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user:General maths

Introduction
''There are two basic rules to mathematics. They are as follow:''

Rule 1 states, any number divide by itself is 1 except zero 

$$\frac{1}{1}=\frac{2}{2}=\frac{3}{3}\cdots=1$$

Rule 2 states any number multiply by itself is itself

$$2\times1=2$$

$$3\times1=3$$

And so on.

Knowing these 2 rules and the technique to transpose formulae it is enough to excel any body in maths

The general form of maths

Symbol = Algorithms = Solution

There are 3 parts to what we are learning together

First part, we will deal with formulae

Second part we will deal with the application

Third part deal with the derivation

GCSE
''There are 4 topics in GCSE. They are as follow:''

Numberacy : Rule 1 and 2 can answer any arithmetic problems

Algebra : Rule 1 and 2 and technique of transposition of formulae can answer any Algebra problems

Geometry : Rule 1 and 2 and Transposition of formulae

Data handling : Rule 1 and 2 and Transposition of formulae

A - Level
Heavily depend upon these 2 rules and technique of transposition of formulae

Degree
Heavily depend upon these 2 rules and technique of transposition of formulae

Lesson 1: Part 1: 27/1/8
fraction

Fraction (means broken)

Example of fractions:

$$\frac{3}{4},\frac{4}{3},2\frac{4}{2}$$

General form of fraction
$$\frac{m}{n}=\frac{m...is...called...numerator}{n...is...called...denumerator}= \frac{chosen...parts}{broken...parts}$$

Explanation
When the numerator is less than the denumerator (Long hand written)

numerator < denumerator (short hand written)

This symbol < means less than

Note: In maths they use symbol to shorten the written work

$$\frac{3}{4}$$

it is called proper fraction.

And when the numerator is greater than the denumerator (Long hand written)

numerator > denumerator (short hand written)

This symbol > means greater than

$$\frac{4}{3}$$

it is called an Improper fraction

The combination of a whole number and proper fraction

$$2\frac{1}{3}=2+\frac{1}{3}$$

it is called mixed fraction

Extra note
Mixed fraction is the result of an Improper fraction 

$$2\frac{1}{3}=2+\frac{1}{3}=\frac{2\cdot3}{1\cdot3}+\frac{1}{3}=\frac{6+1}{3}= \frac{7}{3}$$

Part 2 : Addition and subtraction of fraction
An example

$$1\cdots\frac{2}{5}+\frac{1}{5}$$

$$2\cdots\frac{2}{5}+\frac{1}{5}=\frac{2+1}{5}=\frac{3}{5}$$

The rule of adding fraction:

Fraction can only be added if they all have the same denumerator

Subtracting of fraction

An example

$$1\cdots\frac{2}{5}-\frac{1}{5}$$

$$2\cdots\frac{2}{5}-\frac{1}{5}=\frac{2-1}{5}=\frac{1}{5}$$

The rule of subtracting fraction:

Fraction can only be subtracted if they all have the same denumerator

Lesson 1: Part 2: Homework
Add the following fractions

$$1\cdots\frac{1}{2}+\frac{1}{3}+\frac{2}{5}$$

$$2\cdots\frac{5}{6}+\frac{1}{5}$$

$$3\cdots\frac{5}{8}+\frac{2}{4}$$

Subtract the following fractions

$$1\cdots\frac{1}{2}-\frac{1}{3}-\frac{2}{5}$$

$$2\cdots\frac{5}{6}-\frac{1}{5}$$

$$3\cdots\frac{5}{8}-\frac{2}{4}$$

Mixed exercise
$$1\cdots\frac{1}{2}+\frac{1}{3}-\frac{2}{5}$$

$$2\cdots\frac{5}{6}+\frac{1}{5}-\frac{4}{7}$$

$$3\cdots\frac{2}{8}-\frac{1}{4}+\frac{3}{9}$$

Adding with algebra

$$1\cdots\frac{1}{x}+\frac{1}{y}$$

$$2\cdots\frac{2}{a}+\frac{1}{b}$$

$$3\cdots\frac{3}{2(x-1)}+\frac{2}{y}$$

Subtracting with algebra

$$1\cdots\frac{1}{x}-\frac{1}{y}$$

$$2\cdots\frac{2}{a}-\frac{1}{b}$$

$$3\cdots\frac{3}{2(x-1)}-\frac{2}{y}$$

Part 3 : Indices
indices

$${b^{x}=N}\cdots1$$

x is called the index and b is called the base

Multiplying indices
$$2\cdot2=2^2=4$$

$$2\cdot2\cdot2=2^3=8$$

$$2\cdot2\cdot2\cdot2=2^4=16$$

$$2\cdot2\cdot2\cdot2\cdot2=2^5=32$$

Let examine this one :

$$2\cdot2\cdot2\cdot2\cdot2=2^5$$

$$2^2\cdot2^3=2^{2+3}=2^5$$

'''When a set of same bases are being multiply, with same or different index, just add the indices like the above example'''

Caution : You can not do this

$${2^3}\cdot{3^4}={6^7}$$

General formula of multiplying indices
$$b^{m}{\cdot}b^{m}=b^{n+m}$$

Exercise
$$1\cdots2^3\cdot2^4$$

$$2\cdots2^3\cdot2^4\cdot2^{-2}$$

$$3\cdots{a}^3\cdot{a}^4\cdot{a}^{-2}$$

$$4\cdots5^3\cdot5^3\cdot5^{8}$$

Dividing indices
$$\frac{2\cdot2\cdot2\cdot2\cdot2}{2}=\frac{2^5}{2}=2^{5-1}=2^4=16$$

$$\frac{2\cdot2\cdot2\cdot2}{2}=\frac{2^4}{2}=2^{4-1}=2^3=8$$

$$\frac{2\cdot2\cdot2}{2}=\frac{2^3}{2}=2^{3-1}=2^2=4$$

$$\frac{2\cdot2}{2}=\frac{2^2}{2}=2^{2-1}=2^1=2$$

$$\frac{2}{2}=2^{1-1}=2^0=1$$

'''When a set of same bases are being divided, with same or different index, just subtract the indices like the above examples'''

Special case
$$2^0=1\cdots1$$

This is a very important case

General formula of dividing indices
$$\frac{b^m}{b^m}=b^{n-m}$$

Exercise
$$1\cdots\frac{3^5}{3^3}$$

$$2\cdots\frac{6^{-2}}{6^3}$$

$$3\cdots\frac{a^5}{a^3}$$

The power of indices
$$2^2\cdot2^2\cdot2^2=2^{2+2+2}=\left(2^2\right)^3=2^{2\cdot3}=2^6$$

Examples

$$\left(a^5\right)^4=a^{5\cdot4}=a^{20}$$

$$\left(3^2\right)^5=2^{2\cdot5}=a^{15}$$

General power of indices
$$\left(b^n\right)^m=b^{nm}$$

Exercise
$$1\cdots\left(2^2\right)^5$$

$$2\cdots\left(3^3\right)^{-3}$$

$$3\cdots\frac{\left(4^4\right)^6}{\left(4^4\right)^3}$$

Linear equation
Example of linear equation

$$3x+4=7$$

3x, where x is called the variable and the 3 is called the coefficient of x

4 and 7 are just the constant

Solving linear equation
Solve for x

$$4x-4=16\cdots(1)$$

Solution

$$4x-4=16\cdots(1)$$

$$4x-4+4=16+4\cdots(2)$$

$$4x=20\cdots(3)$$

$$\frac{4x}{4}=\frac{20}{4}\cdots(4)$$

$$x=5\cdots(5)$$

Check the answer

Solve for x

$$4x-4=16\cdots(1)$$

$$4(5)-4=16\cdots(2)$$

$$20-4=16\cdots(3)$$

$$16=16\cdots(4)$$

Exercise
$$5x+5=30\cdots(1)$$

$$-5x-5=5\cdots(2)$$

$$ax+b=c\cdots(3)$$

$$\frac{3}{x}+\frac{1}{2}=\frac{5}{4}\cdots(4)$$

$$3y-4=4y+5\cdots(5)$$

$$\frac{3y}{3}+\frac{1}{2}=\frac{5y}{4}-3\cdots(6)$$

Question number 7

Sam gave to his brother 3 box of sweet, 15 are missing the remaining left in the box 21

Find how many was in the box before 15 was missing

Quadratic equation
Example of quadratic equations

$$x^2-4=0\cdots(1)$$

$$x^2+4x=0\cdots(2)$$

$$x^2+4x+4=0\cdots(3)$$

Ways of solving quadratic equation
There 4 ways of solving this kind of equation, they are:

By formula

$$ax^2+bx+c=0\cdots(1)$$

$$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\cdots(2)$$

$$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\cdots(3)$$

Quadratic equation has two solutions (x1 and x2)

By factorisation 

By Completing the square

By graph

The differences between Linear and quadratic equations
$$ax-b=0\cdots(1)$$

$$ax^2+bx+c=0\cdots(2)$$

In linear equation the variable x is raised to the power of

$$x^1\cdots(1)$$

Where as in the quadratic equation the variable x is raised to the power of

$$x^2\cdots(2)$$

Solving the quadratic by formula
An example

$$x^2+5x+6=0\cdots(1)$$

Step 1:

Where a = 1, b = 5 and c = 6

Step 2:

Formula

$$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}\cdots(2)$$

$$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}\cdots(3)$$

Step 3:

$$x_1=\frac{-5+\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}\cdots(1)$$

$$x_1=\frac{-5+\sqrt{25-24}}{2}\cdots(2)$$

$$x_1=\frac{-5+\sqrt{1}}{2}\cdots(3)$$

$$x_1=\frac{-5+1}{2}\cdots(4)$$

$$x_1=\frac{-4}{2}\cdots(5)$$

$$x_1=-2\cdots(6)$$

step 4:

$$x_2=\frac{-5-\sqrt{5^2-4\cdot1\cdot6}}{2\cdot1}\cdots(1)$$

$$x_2=\frac{-5-\sqrt{25-24}}{2}\cdots(2)$$

$$x_2=\frac{-5-\sqrt{1}}{2}\cdots(3)$$

$$x_2=\frac{-5-1}{2}\cdots(4)$$

$$x_2=\frac{-6}{2}\cdots(5)$$

$$x_2=-3\cdots(6)$$

Steps 5:

The solution of:

$$x^2+5x+6=0\cdots(1)$$

is

$$x_1=-2\cdots(2)$$

$$x_2=-3\cdots(3)$$

Check the answer

$$x^2+5x+6=0\cdots(1)$$

$$(-2)^2+5(-2)+6=0\cdots(2)$$

$$4-10+6=0\cdots(3)$$

$$-6+6=0\cdots(4)$$

$$0=0\cdots(5)$$

$$x^2+5x+6=0\cdots(1)$$

$$(-3)^2+5(-3)+6=0\cdots(2)$$

$$9-15+6=0\cdots(3)$$

$$-6+6=0\cdots(4)$$

$$0=0\cdots(5)$$

Exercise
$$x^2+x-6=0\cdots(1)$$

$$x^2+9x+20=0\cdots(2)$$

$$2x^2-5x-12=0\cdots(3)$$

Basic Factorising of Linear and quadratic equations
Factorise the following expression

Example 1:

$$y=3x+6\cdots(1)$$

Step 1:

$$y=3\cdot{x}+3\cdot{2}\cdots(2)$$

Step 2:

Common 3 and not (x+2)

Step 3:

$$y=3\left(x+2\right)\cdots(3)$$

Example 2:

$$z=xy+y^2\cdots(1)$$

Step 1:

$$z=x\cdot{y}+y\cdot{y}\cdots(2)$$

Step 2:

Common y and not (x+y)

Step 3:

$$z=y\left(x+y\right)\cdots(3)$$

Example 3:

$$T=x\left(a+b\right)+y\left(a+b\right)\cdots(1)$$

Step 1:

$$T=x\left(a+b\right)+y\left(a+b\right)\cdots(2)$$

Step 2:

Common (a+b) and not (x+y)

Step 3:

$$T=\left(a+b\right)\left(x+y\right)\cdots(3)$$

Exercise
Factorise the following expression

$$x=m^2+m^y\cdots(1)$$

$$z=10x^2y+15xy^2\cdots(2)$$

$$y=5x-10\cdots(3)$$

Factorisation of quadratic equation
General quadratic form

$$ax^2+bx+c=0\cdots(1)$$

Special condition

Let a = 1

$$x^2+bx+c=0\cdots(2)$$

The question will be of this form

$$x^2+bx+c=0\cdots(1)$$

The solution will be of this form

$$\left(x+A\right)\left(x+B\right)=0\cdots(2)$$

To find A and B we need to solve these basic 2 equations, they are:

$$A\cdot{B}=c\cdots(1)$$

$$A+B=b\cdots(2)$$

Examples

$$x^2+5x+6=0\cdots(1)$$

Step 1:

Solution of this form

$$\left(x+A\right)\left(x+B\right)=0\cdots(1)$$

Step 2:

Solve these 2 equations

$$A\cdot{B}=6\cdots(2)$$

$$A+B=5\cdots(3)$$

Step 3:

$$2\cdot{3}=6\cdots(2)$$

$$2+3=5\cdots(3)$$

Step 4

Where A = 2 and B = 3 

Step 5

$$\left(x+A\right)\left(x+B\right)=0\cdots(4)$$

Step 6:

$$\left(x+2\right)\left(x+3\right)=0\cdots(5)$$

Step 7:

Let x = -2

$$\left(0\right)\left(x+3\right)=0\cdots(5)$$

$$0=0\cdots(6)$$

Step 8:

Let x = -3

$$\left(-3+3\right)\left(x+3\right)=0\cdots(7)$$

$$0=0\cdots(8)$$

Step 9:

The solution of this quadratic equation:

$$x^2+5x+6=0\cdots(1)$$

is

x = -2 and x = -3

Step 10:

Check the answer

Step 1:

$$\left(-2\right)^2+5\left(-2\right)+6=0\cdots(2)$$

$$4-10+6=0\cdots(3)$$

$$-6+6=0\cdots(4)$$

$$0=0\cdots(5)$$

Step 2:

$$\left(-3\right)^2+5\left(-3\right)+6=0\cdots(2)$$

$$9-15+6=0\cdots(3)$$

$$-6+6=0\cdots(4)$$

$$0=0\cdots(5)$$

Exercise
$$x^2+x-2=0\cdots(1)$$

$$x^2+7x+12=0\cdots(2)$$

$$x^2-6x+8=0\cdots(3)$$

Simultaneous equation
2 by 2 simultaneous equation

Example 1:

$$x+y=5\cdots(1)$$

$$x-y=1\cdots(2)$$

Example 2:

$$2x-3y=3\cdots(1)$$

$$5x+y=16\cdots(2)$$

First method of solving simultaneous equation
The elimination method

Example

$$x+y=5\cdots(1)$$

$$x-y=1\cdots(2)$$

Step 1:

(1) + (2)

$$x+x-y+y=1+5\cdots(3)$$

$$2x=6\cdots(4)$$

$$\frac{2x}{2}=\frac{6}{2}\cdots(5)$$

$$x=3\cdots(6)$$

Step 2:

Substitute x = 3 into (1)

$$x+y=5\cdots(1)$$

$$3+y=5\cdots(2)$$

$$3+y-3=5-3\cdots(3)$$

$$y=2\cdots(4)$$

Answer

The solution of this simultaneous equation

$$x+y=5\cdots(1)$$

$$x-y=1\cdots(2)$$

is

x =3 and y = 2

Check the answer

Step 1:

$$x+y=5\cdots(1)$$

$$x-y=1\cdots(2)$$

Step 2:

$$3+2=5\cdots(1)$$

$$3-2=1\cdots(2)$$

Transposition of Formulae : Introduction
These are the basic rules you need remember when dealing with transposition of formulae 

They are as follow:

Unity

$$\frac{a}{a}=1\cdots(1)$$

Zero

$$a-a=0\cdots(2)$$

Expansion of bracket

$$k\left(a+b\right)=k\times{a}+k\times{b}=ka+kb\cdots(3)$$

Factorisation

$$k\left(a+b\right)=ka+kb\cdots(4)$$

Inversing of fraction

$$a=\frac{x}{y}\cdots(1)$$

''Let inverse it. It become like''

$$\frac{1}{a}=\frac{y}{x}\cdots(2)$$

The most important is the technique of recognition and substitution of formulae

Basic definition:

One term is express in term of other terms

Examples of formulae

One term = Other terms

$$F=ma\cdots(1)$$

$$A=\pi{r^2}\cdots(2)$$

$$Sin\theta=\frac{a}{b}\cdots(3)$$

Technique of transposition of formulae

Example 1:

$$F=ma\cdots(1)$$

Explanation:

1...F is called the subject of this formula

2...The subject (F) is always is on the LEFT HAND SIDE, by itself

3...The subject (F) is defined in term of m and a in the form of

$$m\times{a}$$

If value are assign to m and a. We can find the value of F

Example

m = 4 and a = 3

$$F=m\times{a}\cdots(1)$$

$$F=4\times3=12\cdots(2)$$

Example 2:

Make r the subject of the formula

$$A=\pi{r^2}\cdots(1)$$

Step 1:

The subject is always is on the LEFT HAND SIDE

that

$$A=\pi{r^2}\cdots(1)$$

becomes

$$\pi{r^2}=A\cdots(2)$$

Step 2:

The subject is always is by itself

$$\pi{r^2}={A}\cdots(1)$$

$$\frac{\pi{r^2}}{\pi}=\frac{A}{\pi}{\cdots(2)}$$

Quick note

$$\frac{\pi}{\pi}=1$$

$$r^2=\frac{A}{\pi}\cdots(3)$$

$$\left(r^2\right)^{\frac{1}{2}}=\left(\frac{A}{\pi}\right)^{\frac{1}{2}}\cdots(4)$$

Quick note

$$2\times\frac{1}{2}=1$$

$$r=\left(\frac{A}{\pi}\right)^{\frac{1}{2}}\cdots(5)$$

The subject (r) is defined in terms of A and π in the form of

$$\left(\frac{A}{\pi}\right)^{\frac{1}{2}}$$

... Finally 

$$A=\pi{r^2}\cdots(1)$$

Makes r the subject of the formula

Follow the algorithms ( step 1 to step 5 )

We have

$$r=\left(\frac{A}{\pi}\right)^{\frac{1}{2}}$$

Let check the answer

$$A=\pi{r^2}\cdots(1)$$

π = 3 and r = 4

$$A=3\times{4^2}=3\times4\times4=48\cdots(2)$$

$$r=\left(\frac{A}{\pi}\right)^{\frac{1}{2}}\cdots(1)$$

π = 3 and A = 48

$$r=\left(\frac{48}{3}\right)=16^{\frac{1}{2}}=4\cdots(2)$$

Younger sister Lesson two: 3/2/8
1: Definition of multiplication

2: Property of multiplication : Commutative property 

The order in which two numbers are multiplied does not matter

3: Zero multiplication

4: Long hand multiplication

5: Short hand multiplication

6: Multiplication of indices

7: Multiplication of binomial

8 Multiplication involving shapes: Area of square,rectangle,parallelgram, cirle and Volume of rectangle 

9: Multiplication of fraction

10: Factorial

12: Solving equation with multiplication

13: Multiplication with different signs

14: Square and cube numbers

15: Multiplication of decimal number

16: Multiplication of probability

Defintion of multiplication
Example 1:

$$3\times4$$

Solution

What this mean is that

$$3\times4=4+4+4=12$$

General form of maths

Symbol = Algorithms = Solution

Zero multiplication
Any number multiplied by zero is zero

Example 1:

$$5\times0=0$$

Example 2:

$$554\times0=0$$

Long hand multiplication
Home work

Question 1

$$12\times12$$

$$13\times13$$

$$14\times14$$

Question 2

$$121\times121$$

$$131\times131$$

$$141\times141$$

Question 3

$$234\times12$$

$$123\times14$$

$$241\times8$$

Short hand multiplication
This I will show in the lesson

Home work

Question 1

$$12\times13$$

$$23\times13$$

$$41\times11$$

Question 2

$$121\times103$$

$$203\times113$$

$$141\times101$$

Multiplication indices
Examples

$$1\cdots{2\times2}=2^2$$

$$2\cdots{2\times2\times2}=2^3$$

$$3\cdots{2\times2\times2\times2}=2^4$$

Have you noted that

$$1\cdots{2^4}$$

can be written in a number way, like this

$${2\times2^3}=2^4$$

$${2^2\times2^2}=2^4$$

What this mean is that when you are mutliplying bases raised to a index, you just have to add the indices (plural of index) together

Examples

$$2^3\times2^4=2^{3+4}=2^7$$

$$5^3\times5^4\times5^2=2^{3+5+2}=5^{10}$$

But you can not this is

Examples

$$2^3\times3^3=6^{3+3}$$

You can only multiply with the same base regardless of the index

Home work

$$1\cdots2^5\times2^8$$

$$2\cdots4^{-2}\times4^6$$

$$3\cdots3^4\times3^5\times3^7$$

$$4\cdots{m^1}\times{m^4}\times{m^{-2}}$$

Multiplication of fraction
Multiplication of fraction are like normal multiplication except in fraction form

Example 1

$$\frac{1}{4}\times\frac{2}{5}$$

Solution

$$1\cdots\frac{1}{4}\times\frac{2}{5}=\frac{1\times2}{4\times5}$$

$$2\cdots\frac{1}{4}\times\frac{2}{5}=\frac{2}{20}$$

Example 2

$$\frac{3}{5}\times\frac{2}{6}\times{\frac{1}{2}}$$

Solution

$$1\cdots\frac{3}{5}\times\frac{2}{6}\times{\frac{1}{2}}= \frac{3\times2\times1}{5\times6\times5}$$

$$2\cdots\frac{3}{5}\times\frac{2}{6}\times{\frac{1}{2}}= \frac{6}{60}$$

Home work

$$1\cdots\frac{1}{2}\times\frac{3}{5}\times\frac{4}{7}$$

$$2\cdots\frac{2}{5}\times\frac{3}{9}\times\frac{1}{3}$$

$$3\cdots\frac{3}{8}\times\frac{0}{5}$$

$$4\cdots\frac{x}{2}\times\frac{y}{5}$$

Factorial
Multiplication of integer number

Examples

$$1!=1=1$$

$$2!=1\times2=2$$

$$3!=1\times2\times3=6$$

$$4!=1\times2\times3\times4=24$$

$$5!=1\times2\times3\times4\times5=120$$

Multiplication of binomial
Multiplication of binomial look like this

$$1\cdots{k(a+b)=ka+kb}$$

Example 1

$${4\times(2+3)}$$

Solution

$$1\cdots4\times(2+3)$$

$$2\cdots4\times(2+3)=(2+3)+(2+3)+(2+3)+(2+3)$$

$$3\cdots4\times(2+3)=(3+3+3+3)+(2+2+2+2)$$

$$4\cdots4\times(2+3)=4\times3+4\times2$$

Checking the answer

$$5\cdots4\times(5)=12+8$$

$$6\cdots20=20$$

The short way to the solution

$$1\cdots{4\times(2+3)}=4\times2+4\times3$$

The above example are just for explanation

Example 2

$$3\times(3+5)$$

Solution

$$1\cdots{3\times(3+5)}$$

$$2\cdots{3\times(3+5)}= 3\times3+3\times5$$

Check the anwser

$$1\cdots{3\times(8)}= 9+15$$

$$2\cdots24= 24$$

Example 3

$$4\times(a+5)$$

Solution

$$1\cdots{4\times(a+5)}$$

$$2\cdots{4\times(a+5)}= 4\times{a}+4\times5$$

$$2\cdots{4\times(a+5)}= 4\times{a}+20$$

Check the anwser

Let a = 3

$$1\cdots{4\times(a+5)}= 4\times{a}+20$$

$$2\cdots{4\times(3+5)}= 4\times{3}+20$$

$$3\cdots{4\times(8)}= 12+20$$

$$4\cdots32= 32$$

Example 4

$$a\times(x+y)$$

Solution

$$1\cdots{a\times(x+y)}$$

$$2\cdots{a\times(x+y)}= a\times{x}+a\times{y}$$

Quick note

When you are multiplying by a bracket and letters, just ignore the multiplication sign

Repeat of example 4, without he multiplication sign

Example 4

$$1\cdots{a(x+y)}$$

Solution

$$1\cdots{a(x+y)}$$

$$2\cdots{a(x+y)}= ax+ay$$

Multiplication with different signs
There are only 4 different combination of signs

Example

$$(+4)\times(+5)=+20\cdots(1)$$

''Normally you would not put the plus sign infront of a number, because by convention , it is know to be plus''

$$(-4)\times(+5)=-20\cdots(2)$$

$$(+4)\times(-5)=-20\cdots(3)$$

$$(-4)\times(-5)=+20\cdots(4)$$

Home work

$$(-12)\times(-15)\cdots(1)$$

$$(14)\times(-16)\cdots(2)$$

$$(121)\times(101)\cdots(3)$$

$$(-54)\times(-21)\cdots(4)$$

Square and Cube numbers
Square number are multiply by itself twice

Examples

$$2\times2=2^2=4$$

$$3\times3=3^2=9$$

$$4\times4=4^2=16$$

Home work

Find

$$5^2\cdots(1)$$

$$6^2\cdots(2)$$

$$12^2\cdots(3)$$

Cube number are multiply by itself three times

Examples

$$2\times2\times2=2^3=8$$

$$3\times3\times3=3^3=27$$

$$4\times4\times4=4^3=64$$

Home work

Find

$$5^3\cdots(1)$$

$$6^3\cdots(2)$$

$$12^3\cdots(3)$$

Solving equation with multiplication
Examples

Find x

$$2(x-4)=0\cdots(1)$$

Solution

$$2(x-4)=0\cdots(1)$$

Let x = 4

$$2(4-4)=0\cdots(2)$$

$$2(0)=0\cdots(3)$$

The soulution of

$$2(x-4)=0\cdots(1)$$

is

x = 4 

Home work

Find x

$$5(x+5)=0\cdots(1)$$

$$2(x-3)=0\cdots(2)$$

$$3(2x-1)=0\cdots(3)$$

$$5(20x+5)=0\cdots(4)$$

$$2\left(\frac{10}{x}-5\right)=0\cdots(5)$$

$$8\left(\frac{5}{x}-1\right)=0\cdots(6)$$

Multiplication of decimal number
Examples of decimal numbers

1.5, 1.456 , 54.7454 etc.

1.5

This number has one decimal point

1.456

This number has 3 decimal points

So, to find the number of deciaml point is just to count the number after the point

Example

Whole number multiplication

$$121\times121=14641\cdots(1)$$

Decimal number multiplication

Observe the decimal point carefully

$$12.1\times121=1464.1\cdots(2)$$

$$12.1\times12.1=146.41\cdots(3)$$

$$1.21\times12.1=14.641\cdots(4)$$

$$1.21\times1.21=1.4641\cdots(5)$$

Multiplication of probability
Basic definition of probability:

The likely and unlikely of and event or situation happening

Probability is defined by the fraction

$$P=\frac{Chance}{Outcome}\cdots(1)$$

Example 1:

A dice has 6 sides

So the outcome is 6

Example 2:

A coin has 2 sides

So the outcome is 2

Example 3:

A bag contains 16 marbles

6 are Red

5 are Blue and

5 are Green

So the outcome is 16

Example 1:

A dice has 6 sides

So the outcome is 6

Question 1:

When you throw a dice, what is the chance that the dice land on a side label number 4?

Solution

In a dice we know there are only 6 sides and one side label number 4 on it.

So the chance of it land on side label number 4 are written as

$$P=\frac{chance}{outcome}=\frac{1}{6}\cdots(1)$$

Example 2:

A coin has 2 sides

So the outcome is 2

Question 2:

When you throw a coin, what is the chance is land on head?

Soultion

On a coin we know there are only 2 sides and one side label head on it.

So the chance of it land on head are written as

$$P=\frac{chance}{outcome}=\frac{1}{2}\cdots(2)$$

Example 3:

A bag contains 16 marbles

6 are Red

5 are Blue and

5 are Green

So the outcome is 16

Question 3:

When you pick a marble from the bag, what is the chance that you pick a red marble?

Solution

In the bag we know there are 16 marbles and only 6 of them are red

So the chance of you picking a red marble are written as

$$P=\frac{chance}{outcome}=\frac{6}{16}\cdots(3)$$