module 3 super learning day revision notes november 2012

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Module 3 Super Learning Day Revision Notes November 2012

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Page 1: Module 3 Super Learning Day Revision Notes November 2012

Module 3Super Learning Day Revision

Notes November 2012

Page 2: Module 3 Super Learning Day Revision Notes November 2012

In any exam

Always• Read the question at least twice

• Show ALL your working out

• Check your units eg. cm, cm² etc.

• Read the question again to make sure you have actually answered the question asked

• Check that your answer is sensible

Page 3: Module 3 Super Learning Day Revision Notes November 2012

Revision Notes

Shape and Measurement

Page 4: Module 3 Super Learning Day Revision Notes November 2012

Interior and exterior angles of polygons

• In a REGULAR polygon

• Exterior angles add up to 360°

• 360 ÷ number of sides = exterior angle

• 180 – exterior angle = interior angle

Page 5: Module 3 Super Learning Day Revision Notes November 2012

Regular polygons have

n lines of symmetryRotational symmetry of order n- where n is the number of sides

Page 6: Module 3 Super Learning Day Revision Notes November 2012

Tessellations

A tessellation is a tiling pattern with no gaps

Page 7: Module 3 Super Learning Day Revision Notes November 2012

Learn names of shapes

• Triangles:-Equilateral, isosceles, right angled,and

scalene

• Quadrilaterals:-Square, rectangle, parallelogram, rhombus,

trapezium, kite and arrowhead.

Page 8: Module 3 Super Learning Day Revision Notes November 2012

Perimeter

The perimeter is the distance round the edge of the shape

Page 9: Module 3 Super Learning Day Revision Notes November 2012

Area formulas• Area of a Triangle = base x height ÷ 2

• Area of a Rectangle = length x width

• Area of a parallelogram = length x height

• Area of a Trapezium = (a + b) ÷ 2 x height,

(where a and b are the lengths of the parallel sides)

• Area of a circle = πr²

Page 10: Module 3 Super Learning Day Revision Notes November 2012

Volume Formulas• Volume of cuboid = length x width x height

• Volume of a prism = Area of cross-section

x length

• Volume of cylinder = area of circle x height

Page 11: Module 3 Super Learning Day Revision Notes November 2012

Surface AreaWork out the area of every face separately

then add them togetherNOTE

To find the surface area of a cylinder you need to add together the area of the 2 circles AND the rectangle that

wraps round the cylinder.

The length of the rectangle is equal to the circumference of the circle and the width is the height of the cylinder

Page 12: Module 3 Super Learning Day Revision Notes November 2012

ViewsPlan

Side elevation

Front

Elevation

Plan view

Side view

Front view

Page 13: Module 3 Super Learning Day Revision Notes November 2012

Conversions 1cm² = 100mm²

1cm = 10mm

1cm = 10mm

1m³ = 1 000 000 cm³

1m =100cm

100cm

100cm

Page 14: Module 3 Super Learning Day Revision Notes November 2012

Congruent

Means

Alike in every respect

Page 15: Module 3 Super Learning Day Revision Notes November 2012

Similar

Means

Same shape, Different size( one is an enlargement of the other)

Page 16: Module 3 Super Learning Day Revision Notes November 2012

Metric /Imperial conversions

• 1Kg = 2¼ lbs• 1m = 1 yard (+10%)• 1 litre = 1¾ pints• 1 inch = 2.5 cm

• 1gallon = 4.5 litres• 1 foot = 30 cm• 1 metric tonne = 1

imperial ton• 1 mile = 1.6 Km

or 5miles = 8 Km

Page 17: Module 3 Super Learning Day Revision Notes November 2012

Calculators and time• Beware

When using a calculator to work out questions with time make sure you enter the minutes correctly

e.g. 30 minutes = 0.5 of an hour

15 minutes = 0.25 of an hour

Page 18: Module 3 Super Learning Day Revision Notes November 2012

Density = Mass Volume

e.g. g/cm³

Page 19: Module 3 Super Learning Day Revision Notes November 2012

Speed = Distance Time

e.g. Km/hour

m/sec

Page 20: Module 3 Super Learning Day Revision Notes November 2012

Geometry and Graphs

Page 21: Module 3 Super Learning Day Revision Notes November 2012

Angle Facts• An Acute angle is less than 90°

• A Right angle is equal to 90°

• An Obtuse angle is more than 90°, but less than 360°

• A straight angle is equal to 180°

• A Reflex angle is more than 180°

Page 22: Module 3 Super Learning Day Revision Notes November 2012

BearingsAlways

Measure from the NORTH lineTurn clockwiseUse 3 figures (eg. 30° = 030°)

Page 23: Module 3 Super Learning Day Revision Notes November 2012

Drawing Bearings

To measure a bearing of B from A the North line is drawn at A. This is because the

question says ‘from A’ N

A

B

Page 24: Module 3 Super Learning Day Revision Notes November 2012

Angle Rules• Angles in a triangle add up to 180°

• Angles on a straight line add up to 180°

• Angles in a quadrilateral add up to 360°

• Angles round a point add up to 360°

• The two base angles of an isosceles triangle are equal

Page 25: Module 3 Super Learning Day Revision Notes November 2012

Parallel lines• Look for ‘Z’ angles (Alternate angles)

• Look for ‘F’ angles (corresponding angles)

Alternate angles are equal

Corresponding angles are equal

Page 26: Module 3 Super Learning Day Revision Notes November 2012

Transformations(ask for tracing paper to help you with these)

ReflectionsAlways reflect at right angles to the

mirror lineDiagonal mirror lines are sometimes

called y = xor y = -x

Page 27: Module 3 Super Learning Day Revision Notes November 2012

Rotations

Always check for (or state)

1. The centre of rotation

2. The amount of turn

3. The direction (either clockwise or anti- clockwise)

Page 28: Module 3 Super Learning Day Revision Notes November 2012

Enlargements• Check the scale factor and centre of

enlargement (if there is one)

• Draw construction lines from the centre of enlargement to help you draw the new shape

• Remember a scale factor of ½ will make the shape smaller

Page 29: Module 3 Super Learning Day Revision Notes November 2012

Vector translations A vector translation slides the shape to a

new position

+x

-y

+y

-x

xy

The bottom number y moves the shape up (or down if it is negative)

The top number x moves the shape right (or left if it is negative)

Page 30: Module 3 Super Learning Day Revision Notes November 2012

LociA Locus (more than one are called Loci) is

simply:-

A path that shows all the points which fit a given rule

There are only 4 to remember

Page 31: Module 3 Super Learning Day Revision Notes November 2012

Locus 1

The locus of points which are

A FIXED DISTANCE from a GIVEN POINT

Is simply a CIRCLE

Page 32: Module 3 Super Learning Day Revision Notes November 2012

Locus 2The locus of points which are

A FIXED DISTANCE from a GIVEN LINE

This locus is an oval shape

It has straight sides and ends which are perfect semicircles

Page 33: Module 3 Super Learning Day Revision Notes November 2012

Locus 3

The locus of points which are

A EQUIDISTANT from TWO GIVEN LINES

This is the Angle Bisector (use compasses!)

Page 34: Module 3 Super Learning Day Revision Notes November 2012

Locus 4

The locus of points which are

A EQUIDISTANT from TWO GIVEN POINTS

B

A

This locus is the perpendicular bisector of the line AB

Page 35: Module 3 Super Learning Day Revision Notes November 2012

Pythagoras

Page 36: Module 3 Super Learning Day Revision Notes November 2012

PythagorasThe square on the hypotenuse is equal to the sum of the

squares on the other two sides

Hypotenusea

b

h² = a² + b²

SquareSquareAddSquare root

SquareSquareSubtractSquare root

To find the hypotenuse

To find a short side

Remember

Page 37: Module 3 Super Learning Day Revision Notes November 2012

0 1 2 3 4 5 6 7 8 9 10-9 -8 -7 -6 -5 -4 -3 -2 -1-10 x

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

x and y Coordinates

A graph has 4 different regions

Always plot the x value first followed by the y value

x goes across

y goes up/down

‘in the house and up the stairs’

•(4,2)•(-4,3)

•(-6,-2)

•(7,-5)

x positivey positive

x negativey positive

x negativey negative

x positivey negative

Page 38: Module 3 Super Learning Day Revision Notes November 2012

Midpoint of a line

Midpoint is just the middle of the line!

To find it just add the x coordinates together and

divide by 2

Then add the y coordinates together and divide by 2

You have just found the midpoint

If A is (2,1) and B is (6,3)

Then the x coordinate of the mid point is

(2 + 6) ÷ 2 = 4

And the y coordinate is

(1 + 3) ÷ 2 = 2

So the mid point is (4,2)

For example

Page 39: Module 3 Super Learning Day Revision Notes November 2012

Straight Line Graphs

0 1 2 3 4 5 6 7 8 9 10-9 -8 -7 -6 -5 -4 -3 -2 -1-10 x

y

1

2

3

4

5

6

7

8

9

10

-1

-2

-3

-4

-5

-6

-7

-8

-9

-10

x = -5

y = 2

x= a is a vertical line

through ‘a’ on the x axis

y = b is a horizontal line through ‘b’ on the y axis

Don’t forget that the y axis is also the line x = 0

and the x axis is also the line y = 0

The diagonal line y = x goes up from left to right

and the line y = -x goes

down from left to right

y= x

y= -x

Page 40: Module 3 Super Learning Day Revision Notes November 2012

Straight Line Graph:– y = mx + c

In the equation y = mx + c

The m stands for the gradient and the c is where the line crosses the y axis

Page 41: Module 3 Super Learning Day Revision Notes November 2012

Using y = mx + c to draw a line

1. Get the equation in the form y = mx + c2. Identify ‘m’ and ‘c’ carefully (eg. In the

equation y = 3x + 2, m is 3 and c is 2)3. Put a dot on the y axis at the value of c4. Then go along one unit and up or down by

the value of m and make another dot5. Repeat the last step 6. Join the three dots with a straight line

Page 42: Module 3 Super Learning Day Revision Notes November 2012

Finding the equation of a straight line

1. Use the formula y = mx + c2. Find the point where the graph crosses the

y axis. This is the value of c3. Find the gradient by finding how far up

the graph goes for each unit across. This is the value of m.

4. Now just put these two values into the equation

Page 43: Module 3 Super Learning Day Revision Notes November 2012

Quadratic Graphs

1. Fill in the table of values

2. Carefully plot the points

3. The points should form a smooth curve. If they don’t they are wrong!

4. Join the points with a smooth curve

5. The graph should be ‘u’ shaped

Page 44: Module 3 Super Learning Day Revision Notes November 2012

Simultaneous equations with graphs

1. Do a table of values for each graph

2. Draw the two graphs

3. Find the x and y values where they cross

4. This is the solution to the equations

Page 45: Module 3 Super Learning Day Revision Notes November 2012

Numbers and Algebra

Page 46: Module 3 Super Learning Day Revision Notes November 2012

Special Number Sequences

• Even numbers 2, 4, 6, 8, 10,……

• Odd numbers 1, 3, 5, 7, 9, 11,…..

• Square numbers 1, 4, 9, 16, 25,….

• Cube numbers 1, 8, 27, 64, 125,….

• Powers of 2 2, 4, 8, 32, 64,…..

• Triangle numbers 1, 3, 6, 10, 15, 21,…..

Page 47: Module 3 Super Learning Day Revision Notes November 2012

Number Patterns and SequencesThere are five different types of number sequences1. ADD or SUBTRACT the SAME NUMBER e.g. 2 5 8 11 14 …. 30 24 18 12….

+3 +3 +3 +3 -6 -6 -6

The RULE ‘add 3 to the previous term’ ‘Subtract 6 from the previous term’

2. ADD or SUBTRACT a CHANGING NUMBER

e.g. 8 11 15 20…….. 53 43 34 26…… +3 +4 +5 -10 -9 -8

The RULE

‘Add 1 extra each time to the previous term’. ‘Subtract 1 extra each time from the previous term’

Page 48: Module 3 Super Learning Day Revision Notes November 2012

3. MULTIPLY by the SAME NUMBER EACH TIME e.g. 5 10 20 40…… x2 x2 x2

The RULE ‘Multiply the previous term by 2’

4. DIVIDE by the SAME NUMBER EACH TIME e.g. 400 200 100 50…… ÷2 ÷2 ÷2

The RULE ‘Divide the previous term by 2’

5. ADD THE PREVIOUS TWO TERMS e.g. 1 1 2 3 5 8 + + + + The RULE ‘Add the previous two terms’

Page 49: Module 3 Super Learning Day Revision Notes November 2012

Finding The nth TermTo find the nth term you can use the formula

dn + (a – d)Where ‘d’ is the difference between the terms

And ‘a’ is the first number in the sequencee.g. 3 7 11 15 19…

‘d’ is 4 (because you add 4 to get the next term)and ‘a’ is 3 (that is the first number)

This means that (a – d) is (3 – 4) = -1So the nth term, dn + (a – d) is 4n - 1

Page 50: Module 3 Super Learning Day Revision Notes November 2012

AlgebraTerms

A term is a collection of numbers, letters and brackets, all multiplied /divided together

Terms are separated by + and – signs

Terms always have a + or a – sign attached to the front of them

E.g. 4xy + 5x² - 2y + 6y² + 4 Invisible + sign xy term x² term y term y² term number term

Page 51: Module 3 Super Learning Day Revision Notes November 2012

Simplifying (Collecting Like Terms)

EXAMPLE Simplify 2x – 4 + 5x + 6

x terms number terms 7x +2

So = 7x + 2

1. Put bubbles round each term, making sure that each bubble has a + or –

sign.

2. Then move the bubbles so that LIKE TERMS are together

3. Collect the like terms using the number line to help you

2x – 4 + 5x + 6 +2x +5x -4 + 6=

2x – 4 + 5x + 6

Page 52: Module 3 Super Learning Day Revision Notes November 2012

Multiplying out Brackets

• The thing outside the bracket multiplies each separate term INSIDE the bracket

• When letters are multiplied together they are just written

next to each other e.g. pq

• Remember R x R = R²

• Remember , a minus outside the brackets reverses all the

signs when you multiply

Page 53: Module 3 Super Learning Day Revision Notes November 2012

Expanding Double Brackets

• Remember to multiply everything in the second bracket by each term in the first bracket

( 2p – 4 ) ( 3p + 1 )

= (2p x 3p) + (2p x 1) + (-4 x 3p ) + ( -4 x 1) = 6p² + 2p - 12p -4 = 6p² -10p -4

Page 54: Module 3 Super Learning Day Revision Notes November 2012

Squared Brackets

Example (3p + 5)²

Write this out as two brackets (3p + 5)(3p + 5)

(3p + 5)(3p + 5) = 9p² +15p +15p +25

= 9p² +30p +25

The usual wrong answer is 9p² +25 !!!

Page 55: Module 3 Super Learning Day Revision Notes November 2012

Factorising• This is the exact opposite of multiplying out brackets• Take out the biggest NUMBER that goes into all terms• For each letter in turn take out the highest power that

will go into EVERY term• Open the brackets and fill with all the bits needed to

reproduce each term• E.g. +20 x²y³z -35x³yz²

5x²y (3x² + 4y²z - 7xz² ) Biggest number that Highest powers z wasn’t in all terms so that will divide into of x and y that will it can’t come out as 15, 20, and 35 go into all three terms a common factor

15x4y

Page 56: Module 3 Super Learning Day Revision Notes November 2012

Writing FormulasThese questions ask you to write an equationThe only things you will have to do are:-

Example 1:- to find y, you multiply x by 3 and then subtract 4Start with x 3x 3x – 4 Times it by 3 Subtract 4

Example 2:- to find y, square x, divide it by 3 and then subtract 7 Start with x x² x² x² - 7 3 3 square it divide by 3 subtract 7

Multiply x Divide x Square x (x²) Add or subtract a number

So y = 3x - 4

So y = x² - 7 3

Page 57: Module 3 Super Learning Day Revision Notes November 2012

Formulas from wordsThis time you change a sentence into a formula

Example:- Froggatt’s deep-fry CHOCCO- BURGERS (chocolate covered beef burgers) cost 58 pence each. Write a formula for the total cost, T, of buying n CHOCCO-BURGERS at 58p each.

In words the formula is

Total cost = Number of CHOCCO-BURGERS x 58p

Putting letters in, it becomes:-

T = n x 58

It would be better to write this as T = 58n

Page 58: Module 3 Super Learning Day Revision Notes November 2012

BODMASB BracketsO Other (like squaring )D DivideM MultiplyA AddS Subtract• Remember this is the order for doing the

sums.

Page 59: Module 3 Super Learning Day Revision Notes November 2012

SubstitutionTo substitute numbers into an expression or a formula all

you need to do is replace the letters with their values and

work out either the solution of the equation or the value of

the expression

(don’t forget to use BODMAS)

Page 60: Module 3 Super Learning Day Revision Notes November 2012

Examples of substitutionExample of substituting in a formula

If P = 3Q + 7 what is P when Q is 8?

P = 3Q + 7

P = 3 x 8 + 7

P = 24 + 7

P = 33

Example of substituting in an expression

If a = 3, b = 5and c = 7 what is the value of a² - b + 2c

a² - b + 2c

(3)² - 5 +2( 7 )

9 - 5 + 14

18

Page 61: Module 3 Super Learning Day Revision Notes November 2012

Solving EquationsTo solve equations just follow these 3 rules1. Always do the same thing to both sides of the equation

2. To get rid of something just do the opposite.

The opposite of + is – and the opposite of – is +

The opposite of x is ÷ and the opposite of ÷ is x

3. Keep going until you have a letter on its own

Page 62: Module 3 Super Learning Day Revision Notes November 2012

Examples of Solving Equations 1 Solve 5x = 15

5x means 5 x x so do the opposite and divide both

sides by 5

x = 3

Solve p/3 = 2

p/3 means p ÷ 3 so do the opposite and multiply both

sides by 3

p = 6

Solve 4y – 3 = 17

The opposite of –3 is +3 so add 3 to each side

4y = 20

The opposite of x 4 is ÷ 4 so divide both sides by 4

y = 5

Page 63: Module 3 Super Learning Day Revision Notes November 2012

Examples of Solving Equations 2 Solve 2( x + 3 ) = 11

The opposite of x 2 is ÷ 2 so divide both sides by 2

x + 3 = 5.5

The opposite of + 3 is –3 so subtract 3 from both sides

x = 2.5

Solve 3x + 5 = 5x + 1There are x’s on both sides so

subtract 3x from both sides5 = 2x + 1

The opposite of +1 is –1 so subtract 1 from each side

4 = 2xThe opposite of x 2 is ÷ 2 so

divide both sides by 22 = x (or x= 2)

Page 64: Module 3 Super Learning Day Revision Notes November 2012

REARRANGING FORMULAE

• You do exactly the same for this as for solving equations.

• When you are asked to make a letter the subject of the formula then that letter needs to be by itself on one side of the equation

Page 65: Module 3 Super Learning Day Revision Notes November 2012

REARRANGING FORMULAE

Example: Rearrange the formula 2(b – 3) = a to make b the subject of the formula

We want to get rid of the times 2 outside the bracket and the opposite of times 2 is divide by 2

So b – 3 = a 2

The opposite of –3 is +3 sob = a + 3

2

Page 66: Module 3 Super Learning Day Revision Notes November 2012

Inequalities

There are 4 inequalities and you need to

learn the symbols. > means ‘greater than’ < means ‘less than’ ≥ means ‘greater than or equal to’ ≤ means less than or equal to’

Page 67: Module 3 Super Learning Day Revision Notes November 2012

Inequalities There are 3 types of inequality questions

1. Solving inequalities ( these are just like solving equations)

2. Drawing an inequality – use a number line and remember to fill in the blob if there is ≤ or ≥

3. Writing values that satisfy an inequality. Just use common sense (draw a number line to help you if you need to)

Page 68: Module 3 Super Learning Day Revision Notes November 2012

Trial and ImprovementThis is a good method for finding an approximate answer

to equations that don’t have simple whole number answers

Method1. Substitute two initial values into the equation that give one

answer that is too small and one answer that is too large.2. Choose your next value in between the previous two, and put

it into the equation.3. After only 3 or 4 steps you should have 2 numbers which are

to the right degree of accuracy but differ by the last digit. 4. Now take the exact middle value to decide which is the

answer you want.

Page 69: Module 3 Super Learning Day Revision Notes November 2012

Trial and Improvement

x² x³+3x Comment

3 27 + 9 = 36 Too small

4 64 +12 = 76 Too large

3.5 42.875 +10.5 = 53.375 Too small

3.6 46.656 +10.8 = 57.456 Too large

3.55 44.738875 + 10.65 = 55.388875 Too small

ExampleA solution to the equation x³ + 3x = 56 lies between 3 and 4.

The solution lies between 3.5 and 3.6 but is nearer to 3.6