CCEA Maths M4 🧮

Pythagoras

a² + b² = c²

Trigonometry

• Sinθ = O/H

• Cosθ = A/H

• Tanθ = O/A

  use shift to find angle

Indices

𝑎^m × 𝑎ⁿ = 𝑎^m+n
𝑎^m ÷ 𝑎ⁿ = 𝑎^m-n
(𝑎^m)ⁿ = 𝑎^mn
𝑎⁰ = 1
𝑎^-b = 1/𝑎^b

Interpreting product of primes

• Square number powers are multiples of 2

• Cube number powers are multiples of 3

• Prime numbers only one product

Angles of elevation/ depression

elevation is above horizontal, depression is below

Alternate angles (Z)

equal

Corresponding angles (F)

equal

Vertically opposite angles (X)

are equal

Co-interior angles (C)

add up to 180°

HCF

product of primes numbers in intersection of venn diagram

e.g 3 × 7 = 21

LCM

product of all prime numbers in venn diagram

e.g 5 × 3 × 7 × 2 × 2 = 420

FOIL

(𝑥 + 2)(𝑥 - 5)
*multiply First, Outside, Inside and Last*
𝑥² - 5𝑥 + 2𝑥 - 10

Quadratic

𝑥² + 5𝑥 + 6
*multiply for 6, add for 5*
(𝑥 + 3)(𝑥 + 2)

Quadratic with coefficient

3𝑥² - 13𝑥 - 10
*multiply 3 and - 10 = -30*
*multiply for -30, add for -13*
3𝑥² - 15𝑥 | - 2𝑥 - 10
*factorise each side*
3𝑥(𝑥 - 5) | 2(𝑥 - 5)
(3𝑥 + 2)(𝑥 - 5)

Difference of 2 squares

4𝑥² - 100
*square root 4 = 2, square root 100 = 10*
(2𝑥 - 10)(2𝑥 + 10)

Area of square

𝑥²

Area of triangle

𝑏ℎ ÷ 2

Area of parallelogram

base × perpendicular height

Area of rhombus/ kite

(length × width) ÷ 2 or (𝑑₁ × 𝑑₂) ÷ 2

Area of trapezium

1/2(𝑎 + 𝑏)ℎ

Area of circle

π𝑟²

Area of sector

π𝑟² × θ/360

Circumference

πd

Arc length

πd × θ/360

Volume

area cross section × length

Volume cylinder (cone + sphere given)

π𝑟²ℎ

Volume of frustum

lg cone - sm cone

Surface area of cylinder (cone + sphere given)

2π𝑟ℎ + 2π𝑟²

Speed

distance/ time

Density

mass/ volume

Pressure

force/ area

% Change

change/ orginal x 100

Simple interest

𝑝 + 𝑝𝑟𝘵/ 100

Compound interest

𝑝(1 + 𝑟 ÷ 100)^𝘵

Gradient

(𝑦₂ - 𝑦₁) ÷ (𝑥₂ - 𝑥₁)

Midpt

(𝑥₁ + 𝑥₂)/2 , (𝑦₁ + 𝑦₂)/2

Length of line

√(𝑥₂ - 𝑥₁)² + (𝑦₂ - 𝑦₁)²

Parallel lines

same gradient

Perpindicular lines

negative reciprocal gradient (flip fraction + change sign)

1st Quartile/ Lower quartile (Q₁)

¼ × Σ𝑓

2nd Quartile/ Median (Q₂)

½ x Σ𝑓

3rd Quartile/ Upper quartile (Q₃)

¾ x Σ𝑓

Interquartile range

Q₃ - Q₁

Freq density

freq ÷ cw

Mean

Σ𝑓𝑥/Σ𝑓

Mean from grouped table

• create midpoint (𝑥)

• create 𝑓𝑥 column

• use formula

Median from grouped table

• divide total frequency by 2

• count until you reach n^th value

• median lies in this group

Cumulative frequency

plot of against upper boundary

Box plot

• lowest value

• lower quartile (Q₁)

• median (Q₂)

• upper quartile (Q₃)

• highest value

Histogram

• area of bar represents frequency

• scales are continuous

Median from histogram

• count boxes and divide by 2

• find point on graph and vertical dotted line down

Simplifying algebraic fractions

can only cancel if all terms on top + bottom are multiplying

Add + subtract algebraic fractions

4/(𝑥 + 1) + 3/(𝑥 - 5)
*cross multiply + common denominator*
4(𝑥 - 5)/(𝑥 + 1)(𝑥 - 5) + 3(𝑥 + 1)/(𝑥 - 5)(𝑥 + 1)
*combine into one fraction*
4(𝑥 - 5) + 3(𝑥 + 1)/(𝑥 + 1)(𝑥 - 5)
*expand numerator*
4𝑥 - 20 + 3𝑥 + 3/(𝑥 + 1)(𝑥 - 5)
*simplify*
7𝑥 - 17/(𝑥 + 1)(𝑥 - 5)

Multiplying algebraic fractions

get common denominator, multiply top 2 and bottom 2 then simplify

Dividing algebraic fractions

Flip second fraction + change to multiply

Solving algebraic equations

4/(𝑥 + 1) + 3/(𝑥 - 5) = 2
*add algebraic fractions + simplify*
7𝑥 - 17/(𝑥 + 1)(𝑥 - 5) = 2
*set 2/1 + cross multiply to make equal*
7𝑥 - 17 = 2(𝑥 + 1)(𝑥 - 5)
*expand + simplify*
7𝑥 - 17 = 2𝑥² - 10𝑥 + 2𝑥 - 10
*bring to one side, simplify + equal to 0*
2𝑥² - 15𝑥 + 7 = 0
*use quadratic formula to solve for 𝑥*
𝑥 = 7
𝑥 = 1/2

Stratified sampling

number in group ÷ total × sample size

Why we use stratified sampling

ensure sample accurately reflects proportions of different groups

Angle at centre

is twice angle at circumference
e.g 2 × 50 = 100
e.g 2 × 40 = 80

Angles in same segment

are equal
e.g p = 52
e.g q = 40

Angle at circumference in semicircle

is 90°
e.g STU = 90°

Opposite angles in cyclic quadrilateral

add up to 180°
e.g a = 180 - 60 = 120°
e.g b = 180 - 140 = 40°

Radius and tangent

meet at 90°

Alternate segment theroem

angles are equal

Two tangents off circle

form isosceles triangle
e.g joining CB would form an isosceles triangle

Square numbers

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225

Cube numbers

1, 8, 27, 64, 125, 216

Prime numbers

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31