______ –   5 _________ =      ______________ 2   9h2 – 16 3h2 + 5h – 12 h(3h + 4) + 3(4 + 3h) 3h + 9 – 15h – 20 – 6h + 8 __________________       = 0 (3h + 4)(3h – 4)(h + 3) ∴ –18h – 3 = 0 ∴ h = – 1__6 2 2a = 24 + 7b ............. (1) 3a + 5b = 5 ............... (2)  24 2+ 7b (1): a = ______ 3(24 + 7b) (2):  ________ 2 + 5b = 5 72 + 21b + 10b = 10 ∴ 31b = –62 ∴ b = –2 24 + 7(–2) and a =  _______ 2 = 5 3 3V = 4πR3 – 4πr3 4πR3 = 3V + 4πr3 _______ √ 3 3 R =   3V + 4πr _______ 4π   4 a x2 + 3(2x – 5) = x2 – 2(7 + 3x) 12x = 1 __ 1 ∴ x =  12 b y – 3x = 2x + 2z –6x = 2z – y 2z –y ∴ x = ____   –6 y – 2z or x =  ____ 6 138 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 138 16/01/2012 12:43

Answers c x + 5 + 6 = 6x – 10 + 2x ∴ –7x = –21 ∴ x = 3 d Restrictions: x = ±3 3 _________ +   2 ____ =   1 ____   (x + 3)(x – 3) x + 3 x – 3 ∴ 3 + 2x – 6 = x + 3 ∴ x = 6 e 24x – 8(x – 2) ≥ 36 + 21x ∴ 24x – 8x + 16 ≥ 36 + 21x ∴ –5x ≥ 20 ∴ x ≤ –4 5 7a + 14 + 3b – 15 = 34 7a + 3b = 35 ............. (1) and 3a + 6 – 2b +10 = 8 3a – 2b = –8 ........ (2) (1) × 2: 14a + 6b = 70 .............. (3) (2) × 3: 9a – 6b = –24 . ............. (4) (3) + (4): 23a = 46 a=2 b=7 Chapter 5 1 a In ΔADC: CD ___ tan A =  AD ____ =   15 CD AD =  tan ______ A tan 55° AD = 10,50 cm AB = 2 × 10,50 cm [AD = BD] = 21 cm b In ΔADC: CD sin A =  ___ AD CD ____ AC =   sin A AC =   sin15 _____ 55° AC = 18,31 cm c In ΔACD: ∠ACD = 90° – 55° [∠s of Δ = 180°] = 35° © Via Afrika – Mathematics 139 • Via Afrika Maths Study Guide Gr10 139 16/01/2012 12:43

Answers In ΔDBC: ∠DCB = 35° [ΔADC ≡ ΔBDC] cos ∠DCB =  CD ___ BC BC =   CD ______ ^ cos D C   B BC =  cos15 ______ 35° BC = 18,31 cm [ΔABC is isosceles] d ∠DBC = ∠DAC = 55° 2 a Speed = distance ÷ time Therefore, distance = speed × time speed = 6 000 km/h __hour 2 2 min =  60 __= 200 km 2 Therefore, distance (AD) = 6 000 ×  60 b BE (height after 3 min) = BC (BC = AD = 200 km) + CE CE is opposite ∠EDC, so we need ∠EDC ∠EDC = 90° – 55° = 35° opposite CE sin 35° = ________  hypotenuse = ___   DE Therefore, CE = DE × sin 35° __ 1 DE = 6 000 km/h ×  60 DE = 100 km Therefore, CE = 100 km × sin 35° = 57,36 km Therefore, BE = 200 km + 57,36 km = 257,36 km c AB = DC adjacent cos 35° = ________   DC  hypotenuse = ___DE DC = DE cos 35° and DE = 100 km (from b) Therefore, AB = 100 km × 0,85264 AB = 85,26 km (to two decimal places) 3 You have two right-angled triangles that you can use to solve the problem: ∆ABC and ∆ABD. To determine CD, find the distance BC and subtract it from distance BD: CD = BD – BC. ∠AEC is alternate and equal to ∠C1 (AE and BD are parallel lines) and equal to 25°. The side opposite ∠C1 is AB, and BC is adjacent to ∠C1. So the tan-ratio is suitable. ∠C1 = 25° and AB = 60 ___= tan 25°   AB BC BC =  tan60 ______ 25° BC = 128,6704 140 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 140 16/01/2012 12:43

Answers To determine BD, we use ∆ABD. ∠AED is alternate and = ∠ADB (AE and BD are parallel lines) = 20°. The side opposite ∠ADB is AB and BD is adjacent to ∠ABD. So again the tan ratio is suitable. ∠ADB = 20° and AB = 60 ___= tan 20° AB   BD 60 BD =  _____ tan 20° BD = 164,8486 The distance between the ships C and D = BD – BC = 36,2 metres. 4 a ΔMPN = 180° – 52° – 38° = 90° [Angles in a triangle add up to 180°] opposite b sin 38° = ________  hypotenuse ___ =   MP MN MP = 160 × sin 38° MP = 99 m Or adjacent cos 52° = ________ ___ MP  hypotenuse =   MN MP = 160 × cos 52° MP = 99 m opposite c sin 52° = ________ ___ PT  hypotenuse =   MP PT = MP × sin 52° PT = 98,5 × 0,788 ≈ 78 m 5 a x = 34,75° 0,96 b cos x =  ____ 3 x = 71,34° c 2tan x = 4,2 × 3 tan x = 4,2 × 3 ÷ 2 x = tan–1 (4,2 × 3 ÷ 2) = 80,98° d sin (x + 25°) = 0,813 x + 25° = sin–1 (0,813) = 54,39° x = 54,39° – 25° = 29,39° e tan x = 0,6691 x = 33,79° © Via Afrika – Mathematics 141 • Via Afrika Maths Study Guide Gr10 141 16/01/2012 12:43

Answers __ √3  6 a sin x =  ___2 x = 60° b 3x = 60° x = 20° c 4sin (2x – 10°) = 2 sin (2x – 10°) =  2__4 (2x – 10°) = sin–1  1__2 2x = 30° + 10° = 40° x = 20° cos _____ 30° d x =  tan 60° __ __ √3  ___ ___ =   ÷   √3  2 1 __ √ ___3  ___ =  2 ¿   1__ √3  = 1__2   Chapter 6 1 a The graph is a parabola. b Minimum value: [a = 1 > 0] c y = –4 (Note that just –4 would not be sufficient) d For y = x2 – 4, let y = 0: 0 = x2 – 4 4 = x2 x2 = 4 x = ±2 e x-intercepts: (2; 0) and (–2; 0) minimum value at y = –4 So you have three points (–2; 0), (2; 0) and the turning point (0; –4). The graph is a parabola and a is positive, therefore the “arms” point upwards. 142 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 142 16/01/2012 12:43

Answers y 4 3 2 1 x –5 –4 –3 –2 –1 0 1 2 3 4 5 –1 –2 –3 –4 –5 f y = x2 – 4 + 4 y = x2 g The x-intercepts are equal; they are x = 0. h Domain: all real values of x (x ∈ R), Range: all y-values greater or equal to –4 y ≥ –4, y ∈ R or [–4; ∞) i The y-axis is the axis of symmetry (x = 0). 2 a The general equation is: y = ax2 + q and the turning point is (0; 3). Replace x and y with the values of the coordinates of the turning point: y = ax2 + 3 Substitute(–3; 0) into the equation: 0 = a(–3)2 + 3 0 = 9a + 3 a =  –3 __ =   –1 9 __ 3 y = x2 + 3 b Domain: x ∈ R; Range: y ≤ 3, y ∈ R (or domain: (–∞; ∞) range: [–∞; 3)) c x = 0 (note: “y-axis” is not sufficient, because an equation is asked for) Chapter 7 1. ΔWXZ = ΔWYZ (given) W X ∴ XY || WZ (= Δs on same base; ||) And WX || ZY (given) ∴ WXYZ is a ||m (opp sides ||) Z Y © Via Afrika – Mathematics 143 • Via Afrika Maths Study Guide Gr10 143 16/01/2012 12:43

Answers 2 BG || AE (given) A B ∴ BG || EC CG || DE (given) E G ∴CG || EB ∴ BGCE a ||m (2 pr opp sides ||) And ΔBEC = 90° (diags of rhomb ⊥) D C ∴ BGCE a rect (||m with int ∠ = 90°) 3 DEFG is a parm (both pairs opp sides ||) D E ΔGDO = ΔEFO (DG || EF) ΔDGO = ΔFEO (DG || EF) O DG = EF (opposite sides of parm) ∴ ΔDGO ≡ ΔFEO (AAS) ∴DO = OF and GO = OE G F 4 a In ΔADE and ΔABC: AD : AB = 1 : 2 (AD = DB) ^ ^  A   =  A     (common) AE : AC = 1 : 2 (AE = EC) ∴ ΔADE ||| ΔABC (SAS) ^ ^ b A D   E =       B (ΔADE ||| ΔABC) ∴ DE || BC (corresponding Δs equal) c DE : BC = AD : AB = 1 : 2 (ΔADE ||| ΔABC) ∴ DE =  __12 BC Chapter 8 1 a You need the gradient formula: y –y –1 – 3 ___  xPP – xQQ = _____ mPQ = _______   –4 =   –4 = __ – 6 –10 5   2 b Remember, if lines are parallel, their gradients are equal (mPQ = mSR), therefore: y –y –3 _____ –b __ 2 mSR =  xS – xR =   –4 – 6 =   5 _______ S R –15 – 5b = –20 b = 1 c You need the distance formula: PQ2 = (xp – xq)2 + (yp – yq)2 SR2 = (xs – xr)2 + (ys – yr)2 PQ2 = (–4 – 6)2 + (–1 – 3)2 SR2 = (–4 – 6)2 + (–3 – 1)2 PQ2 = 100 + 16 SR2 = 100 + 16 PQ2 = 116 SR2 = 116 Therefore, PQ = SR 144 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 144 16/01/2012 12:43

Answers d Check the properties of a parallelogram! PQ = PS (just proven) and PS || QR (gradients are the same) ∴ PQRS parallelogram (1 pair opposite sides equal and parallel) xp + xr yp + yr e ( i MPR: _____ )   2   = (   –4   2 ; _____ + 6 –1 _____ 2   2+ 1   )= (1; 0) ; _____ x +x y +y M : ( _____   2   )= (   –4   2 ; _____ + 6 –3 _____   2+ 3   )= (1; 0) ; _____ s q s q ii SQ 2 f They bisect each other at the point (1; 0). g PS2 = (xp – xs)2 + (yp – ys)2 PS2 = (–4 – (–4))2 + (–1 – (–3))2 PS2 = 0 + 4 PS2 = 4 ____ ∴ PS = 2 units and PQ = √116 = 10,77 units ∴ PQ ≠ PS ∴ PQRS is not a rhombus, because the adjacent sides are not equal. 2 aAB2 = (–4 – 2)2 + (–3 – 5)2 = (–6)2 + (–8)2 = 36 ____ + 64 = 100 ∴ AB = √100  ∴ AB = 10 units b K = (–1; 1) y –y _____1 c m =  x2 –x 2 1 = –  8__6 = –  4__3 3 mAB = mDC  57 –– 06 = 5 mDC = ____ ∴ x = 1 and y = –2 4 a AB2 = (5 – 3)2 + (11 – 7)2 = 4___ + 16 = 20 AB = √20 units b BC2 = (6 – 5)2 + (3 – 11)2 = 1 ___ + 64 = 65 BC = √65 = 8,06 units c Midpoint = ( ____  7 +2 3   )   3 +2 6 ; ____ = (4 __12 ; 5) d mAB =  11 _____ 5–3 –7 = 2 and mBC =  11 _____ 5–6 –3 = –8 © Via Afrika – Mathematics 145 • Via Afrika Maths Study Guide Gr10 145 16/01/2012 12:43

Answers  62 –+ 23 = __ 5 mAB = ____   34 mBC =  8k – ____ –2 6 =   k – ____ 6 6 AB ⊥ BC ∴ mAB ¿ mBC = –1 ( ) ∴ (   3__4   )   k – ____ 6 6   = –1   k –8 6 = –1 ∴ ____ ∴ k = –2 6 a HK2 = (1 +___ 5)2 + (1 + 3)2 = 36 + 16 = 52 ∴ HK = √52 units b KD = HG ∴ KD2 = HG2 = (3 + 5)2 + (7 – 1)2 = 100 ∴ KD = 10 units c mGD = mHK 7–y ∴  ____   4 = __ 3 – x –6 ∴ D = (9; 3) d Midpoint of HD = midpoint of KG: (   3 + ____ 2 1 7–   2 3   )= (2; 2) ; ____ 7 a Q = (   –4 +3 0– _____ 2   2 7   )= ( –  1__2 ; –3 1__2   ) ; ____ 1  4 + 3 __ 0+ _____ 7 _____ 2 b mMN =  –4 – 3 = –1 and mPQ =   1__ = 1 7 +     2 c Product of gradients = (1)(–1) = –1 d MN || PQ e PM2 = (7 +____ 4)2 + (4 – 0)2 = 121 + 16 = 137 ∴ PM = √137 units PN2 = (7 –____ 3)2 + (4 + 7)2 = 16 + 121 = 137 ∴ PN = √137 units f The triangle is a right-angled isosceles triangle, because PM = PN and there is an angle of 90° between MN and PQ. 146 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 146 16/01/2012 12:43

Answers Chapter 9 1 A = P(1 + I)n = 70 000(1 + 0,09)6 = R117397,01 2 A = P(1 + in) = 30 000(1 + 0,06 × 8) = R44 400 3 a Cash paid = 20% of R105 000 ___ ×   105 20 =   100 ______ 000 1 = R21 000 Loan amount = R105 000 – R21 000 = R84 000 b A = P(1 + in) = 84 000(1 + 0,145 × 5) = R144 900 Full Amount = R144 900 + R21 000 = R165 900 c Monthly payments =  144 ______ 84 000 = R1 725 d Total amount paid = R780 × 84 + R21 000 + 84 × R1 725 = R231 420 Or Total amount paid = R165 900 + 84 × R780 = R231 420 4 Price of pram in SA currency = R26,09 × 80 = R2 152 5 Amount in British pounds = R530 000 ÷ 11,72 = £45 222 6 A = P(1 + in) = 420 000(1 + 0,125 × 8) = R840 000 7 A = P(1 + i)n 180 000 = P(1 + 0,0725)10 180 ________000 P =  (1.0725)10 = R89 392 8 A = P(1 + i)n = 50 000(1 + 0,04)5 = R60 833 60 833 – 50 000 Percentage growth =  ____________ 50 000 × 100 = 21.67% 9 Amount in $ = 40 000 × $76 = $3 040 000 Amount in R = 3 040 000 × R7,23 = R21 979 200 10 Option 1: Simple growth: A = P(1 + in) = 30 000(1 + 0,125 × 8) = R60 000 Option 2: Compound growth: A = P(1 + i)n = 30 000(1 + 0,125)8 = R76 973,54 He will choose option 2 because he will receive more interest on his investment. © Via Afrika – Mathematics 147 • Via Afrika Maths Study Guide Gr10 147 16/01/2012 12:43

Answers Chapter 10 1 a 197 b KwaZulu-Natal c The province is mainly rural. There are not too many activities in the province over the Easter weekend. Drivers and pedestrians obey the rules of the road. (Or any other meaningful reasons.) 37 ___ d percentage =   197 = 18.78% 2 a The range = 188 cm – 134 cm = 54 cm b The heights of the learners vary considerably (by more than half a metre). 3 a Range value for the marks of female learners = 70% – 48% = 22% b Range value for the marks of male learners = 74% – 22% = 52% c The difference between the range value for female learners tells us that they are of similar ability, whereas the wide spread of marks for the male learners tells us that t here is a wide spread of ability amongst the males. d i Females Males Mark category No. of learners who achieved No. of learners who achieved marks in this category marks in this category 0 – 29% 0 2 30 – 39% 0 1 40 – 49% 1 2 50 – 59% 7 2 60 – 69% 6 3 70 – 79% 3 3 80 – 89% 0 0 90 – 100% 0 1 ii Females Achievment Males Achievment Number of females per Number of males per 8 3 6 2.5 2 4 2.5 2 mark interval mark interval 2 1.5 1 0 0 0 to 30 to 40 to 50 to 60 to 70 to 80 to 90 to 0 to 30 to 40 to 50 to 60 to 70 to 80 to 90 to 29% 39% 49% 59% 69% 79% 99% 100% 29% 39% 49% 59% 69% 79% 99% 100% Mark interval Mark interval 148 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 148 16/01/2012 12:43

Answers iii From the two graphs we notice the following: The female histogram tells us that learners’ abilities are similar. The histogram is fairly compact. The male histogram tells us that learners’ achievements are of a very mixed ability. The histogram is very spread out. iv It looks as though the female learners performed better because the results are grouped together higher up the scale. However, even though the male learners have the highest mark, there is a spread towards the lower end of the scale, which would lower their average. 4 a Learner A: 266 ÷ 6 = 44,33% Learner B: 214 ÷ 6 = 35,67% Learner C: 329 ÷ 6 = 54,83% b Learner A: 69 – 22 = 47% Learner B: 55 – 11 = 44% Learner C: 64 – 45 = 19% c Learner C is the only one of the three with a mean over 50% and is also the most consistent (range = 11%). Therefore, Learner C should receive the bursary. © Via Afrika – Mathematics 149 • Via Afrika Maths Study Guide Gr10 149 16/01/2012 12:43

Answers Chapter 11 1 volumecube = 123 and volumeprism = 18 ¿ 8 ¿ height 3 _____= 12 cm ∴ height of water =  1812 ¿8 2 Surface area = 2 × Area of base + (perimeter of base × height) = Surface area of cube – 2 area of circles on sides + circumference of circle × height of cylinder = (2 ¿ 102 + 40 ¿ 10) – 2 ¿ π ¿ ( __   32   ) + 2 ¿ π ¿ __ 2  32 ¿ 10 = 680,11 cm2 3 Note that measurements are given in mm, but the answer has to be in cm3. Convert: 15 mm = 1,5 cm and 20 mm = 2 cm Volume of material = Outer volume – Inner volume V =  4__3 π(23 – 1,53) cm3 = 19,4 cm3 4 Note that there are two different types of units: mm and cm So, 70 mm = 7 cm V =  1__3 πr2h ___ √_______   3V r = ___πh   r = √_______   3 ¿ 4 224   7π ______ slant height = √h2 + r2  ________ s = √72 + 242  s = 25 cm Total surface area = π ¿ 24(24 + 25) = 3 694,5 cm2 150 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 150 16/01/2012 12:43

Answers Chapter 12 1 a S T R 25 3 14 2 b i n(T ∩ R) = 3 ii P(T) = 28 iii P(T ∪ R) = 42 iv n(Rʹ) = 25 + 2 = 27 v n(S) = 44 vi n(T or R) = 28 + 17 – 3 = 42 2 a There are 12 possible outcomes: (H; 1); (H; 2); (H; 3); (T; 1); (T; 2); (T; 3) b P(heads and 2) = 12__ = 1 2 __ = 0,17 6 c P(tails and an uneven number) = 12 4 __ = 2 __ = 0,33. __ = 1 6 3 © Via Afrika – Mathematics 151 • Via Afrika Maths Study Guide Gr10 151 16/01/2012 12:43

Answers Mid-year exam paper Question 1 a i x > 2 ii x=2 b i 6 – 5 = 6 + 5 and 7 – 6 = 7 + 62 2 2 2 ii If m and n are consecutive integers and n > m, then n2 – m = n + m2. x2 – (x – 1) = x + (x – 1)2, x ∈ Z iii iv LHS = x2 – x + 1 RHS = x + x2 – 2x + 1 = x2 = LHS conjecture proved v (n + 2)2 – (n + 1) = (n + 2) +(n + 1)2 for n ∈ natural numbers Question 2 n –2 n 3 a i   2 .2 .3 .3 ________ 2n.3n =  27 __ 4 16x + 48 – 9x – 48 _____________ __7x ii     12 =   12 b 6x3 +13x2 + 4x – 3 c i (x – 6)(x + 1) ii 4a(3n – 2) + m(3n – 2) = (3n – 2)(4a + m) iii (x + 2)(x2 – 4 ) = (x + 2)(x + 2)(x – 2) = (x + 2)2(x – 2) Question 3 a –2a ≤ 74 ∴ a ≥ –37 b 27 – 2a = 8 = (23) ∴ 7 – 2a = 3 ∴a=2 Question 4 6x + 4y = 10 6x – 9y = 36 ∴ 13y = –26 ∴ y = –2 and x = 3 152 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 152 16/01/2012 12:43

Answers Question 5 2 a x2 – 7x – 8 = 0 b  x – 4x + 3 x + ________ 3–x = ____   2 1 c x = –2 or 0 ∴ (x – 8)(x + 1) = 0 ____ 1 – x =  x + 1 2 ∴ x = –1 or 8 x + 1 = 2 – 2x x =  1__3 Question 6 a V = 9 × 7 × 4 = 252 cm3 b i r = 0, ∴ depth = 6 m If r = 1 m, h = 5 1__4 ii When r decreases by 1 m, h increases by  3__4 m. h = 0 ∴ 6 –  3__4 r = 0 iii ∴ radius = 8 m iv C = 2πr = 2π(8) = 16π = 50,27 m Question 7 _______ ___ a AB = √62 + 32 = √ 45  b Midpoint = ( –  1__2 ; 0 ) c mAB = –2 y =  __12 x + c d ∴ 0 =  1__2 ( –  1__2   )+ c ∴ y =  1__2 x +  1__4 Question 8 a 31 – 2 = 29 b 11 c 13 d 19 e  140 ___= 14 10 Question 9 a 50,5 and 55,5 © Via Afrika – Mathematics 153 • Via Afrika Maths Study Guide Gr10 153 16/01/2012 12:43

Answers b f Mass of Grade 10s 80 70 60 50 40 30 20 10 0 10 20 30 40 50 60 70 80 x i Between 51 and 52,5 ii Between 54,5 and 55,5 154 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 154 16/01/2012 12:43

Answers End-of-year sample exam paper: Paper 1 Question 1 a x = 0,57777777 . .......... (1) 10x = 0,5777777 . ........ (2) (2) – (1): 9x = 5,2 5,2 x =  ___ 9   52 = __ 90   26 = __ 45 b i 125a6 ii 5a6 iii –3a8b3 3 ___ √64 = 4 c 3 ____ √125 = 5 Therefore, between 4 and 5 Question 2 a 17; 20; 23 b Tn = a + (n – 1)d = 5 + 3(n – 1) = 5 + 3n – 3 = 3n + 2 c T49 = 3(49) + 2 = 140 Question 3 a 2 200 000 = 1 005 000(1 + 0,065 × n) n = [   10 2 200 ________ 000 050 000 – 1 ]÷ 0,065 = 18,3 For 19 years b 10 000 = 8 000(1 + i)3  10 (1 + i)3 = __8 __ √__ 3 1 + i =   5__4   i = √__ 3 5    – 1 4 i = 0,0772… i = 7,7% c A = 320 000(1 – 0,2)4 A = R131 072 d 11 200 × 14,70 = R164 640 21 300 × 7,30 = R155 490 The car is cheapest in the USA. © Via Afrika – Mathematics 155 • Via Afrika Maths Study Guide Gr10 155 16/01/2012 12:43

Answers Question 4 a 8d3 – 4d2 – 6d + 27 b i (7n + 1)(2n – 1) ii (x – 1 – 3a)(x – 1 + 3a) (2n – 3)(2n + 3) (2n – 3)(n + 1) c  ___________ 2n + 3 ¿ __________   n+1 = (2n – 3)2 Question 5 a i x2 +3x – 10 = 0 (x + 5)(x – 2) = 0 x = –5 or 2 ii 23x – 1 = 2x + 10 ∴ 3x – 1 = 2x + 10 Therefore x = 11 b i x – 1 ≥ 20 x ≥ 21 ii 7 ≤ 3x < 14 7 __   3  14 ≤ x < __3 iii x6 =  3T __ a ___ √ 6 x = ±   3T __  a c 8x + 6y = 14 . ................ (1) 18x – 6y = –27 ............. (2) (2) + (1): 26x = –13 x = – __12 and y = 3 Question 6 a i y 8 6 4 2 x –4 –2 0 2 4 –2 –4 –6 –8 ii 0 ≤ x ≤ 2 156 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 156 16/01/2012 12:43

Answers iii y = x2 – 4 iv y = –2x – 4 b i y = a.bx 4 = 1.b2 ∴ b = 2 f (x) = 2x ii h(x) = –2x iii y<0 Question 7 a i y 6 5 4 3 2 1 x –6 –5 –4 –3 –2 –1 0 1 2 3 4 –1 –2 ii y=2 b i y 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 4 5 6 7 –1 –2 ii x > 0 iii y =  4__x ; x < 0 © Via Afrika – Mathematics 157 • Via Afrika Maths Study Guide Gr10 157 16/01/2012 12:43

Answers c i 0 ≤ y ≤ 100 ii 2  1__2 100 ¿ 30,48 iii  _________ 100 = 30,48 m iv 5 seconds d i x + y = 15 x + 3 = y ii y 15 10 5 x 5 0 5 10 15 Equation 1: x + y = 15 Equation 2: x + 3 = y iii x = 6; y = 9 158 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 158 16/01/2012 12:43

Answers End-of-year sample exam paper: Paper 2 Question 1 a i Volume = πr2h = π(6)2 ¿ 8 = 904,8 cm3 ii SA = 2πrh + 2πr2 = 2π(6)(8) + 2π(6)2 = 527,8 cm2 iii Entire volume is 3 times bigger. b Volume is 8 times bigger. c i V = (4x ¿ 4x ¿ 50) – πx2 ¿ 50 = (800x2 – 50πx2) mm3 = 50x2(16 – π) mm3 ii V = 800(20)2 – 50π(20)2 = 257 168,15 mm3 Question 2 a i _____ mAD =  –1 –1 __ 2 4 – 1 = –  3  –2 ii mBC = _____–0 2+1 = – __23 iii AD || BC b Midpoint E = (   1 + ____ 2 4 1–   2 1   )= (2,5; 0) ; ____ c  32 mEF = __ ∴ y =  3__2 x + c ∴ 0 =  3__2 (2,5) + c ∴ c = –3 3__4 ∴ y =  3__2 x – 3 3__4 _________________ _________ __ d CD = √(4 – 2)2 +    (–1 + 2)2 = √(2)2 + (1)2 = √5  e No, ABCD cannot be a rhombus as all 4 sides must be equal. Question 3 a Let ∠FAD = x ⇒ ∠FDA = x (FA = FD) ∠EAB = x (vert opp angles) ∠EBA = x (EA = EB) and let ∠ABC = y ⇒ ∠ADC = y (opp angles parm) ∴ ∠EBC = x + y = ∠FDC © Via Afrika – Mathematics 159 • Via Afrika Maths Study Guide Gr10 159 16/01/2012 12:43

Answers b 2x + 60° = 180° x = 60° So both triangles AEB and FAD are equilateral triangles with EB = AB = DC (opp sides parm) FD = AD = BC (opp sides parm) ∠EBC = ∠FDC (from a) ∆EBC ≡ ∆CDF (SAS) Question 4 a i __ cos θ =  12 13 5 __ ii tan θ.cos θ =  13 iii θ = 22,62° b BC = 11 sin 43° = 7,5 cm AB = 8,04 cm Area =  1__2 .AB.BC = 30,18 cm2 c i   BC __= tan 46° 40 BC = 41,42 m 41,42 ii  _____ AC = tan 18° AC = 127,48 m d y (90; 2) 2 1 0 x 60 120 180 240 300 360 –1 –2 (270; –2) Question 5 13 a i  __ =   1__ 52 4 __ ¿   3 4 __ =   1 ___ ii   52 51 221 4 __ ¿   16 __ 4 __ +   16 __ iii   52 51 52 ¿   51 32 ___or 0,0048 =   663 160 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 160 16/01/2012 12:43

Answers b   5__6 Yes Warm   2__3 ¿   5__6 =   10 __ 18  __2 3   1__6 No __ +   3   10 __ =   13 __ 18 18 18   1__2 Yes   1__3 Cold 3   1__3 ¿   1__2 =   18 __   1__2 No Question 6 a i 100 – 18 = 82 ii 41 41 + 50 iii  ______ 2 = 45,5 iv  1____ 151 22 = 52,32 78 – 32 v  ______ 2 = 23 b i y 90 Cumulative frequency 60 30 0 x 140 160 180 200 220 Height ii Median ≈ 171 cm Upper quartile ≈ 177 cm iii 85% (80) = 68th value ≈ 180 cm © Via Afrika – Mathematics 161 • Via Afrika Maths Study Guide Gr10 161 16/01/2012 12:43

Glossary A class interval A group of data values adjacent Next to; in trigonometry, the that fall within a certain range; class adjacent side is the side next to a intervals are useful when working with specified angle in a triangle large numbers of data elements algebraic fractions An algebraic fraction clockwise In a circular direction, moving is a fraction that has an algebraic from right to left expression in the numerator and the coefficient If a number appears in front of denominator a variable, then we call it the coefficient amplitude The amplitude of a graph is of that variable half the distance between the maximum collinear If three points are collinear, and minimum points on the graph then they fall on the same straight line angle of depression An angle of common factors A common factor is one elevation is the angle between a that divides into each term in an horizontal line and the line from the expression observer to some object below the complementary events Two sets are horizontal line complementary when they are mutually angle of elevation An angle of elevation exclusive and between them contain all is the angle between a horizontal line the items in the sample space and the line from the observer to some compound interest With compound object above the horizontal line interest you earn interest on the original apex The highest point, or vertex amount (the principal amount), plus asymptote The points on the x-axis for any interest that you earn during the which a function is undefined investment congruent Two triangles are congruent B if all their corresponding angles are the base An exponent tells you how many same size and their corresponding sides times the base number is used as a are equal in length factor conjecture An opinion or conclusion bearing A method of specifying direction formed on the basis of incomplete using angles, starting from north at 0°; information also called true bearing constant If a number appears on its binomial If a polynomial has two terms, own in an expression, then we call it a we call it a binomial constant C continuous relationship A relationship Cartesian plane A plane divided by two in which all points on a line are possible axes on which all points are described solutions to the function or formula by Cartesian coordinates (x and y) conventional bearing Specifying a circumference The circumference of the bearing using an angle and a compass sphere is a circle around the middle of point as a reference; also called the sphere direction 162 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 162 16/01/2012 12:43

Glossary corresponding angles In similar E triangles, the corresponding angles are elements in a data set A data element is the angles that are in the same relative a single member of a data set positions and are the same size eliminate To remove corresponding sides In similar triangles, event a subset of the sample space the corresponding sides are in the same exchange rate Exchange rates give us a relative positions way of comparing different currencies so cos cosine; the trigonometric ratio that countries can trade with each other adjacent ________   hypotenuse exhaustive A set of events is exhaustive if, between them, they contain all the D possible events in the sample space data Basic or raw values, without any experimental probability The result meaning attached to the values of an experiment is the experimental data elements A single data value in a probability or relative frequency data set exponent An exponent tells you how data points An individual data item, many times the base number is used as which will have a value a factor decreasing function In a decreasing exponential equations Equations in function, the output variable decreases which the variables appear in the in value as the input variable increases exponents of the terms in value exponential function The graph of y = ex dependent variable In a function, is upward-sloping, and increases faster the dependent variable depends on as x increases; the graph always lies the value of the input variable, or above the x-axis but can get arbitrarily independent variable close to it for negative x, which means diameter The diameter is any line that that the x-axis is a horizontal asymptote passes through the centre of a circle or sphere from one side to the other F direction Use a directional compass factorisation Factorisation is the (with points N, E, S and W) and an angle reverse of the process of multiplying to specify direction; north is 0°; also polynomials referred to as conventional bearing first degree A linear equation is an discrete relationship A relationship that equation of the first degree, which results in a set of distinct points, rather means that the highest power of the than a continuous line variable is 1 domain The domain of a function is the first difference the constant difference set of all x-values for which the function between the terms in a sequence is the is defined first difference dual intercept method A method used to frequency distribution A record of how draw a straight line graph by joining the many data elements fall within each points that intercept the x- and y-axes class interval © Via Afrika – Mathematics 163 • Via Afrika Maths Study Guide Gr10 163 16/01/2012 12:43

Glossary frequency table A table that shows independent variable An input variable each class interval and how many data in a function that uniquely determines elements fall within each interval the value of the dependent variable function Associates an input to a index This is another name for an corresponding output according to a exponent rule inflation Inflation is the rate at which the price of goods increases annually G information When data is summarised general term We call the nth term the or transformed in some way to give it general term of the sequence meaning, it becomes information generalise To express an idea in general input A formula takes in an input value, terms, as opposed to using specific changes it according to a rule, and examples produces an output value gradient A gradient is a slope interquartile range The difference gradient intercept method A method between the values of the first quartile used to draw a straight line graph using and the third quartile; the range of the the slope (gradient) of the line middle 50% of the data values H intersection the intersection of the two hemisphere A hemisphere is half of a circles represents the event (A and B) or sphere, cut through the middle in set notation, A ∩ B hire purchase A short-term loan inverse function We use the inverse agreement that lets you buy something functions on a calculator to find the now, but pay if off over a fixed number value of an angle if we know the value of months of the trigonometric ratio (sin–1, cos–1, hyperbola A hyperbola is a curve that tan–1) has two pieces that are mirror images irrational number An irrational number of each other and resemble two infinite is a number that we cannot bows express as a fraction hypotenuse The longest side of a right- isosceles trapezium An isosceles angled triangle trapezium is a trapezium in which the non-parallel sides are equal in length I identity An equation that is true for K every value of the variable is called an kite A quadrilateral with two pairs of identity equation adjacent sides equal in length increasing function In an increasing L function, the output variable increases line segment The distance between two in value as the input variable increases points is the length of the line segment in value between those two points linear In a straight line 164 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 164 16/01/2012 12:43

Glossary linear equation A linear equation is midpoint The midpoint of a line segment an equation of the first degree, which between points A and B is the point half means that the highest power of the way along the line segment variable is 1 minimum The lowest value in a data set linear inequality A linear inequality minimum value The lowest y-value for looks similar to an equation, but which a function is defined contains inequality signs rather than the mode The data value that appears most equals sign: ≥ ≤ < > often in a data set; one of the measures linear number pattern A linear number of central tendency pattern is a set or row of consecutive monomial If a polynomial has a single numbers in which the difference term, we call it a monomial between the numbers remains the same mutually exclusive events two events irrespective of how many numbers there are mutually exclusive if an outcome are in the pattern being part of one event means that it literal equation An equation that can never be part of the other contains more than one variable, N meaning that we need to express one nth term We call the nth term the general variable in terms of the others term of the sequence lowest common denominator (LCD) the lowest number that all the O denominators of a set of fractions can opposite The side opposite a specified divide into angle in a triangle outcome The result of an experiment M output A formula takes in an input value, maximum The highest value in a data set changes it according to a rule, and maximum value The highest y-value for produces an output value which a function is defined mean The result of dividing the sum of P the data elements by the number of data parabola A U-shaped graph formed by a elements; one of the measures of central quadratic function tendency parallel If two lines are parallel, then measure of central tendency A value they have the same slope that reflects a typical or average member parallelogram A quadrilateral with both of a data set pairs of sides parallel to each other measures of dispersion Summary values per annum Per year that show how spread out (dispersed) percentile Percentiles divide a data set the data values are into 10 parts median The data value that falls in the period The period of a graph is the middle of an ordered set of data points; number of degrees in a complete cycle of one of the measures of central tendency the graph © Via Afrika – Mathematics 165 • Via Afrika Maths Study Guide Gr10 165 16/01/2012 12:43

Glossary perpendicular Two lines are range The range of a function is the set perpendicular if they intersect at 90° of all y-values for which the function is polygon A figure enclosed by three or defined more line segments range The difference between the highest polyhedron A polyhedron is a geometric and lowest data values solid in three dimensions with flat faces rational number A rational number and straight edges is a number that we can write as a polynomial A polynomial is an algebraic fraction,  ba__, where a and b ∈ Z, and b ≠ 0 expression that consists of two or more real number The set of real numbers algebraic terms includes all rational and irrational power This is another name for an numbers exponent reciprocal ratio The reciprocal ratios are prime factors the prime factors of a the inverse versions of each of the three number are all of the prime numbers main ratios, sin, cos and tan; they are that will exactly divide the given cosec, sec and cot number rectangle A parallelogram with interior principal amount The amount originally angles all equal to 90° invested or borrowed relative frequency The result of an probability Probability refers to the experiment is the experimental likelihood or chance of an event taking probability or relative frequency place rhombus A parallelogram in which all sides are equal in length Q right circular cone A right circular cone quadrants A Cartesian plane is divided is a 3D shape formed by moving the end into four blocks by the two axes; each of a straight line in the shape of a circle block is called a quadrant while keeping the other end fixed above quadratic A quadratic is a trinomial in the centre of the circle the form ax2 + bx + c right prism A right prism is a polyhedron quadratic equation A quadratic equation with two identical faces, called the is an equation of the second degree, bases which means that the highest power of the variable is 2 S quadrilateral A quadrilateral is a two- sample space The set of all possible dimensional figure with four sides outcomes quartiles Quartiles divide a data set into second degree A quadratic equation is four parts an equation of the second degree, which means that the highest power of the R variable is 2 radius The distance from the centre of a semi-interquartile range The semi- circle or sphere to the edge interquartile range (S-IR) is the middle 50% of the interquartile range 166 © Via Afrika – Mathematics • Via Afrika Maths Study Guide Gr10 166 16/01/2012 12:43

Glossary similar Two shapes are similar if their T internal angles are the same, even if the tan tangent; the trigonometric ratio opposite ______ shapes are different sizes   adjacent simple interest We calculate simple term We call each number in the interest as a percentage of the original sequence a term amount invested or borrowed only trapezium A trapezium is a quadrilateral simultaneous equation Solving two or in which one pair of opposite sides is more equations at the same time, which parallel means finding values for the variables trigonometry Trigonometry is the field that satisfy all of the equations, not just of Mathematics in which we study the one relationship between the three sides and oppsite sin sine; the trigonometric ratio ________  hypotenuse three angles of triangles slant height The length of the side of a trinomial If a polynomial has two terms, cone we call it a trinomial sphere A round 3D solid; the shape of a true bearing We measure the true bearing ball between two points in a clockwise square A rectangle in which all sides are direction starting from north; also equal in length referred to as bearing standard form The standard form of an turning point The point at which the equation is the form in which we usually slope of a graph is 0 write the equation – for example, the U standard form of a linear equation is union The set of elements in one event y = mx + c, and that of a quadratic is or in another event, (A or B), or in set ax2 + bx + c notation, A B substitution To replace one value or expression with another V surd A surd is a root of an integer that we variable A letter used in an expression cannot express as a fraction that can take on different values surface area The surface area of a 3D Venn diagram A Venn diagram is a solid refers to the area of the outside graphical tool that we use to represent a surface of the solid sample space and sets of events symmetrical Two sets of points are vertex A point on a polygon at which symmetrical if they form a mirror image three or more sides meet of each other about a line that runs volume The volume of a solid refers to the between them amount of space inside the 3D solid © Via Afrika – Mathematics 167 • Via Afrika Maths Study Guide Gr10 167 16/01/2012 12:43

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