Elementary Math Addition

Introduction of elementary math addition:

The math is a term which includes mathematical terms such as expressions, arithmetic operations, algebraic expressions and etc. The mathematics is used to increase the technical terms of the each person in their daily life. The samples in math are used to have the great ideas in the daily life. The extended problems in the math are made to work on the sensitive mind.

Elementary Math Addition:

The elementary math addition deals with the simple addition problems which needs a little concentration on the particular topic. The elementary addition is the main basic for the higher classes of the addition. The complicated addition is the formation of the elementary addition. Hence the strong foundation of the elementary addition leads you to excel you on the further addition classes. The natural numbers which includes all the whole numbers. The natural numbers sets are denoted as N. The number systems have the ability to count. The collections of some features are termed with number systems.

The math addition can be varied in to partial sums, normal addition etc. The students can make the addition easier through the teaching type of the addition. The most type of teaching is nothing but the remembering of the number given through the fingers. This is the method which can be absorbed by the children as soon as possible.

Examples of Elementary Math Addition:

The elementary math addition can be started through the single digit, when the student is familiar with the type, next he can move to the two digits and three digits of addition.

1)      Compute the addition of ( 9 +8 )

9 + 8 = 17.

Answer = 17.

2)      Compute the addition of ( 12 + 21 )

12

21 +

33

Answer = 33.

3)      Compute the addition of ( 153 + 765 )

153

765 +

91 8

Answer = 918.

Radicals Explanation Math

Introduction to Radicals explanation math:
Radical explanation math symbol (v), a symbol used to indicate the square root or nth root ,  Radical of an algebraic group, a concept in algebraic group theory  Radical of an ideal, an important concept in abstract algebra  Radical of a ring, in ring theory, a branch of mathematics, a radical of a ring is an ideal of "bad" elements of the ring.                                                      
(Source: wikipedia)

Examples for Radicals Explanation Math:

Addition for radicals explanation math:

Example 1:

Find the value of 25v3 +75 v3

Solution:

=v3 (25+ 75)
=100v3

Here the root is square root, only take the base are same so that we can add the roots easily.

So the result is =100v3

Example 2:

Find the value of 44v11 +77 v23+22v11 +66 v23

Solution:

= v11 (44+22)+ v23 (66+77)

= v11 (66)+ v23 (143)

= 66 v11 + 143 v23

Here the root is square root, the base are same in different terms so that we can add the roots easily.

So the result is =66 v11 + 143 v23

Examples for different radicals explanation math:

Example for square root:

v3 = 1.732

v25 = 5

v7 =  2.64

v12 = 3.46                              

Practice Problem for Radicals Explanation Math:

Multiplication for radicals explanation math:

Example 1:

Solution:

To find multiplication of v3x . v4y2    

= v(12 x y2)            

=  v12x y2   

The result is =  2yv3x   

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.

Example 2:

Solution:

To find multiplication of v(25xy2). v5   

= v(125 x y2)            

= 5yv5 x

The result is 5yv5 x

Both the radicals have an same index of two, so we are multiply mutually under one radical keeping the 2 as its index number.

Factorization for radical radical explanation math:

Example 1:

To find the value of v75

v75 = v5*3*3

= 3v5

The result is v75 = 3v5

Example 2:

To find the value cube root of v125

Cube root of  (v125) = v5*5*5

= 5

The result is v125 = 5

In a cube root of 125, here we take three out of one.

Example 3:

To find the value fourth root of v625

Fourth root of  (v625) = v5*5*5*5

= 5

The result is v625 = 5

In a fourth root of 625, here we take four out of one. Similar to all radicals

Finite Math Problems

Introduction to Finite Math Problems

Finite set:

The number of elements one by one in a set A. If it comes to an end of process, then the set A is said to be a finite set.

Infinite Set:

If we never come to an end in counting the elements, then the set A is said to be an infinite set.

Cardinal Number:

The number of elements in a set A is called the cardinal number of the set A and is denoted by the symbol n(A) which is read as the number of elements in the set A.

if A is a finite set, then n(A) is a whole number.

Finite Sets
Some of the finite sets are show in the following

A = {x | x ?W, x < 5}.
{All schools in Tamil Nadu}.
{Students in IX standard in your school}.


Solution:

1.  x ?W, x < 5 ? x = 0, 1, 2, 3, 4.

? A = {0, 1, 2, 3, 4}. When we count elements of A one by one as 1 for 0, 2 for 1, 3 for 2, 4 for 3, 5 for 4, we come to an end .

Therefore  n(A) = 5 and so A is a finite set.

2. All schools in Tamil Nadu can be counted one by one and we come to an end in the counting process. So the set {All schools in Tamil Nadu} is a finite set.

3. When we count the students of IX standard in your school, we come to an end in the counting process. So the set {Students in IX standard in your school} is a finite set.

Infinte Sets

Some of the infinite sets are show in the following

N.
W.
Z.
The set of all prime numbers.


Solution:

N = {1, 2, 3,…}. When we count elements of N one by one as 1 for 1, 2 for 2, 3 for 3, 4 for 4, we are not able to come to an end in the counting process.

Therefore the set N is an infinite set.

W = {0, 1, 2, 3,…}. When we count elements of Wone by one as 1 for 0, 2  for 1, 3 for 2, 4 for 3, we are unable to come to an end in the counting process. ? the set W is an infinite set.

Z = {0, 1, -1, 2, -2,…}. When we count elements of Z one by one as 1 for 0, 2  for 1, 3 for -1, 4 for 2, 5 for -2 and so on, we are not able to come to an end in the counting process. ? the set Z is an infinite set.

When we write the prime numbers one by one as 2, 3, 5, 7, 11, 13, 17 and so on, we are unable to come to an end in the counting process.

Therefore  the set of all prime numbers is an infinite

Multiplying Matrices

Introduction to multiplying matrices

The rectangular arrangement of some numbers in some rows and columns is called matrix.

Any matrix A is denoted by A_mxn (m by n), where m is the number of rows and n is the number of column in the matrix A and consists of m x n elements.

A_mxn = `[[a_(11),a_(12),a_(13),...,a_(1n)],[a_(21),a_(22),a_(23),...,a_(2n)],[a_(31),a_(32),a_(33),...,a_(3n)],[.,.,.,...,.],[.,.,.,...,.],[.,.,.,...,.],[a_(m1),a_(m2),a_(m3),...,a_(mn)]]`

The number of rows and number of columns m x n (m by n) gives the size of the Matrix.

For example:  A = `[[3,-1,1],[2,3,-4]]`  has 2 rows and 3 columns, so the size of matrix A is 2 x 3 (2 by 3).

Procedure for Multiplying Matrices

The two matrices A_mxn and B_pxq can be multiplied AB only if the number of columns in first matrix  A is equal to the number of rows in second matrix B. The number of columns in matrix A is n and the number of rows in matrix B is p. So, if n = p the the matrix A_mxn can be multiplied by matrix B_pxq and the size of the resultant matrix AB will be number of rows in matrix A (which is m) x number of columns in Matrix B (which is q).

A_mxn X B_pxq = AB_mxq (given that n = p).

Lets take A = `[[a_(11),a_(12)],[a_(21),a_(22)]]` and B = `[[b_(11),b_(12),b_(13)],[b_(21),b_(22),b_(23)]]` , Here the number of columns in Matrix A (which is 2) = number of rows of matrix B (which is 2). Hence we can find A x B.

Assume that A_2x2 x B_2x3 = D_2x3 = `[[d_(11),d_(12),d_(13)],[d_(21),d_(22),d_(23)]]`

To find d11 we use the first row of A and first column of B. Find the product of the elements of this rows and column,

`[[a_(11),a_(12)],[*,*]]` `[[b_(11),*,*],[b_(21),*,*]]` = `[[a_(11)b_(11) + a(12)b_(21),*,*],[*,*,*]]`

To find d12, use the elements of first row of A and second column of B, and obtain their product.

`[[a_(11),a_(12)],[*,*]]` `[[*,b_(12),*],[*,b_(22),*]]` = `[[*,a_(11)b_(12)+a_(12)b_(22),*],[*,*,*]]`

To find d23, we use the elements of second row of A and third column of B, and obtain the product.

`[[*,*],[a_(21),a_(22)]]` `[[*,*,b_(13)],[*,*,b_(23)]]` = `[[*,*,*],[*,*,a_(21)b_(13)+a_(22)b_(23)]]`

Continuing this way, we get

A x B = `[[a_(11),a_(12)],[a_(21),a_(22)]]` x `[[b_(11),b_(12),b_(13)],[b_(21),b_(22),b_(23)]]` = `[[a_(11)b_(11)+a_(12)b_(21),a_(11)b_(12)+a_(12)b_(22),a_(11)b_(13)+a_(12)b_(23)],[a_(21)b_(11)+a_(22)b_(21),a_(21)b_(12)+a_(22)b_(22),a_(21)b_(13)+a_(22)b_(23)]]`

=`[[(first row) x (first coloumn),(first row) x (secound coloumn),(first row) x (third coloumn)],[(second row) x (first coloumn), (secound row) x (second coloumn), (second row) x (third coloumn)]]`

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`[[-1,3],[1,2]]` and B = `[[2],[-1]]` , can we find AB, can we find BA?

AB is possible only when number of columns in A is equal to number of rows in B.

Here number of columns in A = 2

number of rows in B = 2

Therefore, number of columns in A = number of rows in B, so AB is possible

AB = `[[-1,3],[1,2]]` `[[2],[-1]]` = `[[(-1)(2)+(3)(-1)],[(1)(2)+(2)(-1)]]` = `[[-5],[0]]`

BA is possible only when number of columns in B is equal to number of rows in A

Number of columns in B = 1

Number of rows in A = 2

Here number of columns in B `ne` number of rows in A

Therefore, BA is not possible.

Area of Circle Using Circumference

Introduction

                         When the circumference of the circle is given, we can find the area of the circle

Definition of area of circle:

                       The region that is occupied by the circle is called as area of the circle.

                       Formula:

                                   Area of the circle = pi x r² sq.units                

                                             Here “r” the radius of the circle.

Definition of circumference of circle:

                      The length around the circle is called as the circumference of the circle.

                         Formula:

                                 Circumference of circle= 2pi r units.

                                           Here “r” radius of the circle.

Steps to Find the Area of Circle Using Circumference

                    Step 1: From the given circumference of circle, find the radius of the circle

                           Here  Circumference of circle= 2pi r units.

                                                               Radius = (circumference of circle/2pi

                                                                                          (Here pi= 3.14)

                   Step 2: From the radius of the circle, we can find the area of the circle.

                                  Formula for area of the circle = pi x r² sq.units                

                                             Here “r” the radius of the circle.

                                                                                      (Here pi= 3.14)

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Examples on Area of Circle Using Circumference

Ex:1  Find the area of the circle, when circumference of the given circle is  30 cm

                                                                                                 (Here pi= 3.14)

Sol:

                Circumference of the given circle =   30 cm

                                                     2pi r =30

                                           Radius = 30/2pi = 4.7cm

  Area of the circle = pi x r² sq.units

                                             = pi (4.7)²

                                             =69.36 cm²    

Ex:2 Find the area of the circle, when circumference of the given circle is 50 cm

                                                                                          (Here pi= 3.14)

Sol:

                Circumference of the given circle =   50 cm

                                                     2pi r =50

                                           Radius = 50/2pi = 7.9cm

Area of the circle = pi x r² sq.units

                                             = pi (7.9)²

                                             =195.96cm²                                                    

Ex:3 Find the area of the circle, when circumference of the given circle is 100 cm

                                                                                          (Here pi= 3.14)

Sol:

                Circumference of the given circle =   100 cm

                                                     2pi r = 100

                                           Radius = 100/2pi = 15.9cm

Area of the circle = pi x r² sq.units

                                             = pi (15.9)²

                                             =793.82cm²                                                    

Methods of Solving Quadratic Equations

Introduction to methods of solving quadratic equations:

The term "quadratic" comes from quadratures, which is the Latin word for "square" Quadratic equations can be solved by factoring,completing the square,graphing are used in the Quadratic formula.

In general quadratic equation is a polynomial equation of the second degree and Generally it defind as the  ax^2+bx+c=0 in which 'x'a,b,c are the constants with a not equal to zero and the value a=0 then it is a linear equation.The constants a, b, and c, are called respectively, the quadratic Coefficient the linear coefficient and the constant terms or free term. can represents a variable and

A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real

The roots are given by the quadratic formula

x= -b`+-` √ (b2-4ac)/2a 

Steps for solving quadratic equation :

Step 1: Put all terms on one side leaving zero on other side.

Step 2: Factorize.

Step 3: Use the theorem and set each factor to zero.

Step 4: Solve each equation.


Methods for Solving Quadratic Equations :-

Method 1: FACTORING

Method 2:PRINCIPLE OF SQUARE ROOTS.

Method 3:COMPLETING THE SQUARE

Method 4:QUADRATIC FORMULA

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Factoring Method:

Set the equation equal to zero. If the quadratic side is factorable, factor, then set each factor equal to zero.
Example: x2 = −5x − 6
Move all terms to one side x2 + 5x + 6 = 0
Factor (x + 3)(x + 2) = 0
Set each factor to zero and solve x + 3 = 0 x + 2 = 0
x = −3 x = −2

Principle of Square Roots Method:

If the quadratic equation involves a SQUARE and a CONSTANT (no first degree term), position the square on one side and the constant on the other side. Then take the square root of both sides.

Example 1: x2 −16 = 0
Move the constant to the right side x2 = 16
Take the square root of both sides x = ± 4, which means x = 4 and x = −4

Completing the Square Method:

If the quadratic equation is of the form ax2 + bx + c = 0, where a ≠ 0 and the quadratic expression is
not factorable, try completing the square.
Example: (x – 4)2 = 5
x – 4 = ± √(5)
x = 4 ± √(5)
x = 4 – √(5)  and  x = 4 + √(5)

Quadratic Formula Method:

Any quadratic equation of the form ax2 + bx + c = 0, where a ≠ 0 can be solved for both real and
imaginary solutions using the quadratic formula:

x= -b`+-` √ (b2-4ac)/2a