Basic Concepts of Mathematics

Introduction - basic concepts of mathematics:
From the basic concepts of mathematics have a different topic with the accurate solutions of integer to fractions, subtraction, measurement, number sense, multiplication, functions, mixed numbers, division, median problems, place value, ordering decimals. Basic mathematics is essential to the students for the upcoming higher grade.  In this article we shall discuss about basic concepts of mathematics. Is this topic Vertex Calculator hard for you? Watch out for my coming posts.

Example Problems - Basic Concepts of Mathematics:

Example 1:

Adding the positive numbers:` (+31) + (+8)`

Solution: The sum of two positive numbers will be a positive numbers.

` (+31) + (+8) = +39`

Solution is `39.`

Example 2:

Adding the positive numbers: `(+13) + (+14) =?`

Solution: The sum of two positive numbers will be a positive numbers.

` (+13) + (+14) = +27`

Solution is `27.`

Basic concepts of mathematics -Subtraction of negative numbers:

Example 1:

Subtracting the negative numbers: `(1) - (19)`

Solution: while subtracting the two numbers, the result carries sign of larger number

`(1) - (19) = -18`

Solution is `-18` .

Example 2: Subtracting the negative numbers:` (23) - (5)`

Solution: The sum of two negative numbers will be a negative number.

`(23) - (5) = 18`

Answer is `18.`

Example 3: Get the value form the expanded form `9xx 10000 + 5xx 1000 + 5 xx 10`

Solution:  `9xx 10000 + 5 xx 1000 + 5xx 10`

`= 90000 + 5000 + 50`

` => 95050.`

Basic concepts of mathematics - multiplication problems:

Example 1:

Multiply the two numbers: `11 xx7`

Solution:

Here `11` is multiplicand

`7` is multiplier

Step1: Multiply the ones term,

Step 2: Multiply the second term (tens)

Final answer is `77` .

Example 2:

Multiply the two numbers: `12 xx 9`

Solution:

Here `12` is multiplicand

`9` is multiplier

Answer is `108`

Numbers using words - basic concepts of mathematics:

Example problem 1:

Write a number using words: `95`

Solution:

`=95`

`5` is ones place

`9` is tens place.

`=90+5 =` ninety + five

So that `95` is become in a word, ninety five.

Example problem 2:

Write a number using words: `65`

Solution:

`=65`

`5` is ones place

`6` is tens place.

`= 60+ 5=` sixty + five

So that `65` is become in a word, sixty-five. I have recently faced lot of problem while learning free online math help live, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problems- Basic Concepts of Mathematics:

Problem 1:

Sum of positive integers:` (+7) + (+7)`

Result: `12`

Problem 2:

Sum of positive integers: `(+7) + (+1)`

Result: `8`

Problem 3:

Sum of positive integers: `(+7) + (+2)`

Result: `9`

Applied and Computational Mathematics

Introduction to applied and computational mathematics:
In this article we learn about applied and computational mathematics problems. In applied mathematics we solve more advanced math problems. Applied mathematics is advanced method of school level mathematics. Computational mathematics is involving research of some important math topics. Computational mathematics involves areas, volume, symbolic methods and numerical methods. Applied and computational mathematics problems are easily solved by using formulas. Now in this article we solve some applied mathematics problems. Having problem with Computational Formula for Standard Deviation keep reading my upcoming posts, i will try to help you.

Example Problems for Applied and Computational Mathematics

Find the first derivative of the given step function f(x) =`{ (9x + 5 if x gt1), (3x^4 + 4x if x lt 1):}`

Solution:

Given step function is f(x) =` { (9x + 5 if x gt1), (3x^4 + 4x if x lt 1):}`

From the given,

f1(x) = 9x + 5 if x > 1

And

f2(x) = 3x4 + 4x if x < 1

Differentiate the given functions f1(x) and f2(x) with respect to x, we get

f1'(x) = 9

And

f2'(x) = 12x3 + 4

Combining the two functions, we get

f'(x) = `{ (9 if x gt1), (12x^3 + 4 if x lt 1):}`

Answer:

The final answer is f'(x) = `{ (9 if x gt1), (12x^3 + 4 if x lt 1):}`

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Applied and Computational Mathematics Example Problem 2:

Find the double integral value of the given function   `int_0^3int_0^(14)`y dy dx

Solution:

Given function is  `int_0^3int_0^(14)`y dy dx

Here range of x is 3 to 0 and y range is 14 to 0

Integrate the function with respect to 'y'. So, we get

= `int_0^3int_0^(14)`y dy dx

Integrate the coordinates separately, we get

`int_0^(14)`y dy `int_0^3` dx =   `[y^2/2]` 0(14) *  `int_0^3` dx

= `((14^2/2) - 0) xx (x)_0^3`

Substitute the limits in the above equations, in this first substitute the upper limits and then substitute lower limits.

= `(98 - 0)` * `((3) - 0)`

After subtracting, we get

= 98 * 3

After simplifying the above step, we get

= 294

Answer:

The final answer is 294

Essentials of Discrete Mathematics

Definition of Discrete mathematics:

The essential of discrete mathematics is significant one in computer evaluation. It give details the algorithms and theory for computer enhancement.The essential of discrete mathematics is the study of exact structure that are basically discrete relatively than nonstop. In difference to real numbers that have the possessions of varying "effortlessly", the objects are deliberate in essential of discrete mathematics  such as integers, graphs, and statements in logic do not differ smoothly in this way, but have dissimilar, separated values.

Discrete Mathematics Topics

Theoretical computer science:

Calculating area for essential of discrete mathematics is incorporated in theoretical computer science. If we desire to study the result of mathematical computation, use this theoretical science. Model computer systems are utilizing the method algebra.

Information theory:

Quantification of information is anxious with information theory. In this, coding theory is used since it sketch the chart strongly and honest data transmission.

Logic:

Valid reason and assumption philosophy are measured by logic. Truth tables are use in normal logic.

Graph theory:

Graphs and networks are calculated by graph theory. It is used for observe the problem result.

Probability:

The actions are significant the countable samples and dense with discrete probability theory.

Algebra:

Boolean algebra along with relational algebra use the discrete examples. Topology, geometry, calculus, game theory and numeral theory also concern of  the essential of discrete mathematics.

Examples of Discrete Mathematics
Question: 1

Find n.

i) (n+2)!=12*11(n!)

Answer:

(n+2) (n+1)n!=12*11(n!)

(n+2)(n+1)=12*11

`n^2` +3n-130=0

(n-10)(n+13)=0

n=10 or n= -13

n=10(n cannot be negative)

ii) `1/(7!) +1/(8!) = n/(9!)`

`1/(7!) + 1/(8!) = n/(9.8(7!))`

Answer:

`1 + 1/8 = n/(9.8)`

`9/8 = n/(72)`

`n= (9 * 72)/8 = 81`

n =  81

Question: 2

A lady wants to select one cotton saree and one polyster saree from a collection of 8 cotton sarees and 11 polyster sarees. In how many ways can the lady choose the two sarees?

Answer:

Here, there are two events E1 and E2.
E1 = Selection of one cotton saree from 8 cotton sarees.
E2 = Selection of one polyster saree from 11 polyster sarees.
E1 = 8 ways E2 = 11 ways

Question 3

i)`20!+(19!)/(18!)` ii)`30!-(28!)/(26!)` iii) `8!+(6!)/(7!)`

Answer

i)                    `20!+(19!)/(18!)= 19!(20+1)/(18!)=19*(18!)*21/(18!)=19*21=399`
ii)                  `(30!-28!)/(26!)=28!((30*29)-1)/(26!)=28*27*869=656964`
iii)                `(8!+6!)/(7!)=6!(8*7+1)/(7!)=57/7`

Null Meaning

Introduction to Null Meaning

In mathematics, the word null meaning that zero or none. The zero elements in set or value of zero is said to be null set. This is symbolized as F. We are using this symbol for differentiating null from zero. We can also denote the null set as { }. In this article, we will see example problems of null meaning. Is this topic Correlation Coefficient Example hard for you? Watch out for my coming posts.

Example Problems - Null Meaning

Example 1:

Find the intersection of two sets P and Q, where P = {1,2,3,4} and Q = {5,6,7}.

Solution:

Given sets are, P = {1,2,3,4} and Q = {5,6,7}.

We need to find intersection of sets P and Q.

That is represented as PnQ.

PnQ = {The common elements in P and Q}.

Here, There is no common element in P and Q.

Thus, PnQ = { }

The answer is null set.

Example 2:

Consider a set, P = {x: x2 = 16 and 3x = 10} is a null set.

Solution:

Given, P = {x: x2 = 16 and 3x = 10}.

When, x = 4 => x2 = 42 = 16

But, 3x = 3(4) = 12.

Therefore, there is no numerical value that satisfies both conditions.

So, the set P is a null set. I have recently faced lot of problem while learning math help for college students, But thank to online resources of math which helped me to learn myself easily on net.

Example Problems - Null Meaning

Example 3:

Consider the set of rectangles that have five numbers of faces.

Solution:

We know that, the total sides of rectangle = 4.

So, there is no rectangle shape that having 5 sides.

Set of rectangles that have five numbers of faces = { }.

Thus, the given condition is null set.

Example 3:

Consider the set of triangles that have seven numbers of faces.

Solution:

We know that, the total sides of triangle = 3.

So, there is no triangle shape that having seven sides.

Then, Set of triangles that have seven numbers of faces = { }.

Thus, the given condition is null set.

That’s all about null meaning.

Sum of Two Perfect Squares

Introduction to Sum of two perfect squares:

In mathematics, a square number, sometimes also called a perfect square, is an integer that is the square of an integer; in other words, it is the product of some integer with itself. So, for example, 9 is a square number, since it can be written as 3 × 3. Square numbers are non-negative. Another way of saying that a (non-negative) number is a square number is that its square root is again an integer. For example, √9 = 3, so 9 is a square number.

(Source: Wikipedia)

Example for Sum of Two Perfect Squares:

Some of the examples for the sum of two perfect squares,

Example for sum of two perfect squares 1: `9, 100` find the sum of the perfect squares?

i) (9)

Solution:

`= 3^2`

`= 3 xx 3`

`= 9`

Therefore, 9 is a perfect square.

ii) 100 is a perfect square?

`= 10^2`

`= 10*10`

`= 100`

Therefore, `100 ` is a perfect square.

Therefore, sum of the perfect squares are,

`= 9 + 100`

`= 109`

Example for perfect square math 3: sqrt (64) is a perfect square?

`= 8 xx 8`

`= sqrt (64)`

Therefore, `sqrt (64)` is a perfect square.

Example for perfect square math 4: `(x + 6) ^2, (x + 5) ^2`

i) `(x +6)`

Solution: `= (x+6)2`

`= (x+6) (x+6)`

`= x(x+6) +6(x+6)`

`= x^2 + 6x + 6x + 64`

`= x^2 + 12x + 64`

There is another method also by using formula,

`= (x+6)^2 `

`= x^2 + 2 xx x xx 6 + 6^2 `

Therefore, `x^2+ 12x + 64 ` is a perfect square.

ii)` (x + 5)^2`

` = (x + 5) (x + 5)`

`= x(x + 5) + 5(x + 5)`

` = x^2 + 5x + 5x + 25`

` = x^2 + 10x + 25`

There is another method also by using formula,

`= (x + 5)^2`

`= x^2 + 2*x*5 + 5^2`

`= x^2 + 10x + 25`

Therefore, `x^2 + 10x + 25 ` is a perfect square.

Now, sum of those two perfect squares are taken,

`= (x^2+ 12x + 64)+ (x^2 + 10x + 25)`

` = 2x^2 + 22x + 99`

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Practice Problem for Perfect Square Math:

Practice problem for perfect square math 1: `38, 50`

Answer: `3944`

Practice problem for perfect square math 2: `(x+9) ^2, (x + 8) ^2`

Answer: `2x^2 + 34x + 145`

Statements

Introduction for Statements:

The statements are the group of words compiled together to give a complete meaning. The basic types of the sentences are the positive sentences and the negative sentences. In the mathematics the sentences that are doubtful are called as the ambiguous sentences. In the following article we will see in detail about the topic statements.

More about Statements:

Let us start with three sentences:

In 2005, the president of USA was a woman.

Two is greater than one.

There is no rain without clouds

The above sentences of 1st one is false, the 2nd one is true as well as the 3rd one is true. In mathematically, these above three sentences are called statements because these are unambiguous sentences.

In the department of mathematics, The recognized format of the sentences are called statements.  The recognized statements may be true or false and not both. Here, the representation of any statements must be recognized in mathematics.

Let us start with examples are,

Seven minus five is equal to two.

The multiplication of two negative numbers are positive.

All natural numbers are positive whole numbers.

The above sentences are true because all these sentences have no doubtfulness, then these above sentences are called statements.

An ambiguous sentences:

The doubtfulness of the sentences are labeled the ambiguous sentences.

Example:   "The multiplication of a and b is a positive number".

Here, we are not able to decide the above sentence is true or false, unless we know the value of a and b.

For example, the above sentence is true for a = -2 and b= -5 as well as true for a= 3 and b= 7, false for a= -7 and b= 5

Therefore, the above sentence is ambiguous, then it is not a statement.

To modify the above sentence,

“ The multiplication of same sign of numbers a and b is a positive number”, is a statement.

Examples on Statements:

There are 14 months in a year.
This sentence is false, because 12 months in a year. Therefore, this sentence is a statement.

The division of 10 by 2 is 5.
This sentence is true, because (10/2) = 5. Therefore, this sentence is a statement.

Today is a good Friday.
This sentence is doubtfulness, unless we know the particular date of Friday. Therefore, this is not a statement.

The square of a number is an odd number.
This sentence is doubtfulness, unless we know the specific number. Therefore, this is not a statement.

Exponential Equation Definition

Introduction to exponential equation definition:

In this article we are going to discuss about exponential equation.An equation can be defined as it is a mathematical statement which contains constants, variables,operation symbols and an equal sign (Or) it can be also defined as an expression with an equal sign.For example, x +3  = 6.Here ' x ' is a variable, ' 3,6 ' are constants, ' + ' is an operational symbol and ' = ' is an equal sign.The exponent can be defined as how many times the number is to be used in multiplication.And this is also called as power.

Exponential equation:

Exponential equation is an equation which involves exponential functions of a variable.

Here there are two forms of exponential equations.Those can be shown in below

1) In first form of exponential equation, we will get same base on each side of the equation.When it happens there is a rule that says,If the bases are same then the exponents are also same.

For example, a^x = a^y  then x = y and a > 0 and `a!=1` .Here the bases are same as ' a '.

2) In the second form of exponential equation, we will get different bases on both side of the equation.Thenn there are different types to solve the exponential equation.That can be shown in below.

Type 1:

If the exponent equation is 2^x = 64

Here the value can be written in the form of base 2 as

2^x = 26

Here the bases are same,then exponents are also same

x = 6

Type 2:

If the exponent equation is 3^x = 11

Here the value 11 is not able to write in the form of base 2. So here we have solve the exponent equation using logarithm.

log ( 3x ) = log ( 11 )

x log ( 3 ) = log ( 11 )

x = `(log(11))/(log(3))`

x = 2.813

Now we are going to discuss some problems regarding on to the exponential equation. Having problem with online algebra equation solver keep reading my upcoming posts, i will try to help you.

Example Problems on Exponential Functions:

Ex:1 Solve the exponential equation 7^2x - 5 = 77

Sol:

Given exponential equation is 7^2x - 5 = 77

Here the bases are same,so the exponents are also same then

2x - 5 = 7

Add 5 on both sides

2x - 5 + 5 = 7 + 5

2x = 12

Divide 2 on both sides

`(2x)/(2) = (12) / (2)`

x = 6

Answer: x = 6

Ex:2 Solve the exponent equation 9^2x = 6561

Sol:

Given exponent equation is 9^2x = 6561

Here the value 6561 can be written in the form of base 9 as 94, then

9^2x = 94

Here the bases are same,so exponents are also same,then

2x = 4

Divide 2 on both sides

`(2x)/(2) = 4/2`

x = 2

Answer: x = 2

Practice Problems on Exponential Equations:

1) Solve the exponential equation 24x = 7

Answer: x = 0.702

2) Solve the exponential equation 119x - 7 = 12110

Answer: x = 3

Perfect Difference Set

Introduction to perfect difference set:

Collections of an element are called as set. Elements of a set are also called as objects of a set or numbers of a set. Set subtraction is one of the operations of a set. Set subtraction is also called as difference set. If the result of difference set has only positive integers then they are called as perfect difference set.  Let us see perfect difference set in this article.

Perfect Difference Set:

Difference set:

Set subtraction is one of the other names of difference set. Let us consider ‘A’ and ‘B’ are given set. The set of all elements of A which is not present in set B is called as A – B. The elements of difference set are both negative integer and non - negative integer.

Example:

A = {-2, -1, 1, 2, 3, 4, 5}

B = {1, 2, 3}

A – B = {-2, -1, 4, 5}

Perfect difference set:

If the difference set have non-negative integers only – called as perfect difference set.

Example:

A = {1, 2, 3, 4, 5}

B = {1, 2, 3}

A – B = {4, 5}

The result of A – B is not containing the negative integers so the above difference set is a perfect difference set.

Problems for Perfect Difference Set:

Problem 1:

A = {1, 3, 4, 8, 12, 15, 16, 20, 24, 32, 44} B = {1, 4, 8, 15, 20, 22, 30, 41, 44}

Find A – B and B – A. Also state it is a proper difference set or not.

Solution:

Given

A = {1, 3, 4, 8, 12, 15, 16, 20, 24, 32, 44}

B = {1, 4, 8, 15, 20, 22, 30, 41, 44}

A – B = {3, 12, 16, 24, 32}

B – A = {22, 30, 41}

A – B and B – A have only non-negative integers so A – B and B – A are perfect difference set.

Problem 2:

A = {-2, -6, 16, 23, 25, 28, 29, 30} B = {-1, -3, -6, 16, 25, 28, 30}

Find A – B and B – A

Solution:

Given

A = {-2, -6, 16, 23, 25, 28, 29, 30}

B = {-1, -3, -6, 16, 25, 28, 30}

A – B = {-2, 23, 29}

B – A = {-1, -3}

A – B and B – A have negative integers also therefore A – B and B – A are not a perfect difference set.

Computer Solves Math Problems

Introduction to computer solves math problems:

Day-by-day, the technology is improving a lot. The computer solves the math problems in a simpler way. If we run various programs, the computer solves the math problems automatically. Apart from this, with the help of computers, the online tutor solves the math problems manually. In this article computer solves math problems, we are going to discuss few math problems.

Example Math Problems:

Example 1:

Find the perimeter of cube having the side length of 38 cm.

Solution:

Perimeter  =  12 `xx` a

=  12 `xx` 38

=  456 cm

Example 2:

Find the area of a regular hexagon given the side length a = 24 cm

Solution:

Area   =  `3 sqrt (3)/2` `xx`  a2 square units.

=  2.6 `xx`  242

=  2.6 `xx`  576

=  1497.6 cm2

Example 3:

Find the area of a triangle with base of 31 mm and a height of 22 mm.

Solution:

Area of a triangle  =  `1/2` `xx` b `xx` h

=  0.5  `xx` 31 `xx` 22

=  341 mm2

Example 4:

Find the volume of cone given the radius is 18 cm and height is 41 cm.

Solution:

Volume of cone  =  `1/3` `xx` `pi``xx` r2 `xx` h cubic units.

=  0.33 `xx` 3.14 `xx` 182 `xx` 41

=  0.33 `xx` 3.14 `xx` 324 `xx` 41

=  13764.88 cm3

Example 5:

Find the mode of the data set: { 13, 21, 65, 29, 29, 65, 66  }

Solution:

Step 1: The data set can be ordered as { 13, 21, 29, 29, 65, 65, 66 }

Step 2: In the data set, the values 29 and 65 occur two times.

Step 3: Hence, the modes are 29 and 65.

Math Problems to Practice:

1) Find the perimeter of cube having the side length of 30.5 cm.

Answer: Perimeter = 366 cm

2) Find the area of a regular hexagon given the side length a = 28 cm.

Answer: Area = 2038.4 cm2

3) Find the area of a triangle with base of 60 mm and a height of 13 mm.

Answer: Area = 390 cm2

4) Find the volume of cone given the radius is 13 cm and height is 17 cm.

Answer: Volume = 2977 cm3

5) Find the mode of the data set: { 105, 106, 360, 106, 366, 440  }

Answer: Mode = 106

Solve Squared Trigonometric Functions

Introduction to solve squared trigonometric functions:

There are the three basic trigonometric functions from which are the other three functions are derived. The square values of these functions are interrelated. All the trigonometric functions can be related to the squared functions of the other squared trigonometric functions. In the following article we will discuss in detail about the topic Solving Trigonometric Functions.

More about Solve Squared Trigonometric Functions.

The relations between the squared functions of the six trigonometric functions are,

`Si n^2 theta + cos^2 theta = 1`

`Cosec^2 theta - Cot^2 theta = 1`

`Sec^2 theta - Tan^2 theta = 1`

From these relations the values for the squared functions of the others can be derived as,

`Si n^2 theta = 1- Cos^2 theta`

`Cos^2 theta = 1- Si n^2 theta`

`Cosec^2 theta = 1+ Cot^2 theta`

`Cot^2 theta = Cosec^2 theta-1`

`Sec^2 theta = 1+ Tan^2 theta`

`Tan^2 theta = Sec^2 theta-1`

Also the basic relations between the squared trigonometric functions are,

`Si n^2 theta = 1/(cosec^2 theta)`

`Cos^2 theta = 1/(sec^2 theta)`

`Tan^2 theta = 1/(cot^2 theta)`

Using all these relations all the six trigonometric functions are related to each other.

Proof for the trigonometric relations:

In a right triangle let the opposite side = O, adjacent side = A, hypotenuse side = H. And `A^2 + O^2 = H^2` ,

`Si n theta = O /H`

`Cos theta = A/H`

`Tan theta = O /A`

`Si n^2 theta + cos^2 theta = O^2/H^2 + A^2/H^2 = (O^2+A^2)/H^2 = H^2/H^2 = 1`

`Cosec^2 theta - Cot^2 theta = H^2/O^2 - A^2/O^2 = (H^2 - A^2)/O^2 = O^2/O^2 = 1`

`Sec^2 theta - Tan^2 theta = H^2/A^2 - O^2/A^2 = (H^2 - O^2)/A^2 = A^2/A^2 = 1`


Between, if you have problem on these topics math formulas sheet, please browse expert math related websites for more help on Poisson Process.

Example Problem on Solve Squared Trigonometric Functions.
1. Solve `Tan^2 theta - Si n^2 theta` , using the squared trigonometric relations.

Solution:

`Tan^2 theta - Si n^2 theta`

`= (Si n^2 theta)/(Cos^2 theta) - Si n^2 theta`

`= Si n^2 theta (1/(Cos^2 theta) -1)`

`= Si n^2 theta (Sec^2 theta -1)`

`= Si n^2 theta (Tan^2 theta)`

Division with Remainders

Division in math is derived from the term divide; we must have heard when the term is used like divide the money between two brothers or so. What do we mean by that? It simple means that the money given has to be cut or break them into parts, not literally break or cut but to put them in small segments and bifurcate. Let us say I have 4 quarters, and we are just two of us to divide the quarters between us. So notice here there are 4 quarters and we are 2 of us and we need to divide the 4 quarters between us. So, 4 divided by 2. Are we clear on this statement? How 4, as total number of quarters are 4 . And how 2, as total number (number of people) of who will get that quarters are 2. Thus 2 will be at the denominators and 4 will be written on top as numerator. Like 4/2. So as we know 2 times 2 is equals to 4. So, 2 quarters each will be given. This is nothing but 50 cents each. Here the remainder is zero. I like to share this Division Tables with you all through my article.

Let us study more where remainders do come while solving any divison. So what do we mean by remainder? In other words, what is a remainder? Remainder means anything which is left over. In Arithmetic the remainder is the amount left after we have found out the quotient in the divison problem. For example if we divide 27by 5, 5 will go in 27 for 5 times with 2 left over thus here, 2 is the remainder. This was the division with remainders Word problems. Please express your views of this topic continuous random variable by commenting on blog.

Remainder math terms are expressed as r sometimes. Division word problems with remainders are also explained to understand the division term more precisely. So I have 25 candies and 6 bags to fill them in equally, so how do we do that its simple, we divide 25 by 6 ,which is 6 goes 4 times to get 24 and one is left over. So here I mean 4 candies go in each bag and one is left over which is termed as remainder. So we checked it how easy it is to divide or do division with remainders.

We can also do the same by using long division method but it will take some time to solve whereas when we know by the multiplication table, of 6 that it goes 4 times, the calculation becomes easier and saves lots of time.