Common Factoring Fractions

A common factoring fraction is the important topic in algebra.  This method is similar to taking the denominator for comparing two or more fractions.  This method is mainly used for following types of problems. They are

• Comparing fractions
• Adding fractions
• Subtracting fractions

    In this topic we have to discuss about the common factoring fractions with example problems.

I like to share this Least Common Denominator Finder with you all through my article. 

Brief explanation of common factoring fractions – Comparing Fractions

Comparing fractions:

    We can compare two or more fractions; first we can take the common denominator for all fractions. For this we can take the common factors for all denominator values. The type of comparing fractions is mostly used in the following types of problem.

They are

• Ascending order
• Descending order

Example:

Arrange the following fractions in the ascending order `(1)/(2)` , `(1)/(4)` and `(1)/(8)` 

Solution:

     Here the denominator values are 2, 4and 8. They are not equal values. So we can take common factor for all denominator values. The common denominator value is 8.

Consider `(1)/(2)` x `(4)/(4)` = `(4)/(8)`

Consider `(1)/(4)` x `(2)/(2)` = `(2)/(8)`

Consider `(1)/(8)`

     Now the denominator is same. So we can arrange the fractions in the ascending order in the following manner,`(1)/(8)`,`(2)/(8)`,`(4)/(8)`

Brief explanation of common factoring fractions – Adding and Subtracting Fractions

Adding fractions:

    We can add the two or more fractions; first we can take the common denominator for all fractions. For this we can take the common factors for all denominator values. Then we can add the numerator values.

Subtracting fractions:

    We can subtract the two or more fractions; first we can take the common denominator for all fractions. For this we can take the common factors for all denominator values. Then we can subtract the numerator values.

My forthcoming post is on Scatter Plot Graphs and Variance Math will give you more understanding about Algebra.

Example:

Simplify the fraction 1/3 + 1/6 -1/12.

Solution:

     Here the denominator values are 3, 6 and 12. They are not equal. So we can take common factor for all denominator values. That is 12.

Consider `(1)/(3)` x `(4)/(4)` =`(4)/(12)`

Consider `(1)/(6)` x `(2)/(2)` =`(2)/(12)`

Consider `(1)/(12)`

     Now the denominator is same. So we can simplify the numerator values in the following manner,

 `((4+2-1))/(12)`= `(5)/(12)`

     These are the important types of common factoring fractions.

Least Common Multiples by Prime Factorization

The unique factorization theorem says that every positive integer greater than 1 can be written in only one way as a product of prime numbers. The prime numbers can be considered as the atomic elements which, when combined together, make up a composite number. The short term for Least common multiples is LCM. (Source Wikipedia)

Examples for finding Least common multiples by prime factorization:

Example 1:

Find the least common multiples for the numbers 234 and 534 by prime factorization.

Solution:

Prime factorization for the given numbers are

234   :  2 3 3 13

534   :  2 3        89

----------------------------

LCM:  2 3 3 13 89

Least common multiples by prime factorization is 2 * 3 * 3 * 13 * 89 = 20826

Example 2:

Find the least common multiples for the numbers 584 and 564 by prime factorization.

Solution:

584    : 2 2 2 73

564    : 2 2 3 47

------------------------

LCM : 2 2 2 3 73 47

Least common multiples by prime factorization is 2 * 2 * 2 * 3 * 73 * 47 = 82344

Example 3:

Find the least common multiples for the numbers 124 and 164 by prime factorization.

Solution:

124    : 2 2 31

164    : 2 2 41

---------------------

LCM : 2 2 31 41

Least common multiples by prime factorization is 2 * 2 * 31 * 41 = 5084

Example 4:

Find the least common multiples for the numbers 48 and 58 by prime factorization.

Solution:

48      : 2 2 2 2 3

58      : 2            29

------------------------------

LCM : 2 2 2 2 3 29

Least common multiples by prime factorization is 2 * 2 * 2 * 2 * 3 * 29 = 1392

Example 5:

Find the least common multiples for the numbers 45 and 64 by prime factorization.

Solution:

45      : 5 3 3

64      :          2 2 2 2 2 2

-------------------------------------

LCM : 5 3 3 2 2 2 2 2 2

Least common multiples by prime factorization is 5 * 3 * 3 * 2 * 2 * 2 * 2 * 2 * 2= 2880

Practice problems for finding Least common multiples by prime factorization:

Problem 1:

Find the least common multiples for the numbers 56 and 456 by prime factorization.

Least common multiples by prime factorization is 3192

Problem 2:

Find the least common multiples for the numbers 64 and 65 by prime factorization.

Least common multiples by prime factorization is 4160

Problem 3:

Find the least common multiples for the numbers 560 and 4560 by prime factorization.

Least common multiples by prime factorization is 31920

Between, if you have problem on these topics Series Solutions  please browse expert math related websites for more help on Multiple Linear Regression Analysis.

Problem 4:

Find the least common multiples for the numbers 356 and 78 by prime factorization.

Least common multiples by prime factorization is 13884

Problem 5:

Find the least common multiples for the numbers 4562 and 5697 by prime factorization.

Least common multiples by prime factorization is 25989714

Checking Prime Factorization

Prime number is a number that is divisible by 1 and itself. Prime factorization is nothing but the factorization of a prime numbers. To find the prime factorization of a number the number it factored until all its factors are prime numbers.In this topic we are going to see problems in prime factorization and checking the factors are prime or not. Some example problems are solved below.

Prime Factorization and checking – Example problems:

Find the prime factors of 24 and check?
24 ÷ 2 = 12
12 is not a prime number, so factor again,
12 ÷ 2 = 6
6 is not a prime number, so factor again,
6 ÷ 2 = 3
3 is a prime number,
24 = 2 × 2 × 2 × 3
Checking whether 2,2,2,3 are the prime factors of 24
2 × 2 × 2 × 3 = 24
Therefore 2, 2, 2, 3 are the prime factors of 24
Example 2:
Find the prime factorization of 124 and check ?
124 ÷ 2 = 62
62 is not a prime number, so factor again:
62  ÷ 2 = 31 
31 is a prime number.
124 = 2 x 2 x 31
Checking
2 * 2 * 31 = 124
Therefore 2, 2, 31 are the prime factors of 124. 

prime factorization tree.
                      124
                      /    \
                 2         62
                           /    \
                         2    31

Some other examples of prime factorization:

My forthcoming post is on SAS Similarity and Define Box and Whisker Plot will give you more understanding about Algebra.

Example 3:
Find the Prime factor of  81 and check?
Solution:
81 ÷ 9 = 9
9 = 3 * 3
81 = 3 x 3 x 3
Checking
3*3*3 = 81
Therefore the prime factors are 3,3,3
Example 4:
Find the prime factor of 13 and check?
Solution:
39 = 3 x 13
So the prime factor of 19 is 3 and 13.
Checking
3 * 13 = 39
Therefore the prime factors of 31 are 3 * 13.

Precalculus Problem Solution

Calculus is one of the learning about the rates of change and measurement of changing quantities in which it is using symbolic notations in the calculations. In precalculus it identifies the values of the function and it does not involved in the problems of calculus but it explores topics that will be applied in calculus. So that precalculus one of the main division in the calculus

Calculus two types, they are.          

• Differential calculus
• Integral calculus       

      Precalculus functions that are given is, Domain and Range Composition, and Difference of Quotients

Differential calculus:

        Differentiations are been used to determine the rates of change of a function and the larger and smaller values of a function.

        It is a branch of Calculus that deals with derivatives and their applications.

Integral Calculus:

         Integration is the branch of the calculus in which it deals with integrals and its application,  in determining areas, equations of curves, or volumes.                                                                                                                

                          

Some precalculus operations:

• The slope of line passing through points of` (x_1, y_1) ` and` (x_2, y_2)` which is given by,

                        m = `(y2 - y1) / (x2 - x1)`

• slope m has the equations. in which passes through the line through the point (x₁, y₁).

                    y - y₁ = m(x - x₁)

• y-intercept the b has the equation of line y = mx + b.It is the line passes through the slope
• Two lines of the equation with slopes m₁ and m₂.

            If the line of the slope is parallel m₁ = m₂ and

            If the line of the slope is perpendicular m₁m₂ = -1.
                                                      

Precalculus problem solution - Example problems:

Precalculus problem solution - problem 1:

Solve the equation with the  curves y = mx, where, m is arbitrary constant.

Solution:

      We have the equation of the curve

               y = mx ... (1)

     Differentiating either side of equation (1) with respect to x, we get

      `dy / dx` = m

     Substitute the value of m in equation (1) we get

               y = ` (dy / dx)` * x

     or 

            x `( dy / dx ) ` - y = 0

    Hence this is the required differential equation.         
Precalculus problem solution - problem 2:
Find the derivatives for,   (i) Y = `4 / x^3`   (ii) Y = 6`sqrt (x^3)`
   Solution :
                  (i). Y = `4 /x^3 `
                        Y = 4 X ^-3
                     ` dy / dx` = (4-3)x^(-3-1)
    Answer     = `(1 / x^(-4))`
               (ii) Y = ` 6sqrt(x^3)`
                      Y =  `6 x^(1 /3)`
             `dy / dx` = `6 (1/3)` `x ^-(2 / 3)`
           `dy / dx ` = `(6/3)x^-(2/3)`
     Answer = `(6/3)x^-(2/3)`
                                                       

Solve Horizontal Line Test

HORIZONTAL LINE TEST is a method to test whether a function is One-One or Many-One. As the terms HORIZONTAL  and LINE suggests it is a geometrical (Graphical) method.

It tests whether any HORIZONTAL LINE cuts the graph of the function exactly once or not, if the test is positive, means, if there is only one point of intersection then the Function is One – One, otherwise it is Many – One.

e.g.

Case1.

Y = 2x+3

Graph of y = 2x+3

Example of a One-One Function, as no HORIZONTAL LINE intersects the graph more than once

Y = x^2 -3

Graph of y = x2 - 3

Example of a Many-One Function, as  HORIZONTAL LINEs intersects the graph more than once( Twice)

Solving an Equation using HORIZONTAL LINE TEST

The solution to any equation is the values of x for which y = 0, or the X coordinates of the points of intersection of the HORIZONTAL LINE, y =0 with the graph of the corresponding function.

From Case 1 above it can be seen that there is only one solution x = -3/2

And

From Case 2 above it can be seen that there are two solutions x = √3 and x= -√3

HORIZONTAL LINE TEST to find the values of x for any y of the function y = f(x)

Drawing a HORIZONTAL LINE along the required value of y, we can find the corresponding values of x.

eg.

Figure 1 shows the value of x=0, when y = 3  (point of intersection of the HORIZONTAL LINE y =3 and the graph of the function y = 2x +3)



Figure 2 shows the values of x= ± √ 6 , when y = 3  (point of intersection of the HORIZONTAL LINE y =3 and the graph of the function y = x2 -3)


My forthcoming post is on math word problems for 6th grade and cat exam pattern 2013 will give you more understanding about Algebra.


HORIZONTAL LINE TEST- the primary test for the Existance of Inverse

A function has inverse if and only if it is One-One and On-To

HORIZONTAL LINE TEST determines whether the first condition, one- one is satisfied or not , if the HORIZONTAL LINE TEST fails, then there do not exist any inverse for the function.

Volume Translation

Definition of volume:

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains,[1] often quantified numerically using the SI derived unit, the cubic meters. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.

(Source: Wikipedia.)

Definition of volume translation:

Volume translation is the process translating the volume into different types of unit. For example, we take the cylinder and calculate its volume and it can be in the form of cubic meters. It can be translated into different units that are cubic milliliters, Liters, Centiliters etc., Volume translation is the process of calculating the different units.

Different formulas for volume:

Volume formulas:

Cube = side3
Rectangular Prism = side1 * side2 * side3
Sphere = (4/3) * `pi` * radius3
Ellipsoid = (4/3) * `pi` * radius1 * radius2 * radius3
Cylinder = `pi`* radius2 * height
Cone = (1/3) *`pi` * radius2 * height
Pyramid = (1/3) * (base area) * height
Torus = (1/4) *`pi` 2 * (r1 + r2) * (r1 - r2)2


Example problem for volume translation:

Example 1:

Find the volume for the given cylinder with radius = 14 m and height = 24 m.

volume

Solution:

Volume formula for cylinder

V = `pi` r2h

Here `pi` = 3.14 or 22/7

r = 14 m

h = 24 m

V = 3.14 * 142 * 24

= 3.14 * 14 * 14 * 24

= 3.14 * 196 * 25

= 3.14 * 4900

= 15386 m3.

Different translation of volume:

15386 m3 = 1538600000 centiliters

15386 m3 = 153860000 deciliters

15386 m3 = 15386000 L

15386 m3 = 15386000000 cubic milliliters

15386 m3 = 96775.028514 barrels.

Example 2:

Find the volume for the given cone with radius = 8 cm and height = 12 cm.

cone

Solution:

Volume formula for cone

V = (1/3) *`pi` * radius2 * height

Here we use pi = 3.14

= (1/3) * 3.14 * 82` * 12`

` = (1/3) * 3.14 * 8 * 8 * 12`

` = (1/3) * 3.14 * 8 * 8 * 12`

` = 3.14 * 64 * 4`

` = 3.14 * 256`

` = 803.84 `cm3`.`

Different translation of volume:

`803.84 `cm3 = 80.383976 centiliters

`803.84 `cm3 = 8.038398 deciliters

`803.84 `cm3 = 0.80384 L

`803.84 `cm3= 803839.763 cubic milliliters

`803.84 `cm3 = 0.005056 barrels.

Interpretations of Probability

An interpretation of probability is a task of meaning to probability claims. These involve specifying a set of possible cases and defining what a probability claim means in terms of that set.


Understanding Probability Equation is always challenging for me but thanks to all math help websites to help me out.

An interpretation tell us what it means to say that P (p) = r

(Where p can be any sentence and r can be any number between 0 and 1.)

An interpretation must also provide a meaning for conditional probability statements of the form P (p / q) = r.

Interpretations of probability: Over view

Probability ->subjective

->actual degree of belief

->personalist (tempered personalist)

->objective

->classical

->logical

->Frequency

-> Finite freq.

-> Limiting freq.

-> Propensity

Cases of Interpretations of probability:

Classical

Number of possible / total number of possible

Logical

Getting a partially entails, with degree of entailment Finite frequency

Limiting frequency

The limiting frequency in an endless series

Propensity

The limiting frequency of a pair

Interpretations of Probability Example:

What is the probability of getting at least one 6 in two tosses of a die?

Solution:

P (6 on toss 1) = 1/6

P (6 on toss 2) = 1/6

P (6 on both tosses) = 1/36

P (at least one 6 in two tosses) = 1/3

Consider each of the following four bet light (I expect to break even):

1. Pay 36 for [36 if no 6 on 1st toss; 0 if 6]

2. Pay 36 for [36 if no 6 on 2nd toss; 0 if 6]

3. Pay 1 for [42 if 6 on both tosses; 0 otherwise]

4. Pay 12 for [36 if slightest one 6 in two tosses; 0 otherwise]

Algebra is widely used in day to day activities watch out for my forthcoming posts on how to add and multiply fractions and iit jee new pattern 2013. I am sure they will be helpful.

But now consider what happens if I make all four bets:

1st toss        2nd toss      Cost   Winnings     Net gain or loss

6                 6                 73        72                -1

6                 1-5              73        72                -1

1-5              6                 73        72                -1

1-5              1-5              73        72                -1

A guaranteed loss!