Regular Algebra Number Lines

Introduction:

Here we are going to discuss about the regular algebra number lines, generally regular algebra number lines is defined as the graphical representation of the given set of the numbers .It is represents by the form of the straight line in the straight number lines we can mark the values as the 0 first than mark the positive numbers on the right side of the number lines and mark the negative numbers on the left side of the number line .So the regular algebra number lines will be shown as the below diagram.

Uses of the Regular Algebra Number Lines:

It is used for the calculation of the addition ,subtraction ,division and the multiplication .So based on the number lines  we can perform these operation on that .


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Regular algebra number line:

Here we are going to see the problems of addition, based on the regular algebra number lines .First we have to draw the number lines like the above and after that we do the operation as addition ,both the numbers are positive means we have to move the points from the right to the starting point .

Example:

6+3 Use the regular number lines to calculate the addition.

Solution:

First we have to draw the number lines after that mark the starting point ,here the starting point Is 5 so mark the point and both the numbers are positive so move to the right side 3 places so the answer for 6+3=9.The ending point is 9,this is the answer .

here the staring number is 6 it will be marked in the diagram after that it will be moved to the right side in three place

so the answer is the ending point here the ending point is 9 so we add 6+3 we get the answer is 9.

Example 2:

12`-:` 2 use the regular algebra number line to calculate the value?

Solution :

In these problem we are going to draw the regular algebra number line after that we have to mark the number in the line ,the positive numbers are come in the right side of the line .so the answer will be explained through the below diagram,the diagram is based on the regular algebra number lines .

Here we are perform the operation as division ,the division is the simplest form of the subtraction ,here we consider the denominator the value of the denominators is 2 .so we have to split the line with the interval of 2,starting from 0 and it will be end at the denominators value ,here the value of the denominator is 12 so the ending point is 12 ,the answer is the number of the equal interval ,here the number of the equal interval is 6,so the answer is 6

Solving Linear Equations Substitution Method

Introduction :

The linear equations for the substitution methods are one of the methods in the systems of equations in the algebra section. The substitution methods are nothing but the variable can be found by using the equations and the value of one variable can be determined. Then, other variable can be found by substituting the variable value in another equation. Now we see the solving of linear equations by substitution method.

About Solving of Linear Equations by Substitution Method:

Now we are going to see the steps for solving the linear equations by substitution methods are as follows,

The equation should be choose and place one variable in one side and the other variables and values on the other side.

Then, in the second step these equations should be substituted in one equation and the value can be determined.

The other value of the variable can be determined by substituting the value in another equation.

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Problems for Solving the Equations by Substitution Method:

Example 1:

Solve the given two equations by substitution method:

x + y = 2

y + 10 = 2x

Solution:

Choose one of the equations for solving the linear systems as follows,

x + y = 2

x = 2 – y

Now put the value of x in the equation y + 10 = 2x

y + 10 = 2x

y + 10 = 2(2 – y)

y + 10 = 4 – 2y

y + 2y = 4 - 10

3y = -6

y = -2.

Put the value of y in the equation x = 2 - y

x = 2 - y

x = 2 - (-2)

x = 2 + 2

x = 4

The values are x = 4 and y = -2.

Example 2:

The given two equations for solving linear system are:

x + y = 4

y + 8 = x

Solution:

Choose one of the equations for solving the linear systems as follows,

x + y = 4

x = 4 - y

Now put the value of x in the equation y + 8 = x

y + 8 = x

y + 8 = 4 - y

y + y = 4 - 8

2y = - 4

y = -2.

Put the value of y in the equation x = 4 - y

x = 4 - y

x = 4 - (-2)

x = 4 + 2

x = 6

The values are x = 6 and y = -2.

One to One Function Graph

Introduction
Let us study about one to one function graph in mathematics. Functions have the not different component in first and not same component in second place. One to one function means that all the elements of a particular function have at least one element on to another function.  One to one function is a function, in this function of A’s elements should have one element into function B.

One to One Function Graph:
A function P should have one element in function Q. but the function P should not have more than one element in function Q.  And also function P’s all elements must have one element in function Q. It is also said as a function P should produce the corresponding image in function Q at once.

For example:

Consider p1 and p2 any elements of P.
If suppose p1 is not equal topa2. Then the function f(p1) is also not equal to f(p2).
If suppose p1 is equal to p2. Then the function f(p1) is also equal to f(p2).

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Example Problems for One to One Function Graph:
Example 1:Check whether the following function defined by f = {(2 , 5),(8 , 5),(4 , 2),(5 , 9),(9 , 8)} is a one to one function?

Solution:

The given function can be drawn as in below figure:

From the figure we can note that 2 and 8 have the same output.

Therefore the given function f is not a one to one function

Example 2: Find the following function is one to one function or not. f(x) =x^3

Solution:

Given f(x) =x^3

Apply the x= -10 to 10. Draw the graph.

The given function produces that each x value has one unique y value. Therefore this given function is one to one function.

Example 3: Find the following function is one to one function or not. g(x) = | x-2|

Solution:

Given g(x) = | x-2|

Apply the values x= -10 to 10. Draw the graph.


From the above graph we can see that the given function is not a one to one function because 4 and 2 has the same element in function y.

Therefore the given function is not one to one function.

Solving Zero Matrix

Introduction :

The zero matrix is defined as the one which consists of the number zero in all the entries. For example consider 2 x 2 matrix which have 4 elements (2 in each row or column). For zero matrix, the four elements are having zero values. consider 2 x 3 matrix which have 6 elements (3 in each row). For this zero matrix, the six elements are having zero values.

Examples to Explain "solving Zero Matrix "

A matrix O = [oij]m × n is defined to be a zero matrix or said to be null matrix having all the entries zero which is denoted by O    

`O_1` = [ 0 ] ;  `O_2` = `[[0,0],[0,0]]`    `O_3` =   `[[0,0,0],[0,0,0],[0,0,0]] `   `O_4` =  `[[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]] `    `O_5` =  `[[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]] `    and so on   

Multiply this matrix  `[[93,91,91],[92,92,90],[91,92,91]]` with the zero matrix  `[[0,0,0],[0,0,0],[0,0,0]]`   
Solution:

It is given by    `[[93.0+91.0+91.0,93.0+91.0+91.0,93.0+91.0+91.0],[92.0+92.0+90.0,92.0+92.0+90.0,92.0+92.0+90.0],[91.0+92.0+91.0,91.0+92.0+91.0,91.0+92.0+91.0]]`      

`= [[93.0+91.0+91.0,93.0+91.0+91.0,93.0+91.0+91.0],[92.0+92.0+90.0,92.0+92.0+90.0,92.0+92.0+90.0],[91.0+92.0+91.0,91.0+92.0+91.0,91.0+92.0+91.0]]`     

` = [[0+0+0,0+0+0,0+0+0],[0+0+0,0+0+0,0+0+0],[0+0+0,0+0+0,0+0+0]]`        

= `[[0,0,0],[0,0,0],[0,0,0]]`     is the zero matrix obtained here.

Perform the subtraction operation on the matrix  `[[33,32,31],[32,32,30],[31,32,31]]` and     `[[33,32,31],[32,32,30],[31,32,31]]` 
Solution:

These matrix are with same rows (3) and columns (3)

= `[[33-33,32-32,31-31],[32-32,32-32,30-30],[31-31,32-32,31-31]]`       

=  `[[0,0,0],[0,0,0],[0,0,0]]`     is the zero matrix obtained here.

Perform the given operation on  `[[103,102,101],[102,102,100],[101,102,101]]` `-`   `[[92,92,92],[92,92,92],[92,92,92]]` ` -` `[[11,10,9],[10,10,8],[9,10,9]]`  
Solution:

These matrix are with same rows (3) and columns (3)


= `[[103-92,102-92,101-92],[102-92,102-92,100-92],[101-92,102-92,101-92]]`      ` -`     `[[11,10,9],[10,10,8],[9,10,9]]`      

=  `[[11,10,9],[10,10,8],[9,10,9]]`  - `[[11,10,9],[10,10,8],[9,10,9]]`  

= `[[0,0,0],[0,0,0],[0,0,0]]`     is the zero matrix obtained here.

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Problems to Explain "solving Zero Matrix "

Multiply this matrix  `[[93,91,91],[92,92,90]]` with the zero matrix  `[[0,0,0],[0,0,0]]`   
Solution:

It is given by    `[[93.0+91.0+91.0,93.0+91.0+91.0,93.0+91.0+91.0],[92.0+92.0+90.0,92.0+92.0+90.0,92.0+92.0+90.0]]`      

`= [[93.0+91.0+91.0,93.0+91.0+91.0,93.0+91.0+91.0],[92.0+92.0+90.0,92.0+92.0+90.0,92.0+92.0+90.0]]`     

` = [[0+0+0,0+0+0,0+0+0],[0+0+0,0+0+0,0+0+0]]`        

= `[[0,0,0],[0,0,0]]`     is the zero matrix obtained here.

Perform the subtraction operation on the matrix  `[[33,32,31],[32,32,30]]` and     `[[33,32,31],[32,32,30]]`  
Solution:

These matrix are with same rows (3) and columns (3)

= `[[33-33,32-32,31-31],[32-32,32-32,30-30]]`       

=  `[[0,0,0],[0,0,0]]`     is the zero matrix obtained here.

Perform the given operation on  `[[103,102,101],[102,102,100]]` `-`   `[[92,92,92],[92,92,92]]` ` -`   `[[11,10,9],[10,10,8]]`  
Solution:

These matrix are with same rows (3) and columns (3)


= `[[103-92,102-92,101-92],[102-92,102-92,100-92]]`      ` -`     `[[11,10,9],[10,10,8]]`      

=  `[[11,10,9],[10,10,8]]`  - `[[11,10,9],[10,10,8]]`  

= `[[0,0,0],[0,0,0]]`     is the zero matrix obtained here.

Triangular Pyramid Vertices

Introduction

Triangular pyramid is one type of three dimensional shapes. It has four vertices. Three vertices are in the base and one vertex is on top of the pyramid. Vertices are nothing but the corner points of the pyramid. The shape of the triangular pyramid is shown in below. In the figure A, A', B, B', C, C'  are the vertices. Let us see how to calculate the volume and surface area of the triangular pyramid.

Formula Used to Find Volume and Surface Area of Triangular Pyramid:

Surface area of the triangular pyramid (SA) = A + `(p * l) / (2)` square units

A – Base area

p – Perimeter of base

l – Slant height



Volume of triangular pyramid (V) = `1/3` A h cubic units

A – Area of the base

h – Height of the pyramid

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Vertices of Triangular Pyramid– Example Problems:

1. The triangular pyramid has the vertices A (0, 0), B (18, 0), slant height (l) = 15 units, height (h) = 12 units. Find the volume and surface area of the triangular pyramid.

Solution:

Given:

A (0, 0), B (18, 0)

Slant height (l) = 15

Height (h) = 12

Length between two vertices (a) =` sqrt ((y_2- y_1)^ 2 + (x_2- x_1)^2)`

= `sqrt((0-0)^2+(18-0)^2)`

= `sqrt(324)`

a = 18 units

Base area (A):

Base area of the triangle (A) = a2 x `sqrt (3) / 4` square units

= 182 x 0.433

= 324 x 0.433

Base area of the triangle (A) = 140.292 square units

Perimeter of the base (P):

Perimeter of the base (p) = 3 a units

= 3 x 18

= 54 units

Surface area (SA):

Surface area of the triangular pyramid (SA) = A + `(p * l) / 2` square units



= 140.292 + `(54 X 15) / 2`

= 140.292 + 405

Surface area of the triangular pyramid (SA)    = 545.292 square units

Volume (V):

Volume of triangular pyramid (V) = `1/3` A h cubic units

= `1/3` x 140.292 x 12

= `1/3` x 1683.504

Volume of triangular pyramid (V) = 561.168 cubic units

2. The triangular pyramid has the vertices A (0, 0), B (10, 0), slant height (l) = 7.8 units, height (h) = 6 units. Find the volume and surface area of the triangular pyramid.

Solution:

Given:

A (0, 0), B (10, 0)

Slant height (l) = 7.8

Height (h) = 6

Length between two vertices (a) = `sqrt ((y_2- y_1)^2 + (x_2-x1)^2)`

= `sqrt((0-0)^2+(10-0)^2)`

= `sqrt(100)`

a = 10 units

Base area (A):

Base area of the triangle (A) = a2 x `sqrt (3) / 4 ` square units

= 102 x 0.433

= 100 x 0.433

Base area of the triangle (A) = 43.3 square units

Perimeter of the base (P):

Perimeter of the base (p) = 3 a units

= 3 x 10

= 30 units

Surface area (SA):

Surface area of the triangular pyramid (SA) = A + `(p * l) / 2` square units



= 43.3 + `(30 X 7.8) / 2`

= 43.3 + 117

Surface area of the triangular pyramid (SA)    = 160.3 square units

Volume (V):

Volume of triangular pyramid (V) = `1/3` A h cubic units

= `1/3` x 43.3 x 6

= `1/3` x 259.8

Volume of triangular pyramid (V) = 86.6 cubic units

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