Unit Circle

In the previous post we have discussed about one to one correspondence and In today's session we are going to discuss about Unit Circle. Unit Circle is defined as a circle that is having radius value is equal to 1. Let's us see the steps of making a this circle. Steps of making a circle is given as:

Step 1: To construct a circle it is very essential to have a radius value is equal to one always. If it is not so then it is not a unit circle.

Step 2: Now draw a tangent and solve equation with the help of Pythagoras theorem.

In mathematical geometry, Equation of circle (unit) is given by: i2 + j2 = 1, here ‘i’ plot the coordinate value along to x – axis and ‘j’ plot the coordinate value along to y – axis.

Now we will discuss the table based on circle. The table is shown below:

s.no	Ó¨ (rad)	Ó¨0	Sin Ó¨	Cos Ó¨	tanÓ¨ = sinÓ¨ / cos Ó¨
1	0 Л / 6	0 30	√0 / 2 = 0 √1 / 2 = 1 / 2	√4 / 2 = 1 √3 / 2	√0 / √4 = 0 √1 / √3 = √3 / 3
2	Л / 3	60	√3 / 2	√1 / 2 = 1 / 2	√3 / √1 = √3
3	Л / 2	90	√4 / 2 = 1	√0 / 2 = 0	----

Using these value we can solve any problem related to unit circle.

Some properties are also based on circle which are given as:

Distance assess from center of circle to any point on a circle is radius of circle. Radius is always half of diameter.

Line that is passing passes through center of circle is diameter of circle. Diameter of circle is always twice the radius value of circle.

Circumference – Formula to calculate circumference of circle is given by: Circumference of circle = 2 лR.

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one to one correspondence

In the previous post we have discussed about algebra quadratic equations and In today's session we are going to discuss about one to one correspondence. In this blog we will see discuss one to one correspondence. One – to – one correspondence is a process in which a condition in which elements of one set (Let a set A) can be properly (or evenly) matched with elements of second set (other set B). Here the meaning of this word evenly is each element of set 'A' relates to one and only one member of set 'B' and each element of set 'B' relates to one and only one member of set 'A'. It means each element of set 'A' is connect with exactly one element of set 'B' and vice versa. Now we will understand the detail of one to one correspondence. If we understand the terms of order pair (x, y) where 'x' is a element of set 'A' and 'y' is an element of set 'B'. Here two orders are not possible for this condition that has first element same and two order is not correct for same element. If this type of condition stable in a set than it shows one – to – one correspondence between sets A and B.

In other words, if two sets have same cardinality than one – to – one correspondence stable among two sets. Let’s have small introduction about one - to - one function. Basically one - to – one function is taken to check whether one – to – one correspondence stable among infinite sets.

Let's we have given a function and if function is one – to – one then one – to – one correspondence lie among the set of positive integers and set of odd positive integer. We can also calculate one – to – one correspondence between rational numbers and integer numbers, (any number represented as ratio of two whole numbers is called as rational number) but we can not calculate one – to – one correspondence among real numbers and integers.

Pythagorean Triples List contains with three positive integers p, q, and r, such that p² + q² = r². Before attempting the 12th board exam please solve cbse sample papers 12.

algebra quadratic equations

In the previous post we have discussed about Factoring Polynomials and In today's session we are going to discuss about algebra quadratic equations. In mathematics, algebra quadratic equations can be defined as an equation that has highest degree equals to 2. In other words it can be defined as an equation that has highest power is a square not more than two. If any expression has highest power more than 2 then it is not quadratic equation. Quadratic equation can be written as: pt² + qt + r = 0. Formula to solve algebra quadratic expression is given as:

⇨ t = - b + √ (b² – 4ac) / 2a.

Let’s talk about how to write a quadratic equation if roots value are known. Here we will follow some steps to write a quadratic equation.

Step 1: First of all we take two roots of an equation to write quadratic equation. Let we have two roots of an equation that is 5 and -7.

Step 2: Then we have to put roots in given form of q = (p - a) (p – b), here we put one root for variable ‘a’ and other root for next variable ‘b’. Put both roots in given form:

q = (p - a) (p – b), put a = 5 and b = -7,

So, it can be written as:

q = (p - 5) (p + 7),

Step 3: Multiply variable ‘p’ with all value of next pair and apply same procedure for second value. So it can be written as:

On multiplying we get:

q = (p - 5) (p + 7),

q = p² + 7p – 5p – 35.

Step 4: Now we have to combine the same terms if present in equation otherwise this is required solution. So in above expression we have two like terms. On combining the equation we get:

q = p² + 2p – 35. This is required quadratic form. (know more about Quadratic equation, here)

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Factoring Polynomials

In mathematics, any expressions which is join with constants, variables and exponent values is said to be polynomial expression. And also polynomial expression is join with together by mathematical operators like (+, -, *, /). Infinite values are not taken in case of polynomials expression. For example: 12xy² – 4x + 7y³ – 20, this given equation is polynomial equation, in this equation exponents values are 0, 1, 2 and 3. Negative and fraction values are also taken in case of polynomial expression. It is not joined with together by division operator.

Let’s discuss that how to solve Factoring Polynomials. Here we will understand the quadratic to calculate polynomial expressions.

Let we have a polynomial expression 2p² + 4p – 10, we can factorize this polynomial as shown below:

We will find its factor by quadratic formula. Formula to find factors is given by:

P = -b + √ (b² - 4ac) / 2a, here value of 'a' is 2, value of 'b' is 4 and value of 'c' is -10. So put these values in formula. On putting these values we get:

P = - 4 + √ [(4)² - 4(2) (-10)] / 2(2); on moving ahead we get,

P = - 4 + √ (16 + 80) / 4, we can also write it as,

P = - 4 + √ (96) / 4. So, here we get two factor of this expression, one positive and other negative.

P = -4 + √ 24 and P = -4 - √ 24.

These two are factors of above expression. With the help of quadratic formula formula we can find factors of any polynomial expression. (know more about Factoring Polynomials, here)

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