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.