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 edmunds.eu Get StartedI’ve done enough

Algebra Number Shape Back to Start

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 Solve non-linear simultaneous equations by Substitution Changing the subject of a formula Straight line graphs Proportionality Back to main menu

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

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 1.Factorise a.(x - 5 )(x + 3) b.(x – 2)(x – 7) c.(x + 5)(x – 5) 2. Factorise (3x – 2)(x – 1) To Question Back to menu

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

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 a.x 2 – 2x – 24 = 0 ( x – 6 )( x + 4 ) = 0 gives x = 6 and x = -4 b.x 2 – 9x + 18 = 0 (x – 3)(x – 6) = 0 gives x = 3 or 6 c.x 2 – 2x + 20 = 5 x 2 – 2x + 15 = 0 ( x – 5)(x + 3) = 0 gives x = 5 or -3 d.3x 2 – 7x + 2 = 0 (3x -2 )( x – 1) = 0 gives x = 2/3 or 1 To Question Back to menu

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

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

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

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 1 st by 2 and 2 nd 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 To Question Back to menu

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

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 To Question Back to menu

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

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 x 2 + y 2 = 25 and y = x + 1 Substituting the y gives x 2 + (x + 1) 2 = 25 x 2 + x 2 + 2x + 1 = 25 Simplify and bring across 25 gives 2x 2 + 2x - 24 = 0 Check if can divide to give x 2 + 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) To Question Back to menu

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 v 2 =u 2 + 2as to make u the subject To Answer Back to menu

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:v 2 = u 2 + 2as v 2 – 2as = u 2  (v 2 – 2as) = u To Question Back to menu

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) To Answer Back to menu

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 To Question Back to menu

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. a.Find the equation connecting t and a b.Find t when a = 16 ms –2 c.Find a when t = 64 seconds To Answer Back to menu

To Question Back to menu

Pythagoras’ Theorem Angles in Right-Angled TrianglesCircle Theorems Constructions Sides and Angles in Non-Right-Angled Triangles Transformations Vectors Volumes and Surface Areas 3D TrigonometryTransformation of Graphs Loci Back to main menu

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

Key points Only right-angled triangles. Square – Square – add/subtract – Square root Add if finding longest side/subtract otherwise Q = ;  144 = 12cm Q = = 218;  218= 14.8 cm Q3. draw diagram and then dist from = 1300 Dist =  1300 = To Question Back to menu

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

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 ° = To Question Back to menu

Objective Can I find sides and angles in non right-angled triangles To Answer Back to menu 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 Q1 – find the largest angle in a triangle of sides 7cm,8cm and 9cm Q2A C B Angle A = 50°, Angle B = 70° and AC = 12 cm. Find the length of AB

Key points Cosine Rule : a 2 = b 2 + c 2 – 2bcCosA (use for SSS and SAS) Sine Rule: a = b = calso area of Δ = ½ ab SinC (angle SinA SinB SinCbetween) Q1: cosine rule: largest angle opposite largest side 9 2 = 7² + 8² - 2x7x8xCosA gives 81= CosA (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 To Question Back to menu P R Q T 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° 70° 60° 70°

Objective Can I solve 3D Trigonometry Questions To Answer Back to menu ABCD is 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 a.Length EX b.Angle EMX c.Angle ECX

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) = From this AX = ½ of AC = In Δ EAX we now know EA is 13 and AX = 4.95 so we can find EX using Pythagoras again, EX =  (13² ²) = 12.0 cm b.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° c.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° To Question Back to menu

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

Key points 1.tangent/radius meet at 90° 2. Angle at centre = 2x angle at circumference 3.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° To Question Back to menu

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 To Answer Back to menu

Key points 1.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. 2.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 To Question Back to menu

Objective Can I construct Loci To Answer Back to menu 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

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

Transformations Enlargement Back to menu ReflectionRotation Translation

Objective Can I enlarge shapes To Answer Back to menu 1.What is the transformation that takes the shaded shape to shape A 2.What takes B to the shaded shape 3.Enlarge the shaded shape with a scale factor of -2 and a centre of (-1,-1)

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 To Question Back to menu 1. scale factor of 3, centre (-1,-1) 2.Scale factor of ½, centre (-2,5) 3.See diagram with blue triangle

Objective Can I translate shapes To Answer Back to menu 1.What is the translation that moves A to B 2.Move shape C by 4 -3

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

Objective Can I reflect shapes To Answer Back to menu 1.What are the mirror lines that reflect the shaded shape to A,B and C 2.Reflect the shaded shape in the lines x=3 and y=-1

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 To Question Back to menu 1.A : y=3, B : x= -1, C : y = -x (dotted line) 2.Red triangle, green triangle

Objective Can I rotate shapes To Answer Back to menu 1.Rotate the shaded shape 90° clockwise using a centre of (-1,-1) 2.What rotations of the shaded shape produced shape B

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 To Question Back to menu 1.Red triangle 2.90° clockwise centre(0,-2)

Objective Can I transform graphs To Answer Back to menu 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 a.y = f(x) + 6 b.y = f(x + 3) c.y = f(-x) d.y = f(4x)

To transform a graph of a function such as f(x) you use f( x + 1) moves graph 1 leftf(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 values2f(x) doubles all the y values Inside the bracket changes the x values and in the opposite of what you would expect Q1 To Question Back to menu Q2 (a)(2, 18) (b)(-1, 12) (c)(-2, 12) (d)(1/2, 12)

Objective Can I solve vector algebra problems To Answer Back to menu 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

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 Back to menu To Question

Objective Can I solve volume and surface area questions To Answer Back to menu Q1. A cone has a base radius of 5 cm, a height of 12 cm and a slant height of 13cm. a.Find the total surface area of the cone b.Find the volume of the cone c.If the density of the cone is 6 gm/cm 3 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: πr 3 B: a + bC: lwh D: 2  rE: πabF: x 2

You will be given the formulae for cones and spheres but will need to know areas of circles, triangles, parallelograms, trapezia and volumes of cubes, cuboids and prisms. When checking whether a formula is length, area or volume, use the lengths multiplied – one for length, two is area and three is volume – ignore numbers which will include  Q1. a.surface area is the slanted area (  rl) plus the base (  r²) which in this case gives (  x 5 x13) plus (  x 25) for a total of 90  or 283cm² to 3sf b.Vol = 1/3 base area x height = 1/3 x  x 25 x 12 = 100  = 314cm² to 3sf c.Density = mass ÷ volume (see the units) so mass = density x volume which gives 314 x 6 = 1884 gm or Kg Q2. lengths : B, DArea : E, FVolume: A,C Back to menu To Question

Reverse Percentages Compound Interest Standard Form Using Standard Form Bounds Back to main menu

Objective Can I use Reverse Percentages Q1. I bought a shirt in the 15% off sale for £34. What was the original cost Q2. I bought a car two years ago and each year it has lost 20% of its value at the start of the year. If it is now worth £1280 what was it worth when I bought it To Answer Back to menu

Key points 1.Find what percentage it is now 2.Find 1% and then 100% Key is that you are finding the value before Q1: I paid 85%. 1% = 34 ÷ 85 = % = 0.4 x 100 = £40 Q2: lost 20% 1 st year to give 80% Second year lose another 20% of 80% = 16% Now = 64% ; 1% = 1280 ÷ 64 = £20 Original price = 100 x 20 = £2000 To Question Back to menu

Objective Can I use percentages to find Compound Interest Q1. I invest £5000 at 4% per year compound interest. What will I have after 3 years? Q2. A flock of 3000 seagulls loses 4% of the flock every year. How many will there be after 3 years? To Answer Back to menu

Key points 1.Find what percentage it is now 2.Find 1% and then 100% Key is that you are finding the value before Q1: year start interest end of year Or total = start x multiplier period = 5000 x = Q2either way but second gives 3000 x = 2548 seagulls To Question Back to menu

Objective Can I convert between standard form and ordinary numbers Q1. Convert these numbers to standard form and Q2. Convert these standard form numbers into ordinary numbers 3.27 x 10 7 and 4.23 x To Answer Back to menu

Key points Standard form is number between 1 and 10 and a power of ten Power is negative for small number, positive for big Q1: 4.76 x and x 10 5 Q2: and To Question Back to menu

Objective Can I use standard form (with a calculator) Find the value of the following Q x 10 4 x 2.57 x 10 6 Q x 10 4 ÷ 3.57 x 10 6 Q x x 10 6 Q x x 10 6 To Answer Back to menu

Key points As this is a calculator exam you just need to be able to use the standard form buttons. On the Sharp this is the EXP button above the 7. To put in 3.27 x 10 -4, press 3.27 then EXP then -4 Q1: x Q2:4.99 x ( 3 sig figs) Q3: x 10 6 Q4:2.983 x 10 7 To Question Back to menu

Objective Can I find the bounds of numbers and use them Q1. State the upper and lower bounds of 56m when measured to the nearest 2m and 1m Q2. a car takes 57 secs to do a lap to the nearest second. The lap has been measured to 890m to the nearest 10m. What are the bounds for its speed ? What is this in km/h To Answer Back to menu

Key points Lower bound can be equal whereas the upper cannot be reached. The nearest “unit” shows the range either side of the measure. To get the highest when dividing needs highest divided by lowest. Q1: nearest 2m will be 55  length < 57 nearest 1m will be 55.5  length < 56.5 Q2: speed is 56.5  speed < 57.5 distance is 885  distance < 895 fastest speed = 895 ÷ 56.5 = m/s to turn into km/h : x 60 x 60 (to hrs) ÷ 1000(km) = 57 km/h To Question Back to menu

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