Area of Non Right Angle Triangle

Introduction on area of non right angle triangle:

The triangles other than the right triangle may be scalene triangle or isosceles triangle. The area of triangle, when all the sides are given is found by using the Heron’s formula. The other method for finding the area of the polygon is when the base and the height of the triangle are given. In the following article we will calculate the area of the non right angle triangle.

Area of Non Right Angle Triangle:

The formula for finding the area of the triangle with all the three sides known is calculated by using the Heron’s formula,

Heron’s formula = `sqrt [s(s-a)(s-b)(s-c)]`

`s = (a + b + c)/2`

The most common formula for calculating the area of the triangle = bh/2

Where, b is the base length and h is the height of the triangle.

The formula for finding the area of an equilateral triangle = `(sqrt 3* a^2)/2`

The formula for calculating the area when two sides and the opposite angle is known is = (ab sin C)/2

Example Problems on Area of Non Right Angle Triangle:

1. Find the area of the triangle if the two sides are 5cm and 7cm, the opposite angle to these sides is 35 degrees.

Solution:

Area formula = (ab sin C)/2

= (5*7* sin 35)/2

= `(35* sin 35)/2`

= `35* 0.574/2`

= 20.09/2 = 10.04 cm^2.

2. Find the area of the triangle with the sides 6cm, 7cm, 9cm.

Solution:

Area of the triangle =` sqrt [s(s-a)(s-b)(s-c)]`

`s = (a + b + c)/2`

= (6+7+9)/2

= 22/2

= 11

Area = sqrt [11(11-6)(11-7)(11-9)]

= sqrt [11*5*4*2]

= sqrt [11*40]

= `sqrt [440]`

= 20.98 cm^2.

I have recently faced lot of problem while learning Right Triangles, But thank to online resources of math which helped me to learn myself easily on net.

Practice problems on area of non right angle triangle:

1. Find the area of the triangle with height 7cm and base 6cm.

Solution: 21 cm^2.

2. Find the area of the triangle with sides 9cm and 11 cm and the angle opposite to them is 42 degrees.

Solution: 33.12 cm^2.

3. Find the area of a equilateral triangle with the side length of 5cm.

Solution: 21.65 cm^2

Narrator, Narrative and Narration

Narrator, narrative and narration are three of the most important terms in English literature. Let’s understand each of the three terms along with examples in this post.

Narrator:  A narrator is the voice that tells a story. The narrator is not the author except in autobiographies. When the narrator is also a character in the story, the narration is done in first person. The narrator can also be created as a third person narrator by the author. Let’s have a look at two examples for better understanding.

Example 1: “I bought at Fisher Price rocker for my friend’s child. Her child is about to be eight months next year and a Fisher Price rocker was a perfect choice for him. I was really happy for her and her baby. But, unfortunately her husband was not in the town to share the experience of parenthood with her.” – Here, the narrator is a first-person narrator.

Example 2: “Mary has started shopping from online stores and has experienced affordable and quality shopping together. She got Johnson baby kit for her baby as Johnson baby kit is considered as the most trusted of brands. She also bought some dresses from Little Kangaroo India collection. And the best part is she got good discount while buy dress from Little Kangaroo India brand.” – Here, the narrator is a third-person narrator.

Narrative: The narrative is the story or the way the events are prepared and presented. Narrative can be classified into various forms such as narrative through speech, writing, songs, film, television and more. The term narrative is also used as a synonym for story.

Narration: Narration means the way that a story is told. Narration can be presented both from outside and being within it. As mentioned earlier, if the narrator is within the story, it becomes a first-person narrative and the narrator is telling the story from inside. On the other hand, when the narrator is outside the story, it becomes a third-person narrative where the narrator talks from outside the story.

These are the basics about narrator, narration and narrative in a story.  

Random Variable X

Introduction to random variable:

The arithmetical value  of a variable is defined by an event,that variable is known as random variable.Random variable may be discrete or continuous.Discrete random variables contains the positive integer values that may exist between zero and infinity.Common example for discrete random variable is when we toss a coin,the possibility of getting head is the positive integer.In continuous random variable the value may range between zero and one sometimes it may be non-integer value.

Random Variable X
random variable x problem 3:

Select the number at random from 12 to 18. What is the probability for select the number is odd?

Solution

The random variables x of the above experiment is 12,13,14,15,16,17,18

i)Take P(A) is the probability of the odd number occur.

The random variable x of odd numbers are 13,15,17.So n(A)=3

Total outcomes n(S)=7

So P(x)=`(n(A))/(n(S))`

=`3/7` .

random variable x problem 4:

Select the number at random from 12 to 18. What is the probability for select the number is even?

Solution

The random variables x of the above experiment is 12,13,14,15,16,17,18

i)Take P(x) is the probability of the even number occur.

The random variable x of even numbers are 12,14,16,18.So n(A)=4

Total outcomes n(S)=7

So P(x)=`(n(A))/(n(S))`

=`4/7` . Please express your views of this topic Convert Decimals to Fractions by commenting on blog.

Random Variable X

random variable x problem 3:

What is the probability of getting a head when a coin is tossed?

Step 1:

The random variable Y contains two possibilities they are head and tail
Step 2:

The random variable is denoted by p(y)

Step 3:

The probability for getting a head is as follows

p(y)=`(n(A))/(n(S))`

Step 4:

The random variable y gives

`p(Y)= 1/2`

random variable x problem 4:

What is the probability of getting a tail when a coin is tossed?

Step 1:

The random variable Y contains two possibilities they are head and tail

Step 2:

The random variable is denoted by p(y)

Step 3:

The probability for getting a tail is as follows

p(y)=`(n(A))/(n(S))`

Step 4:

The random variable y gives

`p(y) =1/2`

Integer Practice

Introduction of integer practice:

We know that integers have positive and negative numbers. Each negative numbers is matching with a positive number the similar distance from 0 on a number line. Example: -6,-5,-4,-3,-2,-1,0,+1,+2,+3,+4,+5,+6. We can write them down like this: {…,-3,-2,-1, 0,1,2,3}. Here we are going to learn how to practice integers.

Rules:

Rules for adding integers:

Rule 1:  Suppose if add the same sign, add the two numbers and put their common sign.

(+62) + (+14) = +76 (-29) + (-13) = -42.

Rule 2:  Suppose if we add the different signs, we need to find the difference between the two numbers and put the sign of the number.

(+15) + (-8) = +7 (+9) + (-30) = -21

Rules for subtracting integers:

Rule 1: Commonly every subtraction problem can be rewrite as a corresponding addition problem,

Use the follow rule: To subtract an integer, add its reverse.

1. (-8) – (+9) = The opposite of +9 is –9. Change sign to opposite: (-8) + (-9) = -17 using Integer addition rules

Switch rule 1 to subtract signed numbers:

Step 1: First, change double negatives to a positive.

Step 2: Get a sum of terms with same signs and keep the given sign, using the sign in front of the number as the sign of the number.

Step 3: Get the difference when the terms have different signs and use the sign of the larger numeral. Understanding Radian Measure is always challenging for me but thanks to all math help websites to help me out.

Practice Problems for Integers:

Practice adding positive integers:

We know that addition is the basic operations in math. Here we are going see some practice problems of addition.

Example 1:

23+21.

Solution:

Here both the numbers are positive .And then addition process is done here.

The result  is 23+21 = 44.

Practice adding negative integers:

-256 +-24.

Solution:

When we take this above problem, we have both negative numbers. And we need to find the total of these two numbers. And negative 256 plus negative 24 equals negative 280.

The result is -280.

Practice subtraction integers:

Turn the subtraction symbol into addition and get the reverse of the second number. Then take the problem as an addition problem.

Example 1:

Integer subtraction:

-128+18.

Solution:

Here the first number is negative and the next number is positive. The subtraction procedure is finished. The biggest number sign will appear to answer

The result is -128+18= 110.

Unit Circle with Tangent

Introduction:

The Unit circle is used to understanding the sins and cos of angles to find 90 degree triangle. Unit circleis radius is exactly one. The center of circle is said to be origin and its perimeter comprises the set of all points that are exactly one unit from the center of the circle while placed in the plane.It s just a circle with radius ‘one’.

Unit Circle Standard Equation:

In unit circle: The distance from the origin point(x,y) is by using Pythagorean Theorem.

Here radius is one So, The expression should becomes =1

Take square on both sides then the equation  becomes,

X2+y2 =1

Positive angles are found using counterclockwise from the positive x axis

And negative angles are found anti clockwise from negative axis. Please express your views of this topic what is a line segment by commenting on blog.

The Graph of the Unit Circle with Tangent Function:

The  correspondence of  `theta` with tan  ` theta` , Here ` theta` is a real number  so, from the definition of tangent function,

tan `(theta)` = Sin `(theta)` /cos`(theta)`

Values for the tangent function can be created.  Above  is the unit circle.  Remember that the sine, cosine, and tangent are all built from this picture.  Hence, they are called circular functions.

These values are the angle measure (the value closest to the circle) and the tangent values.  So starting with angle 0, notice that the tangent value is 0.

Equivalent Systems of Equations

Introduction:

An equivalent system of equation is a system of equations in which both systems have same solution but they may have different numbers of equations.

Systems of Linear Equations

It is a set of algebraic expressions in the form:


a11x1 + a12x2 + .....................+a1nxn = b1


a21x1 + a22x2 + .....................+a2nxn = b2


am1x1 + am2x2 + .....................+amnxn = bm


xi are the unknowns, (i = 1, 2, ..., n).
aij are the coefficients, (i = 1, 2, ..., m), (j = 1, 2, ..., n).
bi are the independent terms, (i = 1, 2, ..., m).
m, n ; m > n, or, m = n, or, m < n.
The number of equations need not equal the number of unknowns.
aij and bi  .
When n is less, it is usual to assign the unknowns with the letters x, y, z, t, ...
When bi = 0, for all i, the system is called homogeneous.

Equivalent systems of equations are obtained by elimination if:


The value of coefficients is zero.
Two equal rows are present.
Two rows are proportional to each other.
A row is formed by linear combination of others. Looking out for more help on Algebra Mixture Problems in algebra by visiting listed websites.

Equivalence Criteria

The resulting system is equivalent if both members of an equation of a system are added or subtracted by the same expression.
The resultant system is equivalent, if both members of the equations of a system are multiplied or divided by a number other than zero,
The resultant system is equivalent, if an equation of a system is added or reduced by another equation of the same system.
The resultant system is equivalent, if an equation in a system is replaced by another equation that results from adding the equations of a system previously multiplied or divided by nonzero numbers,
If the order of the unknowns of a system or order of the equations is changed, it is another equivalent system.

Want to Learn How to Divide

Introduction of want to learn how to divide:-

In mathematics, especially in elementary arithmetic, division (÷) is the arithmetic operation that is the inverse of multiplication.I like to share this Elementary Row Operations with you all through my article.

Specifically, if c time’s b equals a, written:

c x b = a

Where b is not zero, then a divided by b equals c, written:

`a/b` = c.

In the above expression, a is called the dividend. (Source: Wikipedia)

Step by Step Process of Want to Learn How to Divide 20 by 2:-

In following steps for want to learn how to divide 20 by 2

Step 1:-

-------
2 | 20

In the above equation 2 is divisor and 20 is dividend. In the divisor has two decimal numbers put the value dividend of  20.

Step 2:-

10
-------
2 | 20
20
---------
0
--------

Normally divide the values one by one the right value has 0 means directly divided by two digits. In 2 x 10 = 20 the divisor number 2 is multiplied with 10 to get an answer 20.In 20 is equal to 20. So use the values then subtracts the value and get remainder is 0.Please express your views of this topic trigonometry help by commenting on blog.

Example Problems for Want to Learn How to Divide:-

Problem 1:-

How to divide 54 by 2

Solution:-

In following steps for want to learn how to divide 54 by 2

Step 1:-

-------
2 | 54

In the above equation 2 is divisor and 54 is dividend. In the divisor has two decimal numbers put the value dividend of  5 .

Step 2:-

2
-------
2 | 54
4
---------
14

In 2 x 2 = 4 the divisor number 2 is multiplied with 2 to get an answer 4. In 4 is less than from 5.So use the value then subtract the value and get 1.

Step 3:-

27
-------
2 | 54
4
---------
14
14
--------
0
--------

In 2 x 7 = 14 the divisor number 2 is multiplied with 7 to get an answer 14. In 14 is equal to 14. So use the values then subtracts the value and get remainder is 0.



Problem 2:-

How to divide 48 by 2

Solution:-

In following steps for want to learn how to divide 48 by 2

Step 1:-

-------
2 | 48

In the above equation 2 is divisor and 48 is dividend. In the divisor has two decimal numbers put the value dividend of  48.

Step 2:-

2
-------
2 | 48
4
---------
8

In 2 x 2 = 4 the divisor number 2 is multiplied with 2 to get an answer 4. In 4 is equal to  4. So use the value then subtract the value and get 0. Put the next value 8.

Step 3:-

24
-------
2 | 48
4
---------
8
8
----------
0
----------

In 2 x 4 = 8 the divisor number 2 is multiplied with 4 to get an answer 8. In 8 is equal to 8. So use the values then subtracts the value and get remainder is 0.

Lcm of a Math Number Examples

Introduction for lcm of a math number examples:

The lowest common multiple (LCM) or least common multiple (LCM) of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then LCM(a, b) is defined to be zero.

- Source wikipedia

Examples for Lcm of a Math Number Examples:

Example 1:

Find the lcm of a math number examples 3 and 9.

Solution:

Multiples of the given number,

3      : 3, 6, 9, 12, 15, 18,21,24,27,

9      : 9, 18, 27

Now we have to pick out lest common multiple of the both number examples 3 and 9. Therefore, The LCM of math number examples 3 and 9 is 9

Example 2:

Find the lcm of a math number examples 4 and 12.

Solution:

Multiples of the given number,

4     : 4, 8, 12, 16, 20, 24, 28, 32, 36

12   : 12, 24, 26

Now we have to pick out lest common multiple of the both number examples 4 and 12. Therefore, The LCM of math number examples 4 and 12 is 12

Example 3:

Find the LCM of a math number examples 5 and 6.

Solution:

Multiples of the given number,

5      :  5, 10, 15, 20, 25, 30, 35, 40, 45, 50

6      :  6, 12, 18, 24, 30, 36, 42, 48, 54, 60

Now we have to pick out lest common multiple of the both number examples 5 and 6. Therefore, The LCM of math number examples 5 and 6 is 30.Understanding how to find the variance is always challenging for me but thanks to all math help websites to help me out.

Practice Problems for Lcm of a Math Number Examples:

Problem 1:

Find the LCM of a math number examples 6 and 7.

The LCM of math a number examples 6 and 7 is 42

Problem 2:

Find the LCM of a math number examples 7 and 8.

The LCM of a math number examples 7 and 8 is 56

Problem 3:

Find lcm of a math number examples 8 and 9.

The LCM of a math number examples 8 and 9 is 72

Fractions with Regrouping

Introduction of fractions with regrouping:
The fractions with regrouping are nothing but the fractions which has the formation of the understanding and easy recalling of the particular fractions can be done through the fractions. The regrouping fractions are made along the process that had done in the way which has the computation of the process. The computations of the fractions are done through the regrouping of the fractions.I like to share this Compare Fractions with you all through my article.

Fractions with Regrouping:

Let us have the briefing of the quotient or the ratio. The quotient is nothing but the term that has the 2/3 in which the 2 is the numerator and the 3 is the denominator. The method can be mentioned as 2/3. The ratio is made to have the briefing that gives the situation on it. Hence there are two boys or the every three girls in the team hence in this case two-thirds of the teams are boys.

The regrouping of the fractions are made through the way that happen in the way which gives the understandings through the fraction of summing and the changes that works on the principle of the various changes that would be made useful for the regrouping of the fractions. These regrouping fractions can be made through the various methods like numerical and the x, y terms.Having problem with online math tutor keep reading my upcoming posts, i will try to help you.

Examples for Fractions with Regrouping:

Example 1: Compute the fractions with regrouping in the fractions like “1/80 + 4/80 + 9/80 + 6/80” is made to regrouping as “1/80+9/80 +6/80+4/80” the second terms are considered as simpler one these terms leads to the sum of ten [(1+9)/80, (6+4)/80] this made easier to keep the track off.

Example2: Compute the fractions with regrouping like “3x/80 + 2y/80 + 4x/80 + y/80 = 7” hence the regrouping of the x and y terms which can be made simpler.

This can be regrouped as 7x/80 + 3y/80 = 7.

Multiple Regression Equation

Introduction to multiple regression equation:

Multiple regression analysis is a statistical tool in which a mathematical model is developed to predict a dependent variable by two or more independent variables or in which atleast one predictor is non-linear. The principal advantage of multiple regression is that it allows us to utilize more of the information available to us to fit curves as well as lines.

Multiple Regression Model with Two Independent Variables:

The simplest multiple regression model is one constructed with two independent variables, where the highest power if either variable is one.

The model is given by  y = β0 + β1x1 + β2x2 + ε.
The constants and coefficients are estimated from sample information, resulting in the following model.

Y = b0 + b1x1 + b2x2

Multiple Regression Model Equation:

Multiple regression analysis is similar to simple regression analysis. However, it is more complex conceptually and computationally. The general equation for the probabilistic multiple regression model is given by


y = β0 + β1x1 + β2x2 …+ βkxk +  ε.

Where y = the value of the dependent variable

β0 = the regression constant

β1 = the partial regression coefficient for independent variable 1

β2 = the partial regression coefficient for independent variable 2

.....

.....

βk = the partial regression coefficient for independent variable k

k = the number of independent variables

In multiple regression analysis, the dependent variable, y, is some times reffered to as the responsive variable. The partial regression coefficient of an independent variable, βi represents the increase that will occur in the value of y from a one unit increase in that dependent variable if all other variables are held constant. The partial regression coefficient occur because more than one predictor is included in model.

In actuality, the partial regression coefficients and the regression constant of a multiple regression model are population values and are unknow. In virtually all research, these values are estimated y with sample information.Please express your views of this topic how to cross multiply by commenting on blog.

Y= b0 + b1x1 + b2x2 …+ bkxk

Where Y = the predicted value of y

b0  = the estimate of the regression constant

b1 = the estimate of the regression coefficient 1

b2 = the estimate of the regression coefficient 2

bk = the estimate of the regression coefficient k

k = the number of independent variables.

Determining the Multiple Regression Equation:

The procedure for determining formula to solve for multiple regression coefficients is similar to that of solving for simple regression coefficients. The formulas are established to meet an objective of minimizing the sum of squares of error for the model. Hence, the regression analysis shown here is reffered to as least square analysis. Methods of calculs are applied, resulting in K+1 unknowns for regression analysis with k dependent variables.

For multiple regression models with two independent variables, the result is three simultaneous equations with three unknowns( b0, b1 and b2).



The process of solving these equations is tedious and time consuming. Solving for the regression coefficients and regression constant in a multiple regression model with two independent variable requires `sum` x1, ∑ x2, ∑ y, ∑ x12, ∑ x22, ∑ x1x2, ∑ x1y  and ∑ x2y.

Set Theory Subsets

Introduction to set theory subsets:

The set is represented as the collection of the object. There are different kinds of the set. They are set of set, proper test, power set, and universal set. The set inside of the set is called the subset. The one type of the set is called the subset. Now we see the detailed information about the set theory subset.

Subset and Subset Theory:

Subsets:

The P is the subsets of the set Q, if the set P is inside of the set Q, the connection of single set being a subsets of another set is called the inclusion.

Set theory:

The set theory is the division of the mathematics, which are the group of the object. Some kinds of the object can be grouped into the set, the set theory is applied the majority often to object that are applicable to mathematics.

There are the two set, X and Y.  The Y is the subsets of the X. This is represent by `X sube Y`.


Set Theory of the Subsets:

The group of the object is known as the set. They are various types of the set are included. These are the set of the letter and set of number.

Example:

The set of the letters is {A, B, C, D, ....}.

Odd number set is the {1, 2, 3...}.

Even number set is the {2, 4, 6...}.

Upper case of the letter is set is representing as the {A, B, C, D….}.

The lower case of the letter is set represent as the {a, b, c...}.

The element A is belong to a is represented by A`sube` a.

The given number A is not a n set of the X is denotes as A`sube`X. the A is not the number of the set is S.

Looking out for more help on algebraic word problems in algebra by visiting listed websites.

Examples:

They are different kinds of the set. The one set is the mathematics and other set is the algebra. The algebra is the subset set of the mathematics.

The two specific kinds of the set are class A and class B. The class B is the subsets of the class B. These subsets are represented as the class A `sube` class B.



These are the details about the set theory subset.

Partial Fraction Decomposition

Introduction:

If f(x) and g(x) are two polynomials, then $\frac{f(x)} {g(x)}$ defines a rational algebraic function or a rational function of x.

If degree of f(x) < degree of of g (x), then $\frac{f(x)} {g(x)}$ is called a proper rational function.

If degree of f (x) > degree of g(x) then $\frac{f(x)} {g(x)}$ is called an improper rational function.

If $\frac{f(x)} {g(x)}$is an improper rational function, we divide f(x) by g(x) so that the rational function $\frac{f(x)} {g(x)}$ is expressed in the form F (x) + ?$\frac{f(x)} {g(x)}$ where F(x)and ?(x) are polynomials such that the degree of ?(x)is less than that of g(x)is less than that f(x). Thus, $\frac{f(x)} {g(x)}$ is expressible as the sum of a polynomial and a proper rational functions.

Any proper rational function $\frac{f(x)} {g(x)}$ can be expressed as the sum of rational functions, each having a simple factor of g(x). Each such fraction  is called a partial fraction and the process of obtaining then is called the resolution or decomposition or decomposition of $\frac{f(x)} {g(x)}$ into partial fractions.

How to Find Partial Fraction:

The resolution of $\frac{f(x)} {g(x)}$into partial fractions depends mainly upon the nature of the factors of g(x) as discussed below.

Case:- When denominator is expressible as the product of non-repeating linear factors.I like to share this Cdf of Uniform Distribution with you all through my article.

Let g(x) = (x – a1) (x – a2) … (x – an). Then we assume that

$\frac{f(x)} {g(x)}$ =  A1/ x + A2/x x – a2 + … + An/x – an

where A1, A2, … An are constants and can be determined by equating the numerator on RHS to the numerator o LHS and then substituting x = a1, a2, …, an.

My Previous Blog :- http://wanttolearnmath.blogspot.in/2012/11/multiplying-three-factors.html

Independent Dependent Events

Introduction to independent dependent events:

Let we will discuss about the independent and dependent events in probability. If the two events should be said to be dependent when occurrence or outcome of first event affects the occurrence or outcome the second event. Therefore, their probability will changed in independent and dependent events.

If two events should be called independent when the outcome of first event should not affects the outcome of second event.

Independent Dependent Events-dependent Events:

Let us consider more than two events that are dependent.
When p1 should be probability of first event, p2 be the probability that happens after first event and p3 will be the probability that occurs after first and second events.
Then probability of all events will happen will be the product p1 - p2 - p3.

Please express your views of this topic formula for calculating probability by commenting on blog.

Example problem:

A bag has 6 blue balloons, 4 green balloons and 2 black balloons. In every draw, a balloon is drawn from the bag and not replaced. In three draws, find the probability of obtaining blue, green and black in that order.

Solution:

Given, Blue balloons = 6

Green balloons = 4

Black balloons = 2

Total number of balloons = 6 + 4 + 2 = 12

Here, the three events are dependent.

So the probability = ( 6 / 12 ) × ( 4 / 11 ) × ( 2 / 10 )

= ( 1 / 2 ) × ( 4 / 11 ) × ( 1 / 5 )

=  4 / 110

=  2 / 55

Independent Dependent Events-independent Events:

Two events P and Q are called independent when reality that P occurs should not affect the probability of Q happening.

Example problem:

A die should be tossing two times. What will be the probability of getting 2 or 4 on first toss and 1, 3, or 5 in second toss.

Solution:

Let, probability of getting 2 or 4 is P(E1) and probability of getting 1,3 and 5 will be P(E2).

Now, P(E1) = P (2 or 4) = 2 / 6 = 1 / 3

P(E2) = P (1,3 or 5) = 3 / 6 = 1 / 2

Here, they are independent events.

Therefore, P(E1 and E2) = P(E1) × P(E2)

= 1 / 3 × 1 / 2

= 1 / 6

My Previous Blog :- http://wanttolearnmath.blogspot.com/2012/11/multiplying-three-factors.html

Multiplying three Factors

Introduction to multiplying three factors:

This article we will discuss about multiplying three factors and factorization methods. Any numbers that when multiplied together form a product called as factors. Each number should have a factors each factor is unique it must not zero. For example 2 is a factor of 8 because 2 can be multiply by 4 to give 8 this is called factors. Lets us see about prime factors, and multiplying three factors.Understanding Prime Factors of 72 is always challenging for me but thanks to all math help websites to help me out.

Factors:

Prime factorization:

A Prime number is a whole number that can be divisible by itself. This called prime number. For Example 1, 3, 5, 7 are some of the prime numbers.

Factorization is finding which prime number need to multiply together it gets the original number. Factoring is to express a number as the product is called factors. Factors are numbers.

Multiplication: repeated addition of number is called multiplication for example   4* 3=12 it just add  4 times of  3 like 4+4+4=12 or 3 times of 4 (3+3+3+3=12) Let us see example of multiplying three factors.

Example1: multiplying three factors 20 * 6 * 18

Solution:

Multiply three factors:

Step 1:   20  *

6   

120

we multiply 20 *6 factors we can get 120 now we multiply 120 * 18.

Step2: 120 *

18  

1960

120  

2160

Step 3: Therefore multiplying three factors 20 * 6 * 18 is 2160

Example2: multiplying three factors 10 * 40 * 8

Solution: 

Multiply three factors

Step 1: multiply 10 * 40

10   *

40   

00

40  

400

If we multiply 10 *40 factors we can get 400 now we multiply 400 * 8.

Step 2: multiply 400 * 8

400 *

8 

3200

Step 3: Therefore multiplying three factors 10 * 40 * 8 is 3200.

Multiply three Factors Using Associative Property:

Associative property means the factors are same on either side of the equal sign and when we multiply two or grouping of factors does not change the product. (a * b) * c = a* (b * c) following example for multiply three factors.

solve:2* 6 * 3

Solution:

multiply 2 * 6 * 3

Step 1: 2 * 6 * 3 = 36

Step 2: apply associative property

2 * (6 * 3)

= 2 * (18)

= 36 when we multiply two or grouping of factors does not change the product.

Step 3: therefore multiplying three factors 2 * 6 * 3 is 36.

My Previous Blog :- http://learnmathsrightway.blogspot.in/2012/11/define-random-variable.html

How to Solve Significant Figures

Introduction how to solve significant figures:

The digits, which are used to represent the accuracy of numbers are said to be significant figures. Significant figures can also be called as significant digits. The concepts of significant figures are highly linked with rounding. The term significant figures can be abbreviated as sig figs, sign figs, and sign digs. In this section, we shall discuss how to solve significant figures. In this content we have an overview on identifying and solving significant figures.

Example for How to Solve Significant Figures :

Example problem 1:

How to Identify the number of significant figures from the following.

84, 0.084, 5.8480, 2005, 8400

Solution:

84 – The number 84 is a non-zero number that has two significant figures.

0.084 – Zeroes placed before the other numbers are not significant figures and the number of significant figures in 0.084 is two.

5.8480– Zeroes placed after the other numbers but after a decimal point are significant figures and the number of significant figures in 5.8480 is five.

2005 – The number 2005 has four significant figures because the Zeroes placed between the figures are always significant.

8400:

The number 8400 may have at least two significant figures. It also has three and four significant figures.

That is:

8400 – 8.4 x 103 has two significant figures.

8400 – 8.40 x 103 has three significant figures.

8400 – 8.400 x 103 has four significant figures.

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One more Example for How to Solve Significant Figures :

Example problem 2:

Solve the following addition 5.76 + 4.62 + 31.21 and find the number of significant figures in the solution.

Solution:

The given numbers are 5.76, 4.62, and 31.21.

The number 5.76 has three significant figures.

The number 4.62 has three significant figures.

The number 31.21 has four significant figures.

Adding the numbers, we get:

5. 7 6

4. 6 2

3 1. 2 1   +

4 0. 5 9

Therefore, the solution is 40.59.

40.59 can be rounded to 40.6

Therefore, the number of significant figures in 40.6 is three.

Point Symmetry and Line Symmetry

Introduction to point symmetry and line symmetry:
                Point of symmetry is a special center point for certain types of geometry symmetric objects. If an object or shapes can be rotated at 180 degree about a point P and end up looking an identical to the original, then the P is called as point of symmetry. A line which divides the object into their half and each figure has its mirror image is known as line symmetry.

The Explanation about Point Symmetry:

                   In geometry Point Symmetry is sometimes called as Origin Symmetry, because the "Origin" is the central point about which the figure is symmetrical. If shapes or graphs is rotated about a point by 180 degree and yet looks an identical to its original, that point is called as the Point Symmetry. If shapes have point symmetry, then its order of rotational symmetry must be 2.
Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Variables with Exponents and How do you Find the Degree of a Polynomial. I am sure they will be helpful.
Example of Point of Symmetry geometry:

                In the given shapes, an ellipse or rectangle is rotated 180 degree by a point. The ellipse or rectangle obtained after rotation same with the original rectangle. So, the point with which the ellipse or rectangle geometry shape rotated is called as the point symmetry.

Solving Real Analysis Problems

Solving Real Analysis Problems

Real analysis is a practice where raw record is arranging and organizing such that useful information will be extracting from it. The process of organizing about numbers and to understand what the data does and does not contain. Analysis classified into qualitative and quantitative. Qualitative analysis is involved in interpreting information which can be collected during the course of qualitative research. Quantitative analysis is involved in presenting and interpreting numerical numbers.I like to share this Anova Analysis with you all through my article.

Solving Real Analysis – Solving Example Problems

Solved real analysis example problems

Example 1: A pair of balanced dice is rolled, and what are the probabilities of getting the sum (1) 7 (2) 7 or 8 (3) 9

Solution:-

The sample space S = {(1, 1), (1, 2) … (6, 6)}

Number of possible outcomes n(S) = 36

Let A be the event of getting sum 7.

Let B be the event of getting the sum 8.

Let C be the event of getting the sum 9.

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, n(A) = 6.

B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, n (B) = 5

C = {(3, 6), (4, 5), (5, 4), (6, 3)}, n(C) = 4

(1) P (getting sum 7) = P(A) = n(A) / n(S)

= 6/36 = 1/6

Therefore P(7) = 1/6

(2) P (7 or 8) = P(A or B)

= P (A) + P (B)          (A and B are mutually exclusive i.e. AnB=f)

= 6/36 + 5/36

= 11/36

Therefore P (7 or 8) = 11/36

(3) P (getting sum 8) = P(c) = n(A) / n(S)

= 4/36 = 1/9

Therefore P (8) = 1/9.

Example 2: Find the sum of all integers, from 10 to 1000 inclusive, which are divisible by 10.

Solution:

Sequence of first few elements of integers divisible by 10 are given by 10, 20, 30, 40...

The above sequence has a first element equal to 10 and a common difference d = 10.

We need to know the rank of the term 1000.

We use the following formula for the nth term

an = a1 + (n - 1 )d

1000 = a 1 + (n - 1 )d

Substitute a1 and d by their values

1000 = 10 + 10(n - 1)

Solve for n to obtain

n = 100

1000 is the 100th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s100 = 100 (10 + 1000) / 2 = 50500.
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Solving Real Analysis – Solving Practice Problems

Solve these practice real analysis problems

Problem 1: When a pair of balanced dice is rolled, and what are the probabilities of getting the sum (1) 12 (2) 2 (3) 6 or 7.

Answer: 1) 1/36, 2) 1/36, 3) 11/36

Problem 2: Find the sum of all integers, from 11 to 1100 inclusive, which are divisible by 11.

Answer: 61105

Limit Rules Calculus

Introduction for limit rules for calculus:

In mathematics, the concepts of a “limit” are used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concepts of limit allow one to, in the complete space; define a new point from a Cauchy sequence of previously defined points. Limit is essential to calculus (and mathematical analysis in general) and is used to define continuity, derivatives and integrals.                                                                                                   (Source.Wikipedia)

Limit rules Calculus includes that differential calculus and integral calculus limits is used 

Rules for Limit Calculus:

(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 possessing limits and k is a constant 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)` [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).

(4) `lim_(x->a)`   xn - an/x - a = nan - 1 (a ? 0)

Examples for Limit Rules Calculus:

Example 1:

Evaluate 7`lim_(x->1)` (x3 - 1) / (x- 1)

Solution:

7`lim_(x->1)`x3 - 1 / x- 1

=7* 3(1)3 - 1                                     [limit rules:  `lim_(x->a)` `lim_(x->a)`   xn - an / x - a = nan - 1 (a ? 0)]

= 7*3(1)2

= 21

Example 2:

Find 8`lim_(x->0)`{(1 + x)4 - 1}/x

Solution:

Put 1 + x = t so that t ? 1 as x ? 0 and x= t -1

8`lim_(x->0)`(1 + x)4 - 1/ x = 8`lim_(t->1)`(t4 - 1)4/ (t - 1)           [limit rules:   `lim_(x->a)`   xn - an/x - a = nan - 1 (a ? 0)]

= 8*4(1)4-1 = 4*8=32

Example 3:

calculate the value n so that`lim_(a->2)`an - 2n/a - 2= 32

Solution: We have

`lim_(a->2)`an - 2n/a - 2 = n2n - 1                       [limit rules:   `lim_(x->a)`   xn - an/x - a = nan - 1 (a ? 0)]

? n2n - 1 = 32 = 4 × 8 = 4 × 23 = 4 × 2 4 - 1

Comparing on both sides we get n = 4

Example 4:

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

Solution:

We know that loge (1 + x) =`x/1` - `x^2/2` +`x^3/3` - …

loge `(1 + x) / x ` = 1 - `x/2 ` +`x^2/3` - …

Therefore  3`lim_(x->0)`loge` (1 + x)/x``lim_(x->0)`loge (`(1+0)/ 0` ) 

=3 loge 0

= 3*1=3.

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Calculus Integration Problems

Introduction to calculus integration problems: 

Integration calculus is the definition, properties, and applications of two related concepts, definite integral and indefinite integral. Finding the value of an integral process is called integration Methods. Integral calculus is two related linear operators.Indefinite integral is antiderivative, the inverse operation to the derivative.Definite integral inputs a function and outputs a number, gives the area between graph of input and x-axis. Limit of a sum of areas of rectangles is known as Riemann sum.

Calculus Integration Problems - Importance

Ancient period introduced ideas of integral calculus, have developed these ideas in a rigorous or systematic way. Manipulating very small quantities are usually developed Calculus.Infinitesimals are the first method of the calculus. dx is infinitesimal number could be greater than 0, but less than in the sequence 1, 1/2, 1/3, .... Infinitely small with the Integer multiple of an infinitesimal.    

Provided solid foundations for the manipulation of infinitesimals is introduce of non-standard analysis and smooth infinitesimal analysis.

Objective of integration calculus problems is settle on rate of change.

Main wing of integration calculus problems is

Integral calculus.
Integration calculus problems is decide function of  the rate of change.

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Calculus Integration Problems - Sample Problems

Problem 1:
Find the integral of the given equation

2x + exdx

Solution:

?2x + ex dx=?2x dx + ?ex dx

Integrating the above equation.

We get

2x2 / 2 + eX

x^2 + ex

Problem 2:

Find the integral of the given equation

x2+2x+3 dx

Solution:

?x2+2x+3 dx  = ?x2 dx + ?2x dx +?3 dx

Integrating the above equation

We get

= x3/3 + 2x2/2 + 3x

= x3/3 + x2 + 3x

Problem 3

Find the integral of the given equation

X4+2x2+3x dx

Solution:

?x4+2x2+3x dx  = ?x4 dx + ?2x2 dx +?3x dx
Integrating the above equation

We get

= x5/5+ 2x3/3 + 3x2/2

= x5/5 + 2x3/3+ 3x2/2

Volume of a Sphere Units

Introduction about volume sphere:

A sphere is a perfectly round geometrical object in three-dimensional space, such as the shape of a round ball. Like a circle in two dimensions, a perfect sphere is completely symmetrical around its center, with all points on the surface laying the same distance r from the center point. Let us see how to calculate the volume of sphere.

(Source – Wikipedia)

Examples Problems in Volume of Sphere Calculator:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic unit.

r = radius.

Volume of Sphere Units - Problems:

1. The sphere has the radius 20.3 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 20.3 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (20.3) 3

=`4/3` * 3.14 * 8365.427

Volume of the sphere (V) = 35023.25

2. The sphere has the radius 15.2 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 15.2 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (15.2) 3

=`4/3` * 3.14 * 3511.80

Volume of the sphere (V) = 14702.76 cm3

3. The sphere has the radius 11.25 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 11.25 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (11.25) 3

=`4/3` * 3.14 * 1423.82

Volume of the sphere (V) = 5961.09 cm3

4. The sphere has the radius 3.4 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 3.4 cm

Formula:

Volume of the sphere (V) = 4/3 p r ^ 3 cubic units

= 4/3 * 3.14 * (39.304) 3

=4/3 * 3.14 * 164.552

Volume of the sphere (V) = 164.552 cm3

5. The sphere has the radius 5.6 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 5.6 cm

Formula:

Volume of the sphere (V) = `4/3 ` p r ^ 3 cubic units

= `4/3` * 3.14 * (5.6) 3

= `4/3` * 3.14 * 175.61

Volume of the sphere (V) = 735.22 cm3

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Practice problems in volume of sphere:

1. The sphere has the radius 8.4 cm. Find the volume of the sphere.

Answer: 2481.45408 cm3

2. The sphere has the radius 6.6 cm. Find the volume of the sphere.

Answer: 1203.64992 cm3

Solving Linear Equations Substitution

Introduction for solving linear equations using substitution:

Linear equations substitution is nothing but a process of exchanging a variable in a linear expression with its actual value. In linear algebra the substitution process plays a major role to solve the system of linear equation by substituting the value for the given variable, linear substitution is mainly used to identify a variable in an expression and to find linear relationships between the equations. Here we use the linear substitution method to solve the different types of linear equations.

Having problem with Solving Systems of Linear Inequalities keep reading my upcoming posts, i will try to help you.

Solved Examples on Linear Equation Substitution

Ex 1:

Solve the linear equation by substitution method.

1 / (y - 1) 2 - 4 / (y - 1) + 4 = 0

Solution:

Let s = `1 / (y - 1)` and substitute this term in the given equation.

s 2 – 4s + 4 = 0

By solving the above quadratic equation, we get:
s = 2 and s=2.

Now substitute s by` 1 / (y - 1)` and solve for y
`1 / (y - 1)` = 2

1 = 2(y - 1)

1=2y - 2

3= 2y

Y = `3/2` is the solution for above equation.

Ex 2:

Solve the linear equation by substitution method.

z - 5 `sqrt (z)` = - 6

Solution:

Let s = `sqrt (z)` so that s 2 = z. Substitute z by s and `sqrt (z)` by s 2 respectively to obtain an equation in s.
s 2 - 5 s = - 6

The above equation looks like quadratic form, so rewrite the above term

s 2 - 5 s + 6 = 0

By solving the above equation we get
s = 2 or s = 3

We now substitute s by `sqrt (z)` and solve for z
`sqrt (z)` = 2 or `sqrt ( z )` = 3

z = 4 or z = 9  Is the solution for above equation.

My forthcoming post is on example of a algebraic expression, how to write an algebraic expression in words will give you more understanding about Algebra.

Practice Problems on Linear Equations Substitution for Solving:

1) Solve the linear equation by substitution method.

1 - 2 / (a - c) = 8 at c=1.

Answer:    a = 5/7.

2) Solve the linear equation by substitution method.

1 - 1 / (x - z) = -8 / (x 2 - z 2) at z=4

Answer:    x = -3.

Horizontal Line Segments

Introduction to horizontal line segments:
The line segments are distinct as the distance between two points. The line segments are make clear in another method is, the point that is joined the points of both directions. We can define a line PQ as `bar(PQ)` . Here we are going discuss about the horizontal line segment. Also we shall solve an example problem based on horizontal line segments.

Special Cases of Horizontal Line Segments:
Horizontal line segments have some special properties when compared to normal line segments,

Horizontal and vertical line segments are identified easily, that they have either x- value as similar or y value as same from both the given points of (x1, y1) (x2, y2).

In horizontal line segments, if y-value is similar then the line segment is easily identified as horizontal line segments.

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Example Problem for Horizontal Line Segments:

Plot the given horizontal line segment on the graph (8, 5) (-8, 5).

Solution:

Given: Two points to plot line segments are (8, 5) (-8, 5).

Two pairs of points are (x1, y1) (x2, y2) needed to plot the horizontal line segments in graph.

Here, from the given points x1 is 8 and y1is 5.

From the given points x2 is -8 and y2 is 5.

Now, from the given values of x co-ordinates and the values of the y co-ordinates, we can say that the values of y-coordinates are similar, so the line is horizontal line segments.

According, to the values of the x co-ordinate we have to plot the corresponding x value of the line segment and also on the y co-ordinate, plot the corresponding y value of the horizontal line segments.

These mentioned above, process is done by using graph is shown.

Thus, horizontal line segment is explained clearly and through diagrammatically is explained successfully.

Odd Square Numbers

Introduction to odd square numbers:

Odd square numbers are one of the basis for mathematics. The odd numbers are 1,3,5,7 etc. The formula for representing the odd square numbers are 2M+1, where m value is used to represent the any type of variable function. Simply the odd number can be defined as the number which are not divisible the number two. These numbers are called as the odd number.

Explanation for Odd Square Number

Odd numbers are the alternatives of the even numbers. The number that are not divisible by the number 2, 4, 6 ,8 etc are called as the odd number. For example, the numbers 3, 5, 7, 9, 11 etc are represented as the odd numbers. The diagrammatic representation of odd numbers are shown below,

The odd square numbers are represented by using the formula, (2M+1)2 . The formula can be simplified as 2(m2 + m) +1. By using the this formula the odd square number problems are solved.

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Example Problem for Odd Square Numbers

Pro 1: Find the odd square number for the given value. M = 4.

Sol :  Step 1: The formula for finding the odd square numbers is given below,

Odd square numbers = (2M+1)2 = `4M^2 + 8M + 1` = 4M(M+2) + 1

Step 2: The value of m is given. Therefore substitute the value in the formula.

Step 3: By substituting the formula we get,

= (2(4)+1)2

= 2(42 +4)+1

= 2(20)+1

= 40+1

= 41.

This is the required odd square number.

Pro 2: Find the odd square number for the given value. M = 6.

Sol :  Step 1: The formula for finding the odd square numbers is given below,

Odd square numbers = (2M+1)2 = 4(m2 + 2m) +1.

Step 2: The value of m is given. Therefore substitute the value in the formula.

Step 3: By substituting the formula we get,

= (2(6)+1)2 

= (12+1)2 = 13x13 = 169

This is the required odd square number.

Practice Problem for Odd Square Numbers

Pro 1: Find the odd square number for the given value. M = 8.

Ans : 196

Pro 2: Find the odd square number for the given value. M = 10.

Ans : 221

Slant Height Square Pyramid

Introduction to slant height of a square pyramid:

Pyramid is one of the shapes in geometry. A square pyramid is a general pyramid consist of square base. It is octahedron type.The lateral edge length and slant height, s of a right square pyramid of side length and height are In pyramid, the outer surfaces are triangular and converge at a point. The base of pyramid can be any shape like triangular, square, rectangular or of any polygon shape. Pyramid is classified into Volume of a Pyramid, square pyramid, and rectangular pyramid. All types of pyramid have three triangular faces and a base. Let’s see about basic three shapes of pyramid.

Square Pyramid Figure

Types of pyramid: 

Square pyramid
Triangular pyramid and
Rectangular pyramid.


Formula for Square Pyramid Slant Height:

Formula for finding the slant height of square pyramid:

s² = h² + (b/2)²
here s, slant height
h, height of pyramid
b, base length

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Example Problems in Square Pyramid Slant Height:

Ex 1:

The  square pyramid of height 24 cm. Find the slant height if the base edges are given as 14 cm.

Sol:
Formula for finding slant height will be  
h=24 and b=14 `b/2` =7
s² = h² +` (b/2)^2 `
s= `sqrt(24^2 + 7^2) `

= `sqrt(576 + 49)`

= `sqrt(625)`
= 25 cm

Ex 2:
The height of square pyramid 350 ft. and each side of  base is 646 ft. calculate the slant height length.

Sol:
given h=350 b=646 b/2=323
s² = h² + (b/2)²
s² = 350² + 323²
s² = 226829
s = 426.27 ft.

Ex 3:
The  square pyramid of height 32 cm. Find the slant height if the base edges are given as 12 cm.

Sol:
Formula for finding slant height will be  
h=32 and b=12 `b/2` =6
s² = h² +` (b/2)^2 `
s= `sqrt(32^2 + 6^2) `

= `sqrt(1024 + 36)`

= `sqrt(1060)`
= 32.55 cm
Ex 4:
The  square pyramid of height 16 cm. Find the slant height if the base edges are given as 4 cm.

Sol:
Formula for finding slant height will be  
h=16 and b=4 `b/2` =2
s² = h² +` (b/2)^2 `
s= `sqrt(16^2 + 4^2) `

= `sqrt(256 + 16)`

= `sqrt(272)`
= 16.5 cm

Solve Implicit Function or Relation

Introduction :

The function of implicit function is related to the variables. These two variables are given by an equation. This function has not been solved explicitly. A relation between the variables of function is said to be implicit function. For example   x^2 + y^2 = 100, y is an implicit function of x and x is an implicit function of y. In this article, we shall discuss about solve implicit function or relation.

Solve Implicit Function or Relation - Problems:

Solve implicit function or relation - problem 1:

Calculate the implicit function of x and implicit function of y in the given equation    -x^2 = - 5y  .

Solution:

Given equation is        -x^2 = - 5y.                      --------------(1)

Adding by  x^2 + 5y on both side, So we get

- x^2 + 5y +  x^2 = -5y.+ x^2 + 5y

+ 5y = + x^2                     --------------(2)

Now  divided by 5 on both sides,

`(5y)/(5)` = `( x^2) /(5)` .

Implicit function of x is                    y =  `( x^2) /(5)`.

Take equation (2)    ,         5y =  x^2 

Take square root on both sides,       `sqrt(5y)` = `sqrt(x^2)` .

` sqrt(5y)`   = x .

x = `sqrt(5y)` .

Answer:     y =  `( x^2) /(5)`.  is Implicit function of x .                       

x = `sqrt(5y)` . is Implicit function of y .                               

Solve implicit function or relation - problem 2:

The implicit function function is 10xy^2 - 5y^2  = 5. Evaluate  `(dy/dx)` .

Solution:

Given implicit function is    10xy^2 - 5y^2  = 5.

Now Find the derivative of  xy^2

`d/dx`(10xy^2)    = x 20y `(dy/dx)` + 10y^2 (1).

Find the derivative of 5y^2

`d/dx`(5y^2)    =  10y `(dy/dx)` .

Find the derivative of 5 (constant)

`d/dx`(5)    = 0.

So,                     10xy^2 - 5y^2  = 5.

20xy `(dy/dx)` + 10y^2 - 10y `(dy/dx)` . = 0

Subtract by 10 y^2 on both sides,

20xy `(dy/dx)` + 10y^2 - 10y `(dy/dx)` - 10 y^2 . = 0 - 10y^2

20xy `(dy/dx)` - 10y `(dy/dx)` .=  - 10y^2

Take `dy/dx` in common

` (dy/dx)` (20xy - 10y) = - 10y^2

Divided by (20xy - 10y) on both side so we get,

` (dy/dx)`` ((20xy - 10y)/(20xy-10y))` = `((- 10y^2)/(20xy-10y))`.

` (dy/dx)`  =  `((- 10y^2)/(20xy-10y))`.

Answer:  ` (dy/dx)`  =  `((- 10y^2)/(20xy-10y))`.

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Solve Implicit Function or Relation - Practice Problems:

Solve implicit function or relation - practice problem 1:

Find the implicit function of y in the given equation    x  = 2 - 3xy .

Answer:     Implicit function of y is               x  = `((2)/(1 + 3y))`         ..

Solve implicit function or relation - practice problem 2:

Find the implicit function of x in the given equation    3y  = 9 - 3xy .

Answer:     Implicit function of x is               y  = `((3)/(1 + x))`         ..