Solve Arithmetic Problems Examples

Introduction of solve arithmetic problems examples:

Let us see how to solve arithmetic problems examples. The arithmetic is the major learning of mathematics. The arithmetic can be made by four major operations. These arithmetic procedures are used by everybody in everyday life. The four processes are very simple and easy and also it is used to solve the business calculation and lot of calculations. The common processes are arithmetic addition process, arithmetic subtraction process, arithmetic multiplication process and arithmetic division process.

Arithmetic Operations:

Let us see how to solve arithmetic problems.

Arithmetic process 1: Addition:

This is the straightforward process of the arithmetic. This is the wisdom of addition process. It is demonstrated as “+ “.

Arithmetic process 2: Subtraction:

This is the straightforward process of the arithmetic. This is the wisdom of subtraction process. It is demonstrated as “– “.

Arithmetic process 3: Multiplication:

This is the straightforward process of the arithmetic. This is the wisdom of multiplication. It is demonstrated as “x”.

Arithmetic process 4: Division:

This is the straightforward process of the arithmetic. This is the wisdom of division. It is demonstrated as “/ “.

Please express your views of this topic Exponential Growth and Decay by commenting on blog.

Examples:

Let us see some arithmetic example problem.

Example 1:

In a container there are 28 apples, 42 oranges and 73 pineapples. Solve how many fruits in the container?

Solution:

Number of apples = 28.

Number of oranges = 42.

Number of pineapples = 73.

Total fruits    = 28+42+73.

= 143 fruits.

Total number of fruits is 143.

Example 2: In hotel there are 200 chairs. 132 chairs were occupied. Solve how many chairs were not occupied?

Solution:

Numbers of chairs = 200.

Occupied chairs    = 132.

Remaining chairs  = 200-132.

= 68 chairs.

The remaining chairs are 68.

Example 3: The chocolate cost is $35. Solve the cost of 50 chocolates.

Solution:

The cost of chocolate = $35.

Cost of 50 chocolate   = 35 x 50.

= $1750.

50 chocolates cost is $1750.

Example 4: 43 ice creams cost is $200. Solve the cost of each ice cream.

Solution:

Total ice cream = 43.

Cost of ice cream = 200.

= 200 / 43.

= $4.65.

Each ice cream cost is $4.65.

This is the example problems of the arithmetic.

Weighted Arithmetic Average

Introduction to weighted arithmetic average:

Let us see about weighted arithmetic average. The weighted average is not anything but it’s similar as that of arithmetic average, where as every of the data is donated uniformly with the weighed value. If each the weighted values are identical, then the weighed average is similar to the arithmetic average. The weighted average essentially performs similar fashion to the arithmetic average. The expression weighted average can also be named as weighted arithmetic average or weighted harmonic average. I like to share this What Does Arithmetic Mean with you all through my article.

Formula for Weighted Arithmetic Average:

The weighted average is described as the addition of the experiential values multiplied with the allocated weights which is separated by the addition of the observed values.

Formula of weighted average is defined by,

` barx_w = (sum_(i=1)^n (w_i* x_i) )/ (sum_(i=1)^n (w_i)).`

Description:

`barx_w` is the weighted average variable

`w_i ` is the allocated weighted value

`x_i` is the observed values. Is this topic how to solve a math problem step by step hard for you? Watch out for my coming posts.

Examples of Weighted Arithmetic Average:

Let us see about the weighted arithmetic average problems.

Example 1:

Find the weighted average from the experiential values 25, 30, 35, 40 and the allocated weights are 20, 30, 40, 50.

Solution:

The given `barx_i` = 25, 30, 35, 40

The given ` w_i` = 20, 30, 40,50.

Formula of weighted average is,

`barx_w = (sum_(i=1)^n (w_i* x_i))/ (sum_(i=1)^n (w_i)).`

=  `((20 * 25) +(30*30)+(40*35)+(50*40)) / (20+30+40+50)` .

= `4800 / 140` .

= 34.2857143

Example 2:

Find weighted average from the experiential values 65, 70, 85, 90 and the allocated weights are 60, 70, 80, 90.

Solution:

The given` barx_i` = 65, 80, 85, 90

The given `w_i` = 60, 70, 80, 90

Formula of weighted average is,

`barx_w = (sum_(i=1)^n (w_i* x_i)) / (sum_(i=1)^n (w_i)).`

= `((65*60) + (80*70) + (85*80) + (90 * 90)) / (60+70+80+90)` .

= `24400 / 300` .

= 81.3333333

Example 3:

Find weighted average from the experiential values 5, 10, 15, 20 and the allocated weights are 20, 30, 40, 50.

Solution:

The given `barx_i ` = 5, 10, 15, 20

The given `w_i` = 20, 30, 40, 50.

Formula of weighted average is,

`barx_w = (sum_(i=1)^n (w_i* x_i)) / (sum_(i=1)^n (w_i))`

=`((20 * 5) + (30 * 10) + (40 * 15)) / (20+30+40+50).`

=` 2000 /140.`

`barx_w = 14.2857143. `

Define Discrete Math

Introduction to Definition of  Discrete math:

Discrete mathematics is one of the major topics in mathematics. It defines the study of mathematical formation that is basically discrete rather than continuous. In disparity to real numbers that have the property of changeable "smoothly", the objects studied in discrete mathematics such as integers, graphs, and statements in logic.

Discrete math is branch of mathematics that deals Countable sets. Countable set define that have same cardinality as subsets of integers that comprises rational numbers but not a real numbers.

Dis create Math Application:

Presently researches in discrete mathematics were increased in latter half of twentieth century. The main purpose of study in discrete mathematics is discrete objects and analytic methods from continuous mathematics. Define discrete math contains Theoretical computer science, Information theory, Logic, Set theory, Combinatorial, Graph theory, probability theory, Discretization, Discrete analogues of continuous mathematics and so on.

Some of the main discrete math activities are given in below

Set theory:

Set theory is division of discrete mathematics. These commonly define the sets. Sets mean group of objects. The important development of set theory is infinite sets in outside the scope of discrete mathematics

Graph theory:

Graph theory is part of discrete mathematics. In this Graph theory mostly focus on graph and network. Graph theory is extensive in all the areas of mathematics and science.

Algebra:

Discrete mathematics discrete algebra defines the

Boolean algebra
The Boolean algebra commonly used in logic gates and programs

Relational algebra
In relational algebra commonly used in database side.

Algebraic coding theory
Discrete Semi group
Discrete modeling:

Discrete modeling is some time called as discrete analogue of continuous modeling. Case for Discrete modeling is recurrence relation.

Recurrence relationship formula is

`F(n)-f(n-1)=g(n)`

This above formula is discrete analogue of first order ordinary differential equation.

Example for Define Discrete Math:

Example 1:

Problem:

To solve the all ‘n>10’    n-2< (n^2-n)/12

Solution:

Default property of discrete mathematics is

p (n) is `P(n) ` : `k<=n(k>10 ; n>10)k-2<(k^2-k)/12`

Given Base case `n=15`

`15-2< ((15*15)-15)/12`

`13<210 br="br">
`13<17 br="br">
Using discrete mathematics property to solve the equation.

Basic Arithmetic Operations

Introduction to arithmetic operations:

The arithmetic is considered as the branch of math which deals with the operations of counting the numbers. The study of arithmetic mean  involves the study about the volume of the object. There are four basic arithmetic operations are present to add, subtract, multiply and divide the numbers. Here in this article we are going to see about these basic arithmetic operations.

Basic Arithmetic Operations

Basic operations:

The basic arithmetic operations are given as,

Addition
Subtraction
Multiplication
Division
Addition:

This is the one of the basic arithmetic operation. This is considered as the easiest way to add the two numbers. The addition may be done to the numbers with the infinite series. The addition exhibit the property of associative and commutative. When the addition process is done repeatedly then it is said as summation of the numbers. The addition is identified by using the sign “+”.

Subtraction:

This is the operation which finds the difference between the given two numbers. This is also one of the basic operations. There are two terms used in subtraction they are minuend and subtrahend. Where minuend is the term to be subtracted and subtrahend is the term which subtracts the minuend. When the value of the minuend is greatest then the result is positive. When the value of the subtrahend is greatest then the value is negative.

Multiplication:

This is the operation which finds out the product of the two numbers. The name of the two numbers are multipliers and also the multiplicand. Multiplication is also considered as the basic arithmetic operation. The multiplication also exhibits the properties of commutative and associative. The multiplication also exhibits the identity property and also the inverse property. Please express your views of this topic Properties of Exponents by commenting on blog

Division:

This division is considered as the inverse operation of the multiplication. There are two terms used in division dividend and also the divisor. The dividend is the term to be divided and the divisor is the term which is used to divide the number. Division can or cannot attain the property of associative and commutative.

Example Problems

Problem for addition with single digit:

Add 2 and 5.

2
+  5
_____
7
_____

Problem for addition with two digits:

Add 37 and 12

3 7
+  1 2
_____
4 9
_____

By adding 2 and 7 we get the answer 9 and by adding 3 and 1 we get the answer 4.

Problem for addition with three digits:

Add 352 and 141

3 5 2
+  1 4 1
________
4 9 3
________

Problem for subtraction with single digit:

Subtract 3 from 2.

2
-  3
____
-1
____

Since the value of the minuend is less than the subtrahend the answer is in negative. 2 is less than 3.

Problem for subtraction with two digits:

Subtract 75 and 21.

7 5
-  2 1
____
5 4
____

By subtracting 5 and 1 we get the answer 4 and by subtracting 7 and 2 we get the answer 5.

Problem for subtraction with three digits:

Subtract 690 and 420.

6 9 0
-  4 2 0
______
2 7 0
_______

Multiplication with single digit:

Multiply 2 and 5.

2 * 5 =10

By multiplying 2 and 5 we get the answer 10.

Multiplication with two digits:

Multiply 25 and 12.

2 4
1 2
_______
4 8
2 4 +
_______

2 8 8
_______

First we multiply 24 with 2 and write the answer in first row. Then we multiply 24 by 1 and write the answer in second row. Then we add the two rows and obtain the answer.

Division with single digit:

Divide 6 by 3.

3 ) 6 (2
6
_____
0
Here by multiplying the 3 with 2 we get the answer 6 hence we divided and got the quotient as 2.

Division by two digits:

Divide 72 by 24.

24 ) 7 2 ( 3
7 2
_____
0
_____

Division by three digits:

Divide 484 by 121.

121) 4 8 4 (4
4 8 4
______
0
______

By multiplying 121 by 4 we get the answer 484 thus we get the quotient as 4.

List of Derivatives

This page will help you to learn about the list of derivatives and quality calculus help. Below is given a brief description about derivative and then moving on further, the list of derivatives is provided.

In mathematics, the derivative is a measure of how a function changes as its input changes. The process of finding a derivative is called differentiation. A derivative can be thought of as how much one quantity is changing in response to changes in some other quantity. The reverse process is called anti-differentiation.

Some basic rules and list of derivatives and common derivatives are given below which helps you to learn derivatives. I like to share this Derivative of Tan Inverse with you all through my article.

List of Rules - Common Derivatives

There are the list of rules of common derivatives. In algebra of derivatives, we have established the following list of rules,

`d/dx` (u + v) = `(du)/dx` +` (dv)/dx`

`d/dx` (u - v) = `(du)/dx` - `(dv)/dx`

`d/dx` (uv) = u `(dv)/dx` + v `(du)/dx`

`d/dx` (u/v) = (v `(du)/dx` - u `(dv)/dx` ) / v2

List of Common Derivatives -polynomial and Trigonometric Functions:

The list of common derivatives for polynomial function is given below:

`d/dx` (c) = 0

`d/dx` (x) = 1

`d/dx ` (cx) = c

`d/dx` (xn) = nxn-1

`d/dx` (cxn) = ncxn-1

The list of common derivatives for trigonometric function is given below

`d/dx` (sin x) = cos x

`d/dx` (cos x) =  - sin x

`d/dx ` (tan x) = sec2 x

`d/dx` (cot x) = - cosec2x

`d/dx` (sec x) = sec x tan x

`d/dx` (cosec x) = - cosec x cot x . Is this topic help solving math word problems hard for you? Watch out for my coming posts.

List of Common Derivatives - Inverse Trigonometric and Hyperbolic Functions:

The list of common derivatives for inverse trigonometric function is given below:

`d/dx` (sin-1 x) = `1/sqrt(1 - x^2)`

`d/dx` (cos-1 x) = - `1/sqrt(1 - x^2)`

`d/dx` (tan-1 x) = `1/(1 + x^2)`

`d/dx` (sec-1 x) = `1/(|x|sqrt(x^2 - 1))`

`d/dx` (cosec-1 x) = - `1/(|x|sqrt(x^2 - 1))`

`d/dx` (cot-1 x) = - `1/(1 + x^2)`

The list of common derivatives for hyperbolic function is given below

`d/dx` (sinh x) = cos hx

`d/dx ` (cosh x) = sin hx

`d/dx ` (tanh x) = sec h2x

`d/dx` (sech x) = - tanh x sech x

`d/dx` (cosech x) = - coth x cosech x

`d/dx` (coth x) = - cosech2x

Simple Arithmetic Questions

Introduction to simple arithmetic questions:

Simple arithmetic questions involve the process of solving basic math expressions using arithmetic symbols. The algebraic expressions with arithmetic symbols are referred as simple arithmetic question. The simple arithmetic expressions include addition, subtraction, division and multiplication. The elementary math expressions are also called as arithmetic expressions. The following are the example arithmetic expressions with detailed solutions. All the math problems are solved by using aritnmetic symbols.I like to share this Arithmetic Progression with you all through my article.

Simple Arithmetic Questions Examples:

Simple arithmetic expressions are divided into four sections as basic Math Operation, fractions, integers operations, combining like terms of variables. The example arithmetic problems are discussed below with step by step detailed solution. The solved problems are under arithmetic categories.

Basic math questions:

Simple arithmetic expressions mainly cover addition, subtraction, multiplication, and division problems

1) Solve, 3(4+3) – 9 + 3(4)

Solution:
= 3(7) – 9 + 12
= 21 + 3
= 24

2) Solve, 5 + 8/4 * 5 + 5

Solution:
= 5 + 2 * 5 + 5
= 5 + 10 + 5
= 20

3) Solve, 62 + 72

Solution:

= (6 * 6) + (7 * 7)
= 36 + 49
= 85

Fractions questions:

Fraction is one of the parts of simple arithmetic expressions. Simplify the following (Write in mixed number if any of the answer is an improper fraction)

1) Solve (2/3) + (1/5)

Solution:

Take LCM as 15
= `(10/15) + (3/15)`

Add the above terms.
= `(10+3) / 15`


= `13/15`

2) Solve `(3/5)` – `(1/6)`

Solution:

Take LCM as 30
= `(18/30)` – `(5/30)`

Add the above terms.
` = (18 - 5) / 30`


` = 13/30`

Questions on Integer Operations:

Integer operation deals with performing math operations with integers such as positive and negative integers.

1) Solve, 11(–7)
= – 77

2) Solve,` (-81) / (-3)`

= `(-27) / (-1)`


= 27

3) Solve, –14 – (–3)
= –14 + 3
= –11

Variables – Combining Like Terms:

Variables plays major role in algebra for solving simple arithmetic questions. Simplifying variables by combining like terms helps to calculate the value of the variable.Between, if you have problem on these topics math prime factorization, please browse expert math related websites for more help on class 12 cbse sample papers.

1) Solve, 5x + 3 – 3x
= 5x – 3x + 3
= 2x + 3

2) Solve, y + 2y – 2 + 5
= 3y + 3

Simple Arithmetic Practice Questions:

The following are the simple arithmetic math problems for self practicing.

1) Simplify the arithmetic expressions.

42 + (2 * 4) – 4

Solution: 20

2) Simplify the arithmetic expressions.

(–2)2 + 8 – 10 – 6

Solution: - 4

3) Simplify the arithmetic expressions.

8x + 4y – 8 + 4y – 2x

Solution: 6x + 8y – 8

Solving Arithmetic Mean

Introduction to arithmetic mean:

     Arithmetic mean is the important concept in mathematical statistics.   Arithmetic mean is also referred to as average or mean.  Arithmetic mean is the ratio between the sum of all values and  number of values.  In this  article we have to learn about how to solving arithmetic mean problems through examples.

Formula to find arithmetic mean = Sum of elements / Total number of elements.

Brief Explanation of Solving Arithmetic Mean

Formula for finding arithmetic mean:

     Arithmetic mean = `(Sumof all the values)/(Number of values)`

     It is simply called as A.M.  Arithmetic mean values are usually mentioned in decimal point values.

Having problem with First Order Differential Equation keep reading my upcoming posts, i will try to help you.

Example Problems on Solving Arithmetic Mean

Example 1:

Solving the arithmetic mean for the following values 78, 89, 24,56,77,14, 19

Solution:

The given values are 78, 89, 24,56,77,14, and 19

Formula for finding the arithmetic mean

Arithmetic mean= `(Sumof all the values)/(Number of values)`

Here the sum of values = 78+ 89+ 24+56+77+14+ 19

Adding this we can get,

Sum=357

Here the number of values=7

Therefore arithmetic mean =`(357)/(7)` = 51

This is the arithmetic mean of the given values.

Example 2:

Solving the arithmetic mean for the consecutive four odd numbers u+21, u+23, u+25, u+27

Solution:

The given values are u+21, u+23, u+25, u+27

Formula for finding the arithmetic mean

Arithmetic mean=`(Sumof all the values)/(Number of values)`

Here the sum of values = u+21+ u+23+ u+25+ u+27

Adding this we can get,

Sum=4u+96

Here the number of values=4

Therefore arithmetic mean =`(4u+96)/(4)` = u+24

This is the arithmetic mean of the given values.

Example 3:

Solving the arithmetic mean for the consecutive numbers y, y+1, y+2, y+3, y+4, y+5, y+6

Solution:

The given values are y, y+1, y+2, y+3, y+4, y+5,y+6

Formula for finding the arithmetic mean

Arithmetic mean= `(Sumof all the values)/(Number of values)`

Here the sum of values = y+ y+1+ y+2+ y+3+ y+4+ y+5+y+6

Adding this we can get,

Sum=7y+21

Here the number of values=7

Therefore arithmetic mean =`(7y+21)/(7)` = y+3

This is the arithmetic mean of the given values.

Proportion Questions

Introduction to proportion questions:
Algebra is with the purpose of separation of mathematics in which calculation are made by using any arbitrary characters to stand for the quantities or things considered. In our day-to-day life, by learning ratio and proportion many a times we compare two quantities of the same type. Thus, in convinced situations, comparisons by division make better sense than comparison by taking the difference. The comparison by division is the Ratio. We denote ratio-using symbol ‘:’. If two ratios are the same, we state that they are in proportion and use the symbol ‘:’ or ‘=’ to equate the two ratios.In this article, we are going to discuss about proportion questions.

Proportion Questions – Definition and Types:

Definition of proposition:

If two ratios are not equal, then we state that they are not in proportion. In a statement of learning proposition, the four quantities involved when taken in order are known as respective terms. First and fourth terms are known as extreme terms. Second and third terms are known as middle terms.

a: b = c : d

There are two types of propositions,

1. Simple proposition:

A proposition consisting of just individual subject and one predicate is called a simple proposition.

Example: The following are simple proposition

1. Ram is blind.

2. The flower is not red.

2.  Compound proposition

A proposition consisting of two or further simple propositions in the form of a single sentence is called a compound proposition.Is this topic free online math tutor hard for you? Watch out for my coming posts.

Example: The following are compound propositions,

Quadrilateral is a square and each side of this quadrilateral is 4cm long.

Proportion Questions – Example Problems:

Question 1:

The income and savings of a family are in the ratio 6: 4. If the income of the family is Rs.7, 500. Find how much is being saved.

Solution:

Let the savings be Rs. m.

The proportion is 6: 4 = 7500: m

(Income: Saving) = (Income: Saving)

`6m = 4 xx 7500`

`(6m) / 6 = (4 xx 7500) / 6`

`m = 30000 / 5`

`m = 6000`

The Savings = Rs.6000 .

Question 2:

Jack works as a dental hygienist. Last week, Jack made `$` 720 for 36 hours of work. How many hours must Jack work in order to make `$` 950?

Solution:

Jack works as a dental hygienist

`($720) / (36hours)` = `($950) / (y hours)`

`(36 hours) xx ($950)` = `(y hours) xx ($720)`

34200 = $720 y

`(34200) / (720)` = y

47.5 = x

Jack 47 hours 5 mins work in order to make $950

Question 3:

If the cost of 11m rope is Rs. 80, find the cost of 6m cloth.

Quantity(in m) Cost (in Rs.)

11                    96

6                      ?

Solution:

The proportion is

11: 6 = 96: ?

Product of means = `6 xx 96` = 576

Product of Extremes = `11 xx k`

`11 xx k = 576`

`(11 xx k) / 11` = `576 / 11`

k = `576 / 11`

`k = 52.36 `

The cost of 6 m cloth = Rs. `52.36`

Properties of Arithmetic Mean

Introduction of arithmetic mean properties:

In math and statistics, the arithmetic mean or mean defines group of values is sum of all the values in the group and divide it by the number of items in the group. Based on list or group the name of mean is varied that is if the list defines the statistical population then the mean of the population is known as the population mean. Likewise if the group defines the statistical sample then it is known as sample mean.

Properties of Arithmetic Mean:

Let we see about some of the properties of arithmetic mean.

Arithmetic mean Property 1:

If m1 and m2 are the means of the two lists defined from the values v1 and v2 then the mean m is given by the equation
m = v1m1+v2m2/ v1+v2

Arithmetic mean Property 2:

If each inspection in the data is converted by m, the sum total of all the values unaffected.
That is, m = m1, m2, m3, m4........, mn / v
thus m1,m2,m3,m4........,mn = vm
Replacing each observation by m, we obtains

m+m+m........+m = vm

Arithmetic mean Property 3:

If all value of the variable m is either improved, reduces, divided or multiplied by a constant, the explanation so acquired also improved, reduces, acquire multiplied or acquire divided correspondingly by the similar constant.Please express your views of this topic Solve Equation by commenting on blog.

Arithmetic mean Property 4:

Algebraic sum of the divergence of a group of values from their arithmetic mean is 0.

Example:

Example 1:

Find the mean of the given values 56,35,87,23,63,58,46.

Solution:

We know the formula for find the mean, that is, addition of all the values/ total number of values.

Therefore, 56+35+87+23+63+58+46/7

= 368/7

= 52.571

Example 2:

Find the sum of the deviations of the given different values 5, 10, 15, 20, 25 from their mean.

Solution
Mean of 5, 10, 15, 20, 25 is,

`barx=(5+10+15+20+25)/(5)=(75)/(5)=15`



Therefore the addition of the deviation about mean is 0.

Multiplying Decimals Calculator

Introduction for the multiplying decimals calculator:

Calculator is a device which calculates the required process in that device and also we can substitute the many values in the calculator. Decimals is a number which is indicates the number with the point representation. In decimal number we have the point indication in between the numbers. From left to right we have the point value such as tens, hundreds, thousands etc. For multiplying decimals numbers we have the steps for the multiplication.

Example Problems in Multiplying Decimals Calculator

Example 1:Multiply 0.3 × 0.2

Solution: First enter the two decimal number(0.3 and 0.2) in the first and second box
Now press the enter button
Get the multiplied result(0.006) in the third box.

Multiplier is 0.3 and multiplicand is 0.2
1            
0.3      
× 0.2                      
06                
00  
0.006

Example 2:Multiply 1.256 × 0.32

Solution: First enter the two decimal number(1.256 and 0.32) in the first and second box
Now press the enter button
Get the multiplied result(0.40392) in the third box.
1 1                 
1.256
× 0.32         
1                                    
12512 
3768
0000     
0.40392

Please express your views of this topic Terminating Decimal by commenting on blog.

Example 3:  Multiply 0.21256 × 0.1

Solution:
First enter the two decimal number(0.21256 and 0.1) in the first and second box
Now press the enter button
Get the multiplied result(0.021256) in the third box.

0.21256
× 0.1        
021256             
000000 
0.021256                

Example 4: Multiply 0.589 × 0.4

Solution: First enter the two decimal number(0.589 and 0.4) in the first and second box
Now press the enter button
Get the multiplied result(0.2356) in the third box.

0.589
× 0.4        
02356       
0000 
0.2356

Example 5: Multiply 0.54× 1.1

Solution:         
First enter the two decimal number(0.54 and 1.1) in the first and second box
Now press the enter button
Get the multiplied result(0.594) in the third box.

0.54
× 1.1        
054       
054      
0.594       

A Rhombus is a Regular Polygon

Introduction to a rhombus is a regular polygon:

The regular polygon has different types of shapes. A rhombus is one type of regular polygon. A rhombus is closed with four sides and it is like as diamond. It is also referred as quadrilateral shape. Now we are going to see about rhombus in regular polygon.

Explanation for a Rhombus is a Regular Polygon
Some notes for regular polygon rhombus:

The regular polygon rhombus has four equal length sides. A rhombus is considered as parallelogram.The right angles of rhombus is in square shape.

Properties of rhombus regular polygon:

The rhombus has opposite angles are equal.
The diagonals are vertical.
The rhombus has convex property and isotoxal property.
The congruent triangles are used to prove the rhombus is symmetric. The regular polygon rhombus is include properties of parallelogram.

Rhombus regular polygon area calculation:

A rhombus area calculation is done by three methods. They are,

Base times height method – The area of rhombus is ba.
Diagonals method – The area of rhombus is d1d2/2.
Trigonometry method – The area of rhombus is s2 sin a.
In these methods, the base and length of diagonals are used.

Perimeter of regular polygon rhombus:

The total distance of outside rhombus is called as perimeter. The perimeter of rhombus is 4S. Here, the ‘S’ is length of side.Understanding free tutoring online is always challenging for me but thanks to all math help websites to help me out.

More about a Rhombus is a Regular Polygon

Example problems for regular polygon rhombus:

Problem 1: What is the area of regular polygon rhombus with b = 6 and a = 7.

Answer:

The side length is given as 6 and altitude is 7.

Therefore, the rhombus area is ba = 6 x 7 = 42.

Problem2: Determine the perimeter of regular polygon rhombus with side length 8.

Answer:

The side length is 8.

A rhombus perimeter is 4S = 4 x 8 = 32.

Exercise problems for a rhombus is a regular polygon:

1. What is the area of rhombus with b = 10 and a = 6.

Answer: Area = 60.

2. Find the perimeter of regular polygon rhombus with 4 side length.

Answer: The perimeter is 16.

Probability Math Version

Introduction to Probability Math Version

In math version, probability is a method of expressing knowledge or attitude that an occurrence will happen. In mathematics the idea has been given a right sense in probability theory, that is used broadly in such areas of learn as mathematics, finance, statistics etc. Now we will see the examples of probability.I like to share this Calculating Probability with you all through my article.

Examples - Probability Math Version

The following difficult examples are used to understand the probability concepts.

Example 1

The set A has the numbers from 15 to 23. Find the probability for the following outcomes?

i) Select the numbers below 20.

ii) Select the numbers between 15 to 20.

Solution

The set A={15,16,17,18,19,20,21,22,23}

i) Take P(A) is the probability for select the numbers below 20.

Total numbers n(S)=9

Here the following numbers are the below 20 n(A)={15,16,17,18,19}=5

So P(A)=`(n(A))/(n(S))`

=`5/9` .

ii) Take P(B) is the probability for  select the numbers between 15 and 20.

Total outcomes n(S)= 9

The following numbers are available n(B)= {16,17,18,19}=4

So P(B)=`(n(B))/(n(S))`

= `4/9` .

Example 2

James has 5 white balls, Mercy has 10 blue balls and Tom has the 6 yellow color balls. Find the probability for select James white balls.

Solution

James white color balls n(A)=5

Mercy’s blue color balls n(B)=10

Tom’s yellow color balls  n(C)=6

Total number of balls n(S)=5+10+6

=21.

Take P(A) is the probability for select James white balls.

P(A) =`(n(A))/(n(S))`

=`5/21` .

Example 3

Solve the probability for select the letter ‘R’ from the word ‘ALGEBRA?

Solution

Given word is ALGEBRA.

Total letters n(S)=7

Number of ‘R’ letter n(A)=1

So the probability=`1/7` . 

Please express your views of this topic 8th grade practice math problems by commenting on blog.

Practice Problems- Probability Math Version

1.What is the probability for select the letter ‘E’ from the word ‘SET THEORY’?

Answer:

Probability=`2/9` .

2.What is the probability for select the even numbers from 1 to 6?

Answer:

Probability=`1/2` .

These example problems are used to learn the probability concepts in math version.

One to One Function Calculator

Introduction to One to one function calculator:

In this article we are discussing about one to one function problems solving by calculator.  A function for which each number of the range of the function corresponds to approximately one number of the domain, One to one is often written 1 – 1. In this calculator first enter the problems. Finally press the solve button we get the solutions.

One to One Function Calculator – Example Problems:




Example 1:

Is g(x) = 2x - 1 one to one function?

Solution:

Now we plug if x = p and x = q

g(p) = g(q) implies that  p = q

So,   2p - 1 = 2q - 1

Add 1 on both side of the function. We get,

2p = 2q

Divide by 2 on both sides of the function. Finally we get p = q.

So Thus the given function is one to one function

Example 2:

Is f(x) = 4x + 3 one to one function?

Solution:

Now we plug if x = p and x = q

f(p) = f(q) implies that  p = q

So,   4p + 3 = 4q + 3

Subtract 4 on both sides of the function. We get,

4p = 4q

Divide by 4 on both sides of the function. Finally we get p = q.

So Thus the given function is one to one function.

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

Additional One to One Function Calculator – Example Problems:

Example 3:

Is f(x) = 5x + 4 one to one function?

Solution:

Now we plug if x = p and x = q

f(p) = f(q) implies that  p = q

So,   5p + 4 = 5q + 4

Subtract 4 on both sides of the function. We get,

5p = 5q

Divide by 5 on both sides of the equation. Finally we get p = q.

So Thus the given function is one to one function

Example 4:

Is f(x) = 6x + 5 one to one function?

Solution:

Now we plug if x = p and x = q

f(p) = f(q) implies that  p = q

So,   6p + 5 = 6q + 5

Subtract 4 on both sides of the function. We get,

6p = 6q

Divide by 6 on both sides of the function. Finally we get p = q.

So Thus the given function is one to one function

Plotting Algebraic Equations

Introduction to Plotting Algebraic equations:

In Algebra, We plot the equations by using the ordered pairs that satisfy the given condition. All the ordered pairs are plotted on the 2 dimensional graph and joining those points result in the graph of the given equation. We find the ordered pairs that satisfy the given condition by taking some random  values for x or y and find the other variable.Here the plotting of algebraic equations is done in this article.

The standard algebraic linear equation will be of the form of Ax + By=C.

Example for Plotting Algebraic Linear Equation:

Let the equation be 3x+2y=6

First we find the ordered pairs that satisfy the given condition.

We take some random values for x and find the value of y using the given equation.

In this example, let us take x values as -2,-1,0,1 and 2.

To find the value of y for each value of x, we substitute the value of x in the equation and solving for y.

For x=-2,

3(-2) + 2y=6

-6 + 2y =6

2y=6+6

2y=12

Y=12/2

`=>`  Y=6

Therefore the ordered pair is (-2,6).

Similarly we find all the values of y for different values of x.

After finding the ordered pairs we plot the ordered pairs (-2,6), (-1,4.5), (0,3), (1,1.5), (2,0) on 2 dimensional graph.

Between, if you have problem on these topics straight line, please browse expert math related websites for more help on ellipse equation.

Example of Plotting Algebraic Equations :

We plot all the ordered pairs and plot in 2 dimensional graph sheet.

We get the following graph of the algebraic equation 3x+2y=6

Similarly we can also graph the quadratic, rational, exponential, logarithmic, trigonometric, polynomial equations in a 2 dimensional graph. We can follow the same procedure as the above to graph any equation having 2 variables. Algebraic equations having more than 2 variables cannot be plotted in a 2 dimensional graph sheet. Only the equation having less than 2 variables can be plotted in a 2 dimensional graph sheet.

Dividing Polynomials with Exponents

Introduction to dividing polynomials with exponents:

Dividing polynomials with exponents is nothing but if the polynomials are having the degree we have to expand the polynomials and we have to divide the polynomials. Normally polynomials mean we are having power more than one. We will see some example problems for dividing polynomials. It is better to understand the division of polynomials with exponents. We can use the basic algebraic formulas and using the factorization method we can divide the polynomials.I like to share this Divide Exponents with you all through my article.

Examples for Dividing Polynomials with Exponents:

Example 1: Divide the following polynomials: `((x^2 + 3x + 2)^2) / ((x^2 - 4)^2)`

Solution:

The given polynomials are `((x^2 + 3x + 2)^2) / ((x^2 - 4)^2)`

Here the exponents of the polynomials are 2.

First we have to find the factors of the polynomials of the numerator.

x2 + 3x + 2 = x2 + 2x  + x + 2 = x (x + 2) + (x + 2) = (x + 2) (x + 1)

Next we have to find the factors of the denominators.

x2 – 4 = x2 – 22 = (x +2) (x – 2)

Now we have to plug the values.

=   ` (((x + 2) (x + 1))^2) / (((x + 2) (x - 2))^2)`

=  `((x + 2)^2 (x + 1)^2) / ((x + 2)^2 (x - 2)^2)`

From this we get `(x + 1)^2 / (x - 2)^2`

Example 2: Divide the following polynomials: `((x^2 - 5x + 4)^2) / (x^2 - 1)`

Solution:

The given polynomials are `((x^2 - 5x + 4)^2) / (x^2 - 1)`

Here the exponents of the polynomials in numerator are 2 and the denominator exponents is 1.

First we have to find the factors of the polynomials of the numerator.

x2 – 5x + 4 = x2 - 4x  - x + 4 = x (x - 4) - (x - 4) = (x - 4) (x - 1)

Next we have to find the factors of the denominators.

x2 – 1 = x2 – 12 = (x +1) (x - 1)

Now we have to plug the values.

` = (((x - 4) (x - 1))^2) / ((x +1) (x - 1))`

`= ((x - 4)^2 (x - 1)^2) / ((x + 1) (x - 1))`

From this we get `((x - 4)^2(x - 1)) / (x + 1)^2`

These are some of the examples for dividing polynomials with exponents. First we have to expand the polynomials and we have to divide.

Probability for Math

Introduction to Probability for math:

Probability is a study of math. Probability is number of outcomes divided into total number of events. The probability is also the expected value of the math. These are the two kinds of distribution are used to discrete and continuous distribution.  The general format of the formula for probability math is as follows,

The probability of Event P (A) = `("Number of outcomes n(a)")/("Total Number of events n(s)")`

Probability for Math – Examples:

Probability for math Example 1:

Roll a single dice; find the probability of get number 1.

Solution:

Total Number of possible = n (a) = {1, 2, 3, 4, 5, 6}

n (s) = 6

The number of outcomes n (a) = {6}

n (a) = 1

The probability of getting value = `1/6.`

Probability for math Example 2:

Toss four coins and find the probability of all tails. The possible outcomes are:

Solution:

Step 1:

n (s) = {TTTT, TTTH, TTHT, TTHH, HTTT, HTTH,THHT, HTHH, THTT, TTHH, THHT, HTHH, HTHT, HTHH,HHHT, HHHH }=16

Step 2:

There are 4 tosses with only two tails:

n (a) = { TTTT}=1

Step 3:

Formula:

P (A) = n(a)/n(s)

Answer:

P (A) = `1/16` .

Convert into a decimal 0.0625

The probability of two tails is 0.0625 or Rounded 6%.

Probability for math – Example 3:

The odds in possible of outcomes of an event 9:12.  Find the probability of the outcomes of this event

Solution:

Number of possible outcomes = 9.

Number of non-possible outcomes = 12.

Total number of outcomes = (9+12) = 21.

The probability of Event P (E) = `("Number of outcomes n (a)")/ ("Total Number of events n(s)")`

= `9 / 21.`

Probability for math – Example 4:

A coin is tossed. If the coin ground on top of a container is filled with one black globe and three white spheres. If the coin landed on tails the container is filled with one black sphere and nine white spheres. A sphere is then selected from the container. What is the probability that the sphere selected is black?

Solution:

Let H = Heads, T = Tails and B = Black sphere selected. Then by the law of total probability

P(B)   = P(B|H)P(H) + P(B|T)P(T)

= (0.25)(0.5) + (0.1)(0.5)

P (B)  = 0.175

Between, if you have problem on these topics probability combination formula, please browse expert math related websites for more help on answers to math word problems free.

Probability for Math - Practice Problems:

Probability for math - Practice Problem 1::

If word chosen at random from a writer X has a character missing with probability 0.1, has a character inserted with probability 0.2, and contains both kinds of error with probability 0.05. What are the probabilities that a letters chosen at random.

Answer:

P (missing or inserted, but not both) = 0.25 - 0.05 = 0.2

Ratio Math Problems Tutoring

Introduction to ratio math problems tutoring:

The term ratio math problems tutoring in math the term ratio is used to define the sum or total of the given term. Ratio is processes that are used to differentiate the related or differentiate the given items in the question.  A ratio is represented by two numbers separated by a colon. The term tutoring is the process defines the way of teaching done in maths.

Representation of Ratio:

A ratio that represents two numbers separated by a colon

Colon is represented as       :    

Examples Based on Ratio Math Problems Tutoring:

Example 1:   Problems based on ratio math problems tutoring

A special bag mixture contains rice, wheat and corn in the ratio of 2:3:5. If a bag of the mixture contains 3 pound of rice, how much corn does the bag contain?

Solution:

Given:  The cereal mixture of `"rice, wheat, corn"` in `2:3:5`

Now, we are going to assign the variable:

Let `x="mixture of corn"`

The item is now found to be written in ratio fraction,

`"Rice"/"corn"=2/5=3/x`

Cross multiplication is to be done.

So,     `2*x=3*5`

`2x=15`

`x=15/2`

`x=7.5`

Hence the ratio mixture of `7.5` pound of corn is present in bag.

Example 2:   Problems based ratio math problems tutoring

In a readymade shop they are T shirts and formal shirts, the ratio of T shirts and shirts are found   is 3:6 .If the shop contains 60 formal shirts, Find the total ratio of T shirts present in the readymade shop?

Solution:

Given:  The shop contains `60" formal shirts"` ,

Now, we are going to assign the variable:

Let `x= "T shirts"`

The item is now found to be written in ratio fraction,

`"T shirt"/ "Formal shirt"= 3/6=x/60`

Cross multiplication is to be done.

So,     `3*60=6*x`

`180=6x`

`x=180/6`

`x=30`

Hence the total ratio of `30` `T shirts` is present in the readymade shop.

Problems to Solve Based Ratio Math Problems Tutoring:

In a zoo they are monkeys and parrot the ratio of monkeys and parrot are found in ratio of 9: 6 .If the zoo contains 180 parrots, find the ratio of monkeys present in zoo?
Answer: The ratio of 270 monkeys is present in zoo.

In a flower shop they are yellow and pink color roses, the ratio of yellow and pink color roses  are found in ratio of 2:4 .If the flower shop  contains 100 pink color roses, Find the ratio of yellow color roses present in the flower shop?
Answer: The ratio of 50 yellow color roses present in shop.

Arithmetic Series Functions

Introduction for Arithmetic Series function:

A sequence that is 3 + 7 + 11 + 15 + ··· + 99 or 10 + 20 + 30 + ··· + 1000 that includes an even distinction between terms. The first term known as a1, the general difference is d, and the number of terms is n. The calculation of an arithmetic sequence is established by multiplying the number of terms times the average of the 1st and last terms.

Proof of Arithmetic Series Function:

1) Addition of i as i moves from 1 to n: 1 + 2 + 3 + ... + n function.

2) Adding the calculation of adding the 2nd summation backwards:

1 + ... + n

n + ... + 1

3) The 1st equation as (i), and then the 2nd calculation becomes (n - i + 1)

4) Hence 2 * the equation is (i) + (n - i + 1) or (n + 1) for every item in the sequence.

5) Here there are n terms in the sequence, the 2 * calculation of value is n * (n + 1)

6) For a particular series, we have end up with (n * (n + 1) / 2)

Example Problem for Arithmetic Series Function:

Here we will take more time to assuming about the rule, but the formula for the calculation of a finite arithmetic series, frequently symbolized by sn is

Arithmetic series function Formula: Sn = n (`(a1 + an) / 2` ) or `n / 2` [2a1 + (n - 1) d]

Problem 1:

4 + 8 + 12 + 16 + ··· + 96 has a1 = 4 and d = 4. To find n, use the formula for an arithmetic sequence.

Solution:

We solve 4 + (n – 1) *4 = 96 to get n = 24.

Sum = 24 (4 +` 99 / 2` ) = 1236

Or

Sum = `24 / 2 ` [2* 4 + (24 – 1) *4] = 1236.

Problem 2:

How much will a bank staff receives over his 34 year career if his starting salary is 40,000, and he receives a 1,000 salary increments for every year he works here?

Solution

We have the sequence:

40,000 + 41,000 + 42,000 + ... + 73,000

= `34 / 2` (40,000 + 73,000) = 1,92,1000.

Problem 3:

Find S9 for the series an = 6n – 1

Solution:

The formula tells us we have to be known n, the 1st term, and the nth term n = 9 because we have to find out the sum of the first 9 terms.

a1 = 6(1) – 1 = 6 – 1 = 5

a 9 = 6 (9) - 1 = 54 – 1 = 53

Substitute these values into the equation gives

S9 = 9 (5 + `53 / 2)` = 9 (`58 / 2)` = 9 (29) = 261

Fundamental Theorem of Arithmetic

Introduction to Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic is If (1) p1, p2, p3… pn and q1, q2, q3… qm are primes (both groups written in order of increasing size) and if (2) p1p2p3…pn = q1q2q3…qm, then p1 = q1, p2 = q2 … pn = qm and hence m = n. The Fundamental Theorem of Arithmetic is also called the "unique factorization theorem."  The following example is the fundamental theorem of arithmetic

Example:

3468 = 22 · 3 · 172   or   600 = 23 · 3 · 52

Fundamental Theorem of Arithmetic - Proof

It would be reasonable to prove first that p1 = q1.

Now p1 is a divisor of the product of all the ps (call it N) so it must be a divisor of the product of all the qs (which is also N).

If we knew that this forces p1 to divide at least one of the qs, we could then reason like this: Since p1 divides one of the qs, and since the qs are prime, so p1 must equal the q that it divides.

q1 is the smallest of the qs, so p1 = q1. Identical reasoning about q1 shows that q1 = p1. Thus p1 is equal to q1. So, replacing q1 with p1 in q1q2q3…qm, we get p1p2p3…pn = p1q2q3…qm.

Dividing both sides by p1 we get p2p3…pn = q2q3…qm. This new equation has one less prime on each side.

The same kind of reasoning that proved p1 = q1 also shows that p2 = q2. Step by step we peel away the primes from each side, establishing that pk = qk for all values of k.

Since every p pairs off with one q and vice versa, we see that m = n (the number of ps = the number of qs).

The problem is that we don't know that whenever a prime divides the product of several natural numbers it must divide at least one of them.

Fundamental Theorem of Arithmetic - Examples

Example 1:

216 = 2. 108= 2.2.54 = 2.2.2.27 =2.2.2.3.9 = 2.2.2.3.3.3

That is,

216 = 23. 33

Example 2:

26624 = 8.3328 = 8.8.416 = 8.8.8.52 = 8.8.8.2.26 = 8.8.8.2.2.13

That is

8.8.8.2.2.13 = 83.22.13

Arithmetic Maths

Introduction to arithmetic maths:

Arithmetic or arithmetic is the oldest and mainly basic branch of mathematics, used by approximately everyone, for tasks range from simple day-to-day including to superior science and business computation. It engages the learn of quantity, particularly as the outcome of unite numbers. In general usage, it refers to the easier properties when using the conventional operations of adding, subtracting, multipling and dividing with lesser values of numbers.


Basic Operations for Arithmetic Maths:

The arithmetic maths operations are as follows,

1.    Addition.

2.    Subtraction.

3.    Multiplication.

4.    Division.

Arithmetic Math- Addition:

Calculation of two or more than two numbers is known Arithmetic. The indication for addition is +. We can execute the addition operation for the all kind of values. Addition problems augment the values.

Arithmetic math Addition- Example:

6 + 5 = 11

This is easy addition problem.

Similarly we do the addition operation also more than two digits.

25 + 33 = 58

Arithmetic math- Subtraction:

Subtraction is the one of the essential arithmetic operation. Reduction of two or more than two numbers is known as Subtraction. The sign of subtraction is - (minus). Subtraction problems reduce the values.

The subtraction is finished by the following way,

z = x – y

Where,

z is differentiation.

x is known as minuend value.

y is known as subtrahend value.

Arithmetic maths subtraction- Example:

8-5 = 3.

Between, if you have problem on these topics systems of linear equations in two variables, please browse expert math related websites for more help on help with a math problem.

Arithmetic Maths- Multiplication:

Multiplication is the calculation of two numbers. The indication of multiplication is *. Multiplication also execute by recurrence of calculation. The conditions often used for multiplication are “times”.

Arithmetic maths multiplication- Example:

4 times 8 means 4 multiply with 8.

4 * 8 = 32

The other technique of multiplication recurring addition,

4 times 8 = 32

8 + 8 + 8 + 8 = 32

Arithmetic maths- division:

Division is an annul of multiply operation. The indication of division is /. The major reason of division is reducing the value and it is also recurring calculation.

Appearance of division technique,

`x/y` =z.

Where,

x is called as dividend.

y is called as divisor.

z is called as quotient.

Arithmetic math division- Example:

`21/7` = 3.

Sum of Arithmetic Sequence

Introduction to sum of arithmetic sequence:

Arithmetic sequence or progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 10, 20, 30, 40, 50… is an arithmetic sequence with common difference d = 5. Two types of sequences are arithmetic sequence and geometric sequence.

Formula – Sum of Arithmetic Sequence

The formula for the nth term an of an infinite arithmetic sequence with a common difference d and a first term a1 is given by

an = a1 + (n - 1 )d

The sum sn of the first n terms of an infinite arithmetic sequence is defined by

sn = a1 + a2 + a3 + ... + a n

and is a1 is given by

sn = n (a1 + an) / 2

Examples – Sum of Arithmetic Sequence

Ex 1: Find the sum of all positive integers, from 27 to 4590 inclusive, which are divisible by 27.

Sol : Sequence of first few numbers of positive integers divisible by 27 are given by 27, 54, 81...

The above sequence has a first number equal to 27 and a common difference d = 27.

We need to know the rank of the term 4590.

We use the following formula for the nth term

an = a1 + (n - 1 )d

4590 = a1 + (n - 1 )d

Substitute a1 and d by their values

4590 = 27 + 27(n - 1)

Solve for n to obtain

n = 170

4590 is the 170th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s170 = 170 (27 + 4590) / 2 = 392445.

Ex 2:  Find the sum of all positive integers, from 23 to 3680 inclusive, which are divisible by 23.

Sol :  Sequence of first few numbers of positive integers divisible by 23 are given by 23, 46, 69...

The above sequence has a first number equal to 23 and a common difference d = 23.

We need to know the rank of the term 3680.

We use the following formula for the nth term

an = a1 + (n - 1 )d

3680 = a1 + (n - 1 )d

Substitute a1 and d by their values

3680 = 23 + 23(n - 1)

Solve for n to obtain

n = 160

3680 is the 160th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s160 = 160 (23 + 3680) / 2 = 296240.

Problems in Arithmetic

Introduction to problems in arithmetic:

Directly or indirectly we use arithmetic's in our day-to-day life. The literacy rate is calculated by 3 steps namely ‘Read’, ‘write’ and ‘arithmetic’. From this we can conclude that arithmetic takes an important role in our daily life. For successful life one must know the values of arithmetic. Some arithmetic topics are given below.

1. Ratio

2. Proportion

2. Percentage.

Example Problems on Arithmetic

Ex:1 In a classroom there are 25 boys and 15 girls. What is the ratio between number of boys and number of girls?

Sol:

The ratio between the number of boys and number of girls =25/15
When reduced to lowest form =5/3
It is customary to write this ratio as 5 : 3
Ratio between the number of boys and girls = 5 : 3

Ex:2 The age of Ramu’s father is 40 years and that of Ramu is 10 years. What is the ratio between the father’s age and Ramu's age?

Sol:

Ratio between the father's age and Ramu's age  = 40 : 10
=  4 : 1

Ex:3 The cost of a notebook is Rs. 20 and the cost of a pen is Rs. 15. What is the ratio between the cost of a notebook and the cost of a pen?

Sol:

Ratio between the cost of a note book and the cost of a pen = 20 : 15
=  4 : 3

More Problems on Arithmetic

Ex:1 The ratio of rectangle is 4 : 3. If the breadth is 21 cm, find the length?

Sol:
Breadth = 21 cm
The ratio of length to breadth is 4 : 3
Breadth = 3 parts
3 parts = 21 cm
1 part =21
3 cm = 7 cm
Length = 4 parts
4 parts = 4 × 7 cm = 28 cm
Length of the rectangle = 28 cm

Ex:2 The monthly salary of Meena is Rs. 4000. She spends 80% of her salary every month. How much does she save every month?

Sol:
Meena's monthly salary = Rs. 4000
Expenditure = 80% of 4000
= (80/ 100) × 4000

= Rs. 3200
Savings = 4000 – 3200 = Rs. 800.

Sum of Infinite Arithmetic Sequence

Introduction to Infinite Arithmetic Sequence:
Arithmetic involves the rules for primary operations that are applicable to whole numbers, integers, rational and irrational numbers, in a single word, to all the real numbers. Arithmetic sequence or progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 7, 14, 21, 28, 35… is an arithmetic sequence with common difference d = 5.

Formula for Infinite Arithmetic Sequence
The formula for the nth term an of an infinite arithmetic sequence with a common difference d and a first term a1 is given by                 an = a1 + (n - 1 )d

The sum sn of the first n terms of an infinite arithmetic sequence is defined by

sn = a1 + a2 + a3 + ... + a n

and is a1 is given by

sn = n (a1 + an) / 2

Example Problems on Infinite Arithmetic Sequence

Ex1:   Find the sum of all positive integers, from 13 to 3250 inclusive, which are divisible by 13.

Sol :  Sequence of first few numbers of positive integers divisible by 13 are given by 13, 26, 39...

The above sequence has a first number equal to 13 and a common difference d = 13.

We need to know the rank of the term 3250.

We use the following formula for the nth term

an = a1 + (n - 1 )d

3250 = a1 + (n - 1 )d

Substitute a1 and d by their values

3250 = 13 + 13(n - 1)

Solve for n to obtain

n = 250

3250 is the 250th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s250 = 250 (13 + 3250) / 2 = 407875.

Ex 2:  Find the sum of the first 55 positive integers divisible by 17.

Sol :   The sequence of the first 55 positive integers divisible by 17 is given by 17, 34, 51…

The above sequence has a first term equal to 17 and a common difference d = 4. The nth

term formula to find the 55th term is

an = a1 + (n - 1 )d

a55 = 17 + 17 (55 - 1) = 935

Now we know the first term and last term so we can find the sum of the first 55 terms using the following formula

sn = n (a1+ an) / 2

s55 = 55 (17 + 935) / 2 = 26180.

Define Arithmetic Mean

Introduction :

In day-to-day life, we often came across the math word arithmetic mean or average. The arithmetic mean is defined as the sum of the total number of values divided by total number of values. In general math, the arithmetic mean can be defined easily since it involves simple operation. In this article define arithmetic mean, we are going to dicuss about finding the arithmetic mean of given values. In addition, practice problems are given to understand about arithmetic mean.

Example Problems to Define Arithmetic Mean:

The example problems for learning arithmetic mean are as follows:

Example 1:

Find the arithmetic mean of the given data set: { 12, 10, 23, 32 }

Solution:

Arithmetic mean   =  `( 12 + 10 + 23 + 32) / 4`

=  ` 77/ 4`

= 19.25

Example 2:

Find the arithmetic mean of the given numbers 20, 30, 45, 62, 18, 35

Solution:

Arithmetic mean   =  `( 20 + 30 + 45 + 62 + 18 + 35) / 6`

=  ` 210/ 6`

= 35

Example 3:

Find the arithmetic mean of the given data set: { 26, 14, 35, 44 }

Solution:

Arithmetic mean   =  `( 26 + 14 + 35 + 44) / 4`

=  ` 119/ 4`

= 29.75

Example 4:

Find the arithmetic mean of the given numbers 42, 50, 12, 32, 26, 33

Solution:

Arithmetic mean   =  `( 42 + 50 + 12 + 32 + 26 + 33) / 6`

=  ` 195/ 6`

= 32.5

Example 5:

Find the arithmetic mean of the given data set: { 22, 32, 45, 50 }

Solution:

Arithmetic mean   =  `( 22 + 32 + 45 + 50) / 4`

=  ` 149/ 4`

= 37.25

Example 6:

Find the arithmetic mean of the given numbers 8, 10, 4, 6, 8, 11

Solution:

Arithmetic mean   =  `( 8 + 10 + 4 + 6 + 8 + 11) / 6`

=  ` 47/ 6`

= 7.83

Practice Problems to Define Arithmetic Mean:

The practice problems for learning arithmetic mean are as follows:

1) Find the arithmetic mean of the given data set: { 25, 35, 12, 22, 36 }

Answer: Arithmetic mean = 26

2) Find the arithmetic mean of the given numbers 40, 35, 25, 75, 43

Answer: Arithmetic mean = 43.6

3) Find the arithmetic mean of the given data set: { 30, 26, 45, 28, 44, 55 }

Answer: Arithmetic mean = 38

4) Find the arithmetic mean of the given numbers 100, 150, 200, 350, 260

Answer: Arithmetic mean = 212

5) Find the arithmetic mean of the given data set: { 17, 25, 37, 49, 66 }

Answer: Arithmetic mean = 38.8

6) Find the arithmetic mean of the given numbers 12, 36, 48, 60, 15

Answer: Arithmetic mean = 34.2

7) Find the arithmetic mean of the given numbers 25, 31, 40, 11, 12

Answer: Arithmetic mean = 23.8

Horizontal Number Line

Introduction to horizontal number line:

Generally there are two types of the number lines - horizontal number line and another is the vertical number line. Here we are going to see the horizontal number line it means the lines are parallel to the x axis is called the horizontal number line the line is in the form of y=b here b is the constant, it will be come from the general form of the straight line y=ax we substitute x=0 means we get the horizontal number line.

Pictorial Representation of the Horizontal Number Line:

This is the graphical representation of the horizontal line; it will look like the line parallel to the x axis, generally number line means the image of the straight line we called as the number line so the horizontal number line also look like the straight line shape if it is in the horizontal means we called the number line as the horizontal number line.

Steps for drawing the horizontal number line:

Step 1: draw the graph with the x axis and the y axis

Step 2: then give the points on the both the axis

Step 3: start to point the number

Step 4: we write the positive numbers on the right side for x axis

Step 5: we can write the negative integer on the left side for the x-axis

Step 6: we can write the positive numbers of y axis on the top of the graph from the origin

Step 7: we can write the negative integer of y axis to the down of the graph with the origin

Step 8: then starting to plot y=k like that

Step 9: so the graph is parallel to the x axis

I am planning to write more post on math help online tutor, algebra help factoring polynomials. Keep checking my blog.

Examples Based on Horizontal Number Line :

Ex 1: Draw the horizontal number line:

Y=3

sol:

Step 1: Draw the graph it contains both the and y axis

Step 2:Then start to mark the numbers like -3,-2,-1, 0.1,2,3

Step 3:After that we can draw the graph for the given data.


Ex 2: Draw the horizontal number line:

Y=8

sol:

Step 1:Draw the graph it contains both the x and y axis

Step 2:Then start to mark the numbers like -3,-2,-1, 0.1, 2, 3…… (For both the x and y axis)

Step 3:After that we can draw the graph for the given data.