# data[].attributes.body
Fractions:
A = 2/9
J = 7/9

36 ÷ 9 x 2 = 8
36 ÷ 9 x 7 = 28                        = 8:28

# data[].attributes.body
Addition:

+ 2+7=9
÷ 36÷9=4
x A = 2x4=8
J = 7x4=28

# data[].attributes.body
The block method...

A = ⬛️⬛️ - 8
J = ⬛️⬛️⬛️⬛️⬛️⬛️⬛️ -28

5 blocks = 20
Each block is 4

Now you can work out how many are in total or how many one person has by timesing 4 but the amount of blocks

# data[].attributes.body
Use the bar method to draw out the question in real life.
It then allows you to see the difference and amounts.

# data[].attributes.body
Rotate this triangle by 90 degrees anti-clockwise around A:

# data[].attributes.body
Either a vector is given or you have to find the vector:
Choose a vertice and move ...
(3) - across three
(2) - up two

# data[].attributes.body
y = mx + c
m= gradient
c= y intercept

# data[].attributes.body
3y = 6x + 3

3y ÷ 3 = y
6x ÷ 3 = 2x
3 ÷ 3 = 1

y = 2x + 1

# data[].attributes.body
From the centre of enlargement it is (1,1).
When enlarged you double this so it’s (2,2).

# data[].attributes.body
Congruent: The sand shape and size which can only be rotated or flipped.

Exam Questions ask how to prove congruence:

1: SAS - Two sides and the included angle.
2: ASA - Two angles are the same and a side.
3: SSS - Three sides all the same.
4: RHS - Right-angle, hypotenuse and another side.

# data[].attributes.body
Shapes which are the same just different sizes.
Must be equiangular
Must be enlarged.
Use corresponding angles and sides.

# data[].attributes.body
Use the graph to tell you details about the equation.

# data[].attributes.body
Using algebra name the side and work out ‘x’.

REDUCTION - small ÷ big

ENLARGE - big ÷ small

You must always multiply later though!

AREA SCALE FACTOR = LSF2
New area is LSF2 x the old area.

# data[].attributes.body
Expand and Simplify (2x + 1)²
(2x+1)(2x+1) - 4x² + 2x + 2x + 1
= 4x² + 4x + 1

# data[].attributes.body
Factorise fully 50xy ² - 25xy
50xy² - 25xy ➡️ 25xy(26 - 1)

# data[].attributes.body
Factorise 100x ² - 64
(10x - 8)(10x - 8) ⬇️
You have to squareroot both 100x and 68.

# data[].attributes.body
z = 9x ² - y Find the value of z when x = - 1/3 and y = -1/4
z = 9x² - y
= 9x (-1/3)² - (-1/4)
=9 x 1/9 + 1/4
= 1+ 1/4
= 1and a quarter.

# data[].attributes.body
Work out (1 3/5)2 =
(1 3/5)2 ➡️ 64/25
Then turn into a mixed number: 2 14/25

# data[].attributes.body
Enlarge: LSF = 2
New Volume : 8 or 23

NEW VOLUME: LSF 3 x Old Volume

# data[].attributes.body
1. New Length = LSF1 x Old Length

2. New Area = LSF2 x Old Area

3. New Volume = LSF3 x Old Volume

# data[].attributes.body
Roots that cannot be written as exact decimals are called SURDS. Leaving an answer in sure form means the answer is exact.

# data[].attributes.body
Here are the formulas of how to work out SURDS with an example.

# data[].attributes.body
Factorising is putting something into brackets.

# data[].attributes.body
-An angle measured from the north line
-Measured in a clockwise direction
-Must be written in three figures

# data[].attributes.body
To find a scale factor between two similar figures, find two corresponding sides and write the ratio of the sides.

Smaller number first = less than 1

Bigger number first = more than 1

# data[].attributes.body
The answer must be written in three figures.
Also with the degrees sign.
Eg:  = 058o

# data[].attributes.body
Formula of different shapes ...

# data[].attributes.body
These are when you divide a shape into sections to make it easier to work out the are/volume.
Often it will be an irregular shape split into triangles and squares.

Eg:

# data[].attributes.body
Circumference: πd (perimeter)
Area: πr²
Sector: ¹/2 r2 x (angle given)

# data[].attributes.body
Circumference: πd (perimeter)
Area: πr²
Sector: ¹/2 r2 x (angle given)

# data[].attributes.body
Circumference: πd (perimeter)
Area: πr²
Sector: ¹/2 r2 x (angle given)

# data[].attributes.body
The highest number that can be divided into two or more numbers exactly

Eg: WHATS THE HCF OF 24 and 40??
Factors of 24: 1,2,3,4,6,8,12,24
Factors of 40: 1,2,4,5,8,10,20,40

The highest number which is in both list is eight there fore it is the highest common factor.

=8

# data[].attributes.body
The lowest quantity that is a multiple of two or more given quantities.

Eg: WHAT IS THE LCM OF 6 & 10??
Multiples of 6 are: 6,12,18,24,30,36,42
Multiples of 10 are: 10,20,30,40,50

Therefore the lowest common multiple is 30 because it’s the smallest number in both lists.

=30

# data[].attributes.body
Range: lowest - largest
Mean: average
Mode: difference between them
Median: middle number

# data[].attributes.body
The number infront  of the algebraic term.
-3y ➡️ -3 is the coefficient number

# data[].attributes.body
Putting data from a two collum table into a graph.

Plot according to the data.

Join up with straight lines.

# data[].attributes.body
Find the highest frequency (MODE)
Read off coresponding class interval.

Eg: 10<x>40 or 3-7

# data[].attributes.body
The rule for lower&upper quartile:
EVEN number of items = Find middle of bottom half
ODD number of items = Throw away middle item, find medium of remaining half.

# data[].attributes.body
1. Put the numbers in order.

2. Draw a line of numbers 1-5 down.

3. Put the tens in the front collum and list them on the other side of the line. ( eg 1 | 4 = 14 )

4. Then you put it in order.

5. Write the key.

# data[].attributes.body
To work out the inter Quartile range is the lower Quartile minus the upper Quartile.

# data[].attributes.headline
FREQUENCY DENSITY = FREQUENCY ÷ CLASS WIDTH

# data[].attributes.subject
FREQUENCY DENSITY = FREQUENCY ÷ CLASS WIDTH

# data[].attributes.body
Relative frequency of an event = frequency if an event divided by the number of trials

# data[].attributes.body
If the probability of an event occurring is p then the probability of it not happening is 1 - p.

# data[].attributes.body
Outcomes are mutually exclusive if they cannot happen at the same time.

# data[].attributes.body
In probability ‘or’ means +.
In probability ‘and’ means x.

# data[].attributes.body
Independent events are the outcome if one doesn’t affect the other.

# data[].attributes.body
5! = 5x4x3x2x1 ➡️ 5 factorical

# data[].attributes.body
a(x-b)=j this would then become x=j+ab divided by a.
To make x the subject you have to do the reverse of each part.

# data[].attributes.body
x+a= t ➡️ x =t-a

(Note: t must stay on that side because only a was carried over)

# data[].attributes.body
Three steps of inverting a function.

# data[].attributes.body
You have to make sure that either the y or x’s are the same by multiplying the whole equation by something.