How do Parallel Lines Form

When two lines are placed on a plane in same distance without intersection is called as parallel line. The term parallel in math can be represented as ||. For example line XY || to line AB. The statement states that the line XY is parallel to the line AB. The line XY are parallel to the line AB, if they don’t have any common point or center point.

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How do parallel lines form:

How do parallel lines form:

The parallel lines are never cross each other. When two lines are placed on a plane in same distance without intersection is called as parallel line. The following rules are used to form the parallel lines.

Step 1:

The line AB is drew in the paper or board.

Step 2:

The point P is blotted in the space of plane (other than the line).

Step 3

The next line is drowned through the point P. The new line makes an angle to the line AB. Blot the intersection point as Q in the line AB on the plane.

Step 4:

Draw an arc from the point Q. Among the help of compass sketch an arc width as regards half PQ. This arc must cut cross or intersect the line AB and P.

Step 5:

Acquire the same measurement (similar compass width) and blot an arc in the line P from the point P.

Step 6:

Put the compass width to the lower arc.

Step 7:

Shift the compass to the upper arc. Create off an arc to make a point R.

Step 8:

Draw a straight line. The straight lime must travel through the line PR.

Now the line RP is parallel to the line AB.

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Properties of Parallel Lines:

Properties of parallel lines:

• The parallel lines will be in the same planes.
• Parallel lines are never crossed.
• The parallel lines are moved in same distance in the whole plane. 

Online Trigonometry Equation Solver

In online, many trigonometry solver are available for solving various trigonometry problems. It is very simple to work on the online trigonometry equation solver. If we enter the input, the online trigonometry equation solver automatically generates the solution. For example, if we want to find the solution for the equation sin 3x = 0 using online trigonometry solver, we have to just enter the problem and press calculate button, then the online solver will automatically generate the output as x = `(npi)/3`

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Example problems of online trigonometry equation solver:

Example 1:
Find the solution for the equation cos x = `1/sqrt2`
Solution:
The given equation is cos x = `1/sqrt2`
We know that cos` pi/4` = `1/sqrt2` and cos (2∏ -` pi/4` ) = `1/sqrt2`
Therefore,
cos `pi/4 ` = `1/sqrt2` and cos `(7pi)/4` = `1/sqrt2`
Hence, the solutions are x = `pi/4 ` and x = `(7pi)/4

Example 2:
Find the solution for the equation cos x = - `1/2`
Solution:
       The given equation is cos x = - `1/2`
       We know that cos `pi/3` = `1/2`
       Therefore,
                  cos (∏ - `pi/3` ) = - cos `pi/3` = - `1/2`   and
                  cos (∏ + `pi/3)` = - cos` pi/3` = - `1/2`
                  cos `(2pi)/3` = - `1/2` and cos `(4pi)/3` = - `1/2`
         Hence, the solutions are x = `(2pi)/3` and x = `(4pi)/3`

Example 3:
Find the solution for the equation cosec x + √2 = 0
Solution:
The given equation is cosec x + √2 = 0
The above equation can be written as sin x = - 1/(sqrt2)
                                                                            = - sin `pi/4`
                                                                            = sin (∏ + `pi/4` )
                                                                            = sin `(5pi)/4`
Therefore,
sin x = sin `(5pi)/4`
x = (n∏ + (- 1)ⁿ `(5pi)/4` ), where n ε I
Hence, the solution is x = (n∏ + (- 1)ⁿ `(5pi)/4` ), where n ε I

Practice problems of online trigonometry equation solver:

1) Find the solution for the equation sin x = `1/2`
2) Find the solution for the equation sin x = - `(sqrt3)/2`
3) Find the solution for the equation sin 2x = - `1/2`
Solutions:
1) The solutions are x = `pi/6 ` and x = `(5pi)/6`
2) The solutions are x = `(4pi)/3` and x =` (5pi)/3`
3) x = (`(npi)/2` + (- 1)ⁿ `(7pi)/12` ), where n ε I

Quadrilateral Rectangle

In Euclidean plane geometry, a quadrilateral is a polygon with four sides and four vertices or corners. Quadrilaterals are simple (not self-intersecting) or complex (self-intersecting), also called crossed. Simple quadrilaterals are either convex or concave. The interior angles of a simple quadrilateral add up to 360 degrees. A parallelogram is a quadrilateral with two pairs of parallel sides. Source-Wikipedia.

Quadrilateral rectangle:

Rectangle is one type of regular quadrilateral. It has four sides. The opposite sides are equal in length. It has four vertices. It has four internal angles which are congruent. There are two diagonals in rectangle. The length of diagonals is equal in length. The sum of internal angles of rectangle is 360 degree.

Area of the rectangle (A) = length x width square unit
Area (A) = l x w
Perimeter of the rectangle = 2(length + width)
Perimeter (p) =2(l x w) unit
Length of diagonal (d) =sqrt(length²+width²)
Diagonal length    =sqrt(l²+w²) unit.

Example problems for rectangle:

1. Find the area and perimeter of rectangle, whose length and width are 5meters and 3 meters respectively.
Solution:
Area of rectangle = l x w square unit.
Given:    Length= 5 meters, Width =4 meters
=5 x 4

Area of rectangle = 20 m²                 
Perimeter of the rectangle   = 2(l + w)
=2(5+ 4)
= 2 (9)

\Perimeter of the rectangle = 18 meters
2. Find the area and perimeter of rectangle, whose length and width are 9meters and 5 meters respectively.
Solution:
Area of rectangle = l x w square unit.
Given:    Length= 9 meters, Width =5 meters
=9 x 5

Area of rectangle = 45 m²                 
Perimeter of the rectangle   = 2(l + w)
=2(9+ 5)
= 2 (14)
Perimeter of the rectangle = 28 meters

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3. Find the area and perimeter of rectangle, whose length and width are 12meters and 6 meters respectively.
Solution:
Area of rectangle = l x w square unit.
Given:    Length= 12 meters, Width =6 meters
=12 x 6
Area of rectangle = 72 m²                 
Perimeter of the rectangle   = 2(l + w)
                                               =2(12+ 6)
                                               = 2 (18)

Perimeter of the rectangle = 36 meters

Star Test for Geometry

Geometry is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of practical knowledge concerning lengths, areas, and volumes. The field of algebraic geometry is the modern incarnation of the Cartesian geometry of co-ordinates. The star test geometry example problems and star test practice problems are given below. Source: Wikipedia

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Example problems for star test geometry:

Example problem 1:
Find the supplementary angle of 95°
Solution:
The supplementary angle of  95° = 180° – 95° = 85°.

Example problem 2:
From the figure write

    (a) collinear points.
    (b) Concurrent lines and their point of concurrence.

Solution:
    (a) from the figure, the points A, B, C are collinear.
    (b) The lines AD, BD, CD are concurrent lines. D is the point of concurrence

Practice problems for star test geometry:

Star test question 1:
The sum of the three angles of a triangle is _________ .
              a) 90°
              b) 360°
              c) 180°

Star test question 2:
      In an equilateral triangle, the 3 sides are __________ .
              a) Unequal
              b) Equal
              c) Parallel

Star test question 3:
      The triangle in which, the two sides are equal is called an ________ triangle.
             a) Isosceles
             b) Right
             c) Perpendicular

Star test question 4:
      If a triangle has one right-angle, it is called as _________ triangle.
             a) Perpendicular
             b) Right angled triangle
             c) Isosceles

Star test question 5:
      In a triangle the sum of the measure of any two given sides are _______ than the third side.
             a) Equal
             b) Lesser
             c) Greater

Star test question 6:
      One of the angles of a triangle is 100° and the other two angles are equal. What is the measure of each of these equal angles.
             a) 40°
             b) 80°
             c) 120°

Answer key:
       1. (c) 180°
       2. (b) equal
       3. (a) isosceles
       4. (b) right angled triangle
       5. (c) lesser
       6. (a) 40°

General Quadrilaterals

A general quadrilateral is a polygon with four sides. A general Quadrilateral is the two dimensional plane figures which has four sides so it is a four sided polygon. The total angle of a general quadrilateral is 360 degrees. The number of diagonals for a general quadrilateral are two. When we have two diagonals then both the diagonals will intersect at a common point. The general quadrilateral has four vertices's and four edges. Here in this topic we are going to see about general quadrilaterals.

Types of general quadrilaterals.

Square
Rectangle
Parallelogram
Rhombus
Trapezoid
Kite

General quadrilatel - Figures:

Square
In a square all the sides are equal in length and all the angles are equal to 90 degree. The diagonals of a square are equal to each other.

Rectangle:
In a rectangle, opposite sides are equal to each other and all the angles are equal to 90 degree.

Parallelogram:
In a parallelogram opposite sides are equal in length and opposite angles are also equal to each other.

Rhombus:
Rhombus is a combination of square and parallelogram. In Rhombus all sides of the lengths are equal and the opposite angles are equal to each other
.

Trapezoid:
In a trapezoid one of the pair of opposite sides are parallel.

Kite:
It has two pair of sides and the lengths of the adjacent pair are equal in length.

General Quadrilaterals - Example problems:

1) What is the area of a square having side-length 15 cm?

Solution:

The area of the Square = L²

Where L= 15cm

Length= L*L

= 15*15

= 225cm²

2) What is the area of a rectangle having a length of 8cm and a width of 3.3cm?

Solution:

The area of the rectangle = length * width

= 8 * 3.3= 26.4 cm²

3) What is the area of a trapezoid having the length of the parallel sides as  13 and 7 and the height being 9?

Solution:

The area of the trapezoid = h/2*(b1+b2)

=9/2(13+7)

=4.5(20)

=90 square units

4) What is the area of a parallelogram having a base of 30cm and a corresponding height of14cm?

Solution:

The area of the parallelogram=b*h

= 30*14

=420cm²

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5) What is the area of a square having side-length 5.4cm?

Solution:

The area of the Square =L²

= 5.4 * 5.4

=29.16cm²

Pentagonal Pyramid

Pyramid is a solid whose base is a planar polygon with a triangular face on each side. These triangular faces meet in a point which is not in the plane fo the base. Pentagonal pyramid is pyramid whose base is a pentagon, and having 5 triangular faces on each side.

 Pentagonal Pyramid Formula:
 Let,   a = apothem length, s,b = side, h = height and l = slant height,

• Area of Base(A) = `(5/2)axxs`
• Surface Area of Pyramid = `(5/2) a xx s + (5/2) s xx l = A + (5/2) s xx l`
•  Volume of Pyramid = `(5/6) axxbxxh`

Properties of pentagonal pyramid:

• A pentagonal pyramid has 6 faces. One base and 5 lateral faces
• A pentagonal pyramid has 10 edges
• A pentagonal pyramid has 6 corners

 

Examples on pentagonal pyramid:

1) Find  volume and surface area of a pentagonal pyramid with the given apothem length 7, side 8, height 9 and the slant height 10.
 Solution:   a = 7, s = b = 8, h = 9, l = 10

  Step 1:  Area of the base(A) = `(5/2) axx s`                                 
                                               = 2.5 * 7 * 8
                                               = 2.5 * 56
                                               = 140

  Step 2:  Surface Area of Pyramid = `A + (5/2) s xxl`
                                                       = 140 + ((5/2) * 8 * 10
                                                       = 140 + (2.5 * 80)
                                                       = 140 + 200
                                                       = 340

  Step 4: Volume of Pyramid =`(5/6) a xx b xxh`
                                              =  `(5/6)` * 7 * 8 * 9
                                              = (0.833) * 56*9
                                              = 0.8333 * 504
                                                = 419.98

2) Find the surface area a pentagonal pyramid with the given apothem length 2.5, side 5 and the slant height 9.
 Solution:   a = 2.5, s = b = 5, , l = 9

       Area of the base(A) = `(5/2)axxs`                         
                                          = 2.5 * 2.5 * 5
                                          = 6.25 * 5
                                          = 31.25
    Surface Area of Pyramid = `A + (5/2) s xxl`
                                               = 31.25 + ((5/2) *5 * 9
                                               = 31.25 + (2.5 * 45)
                                               = 31.25+ 112.5
                                               = 143.75

Algebra is widely used in day to day activities watch out for my forthcoming posts on Mean Median Mode Definition and icse syllabus 2013. I am sure they will be helpful.

Pictures of pentagonal pyramid

. 

Sin Angle Formula

Sin is a basic trigonometric function. In this article we are going to deal with sin angle formulas and how to use the sin angle formulas. Sin angle formulas used to calculate the sin values of the angles. Using the sin angle formulas we have to find the side lengths of the triangles normally from a right angle triangle we can say the sin function as
Sin A = `(opposite) / (hypotenuse)` . In this A is the angle of the opposite side.

Once you've gone through these, take a look at our Arc Formula for more refernce.

Sin angle formulas:

           There are five types of sin angle formulas are there.
Sum and angle formula
              Sin(X +Y) = Sin X Sin Y + Cos X Cos Y
              Sin (X – Y) = Sin X Sin Y – Cos X Cos Y
Double angle formula
             Sin 2A = 2 Sin A Cos A
Triple angle formula
             Sin 3A = 3 Sin A – 4Sin³A
Half angle formula
             Sin²(X / 2) = ((1 – Cos A) / 2)
Product and sum formula:
             Sin X Sin Y = Cos(X – y) – Cos (X + y) / 2
             Sin X Cos Y = Sin (X + Y) + Sin (X – Y) / 2

Example problems for sin angle formulas:

 Example 1:
      Find the value of 75^o
Solution:
         Sin 75^o = Sin (30^o+ 45^o)               
             We know the sin angle formula
                                        Sin(X +Y) = Sin X Sin Y + Cos X Cos Y
                                        Sin (30^o+ 45^o) = Sin 30^o Sin 45^o + Cos 30^o Cos 45^o
                                        Sin (30^o+ 45^o) = `(1 / 2)xx (1 / sqrt(2)) + (sqrt(3) / 2) xx (1 / sqrt(2))`
                                        Sin (30^o+ 45^o) = `(1 / (2 sqrt(2))) + (sqrt(3) / (2 sqrt(2)))`
                                        Sin (30^o+ 45^o) = `((1 + sqrt(3)) / (2 sqrt(2)))`

Example 2:
         Find the value of Sin 150^o using double angle formula. Where sin 75^o = 0.9659, Cos 75^o = 0.2588
Solution:
         Given sin 75^o = 0.9659, Cos 75^o = 0.2588
         Sin 150^o = Sin 2 `xx` 75^o
           We have the sine angle formula Sin 2A = 2 Sin A Cos A
                 Sin 150^o = 2 Sin 75⁰ Cos 75^o
                 Sin 150^o = 2 `xx` 0.9659 `xx` 0.2588

                 Sin 150^o = 0.4999        

Addition and Subtraction Facts

A fact is a group of a declaration of  each of which stand for the same meaning. All the declaration has the same meanings. Addition and subtraction facts contain 2 or 3 numbers. The types of facts operations are addition and subtraction facts, multiplication and division facts. In here we shall discuss about addition and subtraction facts. 

Addends:

The facts being added in an addition operation like 4 and 5 in 4 + 5 = 9 or 5 + 4 = 9.

Reverse operation:

Reverse of a process is call as inverse operation of addition and subtraction.

I like to share this Addition and Subtraction Word Problems with you all through my article. 

Sample problem for addition and subtraction facts:

First we take an addition operation like 3 + 5 = 8.
 In the next operation the addends are reversed like 5 + 3 = 8.
Now we can write the inverse operations of both.
8 - 5 = 3
8 - 3 = 5
Ex1: What are the four members of the fact family of 7 + 5 = 12?
Solution: 7 + 5 = 12
Reversing the addends:
       7 + 5 = 12;
Inverse operations:
        12 - 5 = 7;
        12 - 7 = 5;
Ex 2: Write the four facts of the family with numbers 9, 5 and 14.
Solution:9 + 5 = 14;
5 + 9 = 14;
14 - 9 = 5;
14 - 5 = 9;
When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same

Some more examples

Ex 3 :Write the four facts of the family with numbers 11, 5 and 16.
Solution:In the problem we find the facts value for addition and subtraction.
11 + 5 = 16
5 + 11 = 16
16 - 11 = 5
16 - 5 = 11
When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same
Ex 4: Write the four facts of the family with numbers 2, 7 and 9.
Solution:7 + 2 = 9
2 + 7 = 9
9 - 2 = 7
9 - 7 = 2
When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same

Solving Online Vertical Asymptotes

Solving online vertical asymptotes mean we are going to solve the vertical asymptotes of the curve through online. Normally asymptotes mean the distance from the line to the curve tends to zero. Vertical asymptotes mean the distance from the curve to the vertical line which tends to the zero. If a rational functions denominator tends to zero mean we will get the vertical asymptotes equation. We will see some example for solving vertical asymptotes.

Examples for solving online Vertical Asymptote:

Example 1 for solving online vertical asymptotes:
          Find the vertical asymptotes of the given rational function `(x^2 + x + 1) / (x^2 + 3x + 2)`
Solution:
        The given rational function is function `(x^2 + x + 1) / (x^2 + 3x + 2)`
        If we want to find the vertical asymptotes of the given function we have to equal the given rational functions denominator to zero. So we get
                                  x² + 3x + 2 = 0
                                  x² + 2 x + x + 2 = 0
                                  x (x + 2) + (x + 2) = 0
                                  (x + 2) (x + 1) = 0
                                   x + 2 = 0 and x x + 1 = 0
                                   x = -2 and x = -1
            So the vertical line equations of the given rational functions are x = -2 and x = -1. The solutions for the rational functions are except -2 and -1.
And the vertical asymptotes of the rational function are -2 and -1.

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More examples for solving online vertical asymptotes:

Example 2 for solving online vertical asymptotes:
                     Find the vertical asymptotes of the given rational function `(x^2 + 9x + 5) / (x^2 + 2)`
Solution:
        The given rational function is function `(x^2 + 9x + 5) / (x^2 + 2)`
        If we want to find the vertical asymptotes of the given function we have to equal the given rational functions denominator to zero. So we get
                       x² + 2 = 0
                       x²  = -2
                 So there is no solution for x. It means the denominator has no zeros and there is no vertical asymptote for this function.
       If the x value is imaginary there are no vertical asymptotes for this function.

Taylor Series Power

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.

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Definition of taylor series power:

Taylor series power:

Definition:   Taylor polynomial degree of function can be defined as the function of f. For an approximation of a function have a degree of polynomial. It can be differentiate for n times. The degree of Taylor polynomial for centered at a is

Pn(x) = f(a) + f′(a)(x - a) + (f″/ 2!)(x – a) ²+ …… + (f (n) (a)/ n!)(x - a)n

= ∑n k = 0 (f (k) (a) / k!) (x- a) k

Properties of taylor series power approximation:

In the interval (a + r, a - r), the series of converges for x and the result is equal to f(x) which function is analytic.
The power series representation can be used in the simplest form of Euler’s formula. It can be done in algebraic expressions
The result of expansions of Taylor series for cosine, sine and exponential functions are the one of fundamentals fields of harmonic analysis.
In singularity, the Taylor functions are sometimes cannot write as function.
The Taylor series is zero and also the function is not which is not analytic.


Example problem for Taylor series power:

Ex 1:   Solve: f(x) = (1 / (x + 1)), a = 1. Find the Taylor polynomial degree up to two.

Sol :    Given f(x) = (1 / x +1)          f (1) = 0.5

f’(x) = -1 / (x +1)2             f’ (1) = -0.25

f”(x) = 2 / (x +1)3             f’’(1) = 0.125

To find the Taylor polynomial:

General form of the Taylor polynomials is

Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n

Here we calculate up to third degree of Taylor polynomial

Pn(x)   = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!)

= 0.5 + (-0.25) (x - 1) + 0.125((x - 1)2 / 2!)

= 0.5 – 0.25x + 0.25 + (x2 – 2x + 1) (0.0625)

= 0.5 – 0.25x + 0.25 + 0.0625x2 – 0.125x + 0.0625

Pn(x) = 0.0625x2 – 0.375x +0.75.

Ex  2:  Solve: f(x) = (1 / (x - 3)) +1, a = 4. Find the Taylor polynomial degree up to three.

Sol :    Given f(x) = (1 / x - 3) +1          f (4) = 2

f’(x) = -1 / (x - 3)2             f’ (4) = -1

f”(x) = 2 / (x – 4)3             f’’(4) = 2

f’’’(x) = -6 / (x – 4)4          f’’’ (4) = -6

To find the Taylor polynomial:

General form of the Taylor polynomials is

Pn(x) = f (a) + f′ (a) (x - a) + (f″/ 2!)(x – a)2 + …… + (f (n) (a)/ n!)(x - a) n

Here we calculate up to third degree of Taylor polynomial

Pn(x)   = f (4) + f’ (4) (x – a) + f’’ (4) ((x – a)2/ 2!) + f’’’ (4) ((x – a)3/ 3!)

= 2 + - (1) (x - 4) + 2((x – 4)2 / 2!) + (-6) (x – 4)3/ 6

= 2 - x + 4 + (x2 – 8x + 16) - (x3 – 3(x2(4)) + 3(x (16)) + (43)

= 2 - x + 4 + x2 – 8x + 16 - x3 + 12x2 - 48x - 64

Pn(x) = - x3 + 13x2 - 57x – 46.

Taylor Series

                      Taylor series is a infinit number of terms.Taylor series is a  expansion of series and its function about a point.The Taylor series of a real or composite function ƒ(x) that is much differentiable in a neighborhood of a real or composite number a is the power series

             
    that can be written in the more compact sigma notation as 

              
     where n! indicates the factorial of n and ƒ ⁽ⁿ⁾(a) indicates the nth derivative of ƒ calculate at the point a.

I like to share this Taylor Series Expansion Example with you all through my article.

Explanation of taylor series:

Maclaurin Series:
If a=0, then it is said to be maclaurin series.


Derivation:
we define the power series as,


At x=0,


Differentiate the function,


 At x = 0,


Differentiating again will give,



At x=0, we will evaluate the equation as,



Generalizing the equation,we get


Substitute the  values of a_n in the power expansion,



Generalizing  f in a more general form, we have


Evaluating at x = a, we get


Sustitute above equation, we get taylor series.


        Taylor series can be used to evaluate the value of an whole function in each point, if the functional value and its derivatives are identified at single point. Uses of the Taylor series for whole functions are:
        1.The sum of partial series can be used as approximations of the entire function.
        2.The representation of series reduces many mathematical proofs.
       In Taylor series, algebraic functions are indicated using an algebraic equation, and transcendental functions are indicated using properties which holds them, namely differential equation. Example the exponential function is equal to its own derivative and its original value is 1.
       Taylor series are used to identify functions and operators in diverse areas of mathematics. Example: Analytical functions of matrices and operators can be defined as matrix exponential. In formal analysis, we directly work with the power series .

Example:

Example 1:
calculate the taylor series for

Using formula ,

Rewrite the equation,




substitutes


we get,

we get the new series as,


Example 2:

taylor series for f(x) = 1/ ( 1 + x )

the general case is 


hence the taylor series for f(x) is

Composite Functions

Students can learn about Composite Functions and How to Graph Composite Functions. Students can get help with Calculus problems involving Composite function from the online tutors.

In this article about Composite Functions, we will learn about Composite Functions and see about how to draw the graph for composite functions and how to get the composite functions. We shall discuss how to make the composite function and how to draw the resultant function on a graph sheet.

When two functions are f(X) and g(X), the composite function f(g) will be defined as f[g(x)].

Graphing composite functions

Below is given an example explained with graphical representation which will help you to understand the concept of graphing composite functions better:

Example:
Consider the following function f(x)=x+1 find `g(x)=x^2`
Solution:
To answer this f[g(x)]
Substitute the g(x) in f(x) function
`=f(x^2)`
Perform the f(x) function ie f(x)=x+1
here `x=x^2`
so `f(g(x))=x^2+1`

Since the function `y=f(x)=x^2+1`
The value of a y varies with respect to the x-value. Substitute different x-value to the function, the y value will be obtained. After getting y value make the values of x and y value in table format.
`f(x)=x^2+1`
`f(0)=0^2+1=1`
`f(1)=1^2+1=2`
`f(2)=2^2+1=5`
`f(3)=3^2+1=10`

x	0	1	2	3
y=f(x)	1	2	5	10

Assign the scale as per our co-ordinate points.
Scale:
In x-axis 1unit= 1 cm
In y-axis 1unit= 2 cm
Graph for the composite functions

Example Problems

Example: Consider the following function f(x)=2x+1 find `g(x)=x^2`
Solution: To answer this f[g(x)]
Substitute the g(x) in f(x) function
`=f(x^2)`
Perform the f(x) function ie f(x)=2x+1
here `x=x^2`
so `f(g(x))=2(x^2)+1`
Since the function `y=f(x)=2(x^2)+1`

The value of a y varies with respect to the x-value. Substitute different x-value to the function, the y value will be obtained. After getting y value make the values of x and y value in table format.
`f(x)=2(x^2)+1`
`f(0)=2(0^2)+1=1`
`f(1)=2(1^2)+1=3`
`f(2)=2(2^2)+1=9`
`f(3)=2(3^2)+1=17`

x	0	1	2	3
y=f(x)	1	3	9	17

Assign the scale as per our co-ordinate points.

My forthcoming post is on pu board karnataka and cbse syllabus 2012 will give you more understanding about Algebra.

Scale:
In x-axis 1unit= 1 cm
In y-axis 1unit= 3 cm
Graph for the composite functions

Reciprocal Function Tutor

Reciprocal function tutor is nothing but the inverse function tutor. Before knowing about the inverse function we have to know about the one – to – one function. Let us take the domain as X and range as Y of one to one function f. Thus, the reciprocal function of f has the y domain and X range and is represented as  f-1 (y) = x or f(x) = y.

one to one function

Let see about the reciprocal function with example:

Procedure to find the Reciprocal Function Tutor:

To find the formula of the reciprocal function tutor by following the given procedure:

Step 1: Change the given equation in form of function y = f(x).

Step 2: Solve the given equation for x in terms of y.

Step 3: Inter change the x and y variables and therefore y = f-1(x). Thus function of y is equal to reciprocal function of x.

Example Problems – Reciprocal Function Tutor:

Example 1:

Find the reciprocal of the function f(x) = `(3x - 6)/(3x+5)`.

Solution:

Step 1: Let write the given function as y = `(3x - 6)/(3x+5)`..

Step 2: Solve the function of x in term of y.

y = `(3x - 6)/(3x+5)`.

Multiply the denominator of the fraction 3x + 5 to y.

Now, y (3x + 5) = 3x – 6

3xy + 5y = 3x – 6

5y + 6 = 3x – 3xy

5y + 6 = x (3 - 3y)

Now, we get the value of x = `(5y + 6)/(3-3y)`.

Step 3: Now change the variable x in terms of y and vice versa.

Hence y =  `(5x + 6)/(3-3x)`

Thus the reciprocal of the function f(x) =  `(3x - 6)/(3x+5)`. is given by the reciprocal of function f-1 =  `(5x + 6)/(3-3x)`.

Answer: f-1 =  `(5x + 6)/(3-3x)`.

Example 2:

Find the reciprocal of the function f(x) = 7x – 6.

Solution:

Step 1: Let write the given function as y = 7x – 6.

Step 2: Solve the function of x in term of y.

y = 7x – 6

Add 6 on both sides, we get y + 6 = 7x - 6 +6

y + 6 = 7x

Divide 7 on both sides, we get  `(y +6)/(7)`. =  `(7x)/(7)`.

x = `(y +6)/(7)`

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Step 3: Now change the variable x in terms of y and vice versa.

Hence y =`(x +6)/(7)`.

Thus the reciprocal of the function f(x) = 7x - 6 is given by the reciprocal of function f-1 = `(x +6)/(7)`.

Answer: f-1 = `(x +6)/(7)`.

Equivalence Statement

Logical Equivalence in simple word is the comparision of two logical statements. Meaning of the word logic is analysis. Analysis may be granted result or mathematical proof. Two statemnets satisfies the equivalance if the resultant truth table of both the given statements are exactly same. That is, the resultant truth values are the same. Equivalence statements satisfies the "if and only if" or biconditional operations. If A and B are two equivalant statements then one can be proved from another.
Symbol of Equivalence

We use '-= symbol to represent the term logical equivalence in discrete mathematics. Let us see about logical equivalence in this article.

Logical Equivalence Table

If the last column of two given components in the logical equivalence table are the same, then the two components are said to be logical equivalent.

We can  use'-= ' symbol to represent logical equivalences. Other name of logical equivalence is simple equivalence. These are the logical equivalence laws.

Solved Examples

Given below are some of the logical equivalence examples:

Example 1: Show ~ (Xvv Y) = (~X)^^ (~Y) simple equivalence.

Proof:

LHS:

~ (Xvv Y)
X Y Xvv Y
~ (Xvv Y)
T T T F
T F T F
F T T F
F F F T

RHS:

(~X)^^ (~Y)
X Y ~X ~Y (~X)^^ (~Y)
T T F F F
T F F T F
F T T F F
F F T T T

Last column of LHS table is equal to last column of RHS table. And so, it is a simple equivalence.

Example 2: Show   (X^^ Y) = ~ ((~X)vv (~Y)) simple equivalence.

Proof:

LHS:

(X^^ Y)
X Y (X^^ Y)
T T T
T F F
F T F
F F F

RHS:

~ ((~X)vv (~Y))
X Y ~X ~Y ((~X)vv (~Y)) ~ ((~X)vv (~Y))
T T F F F T
T F F T T F
F T T F T F
F F T T T F

Last column of LHS table is equal to last column of RHS table. And so, it is a simple equivalence.

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Example 3: Show (Xvv Y)vv (~(Xvv Y)) =(Xvv (~Y))vv Y simple equivalence.

Proof:

LHS:

(Xvv Y)vv (~(Xvv Y))
X Y Xvv Y
(~(Xvv Y) (Xvv Y)vv (~(Xvv Y))
T T T F T
T F T F T
F T T F T
F F F T T

RHS:

(Xvv (~Y))vv Y
X Y ~Y (Xvv (~Y)) (Xvv (~Y))vv Y
T T F T T
T F T T T
F T F F T
F F T T T

Last column of LHS table is equal to last column of RHS table. And so, it is a simple equivalence.

Multiplying Dividing Decimals

Decimal number is nothing but number with base 10. The multiplying decimals is nothing but to multiply the given number considering it as a whole numbers. At the end of calculation put the decimal point into position. For dividing decimals number by which dividing by, must be a whole number.division by 10 or by powers of 10 is easy in a decimal system. Let us see sample problems for multiplying dividing decimals.

Please express your views of this topic Multiplying Integers Rules by commenting on blog.

multiplying dividing decimals:

Multiplying decimals:

The multiplying decimals is nothing but to multiply the given number considering it as a whole numbers. At the end of calculation put the decimal point into position.


How to work out the position of the decimal point:


Count up how many figures there are after the decimal point in the question. There must be the SAME number of figures after the decimal point in the answer.


Eg: 6.7 x 2

Write this as            67
x 2
-------------
134
--------------
There is one figure after the decimal point in the question, so there must be only one after the point in the answer.
6.7 x 2 = 13.4

Eg: 30.6 x 0.7

Write this as          306
x 7
-----------------
2142
----------------
There are two figures after the decimal point in the question, from both the numbers, so there must be two after the point in the answer.
30.6 x 0.7 = 21.42

Eg: 2.9 x 0.0351
Write this as            351
x 29
--------------
3159
7020 *
-------------
10179
------------

There are five figures after the decimal point in the question so    2.9 x 0.0351 = 0.10179

multiplying dividing decimals:

Dividing decimals:

Division by powers of 10

Like multiplication, division by 10 or by powers of 10 is easy in a decimal system.


A quick way to divide a decimal number by 10 is to move the decimal point ONE place to the LEFT.

←
Eg 24.185 / 10 = 2.4185

dont forget when we are dividing you are making a number smaller.


←
Eg 0.6 / 10 = 0.06

To divide by 100 move the decimal point TWO places to the left.


←
Eg 32.157 / 100 = 0.32157

←
Eg 0.6 / 100 = 0.06

Dividing decimals:

Eg: `4.48 / 0.4`

Procedure:

The number by which dividing by, must be a whole number. To make this as a whole number we multiply it by 10 or 100 or 1000 or a higher power of 10,  if necessary.

Eg: 0.4 x 10 = 4

Whatever you do to the number you are dividing by, do exactly the same to the number which you are dividing.

Eg: 4.48 x 10 = 44.8

Then do the division, putting the decimal point of the answer directly above the other decimal point when you write out the division.

Eg:                11.2
4 ) 44.8

You don’t have to do anything else. The answer is 11.2

Example: `0.36 / 0.6`

First multiply 0.6 by 10 to make it into a whole number: 6

Then do exactly the same to 0.36: 0.36 x 10 = 3.6

Now divide 3.6 by 6:           0.6
6 ) 3.6
The answer is: 0.6

Recurrence Relations Tutor

A recurrence relation is an equation with the intention of recursively describes a series: each term of the series is defined as a function of the above terms. The difference equations refer to a particular type of recurrence relation. Note but the "difference equation" is commonly used to refer to any recurrence relation. Tutoring is method of teaching. Tutoring is to teaching a single student or a group of students.

An example of a recurrence relation is the logistic map:

x n + 1 = r x n (1-xn)

Recurrence relations tutor – Examples:

Recurrence relations tutor  - Example 1:

Find the limiting ratio `lim_(n->oo) (x_{n+1})/(x_n)` , for the recurrence relation `x_n = x_{n-1}+x_{n-2}.`

Solution:

We find what the limit must be, assuming that it exists.

L = `\lim_{n\rightarrow\infty} \frac{x_{n+1}}{x_n} = \lim_{n\rightarrow\infty} \frac{x_n+x_{n-1}}{x_n} = 1 + \lim_{n\rightarrow\infty} \frac{x_{n-1}}{x_n} = 1+L^{-1}`

L = 1 + `L^{-1}`

`L^2=L+1`

`L^2-L-1=0`

L = `\frac{1 \pm \sqrt{ 5 }}{2}` , via the quadratic formula.
Recurrence relations tutor - Example 2:

Let
`x_n = { x_(n-1) + x_(n-2)`     if n > 1
`x_(1) epsi N`                       if n = 1
`x_(0) epsi N`                       if n = 0
0                                if n < 0

Show that `x_n = x_{1}F_{n-1} + x_{0}F_{n-2} \,\! "where" F_n\!` is the n-th Fibonacci number   (F0 = F1 = 1)

Solution:

Using the principle of induction we have:

BASIS: n=`2 \Rightarrow x_2 = x_1 + x_0 = x_{1}F_{1} + x_{0}F_{0}\!`

INDUCTIVE STEP: We have `x_{n-2} = x_{1}F_{n-3} + x_{0}F_{n-4}\,\! and x_{n-1} = x_{1}F_{n-2} + x_{0}F_{n-3}\!`

By definition we have:


`x_{n} := x_{n-1} + x_{n-2} = x_{1}F_{n-2} + x_{0}F_{n-3} + x_{1}F_{n-3} + x_{0}F_{n-4}`

`= x_{1}\(F_{n-2} + F_{n-3}\) + x_{0}\(F_{n-3}+F_{n-4}\) = x_{1}F_{n-1} + x_{0}F_{n-2} \mbox{ } \!`

Recurrence relations tutor – More Problems:

Recurrence relations tutor - Example 1:

Let  .

`x_n := x_{n-1}*x_{n-2}`  if n>1
`x_1`                        if n=1
`x_0`                        if n=0
0                           if n<0 p="">
Show that `x_n = x_{1}^{F_{n-1}}*x_{2}^{F_{n-2}} \!` where `F_n\!` is the n-th Fibonacci number   (F0 = F1 = 1 and F(n < 0) = 0)

Solution:

As before: induction is da way!

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BASIS: For n = 2! we have `x_2 = x_1^{F_1}*x_0^{F_{0}} = x_1*x_0 \!`

INDUCTIVE STEP: `x_{n-2} = x_1^{F_{n-3}}*x_0^{F_{n-4}} \,\! and x_{n-1} = x_1^{F_{n-2}}*x_0^{F_{n-3}} \!` and

so `x_n = x_{n-1}*x_{n-2} = (x_1^{F_{n-2}}*x_0^{F_{n-3}}) * (x_1^{F_{n-3}}*x_0^{F_{n-4}}) `

= `x_1^{F_{n-2}+F_{n-3}}*x_0^{F_{n-3}+F_{n-4}}`

= `x_{1}^{F_{n-1}}*x_{2}^{F_{n-2}} !`

All Kinds of Fractions

A fraction means a part of a whole group or its region.A fraction written as  number  with the bottom part (denominator) showing how many parts the whole is divided into, and the top part (numerator). Fraction in all kinds can be written as many types which depends upon the numerator and denominator by its value.
Fraction in all kinds are

Proper fractions
Improper fractions
Mixed fractions

I like to share this Converting Fractions with you all through my article.

Classification for all kinds of Fractions:

Proper Fractions:
proper fraction is explained as  the denominator shows the number of parts into which whole divided and the numerator expressed  the number of parts which  we taken out. In other words numerator less than denominator.

Example:
`3/10,2/5`
Improper Fractions:
Improper fractions is the fraction whose  numerator is greater  than the denominator are called improper fractions.

Example:
`7/2, 9/7`

Mixed Fractions:
A  mixed fraction is typical fraction which  has  combination of a whole and its part.

Example:
2` 3/4,` 7` 2/9` ,



Addition and Subtraction fractions in all kinds:

Procedure for addition:

For addition,fraction numbers with same denominator, denominator remain same number and we add only the numerator.
For addition with different denominator fraction  we have to take lcm for all denominator and convert different denominator  into like denominator by taking LCM and add.


Example 1:
Add `1/5` and  `3/5`

Solution :
In this proper fraction we have same denominator
`1/5 +3/5` =` (1+3)/5`

=`4/5`
Example 2:

Add `2/3` and `1/5`
Solution:
In this fraction we have different denominator  so,we take LCM
The LCM of 3 and 5 is 15.

Therefore,`2/3+1/5` =`(2xx5)/(3xx5)+(1xx3)/(5xx3)`

=`10/15+3/15`

=`13/15`

Example 3:
Add `1 4/5` and `3 5/6`

Solution:
`1 4/5 +3 5/6` =  `(5+4)/5+(18+5)/6`


Now `9/5+23/6` =`(9xx6)/(5xx6)+(23xx5)/(6xx5) `    since LCM of 5,6 =30

=`36/30+115/30`

=`151/30`

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Rules for subtraction:

For subtraction,fraction numbers with same denominator, denominator remain same number and we subtract only the numerator.



For subtraction with different denominator fraction  we have to take lcm for all denominator and convert different denominator  into like denominator by taking LCM and subtract.


Example 1:

subtract  `1/3` from `2/3`
Solution:
here same denominator
`2/3-1/3`    =`(2-1)/3`

Example 2:
Find fraction for `4 2/5-2 1/3`

Solution :
Taking LCM for 5 and 3 is 15 because of different denominator
`4 2/5 -2 1/3` =  `22/5-7/5`
= `(22xx3)/(5xx3) -(7xx5)/(3xx5)`

=`66/15 -35/15`

=`(66-35)/15`

=`31/15`

Word Problems Review

In mathematics education, the term word problem is often used to refer to any mathematical exercise where significant background information on the problem is presented as text rather than in mathematical notation. As word problems often involve a narrative of some sort, they are occasionally also referred to as story problems and may vary in the amount of language used.(Source - Wikipedia )
In this article of word problems review, we are going to review some of the basic word problems.

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Review on word problems through examples:

Example 1:

An Auditorium has 1020 seats. 875 of them are occupied. What percentage of the seats are occupied?

Solution:

Number of seats in Auditorium  = 1020

Number of seats occupied        =  870

Percentage of the seats occupied  = `<< 870/1020>>`   x 100

=  `<< 87/102>>` x 100

=  85.29

Example 2:

In a Theatre, there are 400 persons consists of men, women, and children.There are four times as many men as children, and thrice as many women as children. How many of children are there?

Solution:

Let 'x' be the number of children.

Then the number of men = 4x and the number of women = 3x

So, the equation is      x + 4x + 3x = 400

8x = 400

x = 400 / 8

x = 50

Example 3:

A train 300 m long is running at a speed of 70 km/hr. What time will it take to cross a 50 m long bridge?

Solution:

In order to cross the bridge, the train will have to cover

distance = ( 300 + 50 )

= 350 m

Speed     = 70 x `<< 5/18>>`

= `<< 175/9>>`

= 19.44 m/s

Required time    = Distance / Speed

= `<< 350/19.44>>`

= 18 sec

Example 4:

A customer deposits an amount of $3000 in his account. The bank offers a interest of 2.5%. Calculate the simple interest for the customer earn in 7 years?

Solution:

Interest rate (r) = 2.5% = 0.025

Amount   (P) = $3000

time (t) = 7 years

Simple interest formula

I = P r t

I = 3000 × 0.025 × 7

= 5000 × 0.15

= $ 525

Practice problems for Review:

1) Adam deposits an amount of $6500 in his account with an interest of 4.5%. Calculate the simple interest that he earn in 5 years?

2) An Auditorium has 950 seats. 855 of them are occupied. What percentage of the seats are occupied?

Answer key:        1) $ 1462.5         2) 90 percent

Mixed Numbers

Mixed numbers the proper numbers is said to be numerator is less than denominator. Improper numbers says numerator is greater than denominator. A number consist of natural number and a fraction is called mixed number.Round mixed numbers as show in the below 2 `1/4` , 3 `4/5` , and 5 `7/8` . These are the mixed number. 2 `1/4` , In this mixed number the numerator is 1 and denominator is 4. 2 is out side of the given mixed number.First we have to find how to round the mixed number, here multiply 4 into the number 2 and add to the numerator, we get `(4xx2+1)/4` = `9/4`.

Problem in round mixed numbers

Problems 1 Round  mixed numbers and Adding mixed number:

(i)                  2 `1/5 ` + 3 `5/6`

Solution:

In the question `1/5` is the improper number because numerator is greater than denominator. Here 2 is out of the number. 2 `1/5` first we have to change in the proper number that is `(5*2+1)/5`

=` 11/5` is the proper number

The second number 3 `5/6`

In this number `(6*3+5)/6 ` we get

=`23/6`

Here we round  mixed numbers.

`11/5` + `23/6`

In this equation we have to take the Least common factor for 5 and 6 is 30

`11/5` +`23/6`

Here LCM is 30

`(11*6+23*5)/30 ` cross multiplication

`(66+115)/30` Adding  the number in the numerator we get

`=181/30.`

This is the answer for the adding mixed number for 2 `1/5 ` + 3 `5/6`

(ii)                2 `2/5` + 3 `1/6`

Solution: In the question `2/5` is the improper number because numerator is greater than denominator. Here 2 is out of the number.

2 `2/5` first we have to change in the proper number that is `(5*2+2)/5`

= `12/5` is the proper number

The second function is  3 `1/6`

In this function `(6*3+1)/6` we get

`=19/6`

Here we round  mixed numbers.

`12/5` + `19/6`

In this equation we have to take the Least common factor for 5 and 6 is 30

`12/5` +`19/6`

Here LCM is 30

`(12*6+19*5)/30` cross multiplication

`(72+95)/30` Adding the numbers in the numerator we get

`=167/30.`

This is the answer for the adding mixed numbers for 2` 2/5` + 3 `1/6.`

Round mixed number and adding mixed number with same denominator:

Problem for round mixed numbers and Add with same denominator

(i)  2 `1/8 ` + 3 `2/8`

Solution:

In the question `1/8` is the improper numbers because numerator is greater than denominator.

Here 2 is out of the number.2 `1/8` first we have to change into the proper numbers

`(2xx8+1)/8` ` = 1 7/8` is the proper numbers

The next fraction 3 `2/8`

The same we have to do this also we get

`(3xx8 +1)/8 = 25/8` is the proper number

Here we round mixed numbers and adding both functions

` = 17/8` + `25/8`

In this step adding mixed numbers is the easier way to add because the denominator is same so we can add directly

`(17+25)/8 =42/8`

So  `48/7` is the answer for the adding mixed fraction of 2 `1/8 ` + 3 `2/8`

Example Problem for round mixed numbers and Add with same denominator

2 `3/8` + 3 `5/8`

Solution:

In the question `3/8` is the improper numbers because numerator is greater than denominator.

Here 2 is out of the number. 2 `3/8` first we have to change into the proper numbers

`(2xx8+3)/8` = `19/8` is the proper numbers

The next fraction is 3 `5/8`

The same we have to do this also we get

`(3xx8 +5)/8 =29/8`

Here we round  mixed numbers

`=19/8` + `29/8`

In this step adding mixed numbers is the easier way to add because the denominator is same so we can add directly.

`(19+29)/8=``48/8`  is the answer for the Adding mixed numbers of 2 `3/8` + 3 `5/8`