Greatest Monomial Factor

Let us make a review on finding gcf of any two numbers before we start learning how to find gcf of monomials.

Introduction to GCF of monomials:

Greatest common factor is abbreviated as GCF. To find gcf of any two numbers, we have to write the prime factors of the numbers and pick out the common terms and find the product.

Ex 1: Consider 4 and 8.

Prime factors of 4 = 2 x 2

Prime factors of 8 = 2 x 2 x 2

Common factors are 2 and 2

The product of common factors are 2 x 2 = 4

GCF of 4 and 8 is 4

Monomial is an mathematical expression that has only one term consisting of constants and variables raised to  any exponents connected by mathematical operation multiplication or division.

Ex: 3x, 2xyz, 6x2y3z4

How to Find the Gcf of Given Monomials:

Consider any two monomials, say 3a3, 9a5

Step 1: Write the factors of the two monomials.

3a3 = 3 x a x a x a

9a5 = 3 x 3 x a x a x a x a x a

Step 2: Pick out the common factors

Common factors are 3, a, a, a

Step 3: Find their product

Product is 3a3

Step 4: Write the product as gcf

GCF of the given monomials is 3a3

By regular practice, one could say the gcf of monomial by seeing the monomials.

Consider x2, x3

The term with lowest power is the gcf

So, x2 is the gcf

Consider a2bc, ab2c, abc2

There are three monomials. Each monomial has the same variables a, b, c

Pick up the variables that has least power in the given monomials

It is a, b and c

So, GCF is abc.

With the above ideas, let us do some problems on GCF .Having problem with what is standard form keep reading my upcoming posts, i will try to help you.

Problems to Find the Gcf of a Monomial:

Ex 1: Find gcf of p8, p9, p7

Sol:

Step 1: Check whether all the terms has same variables

Step 2: Pick out the term has that least power.

It is p7

Step 3: Write the solution.

gcf is p7

Ex 2: Find gcf of the monomials 14m2n, 28mn2, 21m3n3

Sol:

Step 1: Find the gcf of the three numbers.

14 = 2 x 7

28 = 4 x 7

21 = 3 x 7

Greatest common factor is 7

Step 2: The monomials has the same variables m and n.

Step 3: Pick out the variables that hs least power

It is m and n

Step 4: Gcf is 7mn

Method for Solving Angles

Introduction:

Method for solving angles is the important chapter in geometry.  There is more number of properties in geometry. With the help of the properties of the geometry we can solve the angles. Here we have to discuss about complementary angles and supplementary angles and triangle properties with solved example problems. Understanding Scalene Triangle Formulas is always challenging for me but thanks to all math help websites to help me out.

Important Property Methods in Geometry

They are

Method 1: The sum of angles in the triangle is equal to 180 degree.

Method 2: The sum of the complementary angles is equal to 90 degree.

Method 3: The sum of the supplementary angles is equal to 180 degree.

Solving Problems Based on the above Methods
Example 1:

If one of two angles in the triangle have 75 degree and 65 degree. Solve the another angle.

Solution:

We know that the method has

The sum of angles in the triangle is equal to 180 degree.

Here the two triangle measures 75 degree and 65 degree.

Consider another angle be x

Therefore x+ 75 + 65 =180

Adding this we can get,

X+140 =180

Subtracting 140 on both sides we can get

X = 40 degree.

Therefore another angle of the triangle is 40 degree.

Example 2:

If one of the complementary angles is equal to 48 degree. Solve the measure of another complementary angle.

Solution:

We know that the method has

The sum of the complementary angles is equal to 90 degree.

We can consider another angle be x.

Therefore x + 48 = 90

Subtracting 48 on both sides we have to get

X= 42 degree

Therefore 42 is the complement angle of 48.

Example 3:

If one of the supplementary angles is equal to 148 degree. Solve the measure of another supplementary angle. Is this topic Types of Angles hard for you? Watch out for my coming posts.

Solution:

We know that the method has

The sum of the supplementary  angles is equal to 180 degree.

We can consider another angle be x.

Therefore x + 148 = 180

Subtracting 148 on both sides we have to get

X= 32 degree

Therefore 32 is the supplementary angle of 148.

Generalized Permutations and Combinations

Introduction to generalized permutations and combinations:
 
we have 2 formulas  ncr   and  npr
If there are 'n' things, out of which 'r' things are to be selected, the number of ways in which 'r' things out of 'n' things are selected is denoted by nCr. It is called Combination.
But if the selected 'r' things are arranged in a definite order, then we call it npr.  It is called Permutation.
Generalized permutation and combinations are done using the  nCr and  nPr   formulas
Generalized Permutations and Combinations Problems for n things not all  different has the following formula.
Out of 'n' things, if 'p' things are of one kind, 'q' things are of another kind and 'r' things are of a third kind, then
the generalized formula for this permutation is N .p!.q!.r! =  n!    where N = number of permutations
The generalized formula  for this permutation becomes  N =  n!
-------
p!q!r!

Generalized Permutations and Combinations-permutation:-

The number of permutation of n things taken r at a time =npr= n(n-1)(n-2) .............. (n-r+1)
If r = n , then npr = npn
Then we get  npn= n(n-1) (n-2) ............ 3.2.1
This continued product is  denoted by a factorial  n!  Therefore npn= n!    ! sign is called factorial
T  he formula for npr =  `|__` n        L is also called  factorial
------
`|__` n-r
Let us do a couple of problems on permutations.
1) Find 10p4
Answer               10p4=  `|__` 10              `|__` 10         10x9x8x7x6x5x4x3x2x1           10x9x8x7 =  5040
-------  =    ------------ =  -------------------------------  =
`|__` 10-4           `|__` 6                   6x5x4x3x2x1
2) How many 6  digit numbers can be formed using the digits 4,5,6,7,8,8, no digit being repeated in each number
Answer   the permutatio is 6p6  =  6! = 6x5x4x3x2x1 =  720 numbers can be formed
3)  In how many ways  can 3 white balls, 4 red balls and 5 blue balls can be arranged in a row  so as to keep  all the
balls of the same color together?
Answer:-Now we consider the 4 red balls as a unit because they must be together.  Similarly consider the  5 blue balls
as a unit and the  3 white balls as a unit.
Now we get 3 units which can be arranged in  3p3 ways =  3x2x1= 6 ways.
Next  4 red balls can be arranged among themselves in 4p4 ways = 4! = 4x3x2x1= 24 ways
Next  5 blue balls can be arranged in 5p5 ways = 5! = 5x4x3x2x1 =  120 ways
Next  3 white balls can be arranged in 3p3 =  6 ways
The require no of arrangements =  6 x24x 120 x6 = 103,680 ways

Generalized Permutations and Combinations-combinations:-

The formula for nCr =  n!
-------
(n-r)! r!
It can also be written as nCr =  nPr
-------
r!
Let us do a few problems on combinations.
Problem 4     Find 10C4
10C4 =  10!                   10x9x8x7x6x5x4x3x2x1              10x9x8x7
---------             ---------------------------------     =     ------------- =  210
(10-4)! 4!             [ 6x5x4x3x2x1][ 4x3x2x1]            4x3x2x1
Problem 5     A committee of  3 teachers and 2 students  is to be formed from 5 teachers and 10 students.  In how many ways
can this be done?
Teachers     n= 5  and r = 3   so nCr  =  5C3 =  5x4x3
-------   =     10 combinations
1x2x3
Students      n=10        r = 2    so  nCr = 10C2=    10 x9
-------  =   45 combinations
1x2
Total combinations  =  10 x 45 = 450 combinations

Circular Permutation

Circular permutation is an arrangement in which the things are arranged in a circle.
let us arrange  a,b,c,d  in a circular arrangement.
We can arrange them in 4! ways = 24 ways
But when we arrange them in a circle we get 4 circular arrangements .  Hence 24 ÷ 4 = 6 = 3.2.1 = 3!
Thus the generalized  permutation formula for circular  permutation  is   n!
--  =   (n-1)!
n
This circular permutation formula gets modified in case of beads and necklaces.  since there is no difference
in clockwise or anticlockwise arrangement in the case of beads or necklaces of circular permutation, the
formula becomes  nPr  =    (n-1)!
-------
2
Find the number of ways in which 5 beads can be strund in a ring
This is circular permutation of beads.  Hence the formula  is
Number of permutations = (5-1)!         4!        24
-------   =    ----   =  ---   =  12
2             2         2
But when we do  a problem like find the number of ways in which 5 people can be arranged at a round table, I have recently faced lot of problem while learning how to graph an equation, But thank to online resources of math which helped me to learn myself easily on net.
we must remember that the 5 people are not alike then  the number of permutation= (5-1)! = 4! = 24

Generalized Permutation and Combination Problems:-

Let us give the student a few exercise problems
1.  Find 7C2  and 7C5  What do you infer from this?                                   Answer : 7C2 = 21   7C5 = 21Both are  equal
2.  Find 8C3                                                                                                                         Answer : 56
3.  Find the number of diagonals in a decagon                                          Answer ; 35
4.  Find the number of triangles formed by the vertices of a hexagon           Answer : 20
5.  A man has 3 friends.  In how many ways can he invite them to a  dinnerAnswer :   1
6.  Find  5P3                                                                                          Answer : 60
7.  If nP2 = 30   find n                                                                              Answer: n=6
8.  In how many ways can  6 different beads can be strung into a necklace Answer:  60

Standard Deviation of the Mean Calculator

Introduction to standard deviation mean calculator

The standard deviation, also called the residual standard error, of a statistical population, a data set, or a probability distribution is the square root of its variance. Standard deviation is a widely used to measure of the variability or dispersion, being algebraically more tractable though practically less robust than the expected deviation or average absolute deviation..

There are other statistical measures that can use samples that some people confuse with averages - including 'median' and 'mode'. Other simple form of statistical analyses use measures of spread, such as range, inter quartile range, or standard deviation.

Finding Mean Deviation through Calculator:

The followings are the steps to be followed in the mean deviation calculator.

Step 1:

To find the arithmetic mean.

Sum of the given values
Mean =    ----------------------------------------
Total number of values
Step 2:

To find the deviation.

Deviation = mean – given values.

Step 3:

To find the absolute deviation.

Step 4:

To find the sum of the absolute deviation.

Step 5:

To find the mean deviation.

Sum of absolute deviation
=      -----------------------------------------
Total number

Problem on Standard Deviation and Mean

Calculate the sample mean and standard deviation for the given data set.

435 , 235 ,543 , 435, 230

Solution:

Mean: Calculate the sample mean for that the  given set of data.

`sum` (x)
x¯    =____________
n


435 + 235 + 543 + 435 + 230
= __________________________
5


=  1878/5

=375.6

Calculate the sample mean and the standard deviation by the formula.

`sqrt(sum ( x - x))`
s = _________________________
n - 1



`sqrt((435 - 375.6)2 +( 235 - 375.6)2 + (543 - 375.6)2 +( 435 - 375.6)2 + ( 230 - 375.6)2)`

s=     ___________________________________________________________________________________
5 - 1




`sqrt(76044.2.)`
s = ___________
4


s =  `sqrt(19011.05)`

s =   137.8805

The required deviation calcu;ator is 137.8805



Understanding how to make a histogram is always challenging for me but thanks to all math help websites to help me out.

Practice Problem in Standard Deviation Mean Calculator.

Q1:Here the  given   capacity are  44, 45, 44, 48, 47 and 47 and Find the Mean, Median, Mode, Variance, Standard Deviation, Standard Deviation Standard Error.

Mean:    = 46

Standard Deviation,S = 1.67332005

Significant Figures Rounding Rules

Introduction to significant figures rounding rules:

The significant figures are normally those digits in a measured quantity which is known reliably or the ones we are confident of in our measurement plus one additional digit that is uncertain. The number of significant figures in a measurement is directly proportional to the measurement's accuracy. Suppose the time period of the simple pendulum is 1.62 seconds. This digit 1 and 6 are reliable and certain, while the digit 2 is uncertain. So, the time period has significant figures that are three in total. Again, the length of the object measured as 273.6 cm. It has four significant figures. The digits 2, 7 and 3 are certain while the digit 6 is uncertain. Here we discuss the rules for the rounding of the significant figures in any particular measurement.

Significant Figures Rounding Rules:

The significant figures rounding rules are as follows:

(i) If the digit which would be dropped is lesser than 5, in that case, the preceding digit will be unchanged.

(ii) If the digit which would be dropped is bigger than 5, which would be the preceding digit will be increased by 1.

(iii) If the digit which would be dropped is 5 followed by the non zero digits, in that case, the preceding digit will be increased by 1.

(iv) If the digit which would be dropped is 5, in that case, the preceding digit is left unchanged if it is even.

(v) If the digit which would be dropped is 5, in that case, the preceding digit is increased by 1 if it is odd.

Understanding how to find the least common multiple is always challenging for me but thanks to all math help websites to help me out.

Examples for the Significant Figures Rounding Rules:

Round off the following numbers as indicated:

(a) 18.35 upto 3 digits

(b) 143.45 upto 4 digits

(c) 18967 upto 3 digits

Solution

(a) Here the third digit is 3 (odd number) and the next digit is 5 so that the digit 3 is increased by 1. Hence, the rounded figure is 18.4

(b) Here the fourth digit is 4 (even number) and the next digit is 5 so that the digit 4 is remained same. Hence, the rounded figure is 143.4.

(c) Here the third digit is 9 and the next digit is 6 which is more than 5 so that the preceding digit is increased by 1. Hence, the rounded figure is 19000.

Different Properties in Math

Introduction to different properties in math:

In this article we are going to see about the different properties in math. There are different properties in math that are defined in earlier in the field of math. The properties in math are applied for all the branches of mathematics. Among the different properties of math only three properties are used in common they are associative, distributive and commutative property.I like to share this Commutative Property of Addition Definition with you all through my article.

Different Math Properties

Different properties:

The different properties in math are,

Property 1: Associative property.
Property 2: Property of commutative.
Property 3: Property of distributive.
Property 4: Reflexive property.
Property 5: Transitive property.
Property 6: Property of addition.
Property 7: Multiplication property.
Property 8: Identity property
Property 9: Substitution property.

Explanation of Different Properties

Associative property:

The associative property is considered as the property in which the change in the parentheses of of the operation does not affect the result of the operation. This property is common for both the addition and subtraction.

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

(x * y) * z = x * (y * z)

Commutative property:

In this property when we change the order of the operation is changed it does not affect eh result. The commutative property is given as,

x + y = y + x

x * y = y * x

Distributive property:

In this property the process is distributive over the parentheses. The operation is to add or multiply the numbers.

x * (y + z) = x * y + x * z

Identity property:

This identity property may be additive identity or the multiplicative identity. The additive identity is the property in which the zero is added to the number gives the same number as the answer. The additive identity property is given as,

x + 0 = x = 0 + x

In multiplicative identity property we take a number and multiply the number with 1 we get the same number as the answer.

x * 1 = x = 1 * x

Addition property:

When the same non-zero number is added in both the sides then this is referred as the addition property. This addition property is given as,

x + z = y + z

Here z is considered as the non-zero number.

Multiplication property:

Here when the same number which is not equal to zero is multiplied on both sides then it is considered as the multiplication property.I have recently faced lot of problem while learning Calculus Solver, But thank to online resources of math which helped me to learn myself easily on net.

x * z = y * z

Here z is considered as the non-zero number.

Reflexive property:

In this property a number x is equal to its identical number. The reflexive property is given as,

x = x

Transitive property:

When the two numbers are equal to the third number then it is said as transitive. The transitive property is given as,

x = y and z = y hence x = z

Substitution property:

In this property we replace a statement with the other statement which is considered as equivalent.

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.

Solving Equations with Radicals and Exponents

Solving equations with radicals and exponents:

Exponents:      

The term exponent in math is used to find the exponential value of the particular value it may be integer  or fraction . We can easy to get  the power of a  number using online calculator ,Consider  the unknown number , Here x is base and y is the power  of x . Let us consider the  number 23 Here 2 is the base(x)  and 3 is the power of 2(y). We calculate  23= 2 x 2 x 2 =8

Radicals:

Square root of the number is said to be a radical number .A radical equation is an equation in which a variable is under a radical term.Let us discuss about the solving an equations with radical and  exponents,

Example Problems to Solving Equations with Radicals and Exponents:

Example 1: Solve `sqrt(3x^2 +8x)` -3 =0

Solution:

Given `sqrt(3x^2+8x)` -3=0

isolate the radical equation,

`sqrt(3x^2 +8x)=3`

Take square on both sides we get,

`(sqrt3x^2+8x)^2` `=3^2`

3x^2+8x=9

3x^2+8x-9 =0

now you can solve the above equation using factoring method,

Using quadratic formula ,

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

Here b=8 ,a=3 and c=-9

Therefore ,

x= `(-8+-sqrt(8^2-(4)(3)(-9)))/(2(3))`

`=(-8+-sqrt(64+108))/6`

`=(-8+-sqrt(172))/6`

`=(-8+-sqrt(2*2*43))/6`

`=(-8+-2sqrt(43))/6`

`=2([-4+-sqrt(43)])/6`

`=(-4+-sqrt43)/3`

Therefore,

The factors are , `(-4+sqrt43)/3` ,`(-4-sqrt43)/3`

Example 2 : Solve `sqrt( x^2 +5x)-3=0`

Solution:

Isolate the given radical and exponent equation, we get,

`sqrt(x^2+5x)=3`

Take square root on both sides we get,

x^2+5x=32

x^2+5x=9

x^2+5x-9=0

Above equation in the form of ax^2+bx+c.

Therefore  you can find the factors using quadratic formula,

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

Here , b=5,a=1 and c=-9

substitute these values into formula,

`x= ((-5+-sqrt(5^2-4(1)(-9)))/(2(1)))`

`=(-5+-sqrt(25+36))/2`

`=(-5+-sqrt(61))/2`

`=(-5+-sqrt(61))/2`

Therefore ,

The factors are ,

`x= (-5+sqrt61)/2`    ,   `(-5-sqrt61)/2`

More about the Solving of Equations with Radicals and Exponents:

Example3: Solve `sqrt(2x^2+4)=4`

Solution:

Take square on both sides we get,

`(sqrt(2x^2+4))^2` =42

2x^2+4 =16

subtract both sides by 4,we get

2x^2+4-4=16-4

2x^2=12

Divide both sides by 2 ,

x^2 =`12/2`

x^2=6

x=`sqrt6`

Example 4: Solve the equations with radicals  and exponents `sqrt(4x^2+8)=11`

Solution:

Take square on both sides , we get,

4x^2+8 =121

Subtract both sides by 8 we get,

4x^2+8-8 =121-8

4x^2=113

Divide both side by 4,

x^2 =113/4

x=`sqrt(113/4)`

x=`(sqrt113)/2`

Therefore the value of `x= sqrt113/2`

Example 5: Solve the equation with radicals and exponents `sqrt(6x^2+7)` `=8`

Solution:

Take square on both sides we get,

6x^2 +7 = 64

Subtract both side by7,

6x^2+7-7=64-7

6x^2=57

Divide both side by 6,

x^2 =`57/6`

x= `sqrt(57/6)`

Therefore the value of `x = sqrt(57/6)`

Interest Compounded Quarterly

Introduction to interest compounded quarterly:

Interest is a fee paid on borrowed assets. It is the price paid for the use of borrowed money. Compound interest arises when interest is added to the principal, so that from that moment on, the interest that has been added also itself earns interest. This addition of interest to the principal is called compounding (for example the interest is compounded). In this article we shall discuss about interest compounded quarterly

Interest Formula for Compounded Quarterly

The basic formula for Compound Interest is:

FV = PV (1+r)n

PV is the current value or present value

r is the annual percentage rate of interest (percentage)

n is the total number of years the amount is deposit

FV = Future Value (amount of money collect after n number of years, with interest.)

Quarterly compounded interest = P (1 + r/n)nt = (Quarterly Compounding)


Interest Compounded Quarterly Example Problem

Ex 1:Rose deposits $7000 in a bank account, bank paying at the rate of 7% per year, compounded and credited quarterly. Find how much will he have at the end of 5 years?

Here p=$7000, n=4, r=7/100, t=5

Quarterly compounded interest = P (1 + r/4)4(5)

=7000(1+0.07/4)20

= 7000.00 is worth  9,903.45

Ex 2:Jessica deposits $6000 in a bank account, bank paying at the rate of 6% per year, compounded and credited quarterly. Find how much will he have at the end of 3 years?

Here p=$6000, n=4, r=6/100, t=3

Quarterly compounded interest = P (1 + r/4)4(3)

=6000(1+0.06/4)12

= 6000.00 is worth  7,173.71

Ex 3:Joseph deposits $5000 in a bank account, bank paying at the rate of 5% per year, compounded and credited quarterly. Find how much will he have at the end of 4 years?

Here p=$5000, n=4, r=5/100, t=4

Quarterly compounded interest = P (1 + r/4)4(4)

=5000(1+0.05/4)16

= 5000.00 is worth  6,099.45

Ex 4:Jim deposits $4000 in a bank account, bank paying at the rate of 5% per year, compounded and credited quarterly. Find how much will he have at the end of 6 years?

Here p=$4000, n=4, r=5/100, t=6

Quarterly compounded interest = P (1 + r/4)4(6)

=4000(1+0.05/4)24

= 4000.00 is worth  5,389.40

Domain of a Logarithmic Function

Introduction to Domain of a Logarithmic Function
In general, let us consider a function,  y = f(x).

It means, y the value of the function varies depending upon the input variable x and nature of the function.

As long as any value of x makes gives a real value of y, the function said to exist all the time. But this may not be the case with many functions due to the nature and restriction of the functions. In such a case, only for a particular set (or sets) of values of the input variable the function exists.

The set (or sets) of values of the input variable which makes the function exist is called the domain of the function and the corresponding set (or sets) of values of the function is called as the range of the function.

Let study the domain of a logarithmic function.

Description of a Domain of a Logarithmic Function

Before determining the domain of a logarithmic function, let us see what a logarithmic function is.

Mathematicians discovered that the function could be described in the form

f(x) = logbx, which is called as logarithmic function.

Let, n = logbx   Then as per the definition of a logarithmic function, bn = x

Determining the Domain of a Logarithmic Function

To determine the domain of a logarithmic function, let us start from the fundamental concept.

Let us take the simple form of a logarithmic function  y = logbx

Then as per definition   by = x
I am planning to write more post on Definite Integrals, Newton Raphson Method. Keep checking my blog.
If you carefully notice, the value of x becomes closer and closer to 0 as y becomes infinitely smaller and smaller. As ultimate, only when y becomes – infinity, the value of x becomes 0, which is an impossible situation. The following graph shows the variation.

Reversing the above argument, it can be said that a logarithmic function exists only for all values greater than 0.

That is domain of a logarithmic function is x > 0

Please note that for simplicity we assumed f(x) = logbx. In general it could be a logarithmic function of another function of x.

That is, f(x) = logb g(x).

In such a case, the domain of the logarithmic function is determined by solving  g(x) > 0

For example,  if f(x) = logb (x – 1), then the domain of the function is x > 1.

Fraction Decimal Notation

Introduction to fraction decimal notation:

Let us discuss the fraction decimal notation. The fraction notation is defined the part of whole number. The fraction is generally specifying two parts. The first part is numerator and second part is denominator. The numerator part is top part and denominator is bottom part. The decimal notation is same as the fraction notation. Decimal notation has decimal point. The example is 3.16.

Fraction Decimal Notation:

The fraction notation is numerator / denominator. Example is 4 / 6. The top number is known as the numerator and bottom number is known as the denominator.

The 4 is called numerator.
The 6 is called the denominator part.
The example of the fraction number is 9 / 7, 5/ 6 and etc.

The decimal point is very important part of the decimal notation. The example of the decimal notation is 63.147.

The 63 is called  the whole number.
The dot (.) is called  the decimal point.
The 1 is called the tenth place of the number. The meaning is 1/10.
The 4 is called the hundredth place of the number. The meaning is 4/100.
The 7 is called the thousandth place of the number. The meaning is 7/1000.
The fraction notation is written the decimal notation. The example is 5/7 is fraction notation. The decimal notation of 5/7 is 0.71.

I am planning to write more post on double digit multiplication, chat free online. Keep checking my blog.

Example Problem of Fraction Decimal Notation

Problem 1:

Divide the 8 by 10.

Solution:

The 8 is dividend and 10 is divisor

0.8
____
10  ) 8
0
------
80
80
------
0
-------

The 8/10 decimal value is 0.8.

Problem 2:

Divide the 3 by 2.

Solution:

The 3 is dividend and 2 is divisor

0.666
______
3  ) 2
0
------
20
18
------
20
18
-----
2
------

The 3/2 decimal value is 0.66.

Logarithm of Complex Number

Introduction to logarithm of complex number:


In this article we will study about logarithm of complex number. In this foundation is the higher grade mathematics. Logarithm is used to solve many complex problems in easy way. In this content we are going to discuss about logarithm of complex number. The following are the examples involved in logarithm of complex number

Sample Problem for Logarithm of Complex Number:

Logarithm of complex number Problem 1:

Solve the given logarithmic expression: in 3x + in 5 = 3

Solution:                                                                                                                            

Given logarithmic expression: ln 3x + ln 5 = 3

in 3x + 1.6094 = 3                   ( The value of ln 5 = 1.6094)

Subtract by 1.6094 on both side in the above expression.

ln 3x + 1.6094 - 1.6094 = 3 - 1.6094

ln 3x =  1.3906   

log e 3x = 1.3906                        (ln x = loge x)        

(We know, y = logbP       and P = by)

Here, y = 1.3906         P = 3x                b = e

So,    3x = e1.3906   = 4.0172

3x = 4.0172

Divided by 3 on both side in the above equation

=3x/3 = 4.0172/3

x = 1.3390

Answer: The value of x = 1.3390

Logarithm of complex number Problem 2:

Solve the problems `log 10^x = log^5`

Solution:

Step i: given  ` log 10^x = log^5`

Step ii:  simplify the equation using with law (when the log base to the power here we can written like multiplication of log term)

` x.log10=log^5`

Step iii: evaluate for x, here we can dividing each side by log10



`x = (log^5/log10)`

`x = (log^5/log10)`

Or

Step iv: Take log value for the terms

Log 5= 0.69900

Log 10=1

` x =0.69900/1`

Step v: x = 0.699.

Logarithm of complex number Problem 3:

Solve the problems 61 log29+61 log27 = 66 log2 (7x)

Solution:

Logarithmic function 61 log25+61 log27=61 log2 (7x)

61 log2(9*7)  =61 log2(7x)

log2(63)=log2(7x)  by logarithm rules

Equate both sides, s the base are same 2, 7x = 63

Simplification: `x =63/7`

Answer = 9

Algebra is widely used in day to day activities watch out for my forthcoming post on algebraic expressions  I am sure they will be helpful.

Practice Problem for Logarithm of Complex Number:

Get the values of given problems `log^8 + log^4 + log^5`
Answer:log (160)

2. Solve the given logarithmic expression: In 6x + In 5 = 6

Answer: x = 0.163390

Easy Way to Learn Limits

Introduction to Easy way to learn limits:      
In the easy way to learn limits, let f be a function of a real variable x. Let c and l be two unchanging numbers. If f(x)  come within reach of the value l as x approaches c, we say l is the limit of the function f(x) as x tends to c. This is written as

`lim_(x->c)`   f(x)   =  l.

Left Hand and Right Hand Limits:

In thr easy way to learn limits, although defining the limit of a function as x be inclined to c, we consider values of f(x) when x is very close to c (x>c or x
Lf(c) = `lim_(x->c )` _f(x) , provided the limit exists.

Likewise if x gets merely values greater than c, next x is said to tend to c from above or from right, and is denoted symbolically as x ? c + 0 or x ? c+, here the limit of f is then labeled the right hand limit. This is written as Rf(c) = `lim_(x->c+)` f(x). It is significant to reminder that for the survival of`lim_(x->c)` f(x) it is necessary that both Lf(c) and Rf(c) exists and Lf(c) = Rf(c) =`lim_(x->c)` f(x).  As well as the left and right hand limits are labeled as one sided limits.

Fundamental Results for Easy Way to Learn Limits:
The following rules for easy way to learn limits,

(1) If f(x) = k for all x, then
`lim_(x->c) `  f(x) = k.
(2) If f(x) = x for all x, then                                                                                                               
`lim_(x->c)`  f(x) = c.
(3) If f and g are two

functions having limits and k is a invariable then

(i)  `lim_(x->c)` k f(x) = k `lim_(x->c)` f(x)

(ii)  `lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) + `lim_(x->c)`  g(x)

(iii)  `lim_(x->c)`  [f(x) - g(x)] = `lim_(x-> c)`  f(x) - `lim_(x->c)`  g(x)

(iv)  `lim_(x->c)` c [f(x) . g(x)] = `lim_(x->c)` f(x) . `lim_(x->c)` g(x)

(v)    `lim_(x->c)` [`f(x)/g(x)` ] = `lim_(x->c)`  f(x)  / `lim_(x->c)`  g(x)              g(x) `!=` 0

(vi)  If   f(x) = g(x) then

`lim_(x-> c)`  f(x) =  `lim_(x->c)` g(x).

Example of Easy Way to Learn Limits:

Evaluate  `lim_(x->3)` `(x^2+ 7x + 11) / (x^2-9)` .

Solution:
Let f(x) = `(x^2 + 7x + 11)/( x^2-9)`
This is of the form f(x) =`g(x) / (h(x))`  ,
where g(x) = x2 + 7x + 11 and h(x) = x2 - 9. Clearly g(3) = 41 ? 0 and h(3) = 0.
Therefore f(3) = `g(3) /(h(3))`  = `41/ 0`  . Hence `lim_(x->3)`  `(x^2+ 7x + 11) / (x^2-9)` does not exist.

Solving Plane Geometry

Introduction to solving plane geometry:

Geometry is one of the important parts of mathematics deals with properties of shape of geometric objects. Geometry used to say all kinds of shapes and their properties. There are two major classifications in geometry. They are Plane geometry and Solid geometry. Shapes that are drawn at flat surface are called Plane and the study about this is called  Plane geometry. Plane geometry is about the shapes of lines, circles, etc. Plane geometry is study about two-dimensional objects.

Problems on Solving Plane Geometry:

Problem 1:

Solving the area of the triangle with the base of 7cm and the height is 10cm.

Solution:

Given:       Base = 7cm

Height = 10cm

Area = (b * h) / 2                       

= (7 * 10) / 2                            

= 70 / 2

= 35 cm^2

Problem 2:

Solving the area of the circle with 10.5cm radius.

Solution:

Given:     Radius = 10.5cm

Area of the circle =`pi` r2

= 3.14 * 10.5 * 10.5

= 346.18cm^2

Problem 3:

Solving the area of the trapezoid whose length is 10 cm, the width is 5cm and the height is 10 cm.

Solution:

Given:   Length = 10cm

Width = 5cm

Height = 10cm

Area = h (l + w) / 2                                                                                                                

= 10 (10+5) / 2

= 150 / 2

= 75cm3

Problem 4:

Solving the area of the rectangle with length 6.5cm and width 3.5cm. Also find its perimeter.

Solution:

Given:    Length = 6.5cm

Width = 3.5cm

Area = L * W

= 6.5 * 3.5

= 22.75cm^2            

Perimeter = Sum of all sides

= 2L + 2W

= 2 * 6.5 + 2 * 3.5

= 13 + 7

= 20cm^2

Problem 5:

Solving the area and perimeter of the parallelogram whose Height 12 cm, length 14cm, width 16cm.

Solution:

Given:    Length = 14cm

Width = 16cm

Height = 12cm

Area = L * W

= 14 * 16

= 224cm^2

Perimeter = Sum of all sides

= L + W + H

= 14+16+12

= 42cm3


Between, if you have problem on these topics geometric probability formula, please browse expert math related websites for more help on math word problem solver online.

Practice Problems on Solving Plane Geometry:

1.   Find the area of the trapezoid whose length is 3 cm, the width is 1.5cm and the height is 5cm.

Answer:  11.25cm^3

2.  Find the area of the circle with 0.5cm radius.

Answer:  0.785cm^2

3.  Calculate the area and perimeter of the rectangle whose length is 2cm, width is 2.5cm.

Answer:   Area = 5cm^2
Perimeter = 9cm^2