List of Whole Numbers

Introduction for Whole numbers List:

A number system which consist of all types of numbers it can be subdivided into Natural number,whole number,rational number and decimal numbers. number ‘0’ with the natural number gives us the whole number list . 0, 1, 2………smallest whole number is 0 and largest whole is unpredicted but no fractions, no negative and decimal parts in whole numbers. Generally whole number has no proper definition it's start with zero and it has no ends all numbers is whole numbers. Whole numbers list is a set of natural numbers and its represented by W.

W= 0, 1, 2, 3, 4……….

Listing Whole Numbers in Words:

We can have Whole number system which are tabulated Below it represented the numbers and it shows how can we expand and write the values for numbers.All numbers are whole numbers and in large number case we have to separate the numbers with its position left to right for find its value


Any whole number multiplied by zero is zero.

Division by zero is not allowable operation in whole numbers.

The whole numbers for multiples of 2 are said to be even numbers

Example 1:

Write a whole number 54 in words.

Solution:

54 is the number fifty-four.

Example 2:

Write a whole number 225 in words.

Solution:

225 is the number two hundred and twenty five.

Example 3:

Write whole number 7 128 659 in words.

Solution:

7128 659 is the number seven  million, one hundred and twenty-eight thousand, six hundred and fifty-nine.
I like to share this difference between permutation and combination with you all through my article.
Fundamental Operations with Whole Number List:

Addition with Whole numbers:

Example 1:

Add Two whole numbers 22 and 45

Solution:

22

+ 45

67

Example 2:

Add 125 and 221

Solution:

125

+ 221

346

Subtraction  with whole numbers:

Example 1:

Subtract 12 from 46

solution:

46

- 12

34

Example 2:

Subtract 540 from 1020

Solution:

1020

-  540

480

Multiplication with whole numbers:

Example 1:

multiply 15 with 4

Solution:

15

`xx` 4

30

Example 2:

multiply 225 with 2

solution:

225

`xx `  2

550

Division with whole numbers:

Example 1:

Divide 15 by 3

Solution:

5

3 )15

15

0                     Answer:5

Example 2:

Divide 120 by 10

Solution:

12

10 )120

10

20

20

0

Answer:12

Natural Log Laws

Introduction to natural log laws:

The natural logarithms are used for many features of real life, for example zooming, mirroring, rotating images, etc. The natural logarithms are defined by the exponent of the power to which a base number must be raised to equal a given number. The solving natural logarithms are also called the inverse of an exponents. In logarithm, it has two types of logarithms are the common logarithm, also called the base ten logarithms, and the natural logarithm, also called the base e logarithm. Let us see about natural logarithmic law in this article
The natural logarithms are written:

logex or ln x

Where e = 2.71828182846 (base of natural logarithm).

List of Natural Log Laws:

The following laws help in solving natural logarithms : [here, ln x = logex]

Product law:

` ln (x * y) = (ln x) + (ln y)`     `or`      `log_e (x * y) = (log_e x) + (log_e y)`

Quotient law:

`ln (x / y) = (ln x) - (ln y)`         `or`   `log_e (x / y) = (log_e x) - (log_e y)`

Power law:

`ln (x^n) = (n) ln x`        `or`        `log_e (x^n) = (n)log_e x`         `Where` `^nsqrt(x) = x 1/n`

Reciprocal law:

`ln x = 1 / (log_xe)`    `or`      `log_e x = 1 / (log_xe)`            `where` `x` ` represents` ` any` `Numbers.`

Examples for Solving Natural Logarithms by Using the Laws:

Example 1:

Rewrite the following natural logarithms by using the required laws.

`ln` `(a xx b^3 xx c)/(d^2)` .

Solution

Step 1:       `ln` `(a xx b^3 xx c)/(d^2)` .  [Given]

Step 2:      `ln` `(a xx b^3 xx c) - ln(d^2)` .               [Quotient laws]

Step 3:      `ln` `(a) + ln b^3 + ln c - ln(d^2)` .     [Product laws]

Step 4:       `ln` `(a) + (3)ln (b) + ln (c) - 2ln(d)` .    [Power laws]

This is the required rewritten natural logarithms.

Example 2

Rewrite the following natural logarithms by using the required laws.

`ln` `(x^2 xx sqrt(y))/(z^4)` .

Solution:

Step 1:      `ln` `(x^2 xx sqrt(y))/(z^4)` .  [Given]

Step 2:     `ln` `(x^2 xx sqrt(y)) - ln(z^4)` .                [Quotient laws]

Step 3:     `ln` `(x^2) + ln(sqrt(y)) - ln(z^4)` .        [Product laws]

Step 4:     `ln` `(x^2) + ln(y^(1/2)) - ln(z^4)` .        [Radical laws]

Step 5:     `(2)ln` `(x) + (1/2)ln(y) - (4)ln(z)` .  [Power laws]

This is the required rewritten natural logarithms.

Example 3: Solve the following natural logarithms by using the required natural log laws.

`ln 5^(7x-3) = 4` .

Solution:

Step 1: Given   `ln 5^(7x-3) = 4`

Step 2:          `(7x -3)ln 5 = 4` .                [by using the power laws]

Step 3:           `(7x -3) = 4/(ln 5)` .                   [by using the cross multiplications laws]

Step 4:               Addition of  2 on both sides we get.

`(7x -3) +3 = 4/(ln 5) + 3` .

Step 5:            `(7x) = (4 + 3ln(5))/(ln 5) ` .                   [By taking LCM]

Step 6:                    Now we are going to divide 7 on both sides we get.

`(7x)/7 = (4 + 3ln(5))/(7ln 5) ` .

Step 7:          `(x) = (4 + 3ln(5))/(7ln 5) ` .

This is the required solved natural logarithms.

Quadratic Function Roots

Introduction to quadratic function:

A quadratic function, in mathematics, is a polynomial function of the form



The graph of a quadratic function is a parabola whose major axis is parallel to the y-axis.

The quadratic function has the highest degree of 2.

If the quadratic function ax2 + bx + c = 0, then it becomes quadratic equation.

Roots of Quadratic Functions:

If  the coefficients of quadratic function a, b and c are real and complex, then the roots of quadratic function will be

x = `(-b+-sqrt(b^2 - 4ac))/(2a)`

The above formula is called quadratic formula.

The root of quadratic functions are different real numbers, if b2 - 4ac > 0. i.e., The function has two real roots.
The root of quadratic functions are equal real numbers, if b2 - 4ac = 0. i.e., The function has one real root.
The root of quadratic functions are imaginary numbers, if b2 - 4ac < 0. i.e., The function has no real roots.
The expression b2 - 4ac is called the discriminant of a quadratic function.Understanding what is rotational symmetry is always challenging for me but thanks to all math help websites to help me out.

Example Problems to Learn Roots of Quadratic Function:

Example 1:

Find the root of quadratic function f(x) = x2 + 7x - 30.

Solution:

Step 1: Given quadratic function

f(x) = x2 + 7x - 30

Step 2: Rewrite the term ' 7x ', as ' 10x - 3x ', we get

f(x) = x2 + 10x - 3x - 30

Step 3: Take the common terms outside

f(x) = x(x + 10) - 3(x + 10)

f(x) = (x - 3)(x + 10)

Step4: Equate each function to zero to find roots

x - 3 = 0                                                             x + 10 = 0

Add 3 on both side, we get                                       Subtract 10 on both side, we get

x = 3                                                                     x = - 10

Step 5: Solution

The roots of given quadratic functions are 3 and - 10

Example 2:

Find the root of quadratic function f(x) = x2 + 16x + 63.

Solution:

Step 1: Given quadratic function

f(x) = x2 + 16x + 63

Step 2: Rewrite the term ' 16x ', as ' 9x + 7x ', we get

f(x) = x2 + 9x + 7x + 63

Step 3: Take the common terms outside

f(x) = x(x + 9) + 7(x +9)

f(x) = (x + 9)(x + 7)

Step4: Equate each function to zero to find roots

x + 9 = 0                                                             x + 7 = 0

Subtract 9 on both side, we get                                       Subtract 7 on both side, we get

x = - 9                                                                     x = - 7

Step 5: Solution

The roots of given quadratic functions are - 9 and - 7.

Factor Pairs Definition

Introduction to factor pairs definition:

Factorization is a method of splitting up a given numbers into prime or complex numbers. In this factor pair is a pair of numbers when multiply the factors it will give a original number. For example the factor pairs of 20 are 1 * 20 = 20, 2 * 10 = 20, 4 * 5 = 20. The pairs are (1, 20), (2, 10) and (4, 5).

Steps to Find the Factor Pairs:

Step 1: Write down the given number.

Step 2: Split up the given number into complex or prime numbers up to maximum level.

Step 3: Write down the factors in the form of ordered pair. Those ordered pairs are factor pairs.

Example Problems – Factor Pairs Definition:

Example 1 – Factor pairs definition:

Figure out the factor pairs for the number 40.

Solution:

The given number is 40

40 can be rewritten as

1 `*` 40 = 40

2 `*` 20 = 40

4 `*` 10 = 40

5 `*` 8  =  40

These are the factors for the given numbers. To write it in factor pairs the factors should write in the form of ordered pair.

The factor pairs are (1, 40), (2, 20), ( 4, 10) and (5, 8).

Example 2 – Factor pairs definition:

Figure out the factor pairs for the number 55.

Solution:

The given number is 55

55 can be rewritten as

1 `*` 55 = 55

5 `*` 11  =  55

These are the factors for the given numbers. To write it in factor pairs the factors should write in the form of ordered pair.

The factor pairs are (1, 55) and (5, 11).

Example 3 – Factor pairs definition:

Figure out the factor pairs for the number 60.

Solution:

The given number is 60

60 can be rewritten as

1 `*` 60 = 60

2 `*` 30 = 60

3 `*` 20 = 60

4 `*` 15 = 60

5 `*` 12 = 60

These are the factors for the given numbers. To write it in factor pairs the factors should write in the form of ordered pair.

Understanding how to do standard deviation is always challenging for me but thanks to all math help websites to help me out.

The factor pairs are (1, 60), (2, 30), (3, 20), (4, 15) and (5, 12).

Example 4 – Factor pairs definition:

Figure out the factor pairs for the number 24.

Solution:

The given number is 24

40 can be rewritten as

1 `*` 24 = 24

2 `*` 12 = 24

3 `*` 8 =   24

4 `*` 6  =  24

These are the factors for the given numbers. To write it in factor pairs the factors should write in the form of ordered pair.

The factor pairs are (1, 24), (2, 12), (3, 8) and (4, 6).

Practice Weighted Average

Introduction for practice weighted average:

The weighted average is similar to an arithmetic average, where instead of each of the data points contributing equally to the final average, some data points contribute more than others. The notion of weighted average plays a role in descriptive statistics and also occurs in a more general form in several other areas of mathematics.

(Source wikipedia)

Example Problems for Practice Weighted Average:
Example 1:

A pack contains 70 balloons. 40 balloons are sold for $2 and another 30 balloons are sold for $1. Calculate weighted average for all balloons.

Solution:

The sum of the average is (40 * 2) + (30 * 1) = 80 + 30 = 110

Weighted average = `"The sum of the average" / "Total number of balloons"`

= `110 / 70`

= 1.57

Example 2:

A box contains 80 cakes. 50 cakes are soled for $4 and another 30 cakes are sold for $3. Calculate weighted average for all cakes.

Solution:

The sum of the average is (50 * 4) + (30 * 3) = 200 + 90 = 290

Weighted average = `"The sum of the average" / "Total number of cakes"`

= `290 / 80`

= 3.625

Example3:

A class room contains 100 students, the average weight of the 75 students is 68 and another 25 students average weight is 88. calculate weighted average weight of all students.Is this topic Types of Numbers hard for you? Watch out for my coming posts.

Solution:

The sum of the average is (75 * 68) + (25 * 88) = 5100 + 2200 = 7300

Weighted average = `"The sum of the average" / "Total number of students"`

= `7300 / 100`

= 73

Practice Problems for Weighted Average:

Practice Problem 1:

A class contains 100 students. 60 students average score is 85 and other 40 students average score is 78. Calculate the weighted average for the entire class.

Answer: Weighted average = 82.2

Practice Problem 2:

A class contains 75 students. 30 students average score is 80 and other 45 students average score is 75. Calculate the weighted average for the entire class.

Answer: Weighted average = 77

Practice Problem 3:

A class contains 80 students. 50 students average height is 150cm and another 30 students average height is 140cm. calculate the weighted average for the entire class.

Answer: Weighted average = 146.25

Limits of Logarithmic Functions

Introductions to Limits of Logarithmic Functions :
In mathematics, Logarithm is abbreviations of log functions. The logarithmic function contains three parts such as number, base, the logarithmic itself. This limits of logarithmic functions to discover logarithmic functions and their properties, such as domain, range, x and y intercepts and vertical asymptote. This function is called as the limits of logarithmic function. In this article we shall discuss about limits of logarithmic functions.

Definitions:

The logarithmic functions is called as inverse of exponential functions.

bN = A   is equivalent to N=logbA

Where ,        N = number functions,

A = logarithmic functions,

b = Base functions.

Properties for Limits of Logarithmic Functions:
1. Product functions:

`Log_x AB = log_xA+ log_xB`

2. Quotient functions   

`log_x(A/B)= log_xA - log_xB`

3. Power functions           

.`log_xA^B = B log_xA`

4.`LogBA xx logAB=1`

5.`Log_10(1)=0`

6. `Log_BB=1`

7.  `A ^( log_xA)` =x

Limits of Exponential Functions:

`lim_(x->0^+)`    loga x = `-oo` .    if a > 1

`lim_(x->0^+)`  loga x =  `oo`          if a < 1

`lim_(x->oo)` loga x=  `oo`           if a > 1

Having problem with functions and linear equations and inequalities keep reading my upcoming posts, i will try to help you.

Limits of Logarithmic Functions - Problems:

Limits of logarithmic functions - problem 1:

Solve the given limit function `lim_(x->0)` `log(1 + x^3)/cosx` .

Solution:

Given limit function is  `lim_(x->0)` `log(1 + x^3)/cosx` .

= `lim_(x->0)` `log(1 + x^3)/cosx` .

Apply the limit values in the above logarithmic function,   So, we get

=  `log(1 + 0^3)/cos0`.

we know the value of    03   and   cos 0

03   =  0    and   cos 0 = 1

So,                            =  `log(1 + 0)/1`.

= log 1

=  1 .

Answer:   1 .                              

Limits of logarithmic functions - problem 2:     

Solve the given limit function `lim_(x->0)` `log(5 - x^5)cosx` .

Solution:

Given limit function is  `lim_(x->0)` `log(5 - x^5)cosx` .

= `lim_(x->0)` `log(5 - x^5)cosx` .

Apply the limit values in the above logarithmic function,   So, we get

=  `log(5 + 0^5)cos0`.

we know the value of    05   and   cos 0

05   =  0    and   cos 0 = 1

So,                            =  `log(5 + 0)1`.

= log 5

Answer:   0.6989 .                             

Parametric Equations Calculus

Introduction for parametric equations calculus :

In mathematics, parametric equation is a method of defining a relation using parameters. A simple kinematical examples are  when one use a time parameter to determine the position, velocity, and other information about a body in motion. Abstractly, a Parametric Equations define a relation as a set of equation. It is therefore somewhat more accurates defined as a parametric representation. It is part of regular parametric equation representation.Calculus includes that differential calculus and integral calculus is used.(Source.Wikipedia)



Examples for Parametric Equations Calculus:

Example 1 : Prove that the sum of the intercept on the co-ordinate axes of any tangent to the curve x = d cos^4c,  y = d sin4c,   0 = ? =p /2   is equal to d.

Solution :

Take any point ‘C’ as (d cos^4c, d sin^4c, )

Now  `dx/dd ` = – 4d cos^3c sin c ;

And  `dx/dd` = 4d sin3c cos c

? `dy/dx` = –sin^2c/cos2c

Slope of the tangent at ‘c’ is = –sin^2c/cos2c

Equation of the tangent at ‘c’ is (y - d sin4c) =- sin^2c/cos2c(x - d cos^4c)

or x sin^2 c + y cos2 c = d sin^2 c cos2 c

?`x/d` cos2c+`y/d ` sin^2c= 1

sum of the intercepts = x cos2 c + y sin^2 c = d

Example 2 : Find the equations of the tangent and normal at B =p/2 to the curve x = b (B + sin B), y = b (1 + cos B).
Solution : We have

`dx/dB` = b (1 + cosB) = 2b cos2 B

2`dy/dB` = – b sin B = – 2b sin`B/2 ` cos`B/2`

Then dy/dx =`dy/dA` / `dx/dA` =  – tan`B/2`

Slope m = `dy/dx` [B = p/2] = – tanp/4 = –1

Also for B =p/2 , the point on the curve is{ b p/2 + b, b .}

Hence the equation of the tangent at B =p/2 is

y – b = (–1) [x – a(p/2 + 1)]

x + y =`1/2` b p + 2b or x + y –`1/2` b p – 2b = 0

Equation of the normal at this point is

y – b = (1) x – b(p/2+ 1)

x – y –`1/2` b p = 0

Practice Problem for Calculus Parametric Equations:

Find the equations of the tangents and normal to the ellipse x = c cosS, y = d sin S at the point S =p/4

Answer: (cx – dy) `sqrt(2)` – (c2 – d2) = 0.