Elementary Math Work

Introduction to elementary math work:
The elementary math work consists of mathematics topics taught in the primary and secondary schools. The most important in elementary mathematics are arithmetic, geometry, algebra, number work and special functions. In secondary school the topics are trigonometry, Calculus etc. Now we are going to see about the elementary math work.

Elementary Math Work:

1. Number work:  The number work consists of the natural, decimal, fractional, rational numbers etc. Let us see one example for the rational number in number work.

Ex  :  Determine 0.55 in the form of a rational number.

Sol :  The expression can be simplify as,

0.55 = 5 tenths + 5 hundredths

= 5 × (1/10) + 5 × (1/100)

= (5/10) + (5/100)

= 55/100

= 11/20.

2. Arithmetic:   In arithmetic, we can see about the housing finance, speed, time, distance, ratio etc. Let us see an example for the arithmetic.

Ex :  Determine the ratio of 3 Kg to 750 g

Sol : The ratio can be calculated as follows,

3 Kg = 3 × 1000 = 3000 g

Thus the required ratio given as

∴Required ratio = 3000 : 750

= 300 : 75

= 60 : 15

= 12 : 3

= 4 : 1

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Other Elementary Math Works:

1. Algebra:  Algebra is the important branch of mathematics consists of many expressions such as variables, constants and coefficients. It also contains some operations as addition, subtraction, multiplication and division.

Ex :   Expand the expression: (2x + 2y + 2z)^2

Sol :  Let us take the equation with (a + b + c)^2

a = 2x, b =2y, c = 2z

(2x + 2y + 2z)^2 = (2x)^2 + (2y)^2 + (2z)^2 + 2(4xy + 4xz + 4yz)

= 4x^2 + 4y^2 + 4z^2 + 8xy + 8xz + 8yz

2. Geometry:  Geometry is about the learning of various size, shape and properties of the geometrical solid figures. The geometry also consists of some constructional properties for all the solid figures.

Ex : The circumference of a circle is given as 10 centimeters. Determine the diameter of the circle?

Sol : The circumference of the circle can be given as,

C = π × d

10 = 3.14 × d

d = 10/3.14

d = 3.18cm.

The diameter of the circle is 3.18 centimeters.

Pyramid Definition Math

Introduction to pyramid definition math:

In this section we will discuss about pyramid definition math. In math the pyramid is comes under the concept of Euclidean geometry. The faces of pyramid are mostly isosceles triangle. The properties of the pyramid and their images are given below. The pyramid has been named as the basis of the base of the pyramid. Let we see about pyramid definition math.

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Definition – Pyramid Definition Math:

Formulas:

Formula for lateral surface area of the pyramid = `(tP)/(2)`

Formula for total surface area of the pyramid = `(tP)/(2) + B`

Formula for volume of the pyramid = (Bh)/(3)

Where, t = lateral height, B = area of base, P = perimeter of base.

Triangular Pyramid

A triangular based pyramid should be a shape that having the base shape as triangular. Simply a triangular based pyramid is a shape of tetrahedron. A tetrahedron is a polyhedron that self-possessed of four numbers of triangular sides. Three of them should meet at every vertex. It also a kind of pyramid, that having flat polygon shaped base as well as triangular faces that linking base at a common point.

Properties of Triangular based pyramid:

This shape having many properties analogous to those of triangles, include in sphere, circumspect, medial tetrahedron as well as exospheres.
The triangular based pyramid should have respective centers namely, in center, circumference, shrieker center, ex centers as well as points like centroid.

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