MATHSCLINIC_SMARTPREP_GR10_ENG_2023_V2.1.pdf

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VIDEO LESSONS! We have free video lessons for every topic in this book! These videos can only be accessed via Science Clinic's website - follow these steps to access this amazing resource! DOWNLOAD A BOOK Visit our website to download a SmartPrep book. We'll send you an email with download links. VIDEOS WEBPAGE Go to our Videos webpage. Log in with the email address that you used to download a book. SELECT YOUR TOPIC From your videos page, select a topic. You'll be redirected to a playlist with video lessons. ABOUT OUR SMARTPREP BOOKS Our SmartPrep books provide crystal-clear summaries of the essential theory for Grade 8-12 Core Mathematics and Grade 10-12 Physical Science. Our books are available in English and Afrikaans. COPYRIGHT NOTICE AND LICENSE The content in this SmartPrep Book is the original work of Science Clinic (Pty) Ltd. You may distribute this material, provided that you adhere to the conditions in our copyright notice and license: ¢ If you copy and distribute this book electronically or in paper form, you must keep this copyright notice intact. ¢ If you use questions or syllabus summaries from this book, you must acknowledge the author. ¢ This Smart Prep Book is meant for the benefit of the community and you may not use or distribute it for commercial purposes. ¢ You may not broadcast, publicly perform, adapt, transform, remix, add on to this book and distribute it on your own terms. ° By exercising any of the rights to this Smart Prep Book, you accept and agree to the terms and conditions of the license provided on this link: CONTENT ACKNOWLEDGEMENT Thank you to all the amazing people involved in the production, translation and moderation of this book: S Bouwer, E Britz, G Kyle, D Kotze, Q Meades, S Sapsford, S Stevens, G Swanepoel, GM van Onselen, L Vosloo QD ww sciencectinie.coza © exciencectinic EB ew acodook.comisciencectiniesa

TABLE OF CONTENTS CORE THEORY SUMMARIES Paper 1 Algebraic Expressions 2 Equations and Inequalities 5 Exponents 9 Number Patterns 11 Finance, Growth and Decay 12 Functions and Graphs 14 Probability 19 Paper 2 Trigonometry 21 Euclidean Geometry 25 Analytical Geometry 27 Statistics 30 www

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Grade 10 Maths Essentials Algebraic Expressions SCIENCE CLINIC 2022 © TERMINOLOGY: PRODUCTS 1. Monomial by a Polynomial 4. Difference of Two Squares 6. Mixed Questions Numerical Coefficient: the number in front of a Use the distributive law The two binomials are the same except the sign in Follow BODMAS; variable. one is a plus and in the other is a minus. The outer 1. Simplify in brackets if possible EXAMPLE and inners are then additive inverses of each other 2. Square binomial, FOIL or binomial by Trinomial Variable: an algebraic letter(s) used to represent un- 2a 2 (3a 2 + 4a b − a 3c) so answer is only first squared minus last squared. 3. Distribution known(s). 4. Add or subtract like terms. = (2a 2 × 3a 2 ) + (2a 2 × 4a b) + (2a 2 × −a 3c) EXAMPLE Constant: the numerical term = 6a 4 + 8a 3b − 2a 5c (3a − 2b)(3a + 2b) EXAMPLE = − 2a 5c + 6a 4 + 8a 3b 4x (4x y − 16y + 12) − (2x + y)(x − y) = (3a)2 − (2b)2 Algebraic Expression: a mathematical expression = 9a 2 − 4b 2 = 4x (4x y − 16y + 12) − (2x 2 − 2x y + x y − y 2 ) made up of one or more terms separated by addition 2. Binomial by Binomial (+) or subtraction (–). = 16x 2 y − 64x y + 48x − 2x 2 + 2x y − x y + y 2 Use FOIL method (Firsts, Outers, Inners, Lasts) EXAMPLE = − 2x 2 + 16x 2 y + 48x − 63x y + y 2 Polynomial: an algebraic expression where the (5x − 7y 2 )(5x + 7y 2 ) EXAMPLE exponent(s) on the variable(s) are natural numbers. (a − b)(x + y) = 25x 2 − 49y 4 EXAMPLE Monomial e.g. 4 or 2a 2 b c (one term) = 3(a 2 + 3a − 10) − 2(a 2 − 6a + 9) + 2(a 2 − 4) Binomial e.g. 6x + 2y (two terms) = (a × x) + (a × y) + (−b × x) + (−b × y) EXAMPLE = 3a 2 + 9a − 30 − 2a 2 + 12a − 18 + 2a 2 − 8 Trinomial e.g. 6x 2 − 5x + 4 (three terms) F O I L [(a − b) + 5][(a − b) − 5] = 3a 2 + 21a − 56 = a x + ay − bx − by = (a − b)2 − 25 Degree: is the highest value of an exponent of a specific variable in an algebraic expression. = a 2 − 2a b + b 2 − 25 EXAMPLE (e.g. 7x 3 − 3x y + 8x 6 + 4 has the sixth degree in x (2x + y)(3x − 4y) and first degree in y) 5. Binomial by a Trinomial = (2x × 3x) + (2x × −4y) + (y × 3x) + (y × −4y) Multiply each term in the binomial by each term in Like Terms: are terms with the same variable(s) = 6x 2 − 8x y + 3x y − 4y 2 (add like terms) the trinomial and the add like terms. with the same exponents, the coefficients may differ. = 6x 2 − 5x y − 4y 2 1 EXAMPLE (e.g. 6a 2 b and − a 2 b ) (a − 2)(a 2 − a + 1) 2 3. Squaring a Binomial Step 1: Square the first term = a (a 2 ) + a (−a) + a (1) − 2(a 2 ) − 2(−a) − 2(1) Unlike Terms: are terms where the Step 2: Multiple the first term by the second term = a 3 − a 2 + a − 2a 2 + 2a − 2 variables are different. and double it = a 3 − 3a 2 + 3a − 2 (e.g. 2x , 2x 2 and 3x y ) Step 3: Square the last term. EXAMPLE EXAMPLE ( p + q)( p 2 − p q + q 2 ) ( p + 2r)2 SIMPLIFYING ALGEBRAIC EXPRESSIONS = p3 − p2q + p q2 + p2q − p q2 + q3 = ( p)2 + ( p × 2r) × 2 + (2r)2 Follow BODMAS rule but can only add or subtract like = p 3 + q 3 (sum of 2 cubes) = p 2 + 4pr + 4r 2 terms and write answer with variables in alphabetical order and terms in descending order of powers. EXAMPLE EXAMPLE (3a − 2b)(9a 2 + 6a b + 4b 2 ) (3a − 4b)2 EXAMPLE = 27a 3 + 18a 2 b + 12a b 2 − 18a 2 b − 12a b 2 − 8b 3 Simplify the following: = (3a)2 + (3a × −4b) × 2 + (−4b)2 = 27a 3 − 8b 3 (difference of 2 cubes) 1. 6b c a − 7a b c + 4a 2 b c − 3c a b + b c a 2 = 9a 2 − 24a b + 16b 2 = b c a 2 + 4a 2 b c + 6b c a − 7a b c − 3c a b = 5a 2 b c − 4a b c NOTE: the second step in these examples is not usually shown. 2. 6x − 4x 2 − 8x + x 3 − x 2 + 7x − 3x 3 = − 3x 3 + x 3 − 4x 2 − x 2 + 6x − 8x + 7x = − 2x 3 − 5x 2 + 5x Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 2

Grade 10 Maths Essentials Algebraic Expressions SCIENCE CLINIC 2022 © FACTORISATION 3. Difference of Two Squares (DOTS): 5. Trinomials: Factorisation is the inverse operation to products, that is we want Square root the first term minus square root second term in one STEPS: to put the brackets back into the sum. bracket then Square root the first term plus square root second term 1. Put in standard form a x 2 + b x + c in second bracket. 2. Multiply the coefficients of the 1st and 3rd terms (i.e. a × c) STEPS: 3. Find the factors of answer in (2) that add if + c or subtract if −c 1. Look for a common factor first. EXAMPLES to get b 2. If a binomial look for difference of two squares or sum/ Factorise Fully: 4. Write with middle term split into outers and inners difference of two cubes. 1. 4a 2 − 64b 2 Remember to check for HCF 1st 5. Factorise by grouping 3. If a trinomial check if in form a x 2 − b x + c, = 4(a 2 − 16b 2) then factorise. EXAMPLES = 4(a − 4b)(a + 4b) a ×c = 3×2 = 6 4. If 4 or more terms group by looking for patterns first, Factorise Fully: c = + 2; + factors; e.g. difference of squares or perfect square trinomial. 1. 3x 2 + 7x + 2 5. Don’t forget to factorise as far as possible. 2. x 2 (x − k) + y 2 (k − x) Sign change 2×3 ; 2+ 3= 5 = 3x 2 + 6x + 1x + 2 6. Remember terms in brackets can be considered as a variable, = x 2 (x − k) − y 2 (x − k) Take (x − k) out as a common bracket 1×6 ; 1+ 6 = 7 = 3x (x + 2) + (x + 2) = (x − k)(x 2 − y 2 ) 2nd bracket is DOTS = (x + 2)(3x + 1) both terms + as = (x − k)(x − y)(x + y) middle term is + 1. Highest Common Factor (HCF): a × c = 6 × 12 = 72 This is inverse of distribution. 3. (a − b)2 − (2a + b)2 2. 6a 2 − 17a b + 12b 2 c = + 12; + factors; = [(a − b) − (2a + b)][(a − b) + (2a + b)] = 6a 2 − 9a b − 8a b + 12b 2 9 × 8 ; 9 + 8 = 17 EXAMPLES = [a − b − 2a − b][a − b + 2a + b] = 3a (2a − 3b) − 4b (2a − 3b) Factorise Fully: = (−a − 2b)(3a) both terms – as = (2a − 3b)(3a − 4b) 1. 6y 2 + 12y = − 3a (a + 2b) middle term is – = 6y (y + 2) a × c = 3 × 6 = 18 4. Grouping: 3. 3p 2 + 7p − 6 c = − 6; – factors; 2. 3a (2a − b) − a 2 (2a − b) Take (2a − b) out as a common bracket Used if four or more terms. First group then do HCF. Groups can be = 3p 2 + 9p − 2p − 6 9×2 ; 9−2 = 7 = a (2a − b)(3 − a) due to HCF, Difference of two squares or perfect square trinomial. = 3p ( p + 3) − 2( p + 3) biggest factor gets EXAMPLES = ( p + 3)(3p + 2) 3. x (x − y) − 4(x − y)2 Take (x − y) out as a common bracket middle term sign Factorise Fully: 1. 9d + b c − b d − 9c a × c = 1 × 18 = 18 4. = (x − y)[x − 4(x − y)] Simplify 2nd bracket c = − 3; – factors; = (9d − 9c) + (b c − b d ) HCF in each bracket 4. 2x 2 − 6x − 36 = (x − y)[x − 4x + 4y] 6×3 ; 3−6 = −3 = 9(d − c) + b (c − d ) Sign change needed = 2(x 2 − 3x − 18) (HCF) = (x − y)(−3x + 4y) = 9(d − c) − b (d − c) Do HCF = 2(x 2 − 6x + 3x − 18) biggest factor gets = − (x − y)(3x − 4y) = (d − c)(9 − b) = 2[x (x − 6) + 3(x − 6)] middle term sign 2. Sign change: = 2(x + 3)(x − 6) Change of sign in a bracket to make the factors 2. 2a − 3b + 4a 2 − 9b 2 the same. = (2a − 3b) + (4a 2 − 3b 2 ) 2nd bracket DOTS = (2a − 3b) + (2a − 3b)(2a + 3b) Do HCF 5. Perfect Square Trinomial NOTE: = (2a − 3b)(1 + 2a + 3b) a) 4m 2 − 12m n + 9n2 (b + a) = (a + b) but (b − a) ≠ (a − b) 2 = ( 4m 2 − 9n2 ) Do a sign change as follows: 3. 25a 2 − p 2 − 12p q − 36q 2 Group last three terms as they (b − a) = − 1(a − b) = 25a 2 − ( p 2 + 12p q + 36q 2 ) = (2m − 3n)2 make a perfect square trinomial First term and last term are = 25a 2 − ( p + 6q)2 DOTS perfect squares EXAMPLE = [5a − ( p + 6q)][5a + ( p + 6q)] b) 49p 4 + 84p 2 + 36 4a (a − 2b) − 6(2b − a) = (7p 2 + 6)2 = 4a (a − 2b) + 6(a − 2b) = (5a − p − 6q)(5a + p + 6q) = 2(a − 2b)(2a + 3) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 3

Grade 10 Maths Essentials Algebraic Expressions SCIENCE CLINIC 2022 © FACTORISATION (CONTINUED) ALGEBRAIC FRACTIONS 6. Sum or Difference of Two Cubes: EXAMPLES: 2. Simplification of fractions with addition and Subtraction. STEPS: x2 − x − 6 STEPS: 1. Example: 8a 3 + 27 x2 − 9 1. Factorise the denominator(s) (and numerator(s) where necessary) (x − 3)(x + 2) 2. Cancel like factors in each term if any. 1. First bracket (binomial): = 3. Find the Lowest Common denominator (LCD) (x − 3)(x + 3) a. Cube root the 2 terms sign between that of sum 4. Put each term over LCD by creating equivalent fractions. ( 8a + 27 ) 33 3 (x − 3)(x + 2) 5. Carry out the products in the numerator and add like terms. = 6. Factorise numerator if possible and cancel any like factors. (x − 3)(x + 3) = (2a + 3) 3x 2 3 3x 3a − 4 4a (x + 2) 1. − − 2. − = 2 3a 2−a −4 a 2 − 2a − 3 2. Second bracket (trinomial): x −x −6 x −3 x + 2 (x + 3) a. square first term 3x 2 3 3x (3a − 4) 4a = − − = − b. 1st term x 2nd term with opposite sign to 1st (3a − 4)(a + 1) (a − 3)(a + 1) term (x − 3)(x + 2) (x − 3) (x + 2) 12y − 4x −4x 2 + 14x y − 10y 2 1(a − 3) 4a c. add 2nd term squared 2. × 3x 2 − 3x − 6 − 3x 2 + 9x = − 12x − 30y 8x − 24y (2a)2 − (2a × 3) + (3)2 = (a + 1)(a − 3) (a + 1)(a − 3) (x − 3)(x + 2) 2.a. 2.b. 2.c. −4(x − 3y) −2(2x 2 − 7x y + 5y 2 ) 1(a − 3) − 4a = (4a2 − 6a + 9) = × 6x − 6 = 6(2x − 5y) 8(x − 3y) = (a + 1)(a − 3) (x − 3)(x + 2) 8a 3 + 27 = (2a + 3)(4a 2 − 6a + 9) 8(2x − 5y)(x − y) −3a − 3 = 6(x − 1) = 48(2x − 5y) = (a + 1)(a − 3) (x − 3)(x + 2) EXAMPLES: (x − y) −3(a + 1) = = 1. 8h 3 − 125g 3 6 (a + 1)(a − 3) = (2h − 5g )(4h 2 + 10g h + 25g 2 ) −3 = a 2 − a b − 2b 2 a 2 − 4a b + 4b 2 (a − 3) 3. ÷ 2. 24t 3 + 1029 a 2 + 2a b + b 2 a+ b = 3(8t 3 + 343) 3. Restrictions on Fractions = 3(2t + 7)(4t 2 − 14t + 49) (a − 2b)(a + b) (a − 2b)2 All fractions with variables in their denominators will have restrictions, as a denominator = ÷ (a + b)2 (a + b) may not equal zero. If the denominator becomes zero the fraction is undefined. 216 (a − 2b)(a + b) (a + b) EXAMPLE: EXAMPLE: 3. a 3 − = × a3 (a + b)(a + b) (a − 2b)(a − 2b) Determine the value of x for which the Determine the restrictions on the following 6 36 = (a − )(a 2 + 6 + a2 ) 1 fractions will be undefined: fractions: a = a − 2b 7x 4x − 2 1. 2x − b 1. x −1 3. x2 − 1 1. Simplification of a Fraction with multiplication 3x − 2b 5b + 5 6 − 4b 2a + 4a b x −1= 0 and division. 4. × ÷ 3x − 2b = 0 x2 − 1 ≠ 0 2b 2 − b − 3 5b 2 + 10b + 5 2b 2 + 3b + 1 x = 1 2b x = x2 ≠ 1 STEPS: 5(b + 1) −2(2b − 3) 2a (1 + 2b) 3 1. Factorise the numerator(s) and the denominator(s). = × ÷ 3x − 1 x ≠± 1 (2b − 3)(b + 1) 5(b + 1)2 (2b + 1)(b + 1) 2. 2. Cancel like factors. 2x + 1 5(b + 1) −2(2b − 3) (2b + 1)(b + 1) = × × 2x + 1 = 0 3 4x a b (2b − 3)(b + 1) 5(b + 1)(b + 1) 2a (2b + 1) REMEMBER: ÷ 1 2. − b a x = x − 1 (x + 2)(x − 1) −10 2 a a = 10a (b + 1) x −1≠ 0 and x −2≠ 0 = × b b x ≠1 x ≠ −2 −1 = a (b + 1) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 4

Grade 10 Maths Essentials Equations and Inequalities SCIENCE CLINIC 2022 © SIMULTANEOUS EQUATIONS SIMULTANEOUS EQUATIONS: WORD PROBLEMS Determine values that will satisfy both equations simultaneously. Two methods can be used: Elimination or EXAMPLE 1 Substitution. Determine the values of x and y if the quadrilateral is a rectangle. EXAMPLE 1 2x + 14y x − y = 8 and 2x + y = 10 ELIMINATION: SUBSTITUTION: 4y 2x x −y = 8 x − y = 8 ...A Add the equations to eliminate 2x + y = 10 one of the variables 2x + y = 10 . . . B 54 3x = 18 x = y + 8 ...C 4y = 2x . . . A 18 2(y + 8) + y = 10 x = 2x + 14y = 54 . . . B 3 2y + 16 + y = 10 1 x = 6 y = x ...C 3y = 10 − 16 2 1 3y = − 6 2x + 14( x) = 54 x −y = 8 2 Substitute the value into the y = −2 2x + 7x = 54 (6) − y = 8 original equations to find the −y = 8 − 6 y-value x −y = 8 9x = 54 Substitute the y-value into −y = 2 x − (−2) = 8 equation A to find the x-value x = 6 Substitute the y-value into A y = −2 x+ 2= 8 4y = 2(6) x = 6 12 ∴ (6; − 2) y = 4 ∴ (6; − 2) y = 3 EXAMPLE 2 The results of simultaneous equations are the point of intersection if the equations were to be presented graphically (x ; y). The sweet you like is reduced by R2 on a special offer. This means you can get 14 sweets for the same price as you used to pay for 10. What is the usual price? EXAMPLE 2 3x − 2y = 8 3x − 2y = 8 and 4x + 2y = 6 ∴ 3x − 2(−2x + 3) = 8 Usual price: x ∴ 3x + 4x − 6 = 8 Special price: (x − 2) 3x − 2y = 8 . . . A ∴ 7x = 14 4x + 2y = 6 . . . B ∴x = 2 10x = 14(x − 2) 4x = 2y = 6 10x = 14x − 28 2y = − 4x + 6 y = − 2x + 3 Substitute the x-value into 28 = 14x − 10x ∴ y = − 2x − 3 . . . C y = − 2(2) + 3 equation C to find the y-value 28 = 4x y = −4+ 3 7= x y = −1 ∴ (2; − 1) ∴ the usual cost of the sweet is R7. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 5

Grade 10 Maths Essentials Equations and Inequalities SCIENCE CLINIC 2022 © LINEAR INEQUALITIES: EXAMPLE 2 Relationship between expressions that are not equal Inequality Interval Notation for Open Intervals 4(2x − 1) > 5x + 2 x > a: x is greater than a (a , ∞) 8x − 4 > 5x + 2 x < a: x is less than a (−∞, a) 8x − 5x > 2 + 4 x > 2 3x > 6 (2;∞) ∴x > 2 2 x< 2 (–∞;2) 2 2 Inequality Interval Notation for Closed Intervals x ≥a: x is greater than or equal to a [a , ∞) EXAMPLE 3 x ≤a: x is less than or equal to a (−∞, a] x ≥2 1 [2;∞) −2 < 3 − x ≤5 subtract 3 from all terms 2 x ≤2 2 1 −5 < − x ≤2 multiply all terms by -2 2 (–∞;2] 2 10 > x ≥− 4 NOTE: the inequality signs ∴ − 4 ≤ x < 10 had to be REVERSED EXAMPLE 1 Write down the inequality for each of the following: a. –2 –1 0 1 2 3 4 5 6 7 –4 10 b. 0 1 2 3 4 5 6 7 8 EXAMPLE 4 c. 0 1 2 3 4 5 6 7 8 −3 < 2x + 1 < 5; x ∈R subtract 1 from all terms a. x < − 1 or x ≥6 −3 − 1 < 2x + 1 − 1 < 5 − 1 x ∈(−∞; − 1) or x ∈[6; ∞) −4 < 2x < 4 divide all terms by 2 ∴−2< x < 2 b. 3 < x < 6 x ∈(3; 6) c. −∞ < x < ∞, x ≠ 3, x ≠ 6 –2 2 x ∈(−∞; ∞); x ≠ 3, x ≠ 6 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 6

Grade 10 Maths Essentials Equations and Inequalities SCIENCE CLINIC 2022 © LINEAR EQUATIONS (Gr9 Revision) LITERAL EQUATIONS REMINDERS: F Move all the variables to the one side ad the constants to the other on order to Make a specific variable the subject of the equation 1. Linear equation: an equation of degree solve. Linear equations have only one solution. one with at most one solution 2. Distributive law: a (b + c) = a b + a c; EXAMPLE 1 EXAMPLE that is, the monomial factor a is distrib- Solve for p: The surface area of a cylinder is given by A = 2π r(h + r). uted, or separately applied, to each 4(2p − 7) − 8(5 − p) = 3(2p + 4) − 5( p + 7) Use the distributive law A − 2π r term of the binomial factor b + c Prove h = 8p − 28 − 40 + 8p = 6p + 12 − 5p − 35 Group like terms 2π r 3. Like terms: terms that have the same 8p + 8p − 6p + 5p = 28 + 40 + 12 − 35 Simplify variables and powers 15p = 45 A = 2π r(h + r) Manipulate the equation for find h on its own A p= 3 = h+ r 4. Quadratic Equation: an equation of the 2π r second degree A −r = h 2π r 5. Degree of the equation: the exponent of EXAMPLE 2 A r × 2π r the highest power to which that variable Solve for a: − = h Adding fractions (LCD is 2π r) 2π r 2π r is raised in the equation 2a − 1 −a + 5= 0 Find LCD (5) A − 2π r 2 5 = h 6. Constant term: a known, fixed value 2π r 2a − 1 ( 5 ) × 5 − (a) × 5 + (5) × 5 = 0 × 5 Multiply both sides by LCD 7. Coefficient: a number used to multiply a 2a − 1 − 5a + 25 = 0 Group like terms variable 2a − 5a = − 25 + 1 Simplify 8. Inequality: <,> two values that are −3a = − 24 not equal. −24 a = −3 9. Fractions and 0: a = 8 Numerator Denominator 0 x = 0 BUT = undefined x 0 10.Multiplication of signs: (−) × (+ )= (−) (−) × (−)= (+ ) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 7

Grade 10 Maths Essentials Equations and Inequalities SCIENCE CLINIC 2022 © Steps for solving unknowns in the EXAMPLE 1 EXAMPLE 2 denominator: 1. Factorise denominators, apply the 3 x+ 3 6 2x 5x − 3 x + = Change sign to simplify factorisation + = Change sign to simplify factorisation sign-change rule if necessary. x − 2 4 − x2 x+ 2 x −3 9 − x2 x+ 3 2. State restrictions. 3 x+ 3 6 2x 5x − 3 x − = Factorise denominators − = Factorise denominators x − 2 x2 − 4 x+ 2 x −3 x2 − 9 x+ 3 3. Multiply every term by the lowest 2x 5x − 3 x common denominator (LCD). 3 x+ 3 6 − = LCD: (x − 3)(x + 3); x ≠ 3,x ≠ − 3. − = LCD: (x − 2)(x + 2); x ≠ 2,x ≠ − 2. x − 3 (x − 3)(x + 3) x+ 3 4. Solve the equation. x − 2 (x − 2)(x + 2) x+ 2 Multiply every term by the LCD Multiply every term by the LCD 2x 5x − 3 x 3 x+ 3 6 × (x − 3)(x + 3) − × (x − 3)(x + 3) = × (x − 3)(x + 3) × (x − 2)(x + 2) − × (x − 2)(x + 2) = × (x − 2)(x + 2) x −3 (x − 3)(x + 3) x+ 3 x −2 (x − 2)(x + 2) x+ 2 2x (x + 3) − (5x − 3) = x (x − 3) 3 × (x + 2) − (x + 3) = 6 × (x − 2) 2x 2 + 6x − 5x + 3 = x 2 − 3x 3x + 6 − x − 3 = 6x − 12 Simplify x 2 + 4x + 3 = 0 Simplify 6 − 3 + 12 = 6x − 3x + x (x + 1)(x + 3) = 0 15 = 4x (x + 1) = 0 or (x + 3) = 0 15 x = −1 or x = −3 x = 4 BUT x ≠ − 3 ∴ x = − 1 is the only solution QUADRATIC EQUATIONS General Steps for solving Quadratic Equations Cannot Sum factorise 2 a x + bx + c = 0 wh ere a ≠ 0 2 perfect squares Binomial Difference 2. Factor out Highest 1. Standard Form Common Factor Trinomial Factoring EXAMPLE 1: FACTORISATION Factorise Quadratic Formula 5x + 2x 2 = 3 Standard form Complete the EXAMPLE 2: DIFFERENCE OF 2 SQUARES EXAMPLE 2: ALTERNATIVE square 2x 2 + 5x − 3 = 0 Factorise (2x − 1)(x + 3) = 0 Apply the zero-factor law* (x + 5)2 = 9 (x + 5)2 = 9 2x − 1 = 0 or x + 3 = 0 x 2 + 10x + 25 = 9 Standard form (x + 5)2 = 9 2x = 1 or x = −3 x2 + 10x + 16 = 0 Factorise x+ 5= ± 3 *Zero-Factor Law: 1 (x + 8)(x + 2) = 0 Apply zero-factor law* if a × b = 0 then a = 0 or b = 0 x = x + 5 − 3 = 0 or x + 5 + 3 = 0 2 x + 8 = 0 or x + 2 = 0 1 x+ 2= 0 or x+ 8= 0 ∴x = or x = −3 x = −8 or x = −2 2 x = −2 or x = −8 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 8

Grade 10 Maths Essentials Exponents SCIENCE CLINIC 2022 © LAWS OF EXPONENTS NOTES: Laws of exponents only apply to multiplication, division, brackets and roots. NEVER adding or subtracting. For the following: a , b > 0 and m , n ∈Z 1 a 1. a −n = or a ⋅ b −m = 1. 2. am 3. 4. m an bm a m × a n = a m+ n = a m−n (a m ⋅ a n) p = a n×p ⋅ a m×p n am = a n an 2. a0 = 1 MULTIPLICATION AND DIVISION ADDITION AND SUBTRACTION EXAMPLES EXAMPLES NOTE: Simplify, leaving all answers in positive exponential form. Simplify, leaving all answers in positive exponential form. To be able to simplify algebraic expressions of 2 or more terms, 1. [Multiplication bases variables] 3. [Multiplication bases combined] 2n−1 + 2n−2 one must always 1. Expand factorise FIRST 2n − 2n+ 2 (a b ) ⋅ a b 4 2 24 3 1 1 4 (27m ) 3 × (m 3 ) −1 Laws 3 & 4 Change to prime bases 2n ⋅ 2−1 + 2n ⋅ 2−2 (a b 3) ⋅ 3 a −6 4 −1 = Factorise using HCF and Note 1 (81m 2) 2 2n − 2n ⋅ 22 a 8b 4 ⋅ a 4 b 3 1 1 4 2n( 2 + 1 1 ) = Law 1 (3 m ) 3 × (m 3 ) 22 3 −1 a 4 b 12 ⋅ a −2 = Law 3 = Simplify −1 2 (1 − 22 ) n a 12 b 7 = 2 12 ( 34 m 2 2 ) 2+ 1 = ( 4 ) Law 2 a b ÷ (1 − 4) Note 1 −1 4 10 −5 3m 3 × m 3 = a b Note 1 = Laws 1 & 2 3 1 3−2 m −1 = × a 10 4 −3 = −1+ 4+ 1 b5 = 31+ 2 m 3 3 1 = − = 33 m 2 4 = 27m 2 2. [Multiplication bases numerical] 5 ⋅ 22 x−1 + 3 ⋅ 4x+ 1 1 2. Change to prime bases 182n+ 1 ⋅ 9 2 ⋅ 16n−1 22 x − 6 ⋅ 22 x−2 Change to 5 ⋅ 22 x ⋅ 2−1 + 3 ⋅ 22 x ⋅ 22 81n−3 ⋅ 64 n+ 2 prime bases = Change 4 to prime and expand 1 22 x − 6 ⋅ 22 x ⋅ 2−2 (32 ⋅ 2)2n+ 1 ⋅ (32 ) 2 ⋅ (24 )n−1 = Law 3 22 x(5 ⋅ 2 + 3 ⋅ 22) 1 (34 ) n−3 ⋅ (26 ) n+ 2 = Take out common factor 34n+ 2 ⋅ 22n+ 1 ⋅ 3 ⋅ 24n−4 22 x(1 − 6 ⋅ 4 ) 1 = Law 1 34n−12 ⋅ 26n+ 12 5 2 6n−3 ⋅ 34n+ 3 2 + 12 = 6n+ 12 4n−12 Law 2 = 3 Simplify bracketed terms 2 ⋅3 1− 2 = 26n−3−(6n+ 12) ⋅ 34n+ 3−(4n−12) 29 2 = 2−15 ⋅ 315 Note 1 = 1 −2 315 = 29 2 215 = ×− Tip and times 3 15 2 1 = ( ) = − 29 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 9

Grade 10 Maths Essentials Exponents SCIENCE CLINIC 2022 © ADDITION AND SUBTRACTION EXPONENTIAL EQUATIONS EXAMPLES continued EXAMPLES Simplify, leaving all answers in positive Solve for x Equations involving factorisation exponential form. 3. 3 ⋅ 52 x−1 − 3 = 0 2 NOTE: 1. 3x 3 = 12 Divide by 3 3⋅5 2 x−1 = 3 To be able to simplify algebraic 2 expressions of 2 or more terms, x3 = 4 Convert back to surd form 52 x−1 = 1 one must always factorise 3 2 x = 4 Perform inverse operation on both sides ∴ 52 x−1 = 50 Remember that 50 = 1 (Note 2) FIRST ∴ 2x − 1 = 0 ( x ) = (4) 23 33 1 4x − 4 x 2 = 64 x = 3. Change 4x to prime 2 2x − 2 x = ± 8 22 x − 4 Numerator is Diff of = 2 Squares (DOTS) 4. 2x+ 3 = 2x + 28 2x − 2 x x (2 − 2)(2 + 2) 2x+ 3 − 2x = 28 Get powers together = x−1 2 x+ 2 (2x − 2) 2. 2 ⋅ 3 ⋅3 − 4 = 50 Isolate powers 2 x ⋅ 23 − 2 x = 28 = 2x + 2 2 ⋅ 33x+ 1 = 50 + 4 Simplify LHS 2x (23 − 1) = 28 Factorise LHS 3x+ 1 3 = 54 ÷ 2 Divide both sides by 2 (as it has no exponent) x 2 (7) = 28 Simplify bracket & divide 33x+ 1 = 27 Get bases the same by using prime factors x 2 = 4 Change 4 to prime 21026 − 21024 33x+ 1 = 33 If bases same, exponents must be same to be = 2 x = 22 4. HCF and Law 4 22044 ∴ 3x + 1 = 3 ∴x = 2 2 1024 2 (2 − 1) Law 2 and simplify 2 = bracket x = 21022 3 = 21024−1022(4 − 1) = 22(3) = 12 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 10

Grade 10 Maths Essentials Number Patterns SCIENCE CLINIC 2022 © TERMINOLOGY: EXAMPLE EXAMPLE EXAMPLE 1. Consecutive: numbers or terms following directly after i. Given the sequence 3, 8, 13, … Given T6 = 8 and T9 = − 1, determine T5 A shop owner wishes to display cans of food in a each other ii. Determine the next three terms. and T7 . triangular shape as shown in figure. There is one can iii. Determine the general term. in the top row, three in the second row and so on. iv. Determine the value of the 45th term. SOLUTION 2. Common/constant difference: the difference tween two v. Which term has the value of 403? consecutive terms. T5; 8; T7; T8; −1 d = T2 − T1 d = T3 − T2 SOLUTION T7 = T6 + d i. 18; 23; 28 = 8+ d i. Write down the first four terms of this pattern. 3. Terms are indicated by a T and the ii. In what type of sequence are the tins arranged? ii. d = 8 − 3 = 5 or d = 13 − 8 = 5 position or number of the term in the pattern by a sub- iii. Write down a formula for the term of the sequence. T8 = T7 + d script, e.g. term 1 is T1 or term 50 is T50 . c = T1 − d iv.How many cans are needed for the 15th row? = (8 + d ) + d v.In which row will there be 27 cans? = 3−5 4. General term Tn: also referred to as the nth term. = −2 = 8 + 2d SOLUTION • General term for linear patterns: ∴ Tn = d n + c T9 = T8 + d Tn = 5n− 2 i. 1; 3; 5; 7 Tn = d n + c −1 = (8 + 2d ) + d iii. n = 45 −9 = 3d ii. Linear sequence ∴ T45 = 5(45) − 2 −3 = d iii. d = 3 − 1 = 2 or d = 5−3= 2 Linear Patterns: = 223 Sequences with a constant difference between the terms. c = T1 − d iv. Tn = 403 so need to solve for n ∴ T5 = T6 − d T7 = T6 + d = 1−2 Tn = general term = 8−3 and = 8 + (−3) = −1 Tn = d n+ c d = constant difference Tn = 5n− 2 = 11 = 5 n = number of the term 5n− 2 = 403 ∴ Tn = d n + c 5n = 405 Tn = 2n− 1 Steps to determine the nth term: ∴ n = 81 iv. t15 = 2(15) − 1 1. Find the constant difference = 29 d = T2 − T1 = T3 − T2 EXAMPLE EXAMPLE ∴29 cans are needed for the 15th row 2. Find the c-value Mpho is told that a sequence has a nth If the pattern “safesafesafesafe…“ v. Tn = 27 c = T1 − d term of 15n− 2 . She has to find which continues in this way what would the 2n− 1 = 27 term will be equal to 96. She is stuck 263rd letter be? 3. Substitute the c- and d-values to define the nth term. ∴ n = 14 ( ∴ 14th row) because she keeps getting an unexpected answer. Perform the calculations and then SOLUTION explain the answer Quadractic Patterns: Note safe has 4 letters so safesafe has 8 By inspection: and safesafesafe has 12 etc. Tn = general term SOLUTION a = constant difference ÷ 2 ∴ 263 ÷ 4 = 65 remainder 3 2 Tn = a n + c Tn = 96 n = number of the term ∴ 15n− 2 = 96 Thus 65 safe and three more letters, the 263rd letter is f. 15n = 98 8 n= 6 15 ∴ 96 is not a term in the sequence since n ∈ℕ Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 11

Grade 10 Maths Essentials Finance - Simple and Compound Interest SCIENCE CLINIC 2022 © SIMPLE INTEREST COMPOUND INTEREST HIRE PURCHASE LOANS Interest calculated on only the money initially invested Interest calculated on the initial amount and any sub- Short-term loans to buy goods on credit, normally repaid in equal monthly installments. or borrowed. sequent interest that is earned or charged. A = P(1 + in) A = P(1 + in) A = P(1 + i )n EXAMPLE A = accumulated amount P = original amount Tania wants to buy a new TV which costs R9 350. She can’t afford the full amount now and r agrees to buy it on the following hire purchase terms: n = number of years i = interest rate 100 ➡ 12% deposit ➡ 13% interest p.a. ➡ equal monthly installments over 2 years In financial Maths, unless instructed otherwise, always round off your FINAL answer to 2 decimal places a) How much is the deposit amount? EXAMPLE 1 12 % of R9 350 = R1 122 Which investment would be a better option over 6 years? b) How much will Tania repay, including interest? a) 6% p.a. simple interest b) 5,5% p.a. compound interest R9 350 − R1 122 = R8 228 A = P (1 + i n) A = P (1 + i ) n A = P (1 + i n) = P (1 + 0,06 × 6) = P (1 + 0,055)6 = 8 228(1 + 0,13 × 2) = R10 367,28 = 1,36P = 1,38P c) How much will her monthly repayments be? ∴ Option (b) is better R10 367,28 ÷ 24 = R431,97 per month EXAMPLE 2 d) How much would Tania have saved if she could have paid the full amount initially? John wants to have R10 000 available in 4 years’ time for a holiday. How much does he need to invest now if (R1 122 + R10 367,28) − R9 350 = R2 409,28 he can get an interest rate of 8,3% p.a. compounded annually? A = P (1 + i ) n INFLATION POPULATION GROWTH 10 000 = P (1 + 0,083)4 The rising cost of goods and services A = P(1 + i )n 10 000 = P (1,375...) DO NOT round off yet! n A = P(1 + i ) EXAMPLE P = R7 269,20 EXAMPLE A herd of cows was made up of EXAMPLE 3 24 animals in 2014. If the growth A loaf of bread costs R14. If the average inflation rate of the herd was approximately You want to double an investment of R1 200 in five years. What annual interest would yield this return? has been 8%, and assuming it remains constant: 13% p.a., how many cows would a) Simple interest a) Compound interest a) How much will a loaf of bread cost in 5 years? you have expected in the herd in A = P (1 + i n) A = P (1 + i ) n A = P (1 + i ) n 2018? 2400 = 1200(1 + i × 5) 2400 = 1200(1 + i )5 = 14(1 + 0,08)5 A = P (1 + i ) n = R20,57 = 24(1 + 0,013)4 2 = 1 + 5i 2 = (1 + i )5 b) How much did a loaf of bread cost 13 years ago? = 39,13 1 = 5i 5 2 = 1+ i ≈ 39 cows A = P (1 + i ) n ∴ i = 0,2 1 + i = 1,14869... 14 = P (1 + 0,08)13 = 20,00 % p.a. ∴ i = 0,14869... P = R5,15 = 14,87% p.a. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 12

Grade 10 Maths Essentials Finance SCIENCE CLINIC 2022 © FOREIGN EXCHANGE RATES TIMELINES Timelines can be used to visually represent more complicated situations in Financial Maths CURRENCY RATE OF EXCHANGE (OF THE RAND) TYPE 1: MONEY IN AND OUT Pound (ә) 18,23 Thando invests R2 000 in a fixed deposit. Two years later, he adds R1 000. One year after that, he needs to withdraw R650. If the interest rate is 10,25% p.a. compounded annually for the entire period, how much money will Thando have after 5 years? US Dollar ($) 15,42 R2 000 + R1 000 – R650 EXAMPLE a) A South African lady working in London manages to save Y0 Y1 Y2 Y3 Y4 Y5 R4 000 per month. How many pounds does she save in a year? R4 000 ÷ R18,23/ә × 12 i = 0,1025 = ә2 633,02 A = P (1 + i ) n = 2000(1 + 0,1025)5 + 1000(1 + 0,1025)3 − 650(1 + 0,1025)2 b) If she wanted to buy a book from America, on Amazon, for = R3 807,81 $15, how much would she pay in pounds? TYPE 2: CHANGE OF INTEREST RATE $15 × R15,42 /$ Sam deposits R4 300 into a 3 year fixed deposit account. The interest (all compounded annually) is 9,87% p.a. for the first year, and then = R231,30 10,3% p.a. for the remainder of the period. How much will Sam have at the end of 3 years? R4 300 R231,30 ÷ R18,23/ә = ә12,69 Y0 Y1 Y2 Y3 i = 0,098 i = 0,103 A = P (1 + i ) n = 4300(1 + 0,098)1(1 + 0,103)2 MULTIPLY when the rate changes = R5 744,10 TYPE 3: COMBINATIONS Jenny invests R1 500, but two and a half years later, she needs to withdraw half of the initial investment. The interest rate for the first two years is 11% p.a. compound interest and 9% p.a. compound interest for the other 4 years. How much money will Jenny have after 6 years? R1 500 – R750 Y0 Y1 Y2 Y2,5 Y3 Y4 Y5 Y6 i = 0,011 i = 0,09 A = P (1 + i ) n = 1500(1 + 0,011)2 (1 + 0,09)4 − 750(1 + 0,09)3,5 = R1 594,78 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 13

Grade 10 Maths Essentials Functions and Graphs SCIENCE CLINIC 2022 © FUNCTION TYPES OF FUNCTIONS AND THEIR GRAPHS - LINEAR FUNCTION A FUNCTION is a rule by means of which each element 1. LINEAR FUNCTION EXAMPLE: EXAMPLE: of the domain (independent variable or input value(s), (Straight line graph) Determine the equation of the straight Determine the equation of the straight line passing through i.e. x) is associated with one and only one element of y = mx + c or y = ax + q line passing through (–1;–1) and A(–4;3) and B(3;–2). the range (dependent variable or output value(s), i.e. y) • m = a = gradient or slope perpendicular to x − 4y = 4. Functions can be represented in different ways for SOLUTION: y = mx + c y = −mx + c example: SOLUTION: y − y1 y = 2x − 3; y = 2x 2 + 1; x y = 4 or 2 or or m = 2 x − 4y = 4 put in standard form x 2 − x1 or m> 0 m< 0 f (x) = − 5x + 1; g (x) = x 2 − 5 y y 1 −2 − 3 y = x −1 m = 4 3 − (−4) Example 2 is known as function notation and is an easier x x 1 ∴ mf = 5 way of representing the y-value. ∴ y = f (x) m = − y − y1 4 7 m = 2 So we can write y = 3x + 1 as f (x) = 3x + 1 and is read x 2 − x1 5 as follows; the value of the function f at x is equal to ∴= − x+ c Graphs are perpendicular, 7 3x + 1, where f (x) is the range and x is the domain. Thus c = q = y-intercept i.e. (0;c) f (2) will give the output value when 2 is substituted in ∴ m f × mg = − 1 • Domain: x ∈ℝ Range: y ∈ℝ for x, i.e. f (2) = 3(2) + 1 = 7 so ordered pair (2; 7). 5 ∴ mg = − 4 3= − (−4) + c FINDING THE EQUATION 7 EXAMPLE: Remember m can be found in the follow- 1 If g (x) = 3x 2 − 5x, determine; ∴c = ing ways: Thus y = − 4x + c sub into (–1;–1) 7 g (−1) 1. given two coordinates: y − y1 −1 = − 4(−1) + c g (x) = 2 m = 2 5 1 x 2 − x1 −1 − 4 = c ∴y = − x+ 7 7 g (3) − g (−2) 2. Function f parallel to function g ∴ c = −5 m f = mg SOLUTION: 3. Function f perpendicular to function g ∴ y = − 4x − 5 m f × mg = − 1 x is given, solve for y g (−1) EXAMPLE: EXAMPLE: EXAMPLES: = 3(−1)2 − 5(−1) Determine the equation of the straight Determine the equation of the straight Sketch the following graphs on same set of axes showing all = 8 line passing through (–1;3) and(–5;3). line passing through (–1;4) and(–1;–2). intercepts with the axes: y given so solve for x a) f (x) = − 2x + 1 using dual-intercept method. g (x) = 2 SOLUTION: SOLUTION: y-intercept: 3x 2 − 5x = 2 4 − (−2) 3−3 (x = 0 or f (0)) : (0; 1) m = = undefined m = = 0 x-intercept: 3x 2 − 5x − 2 = 0 −1 − (−1) −5 − (−2) 1 (y = 0 or f (x) = 0) f (3x + 1)(x − 2) = 0 ∴ x = − or x = 2 ∴ y = 3 (horizontal line) 0 = − 2x + 1 3 ∴ x = − 1 (vertical line) y 1 4 y 4 ∴x = g (3) − g (−2) 2 = (3(3)2 − 5(3)) − (3((−2)2 − 5(−2)) 2 x 2 b) y = − 3 = 12 − 22 x −4 −2 2 4 = − 10 −4 −2 2 4 y = −3 −2 −4 −2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 14

Grade 10 Maths Essentials Functions and Graphs SCIENCE CLINIC 2022 © TYPES OF FUNCTIONS AND THEIR GRAPHS - QUADRATIC FUNCTION (PARABOLA) 2. QUADRATIC FUNCTION FINDING THE EQUATION SKETCHING THE GRAPHS (PARABOLA) Steps: y = a x 2 + q, a ≠ 0 1. Given turning point (0, q) 2. Given roots (x-intercepts) and 1. Find Turning point or y-intercept. and another point use another point use 2. Find x-intercepts and in none use the table method. • a > 0 or a is +’ve y = a x 2 + q. y = a (x − R1)(x − R2 ) 3. Determine the shape. EXAMPLE: EXAMPLE: Determine equation of Determine equation of EXAMPLE: EXAMPLE: Sketch the following graph showing all Sketch the following graph showing all inter- intercepts with the axes and turning cepts with the axes and turning points: points: f (x) = 2x 2 + 2 g (x) = − 2x 2 + 8 a < 0 or a is –’ve SOLUTION: SOLUTION: Turning point or y-intercept: Turning point or y-intercept: (0;2) (0;8) • q = y-intercept x-intercepts (y=0): x-intercepts (y=0): SOLUTION: • Turning point = (0;q) First sub the turning points 2x 2 + 2 = 0 0 = − 2x 2 + 8 • Domain: x ∈ℝ y-value in for q: SOLUTION: 2x 2 = − 2 has no solution 2x 2 = 8 OR 2(x 2 − 4) = 0 y = a x2 + 4 First sub in the roots –4 and 4 ∴ has no x − intercepts x2 = 4 (x − 2)(x + 2) = 0 • Range: y ∈[ p; ∞) minimum if a > 0 y = a (x − (−4))(x − 4) FOIL out x = ± 4 y ∈(−∞; p] maximum if a < 0 now sub (–2;–8) to find a y = a (x 2 − 16) now sub in (6;10) Shape: x = 2 or x = − 2 • Symmetry: x = 0 −8 = a (−2)2 + 4 a > 0; −16 = 4a 10 = a (62 − 16) Shape: ∴ a = −4 ∴ y = − 4x 2 + 4 10 = 20a Table of data points (on calculator): a < 0; 1 ∴a= x –2 –1 0 1 2 2 1 2 y 10 4 2 4 10 ∴y = (x − 16) 2 1 2 y ∴y = x −8 8 Domain: x ∈ℝ 2 10 y 6 4 Range: y ∈(∞; 8] Domain: x ∈ℝ 2 x 5 Range: y ∈[2; ∞) −3 −2 −1 1 2 3 −2 x −3 −2 −1 1 2 3 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 15

Grade 10 Maths Essentials Functions and Graphs SCIENCE CLINIC 2022 © TYPES OF FUNCTIONS AND THEIR GRAPHS - HYPERBOLA HYPERBOLA SKETCHING THE GRAPHS FINDING THE EQUATION a Steps: Determine the equation of: y= + q, a≠0 x 1. Determine the asymptotes ( x = 0 and y = q ) • a = constant and affects the shape 2. Determine the x-intercepts 3. Determine the shape + a + a y = + q y = −q x x 4. Use table method to plot at least 3 other points a > 0; q > 0 a > 0; q < 0 EXAMPLE: EXAMPLE: y y Sketch the following graph showing all Sketch the following graph showing all inter- a y= + q intercepts with the axes and asymptotes: cepts with the axes and asymptotes: x 3 2 y=q y=q f (x) = − h (x) = + 1 First sub in asymptote y = 2 in place of q x x x x a y= + 2 SOLUTION: SOLUTION: x Asymptotes: Asymptotes: Sub in the coordinate (4; 3) x = 0; y = 0 x = 0; y = 1 a 3= + 2 As y = 0 is an asymptote there are no 2 4 −a −a x-intercepts: 0 = + 1 y = −q y = + q x-axis intercepts. x 12 = a + 8 x x Shape: 0 = 2+ x ∴a= 4 a < 0; q < 0 a < 0; q > 0 a< 0 x = −2 4 y y ∴y = + 2 y Shape: x y a> 0 y=q y=q x x x x Table: Table: • q = constant but shifts graph up or down the y-axis. x –3 –1 1 3 x –2 –1 1 2 • Asymptotes: x = 0; y = 0 y 1 3 –3 –1 y 0 –1 3 2 (values that make the function undefined) • Domain: x ∈ℝ, x ≠ 0 • Range: y ∈ℝ, y≠q • Axis of symmetry: y = x + q or y = − x + q Domain: x ∈ℝ, x ≠ 0 Domain: x ∈ℝ, x ≠ 0 Range: y ∈ℝ, y ≠ 0 Range: y ∈ℝ, y ≠ 1 Lines of symmetry: y = x or y = − x Lines of symmetry: y = x + 1 or y = − x + 1 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 16

Grade 10 Maths Essentials Functions and Graphs SCIENCE CLINIC 2022 © TYPES OF FUNCTIONS AND THEIR GRAPHS - EXPONENTIAL FUNCTIONS EXPONENTIAL FUNCTIONS SKETCHING THE GRAPHS FINDING THE EQUATION y = a ⋅ bx + q a≠0 and b > 0, b ≠ 1 Steps: STEPS: Asymptote: y = q Domain: x ∈ℝ 1. Determine the asymptote (y = q) 1. First sub in the asymptote y = q Range: y ∈(q; ∞) i f a > 0; y ∈(−∞; q) i f a < 0 2. Determine the y- and x-intercepts 2. Sub in y-intercept to find a 3. Determine the shape 3. Sub in other point to find b y = + a ⋅ b x + q, b > 1 y = + a ⋅ b x − q, b > 1 4. Use table method to plot at least 2 other points y y EXAMPLE: EXAMPLE: EXAMPLE: Sketch the following graph showing all Sketch the following graph showing all Determine the equation of: intercepts with the axes and asymptotes: intercepts with the axes and asymptotes: f (x) = 3x 1 x y = ( ) −2 x x SOLUTION: 2 Asymptote: y = 0 SOLUTION: y-intercept: y = 30 = 1 x y = + a ⋅ b + q,0 < b < 1 x y = + a ⋅ b − q, 0 < b < 1 Asymptote: y = − 2 x-intercept: none as y = 0 asymptote 1 Note a = 1 thus only need to find b and q y-intercept: y = ( ) − 2 = − 1 0 y y f (x) = b x − 1 Sub in (2; 15) y 2 Shape: 1 15 = b 2 − 1 x-intercept: 0 = ( ) − 2 x a > 0,b > 1 2 16 = b 2 x Increasing function 2 = 2−x b = ± 4, but b > 0, b ≠1 x x ∴b = 4 ∴ x = −1 Table: ∴ f (x) = 4x − 1 Shape: y x –1 1 y = + a ⋅ bx − q EXAMPLE: y = − a ⋅ b + x + q, b > 1 y = − a ⋅ b + x − q, b > 1 Determine the equation of: 0< b < 1 x y 1 y 3 3 x Table: x –1 1 y 0 1,5 g (x) = a ⋅ b x + 2 4 = a ⋅ b0 + 2 Sub in (0; 4) y = − a ⋅ b x + q, 0 < b < 1 y = − a ⋅ b x − q, 0 < b < 1 2= a g (x) = 2 ⋅ b x + 2 10 = 2 ⋅ b 1 + 2 Sub in (1; 10) 8= 2⋅b Domain: x ∈ℝ 4= b Range: y ∈(0; ∞) or y > 0 Domain: x ∈ℝ Range: y ∈(−2; ∞) or y > − 2 ∴ g (x) = 2 ⋅ 4x + 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 17

Grade 10 Maths Essentials Functions and Graphs SCIENCE CLINIC 2022 © GRAPH APPLICATION DISTANCE NOTATION TRANSFORMATIONS OF GRAPHS Steps for determining VERTICAL DISTANCE • f (x) > 0 1. Reflection in x-axis: y becomes negative (i.e. all signs (above the line y = 0) on right hand side of equation change). 1. Determine the vertical distance y y Vertical distance = top graph − (bottom graph) 2. Reflection in y-axis: all x’s become negative 2. Substitute the given x-value to obtain your answer 3. Reflection in both axes: both x and y become negative x x 4. Horizontal Shift: q changes, if up then add to q and if down subtract from q. A (i.e. where y is positive) B • f (x) < 0 (below the line y = 0) y y EXAMPLE: SOLUTIONS: Steps for determining HORIZONTAL DISTANCE 2 a) y ∈[−1; ∞) Sketched are the graphs of f (x) = x − 1 and 1. Find the applicable x-values x x g (x) = x + 1. b) x = 0 A B = xB − xA (largest − smallest) c) x–intercept ( ∴ y = 0) (i.e. where y is negative) 0 = x2 = 1 0 = (x − 1)(x + 1) A B ∴ x = 1 or x = − 1 • f (x) ⋅ g (x) ≤0 ∴ A(1; 0) d) B is y–intercept ∴ OB = 1 unit NOTE: (one graph lies above y = 0 and one graph lies below y = 0) e) f (x) = g (x) at C • Distance is always positive • Distance on a graph is measured in units x2 − 1 = x+ 1 • f (x) ≥g (x) top bottom x2 − x − 2 = 0 (i.e. f(x) lies above g(x)) (x − 2)(x + 1) = 0 INTERSECTION OF GRAPHS • f (x) = g (x) QUESTIONS: x = 2 or x = − 1 Steps for determining POINTS OF INTERSECTION (point of intersection) a) the range of f (x) y = 2+ 1 n /a 1. Equate the two functions ROOTS & PARABOLAS b) equation of the axis of symmetry of f (x) y = 3 f (x) = g (x) • Equal, real roots c) the coordinates of A ∴ A(2; 3) 2. Solve for x (look for the applicable x-value: A or B) y y d) the length of OB f) OF on left side of x-axis ∴ x = − 4 3. Substitute the applicable x-value into any of the e) the coordinates of C DE = f (−4) − g (−4) two equations to find ‘y’ x x f(x) DE = ((−4)2 − 1) − ((−4) + 1) f) the length DE if OF is 4 units. g(x) DE = 15 − (−3) A • Non-real/ No real roots g) for which value(s) of x is y y DE = 18 units B i) f (x) ≥0 g) i. x ∈ (−∞; − 1] or [1; ∞) x x ii)f (x) ⋅ g (x) < 0 ii. x ∈ (−∞; − 1) or (−1; 1) INCREASING/DECREASING iii)f (x) decreasing alt. x ∈ (−∞; 1); x ≠ − 1 • Real, unequal roots h) Give the equation of h (x) formed if g (x) is reflected iii.x ∈ (−∞; 0) Dec y y asin in the y-axis. h) h (x) = − x + 1 Dec asin g g s s rea i) Give the equation of k (x) formed if f (x) is trans- i) k (x) = x 2 − 1 + 3 rea Inc re Inc x x re ing lated 3 units up. ing k (x) = x 2 + 2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 18

Grade 10 Maths Essentials Probability SCIENCE CLINIC 2022 © TERMINOLOGY Theoretical Probability of an event happening: Complimentary Events: Outcome: Result of an experiment. S = {sample set} Events A and B are complimentary events if they are mutually exclusive and exhaustive. Event: An event is a collection of outcomes that satisfy a certain condition. A = {event A} An event is denoted with the letter E and the number of outcomes in B = {event B} Exhaustive Events: the event with n(E). A ∪ B = {A union B} = in sets A or B Events are exhaustive when they cover all elements in the Dependent events: When the first event (A) affects the other’s outcomes. sample set. E.g. choosing two coloured marbles from a bag, with A ∩ B = {A intersection B} = in sets A and B S replacement, thus, the first choice doesn’t affect the outcome of the second choice. Theoretical Probability of an event happening: A B Independent event: Events that do not affect each other’s outcomes. Cards in a deck: 52 number of possible times event can occur 1 2 5 P(E ) = 3 4 6 number of possible outcomes Suits in a deck: 4 Specific card: 4 n(E ) = Sample spaces: All possible outcomes of the experiment. E.g. rolling a dice n(S) S={1;2;3;4;5;6}. Intersection of sets. Mutually Exclusive: Unbiased: All events are equally likely to happen. E = event S = sample space A and B are mutually exclusive events as they have no elements in common. Complimentary event: Those two mutually exclusive events whose sum of probabilities equal to 1. Relative frequency or Experimental probability: S PROBABILITY number of times the event occured A P(E ) = B Likelihood of an event happening The probability of an event is the ratio be- number of trials done 1 2 5 tween the number of outcomes in the event set and the number of possible 3 4 6 outcomes in the sample space. Equally Addition Rule (OR/+): likely 9 Impossible Unlikely Likely Certain NOTE: 1 1 3 P(A or B) = P(A) + P(B) − P(A and B) or 0 1 P(A and B) = P(A) + P(B) − P(A or B) n(A ∪ B ) = n(A) + n(B ) 4 2 4 0 0,25 0,5 0,75 1 n(A ∩ B ) = 0 P (A ∪ B ) = P (A) + P (B ) 0% 25% 50% 75% 100% Probability using a Venn diagram Inclusive Events: Event E Sample space S Intersecting events: Union of events: The ocean turns A coin dropped into milkshake will land on heads Compliment of an event A B A B Not A A S S S A and B A or B Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 19

Grade 10 Maths Essentials Probability SCIENCE CLINIC 2022 © V ∪M (V ∪ M )′ or V′ ∩ M′ EXAMPLE 1 EXAMPLE 3 EXAMPLE 4 V M V M A dice is rolled 100 times. It lands on 2 six- Questions: Questions: teen times. Calculate the relative frequency Calculate from the Venn diagram for a 120 Gr 12 girls at Girls High where asked about their and compare this to the theoretical probability. grade 6 group in which the number of participation in the school’s culture activities: equally likely ways the events • 61 girls did drama (D) 1 (Reading(R); Sports(S) and Art(A)) can • 29 girls did public speaking (P) S S Probability = = 0,1667 6 occur has been filled in: • 48 girls did choir (C) • 8 girls did all three M V frequency of event Gr 6 • 11 girls did drama and public speaking Relative frequency = number of trials • 13 girls did public speaking and choir V M V M R S • 13 girls did no culture activities 16 = 100 10 97 1. Draw a Venn diagram to represent this information. 20 = 0,16 2. Determine the number of Girls who participate in S S 5 drama and choir only. The more an experiment is repeated the 25 8 3. Determine the probability that a grade 12 pupil closer the relative frequency and the V∩M (V ∩ M )′ or V′ ∪ M′ theoretical probability will be. chose at random will: 3 a. only do choir. V M V M 2 EXAMPLE 2 b. not do public speaking. Elmari carried out a survey in her town to A c. participate in at least two of these activities. establish how many passengers travel in each vehicle. The following table shows her results: 1. P(A⋂R⋂S) S S Solutions: Number of 2. P(R and A and not S) Number of cars 1. n(S) = 120 passengers 3. P(A or R) V′ ∩ M V ∪ M′ 4. P(S or R and not A) n(D) = 61 n(P) = 29 V M V M 0 7 1 11 Solutions: 50–x 3 13 2 6 5 1 8 1. P(A ∩ R ∩ S) = = x 5 3 4 170 34 S S 4 2 25 5 35–x 2. P(R and A and not S) = = 13 M′ What is the probability that a vehicle has 170 34 n(C) = 48 more than two passengers? 70 7 V M 3. P(A or R) = = 2. (50 − x) + 3 + 13 + x + 8 + 5 + (35 − x) + 13 = 120 SOLUTION: 170 17 127 − x = 120 There are 30 vehicles in the survey, so 127 n(S)=30. Let A be the event “cars with more 4. P(S or R and not A) = ∴x = 7 170 S than two passengers”. This means that we ∴ 50 − x = 43 and 35 − x = 28 only count the vehicles with three and four passengers. Therefore, n(A)=4+2=6. 3. 35 28 n(A) a. P(C only) = = = 0,23 P (A) = 120 120 n(S) 120 − 29 91 6 b. P(P′) = = = 0,76 = 120 120 30 3+ 7+ 8+ 5 23 1 c. P(at least 2) = = = 0,19 = 120 120 5 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 20

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Grade 10 Maths Essentials Trigonometry SCIENCE CLINIC 2022 © BASIC DEFINITIONS BASIC CAST DIAGRAM PYTHAGORAS PROBLEMS RECIPROCALS ( ) ( o) Shows the quadrants where each trig ratio is + Steps: 1 r h 90⁰ 1. Isolate the trig ratio = cosecθ or sin θ y 2. Determine the quadrant o 3. Draw a sketch and use Pythagoras sin θ = soh (x) (a) Opposite t h S A 1 r h po 4. Answer the question Hy cos θ = a cah sin + all + = secθ or e h II I EXAMPLE cos θ us 180⁰ 0⁰ en tan θ = o toa III IV 360⁰ If 3sinθ – 2 = 0 and tanθ < 0, determine (y) ( o) a sin2 θ + cos2 θ without using a calculator and 1 x a T C = cotθ or cos + by using a diagram. tan θ tan + θ Step 1: Adjacent Special Angles 270⁰ 3 sin θ − 2 = 0 2 y These are our basic trig ratios. EXAMPLE sin θ = 3 r r = 2 (x; y) 1. In which quadrant does θ lie if tanθ < 0 and cosθ > 0? EXAMPLE (0; 2) Step 2: (1; 3) C o BC tanθ - cosθ + sin θ = = tanθ - h AC cosθ + sinθ + sinθ + ( 2; 2) a AB 90o 60o cos θ = = tanθ - tanθ - o h AC 45 ( 3; 1) ! o BC Quadrant IV o A B tan θ = = ∴ Q u a d r a nt I I 30 a AB 2. In which quadrant does θ lie if sinθ < 0 and cosθ < 0? EXAMPLE (2; 0) P Step 3: 0o cosθ - Q sinθ - x 2 + y2 = r 2 EXAMPLE sinθ - (x;2) cosθ - r=3 x 2 + (2)2 = (3)2 Simplify without the use of a calculator: y=2 x2 = 5 S Quadrant III θ 1. tan 60∘ + cot30∘ x = ± 5 ( 1 ) ( 1 ) 3 3 y On the Cartesian Plane = + R ∴x = − 5 (x;y) = 3+ 3 1. Write down two ratios for cos R y a PR RS sin θ = = 2 3 cos R = = = r Step 4: sin2 θ + cos2 θ h QR PR r y x cos θ = 2 2 (3 ) ( 3 ) r 2 − 5 y = + tan θ = 2. sin2 45∘ − cos2 30∘ (in △ PQR) (in △ PRS) θ x x 4 5 2 2 3 2 ( 2 ) ( 2 ) x = + Remember: 2. Write down two ratios for tan Q 9 9 y = − sin θ = r o PR PS 9 tan Q = = = = and 2 3 a PQ QS Remember: 9 x = − cos θ = r 4 4 • x2 + y2 = r2 (Pythagoras) = 1 1 = − (in △ PQR) (in △ PQS) • Angles are measured upwards from the positive (+) x-axis 4 (anti-clockwise) up to the hypotenuse (r). Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 21

Grade 10 Maths Essentials Trigonometry SCIENCE CLINIC 2022 © TRIG EQUATIONS SOLVING RIGHT-ANLGED TRIANGLES PROBLEM SOLVING For 0∘ < θ < 90∘ * Missing Sides * * Missing Angles * Determine the area of quadrilateral ABCD. EXAMPLES EXAMPLES D Steps: 1. Calculate x 1. Calculate  3 units 1. Isolate the trig ratio 3 units (o) A 2. Reference angle (shift on the calculator) x 5 units (h) 3. Solve for θ (o) 1 units 25º REMEMBER: (a) (h) (a) 6 units Only round off at the END o 5 A sin 25∘ = = h x EXAMPLES 37º ∴ x sin 25∘ = 5 o 3 B C Solve for θ tan A = = 5 a 1 In △ ABC : x = sin 25∘ tan A = 3 * shift tan 3 tan 37∘ = o = 6 1. 3 sin θ − 1 = 0 x = 11,83 units Ref ∠: 60∘ a BC 1 6 sin θ = ∴  = 60∘ BC = 3 tan 37∘ Don’t round off early. Ref ∠: 19,47∘ 2. Calculate y Remember to store ∴ BC = 7,96 units ∴ θ = 19,47∘ (o) answer in calculator 2. Calculate P̂ ∴ AC2 = AB2 + BC2 (Pythag) 2. tan(3θ + 30∘ ) − 1 = 0 y P AC2 = 2 (6) + (7,96) 2 Use the ‘unrounded’ tan(3θ + 30∘ ) = 1 (a) 7 units answer from your (h) 5 units AC2 = 99,3977 calculator Ref ∠: 45∘ 50º (a) (h) ∴ AC = 9,97 units ∴ 3θ + 30∘ = 45∘ a y 3θ = 15∘ cos 50∘ = = Area of △ ABC : Area of △ ACD : h 7 ∘ θ = 5 7 cos 50∘ = y Area = 1 b⋅⊥ h Area = 1 b⋅⊥ h (o) 4 units 2 2 y = 4,50 units 1 1 3. sec2θ = 2 o 4 = (BC)(AB) = (AC)(AD) sin P̂ = = 2 2 1 h 5 1 1 = 2 4 = (7,96)(6) = (9,97)(3) cos 2θ sin P = 2 2 1 5 * shift sin 4/5 = 23,89 units2 = 14,95 units2 ∴ cos 2θ = 2 Ref ∠: 53,13∘ Ref ∠: 60∘ ∴ P̂ = 53,13∘ ∴ 2θ = 60∘ Area of Quad ABCD : ∘ Area = 23,89 + 14,95 θ = 30 = 38,84 units2 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 22

Grade 10 Maths Essentials Trig Graphs SCIENCE CLINIC 2022 © IMPORTANT! When sketching trig graphs, you need to label the following: ! both axes ! x- and y-intercepts Notes for sin x and cos x : Notes for tan x : ! turning points ! endpoints (if not on the axes) ❖ Key points (intercepts/turning pts) ❖ Key points every 45° ! asymptotes (tan graph only) every 90° ❖ Period (1 complete graph): 180° BASICS ❖ Period (1 complete graph): 360° ❖ No amplitude can be defined • y = sin x for x ∈[0∘; 360∘ ] ❖ Amplitude (halfway between min ❖ Asymptotes at x = 90o + k180o, k ∈ℤ and max): 1 y (90°; 1) 180° 360° 0° x AMPLITUDE CHANGE VERTICAL SHIFT (270°; -1) • y = a . sin x or y = a . cos x or y = a . tan x • y = sin x + q or y = cos x + q or y = tan x + q If a > 1 : stretch If q > 0 : upwards (e.g: y = sin x + 1) 0 < a < 1 : compress If q < 0 : downwards (e.g: y = cos x − 2) a < 0 : reflection in x-axis • y = cos x for x ∈[0∘; 360∘ ] EXAMPLES EXAMPLE y 1. y = 2 sin x 2. y = − 3 cos x y = cos x − 1 x ∈[0∘; 360∘ ] (solid line) (solid line) (solid line) y = cos x (dotted line - for comparison) (360°; 1) y = sin x y = cos x (dotted line - for comparison) (dotted line - for comparison) 90° 270° * Amplitude = 1 0° x * Range: y ∈[−2; 0] * Amplitude = 2 * Amplitude = 3 (180°; -1) * Range: y ∈[−2; 2] * Range: y ∈[−3; 3] y • y = tan x for x ∈[0∘; 360∘ ] y (180°;3) y (90°; 2) y 360° 0° x x (45°; 1) 180° 360° (90°;-1) (270°;-1) 0° x 0° 90° 270° xx 0° 180° 360° x (180°; -2) (270°; -2) -3 (360°; -3) x = 90° x = 270° Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 23

Grade 10 Maths Essentials Trig Graphs SCIENCE CLINIC 2022 © MIXED EXAMPLE Solutions: Given f (x) = 3 sin x and g (x) = cos x − 2 1. 360∘ Questions: 2. 3 1. What is the period of f? 3. 1 2. State the amplitude of f? 4. y 3. Give the amplitude of g ? (90°; 3) 4. For x ∈[0∘; 360∘ ] sketch f and g on separate sets of axes. f 5. Use the graphs to determine the values of x for which: a. f (x) > 0 180° 360° b. g (x) is increasing 0° x 6. Explain, in words, the transformation that takes g (x) = cos x − 2 to y = 2 cos x + 2 (270°; -3) y x (0°;-1) (360°;-1) g (180°; -3) 5. a. x ∈(0∘; 180∘ ) Remember: A graph is positive when it b. x ∈(180∘; 360∘ ) is above the x-axis Remember: 6. g (x) = (1)cos x − 2 A graph is increasing when the gradient is positive to y = 2 cos x + 2 double the move 4 amplitude units up ∴ g is stretched by a factor of 2 and translated 4 units up. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 24

Grade 10 Maths Essentials Euclidean Geometry SCIENCE CLINIC 2022 © Grade 8 and 9 Revision CONGRUENT TRIANGLES Remember there are four reasons for congruency and EXAMPLE: EXAMPLE: Deductive logic in geometry the triangles must be written in order of equal parts. Given PQ = PR and circle centre S. Prove Given P Q̂ S = 90∘, Q R = R S and PQ ∥ A R. Prove Working reasoning to conclude next answer that AQ = A S. ̂ . that PS bisects angle Q SR in known as deductive logic. Q P Examples: ΔACR ≡ ΔEGO (SSS) Q 1. If a = b 2. If x = y and b = c and p= q then a = c then x + p = y + q A R 3. If P̂ − Q̂ = 180∘ and Ŝ − Q̂ = 180∘ then: P̂ = Ŝ 1 S ΔDEC ≡ ΔBEA (SAS) P 2 THEORY TO REMEMBER NOTE: S must be the A SOLUTION: INCLUDED angle R In Δ A RQ and Δ A R S AR = AR (common) P ̂ QA = Q ̂ AR = 90∘ (alt . ∠′s, PQ ∥ AR) SOLUTION: QR = RS (given) 1 2 ΔRST ≡ ΔCAB (SAA) In ΔPSQ and ΔPSR ∴ ΔARQ ≡ ΔARS (RHS) B C PS = PS (common) ∴ AQ = AS ( ≡ Δ′s) B̂ = C1̂ (∠'s opp. = sides) PQ = PR (given) Â + B̂ + C1̂ = 180∘ (sum ∠'s of Δ) SQ = SR (radii) △ PSQ ≡ △ PSR (SSS) C2̂ = Â + B̂ (ext. ∠'s of Δ) ∴ Ŝ 1 = Ŝ 2 ( ≡ Δ′s) PROPERTIES OF QUADRILATERALS D EXAMPLE: *For ALL quadrilaterals: sum of interior F ΔPQR ≡ ΔMNR (RHS) Given PQ ∥ R S and PS ∥ Q R, prove: angles is 360° 2 a) PQ = R S 1 2 1. KITE 1 K M 3 b) PS = Q R • Two pairs of adjacent sides are equal. EXAMPLE: • The longest diagonal bisects the angles. E G If A ̂ DC = x, find with reasons, the size of angles • One diagonal is the perpendicular bisector of the other. K2̂ = M̂1 (corres. ∠'s DE//GF) x and y . Show all steps and give all reasons. • One pair of opposite angles are congruent. K2̂ = M̂3 (alt. ∠'s DE//GF) K2̂ + M̂2 = 180∘ (co-int. ∠'s DE//GF) M̂1 = M̂3 (vert. opp. ∠'s) SOLUTION: 2. TRAPEZIUM K2̂ + K1̂ = 180∘ (∠'s on a str. line) • One pair of opposite sides parallel. a) In ΔPQ R and ΔR SP P̂ 1 = R̂ 2 (alt ∠′s, PQ ∥ RS) P PR = PR (common) R̂ 1 = P̂ 2 (alt ∠′s, PS ∥ QR) SOLUTION: ∴ ΔPRQ ≡ ΔRPS (SAA) D̂ 2 = 27∘ (alt . ∠′s DE //CB) ∴ PQ = RS ( ≡ △′ s) R T D̂ 1 = 52∘ (sum ∠′s of Δ) ∴ x = 79∘ b) PS = QR( ≡ Δ′s) P T 2 = PR 2 + R T 2 (Pythag. Th.) y = D̂ 1 = 52∘ (corres . ∠′s, DE //CB) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 25

Grade 10 Maths Essentials Euclidean Geometry SCIENCE CLINIC 2022 © PROPERTIES OF QUADRILATERALS (CONT.) MIDPOINT THEOREM 3. PARALLELOGRAM (parm) EXAMPLE: KLMN is a parallelogram with KM = KL. Mid-Point Theorem: Converse: • The opposite sides are parallel by definition. (mid-pt. Th.) (conv. mid-pt. Th.) (PA ∥ RM & PM ∥ AR) The line segment joining the S The line passing through • The opposite sides are equal. midpoints of two sides of a the midpoint of one side of S PA = RM & PM = AR) triangle, is parallel to the third a triangle and parallel to • The opposite angles are equal. side and half the length of the another side, bisects the ̂ = PMR (PAR ̂ ̂ = M RA) & M PA ̂ third side. Q P third side. The line is also Q P • The diagonals bisect each other. equal to half the length of (PD = DR & AD = DM) Determine the values of a, b and c. the side it is parallel to. • Two pairs of opposite sides parallel and equal. O R O R 4. RECTANGLE (rect) SOLUTION: Therefore if SQ = QO and SP = PR then Therefore if SQ = QO and PQ ∥ OR then • Is a specialised parallelogram so has a = 53∘ (opp ∠′s parm) 1 1 b = 53∘ (∠′s opp = sides) PQ ∥ OR and Q P = OR (mid-pt. Th.) SP = PR and Q P = OR (conv. mid-pt. Th.) all the parallelogram properties. 2 2 M KL̂ = 74∘ (sum ∠′s of Δ) • Diagonals are equal length. (RC = ET) c = 74∘ (alt ∠′s, KL ∥ NM) EXAMPLE • Interior angles are each 90°. In ΔACE, AB = BC, GE = 15 cm EXAMPLE: 5. RHOMBUS (rhom) and AF = FE = ED. M and N are the midpoints of XY and HINTS WHEN ANSWERING • Is a specialised parallelogram so has all the parallelogram properties. XZ. MN is produced its own length to GEOMETRY QUESTIONS • Adjacent sides are equal. D. (RH = HO = OM = MR) Prove that: • Read the given information and mark • Diagonals are perpendicular a) XMZD a parallelogram on to the diagram if not to each other. b) MYZD a parallelogram already done. ( D̂ 1 = 90∘ ) • Diagonals bisect the angles. • Never assume anything. If not given ( R̂ 1 = ̂ R̂ 2; Ĥ 1 = Ĥ 2; Ô 1 = Ô 2; M̂ 1 = M̂ 2 ) or marked on diagram is not true unless proved. 6. SQUARE (squ.) • Is a specialised parallelogram, rectangle and rhombus so has • As you prove angles equal or calculate all their properties. angles mark them on to the diagram Determine the length of CE. and write down statement and reason SOLUTION: there and then. EXAMPLE: In ΔACE: AT = BU = CV = DW. O is the AB = BC and AF = FE (given) • Make sure that by the end of the centre of the circle TUVW and ̂ = 90∘ . AC and BD are SOLUTION: 1 question you have used all the given BOC ∴ BF ∥ CE and BF = CE (mid − pt . Th.) information. a) XN = NZ (given mid pt) 2 straight lines. Prove that ABCD is MN = ND (given) a square. ∴ XMZD a parm . • If asked to prove something, it is true. In ΔDFB For example: if asked to prove ABCD (diag bisect ea . other) FE = ED (given) a parallelogram, it is a b) XY ∥ DZ (opp . sides parm XMZD) BF ∥ GE (proven) parallelogram. If you can’t prove it, SOLUTION: XM = MY (given mid pt.) 1 you can still use it as a ∴ BG = GD and GE = BF (conv . mid − pt . Th.) OT = OU = OV = OW (radii) XM = DZ (opp . sides parm XMZD) 2 parallelogram in the next part of the AT = BU = CV = DW (given) ∴ MY = DZ ∴ BF = 2GE question. ∴ AO = BO = CO = DO ∴ MYZD a parm . ∴ BF = 2(15) = 30 cm AC ⊥ DB (given) (1 pr . opp . sides = & ∥) ∴ ABCD a squ . (diag . = and ⊥ ) CE = 2BF (proven) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 26

Grade 10 Maths Essentials Analytical Geometry SCIENCE CLINIC 2022 © What is Analytical Geometry? (Co-ordinate Geometry): Application of straight line functions in conjunction with Euclidean Geometry by using points on a Cartesian Plane. DISTANCE BETWEEN TWO POINTS MIDPOINT OF A LINE SEGMENT GRADIENT OF A LINE COLLINEAR POINTS The distance between two points (x1; y1) and (x 2; y2 ) The midpoint between (x1; y1) and (x 2; y2 ) is given by: The gradient of a straight line between (x1; y1) and Points on the same line, hence, gradients is given by: x + x1 y2 + y1 (x 2; y2 ) is given by: between the points are equal. M(x ; y) = ( 2 2 ) ; y − y1 m = ( 2 x −x ) d = (x 2 − x1)2 + (y2 − y1)2 2 2 1 EXAMPLE EXAMPLES If T (5; 2), U (7; 4) and V (b; − 5) are EXAMPLES REMEMBER: collinear, calculate the value of b. 1. Determine the midpoint of P (−1; 4) and Q (4; − 2) 1. Determine the length of PQ if P(−1; 4) and • Parallel (∥) lines: m1 = m 2 Collinear ∴ m TU = mUV Q (4; − 2) x + x1 y2 + y1 Midpnt = ( 2 2 ) ; • Perpendicular (⊥ ) lines: m1 × m 2 = − 1 2−4 4+ 5 2 = −1 + 4 4 − 2 • Horizontal (−) lines [y = c]: m = 0 5−7 7−b = ( 2 ) d = (x 2 − x1)2 + (y2 − y1)2 ; 2 9 3 • Vertical (|) lines [x = c]: m is undefined 1= = (−1 − 4)2 + (4 − (−2))2 = ( ; 1) 7−b 2 7−b = 9 = 61 EXAMPLE = 7,81 2. FEGH is a parallelogram. Calculate the b = −2 co-ordinates of G. Given A(2; 3) and B(−3; 1). Remember: 2. If A(1; 2), B(−1; − 5) and C (x ; − 7) and y in a parallelogram A B = BC, calculate x 1. Determine the gradient of the line A B F (4;6) the diagonals y − y1 3−1 2 mAB = ( 2 x 2 − x1 ) ( 2 + 3 ) ( 5 ) AB = (1 + 1)2 + (2 + 5)2 = = bisect each other, x = 53 H (–4;–6) ∴ M is the mid- M BC = (x + 1)2 + (−7 + 5)2 E (14;–10) point of F G and E H. 2. Determine the gradient of the line parallel to = (x + 1)2 + 4 AB but A B = BC G (x;y) 2 For ∥ lines: mAB = m∥ = 5 m1 = m 2 (x + 1)2 + 4 −4 + 14 −6 − 10 Midpnt of E H = ( ) ∴ 53 = ; 2 2 ∴ (x + 1)2 + 4 = 53 = (5; − 8) (x + 1)2 = 49 x+ 4 y+ 6 Midpnt of F G = ( 2 ) 3. Determine the gradient of the line ; x+ 1= ± 7 2 perpendicular to A B For ⊥ lines: x = −1+ 7 or x = −1−7 x+ 4 y+ 6 Flip the fraction and ∴ = 5 and = −8 mAB × m⊥ = − 1 2 2 change the sign x = 6 or x = −8 2 x + 4 = 10 y + 6 = − 16 × m⊥ = − 1 5 x = 6 and y = − 22 5 ∴ m⊥ = − 2 ∴ G (6; − 22) Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 27

Grade 10 Maths Essentials Analytical Geometry SCIENCE CLINIC 2022 © EQUATIONS OF STRAIGHT LINES MIXED EXAMPLE 1 y = mx + c Prove that △ L M N is right-angled 6−2 4 y mL M = = − −2 − 3 5 gradient y-intercept 12 − 2 5 N (11;12) mNM = = 11 − 3 4 EXAMPLE y ∴ mL M × mNM 3. parallel to ST, through point R. 4 5 L (–2;6) = − × 7 5 4 S (–4;2) mST = m∥ = − R (6;2) 6 = −1 M (3;2) 7 ∴y = − x+ c ∴ LM ⊥ MN 6 x x ∴ △ L M N is right-angled Sub in R(6; 2) 7 2 = − (6) + c 6 MIXED EXAMPLE 2 NOTE: T (2;–5) c = 9 There are 5 ways to prove a quad is a parm 7 Quadrilateral PQ R S is given 1. both pairs of opposite sides equal y = − x+ 9 6 2. both pairs of opposite sides parallel Determine the equation of the line: y 3. one pair of opposite sides equal and parallel S (–1;m) 4. perpendicular to ST, through point T. 1. ST 4. diagonals bisect each other 2+ 5 7 mST × m⊥ = − 1 mST = = − 5. both pairs of opposite angles equals −4 − 2 6 7 P (1;1) − × m⊥ = − 1 7 6 y = − x+ c 6 3. If RQ ∥ SP determine the value of m 6 ∴ m⊥ = R (–2;0) x 7 mRQ = mSP Sub in S(−4; 2) (or T) 6 m −1 7 y = + c Q (0;–2) −1 = 2 = − (−4) + c 7 −1 − 1 6 8 Sub in T (2; − 5) 2 = m −1 c = − 6 3 −5 = x+ c ∴m = 3 7 8 7 y = − x− −47 1. Determine the length of PQ 6 3 c = 4. Prove that PQ R S is a parallelogram 7 PQ = (1 − 0)2 + (1 + 2)2 6 47 * You could use methods 1-4 to answer this 2. SR ∴y = x− = 10 question. Let’s use 4 this time (diags bisect) 7 7 Horizontal line 1−2 1−0 1 1 Midpt PR = ( 2 ) ( 2 2) ; = − ; S(−4; 2) and R(6; 2) have the same y-value. 2. Find the gradient of RQ 2 5. perpendicular to SR, through S. −1 − 0 3 − 2 1 1 Midpt Q S = ( 2 ) ( 2 2) ∴m = 0 0+ 2 ; = − ; ⊥ to horizontal line is a vertical line mRQ = = −1 −2 − 0 2 ∴y = 2 through (−4; 2) ∴ Midpt PR = Midpt Q S ∴ x = −4 ∴ Diags bisect each other ∴ PQ R S is a parm Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 28

Grade 10 Maths Essentials Analytical Geometry SCIENCE CLINIC 2022 © MIXED EXAMPLE 3 Parallelogram T U V W is given 4. Prove that △ T U W is isosceles y TU = (0 − 21)2 + (6 − 9)2 = 15 2 T (0;6) U (21;9) ∴ T W = TU ∴ △ T U W is isosceles x V (x;y) W (–15;–9) 5. Hence, what type of parm is T U V W? Give a reason. Rhombus. Parm with adjacent sides equal. 6. Determine the equation of the line perpendicular to U W 1. Determine the gradient of U W and passing through point W 9+ 9 1 mUW = = mUW × m⊥ = − 1 21 + 15 2 1 × m⊥ = − 1 2 2. Determine (by inspection) the co-ordinates of V ∴ m⊥ = − 2 T→U x : 0 → 21 ∴ x + 21 y = − 2x + c TU ∥ W V y :6→9 ∴y+ 3 Sub in W (−15 : − 9) ∴W→V x + 21 : − 15 + 21 = 6 −9 = − 2(−15) + c y + 3:−9+ 3= −6 c = − 39 ∴ V (6; − 6) ∴ y = − 2x − 39 3. Calculate the length of T W (in simplest surd form) 7. If U, R(3; k) and W re collinear, find the value of k TW = (−15 − 0)2 + (−9 − 6)2 Collinear ∴ mUR = mUW k −9 1 = 15 2 = 3 − 21 2 k −9 1 = −18 2 ∴ k −9= −9 ∴k = 0 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 29

Grade 10 Maths Essentials Statistics SCIENCE CLINIC 2022 © REMINDER NB: Always arrange data in ascending order. Discrete data: Data that can be counted, e.g. the number of people. FREQUENCY TABLE STEM AND LEAF PLOTS BAR GRAPH HISTOGRAM Mark Tally Frequency Stem Leaf 7 Continuous data: quantitative data that can be measured, 6 e.g. temperature range. 4 || 2 0 1, 1, 2, 2, 3, 4 5 5 || 2 1 0, 0, 0, 1, 1, 1 4 Measures of central tendency: a descriptive summary of a 6 |||| 4 2 5, 5, 7, 7, 8, 8 3 dataset through a single value that reflects the centre of the data distribution. 7 |||| 5 3 0, 1, 1, 1, 2, 2 2 1 8 |||| 4 4 0, 4, 8, 9 Measures of dispersion: The dispersion of a data set is the 0 9 || 2 5 2, 6, 7, 7, 8 0 10 20 30 40 amount of variability seen in that data set. OR 0<x≤10 10<x≤20 20<x≤30 30<x≤40 10 | 1 6 3, 6 Outliers: Any data value that is more than 1,5 IQR to the left of Q1 or the right of Q3, i.e. MEASURES OF DISPERSION INDICATORS OF POSITION Range Interquartile range Semi-Interquartile range Outlier < Q1 – (1,5×IQR) or Quartiles 1 Outlier > Q3 + (1,5×IQR) range = max value − min value IQR = Q 3 − Q1 semi − IQR = (Q3 − Q1) The three quartiles divide the data into 2 Note: range is greatly influenced Note: spans 50% of the four quarters. Note: good measure of dispersion by outliers data set Q1 = Lower quartile or first quartile for skewed distribution Q2 = Second quartile or median MEASURES OF CENTRAL TENDENCY FOR UNGROUPED DATA Q3 = Upper quartile or third quartile Mean Mode Median Percentiles The mean is also known as the average value. A The mode is the value that appears most fre- The median is the middle number in a set of data that has The pth percentile is the value that p% of disadvantage to the mean as a measure of central quently in a set of data points. been arranged in order of magnitude. The median is less the data is less than. tendency is that it is highly susceptible to outliers. Bimodal: a data set with 2 modes affected by outliers and skewed data than the mean. Q1= 25th percentile Trimodal: a data set with 3 modes 1 Q2= 50th percentile sum of all values Σx position of median = (n + 1) x̄ = = 2 Q3= 75th percentile total number of values n * If all values have the same frequency, there is NO MODE Where; x̄ = mean Σx = sum of all values eg. If the 25th percentile is 12, then 25% n = number of values n = number of values of the data will be less or equal to 12 If n = odd number, the median is part of the data set. EXAMPLE: All other percentiles can be If n = even number, the median will be the average Create a frequency table for the following ungrouped data: calculated using the formula: between the two middle numbers. 1, 5, 3, 1, 2, 3, 4, 5, 1, 4, 2, 4, 4, 5, 1, 4, 2, 4, 2, 2 sum (frequencies × value) p mean = total frequency BOX AND WHISKER PLOT i= (n) 100 Frequency 59 A box and whisker plot is a visual representation of the five where; Streak Number = Chosen-Tally 20 number summary. i = the position of the pth percentile = 2,95 1 4 rem rt i rt i rem e le le e mode : 4 ex t qu a qu a ex t FIVE NUMBER SUMMARY 2 5 Lo Lo ed Up Up median = 1 (n + 1) we we ia n pe pe 1. Minimum value 2 r r M r r 3 2 2. Lower quartile Q1 1 = (20 + 1) 3. Median 2 4 6 = 10,5 4. Upper quartile Q3 5 3 ∴ the median is 3 5. Maximum value 10 20 30 40 50 60 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 30

Grade 10 Maths Essentials Statistics SCIENCE CLINIC 2022 © MEASURES OF CENTRAL TENDENCY FOR GROUPED DATA EXAMPLE: 2. Identify the interval in which each of the following data Estimated mean The mathematics marks of 200 grade 10 learners at a items lies: school can be summarised as follows: 2.1. the median; sum (frequencies × midpoint of interval) mean(x ) = median = 1 (n + 1) total frequency (n) 2 Percentage Number of 1 obtained candidates = (200 + 1) 2 x̄ = estimated mean = 100,5 n = number of values 10 ≤ x < 20 4 median class 50 ≤x ≤60, 20 ≤ x < 30 10 the 100th value is in this class interval Modal class interval The modal class interval is the class 30 ≤ x < 40 37 interval that contains the greatest number of data points. 40 ≤ x < 50 43 2.2. the lower quartile; Median class interval 50 ≤ x < 60 36 Lower quartile = 25th Percentile : The median class interval is the interval 25 60 ≤ x < 70 26 i= (200) that contains the middle number in a set of 100 data points. 70 ≤ x < 80 24 i = 50 1 80 ≤ x < 90 20 ∴ the 25th percentile is in the 30 ≤x ≤40 class interval position of median = (n + 1) 2 n = number of values SOLUTION: 2.3. the upper quartile; If n = odd number, the median is part 1. Calculate the approximate mean mark for Upper quartile = 75th Percentile : of the data set. the examination. 75 If n = even number, the median will i= (200) be the average between the Frequency Midpoint f ×x 100 i = 150 two middle numbers. 20 + 10 4 = 15 15 × 4 = 60 ∴ the 75th percentile is in the 60 ≤x ≤70 class interval 2 GROUPED DATA FREQUENCY TABLES 10 30 + 20 Midpoint = 25 25 × 10 = 250 Class interval Frequency 2 2.4. the thirteenth percentile; uper class barrier + lower class barrier ( f × x) 37 40 + 30 i= 13 (200) (f ) x = = 35 35 × 37 = 1295 100 2 2 i = 26 0 ≤ x < 10 3 10 + 0 3 × 5 = 15 50 + 40 = 5 43 = 45 45 × 43 = 1935 ∴ the 13th percentile is in the 30 ≤x ≤40 class interval 2 2 20 + 10 60 + 50 10 ≤ x < 20 7 7 × 15 = 105 36 = 55 55 × 36 = 1980 = 15 2 2 70 + 60 20 ≤ x < 30 4 30 + 20 4 × 25 = 100 26 = 65 65 × 26 = 1690 = 25 2 2 80 + 70 24 = 75 75 × 24 = 1800 total : 14 220 2 20 90 + 80 = 85 85 × 20 = 1700 Mean: 2 sum (frequencies × midpoint of interval) 200 10 710 mean(x ) = total frequency (n) sum (frequencies × midpoint of interval) 220 mean(x ) = = total frequency (n) 14 10 710 = 200 = 15,71 = 53,55 Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 31

Grade 10 Maths Essentials Statistics SCIENCE CLINIC 2022 © EXAMPLE: EXAMPLE: Examine the following box and whisker diagrams and answer the questions that follow: The following graph indicates the number of iPads sold per week. Class A 12 Maximum value 10 Frequency Class B 8 90 96 100 6 Minimum value 30 66 72 80 84 4 1. Name the value from the five number summary that is the same for both classes. 2 2. For each class, explain if the data is skewed or symmetrical. 1 2 3 4 5 6 SOLUTION: 1. In which week were the sales the highest? 1. The median for both classes are the same. 2. The store has a competition and the winner will be the person who bought their iPad in the middle of 2. Class A: skewed to the left, the data is more dispersed to the left of the median. the sales over the 6 weeks. In which week did the winner buy their iPad? Class B: slightly skewed to the right, the data is more dispersed to the right of the median. 3. What is the mean sales per week over the 6 weeks? EXAMPLE: SOLUTION: SOLUTION: The following stem and leaf diagram represents the scores of 40 people who wrote an exam. 1. mode = 2 mode = 35 Weeks Frequency The total of the scores is: 1544 2. sum of all value 1 5 1 mean = position of median = (n + 1) nr of values 2 Stem Leaf 2 12 1 1544 = (39 + 1) = 2 40 1 7, 7, 8, 8, 9, 9 = 38,6 3 8 = 20 2 0, 2, 4, 4, 6, 7, 7 4 8 ∴ the winner bought their iPad in the 3rd week 1 position of median = (n + 1) 2 3 4, 5, 5, 5, 5, 5, 5, 5, 9, 9 1 5 4 3. = (40 + 1) sum of all value 2 mean = 4 1, 2, 2, 3, 7, 8 nr of values = 20,5 6 2 39 = 6 5 0, 3, 3, 4, 5, 7 ∴ the median is 35 Total 39 = 6,5 6 3, 4, 5, 6, 6 While measures of central tendency are used to estimate "normal" values of a dataset, measures of dispersion are important for describing the spread of the data, or its Calculate the mean, mode and median variation around a central value. for the information provided. Range • Defined as the difference between the largest and smallest sample values. • Depends only on extreme values and provides no information about how the remaining data is distributed, this means it is highly susceptible to outliers. Interquartile Range (IQR) • Calculated by taking the difference between the upper and lower quartiles (the 25th percentile subtracted from the 75th percentile). • A good indicator of the spread in the center region of the data. • More resistant to extreme values than the range. Follow us on Instagram: @science.clinic Visit www.scienceclinic.co.za to download a free copy of these SmartPrep summaries, or to join interactive online Maths & Science classes. Page 32