OpenLLM-France/wikimedia huggingface dataset:
# Arithmetic Course/Number Operation/Multiplication
## Multiplication

Addition is the operation of adding (combining) two quantities which can be represented by
    A x B

## Operation

    a x b = b x a
    a x b x c = a x (b x c) = (a x b) x c

## Rules

Any number multiply by zero is equal to zero
    a + 0 = a
Any number multiply to itself is
    a x a = a$^2$
Any negative number multiply by itself is
    (-a) x (-a) = a$^2$
Any number multiply to its negative is
    a x (-a) = -a$^2$
# Arithmetic Course/Number Operation/Integration/Indefinite Integration
## Indefinite Integration

Mathematic operation on a function to find the total area under the function's curve . Given a function of x f(x) then the Indefinite Integration of function f(x) has a symbol below
    $\int f(x)dx=Lim_{\Delta x\to 0}\Sigma \Delta x[f(x)+{\frac {f(x+\Delta x)}{2}}]$
Result
    $\int _^f(x)\,dx=F(x)+C=\int f(x)dx=f^{'}(x)+C$

## Integration laws

    $\int {\frac {f^{'}(x)}{f(x)}}{\rm {d}}x=\ln |f(x)|+c$
    $\int {UV}=U\int {V}-\int {\left(U^{'}\int {V}\right)}$
       $e^{x}$ also generates itself and is susceptible to the same treatment.
       $\int {e^{-x}\sin x}~dx=(-e^{-x})\sin x-\int {(-e^{-x})\cos x}~dx$
       $=-e^{-x}\sin x+\int {e^{-x}\cos x}~dx$
       $=-e^{-x}(\sin x+\cos x)-\int {e^{-x}\sin x}~dx+c$
       We now have our required integral on both sides of the equation so
       $=-{\frac {1}{2}}e^{-x}(\sin x+\cos x)+c$
* $f(x)=m$
    $\int mdx=mx+C$
* $f(x)=x^{n}$
    $\int {f(x)}dx={\frac {1}{n+1}}x^{n+1}+c$
* $f(x)={\frac {1}{x}}$
    $\int {\frac {1}{x}}dx=\ln x$
# Arithmetic Course/Number Operation/Subtraction
## Subtraction

Subtraction is an operation where one number is deducted from another. It can be represented by
    A - B

## Operation

    a - b = -b + a
    a - b - c = a - (b + c) = (a - b) - c

## Rules

Any number subtract zero is equal to the number
    a - 0 = a
Any number subtract to itself equals zero
    a - a = 0
Any number subtract its negative is doubled
    a - (-a) = 2a
# Arithmetic Course/Number Operation/Ln
## Natural Log

If there is a power of two, 2$^b$ = a, then
    log$_2$ a = b which is equivalent to Ln a = b

## Rules

* $Lna+Lnb=Logab$
* $Lna-Lnb=Log{\frac {a}{b}}$
# Arithmetic Course/Number Operation/Power
## Power

"Power" is the number of times a number is multiplied by itself
    $a^{n}=\underbrace {a\times \cdots \times a} _{n},$
    a is raised to power of n is equal to a times a times a n times

## Rules

* $a^{0}=1$
* $a^{1}=a$
* $a^{-}1={\frac {1}{a}}$
* $(-a)^{-}1=a^{n}$ . n=2m
       $(-a)^{-}1=-a^{n}$ . n=(2m+1)

## Identities

* $a^{n}+a^{m}=a^{n}(1+a^{m-n))$
* $a^{n}-a^{m}=a^{n}(1-a^{m-n))$
* $a^{n}\times a^{m}=a^{n+m)$
* ${\frac {a^{n}}{a^{m}}}=a^{n-m)$
* $(a^{n})^{m}=a^{nm)$
* $({\frac {a}{b}})^{n}={\frac {a^{n}}{a^{m}}}$
# Arithmetic Course/Number Operation/Root
## Root

Root is an arithmatic operation opposite to power . If there exists $a^{n}=b$ Then ${\sqrt {b}}=a$

## Rules

Square root of zero is zero
    ${\sqrt {0}}=0$
Square root of one is one
    ${\sqrt {1}}=1$
Square root of negative one is an imaginary number i
    ${\sqrt {-1}}=i$

## Identities

* ${\sqrt {a}}{\sqrt {-a}}={\sqrt {-a^{2}}}=i{\sqrt {a^{2}}}$
* ${\sqrt {a}}{\sqrt {a}}=a$
* ${\sqrt {a}}{\sqrt {b}}={\sqrt {ab}}$
* ${\frac {\sqrt {a}}{\sqrt {b}}}={\sqrt {\frac {a}{b}}}$
* ${\sqrt {a^{n}}}=a{\sqrt {a^{n-2)}}$
* $({\sqrt {a}})^{n}=a^{n-2)$
# Arithmetic Course/Number Operation/Integration

Integration is an operation on a function to find the area under the curve of a function
* Indefinite Integration
* Definite Integration
# Arithmetic Course/Number Operation/Integration/Definite Integration
## Indefinite Integration

    $\int _^f(x)\,dx$
    $\int _^f(x)\,dx=F(x)+C$
# Basic Algebra/Exponential Functions/Zero and Negative Exponents
## Lesson

There are two very important things you need to know when working with zero or negative exponents. The rules of them.
First, any number the zero power always equals one. For example -500$^0$ = 1
Second, any number to the negative power is the opposite fraction of the exponent. For example 2$^{-2}$ = 1/2$^2$ = 1/4 because 2$^2$ = 4 and when you have 2$^{-2}$ it becomes its reciprocal to get 1/4.
This is a review of section 2 of the polynomials chapter.

## Practice Problems

Easy
Medium
Hard
# Basic Algebra/Factoring/Chapter Review

Lesson 1. Factors of Integers
Lesson 2. Dividing Monomials
Lesson 3. Monomial Factors of Polynomials
Lesson 4. Multiplying Binomials
Lesson 5. Squares of Binomials
Lesson 6. Factoring x^2 + bx + c
Lesson 7. Factoring x^2 - bx + c
Lesson 8. Factoring x^2 - bx - c and x^2 + bx - c
Lesson 9. Factoring by Grouping
Lesson 10. Factoring Cubic Polynomials
# Basic Algebra/Factoring/Chapter Test

Please post your test problems this way:
* Problem:(answer,level of difficulty(easy, medium, hard))
* Example 2x=6: (x=3,easy)
# Basic Algebra/Introduction to Basic Algebra Ideas/Chapter Review
## Lesson 1. Simple Operations

An operation is a thing you do to numbers. You use signs like: +, –, ×, or ÷ for operations. The Equals Sign is not an operation.

### Adding

    Adding is a way to put two numbers together.
* 1 + 2 = 3

### Subtracting

    Subtracting is a way of taking a number out from another number.
* 2 – 1 = 1

### Multiply

    Multiplying is a way of adding a number many times
* 3 × 2 = 6

### Dividing

    Dividing is a way of subtracting a number many times.
* 6 ÷ 2 = 3

### Example Problems

* 2 + 1 = 3
* 8 + 2 = 10
* 8 – 4 = 4
* 5 – 2 = 3
* 6 × 2 = 12
* 2 × 3 = 6
* 12 ÷ 6 = 2
* 4 ÷ 2 = 2
Back to lesson

## Lesson 2. Exponents and Powers

Exponent is the number on the top that shows.
Base is the number to be multiplied by itself.

### Example Problems

* 6$^2$ = 36
* 2$^3$ = 8
* 4$^2$ = 16
* 5$^3$ = 125
* 2$^4$ = 16
Back to lesson

## Lesson 3. Order of Operations

Math problems are done in this order from top to bottom:
* Parenthesis ( )
* Exponent ^
* Multiply ×, Divide ÷ (Left to Right)
* Add +, Subtract – (Left to Right)

### Example Problem

| $2^{2}+(3\times 4)$ $2^{2}+(12)$ $2^{2}+12$ $4+12$ $16$ | Original problem. Do parenthesis first. Do exponent. Add. Answer |
Back to lesson

## Lesson 4. Working With Negative Numbers

A positive number is a number more than zero.
A negative number is a number less than zero.

### Example Problems

* 6 + (–3) = 3
* 3 + (–9) = –6
* –4 × 4 = -16
* 4 × (–9) = -36
* –2 × (–4) = 8
Back to lesson

## Lesson 5. Solving Equations Using Properties of Mathematics

It is very important to show math in the simplest way. For example, 5/10 is the same as 1/2, but 1/2 is better because it is easier to understand. The simplest answer is usually the best.

### Example Problems

    Find $x$ when $y=9$
* $x=8(y\div 3)$
* $x=24$
* $(x-4)=8+y$
* $x=21$
* $(14+x)\div y=3$
* $x=13$
Back to lesson
# Basic Algebra/Factoring/Squares of Trinomials

Special Products Squares of Trinomials Directions: Please factor the following.
Step 1. Verify that the first term and the third term are both perfect squares.
Step 2. Verify that the middle term is twice the product of the square roots of the first and third term.
Step 3. Use the standard form above to write the factored form.
# Basic Algebra/For Contributors

Welcome to the Basic Algebra project!
Each lesson has 5 sections:
* Vocabulary - simple explanations of new terms.
* Lesson - What does this new skill let us do?
* Example Problems - three example problems worked out in detail.
* Practice Games - links to outside games on the internet that reinforce skills needed for this new skill
* Practice Problems - in three sections (easy, medium, hard). Please list the question, then give the answer in parentheses. Example 2x=6: (x=3)
Remember: All textbooks have many, many problems in common. Don't worry if the problem you add is in a copyrighted textbook. No one owns 2x = 16, for example. Be sure that you do not copy lists of problems. Make sure all word problems you give are completely original as far as subject matter is concerned. Please be sure the subject of your word problem is a BE850 word.
We assume that these students are ready for algebra. We do include a review of basic mathematics ideas in order to direct the students towards integrating their prior math knowledge, especially math vocabulary. This is a standard practice in algebra textbooks.
Also, please log in when you contribute so we can contact each other. Thanks!
# Basic Algebra/Factoring/Factoring a^2-b^2 Binomials
## Lesson

Difference of Squares
Any binomial of the form $a^{2}-b^{2}$ may be written as $(a+b)\cdot (a-b)$. That is
$a^{2}-b^{2}=(a-b)\cdot (a+b)$.
Example 1: Factor $x^{2}-9$.
This is clearly seen just take $a^{2}=x^{2}$ and $b^{2}=9$ so that $b=3$. So $x^{2}-9=(x-3)(x+3)$
Example 2:: $32w^{4}-162$.
Here it is unclear where we can use the difference of squares as 32 is NOT a perfect square. However if we look we see that we can factor out a common factor of 2.
$32w^{4}-162=2(16w^{4}-81)$
Now we see we can use the difference of two squares to simplify matters take $a^{2}=16w^{4}$ and $b^{2}=81$:
$2(16w^{4}-81)=2(4w^{2}-9)(4w^{2}+9)$
Now we notice that we can use the difference of squares again in the first factor to get:
$2(4w^{2}-9)(4w^{2}+9)=2(2w+3)(2w-3)(4w^{2}+9)$
This is now completely factored.
This is brings us to our next point that is that $a^{2}+b^{2}$ is NOT FACTORABLE (at least for the purposes of this class).

## Example Problems

Let a = b
Therefore: a^2 = ab
Therefore: a^2 - b^2 = ab - b^2
Therefore: (a + b)(a - b) = b(a - b)
Now divide both sides by (a - b)
Therefore: a + b = b
But since a = b and substituting b for a
Therefore: b + b = b
Therefore: 2b = b
Now divide both sides by b
Therefore: 2 = 1
QED
This doesn't work because in line 5 both sides of the equation are divided by (a-b). Now, we know a = b and therefore, (a - b) = 0. This means that in line 5, both sides of the equation are divided by 0, which is not allowed. There are many ways that unrealities can be shown, if division by zero is permitted which is why this is not possible.

## Practice Problems

Use ^ for exponentiation
# Basic Algebra/Factoring/Squares of Binomials
## Vocabulary

Binomial - An algebraic expression with exactly two terms.
Square - Multiply a number by itself.
FOIL - The product of two binomials is the sum of the products dio of the First terms, the Outer terms, the Inner terms, and the Last terms.
Quantity -Total amount or number.
Please note that the ‘FOIL’ method as well as the shortcut shown below is ONLY for binomial(s).

## Technical Equation

(a+b)$^2$ = a$^2$ + 2ab + b$^2$
(a-b)$^2$ = a$^2$ - 2ab + b$^2$
NOTE: These equations will work if you substitute the first term for a and the second term for b. You may find these equations easier to understand after the concept is learned using the three steps described below.

## Lesson

Lesson 4 has shown you how to multiply binomials. In Lesson 5 we are going to learn how to square binomials. Squaring a binomial can be done using two different methods. The first method uses FOIL (refer to lesson 4). The second method is a shorter alternative to FOIL. The way we use the shortcut is to follow three simple steps.
Step 1: Square the first term of the binomial.
Step 2: Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2).
Step 3: Square the last term of the binomial.

## Example Problems

Here is an example to follow.
Using the binomial x$^2$+12x+36 we will square it creating the problem (x+6)$^2$
Step 1 Square the first term of the binomial.
(x)$^2$=x$^2$
Step 2 Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2).
[(x)*(6)]*2 = (6x)*(2) = 12x
Step 3 Square the last term of the binomial.
(6)$^2$ = 36
Finally we put the three term we have acquired together and get the answer
(x+6)$^2$ = x$^2$ + 12x + 36
Let’s try another problem that may be a bit more difficult.
Let’s square the binomial (x$^2$-4x) giving us (x$^2$-4x)$^2$
Step 1 Square the first term of the binomial.
(x$^2$)$^2$ = x$^4$
Step 2 Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by 2).
*Notice we keep the negative sign with the second term
[(x$^2$)(-4x)]*2 = (-4x$^3$)*(2) = -8x$^3$
Step 3 Square the last term of the binomial.
(-4x)$^2$ = (-4)$^2$(x)$^2$ = 16x$^2$
Our final answer will be the answers from the three steps combined
(x$^2$-4x)$^2$ = x$^4$ -8x$^3$ + 16x$^2$
Problem 3 (2x-6y)$^2$ = 4x$^2$ - 24xy + 36y$^2$

## Online Lesson

You can use this link to view this lesson illustrated by a teacher.
http://www.phschool.com/atschool/academy123/html/bbapplet_wl-problem-431067.html

## Practice Problems

Use ^ for exponentiation.
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/Map/Indexing

A map can be indexed-by-key (a lookup operation). Use either the C-array-operator, the at function, or the find function. These differ in how they deal with a non-existent key.
Additional information about map indexing (includes interactive examples)
# C++ Language/Std/Stl/CollectionClasses/SequenceCollections/VectorOfBool

A special specialization std::vector<bool> compresses each item to a single bit.
For this special specialization, the indexing operator doesn't return bool&. Instead, it returns a reference to an auxillary class that defines both vecyVar[iIndex].flip and a cast-operator for getting and setting its bool value.
Additional information about a vector of bool (includes interactive examples)
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/Map

A "map" is a lookup-dictionary whose items are key/value pairs.
* Construction
* Sort-Class
* Custom Sort-Class
* Size
* Equality
* Indexing
* Removing Items
* Inserting Items
* Assignment
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/Map/CustomSortClass

You could define your own custom sort-class.
Additional information about using a custom sort-class
# C++ Language/Std/Stl/CollectionClasses/SequenceCollections/Vector/Size

STL implements a vector as having an allocation "capacity" (queried by veciVar.capacity) that may be somewhat larger than that collection's current size (queried by veciVar.size).
Additional information about vector size (includes interactive examples)
# C++ Language/Std/Stl/CollectionClasses/SortedCollections

Some STL collection classes are "sorted", in the sense that the class' items are internally stored using a tree data structure. Each item contains a "key", and the tree data structure maintains a property whereby the items are sorted by that key. As a result, client code can still iterate over the data structure from a "beginning" (item having the smallest key) to an "ending" (item having the largest key), even though that data isn't being linearly-stored. To define the sorting of keys, a sorted collection object is associated with a "sort-class" object.
* A Pair as one Item
* Tuple
* Map
* Multimap
* Set
* Multiset
* Hash-Table
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/HashTable

The standard sorted collections each have an alternative implementation that is based on a hash-table. For example, instead of using std::map<char,int>, you might choose to use std::unordered_map<char,int,std::hash<char>>.
Additional information about hash-table-based implementations
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/Map/Assignment

Assignment using the = operator will remove all existing elements.
Additional information about map assignment (includes interactive examples)
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/Map/Construction

Initializing a map from a literal value uses nested-brace-syntax: std::map<char,int> mapciVar = {{'a',0}, {'e',11}, {'i',22}, {'o',33}};
Additional information about map construction
# C++ Language/Std/Stl/CollectionClasses/SortedCollections/Map/Equality

Map-equality depends on the order of items in their internal-representations (due to sort-class), not on the order of items in any literal-specficiation.
Additional information about map equality (includes interactive examples)
# Blender 3D: Noob to Pro/Advanced Tutorials/Advanced Animation/Guided tour/Armature/index

| previous module: "Advanced Tutorials/Advanced Animation/Guided tour/index" | • Table of Contents • Glossary | next module: "Advanced Tutorials/Advanced Animation/Guided tour/Armature/object" |
The Armature Object in itself is a tool for the animator to move an object or group of vertices in a reliable way. An armature is made of bones, which can be parented to each other, or connected to each other. It was built with the idea of a skeleton in mind. And in fact, in the 3D world with animation inclusive, its considered to be the skeleton.
You can add it using the SPACEKEY in 3Dview and selecting Armature. You'll then enter into Edit Mode where you can add or move bones to build your default rig. An armature has 3 states. You can switch using the dropdown menu in the header of the 3Dview or use the TABKEY to switch between Editmode <-> [Objectmode|Posemode] and CTRL-TABKEY to switch between Objectmode <-->Posemode:
* Object Mode: Your armature is like any other Object, you can move it around the scene, scale it, rotate it and edit options in the button window.
* Edit Mode: Your armature is in what we call rest position, you can modify the bones it contains.
* Pose Mode: Your armature is ready to be animated, each bone can be moved, scaled or rotated, constraints get applied, you can pose your character and animate the bones' behavior over time. Take note that Pose mode is now a state of the armature you can switch on/off using CTRL-TABKEY. So when in Pose, you are still in object mode (you can select another object, contrary to the edit-mode)
Note: The following 3 pages of this tutorial contain screenshots and discuss techniques that are only available in Blender 2.40a and later. Refer to the Blender 2.40a release notes on Armature draw types and Armature envelopes.
| previous module: "Advanced Tutorials/Advanced Animation/Guided tour/index" | • Table of Contents • Glossary | next module: "Advanced Tutorials/Advanced Animation/Guided tour/Armature/object" |
# Blender 3D: Noob to Pro/3D View New

| Properties Window | • Table of Contents • Glossary | Continue With Blender 2.49 3D View |
The 3D View looks slightly different in Blender 2.5x:
The obvious visible difference is the addition of shelves: the Tool Shelf on the left, and the Object Properties on the right. You can toggle the visibility of these by pressing T for the Tool Shelf and N for the Object Properties. Alternatively you can drag the inner edge of the shelf to the window edge to make it disappear. Whichever way you do it, when a shelf is hidden, this symbol will appear in the corner of the window, and can be clicked to make the shelf reappear.
Apart from the above, operating in the 3D view is still largely the same.
However, some keyboard shortcuts from older Blender versions may not work any more. They may yet make a reappearance as development of the 2.5x version continues. But in any case, most of the operations are now also accessible from the menus in the window header, so keyboard shortcuts are no longer the only way to get to them.
# Blender 3D: Noob to Pro/Advanced Tutorials/Advanced Animation

| Introduction to Freestyle | • Table of Contents • Glossary | next module: "Introduction" |
This section will show you the Animation system as it is in Blender 3D. Most of the features will be explained and some tutorials will follow. It is assumed that the user has a good understanding of Blender here.

## Index

* Advanced Animation
** Introduction
** Guided tour:
*** Armature Object
**** Armature Object in Object Mode
**** Armature Object in Edit Mode
**** Armature Object in Pose mode
*** Mesh Object
**** Connection between Armature and Mesh
**** Envelope
**** Vertex Groups & Weight Paint
**** Shape Key
*** Constraints
**** Copy Location
**** Copy Rotation
**** Track-To
**** Floor
**** Locked Track
**** Follow Path
**** Stretch-To
**** IK Solver
*** Timeline Window
*** Graph Editor
*** Dope Sheet
*** NLA Editor
**** Introduction To NLA Editor
**** The Stride feature
** Working example: Bob
*** Build The Rig
*** Deform The Mesh
*** Create A Walk Cycle
| Introduction to Freestyle | • Table of Contents • Glossary | next module: "Introduction" |
# Actually Applicable Application Problems and Brainteasers/Wheelchair Ramp Standards
## Overview

Wheelchair ramps cannot be too steep or they will be too difficult for wheelchair users to roll up, and the wheelchair may collect an unsafe amount of momentum in rolling down. The standard is that the slope may be up to ${\frac {1}{12}}$, meaning that for each 1 foot up, the ramp must go at least 12 feet forward. (The same ratio will work with any unit of measurement, so you can use inches instead of feet, or meters, centimeters, etc.) Flatter is OK; steeper is not.
This type of calculation is Actually Applicable for building planners and construction workers, who use this strategy to find out whether existing ramps need to be replaced and check whether a planned ramp meets the standards, as well as for anyone who cares about making sure their community locations are safe for wheelchair users.

## General Method

* Measure the vertical change between the top and bottom of the ramp.
* Measure the horizontal change between the top and bottom of the ramp.
** You may need to get clever to take these measurements, such as by using a plumbline, bubble level, something with a fixed 90 degree angle, etc.
* Divide the vertical change by the horizontal change to find the ramp's slope.
* Compare the slope with the standard to make sure it is not over the limit.

## Problems

* Does a ramp with a vertical change of 8" and a horizontal distance of 6' meet the standard?
* Does a ramp with a vertical change of 2' and a horizontal distance of 30' meet the standard?

### Make Your Own Problem

Measure a ramp in your community and perform the calculations to make sure it is accessible.
Is the ramp in the picture at https://themighty.com/wp-content/uploads/2017/09/img_2394-1280x640.jpg likely to be accessible? Use measurements from a typical staircase to decide.

## Related Problems

Wheelchair Ramp Standards 2 is about using trigonometry to get around a common measurement complication.
# Ad Hoc Data Analysis From The Unix Command Line/Joining The Data with join

Please note - Join assumes that that input data is sorted based on the key on which the join is going to take place.

## Delimited data

In delimited data, elements of a record are separated by a special 'delimiter' character. In the CSV files, fields are delimited by commas or tabs:
Explanation of options:
0 denotes the match (join) field (needed when using "-a") 1.2 denotes field 2 from file 1 ("j1") 2.2 denotes field 2 from file 2 ("j2").
Using the "-a" option creates a full outer join as in SQL.
This command must be given two and only two input files.

## Multi-file Joins

To join several files you can loop through them.
File "J" is now the full outer join of "j1", "j2".
and so on through j4, j5, ...
For many files this is best done with a loop

## Sorted Data Note

join assumes that the input data has been sorted by the field to be joined. See section on sort for details. • Counting Part 2 - sort and uniq
Credits: Some text adapted from Ted Harding's email to the R mailing list.
← Picking The Data Apart With cut · Counting Part 2 - sort and uniq →