Study About Adjacent

Introduction to study about adjacent

Adjacent is an adjective meaning contiguous, adjoining or abutting. In geometry, adjacent is when sides meet to make an angle. In trigonometry the adjacent side of a right angled triangle is the cathetus next to the angle in question. In graph theory adjacent nodes in a graph are linked by an edge. (Source: From Wikipedia)

Here we are going to study about adjacent. I like to share this Adjacent Angles Definition with you all through my article.


Study about adjacent sides

Here we are going to study about adjacent sides of polygons. In a two dimensional shapes, like triangle, square or rectangle, two sides which share a common vertex and angle is called as adjacent sides.

If the adjacent sides of a polygon are in same length, then it is called as a square. In a rectangle, the adjacent sides are not equal but, opposite sides are equal. The angle between adjacent sides in a square and a rectangle is 90 degrees.

If the adjacent sides of a triangle are equal and it is called as an isosceles triangle.

Example problem

If the lengths of adjacent sides of a rectangle is x and x+5 meters respectively. The perimeter of the rectangle is 34 meter. Find the area of the rectangle.

Solution

The perimeter of the rectangle = 2(l + b)

Here the length and breadth of the rectangle is given by it's adjacent sides, x and x+5.

so, 34 = 2(x + x + 5)

34 = 2(2x + 5)

34 = 4x + 10

34 - 10 = 4x - 10

24 = 4x

x = 6

So, the adjacent sides of the rectangle are 6 and 11 meter respectively.

Area of the rectangle = lb square meter

= 6 * 11 square meter

= 66 square meter.

So the area of the given rectangle is 66 square meter. Understanding Find an Angle is always challenging for me but thanks to all math help websites to help me out.


Study about adjacent angles


Here we are going to study about adjacent angles.

Adjacent angles are nothing but, two are more angles which shares a common vertex.
If the sum of two adjacent angles is 90 degree, then the angles are called as complementary angles.
If the sum of two adjacent angles is 180 degree, then the angles are called as supplementary angles.
Example problem

If two adjacent angles are complement to each other, and one angle measures 48 degree, find another angle.

Solution

Given that the adjacent angles are complement, that is the sum of the adjacent angles is 90 degrees.

One angle measures 48 degree.

Measure of another angle = 90 - 48

= 42 degree.

Algebra Study Guide

Introduction to algebra study guide:

Algebra, a branch of mathematics deals with the expressions of various order, their operations, relations and construction and concepts of expressions.Algebra is the combination of numerical numbers, alpha-numeric variables, expressions and etc., the basic aim of the algebra is balancing the algebraic equations on both sides.In algebra study guide you can learn about the various expressions and order of expressions, with their standard forms. I like to share this Adding Polynomials with you all through my article.



General concepts of Algebra – algebra study guide:


Quartic function:


The Quartic function can be defined as the algebraic expression of fourth order. That is the highest power in the equation is equals to 4.

It is an algebraic equation with order number 4.

The general form is given by,

f(x) = ax^4 + bx^3 + cx^2 + e.

thus, the quartic expression from algebra study guide.

Quintic equation:

The Quintic equation can be defined as the algebraic expression of fifth order. That is the highest power in the equation is equals to 5.

It is an algebraic equation with order number 5.

The general form is given by,

ax5 + bx4 + cx3 + dx2 + ex + f = 0.


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Polynomial:

The Polynomial expression can be defined as the algebraic expression of second order. That is the highest power in the equation is equals to 2.

The polynomial equation is of the form,

x^2 – 10x + 7 = 0

Note:

The addition or multiplication of any two polynomial results another polynomial.

These are the various forms of algebraic equations from algebra study guide.

Experiment Learning

Introduction experiment learning:-

In this article we are learning about the experiment concept. A learning probability experiment is a condition where chances affect the outcome (the result of a learning experiment). A coin flip is a learning probability experiment because chance affect whether a coin will ground heads or tails when it is flipped. In learning probability, an experiment is a procedure with an outcome that deepens leading chance. Some learning examples of such experiments are given below. Is this topic Experimental Probability Definition hard for you? Watch out for my coming posts.


Example for learning experiment:-


Learning Experiment 1:-

Toss the coin. But the coin is fair, (that is balanced or unbiased) the chance of in receipt of a head is one chance in two. The chance of in receipt of a tail is also one in two.

Solution:-

In conditions of probability, we say that each outcome is evenly likely to occur and that the probability of tossing a head is one dived by two.

Learning Experiment 2:-

Ten cards number one to ten are placed in pack and then a card is drawn at random. It is even that is some of the ten cards will be selected.

Solution:-

There are four prime numbers and so the chance of choose a card bearing a prime number is four in ten hence, the probability of select a prime number is four dived by ten or two divide by five.

Learning Experiment 3:-

A die is fearful and the number on the highest face in noted. It is uniformly likely that any of the 8 numbers will show on the top. In other words the likely outcomes are 1, 2, 3, 4, 5, 6,7and. The set of these outcomes is called the example space for the experiment.

Solution:-

There are 4 odd numbers and 4 even numbers. so the chance of an odd number being on the peak face of the die is 4 in 8 (as there are 8 numbers in the sample space). So we say that the probability of obtain an odd number on the summit is 4 in 8:   P (odd number) =4/8.

Learning Experiment 4:-

Believe one more random experiment. This time two dice are thrown at once ant the numbers on the top are noted. There are 16 probable outcomes that are; there are 16 elements in the sample space.

Solution:-

By inspecting the following table, we see that there are 4 chances of getting a double. So the probability of getting on the uppermost faces is 4 in 16 P (double) =4/16.

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Practices problem for learning experiment:-


Learning Experiment Problem1:-

During Coin tossing, Toss a fair coin once and ask what the probability of heads is?

Answer:- one dived by two.

Learning Experiment Problem 2:-

A pack contains 124 birth certificates, of which 48 are of male births and 78 of female births. Suppose that a credential is drawn at random from this pack. What is the probability that the certificate tired will be that of a male birth?

Answer:- The probability of drawing a female’s credential is P (F)=78/124.

Learn Harmonic Mean Online

Introduction to learn harmonic mean online:
Harmonic mean is used to find the average of the given terms. Harmonic mean is defined as the number of given value divided by the inverse of the each values given. For calculating the harmonic mean, we are using one of the formula. By using the formula, we are calculating the harmonic mean.


Explanation to learn harmonic mean online

The explanations to learn harmonic mean is as follows,

Online has number of examples about the harmonic mean with suitable solutions.
All the examples given in online are worked out with step by step solutions.
Also the practice problems for the harmonic mean are very helpful to the students.
The formula given for calculating the harmonic mean is as follows,

Harmonic Mean =  `N/( (1/a_1) + (1/a_2) + (1/a_3) + .......(1/a_n))`

Where,

N = Number of terms in the given functions.
`1/a_1`   = Inverse of each terms in the given functions. Understanding Interesting Data Sets is always challenging for me but thanks to all math help websites to help me out.

Example problems to learn harmonic mean online

Problem to learn in online:

Problem 1:  2, 4, 8, 16. What is the value of the harmonic mean?.

Solution:

Step 1: Given:

Data  = 2, 4, 8, 16

N = 4

Step 2: Formula:

Harmonic Mean =  `N/( (1/a_1) + (1/a_2) + (1/a_3)....(1/a_n))`

Step 3: Simplify:

Harmonic Mean =  `N/( (1/a_1) + (1/a_2) + (1/a_3) + (1/a_4) )`

= `4/ ( (1/2) + ( 1/4) + (1/8) + (1/16))`

Step 4: Solve:

=  `4/((8 + 4 + 2 + 1)/16)`

= `64/(8 + 4 + 2 + 1)`

= `64/15`

= 4.2666

Result: Harmonic Mean = 4.2666

Thus, this is the required answer for the above given harmonic function.

Problem 2: 1, 3, 6, 8, 12. What is the value of the harmonic mean?.

Solution:

Step 1: Given:

Data  = 1, 3, 6, 8, 12

N = 5

Step 2: Formula:

Harmonic Mean =  `N/( (1/a_1) + (1/a_2) + (1/a_3)....(1/a_n))`

Step 3: Simplify:

Harmonic Mean =  `N/( (1/a_1) + (1/a_2) + (1/a_3) + (1/a_4) )`

= `5/ ( (1/1) + ( 1/3) + (1/6) + (1/8) + (1/12))`

Step 4: Solve:

=  `5/((24 + 8 + 4 + 3+ 2)/24)`

= `5/((41/24))`

= `120/41`

= 2.9268

Result: Harmonic Mean = 2.9268


Practice problems to learn harmonic mean online

Practice problems to learn in online:

Problem 1: 2, 6, 12, 24, 26. What is the value of the harmonic mean?

Answer: 6.027

Problem 2: 3, 5, 7, 9, 12. What is the value of the harmonic mean?

Answer: 5.742

Tutoring in Math

Introduction to tutoring in math:

Some times some subjects needs an extra effort to score a good grade in them and maths is one of the subject which is always a challenge for students. And tutoring is one of the solution  for that . But there are some limitation of class room tutoring in math like time limitation and limited availability of tutors. But here we have solutions of all these issue in our website tutorvista.com . Here we provide tutoring in maths with the help of a fully developed environment over Internet and experienced tutors. You can take a 10 minutes demo session of our services and in that demo session you can ask any of your doubt. I like to share this online math tutors for free with you all through my article.


Online Solution for math


In today's world where online education is becoming so popular we also provide you with online tutoring in math and in many more subjects. We provide you with so much interactive environment which makes you feel like you are in a class room. Where you can ask any of your doubt. You can have unlimited session of 45 to 60 minutes every day with a constant charges. We have team of expert tutors who are always ready to help you i mean 24 hours.

There other advantages also like you need not to go anywhere else for the tuition , so your time will save , which you can use in studies.  You can get revision of any topic to any number of times until you start feeling comfortable with the topic. Understanding Perimeter of a Square is always challenging for me but thanks to all math help websites to help me out.


Other Advantages of tutoring in math


After subscribing to us you can have any number of session to our website with our tutors. And if you like any of tutors method of teaching then you can also book a session with him. And we assure you that once you will subscribe to us your grades will improve. We have Course material also for students to practice and you can get help on any day whether it's a holiday or any other day, we are always ready to help you with math. we also provide help in other subjects like science, English. With our tutors you can discuss your homework and get solved with a clear understanding of the relevant topic.


All you just need


For having all this advantages you just have to log in to our portal and have a demo session with us and then you can see the difference. and after subscription you can other advantages also like guidance for other competition exams and can learn maths even if the schools are of. We are the most experienced and most professional organization in online tutoring and we assure a balanced return of your investment and your grades will be improved which is more then anything.

Thanks

Set Theory Math

Definition:

Set theory is one of the branches of math that studies sets, which  there are collections of objects. Set theory is applied to objects that are relevant to math. Set theory begins with the binary relation between an item o and a set A. If o is a member or element of the A, since sets are defined as the objects, then the relationship can relate sets as well. Please express your views of this topic Absolute Value Calculator by commenting on blog.



Set theory basic operations

Subset relation is a derived binary relation exists connecting the two sets which is known as the set inclusion. If all the members of sets A are the members of set B, then A is a said to be the subset of B, which is denoted as A U B.

Union: In math, the union of the sets A and B, denoted A U B, is the set of all objects that are a member of A, or B, or can be both. The union of {1, 2, 3} and {2, 3, 4} in this set A U B is  set {1, 2, 3, 4}.
Intersection: In math, the intersection of the sets A and B can be denoted as A ∩ B in the cases that the set of all objects are the members of both AB. Hence the intersection of {1, 2, 3} and {2, 3, 4} is the set {2, 3}. Understanding Ask Math Questions is always challenging for me but thanks to all math help websites to help me out.

Set theory complement

Complement :In math, Complement set A is related to the set U, which is denoted Ac, and that is the set of all members of U, that are not members of A. This expression is most commonly employed when the universal set states as U, as in the study of Venn diagrams. This operation is also mean to be as the set difference of U and A, such that can be denoted by U/A. The complement of the set {1,2,3} relative to {2,3,4} is {4} conversely, the complement of {2,3,4} relative to {1,2,3} is {1} .

Simultaneous Equations

In this page we are going to discuss about simultaneous equations concept. A set of independent equations in two or more variables is called a simultaneous equation.   To find the value(s) of the variables for which the set of equation holds is called solving the equation. There are different methods used to solve simultaneous equations. In this article, we will learn how to solve the simultaneous equations by Substitution and Elimination methods with examples.

The general form of the simultaneous equation can be written as,

AX + BY = C,

DX + EY =  F.


Elimination Method to solve simultaneous equations


Below mare the example on elimination method to solve simultaneous equations -

Examples:

Solve the simultaneous equations examples using Elimination method 4X + 5Y = 12, 3X - 5Y = 9.

Solution:

The given two equations are,

4X + 5Y = 12       -----------------(1)

3X - 5Y =    9       -----------------(2)

Step 1: Here the coefficient of one of the variables numerically equal and also opposite signs. So

Add the two equations for eliminating Y variable

Step 2: Add the like terms, (4X + 3X) + (5Y - 5Y) = 12 + 9.

Step 3: After the addition and elimination of Y, rewrite the equation as

7X = 21

Step 4: Divide 7 on both side of the equation

7X / 7 = 21 / 7

Step 5: After dividing, we get X = 3.

Step 6: Substitute the X value in the first equation

4(3) + 5Y = 12,

12 + 5Y = 12.

Step 7: Subtracting 12 both sides in the equation. We get Y value,

5Y = 12 - 12,

5Y = 0,

Y = 0.

Step 8: The solution for this equation is X = 3, Y = 0. Please express your views of this topic Decimal Multiplication by commenting on blog.


Substitution method to solving simultaneous equations


Below are the example on substitution method to solving simultaneous equations -

Examples:

Solve the simultaneous equations examples using substitution method Y= X + 2, X + Y = 4.

Solution:

Given,               Y   = X + 2  --------------(1)

X + Y = 4         --------------(2)

Step 1: Substitute the value of Y into the second equation. We get,

X + (X + 2) = 4

Step 2: Solve X

2X + 2 = 4    (subtract 2 both sides)

2X = 4 - 2

2X = 2, (divide 2 both sides)

X = 2/2,

X = 1.

Step 5: Substitute the value of X into the first equation.

Y = X + 2,

Y = 1 + 2,

Y = 3.

Step 7: The answer is X = 1 and Y = 3.

Pythagorean Learning

Pythagorean Theorem states that, “In any right angle triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle)”.

Pythagorean Theorem is applied only, when the given triangle is a right angle triangle. I like to share this Pythagorean Identity with you all through my article.


General equation of Pythagorean Theorem

The theorem can be written as an equation,

c^2 =a^2+ b^2

Where c denote the length of the hypotenuse, and a and b represent the lengths of the other two sides.

If we let 'c' be the length of the hypotenuse and 'a' and 'b' is the lengths of the other two sides, the theorem can be expressed as the equation:

c^2 =a^2+ b^2

Or, solved for c:

c=square root of (a^2+b^2)

If c is given, and the length of one of the side must be found, the following equations (which are corollaries of the first) can be used:

c^2 -a^2= b^2

Or

c^2 - b^2 = a^2

This equation gives a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. An overview of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle it will be reduces to the Pythagorean Theorem. Understanding andhra pradesh board of secondary education is always challenging for me but thanks to all math help websites to help me out.


How to solve the right triangle using Pythagorean


Example:

Find the length of a side 'a' in the right triangle, if b = 9 and c=15.

Solution:

We know that formula,

c^2 =a^2+ b^2

We need to find the value of 'a' so,

a^2=c^2- b^2

a^2= (15)2 – (9) 2

a^2 = 225 -81

a^2=144

Take square on both sides,

a =12

The side a value is 12.

Pre Algebre Test

Introduction to Pre algebre test

Algebra is an important part of mathematics and the pre algebra test is the basic operation is involves in mathematics . Here the topic discuss about the usual terms in algebra like addition, subtraction, multiplication and division. The pre algebre test makes very easy for whom studying the algebra.Let we discuss about the pre algebre test problems. Is this topic Polynomial Factorization hard for you? Watch out for my coming posts.

Pre Algebre Test Problems

Pre algebre test 1:

Determine the value of x in the equation x - 6 = 12
Solution:

Add 6 to both sides of the equation:

x – 6 + 6 = 12 + 6

x = 18

The required solution is x = 18.

Example based on Proof:
6(-3x - 2) - (x -5 ) = -6 (3x + 5) + 13

Multiply the factors-18x - 12 - x + 5 = -18x - 20 +13

Group like terms

-18x - 7 = -18x - 7
Add 18x + 7 to both sides 0 = 0
Hence the prooved.

pre algebre test 2 :

Determine: 6(c – 2) + 5 = 8

Solution:

Step 1: Remove the brackets

6c – 12 + 5 = 8

Step 2: Isolate variable a

6c = 12 – 5 + 7
6c = 14
c=16/14 = 3.4

Answer: c = 1.42

pre algebre test 3:

Addition and subtraction
1) Solve:   x + 62 = 262

Solution:

x + 62 = 262

subtract  62 on both sides

x + 62 - 62 = 62 -262

x = 200

pre algebre test 4:

Solve:   y - 82 = 756
Solution:

y - 82 = 756

(Add 82 on both sides)

y - 82 + 82 = 756 + 82

y = 838.

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More Problems in Pre Algebre Test

Pre algebre test 1:

Multiplication and division
Solve:   7y = 49

Solution:

7y = 49

Divide by 7 on both sides

(7y) / 7 = 49 / 7

xy= 7

pre algebre test 2:
Solve:   y / 8 = 64

Solution:

y / 8 = 64

multiply by 8 on both sides

8(y / 8) = 64(8)

y = 512

pre algebre test 3:

Combination of operations
1) Find:   6y - 6 = 36

Solution:

6y - 6 = 36

6y - 6 + 6 = 36 + 6

6y = 42

(6y) / 6 = 42 / 6

y = 7

Pre algebra test 4:

Solve:   4(y + 8) = 32
Solution:

4(y + 8) = 32

[4(y+ 8]/4 = 32/8

y + 8 = 4

y + 8 - 8 = 4- 4

y = 0

Practice Problems in Pre Algebre Test

1) Solve:  x + 72 = 64

Answer: x = 8

2)Solve:   y - 16 = 99

Answer: y = 115

3)Solve:   y / 9 = 82

Answer:   y = 738

4) Find:   2y - 2 = 4

Answer: y = 3

Rounding to the Nearest Cents

Introduction to rounding to the nearest cents:

This article is used to discuss about rounding to the nearest cents.  When we round nearest to cents, we have to consider the third digit after the decimal point. For example if you have the value of  `$` 1.1487 and if we want to round to the nearest cent we can take the third digit after the decimal point. that is 8, so the result is `$` 1.15. Let we see some problems for rounding to the nearest cents. I like to share this Rounding Decimals with you all through my article.

Example Problems for Rounding to the Nearest Cents.

Problem 1:

Round the given value to the nearest cents, $4.7846

Solution:

Here we need to round to the nearest cents

Here we have dollar as =4 and cents as .7846

If we round to the nearest cents, we get

Step1) take the third value after the decimal point,

Step2) we have the third value is 4, here the number 4 is less than 5. so the place value will not change

Step3) The round off value is $4.78

Problem 2:

Round the given value to the nearest cents, $ 0.168

Solution:

Here we need to round to the nearest cents

Here we have dollar as =4 and cents as .168

If we round to the nearest cents, we get

Step1) take the third value after the decimal point,

Step2) we have the third value is 0, here the number 8 is greater  than 5. so the place value will  change to the next value as 7.

Step3) The round off value is $0.17

Problem 3:

Round the given value to the nearest cents, $ 25.567

Solution:

Here we need to round to the nearest cents

Here we have dollar as =4 and cents as .567

If we round to the nearest cents, we get

Step1) take the third value after the decimal point,

Step2) we have the third value is 25, here the number 7 is greater  than 5. so the place value will  change to the next higher value as 7.

Step3) The round off value is $25.57


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Practice Problems for Rounding to the Nearest Cents:

Try to solve the following practice problems

Practice problem 1:

Round to the nearest cents, $564.456

Ans: $564.46

Practice problem 2:

Round to the nearest cents, $35.5766

Ans: $35.58