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Additional-Mathmatics	Angles in Polygon
GCE ‘O’ Level	ADDITIONAL MATHEMATICS TOPIC 2: GEOMETRY AND TRIGONOMETRY
EXAMINER TIPS for OL Mathematics 4024	Formula Booklet 2012
Formulae Sheet	GCSE Mathmatics
Indices	Indices & Standard Form
Inverting Functions	Maths (Transformation Matrices)
Mathematics IGCSE notes	Math Rules
Additional-Mathmatics Subtopic: Matrices	Matrices
Practice Test Maths	Revision Checklist for ‘O’ Level Maths
Rotation	Scalars And Vectors
Solving Simultaneous Equations Algebraically	Set’s Vectors And functions
Square, Square roots, cubes, cube roots	Teach yourself Trigonometry
Trigonometry

Angles in Polygon

Angles in Polygons

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Add Maths (Quick Revision)

The Guide Contains Several Topics:


1:Sets

2: Simultaneous Equations
3: Logarithms and Indices4. Quadratic Expressions and Equations
5. Remainder Factor Theorems
6. Matrices
7. Coordinate Geometry
8. Linear Law
9. Functions
10. Trigonometric Functions
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GCE 'O' Level

GCE 'O' Level Guide

--> Examination Format
--> Mathmatical Formulae
-->Mensuration
--> Trignometry
--> Statistics
--> Vector
--> Algebra
--> Matrices

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Proportion

Proportion

If a "is proportional" to b (which is the same as 'a is in direct proportion with b') then as b increases, a increases. In fact, there is a constant number k with a = kb. We write a ∝ b if a is proportional to b.

The value of k will be the same for all values of a and b and so it can be found by substituting in values for a and b.

Example

If a ∝ b, and b = 10 when a = 5, find an equation connecting a and b.

a = kb (1)

Substitute the values of 5 and 10 into the equation to find k:

5 = 10k

so k = 1/2

substitute this into (1)

a = ½b

In this example we might then be asked to find the value of a when b = 2. Now that we have a formula connecting a and b (a = ½ b) we can subsitute b=2 to get a = 1.

Similarly, if m is proportional to n2, then m = kn2 for some constant number k.

If x and y are in direct proportion then the graph of y against x will be a straight line.

Inverse Proportion (HIGHER TIER)

If a and b are inversely proportionally to one another,

a ∝ 1/b

therefore a = k/b

In these examples, k is known as the constant of variation.

Example

If b is inversely proportional to the square of a, and when a = 3, b = 1, find the constant of variation.

b = k/a2 when a = 3,

b = 1

therefore 1 = k/32

therefore k = 9

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Standard Form

Standard form is a way of writing down very large or very small numbers easily. 10³ = 1000, so 4 × 10³ = 4000 . So 4000 can be written as 4 × 10³ . This idea can be used to write even larger numbers down easily in standard form.

Small numbers can also be written in standard form. However, instead of the index being positive (in the above example, the index was 3), it will be negative.

The rules when writing a number in standard form is that first you write down a number between 1 and 10, then you write × 10(to the power of a number).

Example

Write 81 900 000 000 000 in standard form: 81 900 000 000 000 = 8.19 × 10¹³

It’s 10¹³ because the decimal point has been moved 13 places to the left to get the number to be 8.19

Example

Write 0.000 001 2 in standard form:

0.000 001 2 = 1.2 × 10^-6

It’s 10^-6 because the decimal point has been moved 6 places to the right to get the number to be 1.2

On a calculator, you usually enter a number in standard form as follows: Type in the first number (the one between 1 and 10). Press EXP . Type in the power to which the 10 is risen.

Manipulation in Standard Form

This is best explained with an example:

Example

The number p written in standard form is 8 × 10⁵

The number q written in standard form is 5 × 10^-2

Calculate p × q. Give your answer in standard form.

Multiply the two first bits of the numbers together and the two second bits together:

8 × 5 × 10⁵ × 10^-2

= 40 × 10³ (Remember 10⁵ × 10^-2 = 10³)

The question asks for the answer in standard form, but this is not standard form because the first part (the 40) should be a number between 1 and 10.

= 4 × 10⁴

Calculate p ÷ q.

Give your answer in standard form.

This time, divide the two first bits of the standard forms. Divide the two second bits. (8 ÷ 5) × (10⁵ ÷ 10^-2) = 1.6 × 10⁷