How to Learn Fractions Easy

Introduction to how to learn fractions easy:

In this article we shall discuss about how to learn fractions easy. Here, fractions are also meant through division of a whole. A fraction is can be formation over to a decimal through dividing the upper digit, or numerator, during the lower digit, or denominator. Fractions are instead of as ratios, and significance for fraction which is one of the important math processes. Thus the fractions `3/5` are also used to point out the ratio 3:5 and the fractions 3 ÷ 5 as well.

Example problems based on how to learn fractions easy:

The example problems based on how to learn fractions easy are given below that,

Example 1:

How to learn fractions of 391 divide by 15?

Solution:

Step 1:

Here, 280 divide by 13 is meant by `280/13.`

Step 2:

Now, 280 divide by 13 is given below that,

Here, using long division method. So, long division method is shown given below that,

21.53

13)280             [280 > 13, now divide `280/13` ]

273             [(hint: 21 × 13 = 273)]

70           [7 < 13, so add one zero]

65           [(hint: 5 × 13 = 65)]

50         [5 < 13, so add one zero]

39         [(hint: 3 × 13 = 39)]

11

Step 3:

The final answer is 21.53

Example 2:

How to learn fractions of 228 divide by 7?

Solution:

Step 1:

Here, 228 divide by 7 is meant by `228/7.`

Step 2:

Now, 228 divide by 7 is given below that,

Here, using long division method. So, long division method is shown given below that,

32.57

7)228            [228 > 7, now divide `228/7` ]

224            [(hint: 32 × 7 = 224)]

40          [4 < 7, so add one zero]

35          [(hint: 5 × 7 = 35)]

50        [5 < 7, so add one zero]

49        [(hint: 7 × 7 = 49)]

1

Step 3:

The final answer is 32.57

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Practice problems based on how to learn fractions easy:

The practice problems based on how to learn fractions easy are given below that,

Problem 1:

How to learn fractions of 499 divide by 5?

Answer: The final answer is 99.8

Problem 2:

How to learn fractions of 399 divide by 6?

Answer: The final answer is 66.5

Problem 3:

How to learn fractions of 299 divide by 7?

Answer: The final answer is 42.71

Easy Multiplication Techniques

Introduction to easy multiplication techniques:

Multiplication is the basic operations in mathematics. Multiplication can be done using math tables. Multiplying a small number is a simple task, using math table, but when we consider for a larger number, this process is tedious. We have introduced a lot of easy multiplication techniques to learn as fast as possible. Here we are going to see easy multiplication techniques.

Methods of easy multiplication techniques:

We have a lot of easy multiplication techniques to do fast multiplication. Some of the techniques are,

Methods of easy multiplication techniques for multiplying any two numbers:

To multiply any two numbers, follow the following steps,

Let us consider the example sum, 66 * 64

Take the nearest round value of the given two numbers.
We can write 66 as (65 + 1) and 64 as (65 -1 )
Hence we can compute this as, 65 2 – 1 2.
Hence  we got the answer as  4225 – 1 = 4224.
Thus we got the final right answer as 4224 using easy multiplication techniques.

Note: This technique can be applicable only when the two numbers have the same round values and as well as the distance from the round values is same.

Another example:

To multiply any two numbers, follow the following steps,

Let us consider the example sum, 28 * 22

Take the nearest round value of the given two numbers.
We can write 28 as (25 + 3) and 22 as (25 -3 )
Hence we can compute this as, 25 2 – 3 2.
Hence we got the answer as 625 – 9 = 616.
Thus we got the final right answer as 616 using easy multiplication techniques.

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Methods of easy multiplication techniques for multiplying any two numbers with 11:

To multiply any two numbers with 11, follow the following steps,

Let us consider the example sum, 23 * 11

Take the given number 23.
Add the given number individually.
Hence it becomes 2 + 3 = 5.
Now add 5, in between 2 and 3, we get 253.
Thus we got the final right answer as 253 using easy multiplication techniques.

Another example:

Let us consider the example sum, 75 * 11

Take the given number 75.
Add the given number individually.
Hence it becomes 7+ 5 = 12.
Now add 12, in between 7 and 5.
Here we have carry 1, so the add the carry with 7, we get 825.
Thus we got the final right answer as 825 using easy multiplication techniques.

Dividing Fractions Made Easy

Introduction to dividing fractions made easy:

Fraction is defined as separating the quantities. This is done by using the horizontal line. Also fraction has two parts. The two parts has the numerator function and the denominator function. For example, `9/7` is called as the fraction. In this the number 9 is the numerator part and the number 7 is called the denominator part.

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Explanation for dividing fractions made easy


The explanation for dividing fractions made easy are given below,

For doing the division fraction, we can have many of the rules to be followed,
In the first step for doing division fraction is, we have to take inverse process for the fraction.
In the next step, we have to change the division sign by using the multiplication sign.
Then in the next step, the multiplication rules are followed for doing the division fraction.

Example problem for division fractions made easy


Problem 1:  Solve the given problem by dividing the fractions, `(1)/(6)` and `(4)/(8)`

Solution:

Step 1: Write the given fractions,

`1/6` `-:` `4/8`

Step 2: In the second step, we have to take the inverse process for the given fractions,

= `1/6` `xx` `8/4`

= `1/6` `xx` 2

=`1/3`

This is the required solution for dividing the given fraction.

Problem 2:  Solve the given problem by dividing the fractions, `(2)/(6)` and `(1)/(4)`

Solution:

Step 1: Write the given fractions,

`2/6` `-:` `1/4`

Step 2: In the second step, we have to take the inverse process for the given fractions,

= `2/6` `xx` 4

= `2/3` `xx` 2

=`4/3`

This is the required solution for dividing the given fraction.

Problem 3:  Solve the given problem by dividing the fractions, `(2)/(3)` and `(5)/(2)`

Solution:

Step 1: Write the given fractions,

`2/3` `-:` `5/2`

Step 2: In the second step, we have to take the inverse process for the given fractions,

= `2/3` `xx` `2/5`

= `2/3` `xx` `5/2`

=`5/3`

This is the required solution for dividing the given fraction.

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Practice problem for dividing fractions made easy


Problem 1:  Solve the given problem by dividing the fractions, `(4)/(8)` and `2/3` .

Answer: `6/8`

Problem 2:  Solve the given problem by dividing the fractions, `(3)/(6)` and `1/9` .

Answer: `9/2`

What is Distributive in Math

Introduction to what is distributive in math:

The distributive is basic law in the math for the binary operation. It is used in the algebra and also the ordinary binary number. Distributive is one of the law in the math for perform the basic operations. Distributive law is based on the order of operation. Distributive is used to do the basic operation to get the correct result for the equation. Looking out for more help on factoring polynomials online in algebra by visiting listed websites.


Distributive law in math:

Take a set S and the two binary operations + and also -.

If the operation is left distributive over + and -, then the given elements are x, y and z of S. The distributive law is,

x- (y +z) = (x-y) + (x-z)

If the operation is left distributive over + and -, then the given elements are x, y and z of S. The distributive law is,

(y +z) -x = (y-z) + (z-x)


Examples for distributive in math:


Example 1 for distributive in math:

Solve the equation 3 (x+7) using the distributive law.

Solution:

The given equation is 3 (x+7).

By the distributive law, x- (y +z) = (x-y) + (x-z).

Multiply the number for both x and also 7 to get the simplified value.

3 (x+7) = 3x + 3(7)

3 (x+7) = 3x + 21

The value for the equation 3 (x+7) is 3x+21.

Example 2 for distributive in math:

Find the value for the equation 3x (13+ 7) using the distributive law.

Solution:

The given equation is 3x (13+ 7).

3x (13+ 7) = (3 x 13) + (3 x 7)

3x (13+ 7) = 39 + 21

3x (13+ 7) = 60

The value for the equation 3x (13+ 7) is 60.

Example 3 for distributive in math:

Find the value for the equation 16 x (16- 10) using the distributive law.

Solution:

The given equation is 16 x (16- 10).

16 x (16- 10) = (16 x 16) - (16 x 10)

16 x (16- 10) = 256 - 160

16 x (16- 10) = 96

The value for the equation 16 x (16- 10) is 96.

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Practice problem for distributive in math:


Find the value for the equation 2- (33 +7).
Answer: 38.

Find the value for 5 x (23 -9) using the distributive law.
Answer: 70.

Easy Way to do Division

Introduction to Easy Way to do Division:

A division method can be done by the division symbol ÷.  The number present in the left of the division symbol is dividend and the number present in the right of the division symbol is divisor. The answer you get from the division process is called quotient. The number after dividing process over the remaining number left below the division line is called as remainder. Let us see about an easy way to do division.

Shortest and Easy Way to do Division Problems

Example Problems for Dividing by 9:

Example 1:

Solve the problem using easy way to do division method.

20016 ÷ 9

Solution:

Let us write 20016/9

First find the sum of each digit adding with the next digit and the  total sum got by adding each and every number.

2 (2 + 0) (2 + 0 + 0) (2 + 0 + 0 + 1) last term is (2 + 0 + 0 + 1 + 6)

Write the terms as it is and if the total sum has any carry, it can be add it to the before term.

2 (2) (2) (3) last term is (9)

Divide the last term by the given divisor.

223 + 9/9

223 +1

224

Therefore, the division method for 20016 ÷ 9 is 224.

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Other Example Problem for Easy Way to do Division

Example Problems for Dividing by 9:

Example 2:

Solve the problem using easy way to do division method.

7875 ÷ 9

Solution:

Let us write 7875/9

First find the sum of each digit adding with the next digit and the  total sum got by adding each and every number.

7 (7 + 8) (7 + 8 + 7) last term is (7 + 8 + 7 + 5)

Write the terms as it is and if the total sum has any carry, it can be add it to the before term.

7 (15) (22) last term is (27)

(7 + 10) (5 + 20) (2) Last term is (27)

(80) (70) (2) last term is (27)

(8) (7)(2) Last term is (27)

Divide the last term by the given divisor.

872 + 27/9

872 +3

875

Therefore, the division method for 7875 ÷ 9 is 875.

Algebra Rules Made Easy

Introduction to algebra rules made easy:
As we all know that, algebra consists of constants and variables. There are certain rules and principles for solving algebra problems. Algebra rules really made easy for solving tough algebra problems. Algebra is classified into two categories; they are i) ancient algebra and ii) Modern algebra. There are certain rules to be followed in algebra. These rules are used to solve the problems in algebra. In this article, we are going to see about algebra rules made easy.

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Rules for solving algebra:


The rules for solving algebra is listed below, these rules made easy for solving algebra problems,

Rule 1: All the like terms have to be added/subtracted together on both sides of the given equation.

Rule 2: When there is a multiplication/division operation is to be performed for an equation, then it should be done on both sides.

Rule 3: Change of sign or symbols is applicable in inequalities.

Rule 4: In absolute value, negative sign is altered to positive. Positive value will always remains positive. Negative symbol only changes.

Rule 5: While adding/subtracting polynomials, same degree polynomial is to be added/subtracted in the given problem.

Rule 6: In word problems, translation of word to algebraic expressions is to be done.


Example problems on algebra rules made easy:

Example 1:

Write the equation in slope intercept form,

5y – 4x + 8 = 12

Solution:

Add 4x on both sides,

5y – 4x + 4x + 8 = 12 + 4x

5y + 8 = 4x + 12

Subtract 8 on both sides,

5y + 8 – 8 = 4x + 12 – 8

5y = 4x + 4

Divide 5 on both sides,

5y/y = 4x/5 + 4/5

y = 0.8x + 0.8

The answer is y = 0.8x + 0.8

Example 2:

Solve: (4x ^2 + 9x – 7) + (2x ^2 – 4x + 6)

Solution:

In this problem, we have to add the like terms,

= 4x ^2 + 9x – 7 + 2x ^2 – 4x + 6

= 6x ^2 + 5x - 1

The answer is 6x ^2 + 5x -1.

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Example 3:

Solve 2x + 4 > 10.

Solution:

2x + 4 > 10

Subtract 4 on both sides,

2x + 4 – 4 > 10 – 4

2x > 6

Divide 2 on both sides,

2x/2 > 6/2

x > 3

The answer is x > 3.

Math Help 5th Grade

Introduction to math help 5th grade:
The topics involved in 5th grade math help are number sense, patterns, addition, measurement, subtraction, multiplication, functions,  fractions & mixed numbers, division, algebra, decimals, adding and subtraction of decimals and probability & statistics. In this article we shall discuss about the problems involved in 5th grade math help.

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Math Help 5th Grade Problems:


Example 1:Find the area of a rectangle with base of length is 10 inches and height of 4 inches.

Solution:

Area of a rectangle = length * height

So, Area of a rectangle = 10 *  4

= 40 square inches

Example 2: Find the median of 8, 6, 9, 3, 2, 7

Solution:

Evaluate the total numbers from the given set of values.

From the given set of values, there are totally 6 numbers, Which is even,

Now arrange the total numbers in ascending order,

2, 3, 6, 7, 8, 9

As the given set of numbers is even , Sort out two middle numbers from the above given list of numbers,

Which is  6 & 7 are the two middle numbers,

In order to find the median, add both middle terms and divide by 2,

Adding both numbers `(6+7)/(2)` = 6.5

Therefore median=6.5.

Example 3: Find the value form the expanded form 2 × 10000 + 4× 1000 + 10 × 10 + 8

Solution:  2 × 10000 + 4 × 1000 + 10 × 10 + 8

= 20000 + 4000 + 100 + 8

=> 24108.

Example 4: Find the mode for given set of data. 60, 19, 15, 50, 15, 80, 30

Solution: Mode:The mode is the number that occurs most often in a set of data.

Form the given data set we can say that 15 is the number that has occurred twice.

For the given set of data, Mode = 15.

Example 5: Find the perimeter of a square with length 8.

Solution: The perimeter of a square = 4 * a ( a-> length)

Here length =5, So perimeter of square = 4 * 8 = 32.

Example 6: Find the circumference of a circle with diameter 14.

Solution:  The circumference of a circle = 3.14(pi) * diameter of the circle.

Here diameter = 10, So Circumference of a circle = 3.14 * 14

Therefore the Circumference of a circle =   43.96

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Math Help 5th Grade Practice Problems:


Problem 1:Find the area of a rectangle with base of length is 10 inches and height of 6 inches.

Answer: Area of rectangle =40 square inches.

Problem 2:Find the mode for given set of data. 6, 9, 5, 5, 1, 8, 3

Answer: Mode = 5

Problem 3: Find the value form the expanded form   8× 1000 + 12 × 10 + 10

Answer: 8130

Logic and Set Theory Learning

Introduction to Logic and Set Theory Learning:

Logic theory is a set of sentences in a formal language. The individual sentences of a theory are called as the theorems. A first-order theory is a set of first-order sentences. Many authors require that the theory be closed under logical consequence; a theory with this property can be called a deductive theory. Set theory is the studies of sets, which are collections of objects. Although any type of object can be collected into a set.

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Basic operations of logic and set theory learning


Union in Set theory Learning:
If A and B are different sets then union of A and B denoted as A U B.

For example {3, 4, 5} and {6, 7, 8} is the set {3, 4, 5, 6, 7, 8.

Intersection in Set theory Learning:
If A and B are different sets then intersection of A and B denoted as A n B.

For example {3, 4, 5} and {6, 4, 3} is the set {3, 4}.

Complement in Set theory Learning:
If A and B are different sets then A relative to set b, denoted Ac.

For example complement of {3, 4, 5} relative to {6, 4, 3} is {6}.

Symmetric difference in Set theory Learning:
If A and B are different sets then member of exactly one of A and B.

For example complement of {3, 4, 5} relative to {6, 4, 5} is {3, 6}.

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logic and set theory learning


Negation (NOT) ˜ p in logic learning:
If value of proposition is true then transform into false and vice versa.

Disjunction (OR) p v q in logic learning:
If two propositions are false then the result is false otherwise true.

Conjunction (AND) p ^ q in logic learning:
If two propositions are true then result is true otherwise false.

Conditional (IF…THEN…) p?q in logic learning :
Truth of the proposition p is sufficient to truth of proposition q.

Biconditional (IF AND ONLY IF) p?q in logic learning:
p is sufficient condition for q. q is necessary for p. Unless q not p. Not p unless q. Not p without q.

How to Explain Divide Math

Introduction for explain divide math:

In algebra basic arithmetic operation (addition, subtraction, multiplication, and division) widely used in day to day life. In these articles we are going to see about how to explain divide math. Division can be considered as repeated subtraction or equal distribution.

Simple division teach to find the how frequently one whole number called the divisor, is contain in an additional whole number, called the dividend, or to divide a whole number into some proposed number of equivalent parts, and is a small method of performing frequent subtraction.

The number arise from the procedure is called the quotient; it shows how often the divisor is contain in the dividend, that is into how many equal parts the dividend is divided.

If any thing be over following the division is performing it is called the remainder. The mark for division is ÷ it is name by and shows that the number status previous to the sign is to be divided by the number.


Explain divide math - Definition and steps:


Explain divide math - Definition:

Division is defined as an arithmetic function, which is the opposed process of multiplication. From the process of division, the proportion or ratio of two numbers be capable of be calculated.

Otherwise, the process of decision how many periods of one number is included in a further one. Symbol of division is ‘/’ or ‘÷’. If we divide a number by another number, then

Dividend = (Divisor * Quotient) + Remainder

Explain divide math - Steps:

Step1. Division of two integers by the related signs resolve be positive sign

a) Positive ÷ positive = positive

b) Negative ÷ negative = positive

Step2. Division of two integers by the unlike signs will be negative

a) Positive ÷ negative = negative

b) Negative ÷ positive = negative.


Explain divide math – Example problems:


Problem 1:

The school's Internet connection transferred 646 megabytes of data in 6 seconds. How many megabytes can it transfer in just one second?

Solution:

The school's Internet connection transferred 646 megabytes Data in 6 seconds.

So, therefore total megabytes can it transfer in just one second

= 646 ÷ 6

Step 1: Determine whether the divisor 6 will divide the first digit of the dividend 646. It will since it is not greater than this digit. The result of division is 1, which is placed under the 6.

Step 2: Determine whether the divisor 6 will divide the second digit of the dividend. Since 6 will not divide 4, a zero is placed under the 4.

Step 3: The 4 is now taken with the third digit 6 to become 46. The divisor 6 divides 46 and the quotient 7 is placed under the 7 and remainder 4. The answer is 107.66

= 107.66 megabytes

Problem 2:

There are 48 soft drink machines in the university. They hold 768 cases of soda altogether. How many cases does each machine hold?

Solution:

There are 48 soft drink machines in the university.

They hold 768 cases of soda altogether.

= 768 ÷ 48

= 16

16 cases do each machine hold.

Sixth Grade Math Fractions

Introduction

A fraction 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, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole. Source wikipedia

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Sixth grade math fractions problems:

Sixth grade math problem 1

Add the two fractions `2/3` and `5/3`

Solution:

The given two fractions `2/3` and `5/3`

=`2/3` +`5/3`

Add above the fraction `2/3` and `5/3`

= `(2+5)/3`

We get

=`7/3`

This can be simplified has

=2.33

Answer of the two fraction = `7/3` or 2.33

Sixth grade math problem 2

Add the two fractions `12/3` and `15/3`

Solution:

The given two fractions `12/3` and `15/3`

=`12/3` +`15/3`

Add above the fraction `2/3` and `5/3`

= `(12+15)/3`

We get

=`27/3`

This can be simplified has

=9

Answer of the two fraction = `17/3 ` or 9

Sixth grade math problem 3

Add the two fractions `22/8` and `25/8`

Solution:

The given two fractions `22/8` and `25/8`

=`22/8` +`25/8`

Add above the fraction `22/8` and `25/8`

= `(22+25)/8`

We get

=`47/8`

This can be simplified has

=5.875

Answer of the two fraction = `47/8` or 5.875

Sixth grade math problem 4

Subtract the two fractions `9/3` and `5/3`

Solution:

The given two fractions `9/3` and` 5/3`

=`9/3` -`5/3`

Subtract above the fraction `2/3` and `5/3`

= `(9-5)/3`

We get

=`4/3`

This can be simplified has

=1.33

Answer of the two fraction = `4/3` or 1.33

Sixth grade math problem 5

Multiply the two fractions `2/3` and`5/3`

Solution:

The given two fractions `2/3` and `5/3`

=`2/3` +`5/3`

Multiply above the fraction `2/3` and `5/3`

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

We get

=`10/9`

This can be simplified has

=1.11

Answer of the two fraction = `10/9` or 1.11

Sixth grade math problem 6

Multiply the two fractions `5/3` and `5/6`

Solution:

The given two fractions `5/3` and `5/6`

= `5/3* 5/6`

Multiply above the fraction `5/3` and `5/6`

= `(5*5)/ (3*6)`

We get

=`25/18`

This can be simplified has

=1.38

Answer of the two fraction = `25/18` or 1.38

Learning Sure Event

Introduction to learning sure event:

The word sure event is coming in probability theory.

Probability is the likelihood of occurring of an event. P (A) denotes the probability of an event.

Probabilities of events are always between 0 and 1.

0<=P(A)<=1

We say that an event is a sure event if it is having a probability of occurrence 1.

As we know that the probability varies from 0 to 1.

So if an event has the maximum probability, that is 1;then that event is a sure event.

For example we can consider the case of tossing a coin

In that case the probability of getting a head or tail one.

As one of these event will surely happen. So it is a sure event.

As we talk about sure event we cant avoid the case of impossible events.

Which is the opposite case to the sure event. That is never going to happen.

For example, let us take the example of tossing a die. Here probability of getting an eight is zero

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learning sure event-Examples


1) While drawing a dice,probability of getting one event less than 7

here the sample space is (1,2,3,4,5,6)

Probability of the event is P(E) =1

Here E is a sure event.

2)Suppose a bag contains 4 balls of red color.

Here the probability of getting a red ball is P(getting a red ball)=1

Which is also a sure event.

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learning sure event-Practice problems


1.      Find the probability of getting a green ball, from the bag, which contains 5 green balls

2.      Find the probability of getting a white pearl from the bag of 500 white pearls and one cream pearl.

3.      Find the probability of getting a black ball from the pack of white balls.

Answers:

1.      Here the sample space contains only events that contain the occurring of green balls. There is no chance of getting any other ball.

Here the number of events favorable is 5

And total number of events is also 5

So here the probability of getting a green ball is

P (Getting a green ball)=5/5

So it is 1..

2. This is the chance of almost sure event.

Probability of getting the other color pearl is almost 0 here.

3. This is the case of Impossible events.

Here the probability of occurring the event is 0

As it a not possible to get a black ball from the white ball pack

Array Definition Math

Introduction to array definition math:
Array has a list of data. In math array is a rectangular arrangement of elements. The elements are shown in rows and columns. In math array elements are put in the parenthesis or square brackets. In math the arrays are represented by capital letters for example A, B, C……An array is a symbolic way to represent math facts. Using symbols like circle, square or rectangle is one way of writing math facts in array.

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Definition of Array:

Definition of array:

The symbol m `xx` n is used to represent the array in math. The number of columns is represented by first element in the symbol and number of rows is represented by second element in the symbol. For example 2 `xx` 3 mean that array has 2 rows and 3 columns. The structure of array is rectangle. Array is used to be a symbol of the number of data.

Here

First row elements = a1, a2, a3

Second row elements = b1, b2, b3

Third row elements = c1, c2, c3

First column elements = a1, b1, c1

Second column elements = a2, b2, c2

Third column elements = a3, b3, c3

Symbol representation of above array is 3 `xx` 3

Total rows =3

Total column = 3

Types of Array or Matrix:

Types of array or matrix:

Column matrix
Row matrix
Square matrix
Diagonal matrix
Triangular matrix
Scalar matrix
Identity matrix
Zero matrix
Equality of matrix

Definition of column matrix:

Only one column is present in the column matrix.

Definition of row matrix:

Only one row is present in the row matrix.

Definition of square matrix:

If one matrix has equal number of row and column means called as square matrix.

Definition of diagonal matrix:

Other than diagonal elements in the array is zero means called as diagonal matrix.

Definition of triangle matrix:

Triangle matrix must be a square matrix. All the elements in above the diagonal are zero means called as lower triangle matrix. All the elements in below the diagonal are zero means called as upper triangle matrix.

Definition of unit matrix:

Unit matrix must be a square and diagonal matrix. The diagonal elements are one and the remaining elements in the array is zero means call as unit matrix.

Growth Factor in Math

Introduction to growth factor in math:

Growth factor in math article deals with the definition of growth factor in math and the model problems related to growth factor.

Definition of growth factor in math:

Growth factor is defined, as the constant rate that is multiplied by it self over a certain period.the growth factor is always a positive number. Growth factor cannot be represented in percentage.

Formula to growth factor in math:

When the growth rate is given

Increase in value = P* G*T

P is the principle or original value.

G is the growth rate.

T is the period.
Growth factor is calculating by dividing the increased value by original value

Growth factor = `(("increased value")/("original value"))`

it is always positive value not the negative one.

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Model problems to growth factor in math:

Problem: 1

Find the growth factor of the original value $500 over a period of 7 years with the growth rate of 6%.

Solution:

The original value is $500

Period of time = 7 years

Growth rate is 6%

The increase value is `(500*7*(6/100))`

=(5*7*6)

= 210

The increased value is $710

Now we have to calculate the growth factor:

Growth factor = `(("increased value")/("original value"))`

= `(710/500)`

= 1.42

The growth factor is 1.42

Problem: 2

Find the growth factor of the original value $300 over a period of 9 years with the growth rate of 8%.

Solution:

The original value is $300

Period of time = 9 years

Growth rate is 8%

The increase value is `(300*9*(7/100))`

(3*9*7)

= 147

The increased value is $447

Now we have to calculate the growth factor:

Growth factor = `(("increased value")/("original value"))`

= `(447/300)`

= 1.49

The growth factor is 1.49

Problem: 3 find the growth factor of the numbers behind the numbers 12.5,50, 100, 200, 400.

Solution:

To find the growth, we have to divide the previous value by present value

(50/12.5) = 2
(100/50) =2
(200/ 100)= 2
(400/200)=2

Here the growth factors between the numbers are 2

the growth factor of the numbers is 2

5th Grade Math Functions

Introduction to 5th grade math functions:

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

Is this topic Simple Interest Problems hard for you? Watch out for my coming posts.

5th grade math functions - Example problems:


Here we will discuss about the 5th grade math functions,

5th grade math function - Example problem: 1

Find the simple interest on a sum of Rs. 5000 for 5 years at the rate of 14% per annual.

Solution:

The given principal amount = Rs. 5000

Interest rate = 14% per annual.

Interest on Rs. 100 for 1 year = Rs. 50

Interest on Rs.2500 for 1 year = Rs. 500

Therefore,

Interest on Rs. 2500 for 3 year = 5 * 500

= Rs. 2500

[Answer: The Simple interest value is Rs. 2500]

5th grade math function - Example problem: 2

Write the given number 9564 in expanded form.

Solution:

The given number is 9564

It can be written as,

9564 = 9000 + 500 + 60 + 4

Rearrange the above numbers as,

9564 = (9 * 1000) + (5 * 100) + (6 * 10) + (4 * 1)

Answer: 9564 = (9 * 1000) + (5 * 100) + (6 * 10) + (4 * 1)

5th grade math function - Example problem: 3

Find the sum of the two algebraic functions (x + 52x ^5 - 15) and (11x - 52)

Solution:

The given two functions are (x + 52x ^5 - 15) and (11x - 52)

Add the given two functions, we get

= (x + 52x ^5 - 15) + (11x - 52)

Expand the above functions, we get

= (x + 52x ^5 - 15 + 11x - 52)

= 52x3 + 12x - 67

Answer:  (x + 52x ^5 - 15) + (11x - 52) = 52x ^5 + 12x - 67

5th grade math functions - Practice problems:

1. Find the expand form for the given number 75689.

[Answer: 75689 = (7 * 10000) + (5 * 1000) + (6 * 100) + (8 * 10) + (9 * 1)]

2. Find the simple interest on sum of 6000 for 4 years at the rate of 10%.

[Answer: The simple interest value is 2400]

3. Add the given two algebraic functions f(x) = 89x + 15 and g(x) = 8x2 - 26

[Answer: 8x2 + 89x – 11]

6th Grade Math Problems

Introduction:

This article deals with the math subject topics that are covered under the 6th(sixth) gradewith sample problems. The basics of all topics in mathematics are covered in the 6th grade and the student is prepared to cope with advanced concepts in mathematics in higher classes.

6th grade math topics:

The 6th (sixth) grade math subject includes the topics like basics of the,

Number system (whole numbers, decimal numbers, rational numbers, integers, fractions and ordering the numbers)
Measurement (metric system, temperature, mass, volume, length, time, area, surface area, and perimeter)
Basic geometry (understandings the elementary shapes, ideas and geometric figures)
Basic Algebra
Data handling in statistics and Symmetry
Ratio and Proportion.
These all are the important and basic topics for the 6th grad math topics.

Explanatory:

1. Number system:

In number system they have to know about how to recognize the numbers from 0 to ‘n’ (minimum 1000000) and, to write an order, represent, comparing, locate the numbers and identify the numbers.

And fully know about the decimal, whole, rational numbers, understanding the mixed numbers, fraction numbers from based on the concepts.

2. Measurements:

In measurement topic completely understanding of all the measurements like mm, cm, meter, km, feet, inches, yards, miles and units. And calculate the area, volume an etc…

3. Geometry:

To know about the geometric figures, concepts, ideas, and dimensional shapes and know about how to draw, calculate and understanding.

4. Basic algebra:

Algebra is the basic chapter for understanding all the other chapters and it has covers many topics. It is very useful for create, analyze, identify.

5. Statistics:

This statistics chapter covers the basics about the data, probability, distributions and statistics.

Math problems for 6th grade:

Example 1: Can you instantly find the greatest and the smallest numbers in each row?

1. 352, 856, 89,21,1548, 659, 256.

2. 32, 58, 6985, 20, 568, 125,129.

Solution:

1. 352, 856, 89,21,1548, 659, 256

The greatest number is 1548,

The smallest number is 21.

Solution:

2. 32, 58, 6985, 20, 568, 125,129.

The greatest number is 6985,

The smallest number is 20.

Example 2:  one side of the square is 6cm. Calculate the area of square?

Solution:

One side of the square is 6cm.

Square has totally 4 equal sides,

Area of square formula is = a ^2

We know a = 6cm

So, a ^2 = 6*6

= 36 cm

Answer is = area of square has 36 cm.

These are the problems on 6th grade math problems.

Learning Circles

Introduction About Learning Circles:

By learning, Circle consists of set of points from middle point. The point from the all points on a circle are equidistant is called the center of the circle, and the distance from that point to the circle is called the radius of the circle. illustration of circles is in single letter, its center. Circle has center point C and a radius of length r.


Properties of circle:


Let we see about the properties of learning circles:

The arc of a circle contains two points on the circle and all of the points on the circle that lie between those two points.
Chord is a segment and having end points which is on a circle.
Diameter of circle:
Length of the diameter = 2 × length of the radius.

Circumference of a Circle:
Circumference = 3 × diameter (approx.).

These are the properties of learning circles.


Formula:


Formulas of learning circles:

The area of circle:

Area of circle = 'pi' * r^2

where, r is the radius of the circle.

The diameter of circle:

Diameter of circle = 2 * r.

where,  r is radius of the circle.

The circumference of circle:

Circumference of circle = 2* 'pi' *r (or) 'pi' *d.

where,r is the radius of the circle.

d is the diameter of the circle.

Example 1:

Find the area, diameter and circumference of circle with radius of 4cm.

Solution:

1.The area of circle:

Area of circle = ∏ *r*r.

=(3.14)*4*4.

= 50.24 cm^2.

2.The diameter of circle:

Diameter of circle = 2*r.
= 2* 4.
= 8cm.

3.The circumference of circle:


Circumference of circle = ∏ * d.
= (3.14)* 8.
= 25.12cm.
Example 2:

Find the area, diameter and circumference of circle with radius of 7cm.

Solution:

1.The area of circle:

Area of circle = ∏ *r*r.

=(3.14)*7*7.

= 153.86 cm^2.

2.The diameter of circle:

Diameter of circle = 2*r.
= 2* 7.
= 14 cm.
3.The circumference of circle:

Circumference of circle = ∏ * d.
= (3.14)* 14.
= 43.96 cm.

Learning Articles with Histograms

Introduction to learning with histograms:

A graphical representation, in which learning the articles of histogram were identified as a bar graph for a particular measure of two frequencies and it has format of table that has shown in the bar graph and the graph represents the shapes that are given in the format. The height of a box and the measure is equal to the base side of the frequent data and the interval. A histogram is a graph demonstration of bar graph. Here shown about the types of distribution in the histogram.

learning part in articles with Histogram:

Horizontal X-Axis:

In the horizontal X-axis the bar chart gives the scale valve, which are the dimensions that fit into the data. These dimensions were normally well-known to the periods. Plot the horizontal X-axis points in the bar chart with respect to the values of Vertical Y-axis.

Vertical Y-Axis:

The straight bar or Vertical Y-axis in the bar chart gives the scale value, which shows you the several times the values within a period occurred. In several times it called as "frequency" in Y-axis.

Legend:

The legend provides extra information about the documents where the data came from and how the measurements were gathered.

Bars:

The bars consists of two main characteristics in the histogram, they are height and width. The height represents the several times the values within a period occurred. The width represents the length of the period covered by the bar. It is the same for all bars.

Title:

The title part in the histogram briefly explains the information about the things that are used in the bar chart.

learning Uses of articles with Histogram:

This histogram shows moreover a perfect distribution that covers about the graph, which displays all the brilliancy levels that contained in the scene found from the darkest to the brightest. The values are arranged with the values at the bottom of the graph from left to right. It gives an idea about the histogram and its use.

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Learning Conditions for the articles with Histogram:

The data’s can be marked clearly on the graph by measuring the data values accurately the scale and period.
The scale should contain all the data values. The period divides the scale into equal parts.
Learning Applications of articles with Histograms:

Clear review of total data sets that graphically shown in the bar chart.
Evaluate the measurements to the given specifications.
Share the information to the team.
Assist in administrative.

Find The Difference Math

Introduction to Find the Difference math:

Math uses various operations from which the given expressions can be evaluated. In math, Subtraction is an operation for which the difference between two terms can be manipulated. There are some rules for finding the difference for the given two numbers. In this article, we shall solve the math problems to find the difference between two numbers. Also the math problems are available to find the difference between two numbers with answers.

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Example problems to find the difference math:

Example 1:

Find the difference of 17 and 5.

Solution:

The given sentence can be expressed as 17 – 5

Now subtract 5 from 17,

1 7

–   5

12

We get the result, 12

Therefore the difference of 17 and 5 is 12.

Example 2:

Find the difference of 17 and – 5.

Solution:

The given sentence can be expressed as 17 – (– 5).

Rewrite the expression as 17 + 5, since (– * – = +)

Now add the terms,

1 7

+   5

2 2

We get the result, 22

Therefore the difference of 17 and –5 is 22.

Example 3:

Find the difference of –17 and 5.

Solution:

The given sentence can be expressed as –17 –5

Rewrite the expression, –(17 + 5)

Now add 17 and 5,

1 7

+  5

2 2

Put the minus sign before the answer.

We get the result, –22

Therefore the difference of –17 and 5 is –22.

Example 4:

Find the difference of –17 and – 5.

Solution:

The given sentence can be expressed as –17 – (– 5)

Rewrite the expression as –17 + 5

Now subtract 5 from 17,

1 7

–   5

1 2

Now put the sign of the greater number.

Here 17 is the greater number and its sign is – .

We get the result, –12

Therefore the difference of –17 and –5 is –12.

Practice problems to find the difference math:

Problem 1:

Find the difference of 25 and 10.

Answer: 15

Problem 2:

Find the difference of 25 and –10.

Answer: 35

Problem 3:

Find the difference of –25 and 10.

Answer: –35

Problem 4:

Find the difference of –25 and –10.

Answer: –15

Learning Algebra Questions

Introduction to learning algebra questions:

For learning algebra questions, it is necessary to solve or workout the algebra questions by own. This way of learning makes the students to be good in their subject.  Algebra deals with solving unknown values called variables with the help of known variables. In algebra, we continually use letters to represent numbers. Algebra questions include real numbers, complex numbers, matrices, vectors etc. An algebraic expression contains constants, operating symbols, such as plus and minus signs and constants. The learning method can be explained in the following examples.


Learning algebra Questions examples:


Q 1:  Evaluate the following quadratic equation.   x ^2 - 3x = 0

Sol :    Given
x ^2 - 3x = 0

Take X factor as common
x (x - 3) = 0

So the product x (x - 3) to be equal to zero, then we get

x = 0 or x - 3 = 0

Solve the above simple equations to obtain the solutions.
x = 0
or
x = 3

X= 0 or 3

Q 2:   Evaluate the quadratic equation   x ^2 - 5 x + 6 = 0

Sol :   To factor the expression, we write the equation x ^2 - 5 x + 6 in the form factored:

x ^2 - 5 x + 6 = (x + a)(x + b)

So that the sum of a and b is -5 and their product is 6. The numbers that satisfy these conditions are - 2 and - 3. Hence
x ^2 - 5 x + 6 = (x - 2)(x - 3)

Substitute into the original equation and solve.
(x - 2)(x - 3) = 0

(x - 2)(x - 3) is equal to zero if
x - 2 = 0
or
x - 3 = 0

Solve the above equations to get the solution

x = 2
or
x = 3

x= 2 or 3


Practice for learning algebra questions:


Q 1:   Evaluate the quadratic equation   x ^2 - 6 x + 9 = 0

Ans : x = ±3

Q  2:  Evaluate the following quadratic equation.   x ^2 - 6x = 0

Ans :   x= 0 or 6

Active Learning Theory

Introduction to active learning theory:

Active learning theory is a sunshade word to submit to numerous representation of teaching to center the dependability of learning. Bonwell popularized this advance to training. This buzz word of the 1980s develops into their 1990s account toward the organization used for the Study of Higher Education. This statement they converse a diversity of methodologies for help active learning. Though according toward plan resembling active learning urbanized away of the effort of an former collection of theorists persons support, discovery learning.


Active learning theory:


Though here is rejection problem to apprentice must be occupied through learning with cognitively active, some examine contain well-known to organism performance active through original learning know how to be harmful to schema acquisition.

Alternative activity theory while those appoint with interrelate by their atmosphere, manufacture of apparatus consequences. These apparatus are exteriorized structure of psychological method, with as these psychological procedures are noticeable in apparatus, they develop into further willingly available with transmissible to new group, fetching constructive for public communication.
It contain be optional to apprentice who actively connect by the objects are further probable toward be reminiscent in sequence, except some glowing recognized writer contain dispute this maintain is not glowing maintain through the journalism. Quite than individual behaviorally lively through learning, propose beginner must be cognitively active.


Active learning theory: Learning by teaching


A well organized instructional approach to combine direction by active learning through training. This approach permits apprentice to educate the original satisfied toward every supplementary. They should be perfectly direct through coach. This technique be initiate through the early on 1980s, particularly within Germany, with is currently glowing recognized within every point of the instructive scheme. Learning through education is combination of behaviorism with cognitive also propose a coherent structure for assumption with follow.

Active learning theory and Policy

Strategy might be fulfilled through representative the instructional efficiency of instruction. Rubrics are a high-quality method to estimate active learning foundation training. These instructional apparatus know how to be use to explain the variety of dissimilar behavior of some action. Stipulation known toward the apprentice, they know how to supply additional assistance.

Active Learning Theory

Introduction to active learning theory:

Active learning theory is a sunshade word to submit to numerous representation of teaching to center the dependability of learning. Bonwell popularized this advance to training. This buzz word of the 1980s develops into their 1990s account toward the organization used for the Study of Higher Education. This statement they converse a diversity of methodologies for help active learning. Though according toward plan resembling active learning urbanized away of the effort of an former collection of theorists persons support, discovery learning.


Active learning theory:


Though here is rejection problem to apprentice must be occupied through learning with cognitively active, some examine contain well-known to organism performance active through original learning know how to be harmful to schema acquisition.

Alternative activity theory while those appoint with interrelate by their atmosphere, manufacture of apparatus consequences. These apparatus are exteriorized structure of psychological method, with as these psychological procedures are noticeable in apparatus, they develop into further willingly available with transmissible to new group, fetching constructive for public communication.
It contain be optional to apprentice who actively connect by the objects are further probable toward be reminiscent in sequence, except some glowing recognized writer contain dispute this maintain is not glowing maintain through the journalism. Quite than individual behaviorally lively through learning, propose beginner must be cognitively active.


Active learning theory: Learning by teaching


A well organized instructional approach to combine direction by active learning through training. This approach permits apprentice to educate the original satisfied toward every supplementary. They should be perfectly direct through coach. This technique be initiate through the early on 1980s, particularly within Germany, with is currently glowing recognized within every point of the instructive scheme. Learning through education is combination of behaviorism with cognitive also propose a coherent structure for assumption with follow.

Active learning theory and Policy

Strategy might be fulfilled through representative the instructional efficiency of instruction. Rubrics are a high-quality method to estimate active learning foundation training. These instructional apparatus know how to be use to explain the variety of dissimilar behavior of some action. Stipulation known toward the apprentice, they know how to supply additional assistance.

Define Pie Chart

Pie charts are circular graphs that have different  sections, that are used for organizing a set of data. These pie charts are represented by comparing  the data by using fractions or percentages, with each section  proportional to the fraction or percentage that it represents in the data set . These chart are generally  used in corporate reports, especially to show budget or financial data. These charts  are  so named as "Pie charts" because of its similarity  to a pie that is sliced into pieces.


Defnition of Pie chart


A Pie chart consists of a circle divided into several partitons normally it does not exceed more than 6. Area of each part is called slice .Each part  is of the same percentage of the circle as the component it reperesents is of the whole data set. Also called circle diagram  or sector graph.


Advantages of Pie charts:


Pie charts are easy to read and understand if they are designed appropriately . These pie charts are very effective to show the data , if the intent was to compare one section to another. Understanding Definition of Statistics is always challenging for me but thanks to all math help websites to help me out.

The best way to make these charts more readable is to make them fill with different colors and to display sections in clockwise direction ,from larger to smaller section.

Learning About Calculus

Introduction of learning about calculus:

In calculus there are various process is involves for create many tools of differential theoretical aspects. Geometrical and kinematic significance for first and second order derivatives were also interpreted. Now let us learn some practical aspects of differential calculus. It contains differential calculus and integral calculus. The applications of Calculus are Statistics, Science, Business and Engineering. Having problem with Left Hand Riemann Sum keep reading my upcoming posts, i will try to help you.


Concept of learning differential calculus:


At this level we shall consider about problems concerned with the applications to (i) plane geometry, (ii) theory of real functions, (iii) optimization problems and approximation problems.

Derivative as a rate measure:

If a quantity y depends on and varies with a quantity x then the rate of change of y with respect to x is dy / dx.. Example a rate of changes of current ‘i’ is di / dt and a rate of change of temperature ‘?’ is d? / dt and so on.

Please express your views of this topic Heron's Formula by commenting on blog.

Integral calculus and its applications:


The direct evaluation of definite integrals is about the limit for integral total. The integrands are very simple, direct calculation of definite integrals as the limit of integral total involves great complex. Sometimes this method involves cumbersome computations.

The formula called Second Fundamental Theorem on Calculus about that yields a practical and particular method for calculating definite integrals in case where the anti-derivative of the integrand is known. This method which was discovered by Newton and Leibnitz utilizes ‘the profound relationship’ that exists between integration and differentiation.

In this learning Integral calculus, we have the five sections dealing with the concept and applications of definite integrals:

(i)  To solve simple problems using theorem of calculus.

(ii)  Properties of definite integral.

(iii) Reduction formulae learning

(iv) Area under the curve and volume of solid of revolution about an axis.

(v)  Length of the curve and the surface area of a solid of revolution about an axis.

Surface Area Learning

Introduction to lateral surface learning:
Lateral surface area in a solid is the sum of the surface areas of all its faces without the base of the solid. Or Lateral surface area means the area of the sides only -- without the top and bottom. Lateral surface area is found for an object around its outer area. Lateral surface area are usually expressed in terms of some square units.


lateral surface area learning formula:


Formula for Lateral surface area:

Cube = 4b^2, where b is the base of a cube

Sphere is 4pr^2, where r is the radius of the sphere.

Cone = p × r × l, where r and l are the radius and slant height of the cone

Cylinder = 2prh, where r is the radius and h is the height of the cylinder

Right triangular pyramid = 3 × area of lateral faces

Pentagonal prism = 5 × area of each rectangle

Please express your views of this topic Intersecting Lines by commenting on blog.

Examples for learning of lateral surface area:

Example 1:

Find the lateral surface area of a pentagonal prism, if a = 5 cm and b = 14 cm.

Solution:

Step 1: In the given figure, the base of the prism is a regular pentagon.

Step 2: All five rectangles are congruent.

Step 3: Lateral surface area of the prism = 5 × area of each rectangle

Step 4: = 5 × 5 × 14 [Substitute the values.]

Step 5: = 350

So, Lateral surface area of the pentagonal prism = 350 cm2.

Example 2:

Height and radius of the cone is 5yard and 7 yard.Find the lateral surface area of the given cone.

Solution: Lateral surface area of the cone = prl

Step 1:Slant height of the cone, l =v(25+49)      [l =vr2+h2.]

l = 8.6 yard

Step 2: Lateral surface area = 3.14 × 7 × 8.6   [As r = 7 andl = 8.6]

=

So, the lateral surface area of the cone = 189.03 sq yd.

Practice problem for lateral surface area learning;

1) Find the lateral surface area for of a sphere radius (r) = 3cm ?

Answer:  113.04

Equality Properties of Learning

Equality properties of learning:

Equal properties are the reasonable laws for actual numbers in arithmetic. These properties are used to control, stable the equations. Moreover, shorten the equations. In general, equality is defined as follows,

p = q denotes p is equal to q.
p ≠ q denotes p does not equal q.
Thus, the learning the properties of equality contain the following properties.

Based on balance equation:

a) Addition property

b) Subtraction property

c) Multiplication property

d) Division property

Based on equivalence:

a) Reflexive property

b) Symmetric property

c) Transitive property

Distributive property


1. Balance equation relation property:


The following properties are used for learning the equations with real numbers.

a) Addition property:

For learning the addition property let assume m, n, o are actual numbers. If m =n, then it can be written as m+o = n+o. similar number can add the equation of both side lacking of modifying the result of the equation.

b) Subtraction property:

For learning the subtraction property let assume x, y, z are real numbers. If x =y, then it can be written as x-z = y-z. Equal number can subtract the equation of all side without adjusting the result of the equation.

c) Multiplication property:

Let consider p, q, r are real numbers. If p =q, then it can be written as p*r = q*r. The equation of both sides can be multiplied by similar quantity without adjusting the result of the equation.

d) Division property:

Let consider p, q, r are real numbers(r =/ 0). If p =q, then it can be written as p/r = q/r. The equation of each side can be divided by same nonzero quantity without modifying the result of the equation.


2. Equivalence relation property:


a) Reflexive property:

Let consider ‘m’ is a real number, and then it reflects by itself. That the real number equals itself as, m = m.

b) Symmetric property:

For learning the symmetric property, let consider m and n are real numbers. If m = n, then it can be written as,

n =m. The order of equality is not considered.

c) Transitive property:

Let consider m, n, and o are real numbers. If m = n and n = o, then it can be written as,

m =o. Thus, the two quantities matching to the same extent are identical to each other

3. Distributive property:

From learning of distributive property, let consider p, q, r are real numbers. Then it states that as follows,

p(q+r) = pq+pr

Angle of Depression

Introduction to Angle of Depression Learning:

Angle of depression is a term used mainly in trigonometry where “depression” means “fall” or “drop”. Angle of depression means the angle between the horizontal and the line of sight to an object beneath the horizontal. The angle of depression is mainly used for learning or obtaining the distance of the two objects where we only know their angle and an object’s distance from the ground. I like to share this Pentagon Geometry with you all through my article.

Learning Angle of Depression:

Learning angle of depression plays one of the key roles for human’s day-to-day life. The angle of depression and angle of elevation is used for seeing objects where they are high above us or low below us. The human’s should able to differentiate what is angle of depression and angle of elevation and where to use them. The some examples of angle of depression are finding an angle from top of the building a man seeing a moving car, and a man seeing a stationary car from a moving train. Please express your views of this topic Picture of an Obtuse Angle by commenting on blog.

Example for Learning Angle of Depression:

Consider an example where the distance of a tree and the airplane is to find out where the distance from the ground and the airplane is given and also we know the angle between the tree and the airplane . But the airplane is flying above the tree here we want to find the distance from the airplane and tree. From using the given data’s we can find the angle of depression where the angle for the foot of the tree and airplane’s base is acting as an angle of depression. From the angle and the distance from the ground to the airplane we can find the distance from the tree to the plane using any one of the trigonometric identities. In the case where we have to find the distance between the foot of the ground and the airplane is obtained by the same trigonometric relations.

Math Makes Sense Grade 6

Introduction to Math makes sense grade 6:

Mathematics plays a vital role in the grade of 6. Elementary level of math are easy to learn and simple to solve.  Grade 6 math, consists of algebra, arithmetic calculations, sets, measurements and graphs. In arithmetic we use numerals and variables to represent equation. The grade 6 of mathematics also solves the linear equations and also the multiplication, subtraction, addition and division of the algebra. This grade 6 math sets as basic block for solving aptitude questions. Having problem with Laplace Transform Chart keep reading my upcoming posts, i will try to help you.


Math makes sense grade 6 in algebra sample problems:


On doing the sample problems they can also solve the advanced problems in the mathematics.

Example 1 to Math makes sense grade 6:

Solve the algebraic equation from 3(-3y - 2) - (y - 3) = -10(2y + 2) + 19

Solution:

Step 1:

Given equation is 3(-3y - 2) - (y - 3) = -10(2y + 2) + 19

Step 2:

Multiply the terms

-9 y -6 - y + 3= -20y – 20 + 19

Make them as a group

-10y -3= -20y - 1

-10y + 20y = 3 -1

10y = 2

y = 2/10

y = 0.2

The Answer Y = 0.2

Example 2 Math makes sense grade 6:


Solve the equation of grade 6:

2x + 5 = 4x (3) + 15

Solution:

Eliminate the braces:

2x + 5 = 12x + 15

Subtract -5 on both the sides:

2x +5 -5 = 12 x + 15 – 5

Simplify the equation:

2x = 12x + 10

Subtract – 10 from both the sides:

2x-10= 12x + 10 – 10

On simplifying the equation:

2x-10 = 12x

Subtract – 12x on both sides to get 0 on the right hand side:

2x – 10 – 12x = 12x – 12x

On simplifying the equation we get:

-10 –10x = 0

On adding 10 on both sides:

-10 -10x +10 = 0 +10

On simplifying the equation we get,

-10x = 10

Dividing by -10 on both the sides:

-10x / -10 = 10 / -10

After dividing we get:

X = -1

Thus we got the value of x by simplifying the equation.

Mental Math Techniques

Some people are born great at maths, most of us have to work hard. But why work harder than we need too, when we can often just use a better technique to improve our maths?

Here are some techniques and rules you can use in your maths to improve your maths skills. I like to share this Mental Math Problems with you all through my article.


Estimation

This is one of the most effective techniques to help you work out roughly what sort of answer you should have. So if you need to multiply 305 x 11, then it looks hard. But 300 x 11 is really easy, so do that first to get an idea of the answer - that's 3300. Having problem with Multiplicative Inverse keep reading my upcoming posts, i will try to help you.

Break It Down

Now we know that, we use the breakdown method: 305 x 11 is the same as 300 x 11 + 5 x 11. So to our 3300 we just add 55, to get 3355 as the answer, and we can see we've worked out 305 x 11 really easily!

Multiplying by 0 is always 0

Some sums that look really hard are actually really, check this one out: 26 x 54 - 15 x 576372 x 0 + 57

Looks hard? It's easy - the answer is '0' for any sum that contains multiplying by '0' and once you know that handy little fact such sums become the easiest ones there are!

Ratio Definition Math

Definition for ratio:

In mathematics, a ratio expresses the magnitude of quantities relative to each other. Specifically, the ratio of two quantities indicates how many times the first quantity is contained in the second and may be expressed algebraically as their quotient. I like to share this What is a Ratio? with you all through my article.

Example:

For every Spoon of sugar, you need 2 spoons of flour (1:2)

Source :Wikipedia

Ratio formula:

Let A and B be two given points. Let P be a point on the line segment `barA``barB` or on `barA` `barB` produced. Then P divides `barA` `barB` into two segments `barA` `barP` and `barP` `barB`. The lengths of `barA` `barP` and `barP``barB`  are AP and PB.

These lengths are in some ratio: n; that is AP : PB = m : n or  `(AP)/(PB)` `(m)/(n)`

If P lies inside `barA` `barB` we say that P divides `barA``barB` internally in the ratio m : n. If P lies outside `barA` `barB`, that is, P lies on `barA` `barB`  produced, then we say that P divides `barA``barB` externally in the ratio m : n. With a given ratio m : n, `barA``barB` can be divided either internally or externally. Understanding Dividing Radicals with Variables is always challenging for me but thanks to all math help websites to help me out.

Example for ratio:

Example :

Divide the line segment `barA` `barB` of length 16 units in the ratio 3:5

Solution:

i) Let C be the point inside `barA``barB` such that `(AC)/(CB)`= `(3)/(5)`. Since the numerator is smaller than the denominator, C is closer to A than to B . Then

5AC=3 BC or 5AC=3(AB–AC) or 8AC=3AB =3(16)=48

AC=6 units and so CB=AB–AC =16–6 = 10 units.

Hence C lies inside `barA` `barB` 6 units distance from A and 10 units distance from B. The point C is unique and it divides `barA``barB` internally in the given ratio 3:5.

ii) Let D be the point outside `barA``barB` such that `(AD)/(DB)`=`(3)/(5)`?. Since the numerator is smaller than the denominator, D is closer to A than to B. Now we have

5 AD = 3DB or 5 AD = 3(AD + AB) or 5 AD = 3AD + 3AB

2 AD = 3AB = 3(16) = 48 or AD = 24 Then DB = DA + AB = 24 + 16 = 40

Therefore, D lies outside `barA``barB` 24 units distance from A and 40 units distance from B. The point D is unique and it divides   `barA``barB` externally in the given ratio 3:5.

Solving Mathematics Grind

Introduction for Solving Mathematics Grind:

Well experienced tutor (or) graduate Mathematics student offering Junior and Leaving Certificate grinds around Ireland. Grinds contains various subjects such as Mathematics, French, Accounting etc., in this math grind covers all level of students. They are providing explanation of theory with detailed examples with questions and answers. In this article we shall discuss about solving mathematics grind. The following examples are involved in solving mathematics grind. Is this topic Exponential Function Solver hard for you? Watch out for my coming posts.

Solving Mathematics Grind Example: 1

Solve the sum and find the value of ‘x’

50x + 40 = -200

Subtract 40 from both sides:

50x + 40 – 40 = -200 - 40

Simplify both sides:

50x = -240

Divide both sides by 50:

`(50x)/50` = `-240/50`

Simplify both sides:

x   =   `-24/5`

I have recently faced lot of problem while learning Place Value with Decimals, But thank to online resources of math which helped me to learn myself easily on net.

Solving Mathematics Grind Example: 2

Use Euclid’s algorithms solve the HCF of 2245 and 36548.

Solution:

Since 36548 > 2245, we apply the division lemma to 36548 and 2245, to get

36548 = 2245 × 16 + 628

Since the remainder 628 not equal to 0, we apply the division lemma to 2245 and 628, to get

2245 = 628 × 3 + 361

We consider the new divisor 628 and the new remainder 361, and apply the division lemma to get

628 = 361 × 1 + 267

We consider the new divisor 361 and the new remainder 267, and apply the division lemma to get

361 = 267 × 1 + 94

We consider the new divisor 267 and the new remainder 94, and apply the division lemma to get

267 = 94 × 2 + 79

We consider the new divisor 94 and the new remainder 79, and apply the division lemma to get

94 = 79 × 1 + 15

We consider the new divisor 79 and the new remainder 15, and apply the division lemma to get

79 = 15 × 5 + 4

We consider the new divisor 15 and the new remainder 4, and apply the division lemma to get

15 = 4 × 3 + 3

We consider the new divisor 4 and the new remainder 3, and apply the division lemma to get

4 = 3 × 1 + 1

We consider the new divisor 3 and the new remainder 1, and apply the division lemma to get

3 = 1 × 3 + 0

The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 1, the HCF of 36548 and 2245 is 1.

Notice that 1 = HCF (3, 1) = HCF (4, 3) = HCF (15, 4) = HCF (79, 15) = HCF (94, 79) = HCF (267, 94) = HCF (361, 267) = HCF (628, 361) = HCF (2245, 628) = HCF (36548, 2245).

Meters to Inches

Meters to inches

Meter:

The metre (or meter), symbol m, is the base unit of length in the International System of Units (SI). It is defined as the distance travelled by light in a complete vacuum in 1/299,792,458 of a second.

Inch:

An inch is the name of a unit of length in a number of different systems, including Imperial units, and United States customary units. I like to share this Alternate Exterior Angles with you all through my article.

(Source: wiki)

Let us see how to convert meters to inches in this article.

Formula for meters to inches

1 meter = 39.3700787 inches

Meters to Inches – Examples:
Meters to inches – Example 1:

Convert 5 Meters to inches?

Solution:

Step 1:

Formula for converting meters to inches:

1 meter = 39.3700787 inches

Step 2:

So to find 5 meter

Step 3:

Multiply 5 with 39.3700787 = 196.8503935

Step 4:

Therefore, 5 meter = 196.8503935 inches

Meters to inches – Example 2:

Convert 11 Meters to inches?

Solution:

Step 1:

Formula for converting meters to inches

1 meter = 39.3700787 inches

Step 2:

So to find 11 meter

Step 3:

Multiply 11 with 39.3700787 = 433.070866 inches

Step 4:

Therefore, 11 meter = 433.070866 inches

Meters to inches – Example 3:

Convert 15 Meters to inches?

Solution:

Step 1:

Formula for converting meters to inches

1 meter = 39.3700787 inches

Step 2:

So to find 15 meter

Step 3:

Multiply 15 with 39.3700787 = 590.551181 inches

Step 4:

Therefore, 15 meter = 590.551181 inches

Meters to inches – Example 4:

Convert 26 Meters to inches?

Solution:

Step 1:

Formula for converting meters to inches

1 meter = 39.3700787 inches

Step 2:

So to find 26 meter

Step 3:

Multiply 26 with 39.3700787 = 1 023.62205 inches

Step 4:

Therefore, 26 meter = 1 023.62205 inches

Meters to inches – Example 5:

Convert 0.50 Meters to inches?

Solution:

Step 1:

Formula for converting meters to inches

1 meter = 39.3700787 inches

Step 2:

So to find 0.501 meter

Step 3:

Multiply 0.501 with 39.3700787 = 19.7244094 inches

Step 4:

Therefore, 0.501 meter = 19.7244094 inches

Meters to inches – Example 6:

Convert 0.81 Meters to inches?

Solution:

Step 1:

Formula for converting meters to inches

1 meter = 39.3700787 inches

Step 2:

So to find 0.81 meter

Step 3:

Multiply 0.81 with 39.3700787 = 810

Step 4:

Therefore, 0.81 meter = 31.8897638 inches


Understanding Plane Parallel is always challenging for me but thanks to all math help websites to help me out.

Meters to Inches – Practice Problems

Practice problem -1

Convert 6.5 Meters to inches?

Answer:

255.905512 inches

Practice problem -2

Convert 0.6 Meters to inches?

Answer:

23.6220472 inches

Entrance Exam for College Math

Introduction to entrance exam for college math:

Entrance exam for college evaluates high school students, general educational improvement and their ability to complete college-level work. SAT exam is one of the entrance exams for college math. It is used to find out skills in math, reading comprehension and vocabulary of the students planning to attend college. Now, we are going to see some of the problems on entrance exam for college math.

Entrance Exam for College Math Solved Problems:

Example problem 1:

Solve the quadratic equation by factoring method: x2 - 143x + 142 = 0.

a) 1, 142

b) 1, 132

c) 1, 122

d) 1, 112

Solution:

The given quadratic equation is x^2 - 143x + 142 = 0

Here a = coefficient of x2 = 1

b = coefficient of x = -143

c = constant term = 142

We find a × c = 1 × 142 = 142 = -1 * -142, a + c = (-1) + (-142) = -143 = b.

By splitting the middle term, we get

x2 - 143x + 142 = 0

x2 - 1x - 142x + 142 = 0

x(x - 1) - 142(x - 1) = 0

(x - 1) (x - 142) = 0

x = 1, x = 142

So, the option a is the correct answer.

So, the answer is x = 1, 142. Having problem with word problems in algebra keep reading my upcoming posts, i will try to help you.

Additional Solved Problems-entrance Exam for College Math:

Example problem 2:

Solve the following simultaneous equations using substitution method:

9x – y = 18 ----------Equation (1)

1x + y = 22--------------Equation (2)

a) (2, 25)

b) (3, 27)

c) (5, 27)

d) (4, 18)

Solution:

Step 1: Let us consider the Equation (1)

9x – y = 18

Subtract 9x on both sides of the equation

9x – y –9x = 18 – 9x

-y = -9x + 18

y = 9x - 18---------Equation (3)

Step 2: Substitute the value of y in Equation (2). We get

1x + y = 22

1x + (9x – 18) = 22

10x – 18 = 22

Add 18 on both sides of the equation

10x – 18 + 18 = 22 + 18

10x = 40

Divide by 10 on both sides of the equation

`(10x) / 10 = 40 / 10`

x = 4

Step 3: Substituting this value of x in Equation (3), we get

y = 9x – 18

y = 36 – 18

y = 18

So, the option d is the correct answer.

So, the answer is (4, 18).

Thoughts on Distance Learning

Introduction to thoughts on distance learning:

In this article, we shall discuss about thoughts on distance learning. In mathematics, distance is defined as an object moves from one particular point to another point which is calculated by speed and time. That is, when we multiply the time and speed we will get the distance. Please express your views of this topic Define Permutations by commenting on blog.

Formula for the calculating distance is

Distance = speed `xx` time

Now we shall solve some example problems regarding thoughts on distance learning.

Examples to Thoughts on Distance Learning:

Example 1:

A train is driven at average speed of 3 kilometers per hour to the station 6 hours. If the train is droved at the average speed of 2, how many hour would need to reach the station?

Solution:

3 km/h needs 6 hours

2 km/h needs x hours

This can be written as,

3 = 6 and

2 = x

Also we can written as

3 `xx` 6 = 2x

18 = 2x Also we can written as

2x  = 18 now we have to divide both sided by 2

`(2x)/2`  = `18/2`

X = 9

therefore, if the train is droved at 2 kilometers per hour, it would need 9 hours
Example 2:

Bus and car are leaving from the same place in opposite direction. Bus goes at 3mph and the car goes at 2 mph. how many hours will they need to 22 miles apart?
Solution:

Distance = speed time

Speed of  bus = 3mph

Time of bus = t

Therefore, distance of the bust = 3t

Speed of  car = 2mph

Time of bus = t

Therefore, distance of the bust = 2t

Therefore,

3t + 2t  = 22 miles

5t  =  22   now we have to divide both sides by 5. so we get

`(5t)/5` = `22/5`

t =4.4 hours

Bus and car will reach 22 miles apart in 4.4 hours


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Practices Problems to Thoughts on Distance Learning :

Problem 1:

A train is driven at average speed of 6 kilometers per hour to the station 6 hours. If the train is driven at the average speed of 4, how many hour would need to reach the station?

Answer: 9 hours

Problem 2:

Bus and car are leaving from the same place in opposite direction. Bus goes at 10 mph and the car runs at 6mph. how many hours will they need to 39 miles apart?

Answer: 2.4 hours

Exact Distribution

Introduction to exact distribution:

The exact distribution refers the probability distribution the probability theory. The probability distribution is used to determine the number of possibility for the occurrence of an event. The most commonly used probability distributions are the binomial distribution, geometric distribution, normal distribution and the gamma distribution. These above mentioned distributions are included in the discrete and continuous probability distribution. The major type of the probability distribution is the discrete probability distribution and the continuous probability distribution. This article has the study about exact distribution.

Types of Exact Distribution:

The major types of the probability distribution are

Discrete probability distribution
Continuous probability distribution
Discrete probability distribution:

The probability for a countable number of occurrences for the event is calculated in the discrete probability distribution.
Continuous probability distribution:

The probability values in this are the continuous ranged value it is calculated in the continuous probability distribution.

Examples for Exact Distribution:

Example 1 to exact distribution:

If X is normally distributed the mean value is 1 and its standard deviation is 2. Determine the value of P (0 ≤ X ≤ 7).

Solution:

The given mu value is 1 and the standard deviation is 2.

Z = `(X- mu)/ sigma`

When X = 0, Z = `(0- 1)/ 2`

= -`1/2`

= -0.5

When X = 7, Z = `(7- 1)/ 2`

= `6/2`

= 3

Therefore,

P (0 ≤ X ≤ 7) = P (-0.5 < Z < 3)

P (0 ≤ X ≤ 7) = P (0 < Z < 0.5) + P (0 < Z < 3) (due to symmetry property)

P (0 ≤ X ≤ 7) = (0. 6915- 0.5) + (0.9987 - 0.5)

P (0 ≤ X ≤ 7) = 0.1915 + 0.4987

P (0 ≤ X ≤ 7) = 0.4987

The value for P (0 ≤ X ≤ 7) is 0.4987.

Between, if you have problem on these topics, please browse expert math related websites for more help on cbse 10th sample papers.

Example 2 to exact distribution:

The probability for destroying the target in only one time is 0.37. Compute the probability that it would be destroyed on the third attempt itself.

Solution:

The probability of destroying the target in one trial is p = 0.37.

The value of the q is calculated by q = 1-p

q = 1- 0.37

q = 0.63

By the geometric distribution, the probability for the success is calculated by using the formula

P(X =x) = q x p, the value of x is 0, 1, 2. . .

The target is destroyed at the third attempt, so x = 3.

P(X = 3) = (0. 63) 3 (0.37)

P(X = 3) = (0.25) (0.37)

P(X = 3) = 0.0925

The probability for destroying the target at the third trial is 0.0925.

Easy Way to Learn Statistics

Introduction to Easy Way to Learn Statistics

Statistics is the proper science of creating successful use of mathematical relating to groups of individuals or experiments. It deals with all features of this including not only the collection, analysis and interpretation of such data, but also the planning of the collection of data, in terms of the design of surveys and experiments. Now we will learn the statistics in easy way.

Examples for Easy Way to Learn Statistics

Example 1

What the mean, median,mode and range of the following group of numbers?

11,12,15,17,19.

Solution

The given numbers are 11,12,15,17,19.

Mean

We can find a mean in easy way. Mean is the average of the given number. So find the sum of the given numbers.

Sum of the given numbers are = 11+12+15+17+19

Now divided by 5 (Because 5 is the total given numbers) =74/5

=14.8.

Median

A middle value of the given number series is the median.

The number series is 11,12,15,17,19.

Here the center value is 15.

Therefore 15 is the median.

Mode

It is also very easy way to find. Mode is a duplicate value of the given number series. Here no duplicate value.

Therefore mode is empty or null.

Range

It can find in easy way. Range is the difference between maximum value and the minimum value of the given number series.

Range=19-11

=8.

Example 2

What is the mean, median of the following group in statistics?

14,16,18,20, 22.

Solution

The given numbers are 14,16,18,20 and 22.

Mean

We learn.mean is the average of the given number. We find the sum of the given numbers.

Sum of the given numbers are = 14+16+18+20+22

= 90.

Now divided by 5 (Because 5 is the total given numbers) = 90/5

= 18.

Median

We learn,a middle value of the given number series is the median.

The number series is 14,16,18,20, 22.

The middle value of the above series is 18.

Therefore 18 is the median.

These are the examples and find the solution in easy way.