Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Apply algebraic methods in solving problems Expanding brackets The distributive law is used to expand brackets: a(b + c) = a × b + a × c Expand pairs of brackets using the rule: (a + b)(c + d) = ac + ad + bc + bd The following expansions involving two factors should be familiar. Example Copy correctly Up to 3% of a workbook Copying or scanning from ESA workbooks is subject to the NZ Copyright Act which limits copying to 3% of this workbook. To expand an expression with three factors, select a pair of factors to expand first. This reduces the expression to the product of two factors. Example 1. (x + 5)(x + 4)(x + 3) = (x + 5)(x2 + 3x + 4x + 12) [expanding (x + 4)(x + 3)] 2 = (x + 5)(x + 7x + 12) [simplifying] = x3 + 7x2 + 12x + 5x2 +35x + 60 [(x + 5)(x2 + 7x + 12)] 1. y (2y + 5w) = y2 × 2y3 + y2 × 5w [distributive law] [ y n × y m = y n + m] = 2y5 + 5y2w 2 Externally assessed 4 credits 3 2. –2a(a2 – 7a + 8) = –2a × a2 + –2a × –7a + –2a × 8 [a(b + c + d) = ab + ac + ad] 3 = –2a + 14a2 – 16a 3. (3x + 5)(4x – 7) = 12x2 – 21x + 20x – 35 [(a + b)(c + d) = ac + ad + bc + bd] 2 = 12x – x –35 [simplifying] 4. (5x – 2)2 + x(4 – x) = (5x – 2)(5x – 2) + x(4 – x) = 25x2 – 10x – 10x + 4 + 4x – x2 [distributive law] 2 [collecting like terms] = 24x – 16x + 4 Perfect squares can be expanded more quickly using the following formula. (a ± b)2 = a2 ± 2ab + b2 e.g. (x + 5)2 = x2 + 2 × x × 5 + 52 = x3 + 12x2 + 47x + 60 [collecting like terms] 2 2. –4p(8 – p) = –4p(8 – p)(8 – p) = –4p(64 – 16p + p2) [expanding and simplifying (8 – p)2] = –256p + 64p2 – 4p3 3. (x + 5)3 = (x + 5)(x + 5)2 = (x + 5)(x2 + 10x + 25) = x3 + 10x2 + 25x + 5x2 + 50x + 125 = x3 + 15x2 + 75x + 125 Note: A simple check on an expansion can be done by substituting x = 1 into both the original expression and its expansion. If the answers don’t match, you have made an error. For example, substituting x = 1 in (x + 5)(x + 4)(x + 3) gives (1 + 5)(1 + 4)(1 + 3) = 120. Substituting x = 1 in x3 + 12x2 + 47x + 60 gives 1 + 12 + 47 + 60 which is also 120. The two results agree, so the expansion is most likely to be correct. = x2 + 10x + 25 © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. AS 91261 MATHEMATICS AND STATISTICS 2.6 2 AS 91261 Ans. p. 59 Achievement Standard 91261 (Mathematics and Statistics 2.6) Exercise A: Expanding brackets j. 4(2x – 3)(x – 4) 1. Expand and simplify. a. 8x(3x – 4) b. –6y(1 – x) 2. Expand and simplify the following expressions. a. x(3x – 4) + 2x(x + 1) c. 7x4(x3 – 8) b. 2ab(a + b) – a2(b – a) d. –8p(p2 – 6p – 4) c. 9 – x2(9 – x) e. (7a + 4)(2a –1) d. (w + 11)(w – 2) + 2w(3w – 5) f. (5x + 3y)(xy – 7) g. (5x + 2y)2 e. 12x – (x + 4)2 h. (2x – 7)2 f. i. (3t –7)(3t + 7) – (3t – 7)2 5(x + 1)2 © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. Alternatively x + 2 – 4x 3x + 4 = [subtracting fractions] 6 2 –3x + 2 3x + 4 = 6 2 –6x + 4 = 18x + 24 [cross-multiplying] –24x = 20 –20 –5 or x= 24 6 Use brackets around algebraic expressions in fractions, so that errors with signs are avoided. Example Solve for w: w + 2 1 – w 3w – = 4 2 8 Solution (w + 2) (1 – w) 3w – = 4 2 8 [bracketing expressions] 8(w + 2) 8(1 – w) 8 × 3w [multiplying – = by LCM of 4 2 8 denominators] 2(w + 2) – 4(1 – w) = 3w [cancelling] 2w + 4 – 4 + 4w = 3w [expanding] 6w = 3w [simplifying] 3w = 0 [subtracting 3w (both sides)] w=0 19 Forming linear equations and inequations Read problems carefully to define the unknown variable and identify the correct sequence of mathematical operations (such as adding, dividing, etc.). You will then need to solve your equation. Make sure you check your answer in the context of the question to ensure it applies to the practical situation. Example Miro and Maddie both drive taxis. Miro charges $6.50 flagfall plus $3.20 per kilometre. Maddie charges $4 flagfall plus $3.90 per kilometre. A trip of a certain length costs more with Miro than with Maddie. How long was this trip? Give your answer to the nearest tenth of a kilometre. The SOLV function on a graphics calculator is useful for checking and solving multi-step linear equations. Linear inequations Linear inequations will also involve multi-step reasoning. Remember to reverse the inequality sign when multiplying or dividing by a negative. Example 3 – 4(x + 2) ) 3(x + 1) 3 – 4x – 8 ) 3x + 3 [expanding] –7x ) 8 [rearranging] [dividing by –7 and x* 8 reversing inequality] –7 1 x * –1 [simplifying] 7 Note: Alternatively, collect positive numbers of x on the right side. Rearranging the second line gives –8 )7x. Dividing by 7 gives –8 )x, 7 8 which can be written in reverse as x * – . 7 Solution Let the trip length be x kilometres Miro charges 6.5 + 3.2x Maddie charges 4 + 3.9x 6.5 + 3.2x > 4 + 3.9x [Miro costs more than Maddie] –0.7x + 6.5 > 4 –0.7x > –2.5 –2.5 x< –0.7 [subtracting 3.9x] [subtracting 6.5] [dividing by –0.7 (and swapping signs)] x < 3.5714... The trip was at most 3.5 km (1 d.p.) [rounding down to the nearest tenth] © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. AS 91261 Apply algebraic methods in solving problems 22 Achievement Standard 91261 (Mathematics and Statistics 2.6) AS 91261 4. Robert receives $x per week pocket money. One week he spends $10 then saves half of the remaining money. The next week he spends $6 then saves a third of the remaining money. If he saves more in the second week than the first, solve the inequation x – 10 < x – 6 to find the greatest 2 3 weekly pocket money Robert could receive. 7. At Occasions reception lounge the charge is $480 booking fee plus $85 per guest. Another reception lounge Oasis charges $565 booking fee plus $82 per guest. Set up and solve an inequation to work out the minimum number of guests for which Oasis would be the cheaper option. (Your answer should be a whole number.) 5. Marnie is 32 years younger than her aunt. Marnie worked out that if she added 13 to her age, then tripled it, she would be 1 less than double her aunt’s age. Solve the equation 3(x + 13) = 2(x + 32) – 1 to find her aunt’s age. 8. Hone sits three tests. The first result is 15% less than the second, and the third result is 2.5% above the average of the first two results. If Hone must average at least 65% over the three tests, solve the following inequation x + (x + 15) + 0.5(2x + 15) + 2.5 * 65 3 to work out the minimum mark Hone needs to score in the first test. 6. A journey is travelled at a steady speed of 80 km per hour. If the speed is increased by 10 km per hour, then the time for the journey is reduced by half an hour. Solve the following equation to find the length of the journey. d d 1 = + 80 90 2 © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. 9. Shapes A and B are made from rectangles, with side lengths given in terms of x. (x + 5.5) cm 23 The first activity is 1 km further than the second but only half as long as the third. If all three activities take a total of 1 hour and 45 minutes, how far did she travel altogether? A (x + 2.75) cm 2x cm B x cm 2x cm x cm For what values of x is the perimeter of A at least 5 cm greater than the perimeter of B? 12. The front of an A-frame building is in the shape of an isosceles triangle with base width 5 metres and height 10 metres. The entrance is rectangular and touches the sides of the building. Let the width of the entrance be x metres and the height of the entrance be 5.5 metres. Find the width of the entrance. Diagram is NOT to scale x 10 m 5.5 m 10. A rectangular enclosure is made up of a rectangular garden 2x metres wide and (3x + 4) metres long, surrounded by a path 1 metre wide. If the area of the path is no more than 200 m2, find the maximum area the garden can be. Entrance 5m 11. Sharon records her speed and time for three activities. Activity 1: A walk to the shops and back at 4 km h–1. Activity 2: A jog to and from the post-office at 6 km h–1. Activity 3: A run around the running track at 9 km h–1. © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. AS 91261 Apply algebraic methods in solving problems 32 Achievement Standard 91261 (Mathematics and Statistics 2.6) AS 91261 The quadratic formula A general quadratic formula can be used to solve quadratic equations. This equation is derived by completing the square. x= 12 ± 144 – 12 2 x= 12 + 132 12 – 132 or [± means + or –] 2 2 If ax2 + bx + c = 0 then x= –b ± b2 – 4ac 2a In some cases the quadratic formula can be used as an alternative to factorisation. Example Q. Solve the quadratic equation 2x2 + 7x + 6 = 0 by factorisation and by the quadratic formula. A. By factorisation: (2x + 3)(x + 2) = 0 [factorising] 2x + 3 = 0 or x + 2 = 0 [setting factors to zero] 3 [solving] x = – or x = –2 2 By formula: Comparing 2x2 + 7x + 6 = 0 with ax2 + bx + c = 0 gives a = 2, b = 7, c = 6 Substituting these values in the quadratic formula gives: –7 ± 72 – 4 × 2 × 6 2×2 –b ± b2 – 4ac [x = ] 2a –7 ± 1 [simplifying] x= 4 –7 – 1 –7 + 1 or x = x= 4 4 3 x = –2 or x = – 2 x = 0.255 or 11.745 (3 d.p.) Note: The negative b-value was bracketed to avoid substitution errors. In harder questions, a quadratic equation will be formed from given information. Solutions will need to be interpreted. Example A rectangular photograph is 5 cm longer than it is wide. If the area of the photo is 160 cm2, find the width of the photo. Solution Let width of photo be x, then length is x + 5 x+5 x x= x(x + 5) = 160 [area rectangle = width × length] x2 + 5x = 160 [expanding] x + 5x – 160 = 0 2 [rearranging] Comparing with ax2 + bx + c = 0 gives a = 1, b = 5, c = –160 In other cases, factorisation cannot be used. x= –5 ± 52 – 4 × 1 × –160 2×1 Solution Comparing x2 – 12x + 3 with ax2 + bx + c gives x= –5 ± 665 2 a = 1, b = –12, c = 3 The width of the photo is 10.4 cm (1 d.p.) Example Solve x2 – 12x + 3 = 0, using the quadratic formula. [equating coefficients of x2, x, and constants] Substituting these values in the formula gives x= –(–12) ± (–12)2 – 4 × 1 × 3 2×1 [quadratic formula] [simplifying] x = 10.3938 or –15.3938 (ignore since negative) Graphical calculators can be used to solve quadratic equations of the form ax2 + bx + c = 0. Check how to do it on your calculator. © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. 35 6. Beth enters the number 1 into her calculator, doubles it then subtracts 5. She then squares the result of this calculation and divides by 9. She is surprised to find that the result of these calculations is equal to the number she first entered. She wonders if this is true for any other integer. (2x – 5)2 Use the quadratic formula to solve =x 9 to find out. b. Solve this equation to find the length of the hypotenuse to 1 d.p. 7. A square room is made 3 m longer and 2 m wider. Its new area is 48.75 m2. Solve (x + 2)(x + 3) = 48.75 to find its original area. x x 5. Peter is solving the quadratic equation 5x2 – 14x – 3 = 0 He has made at least one error. Correct his working and find the roots of the equation. x= –14 ± 196 – 4 × 5 × 3 2×5 x= –4 ± 136 10 3 x = 0.766 or –1.566 © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. 2 AS 91261 4. A right-angled triangle has hypotenuse of length (3x + 4) cm. +4 x 3 Its other two sides are of lengths (x + 4) cm and (4x – 1) cm. 4x – 1 a. Show that the relationship between side lengths can be simplified to 8x2 – 24x + 1 = 0. (Hint: Use Pythagoras’ rule – expand and simplify) x+4 Apply algebraic methods in solving problems 38 Achievement Standard 91261 (Mathematics and Statistics 2.6) AS 91261 The nature of the roots of a quadratic equation The graph of the quadratic function y = ax + bx + c is a parabola. 2 Setting y = 0 gives the quadratic equation ax2 + bx + c = 0 to solve. A quadratic equation can have two, one or no solutions, as shown in the graphs below. y Some examples are shown below. Quadratic equation Value of b2 – 4ac Nature of roots 4x2 + 12x + 9 = 0 144 – 4 × 4 × 9 = 0 1 repeated (rational) root x2 + 9x – 2 = 0 81 – 4 × 1 × –2 = 89 2 irrational roots 3x2 + 2x + 6 = 0 4 – 4 × 3 × 6 = –68 No real roots x2 – 7x – 8 = 0 49 – 4 × 1 × –8 = 81 2 rational roots a>0 More complex problems explore the nature of the roots for quadratics with unknown coefficients. x Two solutions (graph cuts x-axis twice) One solution (graph touches x-axis once) For what value(s) of k does the equation 2x2 – kx + k + 2 = 0 have no real roots? No solutions (graph doesn’t cut or touch the x-axis) x y Example Solution For the quadratic 2x2 – kx + k + 2 = 0, a = 2, b = –k and c = k + 2 [comparing with ax2 + bx + c] Substituting in b2 – 4ac, the discriminant is (–k)2 – 4 × 2 × (k + 2) = k2 – 8k – 16 [simplifying] For no real roots, k2 – 8k – 16 < 0. a<0 The discriminant Solving k2 – 8k – 16 = 0 (by calculator or formula) gives k = –1.657 or 9.657. The formula for the solution to ax2 + bx + c = 0 is Since y = k2 – 8k – 16 is an upward-pointing parabola, y –b ± b2 – 4ac x= 2a The expression under the square root sign, b2 – 4ac, is called the discriminant. The value of the discriminant determines the number and nature of the roots (solutions) of a quadratic equation. Discriminant b2 – 4ac Nature of roots Positive • perfect square • not a perfect square Two real roots • 2 rational roots • 2 irrational roots Zero One (repeated) real root Negative No real roots (imaginary roots) –1.657 9.657 x k2 – 8k – 16 is negative between the roots. Thus the solution is –1.657 < k < 9.657. Note: This solution is correct to 3 d.p. The exact 8 ± 128 solution to k2 – 8k – 16 = 0 is 2 so the exact solution to k2 – 8k – 16 < 0 is 8 – 128 8 + 128 <k< which 2 2 simplifies to 4 – 4 2 < k < 4 + 4 2 [since 128 = 64 × 2 = 8 2 ] © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. 8. The value, V dollars, of a motorbike t years after it was purchased is given by the equation: V = 12 500(0.5)0.2t 47 11. 12 600 people visited a tourist attraction in the first month that it was open, but the number of visitors has been falling by 5% per month. When will the value of the motorbike be $8 000? A model for the monthly number of visitors (in thousands) is given by V = A(1 – 0.05) n – 1 where n is the number of months since opening. a. What is the value of A? 9. A population of 12 000 000 is growing by 3% per year. How long does it take for the population to double? 10. The half-life of uranium-235 is 704 million years. This means that 50% of the original radioactivity is present after 704 million years. How long does it take for the radioactivity of uranium to reduce by 10%? b. How many people visited in the 3rd month? Round your answer to the nearest ten. c. If the number of visitors falls below 10 000 in a month, then ticket pricing will be reviewed in the following month’s financial meeting. If the pattern of monthly visitor numbers continues as described above, in which month will the financial meeting to review ticket pricing take place? © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. AS 91261 Apply algebraic methods in solving problems 48 Achievement Standard 91261 (Mathematics and Statistics 2.6) AS 91261 Logarithmic equations Logarithmic equations are converted to exponential equations using the relationship: If logb y = x then y = b x log 52 [dividing by log 3] log 3 y = 3.597 (3 d.p.) [calculator] y= Alternatively, the base-changing formula can be used to evaluate logarithms to any base: Example The logarithmic equation log2 x = 3.5 is equivalent to the exponential equation x = 23.5 So x = 11.314 (3 d.p.) [evaluating by calculator] If the base of the logarithm is unknown, then a power equation must be solved. When dealing with terms of the form logb x , remember that • b > 0 (the base of a logarithm is a positive real number) • x > 0 (logarithms are defined for positive numbers only) Example 1. If logx 3 = 4, then [rearranging using the rule] x4 = 3 4 x = 3 [ignore the negative root (since x > 0)] x = 1.316 (3 d.p.) 2. logx 16 = 9.7 is equivalent to the equation x9.7 = 16 To solve this type of equation, raise both sides to the reciprocal of the power of x. 1 loga x = log10 52 = 3.597 log10 3 Note: log10 is chosen because it can be evaluated by calculator using the log key. It is also possible to use the natural logarithm (loge) using the In key. For example, log3 52 = Other logarithmic equations If two numbers have the same logarithm (to the same base), then the numbers are equal. if logb y = logb x then y = x Example Q. Solve log (5x – 8) = log (2x + 1) A. 5x – 8 = 2x + 1 [if logb y = logb x then y = x] 3x = 9 [rearranging] x=3 The following logarithm laws can be used to rearrange a logarithmic equation, so that it can be solved more easily. log x + log y = log xy x log x – log y = log y n log x = log x n 1 1 ] 9.7 [the product of a number x1 = 16 and its reciprocal is 1] x = 1.331 (3 d.p.) [by calculator] (x 9.7)9.7 = 169.7 [the reciprocal of 9.7 is 1 9.7 To find the logarithm of a number to a given base, form a logarithmic equation and solve it. Example Find log3 52 Solution Let y = log3 52 [equivalent exponential form] 3y = 52 y log(3 ) = log 52 [taking log of both sides] y log 3 = log 52 [logarithm rules] logb x logb a Example Q. Solve 2 log x – log 4x = log (x – 1), for x > 1 A. log x2 – log 4x = log (x – 1) [n log x = log x n] x2 = log (x – 1) [log x – log y log 4x x = log ] y x log = log (x – 1) 4 x =x–1 [if logb y = logb x 4 then y = x] x = 4x – 4 [multiplying by 4] 3x = 4 4 x= 3 [rearranging] © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. Exercise A: Expanding brackets (page 2) 1. 2. 3. 4. 5. = a3 + b3 24x – 32x b. –6y + 6xy 7x7 – 56x4 d. –8p3 + 48p2 + 32p b. a3 – b3 = (a – b)(a2 + ab + b2) e. 14a2 + a – 4 f. 5x2y – 35x + 3xy2 – 21y c. i. (p + 2q)(p2 – 2pq + 4q2) g. 25x +20xy + 4y h. 4x – 28x + 49 ii. (10m – n)(100m2 + 10mn + n2) i. 5x + 10x + 5 j. 8x – 44x + 48 2 2 2 2 2 a. 5x –2x b. a b + 2ab + a c. 9 – 9x2 + x3 d. 7w2 – w – 22 e. –x + 4x – 16 f. 42t – 98 g. 12x – 6x2 h. 20x i. 5y2 – 24y – 5 j. 16x2 + 4x – 31 a. x3 + 12x2 + 41x + 42 b. x3 – 4x2 – 20x + 48 2 2 2 2 3 1. c. 6x – 55x – 6x + 135 d. –18x + 138x – 84x e. x3 – 2x2 – 55x + 200 f. –18x3 – 24x2 – 8x 3 2 3 g. 8x + 12x + 6x + 1 h. 125x – 225x + 135x – 27 –14x2 – 35x + 21 j. –26x3 – 54x2 – 36x – 8 a. i. 2 3 (a + b)(a – b) = a2 – ab + ab – b2 iii. (a + b) = (a + b) (a + b) (a + b) = (a + b)(a2 + 2ab + b2) 2. (a – b)3 = a3 – 3a2b + 3ab2 – b3 d. i. 4x2 + 12xy + 9y2 ii. x3 + 9x2 + 27x + 27 iii. 8x – 12x + 6x – 1 3 4. 3y d. ab–4 e. 7c –3 8 –2b–6 5 1 x4 p3 f. 3. –2 d. p –1 3 1 2c –5 5 y2 2a5 3 4y 3 2c2 f. –w1 or –w h. 3p4 5 a. x5 b. p2 c. w2 d. 4s 2 e. 3w 2 f. x5 g. 8y 3 h. x5 a. 2 b. 3 c. 16 d. 81 e. 8 27 9 or 2 1 4 4 f. 1 000 h. 27 64 a. g. 4. a. 7xy(y2 + 3x) b. 2m2(4m2 – m – 2n) c. 9y(xy – 9z + 4y) d. p6(p – 1) e. a4(1 + a4) f. 2x3(6 + x2) g. 3ab3(3a3 – 2ab3 + 9b) h. x5y8(6x2y2 + 2 – 3y) a. (y + 8)(y + 9) b. (w – 25)(w – 8) c. (x + 6)(x + 2) d. (x – 11)(x + 2) e. 10 e. (x – 3)(x + 10) f. (x – 4)(x – 8) g. a. g. (x – 8)(x + 3) h. (x + 11)(x – 11) 3(x + 7)(x – 2) b. 2(y – 5)(y – 3) c. 8(x + 2)(x – 2) d. –(x – 9)(x + 3) or (9 – x)(x + 3) e. 2(7 + 5m)(7 – 5m) f. x(x + 2)2 g. (ab + 2)(b – 3) h. (3 – a)(a + 13) or –(a – 3)(a + 13) a. (3x + 1)(x + 2) b. (2x – 1)(x + 2) c. (2y – 3)(2y – 1) d. (3p – 4)(p + 1) e. (3m – 1)(2m + 3) f. (h + 2)(4h + 3) g. –(x – 2)(2x + 3) or (2x + 3)(2 – x) h. –(2x – 3)2 or (2x – 3)(3 – 2x) b. g. 9 5 –3 7 2 a. h. 3 Exercise B: Factorising (page 6) 3. c. e. = a3 + 3a2b + 3ab2 + b3 c. 4x–3 c. = a3 + 2a2b + ab2 + a2b + 2ab2 + b3 (a – b)2 = a2 – 2ab + b2 b. g. 3 b. (5 – 4p)(25 + 20p + 16p2) x–5 2 (a + b)(a + b) = a2 + ab + ab + b2 ii. (2x + 3y)(4x2 – 6xy + 9y2) iv. a. 2 i. 3 iii. Exercise C: Indices (page 9) = a2 – b2 2. (a + b)(a2 – ab + b2) = a3 – a2b + ab2 + a2b – ab2 + b3 a. = a2 + 2ab + b2 1. a. c. 2 –2 1 5. a. 10 2 b. 10–1 c. 10 d. 100 f. 10 4 10 2 h. 27x – 17 b. 10 3 –1 1 or x 4 –5 2 3 3 –3 6. c. 5 a4 d. e. b2 216a6 324b a f. g. © ESA Publications (NZ) Ltd, ISBN 978-0-908315-41-3 – Copying or scanning from ESA workbooks is limited to 3% under the NZ Copyright Act. h. – 1 1 x4 1 8x6 9x4y2 16 p 32q5 ANSWERS ANSWERS