Addition of Decimals Methods

In this article we are going to learn about adding decimals concept.Decimal is base 10 number system. A number consists of decimal part and whole number part is known as decimal number. The point which splits the whole number part and decimal part is known as decimal point. The digit after a decimal point is decimal part and the left to the point is whole number part.

Addition of Decimals Methods

Th following are the steps for addition with decimals:

Step 1: Checking the decimal part of given numbers and making them equal in number of digits by adding zeros at the end.

Step 2: Write down the numbers vertically one under the other with the aligned decimal point

Step 3: Add the numbers as normal addition and place the decimal point in the result

Examples

Below are the examples based on Adding decimals:

Example 1:

1)      Add 12.43 with 5

Sol:

Write 5 as 5.00 (as 12.43 has 2 digits after decimal point)

Add 12.43+5.00

12.43

5.00

---------

17.43

Hence the sum is, 12.43 + 5 = 17.43

Example 2:

Add 1.34 + 6 + 12.7.

Sol:

Making decimal part equal ,

6= 6.00, 12.7 = 12.70

1.34

6.00

12.70

--------

20.04

Hence the sum is, 1.34 + 6 + 12.7 = 20.04

Example 3:

Add 0.803+ 3.1+ 12

Sol:

Making decimal part equal,

3.1 = 3.100, 12 = 12.000

0.803

3.100

12.000

------------

15.903

Hence the sum is 0.803+ 3.1+ 12 = 15.903

Example 4:

Add   7.09 +9.20 + 0.36

Sol:

7.09

9.20

0.36

---------

16.65

Hence the result is 7.09 +9.20 + 0.36 = 16.65

Example 5:

Add 45. 56 + 6.7+ 2

Sol:

Making decimal part equal,

6.7= 6.70, 2=2.00

45.56

6.70

2.00

-------------

54.26

Hence the sum is 45. 56 + 6.7+ 2 = 54.26

Addition with decimals Ex 5:

Add   0.5795 + 2.5301

Sol:

0.5795

2.5301

-----------

3.1096

Hence the sum is 0.5795 + 2.5301= 3.1096

Addition with decimals Ex 6:

Add 82.543+322.916

Sol:

82.543

322.916

----------------

405.459

Hence the sum is 82.543+322.916 = 405.459

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Addition with decimals Ex 7:

Add 369.2165 with 100

Making decimal part equal,

100 = 100 .000

369.2165

100 .0000

-------------

469.2165

Hence the sum is 369.2165+100 = 469.2165.

Addition with decimals Ex 8 :

Add 0.00013+3.902+56.7

Sol:

Making decimal part equal ,

3.902 = 3.90200 , 56.7 = 56.70000

0.00013

3.90200

56.70000

------------------------

60.60213

Hence the sum is 0.00013+3.902+56.7 =   60.60213.

Practice problems

Below are the practice problems on adding decimals:

Add 32+4.5+7.03

Answer: 43.53

Add 2.34+5.6+0.007

Answer: 7.947

Add 7.985 with 9.71485

Answer: 17.69985

Add 0.009+7.89+6.0

Answer: 13.899

Add 1.111+4.67+17

Answer: 31.899

Transformations of Functions

In this page we are going to discuss about transformations of functions concept . Function is defined as one quantity (input of transformations functions) associated with another quantity (output of transformations functions). The quantity can be a Real numbers or Elements from any given sets or the domain and the co domain of the function.

For Example: The function is defined as f : C -> D is a relation that assigns to each x belongs to C to y belongs to D. C is Domain of f and D is Range of f.


Types of transformations of functions

There are four classes of transformations,

1. Horizontal translation: The function is transformed along X axis.

g (x) = f (x + c)

It means that the graph is translated c values to the left side for c > 0 or to the right side for c < 0.

2. Vertical translation: The transformation of function is along Y axis.

g (x) = f (x) + k

It means that the graph is translated k values upwards for k >0 or downwards for k < 0.

3. Change of amplitude:

g (x) = b f (x)

It means that the amplitude of the graph is increased by a factor of b if b > 1 and decreased by a factor of b if b < 1, if b < 0, then we get inverted graph.

4. Change of scale:

g (x) = f (cx)

It means that the graph is compressed if c > 1 and stretched out if c < 1. If c < 0 then we get the reflected graph about y axis.

Examples

Below are the examples on transformations of functions -

Examples:

Examples for functions include parabolas, trigonometric curves and polynomial functions.

* f (x) = 2x2 , for all x values are real

* f(x) = y + sin x, x ,y are real.

* f (x) = 1 / (x+1), for all real numbers except -1

*f (x) = x3-4x2+9x , Polynomial function


Even or Odd Functions

A function f:C-->D is said to be even if and only if f (-x) = f (x) for all x belongs to C.

A function is said to be odd if and only if f (-x) = -f(x) for all x belongs to C.

Even function is symmetric about the y-axis; an odd function is symmetric about the origin in the graph.

Example :  * f (x) = 2x2 is an even function.

* f (x) = x + sin x is odd.

Algebra Coefficient Variable

In mathematics, a coefficient is defined as the number in front of the variable. For example 2x is the given expression here the coefficient is 2. The coefficient is usually in numeral with the variable. The expression contains variable and coefficient of the variable. Here in this topic we are going to see about coefficient of variables.

Example problem for the coefficient:

Example 1:

Find the coefficient of the variables in the given algebraic expression:

x + 2y

Solution:

Given that x + 2y

Here x and y is the expression with the coefficient

The coefficient of x is 1

The coefficient of y is 2.

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Example 2:

Find the coefficient of variables in the given algebraic expression:

y2 + 2y + 3xy +1

Solution:

Given that y2 + 2y + 3xy +1

Here y2 , 2y,  3xy, is the expression with the coefficient and 1 is without variable

The 3xy having two variables that two variable consider as a one variable

1 is the constant term of the given algebraic expression

The coefficient of y2 is 1

The coefficient of y is 2

The coefficient of xy is 3

Example 3:

Find the coefficient of variables in the given algebraic expression:

z5 + z + y +6y

Solution:

Given that z5 + z + y

Here z5, z, y, 6y is the expression with the coefficient

The coefficient of z5 is 1

The coefficient of z   is 1

The coefficient of   y is 1

Example 4:

Find the coefficient of  variables in the given algebraic expression:

x3 + y 5 +2xy + 5yx

Solution:

Given that x3 + y 5 +2xy + 5yx

Here x3, y 5, 2xy, 5yx the expression with the coefficient

The coefficient of x3 is 1

The coefficient of y5 is 1

The coefficient of xy is 2

The coefficient of yx is 5.

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Example 5:

Find the coefficient of variables in the given algebraic expression:

100x2 + 2z3 + 102z

Solution:

Given that 100x2 + 2z3 + 102z

Here 100x2, 2z3, 102z the expression with the coefficient

The coefficient of x2 is 100

The coefficient of z3 is 2

The coefficient of z is 102.

Law of Cosines Calculator

Law of cosines calculator is a tool to calculate calculation applying law of cosines easily. First let's understand the concept of law of cosines. In trigonometry, the law of cosines (also known as the cosine formula or cosine rule) is a statement about a general triangle that relates the lengths of its sides to the cosine of one of its angles. The law of cosines states that,

c2 = a2 + b2 – 2ab cos γ,

The law of cosines generalizes the Pythagorean theorem, which holds only for right triangles: if the angle γ is a right angle (of measure 90° or `pi/2` radians), then cos(γ) = 0, and thus the law of cosines reduces to,

c2 = a2 + b2

Formula for Law of Cosine:

The law of cosines is used to solve the third side of a triangle, when other two sides and the angle are known.

By changing the sides of the triangle, one can find the following two formulas also to solve for the law of cosines,

a2 = b2 + c2 – 2bc cos α,

b2 = a2 + c2 – 2ac cos β.

Law of Cosines Using Distance Formula:

To solve for the law of cosines, consider a triangle with a side length of a, b, c, and θ is the measurement of the angle opposite to the side length c.

A = (bcosθ, bsinθ), B = (a, 0), and C = (0, 0).

By using distance formula, we have,

c = `sqrt((a - bcostheta)^2 + (0 - bsin theta)^2)` ,

Now, squaring on both sides, we get,

c2 = (a – bcos θ)2 + (–bsin θ)2

c2 = a2 – 2ab cos θ + b2cos2θ + b2sin2θ,

c2 = a2 – 2ab cos θ + b2(cos2θ + sin2θ)

c2 = a2 + b2 – 2ab cos θ.               [where, cos2θ + sin2θ = 1].

Law of Cosines Using Trigonometry:

To solve for the law of cosines, draw the perpendicular to the side c as shown in the figure,

c = a cosβ + b cosα.

Multiply by c, we get,

c2 = ac cosβ + bc cosα                                                 (1)

By considering the other two perpendiculars, we get,

a2 = ac cosβ + ab cosγ                                                (2)

b2 = bc cosα + ab cosγ                                                (3)

Adding equations (2), and (3), we get,

a2 + b2 = ac cosβ + ab cosγ + bc cosα + ab cosγ

a2 + b2 = ac cosβ + 2ab cosγ + bc cosα                       (4)

Subtracting equation (1) from the equation (4), we get,

a2 + b2 – c2 = (ac cosβ + 2ab cosγ + bc cosα) – (ac cosβ + bc cosα)

a2 + b2 – c2 = 2ab cosγ,

a2 + b2 – 2ab cosγ = c2,

c2 = a2 + b2 – 2ab cosγ.

Examples based on Law of Cosine:

Ex 1: Evaluate the length of side A using law of cosine formula.

Sol:

law of cosine example 1

Step 1:   A² = B² + C² −2(B)(C)cos (`-<`1) ( law of cosine formula)

Step 2:  Plug in the values of B and C

A² = 20² + 13² −2(20)(13)cos(66)

A² = 400 + 169 −520 cos(66)

Step 3: Add and subtract

A² = 569 −211

A² = 358

Step 4:  Take square root

A= √358 = 18.9

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Ex 2:  Evaluate x using law of cosine formula.

Sol:

law of cosine example 2

Step 1:   A² = B² + C² −2(B)(C)cos ( 1)

25² = 32² + 37² −2(32)(37)cos(x)

Step 2:  Solve the equation

625 = 2393 − 2,368 cos(x)

- 1760 = -2, 368 cos (x)

. 7432 = cos x

Cos-1 (.7432) = 42.0 º

Least Common Multiples

In arithmetic and number theory, the least common multiple or lowest common multiple (LCM) or smallest common multiple of two integers a and b is the smallest positive integer that is a multiple of both of a and of b. Since it is a multiple, it can be divided by a and b without a remainder. A few word problems for least common multiples is given below.

(Source: Wikipedia)

Example of word problems for least common multiples:

Word problem 1:

Find the largest number of four digits which when divided by 5, 10, 15, it leaves a remainder 4 in each case.

Solution:

Step 1: Given numbers

5, 10, 15

Step 2: Find least common multiple of 5, 10, 15

Multiples of 5 = 5, 10, 15, 20, 25, 30, 35, 40....

Multiples of 10 = 10, 20, 30, 40, 50, 60, 70, 80...

Multiples of 15 = 15, 30, 45, 60, 75, 90, 105....

From the list of multiples of 5, 10 and 15 the smallest common number in each list 30.

Therefore, the least common multiples of 5, 10 and 15 are 30.

Step 3: Find multiple of 30 which should be slightly less than five digit.

30 * 333 = 9990

So, 9990 is the largest number of four digit which is divisible by 5, 10, 15 and leaves a remainder 0.

Step 4: Find number which leaves a remainder 4.

To get remainder 4, we should add 4 to the obtained number.

Therefore, the required number is 9990 + 4 = 9994

Word problem 2:

Three children John, Nick and Shane run on a round  track. John takes 50 seconds, Nick takes 55 seconds and Shane takes 60 seconds to run a round. If all three of them start together at a point, when do they meet again?

Solution:

Step 1: Find least common multiple of 50, 55, 60

10 |       50       55       60

-----------------------------------------

5  |      5          55       6

-----------------------------------------

1           11       6

Least common multiple = 10 * 5 * 1 * 11 * 6 = 3300

Step 2: Solution

Therefore, they meet after 3300 seconds = 55 minutes.

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Homework of word problems for least common multiples:

1) Find the largest number of three digits which when divided by 7, 14, 28 it leaves a remainder 2 in each case.

2) Two children Joseph and Fleming run on a round track. Joseph takes 75 seconds and Fleming takes 80 seconds to run a round. If both of them start together at a point, when do they meet again?

Solutions:

1) 982

2) 1200 seconds

How to Find Number of Factors

In mathematics, factorization (also factorization in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 × 5, and the polynomial x2 − 4 factors as (x − 2) (x + 2). In all cases, a product of simpler objects is obtained.

How to find number of factor:example problems

Example 1:

Find all number of factors 50.
Solution:

50 = 1x50
= 2x25
= 5x10

So, the factors of 50 are 1, 2, 5, 10, 25 and 50.In the above example for how to find number of factors.

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Example 2:

Find all numbers of factors 80

Solution:

80 = 1x80
= 2x40
= 4x20
= 5x16
= 8x10

So, the factors of 80 are 1, 2,4,5,8,10,16,20,40 and 80.In the above example for how to find number of factors.

More explanation of how to find number of factors:-

Every number greater than 1 have atleast two factors: 1 and itself.

Example,

2 = 1 * 2

3 = 1 * 3

4 = 1 * 4

In two numbers are factors of another number are multiplied.

Note:- In 4 has some other factors besides 1 and 4:

4 = 1 * 4

4 = 2 * 2

A number 36 is a factors 4,

36 =  1 * 36

=  2 * 18

=  3 * 12

=  4 *  9

=  6 *  6

In those factors 1 and 36:

36 = 1 * 36

Divide by the next highest number after 1 and 2 is goes to 36. 18 times, 2 and 18 are a pair of factors:

36 =  1 * 36

=  2 * 18

=  3 * 12

=  4 *  9

Note:- 5 doesn't work, so leave that, and go on to 6:

36 =  1 * 36

=  2 * 18

=  3 * 12

=  4 *  9

=  6 *  6

Again, 7 doesn't work, and neither does 8. But 9 works:

36 =  1 * 36

=  2 * 18

=  3 * 12

=  4 *  9

=  6 *  6

=  9 *  4

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but it would be add on the small  table, because it says the same thing as an earlier entry:

36 = 1 * 36

= 2 * 18

= 3 * 12

= 4 *  9  <-- p="">
= 6 *  6     |  the same factors

= 9 *  4  <-- p="">
Any numbers larger than 6 remaining are smaller than 6. The factors of 36 are 1, 36, 2, 18, 3, 12, 4, 9, and 6.

Numerical Integrals

The process of finding an integral; it may be definite integral or an indefinite integral is referred as integration. Numerical integration is mainly used for finding the numerical value of a definite integral. Numerical integration is also used to finding the numerical solution of differential equations. Numerical integration is also referred to as numerical quadrature.

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Numerical integrals:- types of integrals

1. Indefinite integrations:

An indefinite integration is the family of functions that have a given function as a common derivative. The indefinite integral of f(x) is written ∫ f(x) dx.

2. Definite integrations:

If F(x) is the integral of function f(x) over the interval [a, b] ,i.e., ∫ f(x) dx = F(x) then the definite integral of function  f(x) over the interval [a, b] is denoted by int_a^bf(x)dx and is defined as int_a^bf(x)dx  = F(b) – F(a).

Where 'a' is called the lower limit and b is called the upper limit of integration and the interval [a, b] is called of integration.

Methods of integration:

It is not possible to integrate each integral with help of the following methods but a large number of varieties of the problems can be solved by these methods so, we have the following methods of integration:

1. Integration by substitution

2. Integration by parts.

3. Partial fractions

Example Problems on indefinite and definite integration:

Numerical Integrals Problem:

Example 1:

find ∫ 1 / sin2x cos2x dx.

Solution:

We have ∫ 1/ sin2x cos2x dx

= ∫sin2x+cos2x / sin2x+cos2x dx

=∫ (1/cos2x+1/ sin2x )dx

=∫sec2 x d x+∫cosec 2 x dx

=tan x -cot x+C.

Example 2:

∫ sin x sin (cos x) dx.

Solution:

Let cos x = t

dt = -sin x dx

Therefore we have

∫sin x sin (cos x) dx = - ∫sint dt = cost + C=cos (cos x) + C.

where

t=cos x

Example 3:  int_0^1dx/(1+x2)

Solution:

=[tan-1 x]10

=tan-1 1 - tan-1 0

= /4 -0 = / 4

π/2

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Example 4: Find the value int sin⁷xdx

-π/2

Solution:

Let f (x) = dx.

Then f(-x) = -sin7 x=-f (x)

so, f (x) is odd function.

π/2

π/2

Therefore, int f(x)dx=0 => int sin7x dx = 0.

-π/2                  -π/2

Laws of Exponents for Real Numbers

Let ‘a’ be a positive number. Exponential notation is ax , where x is an integer. When x is a rational number, say p/q with p, an integer and q, a positive integer, we define ax by ax  = root(q)(a) p

Example:

53/8 = root(3)(5)8

laws of exponents for real numbers: Exponents Notation

When x is an irrational number, ax can be defined to represent a real number.

For any a > 0, ax can be defined and that it represents an unique real number u and write u = ax.

The real number u is written in the exponential form or in the exponential notation.

Here the positive number 'a' is called the base and x, the index or the power or the exponent.

The laws of indices which we have stated for integer exponents can be obtained for all real exponents.


laws of Exponents for Real Numbers

We state them here and call them, the laws of exponents for real number:

root(x)(root(y)(a))  = root(xy)(a)


root(x)(a / b)  =  root(x)(a) / root(x)(b)


(root(x)(a) )x = a


axxx ay = ax+y


ax// ay = ax-y


(ax)y = axy


a-x  = 1/ ax


axxx bx = abx


a0 = 1  for a != 0


(a/b)x = ax/bx


a1/x =  root(x)(a)


(root(x)(a) )y/x = (root(x)(a) )


root(x)(a) root(x)(b) = root(x)(ab)



Example sums for Exponents for Real Numbers

Example 1:

Find the exponential form of (25)1/17

Solution of example 1:

Given = (25)1/17

Exponential form  =  root(17)(25)

Example 2:

Simplify 25225 / 53253

Solution of example 2:

Given 25225 / 53253

Now we use the exponents formula for real numbers

By using

(i)axxx ay = ax+y

(ii)axxx bx = abx

We get

= 253 / (5(25))3

= 253 / 1253

= 253 / (5)3(25)3

Cancel the same variables

Now, we get

= 1 / 53

= 1 / 125

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Example 3:

Solve ( root(3)(root(2)(64))92xx 62) / ((43)2+(5/25)2)

Solution of example 3:

Given = ( root(3)(root(2)(64))92xx 62) / ((43)2+(5/25)2)

Now we use the exponents formula for real numbers

By using below laws of exponents formula we can simplify the given data.

(i) root(x)(root(y)(a)) = root(xy)(a)

(ii)axxx bx = abx

(iii)(ax)y = axy

(iv)(a/b)x = ax/bx

= ( root(3)(root(2)(64))92xx 62) / ((43)2+(5/25)2)

Use the (i) formula

We get

= root(6)(49))92xx 62) / ((43)2+(5/25)2)

Use the (ii) formula

We get

= ((root(6)(64))542) / ((43)2+(5/25)2)

Use the (iii) formula

We get

= ((root(6)(64))542) / ((46)+(5/25)2)

Use the (iii) formula

We get

= ((root(6)(64))542) / ((46)+(52/252))

= (2 (542) / ((256) + (25 / 252))

Cancel the same elements

We get

= 2(542) / ((256) + (1 / 25))

= 2(542) / ((6400 + 1) / 25 )

= 2(2916) / (6401/25)

= 2(2916)(25) / (6401)

= 50(2916) / 6401

= 145800 / 6401

= 22.77

What are Perpendicular lines

Two intersecting lines will have four angles formed at the intersection points. If all the four angles are equal, then the two lines are said to be perpendicular to each other. We already know by linear postulate theorem that the two vertically opposite angles are equal. Hence if these two lines are perpendicular, then all four angles are 90 degrees.

Examples of perpendicular lines:

In the graph paper, The X-axis and Y-axis are perpendicular.
In an ellipse two axes, minor axis, and major axis are perpendicular.
For a line segment, any shortest line from a point outside the circle is perpendicular.
Tangent and normal to any curve are perpendicular lines.

Slopes of two perpendicular lines:  In coordinate Geometry, when two lines are perpendicular, the product of the slopes of the lines is -1.  This property has a lot of applications in finding the equation of perpendicular lines, length of perpendicular segment from a point to a given line, etc.

For any curve in a graph with equation y = f(x), the slope of the tangent is defined as the rate of change of y with respect to x at that point. The normal to this curve at this point is perpendicular to the tangent line.

Example:  In a circle, with centre at the origin and radius 3, the equation will be of the form (x)²+(y)² = 3². Take any point say (0,3). To find the tangent, we have to find dy/dx.

Differentiating, 2x+2y  =0

Hence, the slope of the normal is perpendicular to x axis or parallel to y axis.

Example for Perpendicular lines from a point to a line

Let AB be a line with coordinates (1,2) and (3,4).  Measure the length of perpendicular line from (-1,1) to this line segment.

We know that the perpendicular line from (-1,1) has a slope  of -1/slope of AB.

Equation of AB is (x-1)/(3-1) = (y-2)/(4-2)  Or x-1 = y-2 Or y = x+1

Slope of AB passing through (1,2) and (3,4) is 4-2/3-1 =1.

Slope of perpendicular line to AB is -1.

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Since the perpendicular line passes through (-1,1) equation of the perpendicular is y-1 = -1(x+1)  or y =-x -1 +1 or y = -x.

To get the foot of the perpendicular line on AB, we solve the two equations by substitution method.

y = x+1 = -x   This on simplification gives 2x=-1 or x=-1/2.

Since y = -x , we have y = +1/2,

So, foot of the altitude from the point (-1,1) is (-1/2,1/2).

The length of the perpendicular segment is between (-1,1) and (-1/2,1/2) is

√[ (-1/2+1)²+(1/2-1)²] = √(1/4+1/4)   =  √(1/2) = 1/1.414 =0.707 approximately.

Examples of Concave Polygon

The concave polygon has single interior angle which is more than 180 degree. The concave method is drawing few of the straight line only. The concave polygon is used in many interior angles. The three sided polygon denoted the triangle, triangle cannot be a concave polygon. In this article we shall discuss the examples and properties of concave polygon.

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Properties of concave polygons:

The concave polygons do not calculate the exterior angle. The opposite meaning of the concave is called the convex. The polygon is normally declared as concave polygon, but not the convex polygon. The interior angle must be reflex angle of the concave polygon.

The diagonal of the concave polygon is the line through the outside of the polygon. The star polygon is the example of non simple polygon. It must  contain the concave polygon with minimum of four sides. The area of the concave polygon and irregular polygons are same.

Example of the concave polygon

The first example of the concave polygon is declaring the following figure. The concave polygon connected the line of the interior angle side. The important part is declaring interior. There are only  two points selected inside the diagram. Do not select the one point which is inside of the angle and another point is outside of the diagram.

diagram repsent the example of the concave polygon

The inside of the polygon does not enter the line P and Q. The line A and B is entering the inside of the polygon. So that it is said to be concave polygon. The minimum one line segment will not present inside the diagram, this part of the diagram is called the concave.


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The above example is commonly called the concave polygon. The next example of the concave polygon is declaring the following diagram. In figure the blue line segment is declared as two points enter into the inside of the diagram. That line through the outside of the diagram.

Learning Disjunction

Definition:

Logical disjunction is an operation on the two logical values, typically the values of two propositions, that produces a value of false if and only if both of its operands are false. More generally a disjunction was a logical formula that can have one or more literals separated only by ORs. A single literal is frequently considered to be a degenerate disjunction.

Properties:

Associativity :

aV (bVc) = (aVb) V c

In mathematics, associativity was a property of some binary operations. It means that, within the expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. Consider for instance the equation

(5+2) + 1 = 5 + (2+1) = 8

Commutativity:

In mathematics, commutativity is the assets that changing the order of something does not change the end result. It is a primary property of many binary operations, and many mathematical proofs depend on it. The commutativity of easy operations, such as multiplication and addition of numbers, was for many years implicitly assumed and the property was not named until the 19th century when mathematics started to become formalized.

Distributivity:

In mathematics, and in particular to abstract algebra, distributivity is a property of binary operations that generalizes the distributive law from elementary algebra. For example:

2 × (1 + 3) = (2 × 1) + (2 × 3).

Idempotence:

Idempotence is a property of certain operations in mathematics and computer science. Idempotent operations are operations that can apply multiple times without changing the result. The conception of idempotence arises in a number of places in abstract algebra and functional programming.

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Monotonic function:

In mathematics, a monotonic function in which conserve the given order. This concept first arose in calculus, and was shortly generalized to the more abstract setting of order theory.

Symbol:

The mathematical symbol for logical disjunction varies in the text. In addition to the word "or", the symbol "V", deriving from the Latin word vel for "or", is commonly used for disjunction. For example: "A V B " is read as "A or B ". Such a disjunction is false if both A and B are false. In all other cases it is true.

All of the following are disjunctions:

A V B

¬A V B

A V ¬B V ¬C V D V ¬E

Cumulative Frequency Distribution

The total frequency of all classes less than the upper class boundary of a given class is called the cumulative frequency distribution .Relative frequency is the fraction of total number of elements .There are two types of Cumulative  frequency distributions, as follows

1.    Less than cumulative frequency distribution

2.    More than cumulative frequency distribution

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Cumulative distribution function in cumulative frequency distribution

In common terms, the cumulative frequency distribution is the sum of all the frequencies. A cumulative frequency distribution is a sum of a set of data showing the frequency or number of items less than or equal to the upper class limit of each class. The Cumulative distribution function (CDF), describe the Probability distribution of random Variable X

The random variable X is given by

X -> Fx(x) =P(X<=x),

P(X<=x) =The probability that the random variable x takes on a value less than or Equal to x.

FX(b) − FX(a) if a < b.

Where

F for cumulative distribution function

The probability density function f can write as follows:

F(x) = $\int_{-\infty}^x f(t)\,dt$

Where

f(t) means probability density function

Applications and uses of cumulative frequency distribution

Cumulative frequency gives the total number of outcomes that occurred up to some value. Cumulative frequencies are used in risk or reliability analysis. In frequency distribution, every bin has the number of values that lies within the range of values that define the bin. In a cumulative distribution, every bin has the number of data that falls within or below the bin. A graph contains a frequency distribution on the left, and a cumulative distribution of the same data on the right is called a cumulative frequency distribution

The main advantage of cumulative frequency distribution is that one doesn’t need to decide on a bin width or the frequency distribution analysis dialog, can choose among  number of ways to graph the resulting data.


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practice Constant

The values can’t be change. Constants it’s also called variable. In math, constant is a number, But sometimes we can also take the variable as a constant. For example,  In this equation x2+5x+3 = 0, 3 is a constant.  In this equation x+5=-20, 5 and -20 are constants. Now we are doing to practice some constant problems.

Understanding Chain Rule Practice is always challenging for me but thanks to all math help websites to help me out.

Practice constant Problems:

Practice problem 1:

X2+4x+k= 24. If x= 2, solve for k .

Solution:

Step 1: Substitute x value in the given equation.

Step 2: So we get (2)2+4(2)+k=24.

Step 3: Here we need to simplify this.

Step 4: 4+8+k=24.

Step 5: When we add we get 12+k=24.

Step 6: Subtract 12 on both the sides so, 12-12+k= 24-12.

Step 7: Therefore answer is k=12.


Practice problem 2:

X+4y-3z+r= 51...if x=9,y=3 and z=2 ,solve r.

Solution:

Step 1: Substitute x, y and z value in the given equation.

Step 2: So we get, 9+4(3)-3(2)+r= 51.

Step 3: Now we need to simplify this equation.

Step 4: 9+12-6+r=51.

Step 5: When we simplify we get 15+r= 51.

Step 6: Divide using 15 on both the sides.

Step 7: Therefore, the answer is 51/15.


Practice problem 3:

M3+m2+6m+S= 72. If m= 2, solve for S.

Solution:

Step 1: Substitute m value in the given equation.

Step 2: So we get (2) ^3+ (2) ^2+6(2) +S=72.

Step 3: Now we need to simplify this equation.

Step 4: 8+4+12+S=72.

Step 5: When we add we get 24+S=72.

Step 6: Subtract  24 on both the sides so, 24-24=72-24 .

Step 7: Therefore the value of s is 3.

My forthcoming post is on Conditional Probability Venn Diagram and Associative Property of Multiplication Example will give you more understanding about Algebra.


Practice problem 4:

C2+15c+g=195. If c= 3 then find g.

Solution:

Step 1: Substitute m value in the given equation.

Step 2: So we get 32+15(3)+g=195.

Step 3: Here we need to simplify this equation.

Step 4: 9+45+g=195.

Step 5: After addition we get 54+g=195.

Step 6: Divide using 54 on both the sides.

Step 7: So the value of g =195/54.

work out problems:

X2+19x+t=121 if x=12, find the value of t?
A3+4a+z= 147 if a =5, find the z value?
H2+h+m= 12 if h=6, find the m value?
F4+8f+n=546 if f=4, find the n value?

Math Operations

Operator in math is a function which acts on two values ( operands). These operators are known as binar operators as they need two operands. There are four binary operators namely +  ,  -   ,  x   , /. Operator are mainly consists of number of terms. For example, the assignment operator are used for assigning the values to a particular variable.

The four fundamental operations on numbers are Addition, Subtraction, Multiplication and Division. The operator of Addition is ‘+’, Operator of Subtraction is ‘-’, Operator for multiplication is ‘x’ and Operator for division is ‘÷’. Let us see about arithmetic operations with appropriate operator.
Explanation for various types of operator statement:

1. '+' is an  operator statement:

For this operator, the addition operation and the string concatenation operation are included.
Addition operation is used for adding the two values that are given.
String concatenation are used for joining the two statements.

2. ++ is an increment operator:

For this, the values are incremented according to the given values.

3. -- is an decrement operator:

For this, the values are decrement according to the given values.

4. x  is a multiplication operator:

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For this, the multiplication function and the pointer function are included.
Multiplication function is used to multiply the two given values.
Pointer are used for denoting the reference variable.

5. / is a division operator:

For this, the values are divided for getting the remainder and the quotient factor.

6.  ^  exponential operator:

Exponential operator is used for raising the power to a function.

7. -  mathematical operator:

For this, the subtraction and the negation process are included in this operator statement.
Subtraction operation is used for finding the difference of the given two numbers.


Examples using math operations:

Addition Operator ‘+’: For adding two or more numbers we use the ‘+’ operator. This operator is known as plus.Normally additions have two operands and one operator. For example x + y here ‘x’ and ‘y’ is called as operator and the symbol’+’ is called as operand.

Ex: Add 5 and 6.

Sol:

5 + 6 = 11

Where, ‘+’ is the operator of addition.

5 and 6 are operands.

11 is the resultant of the operation of addition.

Subtraction Operator ‘-’: For subtracting two or more numbers, the operator used is ‘-’.  This operator is known as minus.

Ex : Subtract 11 and 5.

Sol:   11 – 5 = 6.

Where, ‘-’ is the operator of subtraction.

11 and 5 are operands.

6 is the resultant of the operation of subtraction.

Multiplication Operator ‘x’: For Multiplying two or more numbers, the operator used is ‘x’.  This operator is known as into.

Ex : Multiply 12 and 5.

Sol:             12 x 5 = 60.

Where, ‘x’ is the operator of multiplication.

12 and 5 are operands.

60 is the resultant of the operation of multiplication.

Division Operator ‘÷’: For dividing two or more numbers, the operator used is ‘÷’.  There is no special name for division operator.

Ex : Divide 10 by 5.

Sol:   10 ÷5 = 2.

Where, ‘÷’ is the operator of Division.

10 and 5 are operands.

2 is the resultant of the operation of division.

Additional Operators:

‘√’:

This operator is known as the square root.

Ex: Find Square root of 4?

Sol: √4 = 2.

Where, ‘√’ is the operator of square root.

4 is the operand.

2 is the resultant of the operation of square root.

Order of operations in math:

The order of operation is square root, division, multiplication, Subtraction and addition.

Ex: Calculate 5 + 6 – [(√4 ÷2) x 3].

Sol:  5 + 6 – [(2 ÷ 2) x 3]

5 + 6 – [1 x 3]

5 + 6 – 3

5 + 3

7

Steps and examples for addition operation in math:

Simple addition uses the following steps to adding the numbers.

Step 1: Add the first place number from the right (first number) first after that take over the carry to next step.

Step 2:   Add the ten’s place number (means second digit) as well as also add the carry from the first step. Now note down the answer and over the carry.

Step 3  :Carry on this process on until reach the end place of the given number.

The followings are the some of the types of addition

Add single digit number with single digit
Add single digit number with double digit
Add three one digit numbers
Add double digit number with three digit number
Add double digit number with double digit number
Add treble digit number with treble digit number etc….

Example for Simple Addition :

Add single digit number with single digit:

Ex: Add 5 with 9

Sol:

5
(+) 9
-----------     14

Ex 2: Adding the following three one digit numbers.

9, 6 and 3

Sol:

Given numbers 9, 6 and 3

Step 1:

Add the first two numbers

9

(+)      6

----------------------

1 5

Step 2:

Add the next number with the previous step sum or total.

Previous step total =15

Third number =3

1 5

(+)        3

----------------------

1 8

Therefore 9 + 6 +3 =18

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Ex 3: Add the following two numbers: 54 with 16

Sol:

Given numbers 54 and 16

Step 1:

Add the first digit (ones place) numbers (4+ 6 = 10) then over the carry (here 1 is the carry)

5 4
(+)   1 6
-----------
0

Step 2:

Add the second digit (tens place) numbers (5 +1 = 6) the add with previous step carry (6 +1 =7)

5 4
(+)   1 6
-----------
7 0

Finally we get 54 +16 = 70

Ex 4: Add the following two numbers: 15948 and 69741

Sol:

Given 15948 and 69741

1 5 9 4 8
(+)     6 9 7 4 1
------------------

Step 1:

Add the one place number then over the carry (8+1=9 carry =0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
9

Step 2:

Add the tens place number then over the carry (4+4 = 8 carry 0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 9

Step 3:

Add the hundreds place number then over the carry (9+7 =16 carry 1)

1 5 9 4 8
+)    6 9 7 4 1
------------------
6 8 9

Step 4:

Add the thousands place number (5+9 = 14) then add with carry from the above step (14 +1 = 15 carry 1)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
5 6 8 9

Step 5:

Add the ten thousands place number (1+ 6=7) then add with previous step carry (7 + 1 =8)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 5 6 8 9

Therefore 15948 + 69741 = 85689

Ex 5: Add 72 with 4.12

Sol:

Given numbers 72 and 4.12

Step 1:

Change the whole number into decimal form

Whole number =72

Decimal form = 72.00

Step 2:

7 2 .0 0

(+)       4. 1 2

--------------------



Start the addition process from the right

Step 3:

Add the first place number from the right and over the carry to the next step (0 +2 =2 carry =0).

7 2 .0 0

(+)       4. 1 2

--------------------

2



Step 4:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 0 +1 =1 carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

. 1 2

Step 5:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 2 +4 =6 carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

6. 1 2

Step 6:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 7 +0 =7, carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

7 6 .1 2

72 + 4.12 =76. 12

Examples on multiplaction operation in math:

Ex 1: Multiply 10 with 5.

Sol:     10 x 5 = 15.

Where, ‘x’ is the operator of multiplication.

10 and 5 are the operands of multiplication.

15 is the resultant of the multiplication operation.



Ex 2: Simplify 10(5) + 20(2) + 9(3) + 4(1).

Sol:     15 + 40+ 27+ 4

86

Ex 3: A train has 9 carriages. There are 42 seats in each carriage. How many seats are there on the train?

Sol:   Number of carriage = 9

Number of seats in each carriage = 42

Multiply number of carriages with number of seats in each carriage.

Number of seats in the train = 9 x 42

= 378.

Ex 4: There are five cupboards in a room. Each cupboard has 7 racks. Find how many racks totally the room has.

Sol:      Number of cupboards = 5.

Number of racks in each cupboard = 7.

To find total number of rack, multiply the number of cupboard with number of racks in each cupboard.

Total number of rack = 5 x 7

= 35.

Ex 5: There are six baskets full of apple in a lorry. 1st basket contains 35 apples, 2nd and 3rd basket contains 40 apples, 4th, 5thth basket contains 50 apples. Calculate the total number of apples in the lorry. and 6

Sol:      35 apples in the 1 basket

40 apples in the 2 basket

50 apples in the 3 basket

Total Number of Apples = 35(1) + 40(2) + 50(3)

= 35 + 80 +150

= 265

Number Percentage Calculator

Percent means ‘for every 100 ’. So, when we say P% .it means P  out of 100 . Thus, P%=P/100 . It is often denoted by symbol “% ”. Any percentage can be expressed as a fraction. For example, 40%=40/100=2/5 .

Percentages are used to find whether one quantity is large or small compared with another quantity. The first term usually represents a part of, or a change in the second term, which should be greater than zero.

Percentages are usually used to express numbers between zero and one, any dimensionless proportionality can be expressed as a percentage.

Please express your views of this topic Percentage of Difference by commenting on blog.

Let us now express some percentages as fractions:

a.       5%=5/100=1/20 .

b.      10%=10/100=1/10 .

c.      25%=25/100=1/4 .

d.     75%=75/100=3/4 .

e.      125%=125/100=5/4 .

f.        175%=175/100=7/4 .

g.       (3 1/8)%=25/800=1/32 .

h.     (6 1/4)%=25/400=1/16 .

i.        (8 1/3)%=25/300=1/12 .

j.        (16 2/3)%=50/300=1/6 .

k.       (66 2/3)%=200/300=2/3 .

l.        (87 1/2)%=175/200=7/8 .



Calculation of Percentage:

Calculation of percentage:

The percent symbol can be treated as being equivalent to the pure number constant 1/100=0.01,  while performing calculations with percentage.

If a number is first changed byP%  and then changed by Q% , then the net change in the number =[P+Q+((PQ)/100)] . Remember that any decreasing value in the formula should be taken as ‘negative’ and increasing value should be taken as ‘positive’.

Similarly, if A’s salary is P%  less than B’s salary, then the percentage by which B’s salary is more than A’s salary is(100P)/(100-P) .

If expenditure also, then percentage change in expenditure or revenue=[P+Q+((PQ)/100)] . Where ‘P’ is the percentage change in price and ‘Q’ is the percentage change in consumption.

Problems on number percentages:

Ex1 : What percentage of 1600 is 40?

Sol:  Let 40  be "P% of 1600.

So, 40=P%   of 1600 =(P/100)(1600)=16P .

Thus, P=40/16=2.5% .

Ex2 : Calculate 40%  of 625 .

Sol: 40%  of a number =2/5  of the number =2/5  of 625=(2/5)(625)=250 .

Ex :3 A number is first increased by 30%  and then decreased by 20% . Find the net change in the number.

Sol:  Let the original number be 100 .

Increasing by 30%, " it becomes " 130 .

Now, if 130  is decreased by 20% , it becomes 104 .

Thus, the net change =(104-100)=4%  increase.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Divide Fractions by Whole Numbers and sample paper of class 9 cbse sa2. I am sure they will be helpful.

Practice problems on number percentages:

Q:1  A’s salary is 25%  more than B’s salary. By what percent is B’s salary less than A’s?

Sol:  Let B’s salary be .

Since A’s salary is 25%  more than that of B, his salary will be Rs. 125 .

Thus, B’s salary is Rs. 25 less than the A’s salary.

So, in percentage: (25125)(100)=20% . Hence, B’s salary is 20%  less than A’s salary.

Multiplication Properties

In this page we are going to discuss about multiplication properties concept. Integer used to measuring and counting in general things. Positive integer and negative integer is the two types of integers. Integer is nothing but using notational symbol represents numbers. Multiplying integer’s mean scaling a number by another.

3 x 4 = 3 + 3 + 3 + 3 = 12

3 x 4 = 4 + 4 + 4 = 12

Properties of multiplication

To study multiplying integers, we have to know five basic properties such as associative, commutative, distributive, and multiplicative identity. These multiplication properties are described as below.

Commutative property
Associative property
Multiplicative identity property
Distributive property
Zero property



Commutative property:

This is the product the two number is same even we change the order. That is, a * b = b * a.

Example: 5 * 4 = 4 * 5

= 20

Associative Property:

This is the product the three more number is same even we change the order. That is, (a * b) * c = a * (b * c).

Example: 5 *( 4 * 3 ) = (5 * 4) * 3

= 60

Identity Property:

This is the multiplication of any number by 1 is that number. That is, a * 1 = a.

Example: 9 * 1 = 9.

Distributive property:

Multiplication of a number with the addition of two numbers is same as addition of two products. That is, a* (b + c)=ab + ac.

Example: 5 * (4 + 3) = 5*4 + 5*3

= 35

Zero property:

Any thing multiply with zero is zero. That is, a * 0 = 0.

Example: 7 * 0 = 0

Multiplication rules

Positive * Positive = Positive

Negative * Negative =Positive

Positive * Negative = Negative

Negative * Positive = Negative

The following examples are to study multiplying integers with multiplication rule

5 * 4 = 20

-5 * -4 = 20

5 * -4 = -20

-5 * 4 = -20

Multiplication grid

Multiplication grid

Methods to multiplying integers

Below are the methods to multiplying integers-

Method 1:

Multiplying integers of 345 * 6

Solution:

Here, 5 is at unit place, so we do    5 * 1       =    5  * 6 =    30

4 is at 10s place, so we do   4 * 10     =  40  * 6 =   240

3 is at 100s place, so we do 3   * 100 = 300 * 6 = 1800
--------
The product result  =  2070
--------

Method 2:

Multiplying integers of 345 * 6

Solution:

345

x 6
--------------
30
240
1800
--------------
2070
--------------

Sample Space

In statistics, learning sample space basically represents the presentation and interpretation of the events of the possible outcomes that occur in a planned learning or scientific investigation. The learning sample space helps to refer all recording of information's are numerical or categorical, as an observation. The learning sample space assists in the following three cases, designed experiments, observational studies, and retrospective studies, the end result was a set of data that of course is subject to uncertainty.


Definition for Sample Space

The set of all favorable combinations of outcomes of a statistical experiment is labeled the sample space and is denoted by the mathematical symbol S.

Each possible outcome of a sample space is labeled a member of the sample space, in other words it is labeled the sample point. The trials of a sample space has a finite number of elements are separated by commas (,) and enclosed in braces ({}).

Thus the: sample space of possible outcomes when a coin is tossed, may be written S = {H, T),

Where, H means that “heads" and T means that “tails,”.


Examples for Sample Space:

Example 1:

Determine the sample space for the event of rolling a die, using the learning sample space.

Solution:

In rolling a die the number that shows on the top face.

The required sample space S1 = {1, 2, 3, 4, 5, 6}.

Example 2:

Determine the sample space for the number is even or odd.

Solution:

The required sample space S2 = {even, odd}.


More than one sample space:

If we know which element in S1 occurs, we can tell which outcome in S2 occurs; however, knowledge of what happens in S2 is of little help in determining which element in S1 occurs. Provides more information than S

Ex:  A coin is tossed twice. What is the Sample space?

Sol: The sample space; for this experiment is

S= {HH, HT, TH, TT}.

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Binary Number Representation

The binary numeral system, or base-2 number system, represents numeric values using two symbols, 0 and 1. More specifically, the usual base-2 system is a positional notation with a radix of 2. Owing to its straightforward implementation in digital electronic circuitry using logic gates, the binary system is used internally by all modern computers.

Understanding Binary Logistic Regression is always challenging for me but thanks to all math help websites to help me out.

The representaion of binary numbers are uses in mathematics are defined as follow,

Binary numbers representation in math:

In the binary number representation consists of octal, decimal, and hexa decimal numbers in the column. We can be represent the binary numbers by use of its operation. In the binary number representation, the decimal is easiest method of understanding binary numbers.

For example we can represent: 4567,

4 is represent the 1000’s

5 is represent the 100’s

6 is represent the 10’s

7 is represent the 1’s

Which means the representaion of 4567 is given as follow,
4567 = 1x1000 + 2x100 + 3x10 + 4x1

Given binary number representation,

1000


= 103 = 10x10x10

100


= 102 = 10x10

10


= 101 = 10

1


= 100 (any number to the exponent zero is 1)

The table above can be represented as the binary numbers,

Such that,

4567 = 4x1000 + 5x100    + 6x10     + 7x1

= 4x103 + 5x102 + 6x101 + 7x100

Examples for binary number representation in math:

The examples of binary number representation in math is given as follow:

Example:1

To determine the decimal number in 10102?

Solution:

Step 1: 1 => 1×2×2×2 = 8

Step 2:  0 => 0×2×2 = 0

Step 3: 1 => 1×2 (=2)

Step 4: 0 => 0

Answer is: 1010 = 8+0+2+0 = 10.

Example 2:

To determine the decimal number in 10112?

Solution:

Step 1:  1=> 1×2×2×2 (=8)

Step 2: 0 => 0×2×2 (=0)

Step 3: 1 => 1×2 (=2)

Step 4: 1 => 1

Answer is: 1001 = 8+0+2+1 = 11.

My forthcoming post is on how to find the prime factorization of a number and neet entrance exam syllabus will give you more understanding about Algebra.

Example 3:

To determine the decimal number in 1.112?

Solution:

Step 1:  1 => 1

Step 2:  1 => 1×(1/2)

Step 3: 1 => 1×(1/4)

Answer is :1.75.

Example 4:

To determine the decimal number in 11.112?

Solution:

Step 1: 1 => 1×2 (=2)

Step 2: 1 => 1

Step 3:  1 => 1×(1/2)

Step 4: 1 =>  1×(1/4)

So, 11.11 is 2+1+1/2+1/4 = 3.75 in Decimal

Answer is: 3.75.

Examples of Poisson Distribution

Definition: A random variable X is  a Poisson distribution if the probability mass function of X is P(X = x) =e−λ  λx / x!,                                       x = 0,1,2, …for some λ > 0

The mean of  Poisson Distribution denoted by λ, and the variance is denoted by λ.
The parameter of Poisson distribution is λ.

The Poisson distribution is a restrictive case of Binomial distribution under the following conditions.
(i)   Number of trials(n) is indefinitely huge(large), that is n → ∞.
(ii)  The constant probability(p) of success in each trial is very less.
ie., p → 0.
(iii) np = λ is finite where λ is a positive real number. When an event occurs rarely, the distribution of such event may be assumed to follow a Poisson distribution. The example problems of poisson distribution is   given below

Examples of Poisson distribution:

Some Examples of poisson distributions are given below

(1) The number of gamma particles emitted by a radio active source in a given time interval.
(2) The number of phone calls received at a telephone exchange in a given time interval.
(3) The number of defective articles in a packet of 250, produced by a good industries limited.
(4) The number of printing errors at each page of a book by a good publication centre.
(5) The number of road accidents reported in a city at a particular time.

Example Problems for Poission distribution:

Example problem 1: If a publisher of  technical books takes a great pain to ensure that his books are free of typological errors, so that the probability of any given page containing atleast one such error is 0.005 and errors are independent from page to page

(i) what is the probability of its 400 page novels will contain exactly one page with error.

(ii) atmost three pages with errors.
[e−2 = 0.1353 ; e−0.2. = 0.819].


Solution :

n = 400 , p = 0.005
np = 2 = λ
(i)  P(one page with error) = P(X = 1) = e−λ λ1/1! = e-2 21/1!
= 0.1363 × 2 = 0.2726
(ii)  P(atmost 3 pages with error) = P(X ≤ 3)
= Σ e−λ λx / x!    [limits 0 to 3]
= `sum` e−2 (2)x  /  x!
= e2 [1 +2/1! + 22/2! + 23/3!]
= e−2 (19/3 )= 0.8569

Learning Simple and Compound Events

Probability of any event  is the ratio of several favorable outcomes of an event to the total number of outcomes of an event. Otherwise if we toss a single dice then it is called as a simple event. If we toss a two dice then it is called as a compound event. Inclusive means when two events are happen at a same time. Mutually exclusive means when two events cannot happen at a same time.

Learning Simple and Compound Events Formulas :

learning  simple and compound events formulas in the probability of an event can be expressed as

P(a) = number of a favorable outcomes / total number of a possible outcomes

learning  simple and compound events formulas in the probability of two independent events(A and B) are multiplied by the probability of the first event by the probability of the second event. Two events are said to be independent. if P(A and B) = P(A) P(B). P(A)and P(B) are are non zero

P(A and B) = P(A) . P(B)

learning  simple and compound events formulas in the probability of two dependent events (A and B ) are multiplied by the probability of A and the probability of B after A occurs.

P(A and B) = P(A) . P(B following A)

learning  simple and compound events formulas in the probability of one or other of two mutually exclusive events (A or B) are added to the probability of the first event to the probability of the second event.Two events are said to be disjoint. if and only if P(A and B) = 0

P(A or B) = P(A) + P(B)

learning  simple and compound events formulas in the probability of one or the other of two inclusive events(A or B) are added to the probability of the first event to the probability of the second event and subtract the probability of both events happening.

P(A or B) = P(A) + P(B) - P(A and B).

practice problems in learning simple and compound events

Example 1:Abraham is going to the Super market to pick a new sports bats. Today, the shopkeeper has 25 cricket bats and 50 Tennis bats are available for sales. If Abraham randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 25 / (25 + 50 )

= 25 / 75

= 1 / 3

= 0.33

if Abraham randomly picks a bat (which is a Cricket bat) having a probability  0.33.

Example 2:Lenin is going to the Super market to pick a new sports bats. Today, the shopkeeper has 20 cricket bats and 40 Tennis bats are available for sales. If Lenin randomly picks the bats, then what is the probability that the bat would be a Cricket bat?

Solution:

Probability(randomly choosing a Cricket bat) = Number of Cricket Bats / Number of cricket bats + number of tennis bats

= 20 / (20 + 40 )

= 20 / 60

= 1/ 3

if Lenin randomly picks a bat (which is a Cricket bat) having a probability  = 0.33

Exponential Growth Formula

A function is said to be Exponential growth that including exponential decay when the growth rate of that mathematical function is proportional to the function's current value. In a discrete domain of definition with equal intervals of the function is called as geometric growth or geometric decay. The exponential growth model is also called as the Malthusian growth model.


Exponential growth formula:

Exponential formula defines the  X as exponentially on time t.

X(t) = a . b(t/r)

"a" denotes the initial value

a = x,

X(0) = a,

b= a

It denotes the positive growth of the factor, t = time required

Example for exponential growth formula:

If a power doubles every 5 minutes, initially there’s only one doubles, how many powers would be there after 2 hours?

Here, a= 1, b= 2, t = 5 minutes.

X(t) = a . b(t/r) = 1 . 2{(120 minutes)/(5 minutes)}

X(2 hour) = 1 . 2 24 = 16777216

After two hours, there would be 16777216 powers.


Exponential growth Problem:

A microbiologist is researching a newly-discovered species of fungi. At time t = 0 hours, he puts one hundred fungi into what he has determined to be a favorable growth medium. Six hours later, he measures 200 fungi. Assuming exponential growth, what is the growth constant "i" for the fungi? (Round i to two decimal places.)

For the given problem, the units on time t will be hours, because the growth is being measured in terms of hours. The starting amount P is the amount at time t = 0, so, for this problem, P = 100. The ending amount is A = 200 at t = 6. The only variable we don't have a value for is the growth constant i, which also happens to be what I'm looking for. So I'll plug in all the values we know, and then solve for the growth constant:
A = Peit
200 = 100e6i
2 = e6i
ln(2) = 6i
`ln(2)/6` = i = 0.11552453

Derivative Step by Step

erivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity.-  Source : Wikipedia

Calculus is used to solve differentiation of any subjective equation and the equivalent output result. we need to find f(x) where,     d/dx f(x) = g(x).The derivative of constant is zero in calculus. The linear equations are also compared using calculus. Integration is considered as one of most important study of calculus in mathematics. British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. The methods that are applied in continous graphs, curves or fucntions  is calculus.

Understanding Derivative of Cot x is always challenging for me but thanks to all math help websites to help me out.

Steps to solve the Derivative function:

Derivative  problem 1:

Find the first derivative and second derivative of

f(x) = x4 + 7x3 - 3x2 - 9x +3

Solution:

Step 1:  First differentiate f(x) = x4 + 7x3 - 3x2 - 9x +3

Step 2:  when we differentiate the first term we get  (1 × 4) x (4-1)

Step 3:  The above equation can above simplified as  4 x 3

Step 4:  Like wise we can differentiate  and simplify all the terms in the equation

Step 5:  Finally after the first derivative we will get  4 x 3 + 21 x2 - 6x - 9

Step 6:  Now we will go for Second derivative
End of the second derivative we will get   12x2 + 42 x - 6


First derivative:

f' = (df )/ (dx)

= (1 × 4) x (4-1) + (7 × 3)x(3-1) - (3 × 2)x(2-1) - (9 × 1)x (1-1)

= 4 x 3 + 21 x2 – 6x - 9

Second derivative:

f '' = (df ' )/ (dx)

= 4 x 3 + 21 x2 – 6x - 9

= 12x2 + 42 x - 6

Answer:

(d^2)/(dx^2)  (x4 + 7x3 - 3x2 - 9x +3) = 12x2 + 42 x - 6

Derivative problem 2:

Find the differential for y = sqrt(2x^3 - 9x)

Solution:

Step 1:  First turn sqrt(2x^3 - 9x)to (2x3 - 9x)1/2
Step 2:  Differentiate in the usual method
Step 3:  Then finally we get,

d/(dx) ((2x3 - 9x)1/2 ) = ½ ( 2x3 - 9x ) ½ -1( 6x2 - 9 )

= ½ ( 2x3 - 9x ) -1/2( 6x2 - 9 )

= ( 6x^2 - 9 ) / (2 sqrt (2x^3 - 9x ))

Answer of given calculus test exam problem is

d/(dx) ((2x3 - 9x)1/2 ) = ( 6x^2 - 9 ) / (2 sqrt (2x^3 - 9x ))

My forthcoming post is on how to do algebra 2 problems step by step and sample papers for class 9 cbse will give you more understanding about Algebra.

Derivative practice problem:

Derivative practice problem 1:

Find the derivative of f(x) = ( x - 5 ) ( 2x + 1 )

Answer:   d/(dx) f(x) = 4x - 9

Derivative practice problem 2:

Differentiate the given equation with respect to t. y =  8t3 + 12t

Answer:  24t2   + 12

Division with Remainders Word Problems

Division is an arithmetic operation which is the opposite of multiplication. The symbol used for division is (÷).

Notation:

Division is often represented by inserting fraction bar which is nothing but the horizontal line between divisor and dividend. Dividend is placed in numerator and divisor are placed in denominator.  For example, a divided by b is written as,

`a/b`

Key terms - division with remainders word problems

Dividend:

The number which is being divided in a division is known as dividend.

Divisor:

The number which divides the dividend is known as divisor.

Quotient:

Quotient is the solution / answer for the problem

Remainder:

The remainder is the amount "left over" after the division

Example problems on division with remainders problems

Division problem 1:

What is division value and remainder of the given fraction 13 / 4

Solution:

When we divide 13 by 4 it gives 3 ( 3 * 4 = 12) as a quotient and 1 is the (13 - 12) remainder value

Quotient = 3

Remainder = 1

Division problem 2:

What is division value and remainder of the given fraction 6 / 4

Solution:

6 / 4 =   There is one 4 in 6

Hence Quotient = 1

Remainder = 2

Solving Division with remainders Word problems:

Division with remainders word problem 1:

A 40 in stick is cut into pieces of 7 in length. How many pieces we can get out of this stick? What is the length of the scrap? Solve the word problem.

Solution:

The problem describes how to divide a 40 in stick into pieces of 7 in each. The remaining piece of the stick is called the scrap.

Therefore, the dividend is 40in., the divisor is 7 in. The quotient of this is the number of pieces and the remainder is the length of the scrap. When we divide 40 by 7, there are fivet 7’s in 40 (5*7= 35) and the remainder is 5.

Hence, the number of pieces of 7 in = 5(quotient)

The length of the scrap = 5 (remainder)

Division with remainders word problem 2:

There are 11 apples and 4 kids . We have to give equal number of apples to each kids. How many apples will each kids get and find the amount of apple left over?

Solution:

Here we have to divide number of apples  10 by number of kids  4.  Hence ,

11 / 4 = 2 with remainder 3

Number of apples each kid will get = 2

The apples left over = 3

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Division with remainders word problem 3:

Consider there are 10 boys in a class. Divide them into 3 groups. How many boys each group will consists and how many boys are left over?

Solution:

Here the dividend is total number of boys 10 and the divisor is 3. We have to divide total number of boys by number of groups to find the number boys each group having .

10 / 3 = 3

Remainder = 1

Hence each group gets 3 boys and 1 boy is left over.

Basic Geometrical Concepts Solve Online

Geometrical is a study of structural mathematics .Basic geometrical has the concept of shapes and structure of an object. It contains two dimensional and three dimensional objects. In this article basic geometrical concepts solve online we are going to see about some basic geometric problems through online.
Generally online help means that here tutor and student should communicate directly with help of online chatting. Student can get more geometrical help and idea from tutor through online with help of chatting. It may be in audio or non audio.


Basic Geometrical Concepts Shapes Solve Online :

Basic geometrical concepts are plane, point, line, angle, ray, two dimensional objects, and three dimensional objects. Main purpose of using geometry is finding area, volume and perimeter of the given shape.

Basic geometrical concepts:

Two dimensional objects:

Square
Rectangle
Triangle
Circle
Parallelogram
Trapezoid

Three dimensional objects:

Cylinder
Cone
Cube
Sphere
Hemisphere
Prism
Pyramid
3d geometry shapes

Solve Example Problems in Basic Geometrical Concepts:

Solve Example problems in basic geometrical concepts:

Example problem 1:

Find the angle of triangle from the given figure?

Geometry-triangle

Solution:

Given data:

Two angles of triangle is 550 and 380

Solve the third angle of triangle

Total angle of triangle=1800

55+38+x=180

93+x=180

X=180-93

X=870

Third angle of triangle is 870

Example problem 2:

Which one is 3D shape of geometrical object?

Geometry- shapes

Solution:

Option (a)-Two dimensional object(square)

Option (b)-Two dimensional objects(Circle)

Option (c)-Three dimensional object(Parallelogram)

Option(d)-Three dimensional object(Cone)

Example problem 3:

How many edges and vertices in the cube?

Cube

Solution:

Cube having 12 edges

Cube having 8 vertices

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So the cube having top portion of four edges

Side portion of four edges

Bottom portion of four edges

So totally have 12 edges

Vertices mean nothing but corner point of the given shapes. When the two edges or lines are meeting It will produce the vertices

Solve Proving Proofs

Mathematical statement considered the sequence of statements by proofs, every statement being adjusted with some definition or a proposition or an axiom that is before launched by the method of deduction using only the permitted logical rules. Repeatedly we prove a proposition straightly from what is given in the proposition. But so many times it is easier to prove a same proposition not proving the proposition by itself,

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Solve Proving Proofs - Types of Proofs:

Solve Proving Proofs Indirect Method:

Solve proving proofs - Indirect method:

Prove the given function f : R  =>  R defined by f(x)  =  2x  +  5 is one – one.

Sol :  A function is one – one if  f(x1)  =  f(x2)  => x1  =  x2.

Using this we have to show that “ 2x1 + 5  = 2x2  +  5” => “  x1  =  x2”.This is of the form p  =>  q, where , p is 2x1  +  5 =  2x2  +  5 and q : x1  =  x2. so this is the direct method proofing.

We can also prove the same by using contrapositive of the statement. Now contrapositive of this statement is ~ q=>  ~ p, is, contrapositive of “if f(x1)  =  f(x2), then  x1  =  x2” is “if x1  =  x2, then f(x1) =  f(x2)”.

Now                 x1  !=   x2

=>                   2x1  !=   2x2

=>                   2x1  +  5  !=   2x2  + 5

=>                   f(x1)  !=   f(x2).
Solve Proving Proofs of Direct Approach:

Solve Proving Proofs -direct Proof:


It  is the proof of  a proposition in which we directly start the proof  with what is given in the proposition.

1)Straight forward approach.

2) Mathematical Induction approach.

3)Proofs By cases or by   exhaustion.

Solve Proving Proofs -Indirect Proof:

Instead of proving the proposition directly, we establish the proof of the proposition through  proving a proposition which is equivalent to the given proposition.

1)Proofs  by contradiction.

2)Proofs by using contrapositive statement.

3)Proofs by a by a counter example.


My forthcoming post is on free math word problems for 2nd grade and ntse 2013 syllabus will give you more understanding about Algebra.


Solving Proving Proofs - some Examples for Proving Proofs:

Solving proving proofs - Direct Method:

Prove the given function f:  R  =>  R

Defined by f(x) = 2x + 5 is one – one.

Sol :  Note that a function f is one – one if

f(x1)  =  f(x2)    x1  =  x2 (definition of one – one function)

Now given that   f(x1)  =  f(x2),

=> 2 x1 +  5 = 2 x2 +  5

=>                 2 x1  +  5  –  5  = 2 x2  +  5  –  5(adding the same quantity on both sides)

=>                 2 x1  +  0  =  2 x2  +  0

=>                 2 x1  =  2 x2 (using additive identity of real number)

=>                 (2)/(2) x1  =  (2)/(2) x2 (dividing by the same non zero quantity)

=>                 x1  =  x2.

Sets and Operations

Sets and relations are two of the most important concepts in mathematics that is taught in middle school and is widely used in practical mathematics. A set is a collection of things such as letters, numbers, words, things etc. For example: {woodcraft construction toys, building blocks, construction sets}. This is a set of educational toys such as woodcraft construction toys and so on. Sets also functions different relationship in mathematics. Let’s have a look at the same in this post.

Union of Sets: When two sets are compared and a final set is created with the elements of set A and set B, the set created is called Union of sets. It is referred as A U B. While creating the union of sets, a single element if occurs in both the sets is used only once in the final set.

For example:
A = { Cloth toys for babies , Wooden play toys for kids , Soft toys for babies, Barbie Doll}
B = {Ludo, Chess, Snakes and Ladder, Barbie Doll}
A U B = {

Cloth toys for babies , Wooden play toys for kids, Soft toys for babies, Barbie Doll, Ludo, Chess, Snakes and Ladder}Intersection of Sets: When two sets are compared and a final set is created using the only common elements of set A and set B, then it is called as intersection of sets. It is referred as A ? B.

For example:
A = {Hair, Eye, Nose, Ear, Lips, Hands, Legs}
B = [Hair, Eye, Nose, Ear, Lips, Teeth, Tongue}
A ? B = {Hair, Eye, Nose, Lips}
Difference of Sets: The difference of sets is nothing but the set that includes the unique elements of set A and B. It is referred as A –B. For example:
A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8, 10}
A – B = {1, 3, 5, 8, 10}
These are some of the basic relations in sets.

Geometric Ray

We know that, a ray is an important topic of line, and geometrical chapter in mathematics subject. In mathematics, a ray is defined as several ways, when we have seen a vector, a ray vector is going to one point to other point, and it is called as rays. In this geometrical part, ray is used in mostly as the lines and it’s relevant. In this article we are going to explain about the geometric ray, and its definitions.

Definition of the Geometric Ray:

Generally a ray is defined as many ways; it is one part of line.
It has a straight line with one end point and other direction (side) extends to infinity.
And a ray is start with end point, and then other point on the ray next.

For example:

Picture representation for geometric ray:

Geometric rays

If suppose the above ray (figure) is vector ray means, a ray is vector (xy) form of point x to point y.
And a ray is start with end point x, and other side goes to infinity on next. It is named ray.

Note:

If suppose the given ray (see figure) is vector ray means, a ray is vector (xy) form of point x to point y.

Ray is` vec (XY)` .

Exact definition for ray:

Ray is straight line that starts at a point and continues outward in another direction. And a ray is start with end point, and then other point on the ray next.

Geometric rays
Examples Pictures and Explain in Geometric Rays:

Example problem 1: Draw a straight line, with two points ray figure.

Figure:

example figure for ray

Example problem 2: Draw the both opposite side rays.

The ray figures:

example figure for ray, and    example figure for ray

These rays are mostly used in the sphere intersection problems.

Example problem 3: Draw the angle figure used on 2 ray figures with 1 common end point:

Angle figure in ray:

example figure for ray

Example problem 4: Draw the parallel line, and use the ray with intersect (cross) the parallel line.

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ray example

These all above the general explanations and example figures are making clear about the geometric ray definition and concept.

Sequence Listing

In this article, we are going to see about the sequence listing. There are two types of sequences

1. Arithmetic sequence and

2. Geometric sequence.

Arithmetic sequence means, the sequence of numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means, the sequence of numbers such that the ratio between two consecutive members of the sequence is a constant. The sequence listing formulas and example problems are given below.



Formula for arithmetic sequence:

nth term of the sequence : an = a1 + (n - 1)d

Series of the sequence: sn = (n(a_1 + a_n))/2

Formula for geometric sequence:

nth term of the sequence: an = a1 * rn-1

Series of the sequence: sn = (a_1(1-r^n))/(1 - r)
Example Problems for Sequence Listing:

Example problem 1:

Find the 15th term of the series 2, 4, 6, 8, 10,. ....

Solution:

First term of the series, a1 = 2

Difference of two consecutive terms, d = 4 - 2 = 2

n = 15

The formula to find the nth term of an arithmetic series, a_n = a_1 + (n-1)d

So, the 15th term of the series 2, 4, 6, 8, 10....... = 2 + (15 - 1) 2

= 2 + 14 * 2

After simplify this, we get

= 2 + 28

= 30

So, the 15th term of the sequence 2, 4, 6, 8, 10,.... is 30

Series of the arithmetic sequence: sn = (n(a_1 + a_n))/2

 s_15 = [15(2 + 30)]/(2)

 s_15 = [15(32)]/(2)

 s_15 = (480)/(2)

After simplify this, we get

 s_15 = 240



Example problem 2:

Find out the 11th term of a geometric sequence if a1 = 4 and the common ratio (C.R) r = 2

Solution:

Use the formula a_n = a_1 * r^(n-1) that gives the nth term to find a_11 as follows

a_11 = a_1 * r^(11-1)

= 4 * (2)10

= 4 * 1024

After simplify this, we get

= 4096

The 11th term of a geometric sequence is

Series of the sequence: sn = (a_1(1-r^n))/(1 - r)

 s_11 = [4(1 - 2^11)]/(1 - 2)

= [4(-2048)]/(-1)

After simplify this, we get

s_11 = 8192

The above examples are helpful study of sequences listing.

Mathematical Symmetry

In mathematical when one shape becomes like another if you rotate over, slide or twist is called the symmetry. In mathematical normally symmetry is classified into three types. Thus three types of symmetry are reflection symmetry, rotational symmetry and point symmetry. Let us see mathematical symmetry in this article.


Mathematical Symmetry:

Symmetry:

When one shape becomes like another if you rotate over, slide or twist it is called as symmetry.

Types of symmetry:

Reflection symmetry
Rotational symmetry
Point symmetry

Definition of reflection symmetry:

Reflection symmetry is the simple type of symmetry. One half of the reflection is the reflection of the other half is known as reflection of symmetry.

Definition of Rotational symmetry:

The shape or image can be rotated on various quantities and it still shows the same is called as rotational symmetry.

Definition of Point symmetry:

If the image is placed the same distance from the starting but in the opposite path then the image has point symmetry.
Brief Explanation about Mathematical Symmetry:

Reflection symmetry:

One half of the reflection is the reflection of the other half is known as reflection of symmetry.

Example:

mathematical symmetry

Here the image (under the line) gives the perfect reflection of the above image.

Rotational symmetry:

The shape or image can be rotated on various quantities and it still shows the same is called as rotational symmetry.

Example:

mathematical symmetry

In above figure the second image represents the rotated structure of first image.

Point symmetry:

If the image is placed the same distance from the starting but in the opposite path then the image has point symmetry.

I am planning to write more post on upsc syllabus 2013 and Fractions and Equivalent Decimals. Keep checking my blog.

Example:

mathematical symmetry

From the above figure we understand the point symmetry. In above figure the images are placed the same distance from the starting but in the opposite path and it similar to rotational symmetry of order 2.