Thursday, September 16, 2010

Transformations in math

Introduction about Transformations in math
Transformation math is planning a function to a set x into a different set or mapping itself. Fundamentally the transformation can be distinct into subsequent types. The first one is rigid transformation and the next one is non- rigid transformation. Rigid transformation when the transformation protects its qualified space then it is called rigid transformation.

It has the subsequent factors. The rigid transformations Composition are rigid. The rigid transformations of contradictory are rigid. The image and pre-image has the specific dimension and outline. To get more help on all study material

Types of Transformations in Math:
  • Translation
  • Rotation
  • Reflection
  • Dilation
  • Glide reflection

Wednesday, September 15, 2010

Rules on Algebra

Introduction:
The Basic algebra rules are,
  1. Commutative property
  2. Associative property
  3. Distributive property
  4. Identities
  5. Order of operation
The most important rule in basic algebra is the order of operations based on these the numerical expression are solved because the numerical expression are contains more than one operation, a rule is needed to analyze which operation we use first, this rule is called the order of operation. This could also help us on liter conversion.

Monday, September 6, 2010

Help with Subtraction Polynomials

Subtracting Polynomials through further three terms have no special names. Polynomials can be used through the articulate geometric relationships.

Types of Polynomial:
  • Monomial – A polynomial of one idiom is defined as monomial.
  • Binomial – A polynomial of two languages is known as binomial.
  • Trinomial – A polynomial of three form is known as trinomial.
  • Linear polynomial – A polynomial of quantity one is called a linear polynomial.
  • Quadratic polynomial – A polynomial of degree two is known as quadratic polynomial.
  • Cubic polynomial – A polynomial of degree three is namely represented as cubic polynomial.
Subtracting Polynomial:
Each appearance is a polynomial. A polynomial is able to have variables, exponents and constants. To subtracting polynomials, you can group like expressions parallel or write them in column form, aligning like terms. This could also help us on quadratic equation calculator

Steps involving the subtracting of polynomials:
Step 1: Write the given polynomials
Step 2: combine like terms and remove zero pairs.
Step 3: Subtract the like terms

Friday, September 3, 2010

Algebra help

Introduction to online free algebra helper:
Learning Online is one of the best tools for students who are studying from their home.In this let me help you with algebra made easy. Through online student can get instant help for their doubts at any point of the world also they can clarify their doubts with the real person at the moment. In concerts with geometry, analysis, topology, combinatorial, and number theory, algebra is one of the main branches of pure mathematics. In this lesson .. we are going to learn about steps by steps explanation for online free algebra helper.

Practice Problems for Online Free Algebra Helper:

Problem 1:

Estimate the given algebraic equation to solve step by step use the elimination method

4a+ 2b = 12,

2b + 4c = 12,

4c + 2a = 12. This could also help us on hands on equations.

Answer: a = 2, b = 2, c = 2. This could also help you on algebra 2 help free

Monday, August 30, 2010

Help with 6th grade math word problems

Introduction:

In this article let me help you on 6th grade math word problems. The first real course in mathematics is free algebra word. Depending on the principles of arithmetic that already have studied, free algebra word uses symbols to create generic solutions which work in various cases.

Free algebra word problem is a type of textbook word problem designed to help students apply abstract mathematical concepts to real-world situations.
Sample Problems:

Problem 1

ax ± b = c. All algebra problems are like the following one.

David spent 24 for shoes, this was 8 less than twice what she spent for a blouse. How much was the blouse?

Solution:

Every word problem has an unknown number. In this problem, unknown number is the price of the blouse. Constantly, let x represent the unknown number. That is, let the x value answer the question. This can also help us on math problem solving questions

Let x, how much he spent for the blouse. Here becomes the equation:

2x – 8 = 24

In the right side, the -8 will become +8.

2x = 24+8

x = 32/2

x = 16

The correct answer is x = 16.

Thursday, August 26, 2010

What is Perimeter?


Introduction to What is perimeter?

In this section let me help you on what is perimeter. The perimeter is the total distances around the outside of a two - dimensional shape (2D). We can calculate it by collating together all the lengths of a shape. The polygon of perimeter is the sum of the lengths of all its sides.
We can also define perimeters as : The distance around a figure is called the perimeter of the figure.
Closed Curve: A curve which begin and ends at the same point is called a Closed Curve.
Simple closed Curve : A closed Curves which doesn't intersect itself is called a Simple closed Curve.This could also help us on percentage increase formula
Unit of Perimeter : Dimensional Perimeter -1 is measured in linear units. Example such as feet or meters.

Tuesday, August 24, 2010

Probability Help

Probability Help:
These solved example problems will help you to understand the probability

Example 1: A box contains 8 sixty watts bulbs and 8 hundred watts bulbs. Five bulbs are taken at random. Find the probability of, five will be of the same watts bulbs?
Solution:
Let S = Sample space
A = the event of taking 5 sixty watts bulbs.
B = the event of taking 5 hundred watts bulbs.
(A or B) = The event of taking 5 sixty watts bulbs or 5 hundred watts balls.
i.e. the event of taking 5 bulbs of same watts.
n(S) = 16C5 = 4368
n(A) = 8C5 = 56
n(B) = 8C5 = 56 . This could also help us on what is the mean
P(A) = n(A) / n(S) = 56/4368 = 1/78
P(B) = n(B) / n(S) = 56/4368 = 1/78
P(A or B) = P(A) + P(B)
= 1/78 + 1/78 = 2/78 = 1/39
P(A or B) = 1/39