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.

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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?