Mathematical Symmetry

In mathematical when one shape becomes like another if you rotate over, slide or twist is called the symmetry. In mathematical normally symmetry is classified into three types. Thus three types of symmetry are reflection symmetry, rotational symmetry and point symmetry. Let us see mathematical symmetry in this article.


Mathematical Symmetry:

Symmetry:

When one shape becomes like another if you rotate over, slide or twist it is called as symmetry.

Types of symmetry:

Reflection symmetry
Rotational symmetry
Point symmetry

Definition of reflection symmetry:

Reflection symmetry is the simple type of symmetry. One half of the reflection is the reflection of the other half is known as reflection of symmetry.

Definition of Rotational symmetry:

The shape or image can be rotated on various quantities and it still shows the same is called as rotational symmetry.

Definition of Point symmetry:

If the image is placed the same distance from the starting but in the opposite path then the image has point symmetry.
Brief Explanation about Mathematical Symmetry:

Reflection symmetry:

One half of the reflection is the reflection of the other half is known as reflection of symmetry.

Example:

mathematical symmetry

Here the image (under the line) gives the perfect reflection of the above image.

Rotational symmetry:

The shape or image can be rotated on various quantities and it still shows the same is called as rotational symmetry.

Example:

mathematical symmetry

In above figure the second image represents the rotated structure of first image.

Point symmetry:

If the image is placed the same distance from the starting but in the opposite path then the image has point symmetry.

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Example:

mathematical symmetry

From the above figure we understand the point symmetry. In above figure the images are placed the same distance from the starting but in the opposite path and it similar to rotational symmetry of order 2.

Adjacent and Opposite Angles

Adjacent angle:

Two angles having a general side and general vertex of that angle is adjacent angles. In adjacent angle, angle that forces never meets. The adj is the short form of the adjacent angle.

Opposite angle:

Intersect two line is framed by four angle.The angles are straightly opposite to other angle is known as opposite angle. The opposite angle is also known as vertically opposite angels.

In this article we see in detail about adjacent and opposite angles.

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Example 1: Adjacent and Opposite Angles:

Following diagram to find the unknown angle a, b , c

example problem for adjacent and opposites angles

Solution:

In this problem given angle is the EFG = 114 degree

Given angle is EOF

Unknown angle is FOG, GOH and HOE

In this diagram angle EOF and angle GOH is equal, since it is opposite angle, written as EOF = GOH

Therefore

Angle EOF = 114

Angle GOH = 114

Angle dimension is 180 degree

EOF+FOG=180

114+c=180

c=180-114

c=66

Angle c and angle a is opposite angle so a=66.

Answer is:

a=66

b=114

c=66
Example 2: Adjacent and Opposite Angles:

Following diagram to find the unknown angle a, b , c

example problem for adjacent and opposite angles

Solution:

In this problem given angle is the EFG = 101 degree

Given angle the angles EOF.

Unknown angle is FOG, GOH and HOE

In this diagram angle EOF and angle GOH is equal since they are opposite angles, written as EOF = GOH

Therefore

Angle EOF = 101

Angle GOH = 101

Angle dimension is 180 degree

EOF+FOG=180

101+c=180

c=180-101

c=79

Angle c and angle a is opposite angle so a=79.

Answer is:

a=79

b=101

c=79

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Example 3: Adjacent and Opposite Angles:

Given complementary angle, angle E is the 74 degree, find the unknown adjacent angle

example for adjacent and opposites angles

Solution:

In complementary angle measurement of the angle is 90 degree

Given the value of angle is 74°

Subtracting form angle measurement of the complementary angle and given angle

90° – 74° = 16°

Answer for this problem is 16 degree.
Example 4: Adjacent and Opposite Angles:

Given complementary angle, angle E is the 71 degree, find the unknown adjacent angle

example for adjacent and opposites angles

Solution:

In complementary angle measurement of the angle is 90 degree

Given the value of angle are 71°

Subtracting form angle measurement of the complementary angle and given angle

90° – 71° = 19°

Answer for this problem is 19 degree.

Running Distance Calculator

How to find a distance covered by a object if it's speed and time  is given,first we will see the formula for running distance calculator.

use the following formula

Distance = Speed x Time

if the speed of the object is running or moving with the speed of 100 km per hr and in 2 hrs that object is going to be cover ------------------------distance?

as we know speed = 100 km per hr and time = 2 hrs

so we know the formula for  distance = speed x time = 100 x 2  = 200 kms

so if the object is moving with 100 km per hr than in 2 hrs it is going to be cover 200 kms distance

when distance covered by the object is asked in miles than use below notation

1 mile =1.6 kilo meters

If the distance i have to find in miles than what i have to do?

As we know 1 mile =1.6 km

so  in 200 km there will be how many miles?

1mile--------------------->1.6km

xmile----------------------->200 km

so by using cross multiplication

x = 200 / 1.6 miles  = 125  miles

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Examples on Running Distance Calculator:

Ex:2 If the speed of a object(plane) is 20 mile per sec than what is the distance traveled by the object in 1 hour.

here speed = 20 miles per sec

so object is traveling 20 mile in 1 sec

so in 1 hour there are 60 sec so it will travel 20 x 60 mile per hour

so it will travel 1200 mile in each hour

if you want to convert this 1200 miles into kilo meters just multiply this 1200 with 1.6

so plane covered the distance of 1200 x1.6 =120 x 16 = 1920 kilo meters

some more examples

example 1: if jhon is driving a car with 90 kilo meter per hour and this is the average speed of his one side journey , for one side he take 5 hours to reach his destination than what is the distance at one side?

answer: jhon is driving with 90 kmph  so speed =90 kmph

and he taken 5 hours of time so time =5 hrs

distance formula is :

distance = speed x time =90 x 5 = 450 km

so his one side journey is 450 kilo meters(answer)

example 2 : if a helicopter is moving with 400 kilo meter per hour  and it traveled about 6 hours than helicopter covered distance in 5 hrs in miles__________?

here helicopter traveled the distance we have to find that to be in miles

for that first we will calculate the distance covered by the helicopter in 5 hrs with the speed of 400 km

than we convert the distance which is in km into miles for that we will use following formula

distance in miles=distance in km/1.6

by this formula distance will change from kilo meters to miles

helicopter  speed = 400 kmph

it taken time about of 5 hrs to reach the destination

so speed =400 kmph and time =5 hrs

distance = speed x time

=400 x 5

=1200 km

so distance in miles =1200/1.6 =12000/16 =750 miles(answer)

so helicopter covered distance of 750 miles in 5 hrs

example 3. pal has to reach his office in exact 15 mins from his home and for that he is moving with his bicycle with the speed if 20 kmph than what is distance  between his house and office?

here bicycle speed is 20 kmph and pal takes 15 mins to reach his office

so speed =20kmph

time=15 mins

here time is in minutes so we have to convert this time in to hours before we substitute time value in formula distance = speed x time , because speed is in km per hours

so distance=20 x15/60 =15/3=5

so distance between pal house and his office =5 kilo meters

example4:  peter average car speed is 60 kmph and he traveled about 3 hours than , distance covered by peter in  this time period=__________in miles

answer: speed = 60 kmph

time = 3 hrs

distance = speed x time = 60 x 3 =180 kilo meters

so peter covered the distance of 180 km but we have to calculate this in miles

so distance =180/1.6 = 1800/16 =900 / 8= 450 / 4 =225 / 2 =112.5 miles

so peter covered distance in miles = 112.5 miles

example 5: robo car is moving with 10 cm/ sec than in 1 minute that robo car is going to cover distance of _________in meters

speed = 10 cm /sec

time = 60 secs

so distance = speed x time = 10 x 60 = 600 cms

but we have to calculate this distance in meters

we know that 100 cms(centi meter)- - - - - - - - -- - - -- - - >1 meter

so 600cms-------------------------------------------------------------->?(take unknown value as X)

so 600 x 1 = 100 x X

X=600/100 meters

distance in meters = 600/100 =6 meters(by cross multiplication)

so fill the blank with the 6 meters

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example 6: distance between two metro stations is _____________in miles, if train is moving with 60kmph and it taken 2 and 1/2 hr to reach the destination

so we have to calculate the distance between two metro stations

here data is given in the question that train is moving with the average speed of 60kmph

and time taken by the train to reach it's destination is 2 and 1/2 hrs

so speed = 60 kmph and time =(2+1/2)= (2x2+1)/2=5/2 hrs

we know distance formula as distance = speed x time

so distance = 60 x 5/2 = 30 x 5 =150 kilo meters

but we have to calculate distance in miles

so convert the kilometers in to miles

we know that 1 mile------------------------------->1.6 km

?       <-------------------------------150 km="" p="">
by cross multiplication

? x 1.6 = 1 x 150

? = 150/1.6

?=150 x 10/16

?=150 x 5/8

?=75 x 5/4

after calculating above expression we will get the distance between two stations as 93.75 miles(this is the required answer)
Problems for Practice on Running Distance Calculator:

some more problems for practice......

1. convert 110 kilometers into miles_________(answer?)

2.car moving with speed 90 kmph and to cover particular distance car taken about 1 and half hours than what is the distance of car starting point to the destination?

3. train is moving with the average speed of 110 kmph and at each stop 5 minutes and total stops between two major stations are 3 and train taken about 3 and 1/4 hr to reach it's destination station , then find the distance between two major stations in miles?

4. peter is ride  his bicycle with average speed of 30 kmph in first 10 minutes of total time and next 5 minutes he ride his bycle with 15 kmph average speed . than he accelerated his speed about 15 kmph and he reached his destination with the same speed in 7 minutes. than find the total distance between his starting and ending point of that race?

5.convert 1200 miles to kilometers?(answer 1920)

Elementary Functions Math

The mathematical concept of a function expresses the intuitive idea that one quantity (the argument of the function, also known as the input) completely determines another quantity (the value, or the output). A function assigns a unique value to each input of a specified type. The argument and the value may be real numbers. (Source: Wikipedia)
Example Problems for Elementary Functions Math :

Elementary functions math - problem 1:

Subtract x3 – 5x2 – 7 from 7x3 + 8x2 – 2x – 9.

Solution:

7x3 + 8x2 – 2x – 9 – x3 + 5x2 +7

6x3 + 13x2 – 2x – 2

Elementary functions math - problem 2:

Find the sum of 6x4 – 7x2 + 9x + 11 and 5x + 7x3 – 6x2 – 9.

Solution:

6x4 + 7x3 – 7x3 – 6x2 + 9x + 11 – 9

6x4  – 13x2 + 9x + 2

Elementary functions math - problem 3:

Factorize: x2 + 10x + 16

Solution:

x2 + 10x + 16

x2 + 8x + 2x + 16

x(x+8) + 2(x + 8)

(x + 2) (x + 8)

Elementary functions math - problem 4:

Simplify 7 * 3 - 2(4)3 ÷ (-6)

Solution:

` (7 * 3 - 2(4)^3 )/ -6`

= =>`(21 - 2(64) )/ -6`

= =>`(21- 128 )/ -6`

= =>`-107/ -6 `

= => `107/6` .

Elementary functions math - problem 5:

Simplify (5y + x)(8y – x)

Solution:

(5y + x)(8y – x)

= = > 40y2 – 5xy + 8xy – x2

= =>40y2 + 3xy – x2

Elementary functions math - problem 6:

Solve the given algebraic equation

3(-5x - 6) - (x - 7) = -8(4x + 7) + 21

Solution:

Given equation is 3(-5x - 6) - (x - 7) = -8(4x + 7) + 21

Multiply the terms

-15 x-18-x+7 = -32x-56+21

Make them as a group

-14x -11= -32x - 35

-14x + 32x = 11 - 35

18x = -24

x = `-24/18`

X = `-12/9` .
Practice Problems for Elementary Functions Math :

1. Subtract x3 – 3x2 – 1 from 3x3 + 6x2 – 4x – 8.

Answer: 2x3 + 9x2 – 4x – 7

2. Find the sum of 5x4 – 8x2 + 7x + 8 and 7x + 6x3 – 3x2 – 1.

Answer: 5x4 – 2x3 – 3x2 + 7

3. Factorize: x2 + 6x + 8

Answer: (X + 2) (X + 4)

4. Simplify 8 * 5 - 5(4)3 ÷ (-8)

Answer: 35.

5. Simplify (4y + x)(4y – x)

Answer: (4y + x)(4y – x) = 16y2 – x2

6. Solve the given algebraic equation  6(-2x - 3) - (x - 2) = -5(2x + 3) + 19

Answer: X = 10.

Parabola Problems and Solutions

In mathematics, the parabola is a conic section, the intersection of a right circular conical surface and a plane to a generating straight line of that surface. Given a point (the focus) and a corresponding line (the directrix) on the plane, the locus of points in that plane that are equidistant from them is a parabola (Source: Wikipedia). In this topic we discuss about parabola problems and solutions, parabola example problems

Parabola Grapher
Parabola Example Problems with Solutions:

Problem: 1

What is the minimum value of the expression 2x2 – 20x + 17?

Solution:Parabola Example 1 Diagram

Consider the function y = 2x2 – 20x + 17. This function is defined by a second degree equation. This xo-efficient of its x2 term is positive. Hence the curve is a parabola opening up ward.

`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` =` (-(-20))/(2(2))` = `20/4` = 5.

For x = 5, y = 2(5)2 – 20(5) + 17 = - 33. Therefore the minimum value of the expression 2x2- 20x + 17 for any value of x is – 33. This minimum value is assumed only when x = 5.
Problem: 2

Find the coordinates of maximum point of the curve y = - 3x2 – 12x + 5, and locate the axis of symmetry.

Solution:

The curve is defined by a second degree equation. The coefficient of x2 term is negative.

`(-coefficient fo X term)/(2.coefficient fo X^2 term)` = `-b/(2a)` = `(-(-12))/(2(-3))` = `12/-6` = -2.

For x = -2, y = -3 (-2)2 – 12(- 2) + 5 = 17. Hence the coordinates of the vertex are (- 2, 17). The curve is symmetric with respect to the vertical line through its vertex, through the point (-2, 17), i.e., the line x = -2


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Problem: 3

If a parabolic reflector is 16 cm in diameter and 4cm deep, find the focus.

Solution:

let POQ be the vertical section of the reflector. Mid - point of PQ is M. Let OX be along OM and OY parallel to MP.

Let the equation of the parabola be y2 = 4ax.

The coordinates of P are (4, 8)

(8)2= 4a (4) or a = 4

Focus = (a, 0) = (4, 0).

Focus coincides with M, the mid-point of PQ

Solve Equations to Vertex Form

Solve equations to vertex form article deals with how to write the standard equation to vertex form and the model problems that helps to understand vertex form.

General form of equation and vertex form:

For solving equation to vertex form, we should have idea in general form

1. The general form of the quadratic function is f(x) =ax2+bx+c

2. The general form of the vertex equation is

f(x) = a(x-h)2 +k

Here (h, k) is the vertex.
Steps in Solving Equations to Vertex Form:

In solving the equation tovertex form, the following steps are very important

Step1: First, we need to factor out the leading coefficient.

Step2:  Add and subtract by common number `(b/2) ^2`

Step 3: simplify the equation to the vertex form.
Model Problems to Solve Equation to Vertex Form:

1. Solve the equation to vertex form f(x) = 2x2 +8x +35.

Solution:

Let f(x) be y

y= 2x2 +8x +35.

Step1: First, we need to factor out the leading coefficient

y = [2x2 +8x]+35
y = 2[x2 +4x]+35

Step2:  add and subtract by common number `(b/2) ^2`

y = 2[x2 +4x + (2)2 -(2)2]+35

Step 3: simplify the equation to the vertex form.

y = 2[(x2 +4x + 4] +35-8
y= 2[(x +2)2] +35-8

y = 2[(x +3)2] +27

Here the vertex form is y =2[(x +3)2] +27 , vertex is (-3,27)

2. Solve the equation to vertex form f (x) = x2+4x+21

Solution:

Let f(x) be y

y= x2+4x+21

Step1: First, we need to factor out the leading coefficient

y = (x2+4x) +21

y = (x2+4x) +21

Step2:  add and subtract by common number `(b/2) ^2`

y = (x2+4x+4) +21-4

Step 3: simplify the equation to the vertex form.

y = (x2+4x+4) +17

y = (x+2)2+17

Here the vertex form is f(x) = (x+2)2+17, vertes is(-2,17)

3. Solve the equation to vertex form y= x2+16x+30

Solution:

y= x2+16x+30

Step1: First, we need to factor out the leading coefficient

y= (x2+16x) +30

y= (x2+16x) +30

Step2:  add and subtract by common number `(b/2) ^2`

y= (x2+16x+64) +30-64

Step 3: simplify the equation to the vertex form.

y= (x2+16x+64)-34

y= (x+8)2-34

Here the vertex form is f(x) = (x+8)2-34, vertex is (-8,-34)

Simple Algebra Calculator

Algebra is one of the most important topics in mathematics. Algebra covers quadratic equation, arithmetic progression, cubic equation, quadratic equation, arithmetic progression, sum of consecutive cubes, equation solver and etc., In this document we shall discus about “simple algebra calculator” with suitable example problems.

Simple Algebraic Equation:

A algebraic equation is the set of the variables, constant and arithmetic operators like addition, subtraction, multiplication, division and equal operators.

To solve the simple algebraic equation using calculator, we have to enter the coefficient of x and y and the constant term. After clicking the calculate button the value of x and y will be displayed on answer box. The algebraic equation calculator is shown below.

Simple Algebraic Equation

Example problem on algebraic equation:

Evaluate the unknown variable for the following algebraic equation.

4x - 5y = 12 ----- (1)

2x+5y = 24 ------ (2)
Solution:

Step 1: Add equation (1) and (2) ,we get
4x - 5y = 12

2x+5y = 24

---------------

6x        = 36

Divide by 6 on both sides,

`(6x)/6 = 36/6`

X = 6
Step 2: Plug x value in second equation;
2x+5y = 24

2 (6) + 5y = 24
Step 3: Expand and simplify the equation:
12 + 5y = 24
12 +5y – 12 = 24 - 12
5y = 12

Divide by 5 on both sides, we get

` (5y)/5 =12/5`

y = 2.4

Algebraic equation of two variable x =6 and y =2.4
Simple Quadratic Equation Calculator:

An equation is collection of  more than one terms are squared but no higher power in terms, having the syntax, ax2+bx+c where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant term

Example: 3x2- 5x + 15

To evaluate the quadratic equation using calculator, we have to enter the value of coefficient of x2 and x and the constant term. After clicking the calculate button the value of x will be displayed on answer box. The quadratic equation calculator

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Example problem of quadratic equation:

Calculate the unknown value in quadratic equation

x2 + 11x - 42

Solution:

Step 1: Multiply the coefficient of x2 and the constant term,

1*-42 = -42 (product term)

Step 2: Find the factors for the product term

-42 --- > 14*-3 = -42 (factors are 14 and -3)

-42 --- > 14-3 = 11 (11 is equal to the coefficient of x)

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Step 3: Separate the coefficient of x

x2 + 11x - 42

x2-3x+14x-42

Step 4: The common term x for the first two terms and 14 for the next two terms are taking outside

x(x - 3) + 14 (x - 3)

(x - 3) (x + 14)

Roots of the quadratic equations (x - 3) (x + 14)