Friday, March 1

Absolutely Continuous Function


DEFINITION

Let [p,q] be the close bounded interval of C. Then a function f:[p,q]→ R will be an absolutely continuous function on [p,q], if for any δ>0 there will be a ε>0 such that the certain conditions which are mentioned below holds good

If (p1q1)..............(pnqn) is a collection which is finite with disjoint open intervals in [p,q] such that

Σni=1 (qi-pi) < ε

and

Σni=1 |f(qi)-f(pi)| < δ



EQUIVALENT DEFINITIONS

The condition on a real-valued function  "  f  "  on the compact interval [ p, q ] are equivalent if

1) f is an absolutely continuous function

2) ' f ' has a derivative f1 almost everywhere which is a Lebesgue integral and

f ( x ) = f ( p ) +  ∫x a  f1  ( c ) dt

for all x on [ p , q ]

3)There exists a Lebesgue integrable function such that g on [ p , q ] such that f ( x ) = f ( p ) + ∫x p  g ( c ) dt

for all x on [ p , q ]

If these conditions are satisfied by the function the definitely g = f' almost everywhere

Properties of the absolute continuous function

PROPERTIES OF ABSOLUTELY CONTINUOUS FUNCTION

1. The sum and difference of two absolute continuous function are also absolutely continuous. The products of two absolute continuous function defined on the bounded interval will also be a absolute continuous function.

2. If an absolutely absolute continuous function is defined on a bounded closed interval and is nowhere zero then its reciprocal is also an absolutely continuous function.

3. Every absolutely continuous function is an uniformly continuous function.

4. If  f: [ p , q ] → R is absolutely continuous, then it will be a function of the bounded variation on [ p,q ]

Thursday, February 28

Learning Geometric Probability


Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.In this case we say probability is 0. If the event is certain to occur, we say probability is likely 1. The probabilities involved in the geometric problem that also called as geometric probability. This geometry probability may be a circle or any polygon from the geometric. It involves the length, area and volume of any one of the geometric shapes.

The definition of probability of an event has shown in below that is depend on their outcomes from the possibilities



Number of successful outcomes

Probability  =        _____________________________

Total number of possible outcomes.

learning geometric probability example problem in triangle:

The figure shows a triangle divided into sectors of different colours. Find the probability of angle for blue sector and orange sector?

learning geometric probability

Solution:

We have to find the Probability for blue color:

Step 1:

Total angle of triangle is 180 degree

Step: 2

Probability finding blue color sector= The blue sector triangle angle/ The total angle of  entire triangle

The blue sector triangle angle=30 degree

=30/360

=1/12

Step 3:

Next we have to find Probability for orange color:

Probability finding orange color sector= The orange sector triangle angle/ The total angle of entire triangle

The orange sector triangle angle=45

=45/360

=9/72

learning geometric probability example in Rectangle:

Example 2:

A rectangle has four sides and that corner having each 4 ball. Find the probability of each corner having 4 balls.?

Solution:

Here the Rectangle has the five corners such as A, B, C, D

We have to find the probability here,

Probability = Number of successful outcomes / total number of possible outcome

Probability  of each corner having the balls=4/4=1

Example 3:

A triangle with area 15 cm2 is inscribed in a circle with radius 3 Cm.Find the probability that a ball thrown fall into the triangle?

Solution:

We have to find the area of the circle and area of the triangle is given. From this we can find the probability of a ball that fall into the triangle easily.

Area of the circle = `Pi` r2

Radius  = 3. So Area of the circle = 3.14 * 3 * 3 = 28.26 cm2.

Area of the triangle  = 15 cm2.

So the probability  = 15 / 28.26 = 0.53

Answer is 0.53.

Wednesday, February 27

free math tutoring live


The best site where you get free math tutoring live  is tutorvista.com . Many sites tell that they provide free math tutoring live  but they don't provide good service.

TutorVista's online Math Help is designed help students to ace their subject.


Get Help with Math

TutorVista provides you all the help you need with Math online with the help of well trained tutors. You can connect to tutor at any time at any place in the world. Get one to one attention and you can see boost in your grades. So for  free math tutoring tutor vista live is the best site.


Math Help available Online Now

To help you learn better and faster our Math tutors are available online 24x7. In fact they not only explain you concepts but also help in completing your home work and explain every concept with examples. TutorVista's tutoring is beyond regular tutoring . We provide math help online every day to help you get better in maths.
You can get  free math tutoring live from the best tutors in the world.


Get Online Math Tutoring from the best Online Math tutor

We have math tutors who are experts in their field.TutorVista tutor certification program is one of the best certification and it will ensure that tutors are properly trained . They are trained to deliver the best and they are familiar with all National and various State Standards required across grades in the US.
in US for free math tutoring live the most visited site is tutorvista.


free math tutoring live:Our portal

Here are some screen shots:

1.) Tutor explaining about irrational numbers.

2.) Tutor explaining commutative property of multiplication:

Diagram

This is how we provide online math homework help. Cant wait any more....

Monday, February 25

How to solve a percent problem


In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Let us see how to solve percent problem.                                                                                                                                - Source from Wikipedia

Formula – How to solve percent problem:

To solve the percent problem convert the fraction else a decimal to a percent and then multiply it by 100:

To get the percent value multiply the fraction into hundred. For example

` 2 / 4` (100) = `2 / 4 ` * 100 = 0.5 * 100 = 50%

To convert the percent value to fraction divide it by hundred.

25% = `25 / 100 ` = `1 / 4`

Example: How to solve percent problem:

Problem 1:

What is the decimal value of 72%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `72 / 100` = 0.72.

Therefore the decimal value of 72% is 0.72.

Problem 2:

What is the decimal value of 86.3%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `(86.3) / 100` = 0.863.

Therefore the decimal value of 86.3% is 0.863.

Problem 3:

What is the decimal value 18% of 60?

Solution:

To find 18 percentages divide 18 by 100 and then multiply it by 60.

= `18 / 100` * 60 = 10.8

Hence 18% of 60 = 10.8

Problem 4:

What is the decimal value for 15.5% of 150?

Solution:

To find 15.5 percent divide 15.5 by 100 and then multiply it by 150.

=` (15.5) / 100` * 150 = 23.25

Hence 15.5% of 150 = 23.25

Problem 5:

The book cost is 5.00 dollars. Now it was 15% increased. What will be the new cost?

Solution:

Multiply the cost by 15%. That is `15 / 100` * 5.00.

= `15 / 100` * 5.00 = 0.75

Add this decimal value with original cost (5.00+0.75= 5.75)

Therefore the original cost of the book is 5.75 dollars.

Problem 6:

In a box there are 85 chocolates. Suppose if a boy took out 60% of chocolates, how many chocolates did he left in the box?

Solution:

The boy took out the chocolates from the box is 60% of 85 or else 60 / 100 × 85

`60 / 100` × 85 = 51%

So the box has 85 chocolates and the boy took out 51 chocolates. Hence the number of chocolates the boy left in the box is 85 – 51 = 34.
Therefore the boy left 34 chocolates in the box.

Problem 7:

A theatre contains 530 chairs. 470 of them are occupied. What percentages of the chairs not occupied?

Solution:

Number of chairs                        = 530

Number of chairs occupied       = 470

Number of chairs not occupied = 530 – 470 = 60

Number of chairs not occupied percentage = `60 / 530` × 100 = 11.32%
Practice problems – How to solve percent problem:

Problem 1:

What is the decimal value of 27%?

Solution:

27% = 0.27

Problem 2:

What is the decimal value 8% of 75?

Solution:

8% of 75 = 6

Friday, February 22

What is the Associative Property


The addition or multiplication of a collection of integer is the same separately from of how the numbers are grouped. A binary operation * is assumed to be associative if any three elements a, b, c of a set a *(b * c) = (a * b) * c. Multiplication of real integers are associative but the division of real integers are not assosiative. And addition is associative but subtraction is not.

What is the Associative Properties

What is the associative property?..The associative property is a property which always involve 3 or more numbers in calculations. The parenthesis indicates the terms that are consider one unit. The gathering (Associative Property) are within the parenthesis. Hence, the figures are 'associated' together. In multiplication, the result is always the same regardless of their combination.

There were two properties are involved in what is the associative property,

They were, Associative property in addition

Associative property in multiplication


Examples for what is the Associative Property (Addition and Multiplication)

Example for what is the Associative property in addition:

Let a, b, c be any real numbers, then

A + (B + C) = (A +B) + C

Let A = 5, B =9, C = 8

L.H.S = 5 + (9 + 8) = 5 + (17) = 22

R.H.S = (5 + 9) + 8 = (14) + 8 = 22

L.H.S = R.H.S

More Examples for addition:

When we change the groupings of addends, the sum does not change:

(8 + 5) + 9 = 22               (or)

8 + (5 + 9)  = 22

(9 + 1) +8 = 18                 (or)

9 + (1 + 8) = 18
Just consider that when the grouping of addends vary, the sum leftovers the same.

Example for what is the Associative property in multiplication:

Let A, B, C be any real numbers, then

A * (B * C) = (A *B) * C

Let, A = 5, B =9, C = 8

L.H.S = 5 * (9 * 8) = 5 * (72) = 360

R.H.S = (5 * 9) * 8 = (45) * 8 = 360

L.H.S = R.H.S

More Examples for Multiplication:

When we change the groupings of factors, the product does not change:

(1 x 9) x 5 = 45          or

1 x (9 x 5) = 45

(2 * 5) * 10 = 100      or

2 * (5 * 10) = 100

Just remember that when the grouping of factors changes, the product remains the same.

Thursday, February 21

Simple Math Multiplication


In mathematics multiplication is used to multiple the numeric values or numbers. Multiplication can also be visualize as counting objects set in a rectangle or as result the area of a four-sided figure whose sides have given lengths. The area of a four-sided figure is not depend on which side is calculated first which illustrates that the order in which values are multiply as one does not matter

Multiplication examples:

Example 1:12 X 8

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 8 = 8.

Step 2: Multiply this 8 by 10 giving 80.

Step 3: Now multiply the 8 by the 2 of twelve giving 16. Add this to 80 giving 84.

Therefore 8 X 12 = 96

Example 2: 18 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 18. So 1 X 8 =1 8.

Step 2: Multiply this 18 by 10 giving 180.

Step 3: Now multiply the 18 by the 2 of twelve giving 36. Add this to 180 giving 216.

Therefore 18 X 12 = 216

Example 3 :12 X 9

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 9 = 9.

Step 2: Multiply this 9 by 10 giving 90.

Step 3: Now multiply the 9 by the 2 of twelve giving 18. Add this to 70 giving 108.

Therefore 9 X 12 = 108

Sample examples:

Example 4: 19 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 9 =1 9.

Step 2: Multiply this 19 by 10 giving 190.

Step 3: Now multiply the 19 by the 2 of twelve giving 34. Add this to 170 giving 228.

Therefore 19 X 12 = 228

Example 5 :23 X 7

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So, 1 X 7 = 7.

Step 2: Multiply this 7 by 10 giving 70.

Step 3: Now multiply the 7 by the 2 of twelve giving 14. Add this to 70 giving 161.

Therefore 7 X 23 = 161

Example 6: 20 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 20. So 2 X 10 = 20.

Step 2: Multiply this 20 by 10 giving 170.

Step 3: Now multiply the 17 by the 2 of twelve giving 34. Add this to 170 giving 240.

Therefore 17 X 12 = 240

Example 7 :12 X 10

Step 1: Multiply the 1 of the 12 by the

number we are multiplying by, in this case 7. So 1 X 10 = 10.

Step 2: Multiply this 12 by 10 giving 120.

Step 3: Now multiply the 10 by the 2 of twelve giving 20. Add this to 70 giving 84.

Therefore 12 X 10 = 120.

Example 8: 30 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 30 =30.

Step 2: Multiply this 30 by 12 giving 360.

Step 3: Now multiply the 30 by the 2 of twelve giving 60. Add this to 170 giving 360.

Therefore 17 X 12 = 360.

Tuesday, February 19

Complex Analysis Problems


Complex analysis is the functions of the complex numbers. Complex numbers have the real and also the imaginary parts. The complex numbers can be represented as x +i y, where x denoted as the real part and y denoted as the imaginary part. The complex numbers comprises the addition of two complex numbers, subtraction of two complex numbers, multiplication of two complex numbers and also the division of two complex numbers. The conception of the complex numbers is the reflection of the fact. This article has the functions of the complex numbers.

Examples for learn complex analysis problems:

Example 1 to learn complex analysis problems:

Compute the value for the complex number f (x) = (75+42i) + (90+72i).

Solution:

The given complex number is f (x) = (75+42i) + (90+72i).

Step 1: (75+42i) + (90+72i) = (75+90) + (42i + 72i)

Step 2: (75+42i) + (90+72i) = 165 + (42i + 72i)

Step 3: (75+42i) + (90+72i) = 165 +114i

The value for the complex number f (x) = (75+42i) + (90+72i) is f (x) = 165 +114i.

Example 2 to learn complex analysis problems:

Resolve the value for the complex numbers f (x) = (13+i) x (11-i).

Solution:

The given complex numbers are f (x) = (13+i) x (11-i).

Step 1: (13+i) x (11-i) = 13(11 - i) + i (11 -i)

Step 2: (13+i) x (11-i) = (143- 13 i) + (11i - i2)

Step 3: (13+i) x (11-i) = 143 - 13i + 11i - (-1) (where i2= -1)

Step 4: (13+i) x (11-i) = (143+1) + (-13i +11i)

Step 5: (13+i) x (11-i) = 144 - 2i

The value for the complex number f (x) = (13+i) x (11-i) is f (x) = 144 - 2i.

Example 3 to learn complex analysis problems:

Calculate the value for the complex number f (x) = (180+53i) - (82+ 7i).

Solution:

The given complex number is f (x) = (180+53i) - (82+ 7i).

Step 1: (180+53i) - (82+ 7i) = (180-82) + (53i - 7i)

Step 2: (180+53i) - (82+ 7i) = 98 +46i

The value for the complex number f (x) = (180+53i) - (82+ 7i) is f (x) = 98 +46i.

Practice problem for learn complex analysis problems:

Compute the value for the complex numbers f (x) = (45+29i)-(2+7i).

Answer: f (x) = 43- 22i

Predict the value for the complex numbers f (x) = (145+88i) + (33+39i).

Answer: f (x) = 178+127i

Compute the value for the complex numbers f (x) = (5+3i) x (2+i).

Answer: f (x) = 7+11i