algebra word problem solver free

In the previous post we have discussed about how to do pre algebra and In today's session we are going to discuss about algebra word problem solver free. We come across many real life problems, which are solved by the help of forming the equation. These equations are formed by considering the variables and then forming the required equations based on the given relations.

We have Algebra Word Problem Solver Free online which can be used to understand the concept how the relations are expressed in the form of the expressions and how the equations are formed. We also use the online Algebra Word problem solver to learn how the formed equation can be solved. The equation solution simply means to find the value of the variable. The value of the variable which we get by solving the equation is the solution to the given equation. This value when placed in the place of the variable in the given equation, will satisfy the equation .

There are different methods used to solve the algebraic equations. One of the very common method of solving the equation is by the Hit and Trial method. By Hit and Trial method we mean that the different values are placed in place of the variable in the given equation. We need to search the values which when placed in the equation, will satisfy the equation. Let's us look at the following problem:

Find the number which when added to 5 results 9.

Here let's us assume that the number is x. We form the equation as follows :

X + 5 = 9.

Now we say that let the value of x = 1, so on putting the value of x = 1, we get :

1 + 5 <> 9.

So it does not satisfy the equation

Similarly we go on trying different values and finally if x = 4, then

4 + 5 = 9

So, x = 4 is the solution to the equation.

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how to do pre algebra

In the previous post we have discussed about how many hours in a year and In today's session we are going to discuss about how to do pre algebra. Pre algebra can be considered as a part of algebra which contains all basic concept of algebra. In school, the study of pre algebra starts from grade 5th that continue to grade 8th. The basic reason behind the study of pre algebra is to make the students ready for handling the mathematical algebra. Sometime students face trouble while they are going to solve algebraic equations. The basic reason behind this problem is that students are not able to clearly understand the concept of algebra, at that pre algebra play their role to make the students ready to face the problem and find the solution of their problem on its own.
Here in this section discussion being held on the topic of how to do pre algebra. As the name of topic describes that how a student is being able to solve a problem of pre algebraic equations. To deal with this topic students need to cover some topic that are study the fundamental of number system and their types like integer, fractional number or the concept of factor of number and properties of operations like commutative, distributive and so on. There are so many other concept that also helps in understanding the concept of pre algebra and after that algebra like:
I ) Understand the order of addend means x + ( y + z ) = (x + y ) + z.
II ) Adding any number to zero gives the same output number.
III ) If two variables ate same to each other then their other tasks are also same.

The concept of Rationalizing the Denominator is a process which is perform when we want to convert root of denominator into their numerator value. The 10th cbse sample paper is a similar kind of board paper that helps the students to evaluate their exam preparation.  

how many hours in a year

The hour and year both can be consider as a unit of time. The Hour which is sometime can be referred by hr. It is a unit of time which is equal to the 1 / 24 part of a day, 60 minutes and 3600 second. Day, minutes and seconds are also a unit of time. Normally, time is a important part of life, on the basis of time we decide our lots of work in a real world life. Generally time is measured on a pattern of two twelve hour segments of a day. The first segment of is denoted by AM and second segment is denoted by PM.

On the other side Year can be describe as a time of the earth which is taken to make one move around the sun. According to various scientist’s, a year is equal to the 365 days and 6 hours but remaining 6 hours can’t be calculated in a year. After each 4 year, one year become the year of 366 days because 6 hours of each year make a one day which is added into every 4thyear. Every 4^th year is equal to 366 days popularly known as leap year.

Here in this section discussion held on how many hours in a year. After discussing about hour and year we can easily elaborate the answer of above question. As we know that one year is equal to 365 days and one day is equal to 24 hours. So, we can easily calculate that how many hours in a year in the below given manner.

Hours in a year = 24 * 365 = 8760 hours

If we want to calculate hours in a leap year then it could be 8784 hours in a year.

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radical calculator

In the previous post we have discussed about How to Graph an Equation and In today's session we are going to discuss about radical calculator. In mathematics, we will see many types of expression such as polynomial expression. Here we will discuss the concept of radical calculator. An expression which is having root values, such as square roots, cube roots are known as radicals. For example: √ (r + s), and 3√ (r + s). The value of 2 means square root, 3 means cube root and so on. Radicals are denoted by the symbol '√'. Now we will see radical calculator.

Radical calculator can be defined as a online machine which is used to add the multiple values within a seconds. Now we will understand some steps that are used to solve the radical values.

Step 1: Let we have to add two radical values then put two root values in two text box.

Step 2: Then enter solve button to get desired result. Now we will discuss it with the help of example:

For example: Let we have to add two unlike radicals. 2 √7 + 4 √9 + √7 + 5 √9.

Solution: Here we need to see some steps so that we can easily add the radical values.

Step 1: Given radical expression is 2 √7 + 4 √9 + √7 + 5 √9, Now we have to add radical values given in the expression.

Step 2: Now we have to find the common term if present in the given radical expression.

In this expression two pairs are same. So we can write them as:

= 2 √7 + 4 √9 + √7 + 5 √9,

= 2 √7 + √7 + 4 √9 + 5 √9,

Now find common term in expression. (know more about radical calculator, here)

= (2 + 1) √7 + (4 + 5) √9, Now add the radicals we get. On adding these values we get;

= 3 √7 + 9 √9. In this way we can easily add radical.

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How to Graph an Equation

Hello friends, in mathematics, we will study different types of equation such as quadratic equation, linear equation. Here we will see How to Graph an Equation. When we are going to plot a graph of the given equation then if we get straight line then the equation is said to be linear equation. For example: y = mx + c; here ‘m’ denotes the slope of a line and ‘c’ denotes Y- intercept where the line crosses the y- axis. Now we will see process of solving systems of linear equations. Here we have to follow some steps to solve the equation. (know more about Graph, here)

Step 1: First we take two equations. Let we have linear equation 2u + v = 12 and 4u – 3v = 6,

Step 2: Then find the value of one variable and put this variable in second equation. Here in this given equation for ‘v’, we cannot solve for ‘u’. So we can write the first equation as:

⇒2u + v = 12;

⇒v = 12 – 2u; so we find the value of ‘v’. Now put the value of ‘v’ in the second equation. On putting the value of ‘v’ in the equation we get:

⇒4u – 3v = 6; put value of ‘v’ to find the value of ‘u’.

⇒4u – 3 * (12 – 2u) = 6, on further solving we get:

⇒4u – 36 + 6u = 6;

Now we add like terms if present in equation;

⇒10u = 42;

⇒u = 4.2.

Now put the value of ‘u’ in 1st equation.

⇒v = 12 – 2u;

⇒v = 12 – 2 * 4.2;

⇒v = 3.6;

So, the value of ‘u’ and ‘v’ is 4.2 and 3.6, if we put these value in the graph then we get a straight line graph. This is how we can solve the equation.

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