text:
User talk:Johanna~enwikiversity

Welcome
Hello Johanna, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( Wikiversity:Chat, wiki blog ) 17:26, 30 January 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Johanna. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Johanna~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:40, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:12, 19 April 2015 (UTC)

title: User talk:Johanna~enwikiversity


text:
User talk:Riikka~enwikiversity

Welcome
Hello Riikka, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( Wikiversity:Chat, wiki blog ) 17:26, 30 January 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Riikka. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Riikka~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:50, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:20, 19 April 2015 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:Riikka~enwikiversity


text:
User talk:B Maags

Welcome
Hello B Maags, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( Wikiversity:Chat, wiki blog ) 17:28, 30 January 2008 (UTC)

title: User talk:B Maags


text:
Materials Science and Engineering/Equations/Magnetism

Force of Charged Particle
When a charged particle moves through a magnetic field B, it feels a force F given by the cross product:
 * $$\vec{F} = q \vec{v} \times \vec{B}$$

Force on Current-Carrying Wire
The formula for the total force is as follows:
 * $$\mathbf{F} = I \mathbf{L} \times \mathbf{B} \,$$

where
 * F = Force, measured in newtons
 * I = current in wire, measured in amperes
 * B = magnetic field vector, measured in teslas
 * $$\times$$ = vector cross product
 * L = a vector, whose magnitude is the length of wire (measured in metres), and whose direction is along the wire, aligned with the direction of conventional current flow.

Magnetic Field from Steady Current
The magnetic field generated by a steady current (a continual flow of electric charge, for example through a wire, which is constant in time and in which charge is neither building up nor depleting at any point), is described by the Biot-Savart law:

 * $$ d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{I d\mathbf{l} \times \mathbf{\hat r}}{r^2} $$

(in SI units), where
 * $$I$$ is the current,
 * $$d\mathbf{l}$$ is a vector, whose magnitude is the length of the differential element of the wire, and whose direction is the direction of conventional current,
 * $$ d\mathbf{B}$$ is the differential contribution to the magnetic field resulting from this differential element of wire,
 * $$\mu_0$$ is the magnetic constant,
 * $$\mathbf{\hat r}$$ is the unit displacement vector from the wire element to the point at which the field is being computed, and
 * $$r$$ is the distance from the wire element to the point at which the field is being computed.

Magnetic Field Inside Coil - Empty Inductor
$$B = \mu_0 n I\;$$

Energy per Unit Volume of Empty Inductor
$$\frac{B^2}{2 \mu_0} = \frac{\mu_0 n^2 I^2}{2}$$

Total Stored Energy in an Empty Inductor
$$\frac{\mu_0 n^2 Al I^2}{2} = \frac{LI^2}{2}$$

Magnetic Field
$$B = \mu_0 n I + \mu_0 M\;$$ $$B = \mu_0 (H + M)\;$$ $$B = \mu_0 \mu_r H\;$$

Relative Permeability of a Material
$$\mu_r = \frac{ \mathbf B}{\mu_0 \mathbf H}$$ $$\mu_r = 1 + \frac{\mathbf M}{\mathbf H}$$ $$\mu_r = 1 + \Chi_m\;$$

Anisotropy Energy
$$E_{an} = K \sin^2 \phi\;$$


text:
User talk:Nevadaq

About your edit; do you have better information? --JWS 19:56, 30 January 2008 (UTC)

title: User talk:Nevadaq


text:
User:Marianna~enwikiversity

American Sign Language

I would like to learn to communicate in ASL, at a basic level at least. I am sure that in a pinch one could use pen and paper or some sort of e-pad, but just as if I were to visit a foreign country, I would like to be better prepared.

Your suggestions are appreciated.

title: User:Marianna~enwikiversity


text:
User:General maths

user:hashim100

Fraction
Addition and subtracting of fraction

Question 1

$$\cdots\frac{1}{5}+\frac{1}{7}$$

The denumerator is not the same in both fractions, so can not be added

To add fraction the denumerator must be the same

Solution:

Step 1:

$$\cdots\frac{1}{5}\cdot{\frac{7}{7}}+\frac{1}{7}\cdot{\frac{5}{5}}$$

Step 2:

$$\cdots\frac{7}{35}+\frac{5}{35}$$

Same denumerator, just add the numerator

Step 3:

$$\cdots\frac{7+5}{35}$$

Step 4:

$$\cdots\frac{12}{35}$$

Step 5:

$$\cdots\frac{1}{5}+\frac{1}{7}=\frac{12}{35}$$

Question 2

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

Solution

Step 1:

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

Step 2:

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

Step 3:

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

Step 4:

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

Question 3

$$\cdots\frac{1}{x-1}+\frac{1}{x+2}$$

Solution

Step 1:

$$\cdots\frac{1}{x-1}\cdot{\frac{x+2}{x+2}}+\frac{1}{x+2}\cdot{\frac{x-1}{x-1}}$$

Step 2:

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

Step 3:

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

Step 4:

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

Step 5:

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

Question 4

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

Solution

Step 1:

$$\cdots\frac{1}{2(x+3)}\cdot{\frac{(x+2)^2}{(x+2)^2}}+ \frac{1}{(x+2)^2}\cdot{\frac{2(x+3)}{2(x+3)}}$$

Step 2:

$$\cdots\frac{(x+2)^2}{2(x+31)(x+2)^2}+\frac{2(x+3)}{(x+3)(x+2)^2}$$

Step 3:

$$\cdots\frac{(x+2)^2+2(x+3)}{(x+3)(x+2)^2}$$

Step 4:

We need to open bracket this one

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

Step 5:

$$\frac{x^2+4x+4+2(x+3)}{(x+3)(x+2)^2}$$

Step 6:

$$\cdots\frac{x^2+4x+4+2x+6}{(x+3)(x+2)^2}$$

Step 6:

$$\cdots\frac{x^2+6x+10}{(x+3)(x+2)^2}$$

Step 7:

$$\cdots\frac{1}{2(x+3)}+\frac{1}{(x+2)^2}=\frac{x^2+6x+10}{(x+3)(x+2)^2}$$

Law of indices
Multiplication of indices

$$1\cdots2\cdot2\cdot2=2^3$$

$$2\cdots2\cdot2\cdot2\cdot2=2^4$$

$$3\cdots2\cdot2\cdot2\cdot2\cdot=2^5$$

$$4\cdots2^3\cdot2^3=2^{2+3}=2^5$$

When the you are multiplying base with different indices, you add the indices raised to the common base

An example

$$1\cdots3^1\cdot3^3\cdot3^5=3^{1+3+5}=3^9$$

General formula of multiplication of indices

$$1\cdots{b^n}\cdot{b^m}=b^{n+m}$$

Home work

Add the following fraction

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

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

$$3\cdots\frac{x}{y}+\frac{1}{x+y}$$

$$4\cdots\frac{2}{x+2}-\frac{1}{x+1}$$

Solve the following indices

$$1\cdots{b^2}\cdot{b^x}$$

$$2\cdots{b^x}\cdot{c^y}$$

$$3\cdots3^a\cdot3^3\cdot3^4$$

$$4\cdots3^{\frac{1}{2}}\cdot3^{\frac{3}{4}}$$

Important arithmetic operation
$$1\cdots+3+3+3=+9$$

$$2\cdots-3-3-3=-9$$

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

Is plus because the 9 is greater than 6

$$4\cdots-9+6=-(9-6)=-3$$

Is minus because the 9 is greater than 6

Younger brother Lesson two : 3/2/8
1 Long addition

2 Addition of like and unlike terms

3 Addition of decimal

4 Addtion of fraction

5 Property of addition

6 Addtion of zero

7 Repeat addition relate to multiplication

Long Addition
Home work

$$1234+4351\cdots(1)$$

$$67464+5647\cdots(2)$$

$$45334+43351\cdots(3)$$

Addition of like and unlike terms
Examples

$$a+a+a=3\times{a}=3a\cdots(1)$$

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

$$a+b=a+b\cdots(3)$$

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

$$xy+xyz=xy+xyz\cdots(5)$$

Home work

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

$$x+y+y+4\cdots(2)$$

$$x^2+y^2+x^2\cdots(3)$$

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

$$(xy)^2+(yx)^2+xy\cdots(5)$$

Addition of fraction
Examples

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

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

Definition of fraction:

Part over whole

$$\frac{part}{Whole}\cdots(1)$$

The part is called the numerator

The whole is called the denumerator

Addition of fraction:

When you are adding fractions you are adding the part of individual fraction together

Rule of addition of fraction states:

You can only add the part together if and only the whole of each fraction are the same

Example will clear the matter

Example 1:

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

This fraction you can add

Solution

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

Example 2:

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

This fraction you can not add

Solution

The soultion will be, you have to use basic rule of arithmetic to change both fractions to a common whole

To change 2 or more fractions to a common whole

Example 1:

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

$$\frac{2}{6}\times{\frac{5}{5}}+\frac{1}{5}\times{\frac{6}{6}}\cdots(2)$$

$$\frac{12}{30}+\frac{6}{30}\cdots(3)$$

Now the whole are the same, just add the parts together

$$\frac{12}{30}+\frac{6}{30}=\frac{12+6}{30}=\frac{18}{30}\cdots(4)$$

Example 2:

$$\frac{1}{2}+\frac{2}{3}+\frac{4}{6}\cdots(1)$$

$$\frac{1}{2}\times{\frac{3}{3}}+\frac{2}{3} \times{\frac{2}{2}}+\frac{4}{6}\times{\frac{1}{1}}\cdots(2)$$

$$\frac{3}{6}+\frac{4}{6}+\frac{4}{6}\cdots(3)$$

Now the whole are the same, just add the parts together

$$\frac{3}{6}+\frac{4}{6}+\frac{4}{6}=\frac{3+4+4}{6}=\frac{11}{6}\cdots(4)$$

Home work
$$\frac{1}{3}+\frac{1}{6}\cdots(1)$$

$$\frac{2}{4}+\frac{5}{6}\cdots(2)$$

$$\frac{2}{4}+\frac{5}{12}+\frac{1}{3}\cdots(3)$$

$$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\cdots(4)$$

Lesson two 3/1/8
In this lesson we will briefly cover the basic of these topics

1: Formulae, see user:hashim100 number:39

2: Trigonometry

3: The circle

4: The areas of triangle

5: Simultaneous equations, see user:hashim100

Trigonometry
We have a triangle ABC

The lenghts of the traiangle ABC will be represent by small case letter

The lenght of a is opposite angle A, b opposite angle B and c opposite angle C

The ratio a over b is defined by

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

The ratio b over c is defined by

$$CosA=\frac{b}{c}\cdots(2)$$

The ratio a over b is defined by

$$TanA=\frac{a}{c}\cdots(3)$$

Application

If given angle A and lenght c, for example

We can find out all the other lenghts and angle, using the above defined ratio

If given lenght c and lenght a, for example

We can find out all the other lenghts and angle, using the above defined ratio

Sin(A) : 30 - 45 - 60 - 90
$$sin(30)=\frac{1}{2}\cdots(1)$$

$$sin(45)=\frac{1}{\sqrt{2}}\cdots(2)$$

$$sin(60)=\frac{\sqrt3}{2}\cdots(3)$$

$$sin(90)=1\cdots(4)$$

Cos(A) : 30 - 45 - 60 - 90
$$cos(30)=\frac{\sqrt{3}}{2}\cdots(1)$$

$$cos(45)=\frac{1}{\sqrt{2}}\cdots(2)$$

$$cos(60)=\frac{1}{2}\cdots(3)$$

$$cos(90)=0\cdots(4)$$

Tan(A) : 30 - 45 - 60 - 90
$$Tan(30)=\frac{1}{\sqrt{3}}\cdots(1)$$

$$Tan(45)=1\cdots(2)$$

$$Tan(60)=\sqrt{3}\cdots(3)$$

$$Tan(90)=Undefine!\cdots(4)$$

Derivation of the above angles are from the following special shapes

Equilateral triangle and a Square

Unity cirlce
Mean that the circle with the radius of 1 unit

This is very important and need to be expalin on the board

The solution and areas of traingles
The area of triangle can be determine in many, depend on the information is given

''If the base and height is given. The area of the triangle can be determine by this formula:''

$$A=\frac{b\times{h}}{2}\cdots(1)$$

''If all the sides are given. The area of the triangle can be determine by this formula:''

$$A=\sqrt{s(s-a)(s-b)(s-c)}\cdots(1)$$

Where s is called the semi-perimeter and is defined by:

$$s=\frac{a+b+c}{2}\cdots(2)$$

Angle C is between lenghts a and b. The area of the triangle can be determin by this formula:

$$A=\frac{a\times{b}\times{SinC}}{2}\cdots(1)$$

Example 1:

Base b = 4 and Height h = 5

$$A=\frac{b\times{h}}{2}$$

$$A=\frac{4\times5}{2}=\frac{20}{2}=10\cdots(1)$$

Example 2:

Lenghts a = 3, b = 4 and c = 5

$$s=\frac{a+b+c}{2}\cdots(1)$$

$$s=\frac{3+4+5}{2}=\frac{12}{2}=6\cdots(2)$$

$$A=\sqrt{s(s-a)(s-b)(s-c)}\cdots(3)$$

$$A=\sqrt{6(6-3)(6-4)(6-5)}=\sqrt{6(3)(2)(1)}=\sqrt{36}=6\cdots(4)$$

Example 3:

Angle C = 30 and lenghts a = 4 and b =5

$$A=\frac{a\times{b}\times{SinC}}{2}\cdots(1)$$

$$A=\frac{4\times{5}\times{Sin30}}{2}\cdots(2)$$

$$A=\frac{20\times{\frac{1}{2}}}{2}\cdots(3)$$

$$A=\frac{10}{2}=5\cdots(4)$$

The Circle
In the cirlce we have r, D , π , s , C , A , A(s) and Φ

r means radius

and it measured from the centre to the circumference

D means diameter and is equal to 2r

$$D=2\times{r}=2r\cdots(1)$$

π is a constant and is equal to roughly 3.14

$$\pi=3.14...$$

and is defined by

$$\pi=\frac{C}{D}\cdots(2)$$

s means the lenght of the sector

and is defined by

$$s=\theta\times{r}=\theta{r}\cdots(3)$$

C means the circumference

and is defined by

$$C=\pi\times{D}=\pi{D}\cdots(4)$$

A means the area

and is defined by

$$A=\pi\times{r^2}=\pi{r^2}\cdots(5)$$

A(s) means the area of the sector

and is defined by

$$A_s=\frac{\theta\times{r^2}}{2}=\frac{\theta{r^2}}{2}\cdots(6)$$

Φ means the angle subtended by the radii (plural of radius) or (the angle in between the radii)

title: User:General maths


text:
User:Hashim100~enwikiversity

Write to me if any problem

E-mail : a.1980@hotmail.co.uk

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

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

title: User:Hashim100~enwikiversity


text:
User:Sisco~enwikiversity

Hello. First let me say that I am NOT a teacher, well at least not an official one. This is a cool site for a nerd like me. I promise to follow the rules.

title: User:Sisco~enwikiversity


text:
User talk:Hashim100~enwikiversity

Welcome
Hello Hashim100, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --mikeu talk 02:07, 31 January 2008 (UTC)

fractions
I'm not sure exactly what you are interested in but there are a couple of pages that you might find helpfull at Primary mathematics:Working with fractions and Primary mathematics:Fractions. Feel free to leave a message on my talk page if you need any help. --mikeu talk 02:32, 31 January 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Hashim100. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Hashim100~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:37, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:10, 19 April 2015 (UTC)

title: User talk:Hashim100~enwikiversity


text:
User talk:General maths

Welcome
Hello General maths, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --mikeu talk 02:11, 31 January 2008 (UTC)

title: User talk:General maths


text:
Introduction to approximate quantum chemistry

Approximate quantum chemistry is a range of approaches and methods that do not aim to solve the quantum mechanical equations in a rigorous manner, but rather to use simple ideas and quantities derived from experiment to obtain insight and rough predictions of chemical quantities. The earliest method, Hückel method, was developed long before digital computers were common and all computatonal methods use few computer resources. Since in computational chemistry the resources used rise, often steeply, with the size of the molecules being studied, these methods can now be applied to very large molecules.

The material will branch out from the links below. The first two sections of this module introduce you to some of the fundamental concepts of molecular orbital (MO) theory. Sections 3 and 4 explore some practical implementations of this theory.

 * 1) Qualitative MO theory
 * 2) Quantitative MO theory
 * 3) Empirical MO methods
 * 4) * Hückel method
 * 5) * Extended Hückel method
 * 6) Semi-empirical MO methods


text:
Qualitative MO theory

Qualitative molecular orbital (MO) theory is a way of building up a simple picture of the molecular orbitals of a molecule using a few basic rules.

The ideas discussed in this section will allow you to make 'back-of-the-envelope' predictions and allow you to interpret the results from more sophisticated calculations.

 * 1) Atomic orbitals to molecular orbitals
 * 2) Molecular orbitals for H2O and CH2
 * 3) Symmetry tutorial
 * 4) MOs for methyl: CH3., CH3+, CH3-
 * 5) Walsh diagrams
 * 6) Qualitative MO theory comprehension test

title: Qualitative MO theory


text:
Quantitative MO theory

The chemist's qualitative picture of molecular orbitals can be rigorously derived from a quantum chemical approach. In this section we give a very brief outline of the theory.

Schrödinger equation
In 1925 Erwin Schrödinger and Werner Heisenberg independently formulated a general quantum theory. Although the two formulations are mathematically equivalent, Schrödinger presented his theory in terms of partial differential equations and, within this framework, the energy of an isolated molecule can be obtained by the solution of the Schrödinger equation. In its time-independent form, this can be written as:

 * $$\hat{H} \Psi = E \Psi $$

where $$\hat{H}$$ is the Hamiltonian operator, $$\Psi$$ is the wavefunction, and E is the energy of the system relative to the state in which the nuclei and electrons are infinitely separated and at rest.

The masses of the nuclei are much larger and their velocities much smaller than those of the electrons, and it is possible to simplify the solution of the Schrödinger equation by separating it into two parts, one describing the motions of the electrons in a field of fixed nuclei and the other describing the motions of the nuclei. This is known as the adiabatic or Born-Oppenheimer approximation. Molecular orbital theory is concerned with finding approximate solutions to the first part, that is, the electronic Schrödinger equation:
 * $$\hat{H}^e \Psi^e = E^e \Psi^e $$

Each quantity is implicitly a function of the nuclear co-ordinates.

The orbital approximation
The orbital approximation simplifies the above equation by assuming that each electron is associated with a separate one-electron wavefunction or spin orbital, [chi]. Thus, Hartree proposed that the wavefunction could be expressed simply as a product of spin orbitals, one for each electron:

 * $$ \psi = \chi_1 (1) \chi_2 (2) ... \chi_n (n)\,\! $$

The LCAO approximation
Each spin orbital is actually a product of a spatial function, $$\chi_i (x,y,z)\,\!$$, and a spin function, α or β. The spatial molecular orbitals, $$\Phi_i\,\!$$, are usually expressed as linear combinations of a finite set of known one-electron functions. In the simplest case where these functions take the form of the atomic orbitals of the constituent atoms, this expansion is called a linear combination of atomic orbitals (LCAO):

 * $$\Phi_i = c_{1i}\chi_1 + c_{2i}\chi_2 + ... c_{Ni}\chi_N\,\!$$

Qualitatively, this is like saying that the two molecular orbitals in H2 are linear combinations of the 1s atomic orbitals:

Φ2 = 0.5χa - 0.5χb Φ1 = 0.5χa + 0.5χb

One-electron approximations
The simplest way to solve the Schrödinger equation is to treat each electron in isolation from the rest. This "one-electron approximation" leads to Hückel theory and Extended Hückel theory.

title: Quantitative MO theory


text:
Empirical MO methods

Explanation of the empirical MO approach
Empirical methods are differentiated from semi-empirical methods in general by the choice of Hamiltonian operator. Empirical methods use a Hamiltonian which is simply a sum of one-electron terms. The two-electron electron repulsion term is in effect averaged into the one-electron terms. In contrast, semi-empirical methods use the correct Hamiltonian with both one- and two-electron terms. In both cases the resulting integrals that arise from the terms in the Hamiltonian are often approximated, neglected or fitted to experimental data.

All the methods we will discuss, both empirical and semi-empirical, use the molecular orbital single determinant method, although, in general, this is not a restriction. The PCILO (perturbative configuration interaction using localised orbitals) method, for example, uses an approach which goes beyond the molecular orbital approach while still using the same approximations as other semi-empirical methods.

The Hückel method
The earliest empirical method is the Hückel method which is applicable to planar aromatic or conjugated systems. In fact, the Hückel method, in 1930, was the first application of quantum mechanics to a polyatomic system.

The earliest and simplest method is restricted to hydrocarbons and we will start with that approach.

Later the method was extended to include so-called heteroatoms (i.e. not carbon) in order to study heterocyclic aromatic systems.

The Hückel method calculates the π electron molecular orbitals only. This is often described by the term σ-π separability. Another term that is often used is the π electron approximation.

Now we can start looking at Hückel approximations in detail and then do some actual calculations.

Go to Hückel method - Go to Extended Hückel method

title: Empirical MO methods


text:
Sigma - pi separability

Sigma - pi separability
There is nothing magical about the separation of σ and π electrons. It is a simple consequence of symmetry. If we have a planar compound, then reflection in the plane that contains all the atoms is a symmetry operation in the point group that the molecule belongs to. For example, ethene (ethylene) has D2h symmetry and reflection in the plane of the 6 atoms is one of the 3 reflection operations of the D2h point group.

All molecular orbitals must transform like one of the irreducible representations of the group. This jargon simply indicates that for these planar compounds the molecular orbitals must be symmetric or antisymmetric with respect to reflection in the plane. Molecular orbitals which are symmetric, i.e the same above and below the plane, are called σ orbitals. Those that are antisymmetric, i.e. they change sign going through the plane, are called π orbitals. σ molecular orbitals must be formed from symmetric atomic orbitals. This means that only s orbitals and the px and py orbitals (we specify the xy axis as being in the plane and the z axis being perpendicular to the plane) can contribute to the σ orbitals. Similarly only antisymmetric atomic orbitals can contribute to the π molecular orbitals.

This means that only the 2pz orbital contributes to the π molecular orbitals. Note that in these empirical methods we are only using atomic orbitals that are occupied in the free atom. This is called a minimal basis set. We are not using d orbitals.

The simple conclusion from this is that the π molecular orbitals are built from just one atomic orbital (the 2pz orbital) on each heavy atom (carbon for hydrocarbons). The hydrogen atomic orbitals only contribute to the σ orbitals.

In a proper treatment, the form of the π orbitals is determined by the nature of the σ orbitals and vice versa because of the repulsion between the π and σ electrons. However, in an empirical method, we simply fudge in the repulsion from the σ electrons into the terms we include in the calculation of the π orbitals.

Return to Empirical MO methods

title: Sigma - pi separability


text:
User talk:Marianna~enwikiversity

Welcome
Hello Marianna, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Gbaor 07:03, 31 January 2008 (UTC)

American Sign language
Hi Marianna, there seems to be a decent list of resources at the bottom of the Wikipedia article Sign language (I haven't looked at American sign language), and there's also a very much work-in-progress Wikibook: American Sign Language. Following this link: American sign language will take you to creating a new page on Wikiversity - you can form this around a set of learning goals for example - I can help if you like. Cormaggio talk 10:43, 31 January 2008 (UTC)

Thank you so much for the information and your suggestions. I learned the alphabet already! As soon as I have a draft for Learning Goals I will post the link here, so that you and I can discuss and receive feedback from others, as well as their contributions. Marianna 19:03, 8 February 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Marianna. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Marianna~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:42, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:15, 19 April 2015 (UTC)

title: User talk:Marianna~enwikiversity


text:
User talk:Sisco~enwikiversity

Welcome
Hello Sisco, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Gbaor 07:03, 31 January 2008 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called Sisco. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name Sisco~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:43, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:22, 19 April 2015 (UTC)

title: User talk:Sisco~enwikiversity


text:
Hamiltonian operator

Hamiltonian operator tutorial
The Hamiltonian contains one- and two-electron terms. The two-electron terms (summed over i and j) are just the repulsion potential energies between all pairs of electrons. Thus:

 * $$\hat{H} = \sum_i\hat{h}_i + \sum_i\sum_j \frac{1}{r_{ij}} $$

$$1/r_{ij}\,\!$$ is the repulsion between a pair of electrons (distance $$r_{ij}\,\!$$ apart).

The one-electron terms (summed over i) are more varied. For each electron, there is a kinetic energy term and a sum of attractive potential energy terms for each nucleus, A, in the molecule.

 * $$\hat{h}_i = -\frac{1}{2} \nabla_i^2 - \sum_A \frac{Z_A}{r_{Ai}} \,\!$$

$$-1/2 \nabla_i^2\,\!$$ is the kinetic energy term with:

 * $$\nabla_i^2 = \frac{\partial^2}{\partial x_i^2} + \frac{\partial^2}{\partial y_i^2} + \frac{\partial^2}{\partial z_i^2}\,\!$$

$$Z_A/r_{Ai}\,\!$$ is the coulombic attraction between electron i and nucleus A.

$$Z_A\,\!$$ is the nuclear charge (atomic number) of atom A and $$r_{Ai}\,\!$$ is the distance between electron i and nucleus A.

Return to Quantitative MO theory - Return to Empirical MO methods

title: Hamiltonian operator


text: Template:Computational Chemistry

title: Template:Computational Chemistry


text:
User talk:Passawuth

Welcome
Hello Passawuth, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Gbaor 12:16, 31 January 2008 (UTC)

title: User talk:Passawuth


text:
User talk:Padraigdoran

Welcome
Hello Padraigdoran, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Gbaor 12:18, 31 January 2008 (UTC)

title: User talk:Padraigdoran


text:
Radio Discussion

Welcome to the Radio Discussion resource! Participants in Radio Discussions create online reading materials to compliment radio programs, which serve as "lectures". A number of radio program sources can be used, such as shows from Public Radio and other media outlets that host in-depth stories and discussions.

Science

 * /The Science Show/: The Science Show (ABC), about science
 * /Living on Earth/: Living on Earth (PRI), about environmental issues

Religion

 * /Speaking of Faith/: Speaking Of Faith (PRI), about religion and spirituality

Media Studies

 * /On the Media/: On The Media (NPR), about the news media

title: Radio Discussion


text:
Radio Discussion/Living on Earth

Living on Earth is a weekly 1-hour show from PRI which discusses environmental issues. Podcasts are available both through the website (http://www.loe.org) and on iTunes.

January, 2008

 * Show broadcast January 18, 2008

February, 2008

 * Show broadcast February 1, 2008

title: Radio Discussion/Living on Earth


text: Category:Radio Discussion

title: Category:Radio Discussion


text:
Talk:Radio Discussion/Living on Earth

Short note about organization of the page on the example of "Biofuels": I think we shoud reorganize the page and make it more simple. There are plenty of titles but no content and extremely long table of contents. I know it is just the beginning and the page is about an hour old... My suggestion is this: Keep just the title ("Biofuel" in this case) and a short introduction whit the WP link as it is done now. Under it simple a "Discussion" subtitle. At the beginning we could encourage the readers/participants to add photos, ext links and other stuff. After it we could organize each section if necessary. --Gbaor 14:09, 31 January 2008 (UTC)

title: Talk:Radio Discussion/Living on Earth


text:
Radio Discussion/Living on Earth/2008.1.18

This show was first broadcast on January 18, 2008. Audio downloads and a transcript are available on this website:
 * http://www.loe.org/shows/shows.htm?programID=08-P13-00003

The links for this week's show were provided by:
 * User:SB_Johnny
 * User:Gbaor
 * User:Mu301

Participant feedback will be in Category:Radio Discussions feedback/Living on Earth/2008.1.18

Biofuel the Best Bet?
Biofuels are controversial because they use the same resources as the human food supply.

A Spoonful of CAFÉ
Corporate Average Fuel Economy regulations in the United States.

Waste Not....Here
Yucca Mountain (radioactive waste storage site)

Cloned Little Piggies Going to Market?
Meat from cloned animals.

From Toilet to Tap
Recycled drinking water.

Water Calculator
Water use calculator.

Cool Fix For a Hot Planet
Tips for reducing your carbon footprint and conserving water.


text:
Materials Science and Engineering/Diagrams/Device Fabrication Processes

Schematic of MBE growth system
Any evaporation system includes a vacuum pump. It also includes an energy source that evaporates the material to be deposited. Many different energy sources exist:

 * In the thermal method, the source material is placed in a crucible, which is radiatively heated by an electric filament. Alternatively, the source material may be hung from the filament itself (filament evaporation).
 * Molecular beam epitaxy is an advanced form of thermal evaporation.
 * In the electron-beam method, the source is heated by an electron beam with an energy up to 15 keV. See Electron beam evaporation.
 * In flash evaporation, a fine wire of source material is fed continuously onto a hot ceramic bar, and evaporates on contact.
 * Resistive evaporation is accomplished by passing a large current through a wire or foil of the material that is to be deposited.

Some systems mount the substrate on an out-of-plane planetary mechanism. The mechanism rotates the substrate simultaneously around two axes, to reduce shadowing.

 * Heat shielding
 * Mass spectrometer
 * Ionization gauge
 * Electron gun
 * Thermocouple
 * Quartz-crystal thickness monitor
 * Hot plate
 * Substrate
 * Holder
 * View port
 * Mechanical shutter
 * Sb effusion cell

Design, process, and control parameter
Design parameters
 * Hearth design
 * H

Process parameters
 * Pback
 * Tsor
 * Tsub

Control parameters
 * film thickness
 * impurity concentration
 * defect density

Traveling from source to substrate

 * Evaporant flux at substrate
 * Background gas flux at substrate
 * Mean free path
 * Evaporation
 * Evaporant travels at line of sight

Ion Implantation

Ion implantation is a materials engineering process by which ions of a material can be implanted into another solid, thereby changing the physical properties of the solid. Ion implantation is used in semiconductor device fabrication and in metal finishing, as well as various applications in materials science research. The ions introduce both a chemical change in the target, in that they can be a different element than the target, and a structural change, in that the crystal structure of the target can be damaged or even destroyed.

Ion implantation equipment typically consists of an ion source, where ions of the desired element are produced, an accelerator, where the ions are electrostatically accelerated to a high energy, and a target chamber, where the ions impinge on a target, which is the material to be implanted. Each ion is typically a single atom, and thus the actual amount of material implanted in the target is the integral over time of the ion current. This amount is called the dose. The currents supplied by implanters are typically small (microamperes), and thus the dose which can be implanted in a reasonable amount of time is small. Thus, ion implantation finds application in cases where the amount of chemical change required is small.

Typical ion energies are in the range of 10 to 500 keV (1,600 to 80,000 aJ). Energies in the range 1 to 10 keV (160 to 1,600 aJ) can be used, but result in a penetration of only a few nanometers or less. Energies lower than this result in very little damage to the target, and fall under the designation ion beam deposition. Higher energies can also be used: accelerators capable of 5 MeV (800,000 aJ) are common. However, there is often great structural damage to the target, and because the depth distribution is broad, the net composition change at any point in the target will be small.

Atomic Layer Deposition
Atomic Layer Deposition (ALD) is a gas phase chemical process used to create extremely thin coatings. The majority of ALD reactions use two chemicals, typically called precursors. These precursors react with a surface one-at-a-time in a sequential manner. By exposing the precursors to the growth surface repeatedly, a thin film is deposited.

ALD is a self-limiting, sequential surface chemistry that deposits conformal thin-films of materials onto substrates of varying compositions. ALD is similar in chemistry to chemical vapor deposition (CVD), except that the ALD reaction breaks the CVD reaction into two half-reactions, keeping the precursor materials separate during the reaction. ALD film growth is self-limited and based on surface reactions, which makes achieving atomic scale deposition control possible. By keeping the precursors separate throughout the coating process, atomic layer control of film grown can be obtained as fine as ~ 0.1 angstroms per monolayer.

ALD has unique advantages over other thin film deposition techniques, as ALD grown films are conformal, pin-hole free, and chemically bonded to the substrate. With ALD it is possible to deposit coatings perfectly uniform in thickness inside deep trenches, porous media and around particles. The film thickness range is usually 1-500 nm.

ALD can be used to deposit several types of thin films, including various ceramics, from conductors to insulators.

Reactive-Ion Etching
Reactive ion etching (RIE) is an etching technology used in microfabrication. It uses chemically reactive plasma to remove material deposited on wafers. The plasma is generated under low pressure (vacuum) by an electromagnetic field. High-energy ions from the plasma attack the wafer surface and react with it.

Equipment
A typical (parallel plate) RIE system consists of a cylindrical vacuum chamber, with a wafer platter situated in the bottom portion of the chamber. The wafer platter, which is usually grounded, is electrically isolated from the rest of the chamber. Gas enters through small inlets in the top of the chamber, and exits to the vacuum pump system through the bottom. The types and amount of gas used vary depending upon the etch process; for instance, sulfur hexafluoride is commonly used for etching silicon. Gas pressure is typically maintained in a range between a few millitorr and a few hundred millitorr by adjusting gas flow rates and/or adjusting an exhaust orifice.

Other types of RIE systems exist, including inductively coupled plasma (ICP) RIE. In this type of system, the plasma is generated with an RF powered magnetic field. Very high plasma densities can be achieved, though etch profiles tend to be more isotropic.

A combination of parallel plate and inductively coupled plasma RIE is possible. In this system, the ICP is employed as a high density source of ions which increases the etch rate, whereas a separate RF bias is applied to the substrate (silicon wafer) to create directional electric fields near the substrate to achieve more anisotropic etch profiles.

Deep Reactive Ion Etching

A special subclass of RIE which continues to grow rapidly in popularity is deep RIE (DRIE). In this process, etch depths of hundreds of micrometres can be achieved with almost vertical sidewalls. The primary technology is based on the so-called "Bosch process", named after the German company Robert Bosch which filed the original patent, where two different gas compositions are alternated in the reactor. Currently there are two variations of the DRIE. The first variation consists of three distinct steps (the Bosch Process as used in the UNAXIS tool) while the second variation only consists of two steps (ASE used in the STS tool). In the 1st Variation, the etch cycle is as follows: (i) SF6 isotropic etch; (ii) C4F8 passivation; (iii) SF6 anisoptropic etch for floor cleaning. In the 2nd variation, steps (i) and (iii) are combined.

Both variations operate similarly. The C4F8 creates a polymer on the surface of the substrate, and the second gas composition (SF6 and O2) etches the substrate. The polymer is immediately sputtered away by the physical part of the etching, but only on the horizontal surfaces and not the sidewalls. Since the polymer only dissolves very slowly in the chemical part of the etching, it builds up on the sidewalls and protects them from etching. As a result, etching aspect ratios of 50 to 1 can be achieved. The process can easily be used to etch completely through a silicon substrate, and etch rates are 3-4 times higher than wet etching.

Chemical Vapor Deposition
Chemical vapor deposition (CVD) is a chemical process used to produce high-purity, high-performance solid materials. The process is often used in the semiconductor industry to produce thin films. In a typical CVD process, the wafer (substrate) is exposed to one or more volatile precursors, which react and/or decompose on the substrate surface to produce the desired deposit. Frequently, volatile byproducts are also produced, which are removed by gas flow through the reaction chamber.

Physical Vapor Deposition

Physical vapor deposition (PVD) is a general term used to describe any of a variety of methods to deposit thin films by the condensation of a vaporized form of the material onto various surfaces (e.g., onto semiconductor wafers). The coating method involves purely physical processes such as high temperature vacuum evaporation or plasma sputter bombardment rather than a involving a chemical reaction at the surface to be coated as in chemical vapor deposition.

Variants of PVD include, in order of increasing novelty:

 * Evaporative deposition -- In which the material to be deposited is heated to a high vapor pressure by electrically resistive heating in "high" vacuum.
 * Electron Beam Physical Vapor Deposition -- In which the material to be deposited is heated to a high vapor pressure by electron bombardment in "high" vacuum.
 * Sputter deposition -- In which a glow plasma discharge (usually localized around the "target" by a magnet) bombards the material sputtering some away as a vapor.
 * Cathodic Arc Deposition - In which a high power arc directed at the target material blasts away some into a vapor.
 * Pulsed laser deposition - In which a high power laser ablates material from the target into a vapor.

LOCOS

LOCOS, short for LOCal Oxidation of Silicon, is a microfabrication process where silicon dioxide is formed in selected areas on silicon wafer having the Si-SiO2 interface at a lower point than the rest of the silicon surface.

This technology was developed to insulate MOS transistors from each-other. The main goal is to create an silicon oxide insulating structure that penetrates under the surface of the wafer, so that the Si-SiO2 interface occur at a lower point than the rest of the silicon surface. This cannot be easily achieved by etching field oxide. Thermal oxidation of selected regions surrounding transistors is used instead. The oxigen penetrates in depth of the wafer, reacts with silicon and transforms it in to silicon oxide. In this way, an immersed structure is formed.

The immersed insulating barrier limits the transistor cross-talk.

Typical process steps are the following:

 * 1) Preparation of silicon substrate (layer 1)
 * 2) CVD deposition of SiO2, pad/buffer oxide (layer 2)
 * 3) CVD deposition of Si3N4, nitride mask (layer 3)
 * 4) Etching of nitride layer (layer 3) and silicon oxide layer (layer 2)
 * 5) Thermal growth of silicon oxide (structure 4)
 * 6) Furhter growth of thermal silicon oxide (structure 4)
 * 7) Removal of nitride mask (layer 3)

There are 4 basic layers/structures:

 * Si, silicon substrate, wafer
 * 1) SiO2, buffer oxide (pad oxide), chemical vapor deposition silicon oxide
 * 2) Si3N4, nitride mask
 * 3) SiO2, insulation oxide, thermal oxide