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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Algebra

Number

Shape

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Factorise Quadratic Expressions

ALGEBRASolve 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

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Objective

Can I factorise a quadratic expressionGrade : B / A (if x2 is more than 1)

1. Factorise

a. x2 – 2x – 15b. x2 – 9x + 14

c. x2 – 25

2. Factorise 3x2 – 5x + 2

To AnswerBack to menu

Key points

Use signs at the end and in the middle to put signs in bracketsFactors 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 QuestionBack to menu

Objective

Can I solve a quadratic equation by factorisingGrade : B / A (if x2 is more than 1)

a. x2 – 2x – 24 = 0b. x2 – 9x + 18 = 0c. x2 – 2x + 20 = 5d. 3x2 – 7x + 2 = 0


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. x2 – 2x – 24 = 0( x – 6 )( x + 4 ) = 0 gives x = 6 and x = -4

b. x2 – 9x + 18 = 0(x – 3)(x – 6) = 0 gives x = 3 or 6

c. x2 – 2x + 20 = 5x2 – 2x + 15 = 0

( x – 5)(x + 3) = 0 gives x = 5 or -3d. 3x2 – 7x + 2 = 0

(3x -2 )( x – 1) = 0 gives x = 2/3 or 1


Objective

Can I solve a quadratic equation by using the formulaGrade : B

Solve giving your answer to 2 decimal places

3x2 – 2x – 24 = 0

6x +15x + 2 = 0


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 -2.52Second example

answers = - 15 ± 129/12 = -2.20 or -0.30


Objective

Can I solve simultaneous equations by elimination

Solve

3x + 2y = 132x + 3y = 12

2x – 3y = 76x + 2y = -1


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 = 132x + 3y = 12

Multiply 1st by 2 and 2nd by 36x + 4y = 266x + 9y = 36

Subtract gives 5y = 10 and so y =2Into first equation gives 3x + 4 = 13 so x = 3

Check in other 2x3 + 3x2 = 12Second example x = ½ and y = -2


Objective

Can I solve simultaneous equations by substitution

Solve

3x + 2y = 12y = x + 1

5x + 4y = 35Y = 2x - 1


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 = 12y = x + 1

Substituting the y gives 3x +2(x + 1) = 125x + 2 = 12

So x = 2 From the y= equation y = 3

Second example x= 3, y = 5


Objective

Can I solve non-linear(x2) simultaneous equations by substitution

Solve

x2 + y 2 = 25y = x + 1

x2 + y 2 = 29Y = 2x + 1


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 + 1Substituting the y gives x2 + (x + 1)2 = 25

x2 + x2 + 2x + 1 = 25Simplify and bring across 25 gives 2x2 + 2x - 24 = 0

Check if can divide to give x2 + x -12 = 0Factorising 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)


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


Key points

Move every unwanted letter or number to the other side changing sign as it moves

Q1: v = u + atv – u = atv – u = t a

Q2: v2 = u2 + 2asv2 – 2as = u2

(v2 – 2as) = u


Objective

Can I solve problems involving straight line equations

Q1: In the equations y = 3x + 4 , what do the 3 and 4 representQ2: 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)


Key pointsAny straight line is y = mx + c where m is gradient and c is y intercept

Lines are parallel if gradient the sameLines 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=6So y = - x + 6


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 ab. Find t when a = 16 ms–2

c. Find a when t = 64 seconds


Key pointsWrite 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 b. When a = 16 , t = 16 x 1/16 = 16 x ¼ = 4c. When t = 64 , 64 = 16 x which rearranges to a = 16/64 which

is ¼ so a = 1/16


Pythagoras’ Theorem

GEOMETRY

Angles in Right-Angled Triangles Circle Theorems

Constructions

Sides and Angles in Non-Right-Angled Triangles

TransformationsVectors

Volumes and Surface Areas

3D Trigonometry Transformation of Graphs

Loci

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Objective

Can I use Pythagoras Theorem to find lengths in right-angled triangles


13cm5cm

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 pointsOnly right-angled triangles.

Square – Square – add/subtract – Square rootAdd 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 = 1300Dist = 1300 =


Objective

Can I use SOHCAHTOA with angles in right-angled triangles


13cm5cmx

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 pointsSOCAHTOA : 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


Objective

Can I find sides and angles in non right-angled triangles


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

Q2 A

C B

Angle A = 50°, Angle B = 70° and AC = 12 cm. Find the length of AB

Key pointsCosine Rule : a2 = b2 + c2 – 2bcCosA (use for SSS and SAS)Sine Rule: a = b = c also area of Δ = ½ ab SinC (angle

SinA SinB SinC between)

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


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=41.1kmUsing 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


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. Finda. Length EXb. Angle EMXc. 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) = 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

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°


Objective

Can I solve angles in circles


Key points1. tangent/radius meet at 90° 2. Angle at centre = 2x angle at circumference3. Angle in semi-circle is 90° 4. Angles from same chord are equal5. 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°


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


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


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 Lb. 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


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 pointsLoci are points or areas set by a rule

a. Same distance from a point = a circleb. Same distance from 2 points = perpendicular bisector

c. Same distance from a line = parallel linesd. Same distance from two lines that meet = line that bisects that the angle

A

B

Transformations

Enlargement

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ReflectionRotation

Translation

ObjectiveCan I enlarge shapes


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


1. scale factor of 3, centre (-1,-1)

2. Scale factor of ½, centre (-2,5)

3. See diagram with blue triangle

ObjectiveCan I translate shapes


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


1. A to B is -1 -7

2. Blue triangle

ObjectiveCan I reflect shapes


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


1. A : y=3, B : x= -1 , C : y = -x (dotted line)

2. Red triangle, green triangle

ObjectiveCan I rotate shapes


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


1. Red triangle

2. 90° clockwise centre(0,-2)

ObjectiveCan 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 transformationsa. y = f(x) + 6b. 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 left f(x) + 1 moves up 1f( -x) reflects in the y axis -f(x) reflects in the x axisf(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)

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

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 – aL 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 getOY + 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

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. a. Find the total surface area of the coneb. Find the volume of the conec. 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: 2r

E: πab

F: x2

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 3sfb. Vol = 1/3 base area x height = 1/3 x x 25 x 12 = 100 = 314cm² to 3sfc. Density = mass ÷ volume (see the units) so mass = density x volume

which gives 314 x 6 = 1884 gm or 1.884 Kg

Q2. lengths : B, D Area : E, F Volume: A,C

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Reverse Percentages

NUMBER

Compound Interest

Standard Form

Using Standard Form

Bounds

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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


Key points1. 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 = 0.4100% = 0.4 x 100 = £40

Q2: lost 20% 1st year to give 80%Second year lose another 20% of 80% = 16%

Now = 64% ; 1% = 1280 ÷ 64 = £20Original price = 100 x 20 = £2000


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?


Key points1. 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 1 5000 200 5200 2 5200 208 5408 3 5408 216.32 5624.32

Or total = start x multiplierperiod = 5000 x 1.043= 5624.32Q2 either way but second gives

3000 x 0.964 = 2548 seagulls


Objective

Can I convert between standard form and ordinary numbers

Q1. Convert these numbers to standard form0.0000476 and 347812

Q2. Convert these standard form numbers into ordinary numbers

3.27 x 107 and 4.23 x 10-5


Key pointsStandard 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 10-5 and 3.48712 x 105

Q2: 32 700 000 and 0.000 042 3


Objective

Can I use standard form (with a calculator)

Find the value of the following

Q1. 3.24 x 104 x 2.57 x 106

Q2. 1.78 x 104 ÷ 3.57 x 106

Q3. 3.24 x 104 + 2.57 x 106

Q4. 3.24 x 107 - 2.57 x 106


Key pointsAs 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: 8.3268 x 1010 Q2: 4.99 x 10-3 ( 3 sig figs)Q3: 2.6024 x 106

Q4: 2.983 x 107


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


Key pointsLower 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 < 57nearest 1m will be 55.5 length < 56.5

Q2: speed is 56.5 speed < 57.5 distance is 885 distance < 895fastest speed = 895 ÷ 56.5 = 15.84 m/sto turn into km/h : x 60 x 60 (to hrs) ÷ 1000(km)

= 57 km/h


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Welcome to the Higher Unit Three Summary

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