Reflex Measure of an Angle

Introduction to reflex measure of an angle.

Let x be the given angle. Then we can define three important angles as follows:

1. The complimentary angle of x is 90 – x

2. The supplementary angle of x is 180 – x.

3. The reflex angle of x is 360 – x.

The above difference of angles must be known clearly.

Example: if x = 400, Then the reflex angle of x is given by 360 – x = 360 - 40 = 3200.

The above diagram, clearly shows that, the reflex angle of any angle is the remaining value of 360 beside the given angle x.

Now let us see some examples to understand and to practice these concepts in the following article.

Example Problems on Reflex Measure of Angle.
Ex 1: What is the reflex angle of  800?

Solution: The reflex angle of 80 is 360 - 80 = 2800.

Ex 2: Find the reflex of A is the given diagram Here XY is a straight line.

Solution: In the given diagram, A+15 = 180.

Therefore A = 180 -15 = 1650.

Therefore the reflex angle is given by 360 - A = 360 – 165 = 1950.

Ex 3: Find the reflex angle of B in the given diagram. Here XY is a straight line. Angle POY is 900.

Solution:

Since POY = 90

POY = 35+B = 90 implies B = 90-35

Therefore B = 550.

Therefore the reflex angle of B is 360 – B = 360 -55

= 3050.

Ex 4: Find the reflex angle of P in the given diagram. Here XY is a straight line.

Solution: Since XY is a straight line, Angle XOY = 180

implies 25+P = 180

implies P =180-25 =1550

Therefore the reflex angle of P is 360 – P = 360 -155 = 2050.



Ex 5: The reflex angle of P is 2250 . Find P.

Solution: Given: The reflex angle of P is 225 implies 360 - P = 225.

implies –P = 225 - 360 = -1350.

Therefore P =1350.

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Practice Problems on Reflex Measure of Angle

1. Find the reflex angle of 300.

[Ans: Reflex angle of 300 is 3300]

2. If the reflex angle of X is 1500, find the angle X.

[Ans: X = 2100]

I hope these examples would have helped you to find and measure the reflex angle for the given angle.

Introduction to Practice Number Line

Introduction :                                                                                           

A number line is a line on which real numbers are placed according to their value. For example, the number 3.5 or 3 1/2 corresponds with the point on a number line that is halfway between the numbers 3 and 4. Each point on a number line corresponds to a real number, and each real number has a unique point on the Number line that corresponds only to that number. Number lines are valuable tools which are easily used to illustrate the mathematical concepts such as subtraction, addition and positive and negative numbers.          

The number line is usually represented as a horizontal line. An arrowhead on either end of the line is suggests that the line  continues indefinitely in the positive and negative real number's which are denoted by ‘R’. The real numbers may be irrational and rational numbers, integers, whole numbers or also natural numbers.      

Steps to Draw a Number Line:

In order to draw a number line,

Draw a straight line with 1 inch divisions.
Position the bottom edge of your ruler at the start of the line. Mark each 1inch measurement divisions on the line until you reach the end of the line. Draw an arrow head (i.e., "<" and ">") on the left and right ends of the line.
Now, draw a perpendicular line at the centre and label it "0” where both the lines meet.
The divisions on the right hand side of the horizontal line will be the positive divisions i.e., numbers are prefixed with “+” sign and written as +1   +2   +3   +4   +5   +6……….

·       The divisions on the left hand side of the horizontal line will be the negative divisions i.e., numbers are prefixed with “-” sign and written as -1   -2   -3   -4    -5   -6………..

The points on a number line are called as the coordinates and the zero point is called the origin and the numbers on the right side of the origin are the positive numbers and those on the left side of the origin are the negative numbers. Hence the number line is divided as two symmetric halves by origin.

My forthcoming post is on factoring polynomials problems, change fractions to decimals will give you more understanding about Algebra.

Examples for Practice Based on Number Line:

Below are the examples on how to perform addition and subtraction using Number line

Ex 1: +6  - +2

Sol: +6 - +2 = +6  - 2 = +4

Ex 2: +8  +  +2

Sol: +8  +  +2 = +10

similarly,

+7 - +10 =    +7 - 10 =    -3
+7 + -10 =    +7 - 10 =    -3
+7+ +10 =    +7 + 10 =    +17
-7 - -10 =    -7 + 10 =    +3

Factor the Trinomial Calculator

Introduction of factor the trinomial calculator :
Any expression which have three non zero terms is known as trinomial.

Ex :

Ax2+ Bx + C  is a trinomial because it is having three non zero terms ( Ax2 , Bx and C)

While cx + d is not a trinomial because it has only two non zero terms ( cx and d ) hence it is known as binomial.

Factor of trinomial- By factoring trinomial we mean that to express the trinomial as a product of two binomials.

Ex : Suppose we have to factorize 35 .Then we have to find two prime numbers whose product is zero so factors of 35 will be 5 and 7 . In the same way we will write each trinomial as a product of two binomials.

Example for Factor the Trinomial Calculator

We have to follow the following steps for factoring type of trinomials

1. Look for the sign of third term.

2. if the sign of third term is positive then we have to divide the coefficient  of middle term in two parts  in such a way that sum of these two part is equal to the middle terms and product of these two parts is equal to the coefficient of third term.

3. Take the common parts out and solve.

Ex : factorize  x2 + 14x + 24

Sol :

Step1. Sign of third term is positive

Step 2. Coefficient of middle term is 14. So we will divide in two parts such that the sum of these two parts is 14 and product of these two parts is 24. Clearly these two parts will be 12 and 2.

Step3.

x2 + 14x + 24

= x2 +  (12 + 2 )x + 24

= x2 +  12x + 2x + 24

= x ( x + 12 ) + 2 ( x +12 )  [take out the common terms and solve]

= (x + 12 ) (x + 2 )

So x2 +  14x + 24 = (x +12 ) (x +2)

I am planning to write more post on list of all prime numbers, translate to an algebraic expression. Keep checking my blog.

Explaination for Factoring the Trinomial of Type X^2 + Bx - C

We have to follow following steps.

1. Look for the sign of third term.

2. If the sign of third term is negative then we have to divide the coefficient  of middle term in two parts  in such a way that difference of these two part is equal to the middle terms and product of these two parts is equal to the coefficient of third term.

3. Take the common parts out and solve.

Example: Factorize  x2 -5x -6

Solution:

Step1. Sign of third term is negative

Step 2. Coefficient of middle term is 5. So we will divide in two parts such that the difference of these two parts is 5 and product of these two parts is 6. Clearly these two parts will be 6 and 1.

Step3.

x2 - 5x – 6

= x2 – 6x  + x – 6

= x ( x -6 ) + 1 (x -6 )

= (x +1 ) (x – 6)

So x2 - 5x – 6 = (x +1 ) (x – 6)

Rules for Factoring the Trinomial of Type Ax^2 + Bx - C and Ax^2 + Bx +c

We have to follow following steps.

1. Look for the sign of third term.

2. Multiply the coefficient of first term and last term,say the product is A

3. If the sign of the last term is positive then divide the A in two parts such that addition of those parts equal to the middle term and multiplication of those two parts equals A.If the sign of the last term is negative then divide the A in two parts such that subs traction of those parts equal to the middle term and multiplication of those two parts equals A.

4. Take out the common factor and solve.

Sum of Positive Integers

Introduction about sum of positive integers:

The integers are formed by the natural numbers including 0 (0,1,2,3…..) together with the negatives  of the non-zero natural numbers (-1, -2, -3, ...) The positive integers are 0,1,2,3…..

To find the sum of consecutive positive integers we have to use the following formula,

                 `sum_(i=1)^n`n = `(n (n+1))/(2)`  where n is the positive integer

Properties of integer

Addition (or) sum: If we have added two positive integers the result will be a positive integer. If negative integer involved in addition means the result is based on the biggest number.

Commutative property: a+b= b+a
Associative property:a+(b+c)=(a+b)+c
Identity property: a+1= 1+a

Algebra is widely used in day to day activities watch out for my forthcoming posts on how to solve quadratic inequalities and algebra linear equations. I am sure they will be helpful.

Lets Solve some Examples on Sum of Positive Integers

Pro 1 :Find the sum of consecutive positive integers from 1 to 25

Solution:From the formula  `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =25  `sum_(i=1)^25`25  = `(25 *(25+1))/(2)`

                                       = `(25 * 26 )/(2)`

                                       =`(650)/(2)`

                                       = 325

The answer for sum of consecutive integer from 1 to 25 is 325

Pro 2:Find the sum of consecutive positive integers from 1 to 50

Solution:From the formula  `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =50  `sum_(i=1)^50`50  = `(50 *(50+1))/(2)`

                                       = `(50 * 51)/(2)`

                                       =`(2550)/(2)`

                                       = 1275

The answer for sum of consecutive integer from 1 to 25 is 1275

Pro 3:Find the sum of positive integer 5 + 10.

The symbol plus +, represents the positive direction.
So, To find 5+10, number line starts from 5,  go to the  positive direction and move 10 units frontward we will get the 15

5+10=15

The answer is 15

Some more Problems on Sum of Positive Integers

Pro 4: Find the sum of consecutive positive integers from 1 to100

Solution:

From the formula  `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =100 =  `sum_(i=1)^100`100 = `(100 *(100+1))/(2)`

                                       = `(100 * 101)/(2)`

                                       =`(10100)/(2)`

                                       = 5050

The answer for sum of consecutive integer from 1 to 100 is 5050

Pro 5:Find the sum of consecutive positive integers from 1 to 45

Solution:From the formula `sum_(i=1)^n`n  = `(n (n+1))/(2)`

Here n =45  `sum_(i=1)^45`45  = `(45 *(45+1))/(2)`

                                       =`(45 * 46)/(2)`

                                       =`(2070)/(2)`

                                       = 1035

The answer for sum of consecutive integer from 1 to 45 is 1035

Introduction of add polynomials

Introduction of add polynomials:

          A polynomial is an expression formed by variables and constant, using the  fundamental operation addition, subtraction, multiplication, and non-negative, number exponents. For example, x 2 - 4x + 5 is a polynomial. An algebraic expression, ax 2 is called a monomial in x, x is a variable a. The number a is called the coefficient of x^2 , the degree of the monomial. The sum of two monomials is called a binomial. The sum of a limited number of monomials in x is called a polynomial in x.

Please express your views of this topic solve a math problem by commenting on blog.

Polynomials and Methods for Adding

The monomials in a polynomial are called the terms and the highest degree of the terms, the degree of the polynomial. The coefficient of the highest power of x in a polynomial is called the leading coefficient of the polynomial. For example, 3x – 1 + 5x^2 + 6x^5 – 4x3 is a polynomial of degree 5 with terms 2x, –1, 5x^2, 6x^5 and –4x3 and most important coefficient 6. The terms of a polynomial are real numbers, they are rearranged using the property of real numbers

A first degree polynomial is called a linear polynomial. A second degree polynomial is called a quadratic polynomial. A degree polynomial is third called as cubic polynomial and a degree polynomial is four called as bi-quadratic polynomial. Since the terms of a polynomial are numbers, all polynomials represent numbers. So we can add, subtract, multiply and divide polynomials. The following method used to add the polynomials.

Adding Polynomials Examples

Ex 1: Find the sum or add of 3x^4 – 4x^2 + 5x + 3 and 4x + 6x3 – 6x^2 – 2 polynomials.

Sol :  Using the properties of real numbers, we add the polynomial

(3x^4 – 4x^2 + 5x + 3) + (6x3 – 6x^2 + 4x – 2) = 3x^4 + 6x3 – 4x^2 – 6x^2 + 5x + 4x + 3 – 2

= 3x^4 + 6x3 – (4+6)x^2 + (5+4)x + 1

= 3x^4 + 6x3 – 10x^2 + 9x + 1.

The following method is useful in add two polynomials

3x^4 + 0x3 – 4x^2 + 5x + 3

0x^4 + 6x3 – 6x^2 + 4x – 2

_____________________

3x^4 + 6x3– 10x^2 + 9x + 1

_____________________

Graphing Linear Equations and Functions

Introduction

Linear equations are relation between two variables. Generally in form of ax+by+c = 0 or ax+by = c. We can see linear functions also. These are nothing but similar to linear equations. Generally a function can be in form of f(x) = ax + b, we can write this function as y = ax + b also.

Now graphing linear equations and functions are very simple. Since these are a straight lines, you have to just find two points through which these lines passes and once you find the point you can draw a line meeting both the point. Here are steps for graphing equations and functions.

Steps for Graphing Linear Equations and Functions

Step1: Put x=0, you will get value of y, lets say y = b, first point will be (0,b)

Step2: put y=0, you will get value of x, lets say x = a, second point will be (a,0)

Step3: Plot both the points and draw a line meeting both the points and extend this line beyond these points.

In this way you will get required graph of linear equation and functions.

Following examples help you in understanding these steps and similar problems.

Examples for Graphing Linear Equations and Functions

Ex:1 Graph the linear equation: 2x + y = 6

Put x=0, we get y =6

first point wil be (0,6)

Put y=0, we get x=3

second point will be (3,0)

Now plot these points and draw the required graph of this linear equation.

Understanding algebra 2 help online is always challenging for me but thanks to all math help websites to help me out.

Ex:2 Graph the function f(x) = 4x + 4

put in the form of y = 4x + 4

Put x = 0, we get y = 4, First point will be (0,4)

Put y = 0, we get x=-1, second point will be (-1,0)

Now plot the points and draw the graph.

Introduction of sample proportion

Introduction of sample proportion:

          Two equal ratios form a proportion. Thus, if a:b=c:d then a,b,c,d are said to be sample proportional. We write this as, a:b::c:d. This is read as “a is to be is as c is to d”A and b are called the extremes and b an c are called the means. The terms d is called the fourth sample proportional to a,b,c.

Types of Sample Proportions:

Continued proportion: quantities a,b,c,d,…. Are said to be in continued proportion if, a/b=b/c=c/d….. i.e., a:b=b:c=c:d=……………….

Mean proportional:  consider a,b, and c are in continued proportion, Proportion will be represented as a:b::b:c. Here, b is called the mean proportional of a and c..

Also c is called the third proportional to a and b.. If a,b, c are in continued proportion, then ac:b² . Hence, if three quantities are in continued proportion, then the product of extremes is equal to the square of the mean.

Note: The product of the extremes is equal to the product of the means.

Inadvertent: If a:b::c:d, then, b:a::d:c. for, a/b=c/d gives a/c=b/d. This operation is called inadvertent. It says that if quantities are proportion they are also proportional when we take it inversely.

Alternendo: if a:b::c:d then a:c::b:c.

For, a/b=c/d gives a/c=b/d. This operation is called alternendo. It says that if four quantities are proportional, they are also proportional when taken alternatively.

Componendo: If a:b::c:d, then (a+b):b:(c+c):d.

Given that, a/b=c/d,

So,(a/b)+1=(c/d)+1

Or, (a+b)/b=(c+d)/d

This operation is called componendo.

Dividendo: If a:b::c:d,

Then (a-b):b::(c-d):d.

Given that, a/b=c/d,

So,(a/b)-1=(c/d)-1

Or, (a-b)/b=(c-d)/d.

This operation is called dividendo.

Componendo and Dividendo:

If a:b::c:d,

Then (a+b):(a-b)::(c+d):(c-d).

Using componendo and dividendo respectively, we get

(a+b)/b=(c+d)/d   -------------------------1

(a-b)/b=(c-d)/d.    -------------------------2

Dividing the respective sides of 1 by those of 2 we get

(a+b)/(a-b)=(c+d)/(c-d)

This operation is called componendo and dividendo.

Examples for Finding the Sample Proportion:

Ex 1  : 2/5=4/10

Sol :  2/5=4/10

If we do the cross multiplication, we will get

5*4 = 2 *10

20 = 20 so, : 2/5=4/10 is a proportion.

Ex  2:  2/5=4/10

Sol :    1/5=4/10

If we do the cross multiplication, we will get

5 X4 = 1 X 10

20 = 10 so, : 1/5=4/10 is not a proportion.