Define Mode in Math

Introduction to define mode in math:

In mathematics, mode is one of the important topics in statistics. Mode defined as most common value of the data. The mode is never a necessarily single, because the similar most frequency may be getting at number of values. In this article, we are going to discuss about math mean median mode tutorial with suitable example problems.

Example Problem for Define Mode in Math:

Problem for define mode in math:

1. Solve the statistics data problem using the given data base, find the mode of the given data.

5, 3, 2, 9, 8, 4, 6, 1, and 8

Solution:

Rearrange the given data for ascending order.

1, 2, 3, 4, 5, 6, 8, 8, 9

Mode:

The most common value in given data is called mode.

Most common value is 8.

Therefore, mode is 8.

2. Solve the statistics data problem using the given data base, find the mode of the given data.

15, 13, 12, 19, 18, 14, 16, 11, and 15

Solution:

Rearrange the given data for ascending order.

11, 12, 13, 14, 15, 15, 16, 18, 19

Mode:

The most common value in given data is called mode.

Most common value is 15.

Therefore, mode is 15.

3. Solve the statistics data problem using the given data base, find the mode of the given data.

85, 83, 82, 89, 88, 84, 86, 84, and 84

Solution:

Rearrange the given data for ascending order.

82, 83, 84, 84, 84, 85, 86, 88, 89

Mode:

The most common value in given data is called mode.

Most common value is 84.

Therefore, mode is 84.

I like to share this Composite Functions with you all through my article.

More Example Problem for Define Mode in Math:

4. Solve the statistics data problem using the given data base, find the mode of the given data.

4, 7, 5, 2, 9, 7, and 1

Solution:

Rearrange the given data for ascending order.

1, 2, 4, 5, 7, 7, 9

Mode:

The most common value in given data is called mode.

Most common value is 7.

Therefore, mode is 7.

5. Solve the statistics data problem using the given data base, find the mode of the given data.

394, 387, 394, 392, 389, 390, and 391

Solution:

Rearrange the given data for ascending order.

387, 389, 390, 391, 392, 394, 394

Mode:

The most common value in given data is called mode.

Most common value is 394.

Therefore, mode is 394.

Preposition and its types

Preposition is one of the parts of speech in English grammar. Preposition is a part of speech that expresses the relation between noun or pronoun and other words in a sentence. In, of, at, on are some commonly used prepositions. For example: There are wide ranges of Hot Wheels India toys in online kids’ stores. There are five different types of prepositions. Let’s have a look at the same along with examples in this post.
Simple Prepositions: In, on, at, under, over, off, for, from, to, of etc. are examples of simple prepositions.

\For example:
• She got a toy for her child from Disney store . (For, from)
• Do you get Hot Wheels India collection at online stores? (At)
• I came from my cousin’s place last evening. (From)
Compound Prepositions: Within, without, inside, outside, into, beside, behind, between etc. are examples of compound prepositions. For example:
• There is a Disney store behind the red building. (Behind)
• Fisher Price bouncer is really comfortable traveling with baby outside the house. (Outside)
• There is a gap between the two rods. (Between)
Double Prepositions: Out of, from behind, from beneath etc. are examples of double prepositions. For example:
• She got the toy from beneath the lane. (From beneath)
• Fisher Price bouncers are one of the best bouncers out of all in the market. (Out of)
• She came from behind the stage. (From behind)
Participle Prepositions: Concerning, notwithstanding, pending, considering etc. are examples of participle prepositions. For example:
• She got her a Barbie doll considering her fondness for dolls. (Considering)
• Her mobile dues for the month are still pending. (Pending)
• There was little hope, notwithstanding they moved ahead. (Notwithstanding)
Phrase Prepositions: By means of, with regards to, instead of, for the sake of etc. are examples of phrase prepositions. For example:
• She got a pair of shoes instead of floaters for her child. (Instead of)
• I am pursuing MBA for the sake of my parents. (For the sake of)
• She earns her living by means of designing clothes. (By means of)
These are the types of prepositions.

Probability Worksheets Elementary

Introduction to elementary probability worksheets

Elementary probability is a branch of mathematics. Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory that is used extensively in such areas of study as mathematics, statistics, finance, gambling, science, and philosophy to draw conclusions about the likelihood of potential events and the underlying mechanics of complex systems. In this article we shall discuss elementary probability worksheets. I like to share this Elementary Statistics Problems with you all through my article.
(Source: wikipedia).

Elementary Probability Worksheets Example Problem

Following are some example worksheets problems

Example 1:

If three coins are tossed, then what is the sample space?

Solution: Sample space S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.

Example 2:

Four coins are tossed simultaneously, then what is the sample space?

Solution:

Toss four coins at the same time the sample space are:

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THTH, TTHH,

THHT, HTTT, THTT, TTHT, TTTH, TTTT}

Example 3: Two dice are thrown. What is the sample space?

Solution: The sample space in throwing two dice

S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.

Example 4:

Two die is tossed once. Find the probability of getting the number 3.

Die is tossed once, the sample space is S= {1, 2, 3, 4, 5, 6} and, therefore, n(S)=6..

Let E1 = event of getting the number 3.

E1= {3} and, therefore, n (E1) =1.

Therefore P (getting 3) = P(E1) =  n(E1) / n(S) = `1 / 6`

Example 5:

Two die is tossed once. Find the probability of getting the number 4.

Get a number less than 4.

Let E1 = event of get a number less than 4. Then,

E1= {1, 2, 3} and, therefore, n (E3) =3.

Therefore P (get number less than 4) = P (E1) = n(E1) / n(S) = `3 / 6 ` = `1 / 2` . Please express your views of this topic define irrational number by commenting on blog.

Elementary Probability Practice Worksheets Problem

Problem 1:

One coin is tossed. What is the sample space?

Answer:

Sample space= {HT, TH}

Problem 2:

Two coins are tossed simultaneously. What is the sample space?

Answer:

The sample space is S = {HH, HT, TH, TT}

Problem 3:

Two die is tossed once. Find the probability of getting the number 3.

Answer:

`1 /3`

Sum of Uniform Distributions

Introduction to Sum of Uniform Distributions:

Uniform distributions is the part of the probability distributions, It is one of the simplest way of distributions. The Sum of uniform distributions is also known as uniform random distributions. Uniform random distributions is analysis numerical data value.That datas are arranged in systamatically.In this article we see the types of uniform distributions, and some example problems of the sum of uniform distributions.

Types of Uniform Distributions:

Sum of  Uniform Distributions has three types ,there are

Continues Uniform Distributions
Discrete Uniform Distributions


1.Continues Uniform Diostributions:

The continues uniform distributions is the density function of the random variable. Continues interval between a and b. The density function of continues uniform distribution is ,

` [ f(x) = {( 1/ (b-a) when a<= x <= b) , ( 0 when xb):} ]`

Now see the graph of continues uniform distributions with a=1, b=3

2.Discrete Uniform Distributions :

Discrete distributions are also known as statistical distributions, It is the simplest way of Probability distribution functions.

Discrete distribution   of Probability distributions functions of p(Xm) is defined  over m= 1,2,….,N ,

The discrete distribution function as

D(Xn)= ∑ p(Xm)

Uniform Random Distribution General Formula:

The general formula of probability density function of the uniform random distribution function is defined as follows:

f(x) = 1 /b-a               for  a` <=` x `<= ` b

Where a is the position parameter and (b-a) is the scale parameter. In case where a = 0 and b = 1 is called the standard uniform random distribution.

The equation of the standard uniform random distribution is

f(x) = 1          for   0 ` <=` x `<= ` 1.

These all are important in the Uniform random distributions. Is this topic how do you determine if a polynomial is the difference of two squares? hard for you. Watch out for my coming posts.

Examples Problems for the Sum of Uniform Distributions

Example 1:

f(x) = 6x^3+8 for 0 x  1. Find the expected value continuous of given f(x).

Solution:

E (X) = X’ =  x f(x) dx

=   x (6`x^3` +8) dx

=    (6`x^3` +8x) dx

= [6`x^4` /4 + 8`x^2` /2]

=6/4+8/2 – 0 -0

E(X) = 22/4 = 5.5

Example 2:

f(x) = x^3+4x^2 for 0 x  1. Find the expected value continuous of given f(x).

Solution:

E (X) = X’ =  x f(x) dx

=   x (`x^3` +4`x^2` ) dx

=    (`x^4` +4`x^2` ) dx

= [`x^5` /5 + 4 `x^3` /3]

= 1/5 +4/3 -0 -0

E(X) = 23/15 = 1.5333

Exercise Problems for  the Sum of Uniform distributions

Problem -1: f(x) = 8x for 0 x  1. Find the expected value continuous of given f(x).

Answer:  2.6667

Problem -2: f(x) = `x^2` (2`x^4` ) for 0 x  1. Find the expected value continuous of given f(x).

Answer: 0.25

Problem -3: f(x) = (`4x)^3` for 0 x  1. Find the expected value continuous of given f(x).

Answer:  12.8

Set Theory Problems and Solutions

Introduction to set theory problems and solutions:
A set theory should be one of the major concepts in mathematics. All the operations in set theory could be based on sets. Set should be a group of individual terms in domain. The universal set is set that having every element of domain. Finding the solutions for set theory problems are very easy. Here we will see about the example problems with solutions in set theory.

Example Problems:

Let we see few set theory problems and solutions.

Example 1:

If the sets, P, Q and R are, P = {1,2,3,4,5,6}, Q = {2,5,7,9} and R = {2,4,7,10}. Find,

i) P?(QnR)

ii) Pn(QnR)

Solution:

Given, sets P, Q and R are,

P = {1,2,3,4,5,6}

Q = {2,5,7,9}

R = {2,4,7,10}

i) P?(QnR):

First we need to find QnR.

QnR = {2,7}

Then, P?(QnR) = {1,2,3,4,5,6} ? {2,7}

= {1,2,3,4,5,6,7}

ii) Pn(QnR):

First we need to find QnR.

QnR = {2,7}

Now, Pn(QnR) = {1,2,3,4,5,6} n {2,7}

= {2}

Example 2:

Prove the associative laws for the sets A = {a,b,c,d,e,f}, B = {a,c,g,h} and C = {b,c,g,h,k}.

Solution:

Given sets are,

A = {a,b,c,d,e,f}

B = {a,c,g,h}

C = {b,c,g,h,k}

Associative laws are,

i) A?(B?C) = (A?B)?C

ii) An(BnC) = (AnB)nC

i) A?(B?C) = (A?B)?C:

Find A?(B?C):

B?C = {a,b,c,g,h,k}

A?(B?C) = {a,b,c,d,e,f,g,h,k}

Find (A?B)?C:

A?B = {a,b,c,d,e,f,g,h}

(A?B)?C = {a,b,c,d,e,f,g,h,k}

Therefore, A?(B?C) = (A?B)?C.

ii) An(BnC) = (AnB)nC:

Find An(BnC):

BnC = {c,g,h}

An(BnC) = {c}

Find (AnB)nC:

AnB = {a,c}

(AnB)nC = {c}

Thus, An(BnC) = (AnB)nC.

Hence, the associative laws are proved. Please express your views of this topic how to round numbers by commenting on blog.

Example Problem 3:

The sets A = {1,2,3,4}, B = {2,3,5,6} and C = {1,2,6,7,8}. Find,

i) A-B

ii) A-(BnC)

Solution:

Given sets are,

A = {1,2,3,4}

B = {2,3,5,6}

C = {1,2,6,7,8}

Find A-B:

A-B means that the set of numbers that are in A not in B. That is,

A-B = {1,4}

Find A-(BnC):

BnC = {2,6}

A-(BnC) = {1,3,4}

Answer:

i) A-B = {1,4}

ii) A-(BnC) = {1,3,4}

These are few set theory problems with solutions.

Function Rules in Math

Introduction to function rules in math:

An association f from x to y will be called as a function if every element x is associated to a unique element.  Here x and y are two non-empty sets.  x is called the domain of f and the set y is called the range of f.
Understanding Elementary Statistics Problems is always challenging for me but thanks to all math help websites to help me out. Now let us see more rules to satisfy a particular definition of a function.

More Definitions on Function Rules in Math:

Definition 1: A function f from x to y is called an into function, if some elements of co-domain are not images.

Example: Let x = {1, 2, 3, 4} and y = {a, b, c}.

Therefore By f: x `|->`  y, we have {(1, a), (2, b), (3, b), (4, a)}

This is a function but ‘c’ is not associated.  Hence it is called an into function.

Definition 2: A function f from x to y is called an onto function if all the elements of y co-domain y are images.

That is f(x) = y.

Ex: Let A = {1, 2, 3}, B = {4, 5, 6}.

Then f: A `->` B are given by {(1, 4), (2, 5), (3, 6)}.

implies f (A) = B.

Therefore f is called as an onto function.Is this topic adding exponents hard for you? Watch out for my coming posts.

More Definitions on Function Rules in Math:

Definition 3:  A function f from x to y is called a one – one function of if different elements of domain x will have different images.  That is x1? x2, f(x1) ? f(x2).

Ex:  Let x = {1, 2} and y = {a, b}.

Then f: x`->` y are given by {(1, a), (2, b)}.

Here 1 ? 2 and their images a ? b.

Hence it is called an one – one function.

Definition 4: A function f from x toy is called a one – one function if two or more elements of domains x are associated with one element of co-domain y.

Definition 5: A function f from x to y is called a bisection if it is both one – one and onto.

Elementary Calculus Problems

Introduction to elementary calculus problems:

Elementary calculus problems deal with solving basic problems in calculus which help to explain the elementary calculus clearly. Calculus is the defined as the process of finding the rate of change of the given function with respect to the change in the given function. In order to find the rate of change, calculus is divided into differential calculus and integral calculus. The following are the example problems in elementary calculus.Please express your views of this topic Mean Value Theorem Example by commenting on blog.

Elementary Calculus Example Problems:

Example 1:

Solve the elementary function by differentiation.

f(n) = 2n 3 – 5n 6  + 7n

Solution:

The given function is

f(n) = 2n 3 – 5n 6  + 7n

Perform differentiation operation for the given function

f '(n) = 2(3 n 2) – 5(6n 5 ) + 7

Solving the above terms we get,

f '(n) = 6n 2 – 30n 5 + 7   is the answer.

Example 2:

Solve the elementary function by differentiation.

f(n) = 4n2 – 7n 3 – 4n 6  + 5

Solution:

The given function is

f(n) = 4n2 – 7n 3 – 4n 6  + 5

Perform differentiation operation for the given function

f '(n) = 4(2n) – 7(3n 2 ) – 4( 6n 5) + 0

Solving the above terms we get,

f '(n) = 8n – 21n 2 – 24n 5  is the answer.

Example 3:

Solve the elementary function by integration.

`int ` f(n) dn = 4+4n2+ 6n3 + 5n4 dn

Solution:

The given function is

`int ` f(n) dn = 4+4n2+ 6n3+ 5n4 dn

`int ` f(n) dn = `int ` (4+4n2+ 6n3 + 5n4) dn

`int ` f(n) dn = `int ` 4dn + `int ` 4n2 dn+ `int ` 6n3 dn + `int ` 5n4 dn

Perform integration operation for the given function

We get

`F(n) = (4n) + (4n^3)/3 + (6n^4)/4 + (5n^5)/5`

Solving the above function we get

`F(n) = 4n+ (4n^3)/3 + (3n^4)/2 + n^5` is the answer.

Example 4:

Solve the elementary function by integration.

`int ` f(n) dn = n5+ 3n2+ 4n dn

Solution:

The given function is

`int ` f(n) dn = n5+3n2+4n dn

`int ` f(n) dn  = `int ` (n5 + 3n2+ 4n) dn

`int ` f(n) dn = `int ` n5 dn + `int ` 3n2 dn + `int ` 4n dn

Perform integration operation for the given function

We get

`F(n) = (n^6)/6 + (3n^3)/3 + (4n^2) /2`

Solving the above function we get

`F(n) =(n^6)/6+ n^3 +2n^2` is the answer.Is this topic how to solve math problems hard for you? Watch out for my coming posts.

Elementary Calculus Practice Problems:

1) Solve the elementary function by integration.

`int ` f(n) = 4n2+ 6n + 5 dn

Answer: `F(n) = (4n^3)/3 + 3n^2 + 5n`

2) Solve the elementary function by differentiation.

f(n) = 13n +5n 3 – 7n 4

Answer: f '(n) = 13 + 15n 2 – 28 n 3

Elementary Math Tutorial

Introduction to elementary math tutorial:

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. In elementary math tutorial, the basic operations and problems are clearly taught by the tutors with the examples solved in steps. Now we are going to see about elementary math tutorial.

About Elementary Math Tutorial:

Now we see the elementary mathematics with the help of the tutorial problems.

1. Number system:

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

Tutorial Example:

Find the fraction for the given decimal number 0.66

Solution:

The given decimal can be simplified as follows,

0.66 = 6 tenths + 6 hundredths

= 6 × `(1/10)` + 6 × `(1/100)`

= `(6/10)` + `(6/100)`

= `66/100`

= `33/50` .

2. Arithmetic:

In arithmetic, we can see about the housing finance, speed, time, distance, ratio etc.

Tutorial example:

Find the ratio of 5 Kg to 850 g

Solution:

The ratio can be determined and explained below

5 Kg = 5 × 1000 = 5000 g

The ratio can be given as

Required ratio = 5000: 850

= 500: 85

= 100: 17

Thus, the ratio is given as 100: 17

Other Elementary Math Tutorials:

1. Algebra:

Algebra is the part of mathematics where it consists of many expressions such as variables, constants and coefficients. Now we see an example

Tutorial example:

Determine and expand the given expression: (2a + 2b + 2c)2

Solution:

Let us take the equation with ( x + y + z)2

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

(2a + 2b + 2c)2 = (2a)2 + (2b)2 + (2c)2 + 2(4ab + 4ac + 4bc)

= 4a2 + 4b2 + 4c2 + 8ab + 8ac + 8bc

2. Geometry:

Geometry is about the learning of various size, shape and properties of the geometrical solid figures.

Tutorial example:

Find the diameter of the circle where the circumference of circle is about 8 cm.

Solution:

The circumference of the circle determined as follows,

C = π × d

8 = 3.14 × d

d = `8/3.14`

d = 3.18cm.

The diameter of the circle is determined as 2.54 cm.

Ratio Problems with Solutions

Introduction to ratio problems with solutions.
Two values can be compared by (i) Subtraction (ii) Division.

If the weights of two quantities A and B are given by 8 kg and 32kg, then the weight of B is 32 – 8 = 24 kg more than the weight of A.

Also `32 / 8` = 4. The weight of B is in 4 times the weight of A.

Therefore, ratio is a relationship between two quantities which expresses how many times one quantity is of the other.

The ratio of two quantities x and y can be expressed as follows:

(i) as a fraction `x / y` .

(ii) as a division x ÷ y

(iii) also as x : y

The ratio can be obtained by dividing the first quantity by the second quantity.

Now let us see few problems on this topic ratio problems with solutions.

Example Problems on Ratio Problems with Solutions.

Ex 1: A plot is 100 m long and 20 m wide. Find the ratio between the following:

(i) Wide and length

(ii) Length and perimeter

Soln: Here wide = 40m, Length = 100 m

Therefore perimeter = 2 (wide + length)

= 2 (40 + 100) = 280 m.

(i) The ratio of wide and length = `40 / 100` = `2 / 5`

Therefore the ratio is 2 : 5.

(ii) The ratio of length and perimeter = `100 / 280` = `5 / 14` .

Therefore the ratio is 5 : 14.

Ex 2: Paul secured 45 marks out of 50 in English and 25 out of 30 in French. In which subject did he perform better?

Soln: Let us find the ratios of the marks.

In English, it is `45 / 50` = `9 / 10` `=>` 9 : 10

In French, it is `25 / 30` = `5 / 6` `=>` 5 : 6.

Hence Paul performed well in English as 9 : 10 is greater than 5 : 6

I like to share this How to do Ratios with you all through my article.

More Example Problems on Ratio Problems with Solutions.

Ex 3: Divide 1250 dollars between two persons in the ratio 12 : 13.

Soln: Here the total ratio is 12 + 12 = 25.

Therefore the value of 1 ratio = `1250 / 25` = 50

Therefore two persons will get 12 * 50 = 600 dollars and 13 * 50 = 650 dollars.

Ex 4: Two numbers are in the ratio 2 : 7. If the second number is 378, find the first number.

Soln: Given: `(First number) / 378` = `2 / 7` `=>` First number = `2 / 7 xx 378` = 108

Therefore the first number is 108.

Real Valued Function

Introduction to real valued function:

A function whose range is within the real numbers be assumed to be a real function, moreover called a real-valued function. During math, a real-valued function is a function to associates near each part of the domain a real number within the image. f be a function as of set A toward a set B but all element x within A be able to be related through a unique element within B. It can be written as, `f:A->B`

Examples of Real Valued Functions:

A real valued function f : A to B or just a real function 'f ' be a rule which associates near all feasible real number xε A, a unique real number f(x)εB, while A with B are subsets of R, the set of real numbers.

In further words, functions whose domain with co-domain be subsets of R, the place of real numbers, be call real valued functions.Is this topic Properties of Real Numbers hard for you? Watch out for my coming posts.

Problem 1 for real valued function

Solve the field of the real valued square root function f known .

f(x) = sqrt(4x)

Solution :

The known function be as follow

f(x) = sqrt(4x)

This collected square root function. The domain be establish through Finding  the inequality

4x > = 0.

The answer set for the exceeding dissimilarity be the domain along with is specified through the interval

(0 , +infinity)

Problem 2 for Real Valued Function :
The known function is a rational

f(x) = (x - 1) / (x - 7)

Its field is the put of all real numbers except for individual’s value of x to create the denominator zero. Therefore the field is known through the interval

(-infinity , 7) U (7 , +infinity)

Comparing Fractions Answers

Introduction for Comparing Fractions Answers:

A fraction (from the Latin fractus, broken) is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (½, ?, ¾, etc.) and which consist of a numerator and a denominator.

Source – Wikipedia.

Comparing Fractions Answers - which are Smaller:

The examples show the fractions compare the smaller.

Comparing  fractions which are smaller: `1/7` or `2/9` ?

Solution:

The LCD for the denominators 7, 9 is 63.

Multiply the numerator `(1*9)/63` and `(2*7)/63` .

The solution for the fractions is `9/63` and `14/63`

When comparing the answers `1/7` is smaller.



Comparing the fractions which are smaller: `4/3` or `12/7` ?

Solution:

The LCD for the denominators 3, 7 is 21.

Multiply the numerator `(4*7)/21` and `(3*12)/21` .

The solution for the fractions is `28/21` and `36/21`

When comparing the answers `4/3 ` is smaller.

Between, if you have problem on these topics how to solve proportions, please browse expert math related websites for more help on how to find the volume of a cube.

Comparing the fractions which are smaller: `17/9` or `21/8` ?

Solution:

Convert the given fractions in to decimal form.

Divide the fractions as (17÷9) and (21÷8).

The solution for the fractions is `17/9` = 1.88 and `21/8` = 2.62

When comparing the answers `17/9` is smaller.

Comparing Fractions Answers - which are Greater:

The examples show the fractions compare the greater.

Comparing the fractions which are greater: `13/7` or `5/3` ?

Solution:

The LCD for the denominators 7, 3 is 21.

Multiply the numerator `(13*3)/21` and `(7*5)/21` .

The solution for the fractions is `39/21` and `35/21`

When comparing the answers `13/7` is greater.

Comparing the fractions which are greater: `9/3` or `12/5` ?

Solution:

The LCD for the denominators 3, 5 is 15.

Multiply the numerator `(9*5)/15 ` and `(3*12)/15` .

The solution for the fractions is `45/15` and `36/15`

When comparing the answers `9/3` is greater.

Comparing the fractions which are greater: `23/8` or `33/6` ?

Solution:

Convert the given fractions in to decimal form.

Divide the fractions as (23÷8) and (33÷6).

The solution for the fractions is `23/8` = 2.875 and `33/6` = 5.5

When comparing the answers `33/6 ` is greater.

Definition of Mathematical Function

Introduction to definition of mathematical function:

A mathematical function is equations that assigns  only answer to every possible query. A variable indicates to unknown value in the function's equation. The function's key variable is generally denoted by an x. This is the self-determining variable to the purpose. The function's output is a variable generally denoted by y. Let us see about the topic definition of mathematical function are given some example with solution are shown below content.

Example Problem for Definition of Mathematical Function

Example problem 1:-Using definition of mathematical function
To calculate the equation x =16y+1

Solution:

Step: 1 x is a function of y, because of every value of y; there is only one value of x.

Step: 2 If we substitute y=4, we get x=65 and no other value.

Step: 3 The values of x only depend on the values selected for y.

Step: 4 So from this we obtain, independent variable is y and needy variable is x.

Example Problem 2:-using Definition of Mathematical Function

To calculate the equation A =9B+4

Solution:

Step: 1 A is a function of B, because of every value of B; there is only one value of A.

Step: 2 If we substitute B=2, we get A=22 and no other value.

Step: 3 The values of A only depend on the values chosen for B.

Step: 4 So from this we obtain, independent variable is B and dependent variable is A.

Between, if you have problem on these topics functions and linear equations and inequalities, please browse expert math related websites for more help on math answers to problems.

Example Problem 3:-using Definition of Mathematical Function

To calculate the given equation 4x2+6x+4

Solution:

Step 1: Given 4x2+6x+4

Step 2: It is a different example and it can be write in a function like,

Step 3: f(x)= 4x2 + 6x + 4

Step 4: Function is nothing but just substitution.

Step 5: Simply, f(x) =f (a) when x=a

I.e if x=2 means like

y=4(22) +6(3) +4

f (2)=4(4)+18+4

=16+18+4

=38.

So when x=2, the value of the function f(x) =38.

Non Negative Real Numbers

Introduction to non negative real numbers:Let us study about non negative real numbers. Real numbers are commonly defined as the combined form rational and irrational numbers.
Smallest numbers in the ratio form as `a/b` , where ‘b’ should not be equal to ‘0’ at any situation are termed to be as rational numbers.
Numbers that cannot be represented in the ratio forms are termed to be as irrational numbers. Examples for non negative real numbers are below.

Non Negative Real Numbers:

Example 1:

Joseph is having 3 fruits, `1/3` chocolates and` 2/7` nuts, Jennifer is having `1/3` fruits, 3 chocolates and `3/5 ` nuts and Nimmi is having `2/5` fruits, `2/3` chocolates and 2 nuts. Find the total number of fruits, chocolates and nuts?

Solution:

Given:
Joseph having 3 fruits, `1/3` chocolates and `2/7` nuts

Jennifer is having `2/3` fruits, 3 chocolates and `3/5` nuts

Nimmi is having `1/3` fruits, `2/3` chocolates and 2 nuts

To calculate the total number of fruits, chocolates and nuts follow the steps as below:
Fruits:
= 3 + `2/3` + `1/3` (add all the fruit numbers from three of them)

= 3 `(3/3)` + `2/3` + `1/3 ` (multiply and divide ‘3’ with the number 3 in order to make it into fraction)

=` 9/3` + `2/3` + `1/3`

= `(9 + 2 + 1)/3` (take the number ‘3’ as common divisor)

= `12/3`

= 4 (total non negative real numbers of fruits)

Chocolates:
= `1/3 ` + 3 + `2/3`

= `1/3` + 3 `(3/3)` + `2/3` (multiply and divide ‘3’ with the number 3 in order to make it into fraction)

= `1/3 ` + `9/3` + `2/3`

= `(1 + 9 + 2)/3` (take the number ‘3’ as common divisor)

= `12/3`

= 4 (total non negative real numbers of chocolates)

Nuts:
= `2/7 ` + `3/5` + 2

= `(2/7)` `(5/5)` + `(3/5)` ` (7/7)` + 2` (35/35)` (multiply and divide the number ‘5’ with `2/7` , ‘7’ with `3/5` and ‘35’ with 2 in order to get the common divisor as ‘35’ for evaluation)

= `10/35 ` + `21/35` + `70/35`

= `(10 + 21 + 70)/35` (take the number ’35’ as common divisor)

= `101/35` (total non negative real numbers of nuts)

Having problem with One to One Correspondence keep reading my upcoming posts, i will try to help you.

Example 2:

If sita is completing 7 puzzle squares per hour then how hour she will take to complete 119 squares?

Solution:

7 puzzles 1 hour
Then to calculate the total hours required to complete 119 squares are as follows:
=`119/7`

= 17 hours (total non negative real numbers hours to complete 119 squares)


Exercises:

Rani is having 3 meters of fabrics, Srikanth is having `1/2 ` meters of fabrics and Mani is having `1/5` meters of fabrics. Find the total meters of fabrics they have with them? (Answer: `37/10 ` meters)
If Meena is having `2/5` liters of coke and Nakul is having `3/5` liters of coke. Find the total liters of coke they have with them? (Answer:1 liter)

Study Online Algebraic Expressions with Variables

Introduction to study online algebraic expressions with variables:

Algebraic expressions are the mixture of numbers, letters and the basic arithmetic operations. The Variables are the symbols used in algebraic expressions to represent letters, and the constants are symbols used in algebraic expressions to represent variables.

Online Math problem is expected to be one of the smart trials, as it supplies the part an excellent chance to work out. The event is all about trying the originality and logical skills of a person and making mathematics fun.

In this session we are going to discuss the online study of algebraic expressions with variables.

Example Problems:

Study online algebraic expressions with variables problem 1:

Write an algebraic expression for the following statement:

Nine times a certain number

Solution: 

Let us represent ‘a’ be the ‘the number’.

Nine times a certain number = 9  `xx`  a certain number = 9 `xx` a = 9a

Study online algebraic expressions with variables problem 2:

Write the algebraic expression for the following phrase:

Eight more than twice a number

Solution:

Let us choose ‘m’ be ‘the number’.

First we should understand the meaning the given phrase.

The word ‘more than’ stands for the operation ‘addition’.

The word "twice" means ‘two times’.

Twice a number = 2 `xx` a certain number = 2 `xx` m = 2m

Therefore,

Eight more than twice a number = 2m + 8.

Study Online Algebraic Expressions with Variables Problem 3:

Write a real life situation for the following algebraic expression.

8 + n

Solution:

8 are added with a certain number ‘n’.

The given algebraic expression has the following situation.

John has 8 new shirts. His brother gives him some more.

The algebraic expression 8 + n represent the number of new shirts that john has in all.

Study online algebraic expressions with variables problem 4:

Evaluate 4n + 5 `xx` 6 if n = 5.

Solution:

4n + 5 `xx` 6                    (given)

= 4 `xx` 5 + 5 `xx` 6 (n = 5)

= 20 + 5 `xx` 6

= 20 + 30

= 50

Hence the solution is 50.

Study Online Algebraic Expressions with Variables Problem 5:

Evaluate 3x + 10y + 1 if x = 2 and y = 5.

Solution:

3x + 10y + 1                            (Given)

= 3(2) + 10y + 1          (x = 2)

= 6 + 10(5) +1             (y = 5)

= 6 + 50 + 1

= 57

Hence the solution is 57.

Study online algebraic expressions with variables problem 6:

Find the multiplication for the following expressions:

`(3x + 5)` and `(2x^2 + x + 4)`

Solution:

`(3x + 5) xx (2x^2 + x + 4)`          (Given problem)

= `(6x^6 + 3x^2 + 12x) + (10x^2 + 5x + 20)`

= `6x^6 + 3x^2 + 12x + 10x^2 + 5x + 20`

= `6x^6 + 12x^2 + 17x + 20`

Hence the solution is `6x^6 + 12x^2 + 17x + 20`

Practice Problems:

Find the multiplication of the expression (9x + 2) (3x - 1).
Evaluate 9x + 3 if x = 2


Answer key:

`27x^2 -6x -2`
21

Fractional Powers in Math

Introduction to Fractional powers in math:

In math, when the power or exponent is a fraction number then the number said to be having the functional powers in math. The example for the fractional power is `2^ (2/3)`

Rules:

Rules for the various operations on fractional powers in math.

For multiplication:

`m^ (a/b)` *`m^ (c/d)` = `m^ ((a/b) + (c/d))`

For division:

`m^ (a/b)` /`m^(c/d)` =` m^((a/b)-(c/d))`

`(m^ (a/b))^c` = `(m^(c/b))^a`

Model Problems for Multiplication of Fractional Power in Math:

Q 1:  Find the value of the fractional power terms?

` 2^ (1/2)` * `2 ^ (3/2)`

Solution:

When we multiply the two terms with fractional powers. we need the fractional powers.

The rule is

`m^ (a/b)` *`m^ (c/d)` = `m^ ((a/b) + (c/d))`

here,

` 2^ (1/2)` * `2 ^ (3/2)`   = `2 ^(1/2+3/2).`

= `2^((1+3)/2)`

= `2^(4/2)`

= `2^2`

= 4

Q 2 : Find the value of the fractional power terms?

` 2^ (5/2)` * `2 ^ (3/2)`

Solution:

When we multiply the two terms with fractional powers.we need to add the fractional powers.

The rule is

` m^ (a/b)` *`m^ (c/d)` = `m^ ((a/b) + (c/d))`

here

`2^ (1/2)` * `2 ^ (3/2)` =` 2 ^(5/2+3/2).`

= `2^((5+3)/2)`

= `2^(8/2)`

= `2^4`

= 16

Answer is 16.

Model Problems for Division of Fractional Power in Math:

Q 1: Find the value of the fractional power terms?

`2^ (5/2)` / `2 ^ (3/2)`

Solution:

When we divide the two terms with fractional powers. we need to subtract the fractional powers

The rule is

`m^ (a/b)` /`m^ (c/d) ` = `m^ ((a/b) - (c/d))`

here

`2^ (1/2)` / `2 ^ (3/2)`   = `2 ^(5/2-3/2)` .

= `2^((5-3)/2)`

= `2^(2/2)`

= `2^1`

=  2

Answer is 2

Q 2 : Find the value of the fractional power terms?

`2^ (5/2)`   / `2 ^ (3/4)`

Solution :

When we multiply the two terms with fractional powers. we need to subtract the fractional powers

The rule is

`m^ (a/b)` /`m^ (c/d)` = `m^ ((a/b) - (c/d))`

here

`2^ (1/2)` / `2 ^ (3/2)` = `2 ^(5/2-3/4).`

= `2^((10-3)/4)`

= `2^(7/4)`

Answer : 274

Q 3 : Find the value of the `(5^ (4/5)) ^10.`

Solution:

To find the value we use

` (m^ (a/b))^c` = `(m^(c/b))^a`

`(5^ (4/5)) ^10 ` = `(5^ (10/5)) ^4.`

= `(5^ (2)) ^10.`

= `5^ (2*10)`

= `5^20`

Answer is `5^20`

Radical Form Math

Introduction to Radical form Math:

In mathematics subject there are so many chapters and so many topics are there. A radical is one of the important topics in that. A radical number is known as, which is the factor value of under the root symbol, and radical expression is containing a square root. Radical is denoted by the symbol of v (Sqrt).it has two forms, one is radical form, and other one is exponential form. Now we are going to see about radical form in math.

Radical Form Math:

Radical form is also called as Root form. it is nothing but to change the order of the exponential form into radical (root) form.

Important steps to remember how to change radical form:

Radical form into exponential form changed by to the power of exponent. For Example: 3v x = (x) 1/3 .
2.  Exponential form into radical form is change the exponent to under the radical. For example: (y) 1/2 = 2v y.

3.  Xn is called the nth root of x and is written as nv x or n Sqrt (x).

4.  If x and y are positive rational numbers and n is any rational exponent, Then xn * yn = (x * y) n. 

5.  If y = q Sqrt (x) or qv x = x 1/q then, x1/q is the exponential form and qv x or q Sqrt (x) is the radical form of x. (Where, q is the index of the radical, x is the radicand).

6.  If x and y are positive rational numbers and n is any rational exponent, Then xn * yn = (x * y) n.

Example Problems in Radical Form Math:

Radical to exponential form:

Example problem 1:

To change the given radical form into the exponential form 3v (11)

Solution:

Given: 3v (11)

= (11)1/3

The required exponential form is (11)1 / 3.

Exponential form to Radical form:

Example problem1:

To change the given exponential form into the radical form (17) - 1/2

Solution:

Given: (17) -1/2

= (17) - 1/2 = (1 / 17)1/2

The required radical form is 2v 1 / 17).

Example problem 2:

To change the given exponential form into the radical form (27 / 220) - 1/4

Solution:

Given: (27 / 220) -1/4

= (27 / 220) - 1/4 = (220 / 27)1/4

The required radical form is 4v (220 / 27).

Non Real Roots

Introduction to Non real roots:
Non real roots are the imaginary roots. A complex number is a number that consists of real part and imaginary part.   We write it as a + bi where i is the imaginary  unit.  i = v-1
If z = a + bi then a is the real part and  b is called the imaginary part.  a is denoted as Re(z) and b is denoted as Im(z)
Roots of a complex number:-
Non real roots are roots of a complex number.
A number is called the nth root of a complex number z if ?n = z  and ? = z1/n
To find the non real roots  of a complex number, we follow the following steps.
Step1 :  Write the given number in polar form.
Step2 :  Add 2kp to the argument..
Step3 :  Apply De Moivre's theorem.
Step4 :  Put k= 0,1 ...... up to infinity.

Non Real Roots-the Nth Root of Unity:

The nth root of unity:-
Take the number 1
Step1 :  1 =  1(cos 0 + i sin 0)
step 2:         1( cos (0 + 2kp) + i sin)0 + 2kp) =  cos 2kp + i sin 2kp
Step 3: nth root of unity = 11/n = (cos 2kp + i sin 2kp)1/n where k = 0,1 .......(n-1)
=  [ cos 2kp/n + i sin 2kp/n ] where k = 0,1   ... ..(n-1)
Step 4 :when k=0 the nth root is cos 0 + i son 0
when k = 1 the nth root is cos 2p/n + i sin 2p/n,...
So ? = cos 2p/n + i sin 2p/n  which is written as ei 2p/n   

Hence the nth root of unity is e0, ei2p/n, ei4p/n etc  which is written as 1,?,?2, ?3 .......... ?n-1
From the above, we note that the non real roots are in geometric progression and the sum of the non real roots  is zero.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Rationalize a Denominator and Graphing Quadratic Functions in Standard Form. I am sure they will be helpful.

Non Real Roots - Cube Root of Unity:

Cube root of unity:-
Let us find the non real roots of 11/3
Let x = 1 1/3
Then x3 = 1
x3 = cos 0 + i son 0 = cos 2kp + i sin 2kp
x  =  (cos 2kp + i sin 2kp)1/3   =   cos 2kp/3  + i sin 2kp/3  ......... k = 0, 1, 2 ......
Therefore the cube roots are cos 0 + i sin 0  = 1
cos 2p/3 + i sin 2p/3 =  ?
cos 4p/3 + i sin 4p/3 = ?2
The values of the non real roots  ? =   -1 + i v3
2
The value of the non real root  ?2 =  -1-iv3
2
When the points are plotted on garland diagram, we see them lying on the circle  of unit radius.
The roots 1,?,?2 are in geometric progression and the sum of the roots = 0                                                                 

Congruent Figures

Congruent Figures:

Two figures are said to be congruent to each other, if on placing one over the other, they exactly coincide. Congruent figures are same in size and shape both. Two circles are always congruent to each other. In the congruent triangles, the sides and the angle that coincide by superposition are called . If two line are same in size they are congruent.

In a congruent triangle corresponding sides and corresponding angles are equal. Two figures are congruence if there is an involvement with their vertices such that coupled angles are the comparable to and the comparable sides are interrelated. It is called as word like the one congruence.

SAS Property:

If two sides and the included angle of one triangle are congruent to the corresponding two sides and the included angle of another triangle, then the two triangles are congruence.

S.S.S Property: 

If the all three sides of one triangle is equal to the three sides of other triangle, both triangles are know as congruent triangles.

ASA Property:

If two angles and the included side of one triangle are congruence to the corresponding two angles and the included side of another triangle, then the two triangles are congruence.

SAA or AAS Property:

If two angles and a non- included side of one triangle are congruence to the corresponding parts of another triangle, then the triangles are congruence.

R.H.S Property:

In two triangles if one of  angles are right angles and hypotenuse and one side of first triangle is equal to hypotenuse and one side of another triangle, triangles are know as congruent triangles.

The L-L Property (The Leg – Leg Property)

If two legs of one right triangle are congruence to the corresponding legs of another right triangle, then the two triangles are congruence.

The L-AA Property (The Leg-Acute Angle Property)

If a leg and an acute angle of one right triangle are congruence to the corresponding leg and an acute angle of another. Then the two triangles are congruence.

The H-AA Property (The Hypotenuse – Acute Angle Property)

If the hypotenuse and an acute angle of one right triangle are congruence to the corresponding hypotenuse and an acute angle of another. Then the two triangles are congruence.

The H – L Property (The Hypotenuse – Leg Property)

If the hypotenuse and leg of one right triangle are congruence to the corresponding hypotenuse and leg of another, then the two triangles are congruence.

Examples of Congruent Figures:

Statement:

In SSS proof, if the corresponding sides of two triangles are proportional, then they are similar.

The proof of the above property is discussed in the form of the following solution.

Solution:

Given:    

In `Delta`ABC and `Delta`DEF, we are given that

`(AB)/(DE)` = `(BC)/(EF)` = `(AC)/(DF)`   

  
To prove:

`Delta`ABC ~ `Delta`DEF.

Construction: mark point P on DE and the point Q on DF such that DP =  AB, DQ = AC. Join P and Q.

Proof:          

Since  `(AB)/(DE)` = `(AC)/(DF)`                 (given)

`(DP)/(DE)` = `(DQ)/(EF)`                   (construction)

By the converse of Thales theorem, PQ || EF.

`anlge`DPQ = `angle`E                       (corresponding angles)

`angle`DQP = `angle`F                       (corresponding angles)

By AAA similarity ( or by AA similarity)

We get             `Delta`DPQ ~ `Delta`DEF

`(DP)/(DE)` = `(PQ)/(EF)` `rArr` `(AB)/(DE)` = `(PQ)/(EF)`                        (given)

But                  `(AB)/(DE)` = `(BC)/(EF)`                 (given)

BC = PQ and AB = DP, AC = DQ                (construction)

`Delta`ABC `~=` `Delta`DPQ                                    (SSS)

Since               `Delta`DPQ ~ `Delta`DEF

`Delta`ABC ~ `Delta`DEF.

Hence, the required result is proved.

Between, if you have problem on these topics greatest integer function, please browse expert math related websites for more help on mathematical induction.

Properties of Congruent Figures

Congruent figures have following properties:


1. The corresponding part of congruent figures are also congruent.


2. The corresponding sides lie opposite to the equal angles and corresponding angles lie opposite to the equal sides.


3. All squares can be similar but not congruent. If squares have equal size  (length and width ) they are know as congruent squares.


4. All the rectangles can be similar but Congruent rectangle are equal in size and shape.

Example : Find the missing sides of congruent triangles:

AB =  5cm

AC = 3 cm

EG = 3 Cm

EG = 4 Cm

Find BC and EF?

Sol :

Both triangles are congruent ( Given )

So All the corresponding sides are equal

BC = FG = 4cm

EF = AB = 5cm

Solve Linear Equations Graphically

There are various methods to solve linear equations like elimination method, substitution method etc. There is another method by which we can solve linear equations, called graphical method. In this method we need to draw graphs of given equations and the point of contact or the point where both the equation intersect each other becomes the solution of the linear equation. Now understand the steps to solve linear equation graphically.

Steps to Solve Linear Equation Graphically

Step1: Take first linear equation and put x = 0, you will get value of y. lets say y = a, it will give you first point (0,a).

Step2: Now put y = 0, it will give you value of x, lets say x = b, it will give you second point (b,0)

Step3: Plot both points and draw a line between them and extend this line beyond these points, This will be graph of first linear equation.

Step4: Repeat step 1,2,3 for second linear equation, you will get graph of second linear equation.

Step5: Find out the point where these lines intersect each other, lets say point is (u,v)

the solution will be x = u, y = v

Note: In case, both lines does not intersect extend both the lines to find intersecting point and once you get intersecting point, it will be required solution. Only parallel line does not intersect each other.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Online Algebra Calculator and algebra 1. I am sure they will be helpful.

Example to Solve Linear Equation Graphically

The following example will make more clear to you how to solve linear equation graphically.

Question: Solve: x + y = 1 and 2x + y = -2

Solution: x + y = 1

put x = 0, w get y = 1, so first point will be (0,1)

put y = 0, we get, x = 1, so second point will be (1,0)

2x + y = -2

put x = 0, we get, y = -2, so first point will be (0, -2)

put y = 0, we get, x = -1, so second point will be (-1, 0)

Now drawing both the lines on graph we get,

We can see that these lines cut each other at point (-3, 4)

Hence Solution will be x = -3 and y = 4.

Elementary Statistics Problems

Introduction to elementary statistics problems: 

Elementary statistics problems were done normally with data collected for specific purposes. In elementary statistics problems, we make decisions about the data by analyzing and interpreting it. There are several methods in elementary statistics for representing data graphically and in tabular form.

The method used in elementary statistics for finding a representative value for the given data are called the measure of central tendency. The three measures of central tendency used in elementary statistics problems are  Arithmetic Mean, also median and mode.

Examples on Elementary Statistics Problems

Example 1:

Find the mean deviation of the mean for given data:

8,5,15,9,13,2,6,14

Solution:

Step 1 Mean of the given data are  `barx`


`barx`  = `(8+5+15+9+13+2+6+14)/8` = `72/8` = 9

Step 2 The deviations of the respective observations from the mean x, i.e., xi– x are
8– 9,5–9,15–9,9–9,13–9,2–9,6–9,14–9      ( or )   –1,–4,6,0,4,–7,–3,5

Step 3 The absolute values of the deviations, i.e.,|xi - x |are  -1,-4,6,0,4,-7,-3,5


Step 4 The required mean deviation about the mean is

M.D. (`barx` ) = `sum` 8 i-1 |xi-x| / 8

= `(1+4+6+0+4+7+3+5)/8 ` = `30/8` = 3.75

Example 2:

Find the mean deviation of the mean for given data :

13, 4, 19, 18, 5, 10, 18, 20, 21, 8, 15, 18, 2, 3, 16, 11, 3, 1, 10, 5

Solution:

find the mean ( `barx` ) of the given data

`barx` = `1/20` `sum` 20i-1  xi = `220/20` = 11


The respective absolute values of the deviations from mean, i.e.,|x-`barx` | are

2,7,8,7,6,1,7,9,10,3,4,7,9,8,5,0,8,10,1,6

Therefore
`sum` 20i-1 |xi - `barx` | = 118

and M.D. ( `barx` ) = `118/20` = 5.9

Examples on Elementary Statistics Problems

Example 3:

Find the mean deviation of the median for the following data: 5,11,7,5,14,12,20,6,8,21,23.

Solution :

Here the total number of observations is 11 which is odd. Arranging the data into ascending order,

5 , 5 , 6 , 7 , 8 , 11 , 12 , 14 , 20 , 21 , 23

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi - M| are

6 , 6 , 5 , 4 , 3 , 0 , 1 , 3 , 9 , 10 , 12
Therefore

? 11i-1 |xi - M| = 59

M.D.(M) = `1 / 11` ? 11i-1 |xi - M| = `(1 / 11) * 59`   = 5.36

Example 4:

Find the mean deviation of  the median for the following data: 10,5, 6, 3, 12, 11, 18, 4, 7, 18, 22.
Solution:

Here the total number of observations is 11 which is odd. Arranging the data into

ascending order, we have 3 , 4 , 5 , 6 , 10 , 11 , 12 , 18 , 18 , 19 , 22

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi - M| are

8 , 7 , 6 , 5 , 1, 0 , 1, 7 , 7 , 8 , 11
Therefore
`sum` 11i-1 | - M| = 61

and  M.D.(M) = 1/1111i-1 |xi - M| =` 1/11*61 ` =5. 545. 

Lcm Prime Factorization

Introduction to LCM prime factorization:

Prime factorization is defined as the two factors of number is must be a prime number.

For example:

The prime factorization of the number x is y and z. Then the factors of the number y and z.

Y must be 1 and y similarly the factor of the number z must one and z. In this article we shall discuss about LCM prime factorization with suitable example problems

I like to share this What is LCM with you all through my article.

Example Problem for Lcm Prime Factorization:

LCM prime factorization problem: 1

find the LCM prime factorization for 24, 12.

solution:

Given that 24 and 12. Find the Factors of two numbers individually

Take the 24 and divide through small number.

`24 / 2` = 12

`12 / 2 ` = 6

`6 / 2` = 3

`3 / 3` = 1

24 = 2 * 2 * 2 * 3

`12 / 2` = 6

`6 / 2` = 3

`3 / 3` = 1

12 = 2 x 2 x 3

prime factorization of 24 = 2 x 2 x 2 x 3

prime factorization of 12 = 2 x 2 x 3

LCM of two numner 12 , 24 is 24.

So the LCM prime factorization of 24 and 12 is 2 x 2 x 2 x 3 = 24 .

LCM prime factorization problem: 2

find the LCM prime factorization for45 ,  25.

solution:

Given that 45 and 25. Find the Factors of two numbers individually

Take the 45 and divide into small number.

`45 / 3` = 15

`15 / 3` = 5

`5 / 5 ` = 1

45 = 3 x 3 x 5

`25 / 5` = 5

`5 / 5 ` = 1

25 = 5 x 5

prime factorization of 45 = 3 x 3 x 5

prime factorization of 25 = 5 x 5.

LCM of 45, 25 = 225

So LCM prime factorization of 45 and 25 is 3 x 3 x 5 x 5 = 225

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Example Problem for Lcm Prime Factorization:

Find the LCM prime factorization for 54, 50

solution:

Given that 54, 50. Find the Factors of two numbers individually

Take the 54 and divide throughout small number.

`54 / 2` = 27

`27 / 3` = 9

`9 / 3` = 3

`3 / 3` = 1

54 = 2 * 3 * 3 * 3

`50 / 2` = 25

`25 / 5` = 5

`5 / 5` = 1

50 = 2 x 5 x 5

prime factorization of 54 = 2 x 3 x 3 x 3

prime factorization of 50 = 2 x 5 x5

LCm of 54 and 50 = 1350

So LCM prime factorization of 54 and 50 is 2 x 3 x 3 x 3 x 5 x 5 = 1350

Example 4:

find the LCM prime factorization for 96, 18

solution:

Given that 96, 18. Find the Factors of two numbers individually

Take the 96 and divide through small number

`96 / 2` = 48

`48 / 2` = 24

`24 / 2` =12

`12 / 2` = 6

`6 / 2` = 3

`3 / 3 ` = 1

96 = 2 * 2 * 2 * 2 * 2 * 3

`18 / 2` = 9

`9 / 3` = 3

`3 / 3` = 1

18 = 2 x 3 x 3

prime factorization of 96 = 2 x 2 x 2 x 2 x 2 x 3

prime factorization of 18 = 2 x 3 x 3

LCM of 96 and 18 = 268

So LCM prime factorization of  96 and 18 is 2 x 2 x 2 x 2 x 2 x 3 x 3= 288

Roman Numerals Adding Answers

Introduction to Roman numerals:
Roman numerals are a numerals system of ancient Rome based on letters of the alphabet, which are combined to signify the sum (or in some cases, the difference) of their values. The first ten Roman numerals are as follows:

I, II, III, IV, V, VI, VII, VIII, IX, and X.

In this article we shall discuss the adding Roman numerals with answers.

(Source: Wikipedia)

Letters in Roman Numerals:

There are seven letters in Roman numerals. These letters are used for adding numerals.

They are the following:

I in decimal is one.

V in decimal is five.

X in decimal is Ten.

L in decimal is fifty.

C in decimal is hundred.

D in decimal is five hundred.

M in decimal is thousand.

Where I, V, X, L, C, D, M.

The rule for adding Roman numerals is:

In order to get answers for adding Roman numerals we must follow the following rules.

Add the amount of the Roman value if one or more letters are located after another letter of larger value.

Subtract the amount of the Roman value if one or more letters are located before another letter of larger value.

I am planning to write more post on calculus problem solver step by step, fractions to decimal converter. Keep checking my blog.

Example Problems for Adding Roman Numerals with Answers:

Example 1:

Add the Roman numerals XV and IX.

Solution:

The given Roman numerals are XV and IX.

XV = 10+5 = 15.

IX = 10-1 = 9

Then adding the Roman numerals XV and IX we get

15+9 = 24

19 can be written as 10+5+ (5-1)

10 in Roman numerals as X

5 in Roman numerals as V

1 in Roman numerals as I

24 in Roman numerals as XXIV.

Finally the answers for adding XV and IX is  XXIV.

Example 2:

Add the Roman numerals XIV and CX.

Solution:

The given Roman numerals are XIV and CX.

XIV = 10+5-1 = 14.

CX = 100+10 = 110

Then adding the Roman numerals XIV and CX we get

14+110 = 124

124 can be written as 100+10+10+ (5-1)

100 in Roman numerals as C

10 in Roman numerals as X

5 in Roman numerals as V

1 in Roman numerals as I

124 in Roman numerals as CXXIV.

Finally the answers for adding XIV and CX is  CXVIV.

Line Segment Chart

Introduction to line segment chart

A line chart or line graph is a type of graph, which displays information as a series of data points connected by straight line segments. It is a basic type of chart common in many fields. It is an extension of a scatter graph, and is created by connecting a series of points that represent individual measurements with line segments (Source: Wikipedia)

Line Segments Chart

Line segments is a division of a line to be surrounded through two finish points. Line contains each point lying on the line among its finish points. Instance of line segments contain the faces of a triangle otherwise square. More usually, while the finish points are equally vertices of a polygon, the line part is any an edging if they are diagonal.

A line chart is over also over again utilized to imagine a development within data in excess of intervals of time; therefore the line is frequently haggard chronologically.Within the experimental sciences, data composed as of testing are frequently imagine through a graph to include an overlay arithmetical meaning representing the finest fit trend of the sprinkled data graph.

For instance, if one were to accumulate data lying on the speed of a body on assured points within time, one could visualize the data:

Example for Line graph

Time: 0 1 2 3 4 5

Speed: 0 4 8 12 14

Time can be represented within x-axis.

Time can be represented within y-axis.

Examples for Line Segment

Example 1

Within a line segment one endpoint is (-2, 6) also the midpoint of the line is (- 6, 10). Compute the co-ordinates of the other endpoint are (x1, y1).

Solution:

`(-6,10)=((x_(1)+(-2))/2),(y_(1)+6)/2`

To find for x1,

`-6=((x_(1)+(-2))/2)`

-12= x1-2

-10 = x1

To compute for x2,

`10=((x_(2)+(6))/2)`

20 = y1 + 6

Y1 = 14

So, the next endpoint of the line is (-10, 14)


Example 2

Within a line segment one endpoint is (-4, 8) also the midpoint of the line is (- 6, 8). Compute the co-ordinates of the other endpoint are (x1, y1).

Solution:

`(-6,8)=((x_(1)+(-4))/2),(y_(1)+8)/2`

To compute for x1,

`-6=((x_(1)+(-4))/2)`

-12= x1-4

-8 = x1

To solve for x2,

`8=((x_(2)+(8))/2)`

16 = y1 + 8

y1 = 8

So, the next endpoint of the line is (-8, 8)

Elementary Matrices

Introduction to Elementary Matrices:

Elementary matrix operation play an significant role in many matrix algebra application, such as finding the inverse of a matrix and solving instantaneous linear equations.

Elementary Operations

The three kind of basic matrix operations.

Interchange two columns.

Multiply each element in a column by a non-zero number.
Multiply a column by a non-zero number and add the result to another column.
While these operation are performed on rows, they are called basic  row operations; and when they are perform on columns, they are called basic column operations.

Types of Elementary Matrices Operations

There are three types of elementary matrices, which match to three types of row operations .

Row switching                RI—>RJ
A row within the matrix can be switched with a new row.

Row multiplication           K RI—>RJ, where k?0            
Each element in a row can be multiplied by a non-zero even.

Row addition              RI+ KRj—> RI
A line can be replaced by the sum of the  row and a multiple of a new row.

The elementary matrix for any row operation is obtained by execute the operation on an identity matrix.

Elementary Operation Notation

In many reference, you resolve encounter a compacted notation to describe elementary operations. That notation is given by

Row operations

1. Interchange rows i and j Ri <--> Rj

2. Multiply row i by s, where s ? 0 sRi --> Ri

3. Add s times row i to row j sRi + Rj --> Rj

Column operations

1. Interchange columns i and j Ci <--> Cj

2. Multiply column i by s, where s ? 0 sCi --> Ci

3. Add s times column i to column j sCi + Cj --> Cj

Rules on Elementary Matrices

Elementary Matrices

Now, imagine R has 1 ,n>n elementary square matrix obtained from the identity matrix by performing arts an basic operation. A result, we have three types of elementary matrices, every subsequent to a type of elementary operations:

Transposition matrix Tij : an matrix obtain from In with rows i and j switch,

Basic diagonal matrix Di(r) : a diagonal matrix whose entries are 1 except in cell (i' > i)  , whose entry is a non-zero element r of R

Row substitution matrix Eij(s) : In+sUij , where s > R and Uij is a matrix unit with i=j  .



Examples based on elementary matrices:

Ex 1: Construct a 3 × 2 matrix whose entries are given by aij = i - 2j

Sol : The general 3 x 2 matrix is of the form,

A= [aij]  =`[[a_11,a_12],[a_21,a_22],[a_31,a_32]]`

where ,i= 1,2,3(rows),j= 1,2 (columns)

it is given that aij = i-2j

a11 =1-2 = -1

a12 =1-4 = -3

a21  =2-2 =0

a22 =2-4 = -2

a31 = 3-2 =1

a32 = 3-4 = -1

Therefore , The required matrix  is  A = `[[-1,-3],[0,-2],[1,-1]]`

Ex 2: Elementary column operation on the term “A”, where the identity is used in matrix   `A= [[1,6], [4,1], [0,1]]` and identity matrix `x = [[1,0], [0 ,1]]`

Sol:

Given: Identity matrix `x = [[1,0], [0 ,1]]`

`A= [[1,6], [4,1], [0,1]]`

Step 1:  The given identity matrix x is interchanged of first and second column and the finding result is named it   (x1).

Step 2: To do the elementary column operation on given `A= [[1,6], [4,1], [0,1]]`

Step 3: The operation in matrix is done by [row *column]

`X= [[1,0], [0,1]]`

The first and second column is interchanged and the result is,

`X_1= [[0,1], [1,0]].`

Now, the found X1 is calculated with given “A”

`A= [[1,6], [4,1], [0,1]]`

`X_1= [[0,1], [1,0]]`

`A*X_1` for calculation elementary column operation on

`A*X_1`

` [[1,6], [4,1], [0,1]] * [[0,1], [1,0]]`

` [[1*0+6*1 , 1*1+6*0],[4*0+1*1 , 4*1+1*0],[0*0+1*1, 0*1+1*0]]` 

`[[0+6, 1+0], [0+1,4+1], [0+1, 0+0]]`

`[[6,1],[1,4],[1,0]]`

Hence ,the elementary column operation for the given value `A=[[6,1],[1,4],[1,0]]`

Practice arithmetic word problems

Learning arithmetic/arithmetics (from the Greek word ἀριθμός = number) is the very oldest and most basic branch of mathematics, used by almost everyone, for tasks collection from simple day-to-day counting to highly developed science and business computations. It refers to the simpler properties while using the conventional operations of addition, subtraction, multiplication and division with lesser values of numbers. Let us see practice arithmetic word problems in this article.

Practice Arithmetic Word Problems - Addition and Subtraction:

Addition Problem :

Abraham has 20 ice creams; Antony has 15 ice creams. Find out the total number of ice creams from both of them?

Solution:

Abraham has 20 Ice creams

Antony has 15 ice creams

Totally 20 + 15 = 35 ice creams

Subtraction Problem :

There are 40 students in a class room. 32 students got pass in computer science. How many students failed?

Solution:

Total students in the class room = 40

Pass in computer science students = 32

Therefore failed students = 40 – 32 = 8.

Practice arithmetic word problems - Multiplication and Division:

Multiplication Problem :

Victoria delivers 80 newspapers in a day. How many newspapers does she deliver in 22 days?

Solution:

Per day Victoria delivers newspapers = 80

If she delivers for 22 days, then 80 * 22 = 1760

Hence she can deliver 1760 newspapers in 22 days.

Division Problem:

Davina has 72 sweets. If she shares with her friends 4 sweets each, how many friends can share her sweets with?

Solution:

Davina has 72 sweets.

Each friend shares 4 sweets.

Total friends shares with her sweets = `72 / 4` = 18.

Hence 18 friends can share her sweets with.

Practice Problems – Practice Arithmetic Word Problems:

Problem 1:

There are 14 apples and 20 oranges in a fruit basket. Totally how many fruits are there in the basket?

Answer:

Total fruits in the basket = 34 fruits

Problem 2:

There are 67 students in a class. 12 students were not wear uniform. How many students are wearing uniform?

Answer:

Number of students wearing uniform = 55

Problem 3:

A balloon cost is $2.What is the cost of 12 balloons?

Answer:

Cost of 12 balloons = $24

Problem 4:

5 toys cost is $25. What is the cost of 1 toy?

Answer:

Cost of 1 toy = $5

Basics about the ogive statistics

We consider the following types of graphical representation of frequency distribution in statistics.
(1) Histogram or Area diagram
(2) Frequency polygon
(3) Ogive or cumulative frequency distribution graph
We shall discuss a bit in detail about the ogive graph in this article. The curve that is obtained by plotting cumulative frequencies against their respective class boundaries is called an cumulative line graph. As such there are two types of cumulative line graph – the less than the cumulative line graph and the more than cumulative line graph. The less than cumulative line graph is obtained by using the less than type cumulative frequency on the vertical axis to plot the cumulative line graph. The more than cumulative line graph is obtained by using the more than type of cumulative frequency on the vertical axis to plot the cumulative line graph. The points thus plotted are then joined successively by short line segments. The graph thus obtained is called an ogive graph.

Ogives may be considered useful for obtaining quartiles graphically. The more than type and the less than type cumulative line graph intersect in a point. If a perpendicular line is drawn from the point of intersection of the two cumulative line graph (more than type and less than type) on the horizontal axis, then the x co ordinate of this point gives us the value of the median, also called the second or the middle quartile.
Q gives are very useful in making short term projections.

Example:
The ages of students in the anatomy class are shown in the following frequency table.
Age Interval(yrs) Frequency
15 - 17                    13
18 – 20                    15
21 – 23                    20
24 – 26                    10
27 – 29                     8
30 – 32                     4
Draw an ogive for the above data.
Solution: First lets add a cumulative frequency column to that table:
Age Interval(yrs) Frequency Cumulative frequency
15 - 17                     13                    13
18 – 20                     15                    28
21 – 23                     20                    48
24 – 26                     10                    58
27 – 29                      8                            66
30 – 32                      4                            70

The cumulative line graph can now be plotted using the upper limit of class interval on the x axis and the cumulative frequency on the y axis as follows:

Ogive Graph

The above graph represents the cumulative line graph of the given example problem question.

Introduction to 4th grade geometry problems

In general, 4th grade geometry problems includes types of polygons, area and perimeter of the polygons. Here we are going to see about  4th grade geometry problems on area and perimeter.  Area (A) is a two-dimensional measure. The area is measured in terms of square units such as square inches, square feet and square centimeters. Hectares and acres are also considered in some special cases. The entire boundary of a figure or the distance around the figure is called as the perimeter. For circle it is called as the circumference, such lengths are measured with inches, feet, and centimeters.

4th Grade Geometry Problems in Square:

Formula to find area and perimeter:

Area of square = side x side square unit.

Area of square (A) = a^2 square units,   (a is the side length of the square)

Perimeter of the square = 4 x side length.

Perimeter of the square (P)=4 x a.

1. Find the area and perimeter of the square, whose side length is 5 meters.

Sol:

Area of square =a^2

= 5 x 5

Area of square =25m^2

Perimeter of the square = 4 x a

=4 x 5

Perimeter of the square  = 20 meters

2. Find the area and perimeter of the square, whose side length is 12 feet.

Sol:

Area of square   =a^2

= 12 x 12

Area of square  =144ft^2

Perimeter of the square    = 4 x a

=4 x 12

Perimeter of the square = 48 feet ft.

4th grade geometry problems in Rectangle:

Formula to find the Area and perimeter of the rectangle:

Area of the rectangle (A) = length x width

Area (A) = l x  w (l is the length and w is the width of the rectangle )

perimeter of the rectangle = 2(length + width)

perimeter (p) =2(l x w)

1. Find the area and perimeter of rectangle, whose length and width are 12meter and 6 meter respectively.

Sol:

Area of rectangle = l x w  square unit.

Given:    Length= 12 meters, Width =6 meters

=12x6

Area of rectangle  = 72 m^2                

Perimeter of the rectangle   = 2(l + w)

=2(12 + 6)

= 2 (18)

Perimeter of the rectangle  = 36 meter

2. Find the area and perimeter of rectangle, whose length and width are 8.5meter and 3 meter respectively.

Sol:

Given: Length= 8.5 meters, Width =3 meters

Area of rectangle   = l x w square unit.

=8.5x3

Area of rectangle = 25.5m^2                

Perimeter of the rectangle    = 2(l + w)

=2(8.5 + 3)

= 2 (11.5)

Perimeter of the rectangle = 23 meter

4th Grage Geometry Problems in Circle:

Formula to find the area and circumference of the circle:

Area of the circle = πr^2

( r is the radius of the circle)

Circumference of the circle = 2πr


1. The radius(r) of a circle is 5 inches. Find the area and circumference of that circle?

Sol:

Given: r = 5inches

Area of the circle= π x r^2

π = 3.14

A = 3.14 x (5)^2

=3.14 x 25 inch^2.

Area  = 78.5 in^2

Circumference of the circle = 2πr.

π=3.14, r = 5 inches

Circumference of the circle = 2 x 3.14 x 5

= 31.4 inches


2. The radius(r) of a circle is 7 inches. Find the area and circumference of that circle?

Sol:

Area of the circle= π x r^2

Given:  r= 7inches,            

value of π=3.14

Area = 3.14 x (7)^2

= 3.14 x 49 inch^2.

Area  =  153.86 in^2

Circumference of the circle = 2πr.

π=3.14 , Given r = 7 inches

=2 x 3.14 x 7

Circumference of the circle = 43.96inches

Introduction to Word Problems based on Ratio and Proportion

A ratio is the comparison of two quantities by division. It is a relation that one quantity bears to another with respect to magnitude. If A and B are two numbers, than the ratio of A to B is A/B and is denoted by A:B. The ratio does not have any unit.

The equality of two ratios is called Proportion. If (A/B) = (C/D) , then A,B,C,D are said to be in proportion and can be written as

A : B :: C:D

The Ratio and Proportion world problems are a kind of problems in which , the relation between different quantity has to be determined and then using the definition of ratio and proportion, the unknown has to be calculated.

Here are some of the ratio and proportion word problems:

Ratio and Proportion: Word Problems.

Problem: Find the value of k that must be added to 7, 16, 43, 79 so that they are in proportion. (Answer: 5)

Problem: Find the fourth proportional to the numbers 60, 48, 30. (Answer: 24)

Problem: The Income of Alex and Bob are in the ratio of 3:2 and their expenditure in the ratio of 5:3. Find the income of Alex if each saves dollars 1000. (Answer:  $6000)

Problem: A mixture contains alcohol and water in the ratio of 12:5. On adding 14 litres of water, the ratio of alcohol to water becomes 1:1. Find the quantity of alcohol in the mixture. (Answer: 24 litres)

Ratio and Proportion: Multiple Choice Word Problems:

Problem 1: If the ratio of ages of Alex and Bob is 6:5 at present and fifteen years from now, the ratio will get changed to 9:8, then find Alex's age.

(A). 24 years

(B) 30 years

(C) 18 years

(D) 33 years

(Answer: (B) 30 years)

Problem 2: If dollars 58 is divided among 150 children such that each girl and each boy gets 25 dollars and 50 dollars respectively. Then how many girls are?

(A) 52

(B) 54

(C) 68

(D) 62

(Answer: (C) 68)

Problem 3: The number that must be added to each of the numbers 8, 21, 13 and 31 to make the ratio of first two numbers equal to the ratio of last two numbers is

(A) 5

(B) 7

(C) 9

(D) None of these.

(Answer: (A) 5)

Unit conversion chart

Introduction: Units may be written in full or using the agreed symbols, but no other abbreviation may be used. The letter ‘s’ is never added to symbols to indicate the plural form. A full stop is not written after symbols for units unless it occurs at the end of a sentence. When unit symbols are combined as a quotient

e.g. metre per second, it is recommended that they may be written as m/s or better still as ms^-1. Three decimal signs are used commonly internationally and they are the full point, followed by the mid-point and lastly the comma. The full point sometimes is used as a sign of multiplication and the comma for spacing the digits in large numbers; we shall be on the safe side if we use the mid-point for decimals.

Basic units of measurement: the unit of measure for length is metre and it is denoted by the letter ‘m’. The unit of measure for mass is kilogram and it is denoted by the symbol ‘kg’. The unit of measure for time is second and it is denoted by the symbol‘s’.

Unit conversion chart: 
1. Units of weight: SI unit for weight is kg and let us understand unit weight through weight conversions  chart. Following are the units of weight and weight  conversions chart
(i) 10 milligrams = 1 centigram 10 mg = 1 cg
(ii) 100 centigram = 1 gram 100 cg = 1 g
(iii) 100 grams = 1 kilogram 1000 g = 1 kg
(iv) 100 kilograms = 1 quintal
(v) 10 quintals = 1 metric tonne
2. Units for volume: The SI unit of volume is cubic metres (m^3). Litre is also sometimes used as the unit of volume. Following are the units of volume and volume conversions:
(i) 1 cm^3 = 1 ml = 1 cm × 1 cm × 1 cm = 10 mm × 10 mm × 10 mm = 1000 mm^3
(ii) 1 litre = 1000 ml = 1000 cm^3
(iii) 1 m^3 = 1 m × 1 m × 1 m = 100 cm × 100 cm × 100 cm = 10^6 cm^3 = 1000 litre = 1 kilolitre
(iv) 1 dm^3 = 1000 cm^3
(v) 1 m^3 = 1000 dm^3
(vi) 1 km^3 = 10^9 m^3
3. Units of length: The SI unit of length is metre.Following are the unit of length and length conversions:
(i) 10 millimetres = 1 centimetre 10 mm = 1 cm
(ii) 100 centimetres = 1 metres 100 cm = 1 m
(iii) 1000 metres = 1 kilometre 1000 m = 1 km
4. Square and cubic units
1 cm^2 = 100 mm^2 1 cm^3 = 1000 mm^3
1 m^2 = 10000 cm^2 1 m^3 = 1000000 cm^3
1 litre = 1000 cm^3
1 m^3 = 1000 litres
1 hectare = 10000 m^2.