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Welcome to the Higher Unit Three Summary

Welcome to the Higher Unit Three Summary On the next pages you will find a list of the topics that are covered in unit 3. For each there is one or more sample questions and then a screen with answers and a brief description of what you need to know for this topic

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Welcome to the Higher Unit Three Summary

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  1. Welcome to the Higher Unit Three Summary On the next pages you will find a list of the topics that are covered in unit 3. For each there is one or more sample questions and then a screen with answers and a brief description of what you need to know for this topic Any problems please let me know on kgrimsley@st-edmunds.eu Get Started I’ve done enough

  2. Algebra Number Shape Back to Start

  3. ALGEBRA Factorise Quadratic Expressions Solve Quadratic equations by factorising Solve Quadratic equations by using the formula Solve Simultaneous equations by elimination Solve Simultaneous equations by Substitution Changing the subject of a formula Solve non-linear simultaneous equations by Substitution Straight line graphs Proportionality Back to main menu

  4. Objective Can I factorise a quadratic expression Grade : B / A (if x2 is more than 1) Factorise x2 – 2x – 15 x2 – 9x + 14 x2 – 25 2. Factorise 3x2 – 5x + 2 Back to menu To Answer

  5. Key points Use signs at the end and in the middle to put signs in brackets Factors of end have to sum to the numbers in the middle Factorise (x - 5 )(x + 3) (x – 2)(x – 7) (x + 5)(x – 5) 2. Factorise (3x – 2)(x – 1) Back to menu To Question

  6. Objective Can I solve a quadratic equation by factorising Grade : B / A (if x2 is more than 1) • x2 – 2x – 24 = 0 • x2 – 9x + 18 = 0 • x2 – 2x + 20 = 5 • 3x2 – 7x + 2 = 0 Back to menu To Answer

  7. Key points Factorise into brackets and then find the value that makes each bracket zero. If one side is not zero move until it is • x2 – 2x – 24 = 0( x – 6 )( x + 4 ) = 0 gives x = 6 and x = -4 • x2 – 9x + 18 = 0(x – 3)(x – 6) = 0 gives x = 3 or 6 • x2 – 2x + 20 = 5x2 – 2x + 15 = 0( x – 5)(x + 3) = 0 gives x = 5 or -3 • 3x2 – 7x + 2 = 0 (3x -2 )( x – 1) = 0 gives x = 2/3 or 1 Back to menu To Question

  8. Objective Can I solve a quadratic equation by using the formula Grade : B Solve giving your answer to 2 decimal places 3x2 – 2x – 24 = 0 6x +15x + 2 = 0 Back to menu To Answer

  9. Key points Set values of a,b,c which are number of x2, x and numbers. Formula is on your sheet • X = - b ±  (b2 – 4ac) • 2a • Example 1 : a= 3, b= -2, c = -24 • X = +2 ±  (-2x-2 – 4 x 3 x -24)/6 • = 2 ± ( 302)/6 • = (2 + 17.09)/6 or (2 -17.09)/6 • = 3.18 or -2.52 • Second example • answers = - 15 ± 129/12 = -2.20 or -0.30 Back to menu To Question

  10. Objective Can I solve simultaneous equations by elimination Solve 3x + 2y = 13 2x + 3y = 12 2x – 3y = 7 6x + 2y = -1 Back to menu To Answer

  11. Key points Pick a letter to eliminate and multiply each equation by how many of that letter are in the other equation. Add or subtract to eliminate. Put value back into an equation to find other letter. Check with other equation 3x + 2y = 13 2x + 3y = 12 Multiply 1st by 2 and 2nd by 3 6x + 4y = 26 6x + 9y = 36 Subtract gives 5y = 10 and so y =2 Into first equation gives 3x + 4 = 13 so x = 3 Check in other 2x3 + 3x2 = 12 Second example x = ½ and y = -2 Back to menu To Question

  12. Objective Can I solve simultaneous equations by substitution Solve 3x + 2y = 12 y = x + 1 5x + 4y = 35 Y = 2x - 1 Back to menu To Answer

  13. Key points From one equation find a simple form for y= or x = . Replace this in the other equation and then find the letter left. Use other equation to find the other letter 3x + 2y = 12 y = x + 1 Substituting the y gives 3x +2(x + 1) = 12 5x + 2 = 12 So x = 2 From the y= equation y = 3 Second example x= 3, y = 5 Back to menu To Question

  14. Objective Can I solve non-linear(x2) simultaneous equations by substitution Solve x2 + y 2 = 25 y = x + 1 x2 + y 2 = 29 Y = 2x + 1 Back to menu To Answer

  15. Key points From one equation find a simple form for y= or x = . Replace this in the other equation and then find the letter left by solving the quaddratic. Use other equation to find the other letter x2 + y2 = 25 and y = x + 1 Substituting the y gives x2 + (x + 1)2 = 25 x2 + x2 + 2x + 1 = 25 Simplify and bring across 25 gives 2x2 + 2x - 24 = 0 Check if can divide to give x2 + x -12 = 0 Factorising gives ( x + 4)(x – 3) = 0 so x = -4 or 3 and y is -3 or 4 (-4,-3) or (3,4) Second example (2,5) and (-2.8,-4.6) Back to menu To Question

  16. Objective Can I change the subject of a formula Q1: change the formula v = u + at to make a the subject Q2: change the formula v2=u2 + 2as to make u the subject Back to menu To Answer

  17. Key points Move every unwanted letter or number to the other side changing sign as it moves Q1: v = u + at v – u = at v – u = t a Q2: v2 = u2 + 2as v2 – 2as = u2 (v2 – 2as) = u Back to menu To Question

  18. Objective Can I solve problems involving straight line equations Q1: In the equations y = 3x + 4 , what do the 3 and 4 represent Q2: Find the line parallel to that in Q1 that passes through (2, 11) Q3: Find the line perpendicular to that in Q1 that passes through (6, 4) Back to menu To Answer

  19. Key points Any straight line is y = mx + c where m is gradient and c is y intercept Lines are parallel if gradient the same Lines are perpendicular if gradients multiply to -1; if you know one then FLIP, FLIP to find the other Q1: gradient is 3, crosses y axis at 4 Q2: as parallel, line is y = 3x + c ; as it goes through (2,11) then 11= 3x2 + c which makes c=5 to give y = 3x + 5 Q3: gradient is - (flip, flip) so y = -x + c and as goes through (6,4) then 4 = -1/3 x 6 + c which gives c=6 So y = - x + 6 Back to menu To Question

  20. Objective Can I solve problems involving proportionality • The time, t, taken to travel a fixed distance from a standing start is inversely proportional to the square root of the acceleration, a. • When a = 4 ms–2, t = 8 seconds. • Find the equation connecting t and a • Find t when a = 16 ms–2 • Find a when t = 64seconds Back to menu To Answer

  21. Key points Write connection using proportion ( Replace with k x Using values find k to get equation Use equation to solve rest of question • a. t • t = k x • When a = 4, t = 8 so 8 = k x ½ so k = 16 which gives t = 16 x • When a = 16 , t = 16 x 1/16 = 16 x ¼ = 4 • When t = 64 , 64 = 16 x which rearranges to a = 16/64 which is ¼ so a = 1/16 Back to menu To Question

  22. GEOMETRY Pythagoras’ Theorem Sides and Angles in Non-Right-Angled Triangles Angles in Right-Angled Triangles Circle Theorems Constructions Loci Transformations Vectors 3D Trigonometry Transformation of Graphs Volumes and Surface Areas Back to main menu

  23. Objective Can I use Pythagoras Theorem to find lengths in right-angled triangles Q1 13cm Q3. A ship leaves a port and sails 20 km east and then 30 km south. How far is it from the port? 5cm x x Q2 7cm 13cm Back to menu To Answer

  24. Key points Only right-angled triangles. Square – Square – add/subtract – Square root Add if finding longest side/subtract otherwise Q1. 132 - 52 = ; 144 = 12cm Q2. 72 + 132 = 49 + 169 = 218; 218= 14.8 cm Q3. draw diagram and then dist from 202 + 302 = 1300 Dist= 1300 = Back to menu To Question

  25. Objective Can I use SOHCAHTOA with angles in right-angled triangles Q1 Q3. A ship leaves a port and sails 20 km east and then 30 km south. What bearing is the ship from the port 13cm 5cm x 20km Q2 x 30km 13° 7cm Back to menu To Answer

  26. Key points SOCAHTOA : Sin = opp/hyp; Cos= adj/hyp Tan = opp/adj Use the -1 to find and angle, button to find length Q1: Opp =5; hyp =13  Sin-1 (5/13) = 22.6° Q2: angle = 13° , adj = 7 ; hyp = 7 ÷ Cos 13° = 7.18cm Q3: opp = 30; adj = 20 : angle = Tan-1(3/2)= 56.3° Bearing is 90 + 56.3° = 146.3 Back to menu To Question

  27. Objective Can I find sides and angles in non right-angled triangles Q1 – find the largest angle in a triangle of sides 7cm,8cm and 9cm Q3. A ship leaves a port and sails 20 km on a bearing of 070 and then 30 km on a bearing 120. How far from the port is it and on what bearing is the ship from the port A Q2 B C Angle A = 50°, Angle B = 70° and AC = 12 cm. Find the length of AB Back to menu To Answer

  28. Key points Cosine Rule : a2 = b2 + c2 – 2bcCosA (use for SSS and SAS) Sine Rule: a = b = c also area of Δ = ½ abSinC (angle SinASinBSinCbetween) Q1: cosine rule: largest angle opposite largest side 92 = 7² + 8² - 2x7x8xCosA gives 81= 113-112CosA (not CosA) Move 113 to give -32=-112CosA so CosA = (-32÷-112) so A=73.4° Q2: Use Sine rule but note C is 60° so AB ÷ Sin60= 12÷ sin 70 so that AB = 12 x Sin 60 ÷ Sin 70 = 11.1 cm Q3. from diagram From question angle PQT=70 and RQT=60 making PQR=130. Using cosine rule PR² = 20² + 30² -20x30xcos130 = 1686 so PR=41.1km Using Sine Rule Sin P = 30x Sin 130 ÷ 41.1 so P= 34° so the bearing of the ship is 104° Q 70° 70° 60° P T R Back to menu To Question

  29. Objective Can I solve 3D Trigonometry Questions • ABCDis a square of side 7 cm and X is the midpoint of ABCD. M is the midpoint of AD and E is directly above X. Find • Length EX • Angle EMX • Angle ECX Back to menu To Answer

  30. Key points Solve all problems by finding 2D triangles and solving them usually using Pythagoras and SOHCAHTOA a. The first step in finding EX is to find AC using the right-angled triangle ADC which will give AC as (7² + 7²) = 9.90. From this AX = ½ of AC = 4.95. In Δ EAX we now know EA is 13 and AX = 4.95 so we can find EX using Pythagoras again , EX = (13² - 4.95²) = 12.0 cm In Δ EMX for angle EMX, we now know that EX(Opp) is 12.0 and MX(Adj) is 3.5 ( ½ of 7) so that angle is tan-1( 12 ÷ 3.5) = 73.7° In Δ ECX for angle ECX , EC(Hyp) = 13 cm and CX(Adj) = 4.95 from part a. This gives us that ECX = Cos-1 ( 4.95 ÷ 13) = 67.6° Back to menu To Question

  31. Objective Can I solve angles in circles Back to menu To Answer

  32. Key points tangent/radius meet at 90° 2. Angle at centre = 2x angle at circumference Angle in semi-circle is 90° 4. Angles from same chord are equal 5. Angle in alternate segment equal to angle between chord and tangent Angle a = 66° (angle at centre theorem) Angle x = 45° ( two unmarked angles are 90 because they are tangents meeting radii; shape is a quad and so they must add up to 360) Angle h : h and 32 make up to 90 as a semi-circle so 58° Angle i : 32° (angle from the same chord across base) Angle k : top left angle next to i is 22 as from same chord and top triangle has 22 and 58 so k must be 100° Back to menu To Question

  33. Objective Can I construct bisectors Q1: Draw an angle of 140° and bisect it Q2: draw two points 8 cm apart and construct their perpendicular bisector Back to menu To Answer

  34. Key points Bisect an angle : draw arc from vertex that cuts both lines. From where they meet the line draw two equal arcs. From where these meet draw back to vertex. Set compass to more than distance between the points. Draw two equal arcs from each end. They will cross twice – join these up to make perpendicular bisector Back to menu To Question

  35. Objective Can I construct Loci • Two mobile phone masts are at L and M. Another mast is to be erected. It must be • a. at least 4 km from L • b. within 6 km of M • c. the same distance from L and M • Show the possible positions of the new mast. • The scale of the diagram is: 1 cm represents 1 km Back to menu To Answer

  36. Key points Loci are points or areas set by a rule Same distance from a point = a circle Same distance from 2 points = perpendicular bisector Same distance from a line = parallel lines Same distance from two lines that meet = line that bisects that the angle • at least 4 km from L outside a circle radius 4 cm from L • within 6 km of M inside a circle radius of 6cm from M • the same distance from L • and M on the perpendicular bisector of L and M • This is on the red line at A and B A B Back to menu To Question

  37. Transformations Enlargement Translation Rotation Reflection Back to menu

  38. Objective Can I enlarge shapes What is the transformation that takes the shaded shape to shape A What takes B to the shaded shape Enlarge the shaded shape with a scale factor of -2 and a centre of (-1,-1) Back to menu To Answer

  39. To Enlarge, count move to each corner from centre and then multiply this by the Scale Factor. Do each corner and connect. To go back, find scale factor by comparing matching sides; find the centre by drawing rays from matching corners-where they meet is the centre scale factor of 3, centre (-1,-1) Scale factor of ½, centre (-2,5) See diagram with blue triangle Back to menu To Question

  40. Objective Can I translate shapes What is the translation that moves A to B Move shape C by 4 -3 Back to menu To Answer

  41. Translation moves left/right and/or up/down with negative numbers meaning left and down A to B is -1 -7 Blue triangle Back to menu To Question

  42. Objective Can I reflect shapes What are the mirror lines that reflect the shaded shape to A,B and C Reflect the shaded shape in the lines x=3 and y=-1 Back to menu To Answer

  43. To reflect just count from each corner to the mirror and put the new corner the same distance the other side. If the mirror line is at angle count up or across until you hit the mirror and then change through 90° and count the same distance A : y=3, B : x= -1 , C : y = -x (dotted line) Red triangle, green triangle Back to menu To Question

  44. Objective Can I rotate shapes Rotate the shaded shape 90° clockwise using a centre of (-1,-1) What rotations of the shaded shape produced shape B Back to menu To Answer

  45. To rotate a shape. Put tracing paper over shape, mark centre of rotation and draw shape. Turn shape using specified angle and direction. To find a rotation , the angle and direction can be found by looking at it or drawing using tracing paper; centre is best found using trial and error but try (0,0) first Red triangle 90° clockwise centre(0,-2) Back to menu To Question

  46. Objective Can I transform graphs Q1. The sketch shows a graph y=f(x). What will f(-x) and f(x+2) look like Q2. the second sketch has a maximum at the point (2,12) . Where does this move to when we do the transformations y= f(x) + 6 y = f(x + 3) y= f(-x) y= f(4x) Back to menu To Answer

  47. To transform a graph of a function such as f(x) you use f( x + 1) moves graph 1 left f(x) + 1 moves up 1 f( -x) reflects in the y axis -f(x) reflects in the x axis f(2x) halves all the x values 2f(x) doubles all the y values Inside the bracket changes the x values and in the opposite of what you would expect Q1 Q2 (a) (2, 18) (b) (-1, 12) (c) (-2, 12) (d) (1/2 , 12) Back to menu To Question

  48. Objective Can I solve vector algebra problems • OYX is a triangle. M and L are the mid-points of OY and MX respectively. • N is the point of YX such that YN:NX = 2:1. • OX= a. OM = b • Find • a. OY b. MX c. OL Show that O, L and N are on a straight line Back to menu To Answer

  49. The key to vector problems is to use the triangle of vectors e.g AC= AB + BC Vectors are parallel if they are multiples of each other OY = 2a OM + MX = OX  a + MX = b so MX = b – a L is half way along MX and so ML = ½ (b – a) OL = OM + ML = a + ½ (b – a) = ½ ( a + b) To find ON we need to know YN which uses a similar method to get OY + YX = OX so that YX = OX – OY = b – 2a From the ration of YN:NX as 2:1 then YN must of YX = (b – 2a) ON is then found as ON=OY+YN = 2a +(b – 2a) = (a + b) As ON and OL are both multiples of (a + b) they must be parallel and as they both go through the same point O, they are on the same line Back to menu To Question

  50. Objective Can I solve volume and surface area questions • Q1. A cone has a base radius of 5 cm, a height of 12 cm and a slant height of 13cm. • Find the total surface area of the cone • Find the volume of the cone • If the density of the cone is 6 gm/cm3 • Q2. Below are six formulae. Two of the formulae are lengths, two of them are areas and two of them are volumes. Write down which are which. • A: πr3 B: a + b C: lwh • D: 2r E: πab F: x2 Back to menu To Answer

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