Exponents to Write Prime Factors

Introduction of exponents to write prime factors:
In mathematics a number or an expression can be factored. A number can be factored only by prime numbers or by a non – prime number or by a combination of prime numbers and a non prime number. A prime number is a number which should divide by itself or by 1. When factorization of a number takes place it is written as exponents finally. Prime factoring is done from the smallest prime and goes upwards.


Exponents to Write Prime Factors Example 1

Write the number 10000 in prime numbers with exponents.

Solution: Given 10000,

Let’s start from the lowest prime 2.

10000 can be written as 2 * 5000.

Now take 5000 and write it in prime factors

5000 is same as 2 * 2500

Take 2500 and write it in prime factors.

2500 is equal to 2 *1250.

Take 1250 and write it in prime factors.

2* 625.

Now take 625 which cannot be factor with 2.

So, take the next prime 3, which is also not possible.

Now take next prime 5 and write 625 in factor form with 5

625 = 5 * 125.

Take 125 and prime factor it now and start from the lowest prime

125 = 5 * 25

Take 25 which is 5 * 5

Take 5 which is 5 * 1.

So the prime factors of 10000 is

=10000

=2 * 5000

=2 * 2 * 2500

=2 * 2 * 2 *1250

=2 * 2 * 2 * 625

=2 * 2 * 2 * 5 * 125

=2 * 2 * 2 * 5 * 5 * 25

=2 * 2 * 2 *5 * 5 * 5 * 5

=23*54, which is the exponents to write prime factors.

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Exponents to Write Prime Factors Example 2 and 3

Write the number 75 in prime numbers with exponents.

Solution: Given 75,

To factor using prime numbers take the least prime 2 first which is not possible.

So take the next prime 3 and write the 75 in terms of 3

75 = 3 * 25.

Now factor the 25 with the prime numbers.

25 cannot be factor in terms of 3. So take the next prime number 5 and write 25 in terms of 5

25 = 5 * 5

=75

=3 * 25

=3* 5 * 5

=31 * 52, which is the exponents to write prime factors.

Write the number 175 in prime number with exponents.

Solution: Given 175,

175 can be written as 5 * 35

Now factor 35 with prime numbers.

35 can be written as 5 * 7

7 can be factor as 7* 1

=35

=5 * 35

=5 * 5 * 7

=5 * 5 * 7 *1

=52 * 71

Which is the exponents to write prime factors.

Volume of Sphere Calculator

Introduction about volume sphere calculator:

The space occupied by an object is known as volume of that object. The volume of the sphere can be calculated easily by using online calculator. If the radius of sphere is given the volume of the sphere can find using online calculator. By using that calculator the operation is carried out robotically once the value is entered. The sphere calculator makes the calculation very easy.  In this article we are going to disuse about how to calculate the volume of sphere formula.

Examples Problems in Volume of Sphere Calculator:
Volume of the sphere = 4/3 p r 3

r = radius.

Problems:

1. The sphere has the radius 7 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 7cm

Formula:

Volume of the sphere= 4/3 p r 3

= 4/3 * 3.14 * (7) 3

=4/3 * 3.14 * 343

= 1436.0266

Volume of the sphere=1436.0266 cm2

2. The sphere has the radius 4 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 4cm

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (4) 3

=4/3 * 3.14 * 64

Volume of the sphere = 268.08cm3

3. The sphere has the radius 11 m. find the volume of the sphere.

Solution:

Given:

Radius (r) = 7m

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (11) 3

=4/3 * 3.14 * 1331

Volume of the sphere = 5575.27976m3

4. The sphere has the radius 2.5 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 2.5cm

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (2.5) 3

=4/3 * 3.14 * 15.62

Volume of the sphere = 65.44985cm3

5. The sphere has the radius 3.7 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 3.7 cm

Formula:

Volume of the sphere = 4/3 p r 3

= 4/3 * 3.14 * (3.7) 3

=4/3 * 3.14 * 50.653

Volume of the sphere  = 212.174cm3

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Practice Problems in Volume of Sphere Calculator:

1. The sphere has the radius 5m. Find the volume of the sphere.

Answer:  Volume of the sphere =523.59878 m3

2. The sphere has the radius 6m. Find the volume of the sphere.

Answer: Volume of the sphere = 904.77868 m3

3. The sphere has the radius 7m. Find the volume of the sphere.

Answer: Volume of the sphere = 1436.75504 m3

4. The sphere has the radius 8m. Find the volume of the sphere.

Answer: Volume of the sphere = 2144.66058 m3

The Product of Two Consecutive Integers

Introduction to Product of two consecutive integers:

Consecutive integer is the number after the same number. Let take the a number a  and its consecutive number can be said as      a + 1. 

Example: (4,5), (6,7) (5,6), (21, 22)

Product of two consecutive integers is nothing but multiplication of the number and its consecutive number. Thus the result of the multiplication of the number and its consecutive is called as product of two consecutive integers. In this article, we see about the product of two consecutive integers.

Product of Two Consecutive Integers:

Formula: multiplication of integer and its consecutive

Let take the number as ‘’a’’ and its consecutive is ‘’a + 1’’.

Step 1: Multiplication of the number ‘’a’’ and its consecutive

a * (a + 1) = a + a2

Thus the product of the integer and its consecutive is equal to the sum of the integer and its square of the same integer.

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Example Problems – Product of Two Consecutive Integers:

Example problem 1:

What is the product of the integer 10 and its consecutive?

Solution:

Formula: a * (a + 1) = a + a2

Here a = 10

10 * (10 + 1) = 10 + 102

Square of 10 =100

10 + 100 = 110

Answer: 110

Check the answer:

Step 1: The consecutive integer of 10 is 11

10 * (11) = 110

Therefore the answer is correct.

Example 2:

The product of two consecutive positive integers is 420. Find the integer and its consecutive integer.

Solution: Let take the integer as ‘’a’’ and its consecutive is ‘’a + 1’’

Formula: a * (a + 1) = a + a2

a * (a + 1) = a + a2 =  420

a2 + a - 420 = 0 --- > equation 1

Step 1: Factoring the equation1, we get the integer.

a2 + a - 420 = 0

a2 -20a + 21a - 420 = 0

a (a - 20) + 21 (a - 20) = 0

(a -20) (a + 21) = 0

Possibilities of ‘’a’’ value is

a = 20

a = -21

Note: in given problem the integer is positive. Hence the value of the integer is 20 and its consecutive is 20 +1 = 21

Answer: 20 and 21

Practice Problem – Product of Two Consecutive Integers:

1. What is the product of the integer -5 and its consecutive?

Answer: 20

2. What is the product of the integer 5 and its consecutive?

Answer: 30

Converting Quadratic Equations

INTRODUCTION

Any equation which has been written in the form  ax 2+bx + c = 0 is called as quadratic equations. In quadratic equations the numbers a, b, and c are real numbers and a ? zero. If p(x) is an equation and which is the polynomial having the degree of 2 and in the form of p(x)= 0 , then the equation is  quadratic.

Steps for Converting Quadratic Equations:

The General form of quadratic equation is: y=ax2+bx+c

The Vertex form   is: y= a(x-h) 2 +k.

The following steps are used for converting quadratic equations,

Step 1: The initial step is to factor out the leading co-efficient from the first two terms of the given equation.

Step 2:  Complete the square to which we have factored out (First two terms).

Step 3: Balance the constant term we get after completing the square, by multiplying it with the co-efficient.

Step 4: Add the opposite to the constant term of given equation.

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Example Problems on Converting Quadratic Equations :
Converting quadratic equations Example 1:

Find the vertex form of the given quadratic equation y = 4x² - 16x + 6

Solution:

Step 1: Factor out the leading co-efficient from the first two terms of the given equation.

Y= 4 (x2-4x) + 6

Step 2:  Complete the square to which we have factored out (First two terms).

Y= 4 (x2-4x) + 6

(x2-4x) = (x-2)2   { (a-b)2 = a2-2ab+b2  , here a= x, b=2 }

= x2-(2)(x)(2) +22

y = x2-4x+4

Step 3: Balance the constant term 4, by multiplying it with the co-efficient 4.

y  = 4(x2-4x+4)  +6 +4(4)

Step 4: Add the opposite to the constant term of given equation.

y= 4(x-2)2   +6-16

Vertex form is   :  y= 4(x-2)2   -10

Vertex            :   (h=2, k= -10).

Converting quadratic equationsExample 2:

Convert the quadratic equation y = -5x2+30 into vertex form.

Solution:

Step 1: Factor out the leading co-efficient from the first two terms of the given equation.

y = -5x2+30

y= -5(x2-6x)

Step 2:  Complete the square to which we have factored out

y= -5(x2-6x)

(x2-6x) = (x-3)2   { (a-b)2 = a2-2ab+b2   , Here a= x, b= 3}

= x2-(2)(x)(3)+ 32

y= x2-6x+9

Step 3: Balance the constant term 9, by multiplying it with the co-efficient -5

y= -5(x2-6x+9) + (-5)(9)

y  = -5(x2-6x+9)-45

Step 4: Add the opposite to the constant term of given equation.

Here there is no constant term .So just change the sign of -45.                                 

y= -5(x-3)2 +45

Vertex form is   :  y= -5(x-3)2 +45

Vertex      :  ( h=3, k=45)

Kinds of Polygons According to Sides

Introduction to Kinds of polygons according to sides

One of the all encompassing shapes in geometry is polygon. it has at least three sides that consists of a line segment in a closed shape. It does not intersect other than its vertices. The sum of the interior angle is 180 degree for n sides. And the sum of the exterior angle is 360 degree’s polygon which has all its sides and angles are equal than it is called regular polygon.

Different Kinds of Polygons According to Sides:

These are classified according to their convexity, angle s, and number of sides so on… A few types of polygons are

· Regular polygon

· Irregular polygon

· Congruent Polygon

· Concave polygon

· Crossed polygon

· Equiangular polygon.

Regular kinds of  polygons according to sides:

All its sides and interior angles are equal. It has the following types

· Triangle- it has 3 sides and 3 angles

·Quadrilateral- it has 4 sides and 4 angles.

· Pentagon- it has 5 sides and 5 angles

· Hexagon- it has 6 sides and 6 angles.

· Heptagon- it has 7 sides and 7 angles.

· Octagon- it has 8 sides and 8 angles.

· Nonagon- it has 9 sides and 9 angles.

· Decagon- it has 10 sides and 10 angles.

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Irregular Kinds of Polygons According to Sides:

These are classified according to their convexity, angle s, and number of sides so on… A few types of polygons are

· Regular polygon

· Irregular polygon

· Convex Polygon

· Concave polygon

· Crossed polygon

· Equiangular polygon.

Regular kinds of  polygon sides with their angles:

All its sides and interior angles are equal. It has the following types

· Triangle- it has 3 sides and 3 angles

·Quadrilateral- it has 4 sides and 4 angles.

· Pentagon- it has 5 sides and 5 angles

· Hexagon- it has 6 sides and 6 angles.

· Heptagon- it has 7 sides and 7 angles.

· Octagon- it has 8 sides and 8 angles.

· Nonagon- it has 9 sides and 9 angles.

· Decagon- it has 10 sides and 10 angles.

Irrational Factors

INTRODUCTION 

In mathematics,an irrational factor is a number that cannot put in the form of P/Q (i.e fraction of two integers), where P and Q are integers and Q is not equal to zero.Now such numbers that are not rational are termed as irrational.Irrational number (or factor)  is a set of real numbers.

Example:-  ,where x= 2,3,5 etc.

Any irrational factor may be represented as a non-terminating decimal or as a non-recurring decimal(a number in  which not allow repeated number up to infinity)  and conversely .This means, it has infinite expansion  which cannot be put as a fraction

.e.g:- .3333333........................   is a recurring infinite expansion  as 3 repeats itself infinitely.Now .333...........   can be put as a fraction ; hence it is a rational number.

Any root of an arithmetical numbers whose value cannot be accurately determined is called a surd.All surds are irrational numbers,but all irrational numbers are not surds.

Example:-  The base of logarithm e is irrational but not surd.The value of e =2.71828..... which is non-terminating and non-recurring decimal.

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Irrational Factors-rationalising Factors

Some times when we multiply two irrational number then there product becomes rational number.In this case each of the irrational number is called a RATIONALISING FACTOR to the  other.

Example:(i)(√3 ) x (√3 ) = [(√3)2] = 3,which is rational number.So,we can say that ,√3  is a rationalising factor of a √3 . 

(ii) (√3 + √2) x (√3 - √2) = [(√3)2 - (√2)2] = 3 - 2 = 1 which is a rational number.So we can say that , (√3 + √2) and (√3 - √2) are rationalising factors of each other. 

Properties of Irrational Factors

There are mainly three types of properties in  irrational factors.

1. ORDER Properties:-  For any irrational factors a,b any one of the following relationals is true

a>b,a<b

Again for a>b>c,we have a>c

or for                     a<b<c,we get a<c

2. ADDITIVE Properties:-For any irrational factors a,b,c all these relationals is true

(i) Commutative :-  a+b=b+a

(ii) Associative :- a+(b+c)=(a+b)+c

(iii) Identity(zero) :- a+0=0+a=a

(iv )Inverse :- a+ (-a)=0,Here (-a) is additive inverse of a.

(v) Cancellation :- If a+c=b+c then a=b

3. MULTICATIVE Properties:- For any irrational factors a,b,c all these relationals is true

(i) Commutative : a.b= b.a

(ii) Associative :- a.(b.c)=(a.b).c

(iii) Identity :- a X1=1X a=a

(iv) Inverse :- a.1/a = 1

(v) Cancellation :- If a.c = b.c then a=b