Trigonometry Radian Measure

Introduction to trigonometry radian measure:

An angle is determined by rotating a ray about its endpoint. One way to measure angles is in radians. To signify that a given angle is in radians, a superscript c, or the abbreviation rad might be used. If no unit is given on an angle measure, the angle is assumed to be in radians. `(3pi^c)/2-=(3pi)/2 rad.-=(3pi)/2`                                                                                                                            (Source: Wikipedia)

Trigonometry Radian Measure

Middle angles of a circle contain an angle measure of 1° if it subtends an arc to be 1/360 of the boundary of the circle. These appearances of angle determine is reasonably general. Another appearance of angle determine namely in utilize is radian measure. If a middle angle subtends an arc i.e. identical toward the radius of the circle after that the middle angle contain a computing of one radian.

If a middle angle ? of a circle through radius r subtends an arc of length s, subsequently their radians establish is describing since  `theta= s/r`

Given, radius is 4 cm, and length of arc is 60 cm.

We know that the formula for radian measure of `theta=s/r` .

As a result,`theta = 60/4`

?=15

The angle of the arc is 15 radians.
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Examples for Trigonometry Radian Measure

Example 1 for trigonometry radian measure

Calculate the angle of the arc, if the radius is 6 cm, and length of arc is 120 cm

Solution:

Given, radius is 6 cm, and length of arc is 120 cm.

We know that the formula for radian measure of `theta=s/r` .

As a result,`theta = 120/6`

?=20

The angle of the arc is 20 radians.

Example 2 for trigonometry radian measure

Calculate the angle of the arc, if the radius is 8 cm, and length of arc is 135 cm

Solution:

Given, radius is 8 cm, and length of arc is 135 cm.

We know that the formula for radian measure of `theta=s/r` .

As a result,`theta = 135/8`

?=17

The angle of the arc is 17 radians.

Example 3 for trigonometry radian measure

Convert 1650 into radians

Solution:

We know that,

`(radians)/pi=(degrees)/180^0`

`radians=degrees pi/(180^0)`

`Given radians = `1650

`radians=165 pi/(180)`

Therefore,`radians=(11pi)/(12)`

Percent Discount Formula

Introduction to percent discount formula:

The word percent means out of hundred or per hundred. Percentage is a fraction with its denominator 100. The numerator of such a fraction is called rate percent.

Ex: (i) `3 / 5` = `3 / 5` `xx` 100% = 60 %

(ii) 0.25 = 0.25 `xx` 100% = 25%

(iii) 15% of 80 = `15 / 100` `xx` 80 = 12.

Now let us see few problems on this topic percent discount formula.

Example Problems on Percent Discount Formula.

Ex 1: What is the single discount which is equivalent to successive discounts of 20%, 15% and 10%?

Soln: Let us consider the price of an article be 100 dollars.

Therefore After the first discount 20%, it will cost = 100 – 20 = 80 dollar.

Second discount = 15%

Therefore 15% of 80 = 15 / 100 * 80 = 12

Therefore the new price = 80 – 12 = 68 dollars.

Third discount = 10%

Therefore 10% of 68 = `10 / 100` `xx` 68 = 6.8

Therefore the new price = 68 – 6.8 = 61.20 dollars

Therefore the single discount is 100 – 61.20 = 38.80%

Here one important thing is, we can get the single discount by adding all the discounts as 20 + 12 + 6.8 = 38.8%.

Ex 2: An article price is 450 dollars. It is sold at a discount of 20%. Find:

(i)                  The discount given

(ii)                The selling prize of the article.

Soln: Given: The price of the article is 450 dollars,

(i) The discount given = 450` xx 20/100` = 90 dollars.

(ii)  The selling price  = 450 – discount

= 450 – 90

= 360 dollars

Ex 3: In a sale, a shopkeeper allows 10% discount on his article. What price must he mark on an article, which costs his 750 dollar, to make a profit of 20%?

Soln: Given: Cost price = 750 dollars.

Therefore to give 10% discount and to gain 20%, his selling price should be as follows:

750 + 20% of 750

= 750 + `20 / 100``xx` 750 = 900 `=>` 90% (x) = 900

`=>` x = 900 `xx` `100 / 90` = 1000

Therefore the marked price = 1000 dollars.

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Practice Problems on Percent Discount Formula.

Find a single discount for 30%, 20% and 10%
[Ans: Single discount = 50.4%]

2. A dealer is selling an article at a discount of 5% on the marked price. What is the selling price if it is marked 140 dollar?

[Ans: Selling price = 133 dollars]

Descriptive Set Theory

Introduction to descriptive set theory:
A set is a well-defined collection of things

The things belonging to a set are called it's members or elements. These elements may be objects, persons, letters, numbers or any items at all.
Well-defined means that given a member, there must be no doubt in deciding whether it belongs to, or does not belong to the given set. Example:-
A set of apples weighting 200 g or over is a well-defined.
A collection of heavy apples is not well-defined and so it not a set.
All elements are separated by a comma.
The elements should be enclosed in braces (”{ }”).
Sets are usually represented by capital letters.
The order in which the members appear is not important.
If one or more elements are repeated, the set remains the same.
A set may consist of a single member like, B = {National bird of India} = {Peacock}. If 'A' is a set and 'a' is an elements of this set, then “a belongs to A”
This article is about descriptive set theory and the topics in that.

Descriptive Set Theory-representation of a Set (notation):

A set is determined by its members or elements. This determination is brought out in the following three ways.

Roster method :-

The method of listing the elements inside the braces is called Roster method. Example :- A set of alphabets of the English language represented by P is written as :
P = {A , B , C , D , E , F , ….}

Description method :-

The method of listing a set by a well-defined statement or description is called description method.
Example :- {Vowels in English alphabet}

Set builder form :-

The method of listing the elements inside the braces by stating the rule or property or formula is known as Set builder form. Example :- If A is a set of a natural numbers less than 80, then it can be represented as, A = {x : x is a natural number and is less than 80} or as

A = {=x : x E N, x<80 br="br">
Types of Sets in Descriptive Set Theory:

Finite set

The elements are limited. Example :-
Natural number less then 60.
A = {1,2,3,4,5,6,7,.....}
Days of a week
W = {Monday, ….. , Sunday}

In finite set

Unlimited numbers of elements Example :-
Set of odd numbers
N = {1 , 3 , 5 , 7....}
Set of even numbers

E = {2 , 4 , 6 , 8....}

Empty set

No elements is represented
Set of odd numbers between 5 and 6
A = { }
30th day of February

D = { }

Anti Derivative of Log X

Introduction to anti derivative of log x:

A function `Phi (x) ` is called a primitive or an antiderivative of a function f(x) if `Phi '(x)`  = f(x).

For example , `(x^4)/4` is an antiderivative of x 3, because `d/dx (x^4)/4`  = x 3.

To find the antiderivative of log x we use a special form of antiderivatives known as uv form (antiderivative by parts).

Theorem : If u and v are two fucntions of x, then

`int` u v dx  = u `int` vdx - `int { du/dx int v dx } `  dx

Finding Antiderivative of Logx

We use the above said formula of uv form to find the antiderivative of logx.

We need two functions for that logx can be written as 1 * logx.

We take logx = u , as differentiation of log x is easy to find and v = 1.

`(du)/dx = (d(logx))/dx`  = `1/x`

and `int` v dx =  `int` 1 dx = x

We use all these values and plug it in the formula

The uv form is `int` uv dx = u `int` v dx - `int { (du)/dx int v dx }` dx

`int` 1 logx dx = logx `int` 1 dx  -  `int { (du)/dx int 1 dx }` dx

= logx * x -  `int 1/x xx x dx`  dx

= x logx - `int 1 dx`

= xlogx - x  +c

= x(logx -1) +c

The antiderivative of logx is x(logx-1) +c

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Solved Problem on Antiderivative of Logx

Find the antiderivative of xlogx   

Solution :  The function is x logx

Let v =x and u = logx, as log x is easily differentiable and the integral of x is easy to find.

Now `int` v dx  = `int` x dx = `(x^2)/2`

and `(d)/dx = (d(logx))/dx = 1/x`

Plugging in all the values in the formula

`int` uv dx  =  u `int` vdx - `int { (du)/dx int vdx }` dx

`int` x logx dx = logx `int` x dx - `int {(d(logx ))/dx int x dx}`  dx

= logx `x^2/2`  - `int` `1/x xx x^2/2` dx

= `x^2/2`  logx - `1/2 int x` dx

= `x^2/2` logx - `1/2 * x^2/2` +c

= `x^2/2`    ( logx - `1/2` ) +c

Antiderivative of xlogx is  `x^2/2`    ( logx - `1/2` ) +c

Odd Number Theorem

Introduction to odd number theorem:

The strong gravitational lensing is use the odd number theorem and it is derived from differential topology. The odd number theorem derive the odd numbers by using formula. The integer is determined by normal form of odd number theorem and converse the given integer for odd number theorem.

Explanation for Odd Number Theorem

Statement of odd number theorem:

The odd number theorem illustrated as the square number is equal to the sum of first ‘n’ odd numbers. This statement is expressed as `sum_(k=1)^n` (2k -1) = n2.

The converse is performed in this theorem and defined as the nth odd number is equal to subtraction of  squares of nth and (n – 1)st value. This can be expressed as n2 – (n – 1)2 = 2n – 1.

Odd number theorem output:

The odd numbers are derived from the odd number theorem. The odd number means if the number is divided by two,the number have a remainder. The odd number is an integer and the form that integer is n = 2k + 1. The series of odd numbers are 1, 3, 5, 7, 9 and so on. The odd numbers are also called as gnomonic numbers by odd number theorem.

The parity value of odd number is 1 and the odd  number’s generating function is `(x(1 +x))/(x-1)^2` .

More about Odd Number Theorem

Examples for finding the odd numbers by using odd number theorem:

Problem 1: Find t out the odd numbers from given set of data by using function:

12, 14, 6, 21, 16, 24, 30

Answer:

The odd number of  given set of data is 21. Because it give the remainder as 1 when it divided by 2.

Problem 2 : Choose the odd number from given set of numbers.

7, 4, 2, 8, 10, 12, 14

Answer:

The odd number of given set of number  is 7.Because the remainder of  that number is 1 when divided by 2.

Exercise problem for finding the odd number by using odd number theorem:

1. Find out the odd number from given set of number:

54, 26, 38, 41, 52, 46

Answer: The odd number is 41.

2. Choose the odd number from series of data:

25, 24, 32, 44, 58, 60

Answer: The odd number is 25.

Graph Parabola Equation

Introduction to graph parabola equation:

The parabola is a member of conic section which means the intersection of a cone with a plane.
The general equation of a parabola is given as  (Ax + By)2 + Cx + Dy + E = 0.
It is very simple to graph parabola equation. To graph parabola equation, we have to find vertex, x  and y intercepts. The steps for graphing parabola equations are given below with an example problem.

Steps to Graph Parabola Equation:

Step 1: Allocate the variables a, b and c from the given equation

Step 2: Check and determine whether the parabola opens upwards or downwards.

If a > 0, then the parabola opens upwards (U-shaped).

If a < 0, then the parabola opens downwards (n-shaped)

Step 3: To find vertex:

The next step is to find vertex. To find vertex, we have to find the x-coordinate of the maximum point (or minimum point) by

X = - `b/(2a)`

By substituting this x-value into the given quadratic function i.e., the y expression, we obtain y-coordinate.

The obtained coordinate (X, Y) is the vertex of the parabola.

Step 4: To find the coordinates of the y-intercept.

By substituting x = 0 in the expression, we can find the y-intercept coordinate.

Step 5: To find the coordinates of the x-intercept.

By substituting y = 0 in the expression and solving the quadratic equation, we can find x-intercept coordinate.

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Example Problem to Graph Parabola Equation:

Graph the equation y = x2 - 7x +12.

Step 1:  Allocate the variables a, b and c from the given equation

a = 1

b = - 7

c = 12

Step 2: In the given equation, a is greater than 0 ( a = 1), therefore the graph of given equation is a parabola opens upwards (U-shaped) i.e., it has minimum point.

Step 3:  To find the vertex

x = - `b/(2a)` .

=  -`(-7)/(2(1))` .

= `7/2` .

= 3.5 

Y has the minimum point. So,

y = (3.5)2 - 7(3.5) + 12

= 12.25 - 24.5 + 12

= -0.25

Therefore, the minimum point obtained is (3.5, - 0.25)

Step 4: To find y-intercept.

By substituting x = 0 in the given equation y = x2 - 7x + 12, we get

y = (0)2 - 7(0) + 12 = 12

So, y intercept = (0, 12)

Step 5: To find x-intercept.

By substituting y = 0 in the given equation y = x2 - 7x + 12, we get

0 = x2 - 7x + 12

By factoring the above equation, we get

x = 3,  x = 4

The graph is drawn below with the help of the above information.  

   

Properties of Absolute Value

Introduction to properties of absolute value:

Normally absolute value is nothing but if any value regards to its sign in math. For example let us consider the numbers 8 and -8. Absolute value of these numbers is 8. |8| = +-8. Here we are going to learn about the properties of the absolute value. If we know the properties of the absolute value it is easy to do the operations on the absolute vales.

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Properties of Absolute Value:

Non negativity property:

The absolute value of any numbers is greater than 0. There is no negative numbers in this.

For example take any value -9. So the absolute value is 9. Because|9| = +-9. So for -9 and +9 the absolute value is +9.

|x| >= 0

Positive definiteness:

The absolute value of 0 is always 0. |x| = 0 then x = 0 (always)

Normally for an absolute value we have two values. Here the absolute value is 0. So value of this is 0. There is no sign for the value 0.

Multiplicative property:

Multiplication of any two absolute values same as the individual Absolute Value Equations. This mean

|x `xx` y | = |x| `xx` |y|

Example:

|-2 X 3| = |-6| = + 6

|-2| `xx` |3| = +2 `xx` +3 = + 6

So both the values are equal.

Subtraction addition property:

Addition of any two absolute values is always less than its individual addition.

|x + y| `lt=` |x| + |y|

Example:

|5 + -3| = |2| = 2

|5| + |-3| = 5 + 3 = 8

2 `lt` 8

Other Properties of Absolute Value:

Symmetry property:

Symmetry property is nothing but |-x| = |x|

Absolute value of –x and absolute value of x is always equal.

Identity of indiscernible property:

If the subtraction of any two absolute values is 0 then these two absolute values are equal.

|x - y| = 0 then x = y

Preservation of division:

The division of any two absolute values same as individual absolute value division.

|x / y | = |x| / |y| (where y `!=` 0)

These are the some basic properties of the absolute values.

Least Common Multiple and Greatest Common Divisor

Introduction to least common multiple and greatest common divisor
Least common multiple

In mathematics, the 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.

Greatest common divisor

In mathematics, the greatest common divisor (gcd), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder. (Source: From Wikipedia).

Example Problems to Find the least Common Multiple and Greatest Common Divisor

Example problems to find the least common multiple of two numbers

Example 1

Find the least common multiple of 3 and 7.

Solution

To find the least common multiples of 3 and 7, first we have to find the multiples of 3 and 7.

Multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Multiples of 7 = 7, 14, 21, 28, 35, 42, 49, 56, 63, 70

Here, 21 is the lowest common number in the multiples of 3 and 7.

So, 21 is the least common multiple of 3 and 7.

Example 2

Find the least common multiple of  5, 11

Solution

The multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70.

The multiples of 11 = 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132.

Here, 55 is the lowest common number in the multiples of 5 and 11.

So, 55 is the least common multiple of 5 and 11.

Example problems to find the greatest common divisor of two numbers.

Example 1

Find the greatest common divisor of  54, and 40

Solution

The greatest common divisor of 54 and 40 can be found by the prime factors of the given numbers.

Prime factorization of 54 = 2 * 3 * 3 * 3

Prime factorization of 40 = 2 * 2 * 2 * 5

The common factors in the prime factorization is 2

So, 2 is the greatest common divisor of 54 and 40.

Example 2

Find the greatest common divisor of 21 and 35

Solution

Prime factorization of 21 = 3 * 7

Prime factorization of 35 = 5 * 7

Here, 7 is the common term.

So, 7 is the greatest common divisor of 21 and 35.

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Practice Problems to Find the least Common Multiple and Greatest Common Divisor

Problem 1

Find the greatest common divisor of 3, 5, 7

Answer: 1

Problem 2

Find the greatest common divisor of 54, 27

Answer: 27

Problem 3

Find the least common multiple of 21 and 3

Answer: 21

Problem 4

Find the least common multiple of 51 and 16.

Answer: 816