Information for Line Segment

Information for line segment, the division of a line with two end points is called a line segment. Line segment FG which we denoted by the symbol `bar(FG)` .

Note: We shell denote a line segment `bar(FG)` by FG only.

From the above figure, we call it a line segment FG. The points F and G are called end-points of the line segment FG.

We can also name it as line segment FG.

A line segments:

(a) A line segment has a definite length.

(b) A line segment has two end-points
Information for Drawing a Line Segment by Using Ruler:

Example:

Describe the information about to draw a line segment FG of length 11.5 cm?

Solution:

Given:

length of  line segment FG = 11.5 cm.

Steps to draw a line segment:

1. Place the ruler on a drawing paper.

2. Mark a point at the zero of the ruler. Here, we mark point F at the zero of the ruler.

3. Count the divisions in the ruler till we reach the required length. Here, we count the divisions in the ruler till we reach 11.5 cm.

4. Mark the required end point. Here, we mark the end point G.

5. Join points F and G.

information for line segment

Therefore, FG is the required line segment of length 11.5 cm. The required line segment is represented by `bar(FG)`
Information for Line Segment on a Polygon:
Describe the information to find the line segments of the given polygon. The polygon shown below figure,

information for line segment

Solution:

Given:

Polygon Triangle  EFG.

To find the line segments on a polygon:

We know that the line segments are consisting of two end points. Here, the triangle has three end points, such as E, F, and F. In the given polygon, the three end points to form the line segments on a triangle by connecting these end points consecutively, such line segments are EF, FG, and GE. These line segments are represented by `bar(EF)` , `bar(FG)` , and `bar(GE)` . Therefore, the given polygon triangle has three-line segments.

Simple Data Format

Any collection of information that makes a form to giving the required information is called data. Simple data format are used to compare the collection of data. Graphs are helping to analysis the various types of data. For example data analysis is statistics, bar graphs, histogram graphs, pie charts, and line graphs are used to simply analysis the data. In this article, we are going to discuss about simple data format with suitable example problems.
Example Problem for Simple Data Format:

1. Analyze the statistical data using the given data, find the mean, mode, median, and range of the given data.

14, 7, 13, 12, 10, 10, and 11

Solution:

Rearrange the data for ascending order.

7, 10, 10, 11, 12, 13, 14

Mean:

Mean is the sum of data divided by the number of data in the given data.

Mean = `("sum of data")/ ("number of data")`

Mean = `77/7`

= 11

Mode:

Mode is the most common value in given data.

Most common value is 10.

Therefore, mode is 10.

Median:

Median is a middle number else we have two middle value means; we find the average of their two values is median. We need to find it, we should make to arranged in ascending order from the given data.

Median:

In this problem we have a middle number.

Therefore, Median is the middle number 11.

Range:

Range is difference between maximum and least values in the given data.

Range = maximum value – minimum value

= 14 - 7.

Range = 7.
More Example Problem for Simple Data Format:

2. Analyze the data and create histogram the given simple data.

Solution:

histogram

The above histogram analyzed the given data that which one is high range and low range. This is the needed histogram.

Explanation:

Step 1:

First represented, the class intervals in the x-axis with the value 0 to 10, 10 to 20, 20 to 30, 30 to 40, and 40 to 50

Step 2:

Then represented the frequencies in the y-axis, which ranged as 250, 100, 50, 175, and 75

Given frequencies are represents to each area of rectangle in the histogram.

Step 3:

From the table values according to the class intervals and the frequencies are drawn. This is the required histogram for the simple data.

Total Property Solutions

This article is about total property solutions. Total property solution is nothing but it is some properties which helps to solve some math problems. There are different math properties we use in different chapters of mathematics. With the help of the total property solution we can complete many different problems. Tutor vista is the best website to help with total property solutions. There will be online tutors in these websites to help the students anytime about the different topics in mathematics. Below we can see some total property solutions.


Total Property Solutions

Here we can work total property solutions under the topic of binary operations.

Closure property

An operation * on a nonempty set S is said to satisfy the closure property, if

a in S, b in S rArr  a*b in S for all a, b in S

Also, in this case, we say that S is closed for *.

An operation * on a nonempty set S, satisfying the closure property is known as a binary operation

Properties of a binary operation

(i) Associative law : A binary operation * on a non empty set S is said to be associative, if

(a*b)*c = a*(b*c) for all a,b,c in S

(ii) Commutative law   A binary operation * on a nonempty set S is said to be commutative if

a*b = b*a for all a, b in S

(iii) Distributive Law   Let * and . be two binary operations on a nonempty set S. We say that * is distributive over(.), if

a*(b.c) = (a*b).(a*c) for all a, b, c in S
Total Property Solutions

Here we see some example using of total property solutions

Example 1: (i) Addition on the set N of all natural numbers is a binary operation, since

a in N, b in N rArr a+b in N for all a,b in N

(ii) Multiplication on N is a binary operation, since

a in N, b in N rArr a x b in N for all a,b in N

Similarly, addition as well as multiplication is a binary operation on each one of the sets Z, Q, R and C of integers, rationals, reals and comples numbers respectively.

Example 2 : Let R be the set of all real numbers. Then,

(i) Addition on R satisfies the closure property, the associative law and the commutative law,

(ii) Multiplication on R satisfies the closure property, the associative law and the commutative law,

(iii) Multiplication distributes addition on R, since

a. (b+c) = a.b+a.c for all a, b, c in R

Solving Vertical Reflection

The vertical reflection is usually a transformation that can be performed over a point or a line.

In vertical reflection we transform all points of an object to an another point which is to the equal length of the opposite side of a vertical line.

This produce a general rotation of straight angle (180°) that is  half turn along the axes.

Solving Vertical Reflection : Description

The vertical reflection flips the image or a given object across a given line x = y + c, c=variable. The new object is a vertically reflected object of the original given object.

Solving vertical reflection follow the given steps :

Step 1: In this step, we have to determine the distance from the coordinates of the object to the given horizontal line (y = x + c)

Step 2: To plot the  reflected coordinates on the vertically opposite side of the given horizontal line from the equal distance.

Step 3: Join all the new coordinates to get the new object .
Solving Vertical Reflection : Examples

Given is a triangle whose coordinates are P(x, y), Q(x, y), R(x, y).It is transformed with the help of vertical reflection, that is with respect to the horizontal line.

Solution:

Step 1: Draw the triangle object as per the given co-ordinates are P, Q, R in the X-Y plane.


Step 2:   Draw the horizontal line of that triangle PQR, and draw the vertical line from each co-ordinates of  given triangular object. Measure the distance from each point of each vertical lines are y1, y2, y3.


Step 3: Plot the vertically reflected  points are A', B', C', are at the  same vertical distance from the horizontal  line


Step 4:  Now, join the vertically reflected points  A', B', C'. We get the vertically reflected triangle.


Step 5:  This is the vertical reflection of the given triangle.

Solving Vertical Reflection : Practice Problems

Problem  : Perform vertical reflection to a given square having points A( x1,y1) B(x2,y2) c(x3,y3) and d(x4,y4). specify the new coordinates of the square after reflection.

Average Deviation Equation

The average deviation is also called as average absolute deviation. Average deviation is  the deference between the individual values and average of the all values. The formula for the average deviation is

Average Deviation = (sum ( |x-barx|) )/ n

Where x is value appeared in given set of values and barx is mean of given set of values.

barx = "sum of all value"/"total number of values" .

Example for Average Deviation Equation:

Example:

Find the average deviation for the given set of values by using the equation 44, 59, 76, 48, 31.

Solution:

barx = "sum of all value"/"total number of values" .

barx = (44 + 59 + 76 + 48 + 31)/5

barx = 258/5

barx = 51.6

Average Deviation = (sum ( |x-barx|)) / n

Average Deviation = 63.6 / 5

Average Deviation = 12.72
Practice Problem for Average Deviation Equation:

Problem 1:

Find the average deviation for the given set of values by using the equation 25, 50, 75, 100, 150.

Solution:

Average Deviation = 40

Problem 2:

Find the average deviation for the given set of values by using the equation 12, 6, 7, 3, 15, 10, 18, 5.

Solution:

Average Deviation = 4.25

Problem 3:

Find the average deviation for the given set of values by using the equation 15, 19, 17, 18, 16.

Solution:

Average Deviation = 1.2

Problem 4:

Find the average deviation for the given set of values by using the equation 3, 5, 9, 7, 4, 10, 12, 2.

Solution:

Average Deviation = 3

Problem 5:

Find the average deviation for the given set of values by using the equation 39, 35, 34.

Solution:

Average Deviation = 2

Solving Proportionality Rule

In general solving proportionality rule; categorization of proportionality for geometric ratios is a essentially serious way of say what is fundamentally obvious in the specific state of proportional ratios of numbers.  In accumulation, it may be well-known that, given incommensurable magnitudes p and q, this explanation in effect split the field of rational numbers m/n into two disjoint sets: the set L of those for which holds, or m : n < p : q, and the set U of those for which holds, or m : n > p : q.


Rule for Solving Proportionality Rule:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio for solving proportionality rule.

Solution:

Given:

In preparation for basic proportionality theorem, a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

Rule for Proportionality

To Prove:

(AD)/(DB) = (AE)/(EC)

Construction:

Join BE and CD and then draw DM _|_ AC and EN _|_AB.

Proof:

Area of DeltaADE = (((1)/(2)) base xx height)

So,                   area (DeltaADE) = 1/2AD xx EN

In addition,     area (DeltaBDE) = 1/2 DB xx EN.

Similarly,         area (DeltaADE) = 1/2 AE xx DM and area (DeltaDEC) = 1/2 EC xx DM.

Therefore,        (area ( Delta ADE))/(area ( Delta BDE)) = ( (1/2) AD xx EN )/( (1/2) DB xx EN ) = (AD)/(DB) rArr (i)

And                 (area ( Delta ADE))/(area ( Delta DEC)) = ( (1/2) AE xx DM )/( (1/2) EC xx DM ) = (AE)/(EC) rArr (ii)

Now the DeltaBDE and DeltaDEC is on the same base DE and between the same parallel lines BC and DE.

So,       area (DeltaBDE) = area (DeltaDEC)         rArr (iii)

Therefore, from (i), (ii) and (iii) we have,

(AD)/(DB) = (AE)/(EC).

Hence, the theorem is proved in preparation for solving proportionality rule.


Example for Solving Proportionality Rule:

In solving proportionality rule, figure, AB is parallel to CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, determine x.

Example for Proportionality

Solution:

Given:

In a quadrilateral ABCD, AB || CD

OA = 3x – 19, OB = x – 4, OC = x – 3, OD = 4

To find:

x = ?

AB || CD

ABCD is a trapezium [The diagonals of trapezium divide each other proportionally]

(3x - 19)/(x - 3) = (x - 4)/(4)

rArr   (x – 4)(x – 3) = 4(3x – 19)

rArr   x^2 - 3x - 4x + 12 – 12x + 76 = 0

rArr   x^2 - 19x + 88 = 0                rArr   x^2 - 11x – 8x + 88 = 0

rArr   x(x – 11) – 8(x – 11) = 0         rArr   (x – 11)(x – 8) = 0

rArr   x – 11 = 0                                or                     x – 8 = 0

x = 11 units or 8 units.

Equality of Condition

Equality of condition states that the given conditions have two or more options that have an equal out come. .

Step  1:Check the given equation of left hand side is equal to right hand side

Step 2:If the left hand side is equal to right hand side the equality condition satisfies.Otherwise it is not a equality of condition(inequality condition)
Solved Examples on Equality of Condition

Example 1: Prove that the given equation satisfies "equality of condition"

3x+7=19

Solution: The Given equation is 3x+7=19

Step 1: Subtract both sides by 7.

3x+7-7=19-7

3x=12

Step 2: Divide both sides by 3

3x/3=12/3

x=4

Step 3:When x=4 only the equality of condition satisfies

Example 2:Check the given equation satisfies "equality of a condition"

(x+5)-(y+9)=4 at x=12 and y=4

Solution: The Given equation is (x+5)-(y+9)=4

Step 1:  Substitute x=12 and y=4 in the equation.

(12+5)-(4+9)=4

17-13=4

4=4

Step 2:So x=12 and y=4 satisfies the equation,this is a equality of a condition.

Example 3:Prove that the given equation satisfies "equality of condition"

4x+5=21

Solution:The Given equation is 4x+5=21

Step 1:Subtract both sides by 5

4x+5-5=21-5

4x=16

Step 2:Divide both sides by 4

4x/4=16/4

x=4

Step 3:When x=4 only the "equality of a condition" satisfies our equation.

Example 4:Check the given equation satisfies "equality of a condition"

(x+2)+(y+1)=9 at x=5 and y=1.

Solution:The given equation is (x+2)+(y+1)=9

Step 1:Substitute x=5 and y=1 in the equation.

(x+2)+(y+1)=9

(5+2)+(1+1)=9

7+2=9

9=9

Step 2:The x and y values satisfies at x=5 and y=1 ,so this is a equality of condition at x=5 and y=1.
Practice Problems on Equality of Condition

Problem 1:Check the given equation is equality of a condition or not?

6x+9=15 at x=1

Problem 2:Check the given equation satisfies "equality of a condition"

(x+3)-(y+1)=9 at x=7 and y=0

Learn Real Roots

In this article we are going to learn about real roots .Roots are the value of the variable that satiafies the given equation.It is also called as sollutions of the equations.The solution may be positive,negative or imaginary numbers.Whenthe roots are real values then its called as real roots. An equation which contains more than one terms are squared but no higher power in terms, having the syntax, ax2+bx+c where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant term

Example: 4x2+7x+18
Identify the Real Roots in Quadratic Equation

A quadratic equation is in the form of ax2+bx+c,

First we need to find the discernment d = b2- 4ac

If d > 0, the roots are real roots and unequal

If d = 0, the roots are real and equal

If d< 0, the roots are imaginary.
Example of Real Roots

Below you can see the example of real roots -

Example1: Solve the quadratic equation

x2+21x+20

Solution:

The equation is in the form of  ax2+bx+c

Where a=1 and b= 21 and c= 20

Find the discernment d = b2- 4ac

d = (21)2 - 4* 1* 20

d = 441 -80

d = 361 > 0

If d > 0, the roots are real roots and unequal

To find the roots, the following steps are used

Step 1: Multiply the coefficient of x2 and the constant term,

1*20 = 20 (product term)

Step 2: Find the factors for the product term

20--- > 1*20 = 20 (factors are 1 and 20)

20 --- > 1 + 20 = 21 (21 is equal to the coefficient of x)

Step 3: Separate the coefficient of x

x2 + 21x+20

x2 + x + 20x + 20

Step 4: The common term x for the first two terms and 20 for the next two terms are taking outside

x(x + 1) + 20 (x + 1)

(x + 1) (x + 20)

Set this value equal to zero

(x +1) = 0; x = -1

(x+20) = 0; x =-20

Roots of the quadratic equations x =-1 and x= -20

Example2: Solve the quadratic equation

x2+4x+4

Solution:

The equation is in the form of  ax2+bx+c

Where a=1 and b= 4 and c= 4

Find the discernment d = b2- 4ac

d = (4)2 - 4* 1* 4

d = 16 -16

d = 0

If d = 0, the roots are real roots and equal

To find the roots, the following steps are used

Step 1: Multiply the coefficient of x2 and the constant term,

1*4 = 4 (product term)

Step 2: Find the factors for the product term

4--- > 2 * 2 = 4 (factors are 2 and 2)

4 --- > 2 + 2 = 4 (4 is equal to the coefficient of x)

Step 3: Separate the coefficient of x

x2 + 4x+4

x2 + 2x + 2x + 4

Step 4: The common term x for the first two terms and 2 for the next two terms are taking outside

x(x + 2) + 2 (x + 2)

(x + 2) (x + 2)

Set this value equal to zero

(x + 2) = 0; x = -2

(x + 2) = 0; x =-2

Roots of the quadratic equations x =-2.

Problem Solving Situations

The problem solving situations is defined as the process by which the new circumstances situations are collected and resolved. It begins with an solving of all the aspects of the situations problems are solved and ends when a satisfactory answer has been found. The problem solving situations are also functioned in every one life .And now let us see about the problem solving situations.



Problems Based on Problem Solving Situations in Algebra

Determine the given Factor: 144x2 – 81 in term using problem solving situations

Solution:

Step1: The given factor is 144x2 – 81



Step2 : The problem solving situation here is to make squares on both sides means we get the values of 144 = 12 and 81 = 9.

Like 144x2 =81x2

144 = 122

81 = 92

Step3: Arrange the terms 144x2 – 81 = (12x)2 – (9)2      this produced by the different of squares formula

Like, (a + b) (a - b) = a2 + b2 where a is denoted by 12x and b is denoted by  9

Step 4: Therefore the factor of  144x2 – 81 are (12x + 9) (12x – 9).

Find:  8x - 1 = 33  in term using problem solving situations

Solution:

We know about the problem solving situation is to add one on both side we get

8x - 1 + 1 = 31 + 1

After simplifying, 8x = 32.

The required solution are x = 4

One more Problem in Problem Solving Situations Using Algebra

In term using problem solving situations

3x - y = -16---------------------- (1)

2x+8y =-28---------------------- (2)

The above problem handles the problem solving situation as rearranging the equation one means we can get the value for  y .

Step 1: The equation 1 can be written as,

3x - y = -16

y = 3x + 16--------------------- (3)

Now we can assume the value of y in equation 2 means and we get answer for x .

Step 2: Assume the y values in equation (2)

2x + 8(3x + 16) = -28

Step 3: Determine the above equation:

2x + 24x + 128 = -28

26x = -156

x = -6

Step 4: Substitute x values in 1st equation

3(-6) - y = -16

-18-y=-16

Y = -2

The required answer are x = -6, y = -2

The problems can be in term using problem solving situations .

Trigonometric ratios of compound Angles

Definition :

The algebraic sum of two or more angles is called a ' compound angle ' .

If A , B , C are angles , then A + B , A - B , A + B - C  , A - B + C , A + B + C , etc., are compound angles .

Formulas

cos(A + B)  =  cosA cosB  -  sinA sinB  for all A , B in  R .
cos(A - B)   =  cosA cosB  + sinA sinB  for all A , B  in  R .
sin(A + B)   =  sinA cosB  +  cosA sinB  for all  A , B in  R.
sin(A - B)    =  sinA cosB   -  cosA sinB  for all A , B  in  R.
tan(A + B)   =  (tanA+tanB)/(1-tanAtanB)   if  none of A , B , A + B is an odd multiple of (pi)/2  .

cot(A + B)   =  (cotBcotA-1)/(cotB+cotA)   if  none of A , B , A + B is an odd multiple of   (pi)/2 .
tan(A - B)    =   (tanA-tanB)/(1+tanAtanB)    if  none of A , B , A - B is an odd multiple of (pi)/2.
cot(A - B)    =   (cotBcotA+1)/(cotB-cotA)    if  none of A , B , A + B is an odd multiple of  (pi)/2.
sin(A + B) sin(A - B)   =  sin2A - sin2B    =    cos2B  -  cos2A .
cos(A + B) cos(A - B)   =  cos2A - sin2B   =   cos2B  -  sin2A


Solved Problems on Trigonometric Ratios of Compound Angles

1) Find the value of sin 105o and cos 165o .

Sol: sin 105o   =   sin (45o + 60o)

=    sin45o  cos60o   +  cos45o sin60o

=   (1/sqrt(2)) (1/2)   +  (1/sqrt(2)) (sqrt(3)/2)

sin105o   =   (sqrt(3)+1)/(2sqrt(2))

Now , cos165o   =   cos (180o - 15o)    =   - cos15o

=   - cos(45o - 30o)

=   - [ cos45o cos30o + sin 45o sin 30o]

=   -[(1/sqrt(2))(sqrt(3)/2)+(1/sqrt(2))(1/2)]

cos 165o   =   -(sqrt(3)+1)/(2sqrt(2))

2) Show that   cos42o + cos78o + cos162o = 0 .

Solution :  cos42o + cos78o + cos162o

=   cos (60o - 18o)  +  cos(60o + 18o)  +  cos(180o - 18o)

=   (cos60o cos18o + sin60o sin18o)  +  (cos60ocos18o - sin60o sin18o)  -  cos18o

=   2 cos60ocos18o  -  cos18o

=   2(1/2) cos18o  -  cos18o  =  0


More Problems on Trigonometric Ratios of Compound Angles:

1)Find the values of sin 15o , cos 15o , tan15o

Sol:

sin15o   =  sin (60o - 45o)   =  sin 60o cos45o  -  cos60o sin45o

=  (sqrt(3)/2)(1/sqrt(2)) -  (1/2)(1/sqrt(2))

sin 15o  =  (sqrt(3)-1)/(2sqrt(2))

cos 15o   =   cos(60o - 45o)   =   cos60o cos45o  +  sin60o sin45o

=   (1/2)(1/(sqrt(2)))   +  (sqrt(3)/2)(1/sqrt(2))

cos15o     =   (sqrt(3)+1)/(2sqrt(2))

tan15o   =   tan(60o - 45o)   =   (tan60^o-tan45^o)/(1+tan60^otan45^o)

=    (sqrt(3)-1)/(2sqrt(2))

tan15o   =   2 -  sqrt(3)

2) Show that  cos100o cos40o  +  sin100o sin40o   =  1/2

Solution :  cos100o cos40o  +  sin100o sin40o

The above equation is in the form of  cosA cosB  +  sinA sinB  which is equal to cos(A - B)

Here A = 100o  and  B = 40o

cos(A - B)  =  cos(100o - 40o)

=  cos(60o)

=  1/2

Hence  cos100o cos40o  +  sin100o sin40o  =  1/2

Solve Two Pairs of Angles Problems

Let us see about solve two pairs of angles problems in this article. Two lines are intersecting to make a two pair of angle. Two lines sharing the common end point is representing as angle. The word angle is derived form the Latin word. In Geometry angle is the one of the figure. In two pair of angle opposite angle are same and parallel.
Two Pairs of Angle:

There are several types of two pairs of angles is available for the geometry. These are

Complementary angle
Supplementary angle
Vertical angle

Example Diagram for Solve Two Pairs of the Angles Problems:

Pair of complementary angle:

The sum of the two angle measurement is equal to 90 is called as complementary angle.

example figure for two pair of the angles

Pair of supplementary angle:

The sum of the two angle measurement is equal to 180 is calling as complementary angle.

example figure for supplementary pair of angles

Pair of vertical angle:

One line crossed by another line is formed by the intersecting is said to be vertical angle.

example figure for vertical pair of angles
Example 1: Solve Two Pairs of Angles Problems:

In the given diagram find the unknown value of the a, b, c.

example for two pair of angle diagram

Solution:

Step 1: angle x is supplement of 80°

Step 2: a + 80° = 180°

a = 180° - 80°

a = 100°

Step 3: In this diagram angle c and 100° are vertical angles.

So, angle c = 100°

Step 4; In this diagram angle b and 80° are vertical angle

So, angle b = 80°

Step 5:

a = 100°

b = 80°

c = 100°


Example 2: Solve Two Pairs of Angles Problems:

In complementary angle, angle A is 37 degree calculate the unknown angle

Solution:

Step 1: Total angle measurement of the complementary angle is 37°

Step 2: Given angle is A = 37°

Step 3: Subtracting given angle measurement from total angle measurement of complementary angle. 90° - 37° = 53°

Step 4: Unknown angle is 53°

Problem Solving Training

In this article we are giving problem solving training and we can understand how to solving problems. It is very helping you to improve your problem solving skills. In mathematical terms we can see different types of problem solving. Here you can learn math terms, Exam practices test, problem solving and online test and our tutor will helps you and you can get free online tutor. Let us see Problem solving training.
Problem Solving:

Let us see few problems and their solving methods.

Problem solving training 1:

Solve:  `12 / 6 + 5 / 3`

Solution:

Above problem is showing fraction addition. So we should solve this problem in fraction addition operation.

Step 1: `12 /6 + 5 / 3`

Here numerator values are same, but denominators are different. So we should take LCM then only we can add both values. (We can take LCM if denominators are different).

Step 2:  `12 / 6 + 5 / 3`

Take LCM 6, 3 (therefore LCM is 6)

we have to change denominators values like 6.

Step 3: `(12*1)/(6*1) = 12/6 , (5*2)/ (3*2) = 10 / 6`

`12/6 + 10 / 6`

Now denominators are same so add both values

Step 4: `12 / 6 + 10 / 6`

`22 / 6`

Therefore `12/ 6 + 5 / 3 = 22 / 6.`

Problem solving training 2:

Solve:    20 ___ 5   = 25

Solution:

Step 1: Given 20 ___ 5   = 25.

Step2: here we find the symbols which are need for this operation.

Step 3: if we put the - (minus) symbols like 20 - 5 = 15 we can get 15.

So minus operation is not accept

Step4: + (plus) is correct operation for this problem.

20 + 5 = 25

Step 5: Therefore + (plus) symbol is making the number sentences true.

Problem solving training 3:

Divide the two fractions `40 -: 1/ 5`

Solution:

Above problem is showing fraction division. So we should solve this problem in fraction division operation.

Step 1: given` 40 -: 1/ 5`

Step 2: It denoted by `40 -: 1/5`

The right hand side denominator will be change like as 5/1 so

=   `40 * 5/1`

= `200`

Step 3: Therefore answer is 200.
Practices Problems:

1) Solve this fraction `40/ 20`      answer: 2

2) Add `2 / 3 + 3/ 2 `                      answer: `13/6`

3) Find missing number 4 , 8  12 , ___ , 20 , 24 , ____ , 32      answer: 16 , 28

Exponential Growth Graph

The exponential growth graph is nothing but the exponential function occurs if the rate of growth is proportional to the functional value and the current value in the functional part. The exponential graph contains some of the equal intervals and can be called as the exponential growth or exponential decay. Now we are going to see about the exponential growth graph.
Problems on Exponential Growth Graph

Determine the exponential growth for the function $3000 and to double at 21/2 % continuously

Solution:

The exponential growth function can be calculated as,

A = Pert

The rate value which can be taken as 0.105

6000 = 3000 e 0.105t

We have to take natural log on both sides we get,

2 = e0.105t

ln 2 = ln e 0.105t

ln 2 = 0.105t (ln e)

ln 2 = 0.105t

By using the calculator the value can be found as,

0.693147 = 0.105t

t = 6.666

Thus it takes 6.66 years to double the money.

Graph:Graph
More Problems on Exponential Growth Graph:

Example 1:

Determine the value of ‘r’ where A = 50 at t = 6 years and P = 10

Solution:

The exponential growth can be calculated as,

A = Pert

50 = 10 e6r

5 = e6r

The logarithm for the above equation given as,

ln 5 = 6r

ln (5)/6 = r

r = 1.6094/6

r = 0.2682

The growth of exponential is 0.2682.

Example 2:

Determine the exponential growth for the function $4000 and to double at 24/2 % continuously

Solution:

The exponential growth function can be calculated as,

A = Pert

The rate value which can be taken as 0.12

8000 = 4000 e 0.12t

We have to take natural log on both sides we get,

2 = e0.12t

ln 2 = ln e 0.12t

ln 2 = 0.12t (ln e)

ln 2 = 0.12t

By using the calculator the value can be found as,

0.693147 = 0.12t

t = 5.77

Thus it takes 5.77 years to double the money.

Graph:

Graph

Relational Algebra Notation

The algebra is a mathematical system that consists of operands and operators. The operand is a variable or value from which new value can be constructed. The operator is a symbol denoted the procedures of that construct new value from given values. The relational algebra is the operand variable that represented relations and operator is designed to relation in the database. In this below details about the relational algebra notation.

Symbolic Notation in Relational Algebra Notation:

The relational algebra is whose operand is variable that represent in the relations and the operator is designed to do the most common things that we need to do with relations in a database. The algebra result can be used to query language for relations. There are following as relational algebra notation.

Symbolic notations are,

SELECT,

PROJECT,

PRODUCT,

JOIN,

UNION,

INTERSECTION,

DIFFERENCE,

RENAME,

Usage of notations in relational algebra notation:

1.      SELECT:

The SELECT notation represent in symbol as σ (sigma).The SELECT operation used to select the particular data details get in the database.

2.      PROJECT:

The PROJECT notation represent in symbol as π (pi). The PROJECT operation used to can provide itself to concision in the particular database.

3.      PRODUCT:

The PRODUCT is representing in symbol as x (times).The PRODUCT is performed combination of tuples in the data base.

4.      JOIN and UNION:

The JOIN is representing in symbol as |x|(bow-tie) and UNION is representing in symbol as U (cub). The JION   used to get particular data in that it has two tuples in restriction of their cartesian product based on the conditions.  The UNION is used to set the rotational attributes are same means at that data provided.

5.      INTERSECTION and DIFFERENCE,

The INTERSECTION and DIFFERENCE have symbols as ∩ and – (minus). The intersection is used to intersection of two relations and provides the common tuples in the relations. The difference used to when applied to relations which first relation has in the tuple but in the second relation.
Example of Relational Algebra Notation:

To find all employees in department cs.

SELECT depname = cs (employee), becomes σdepname = cs (employee).

To find all employees in department cs called jack.

SELECT depname = cs ^ jack = jack(employee), becomes σdepname = cs  ^ jack= jack(employee).

Volume Paraboloid

Paraboloid is defined as one of the most important mathematical shape. Paraboloid has three dimension shape. Paraboloid is defined as the combination of ellipse and parabola.  All the sections present in the paraboloid parallel to one coordinate plane is parabola and all the sections present in paraboloid parallel to one coordinate is ellipse. In this section, we are going to see about the volume of paraboloid in detail.
Explanation to Volume Paraboloid

The explanation to volume of paraboloid is given below the following section,

Formula:

Volume of paraboloid = `1/2` `Pi` a2 h

where,

h = height of the paraboloid

Example Problem to Volume Paraboloid

Problem 1: Find the volume of paraboloid, where, h = 10, a = 5.

Solution:

Step 1: The given values for finding the volume of paraboloid is as follows,

h = 10,

a = 5.

Step 2: To find:

Volume of Paraboloid

Step 3:The formula given for finding the volume of paraboloid is as follows,

Volume of Paraboloid = `1/2` `Pi` a2 h

Step 4: By substituting the values in the volume formula,

Volume of Paraboloid = `1/2` `Pi` a2 h

= `1/2` (3.14) (52 ) (10)

= `1/2` (3.14) (25) (10)

= `1/2` (3.14) (250)

=  `1/2` (3.14) (250)

= `1/2` (785)

= 392.5

Result: Volume of paraboloid = 392.5

Thus, this is the require answer for solving the volume of paraboloid.

Problem 2: Find the volume of paraboloid, where, h = 15, a = 6.

Solution:

Step 1: The given values for finding the volume of paraboloid is as follows,

h = 15,

a = 6.

Step 2: To find:

Volume of Paraboloid

Step 3:The formula given for finding the volume of paraboloid is as follows,

Volume of Paraboloid = `1/2` `Pi` a2 h

Step 4: By substituting the values in the volume formula,

Volume of Paraboloid = `1/2` `Pi` a2 h

= `1/2` (3.14) (62 ) (15)

= `1/2` (3.14) (36) (15)

= `1/2` (3.14) (540)

= `1/2` (1695.6)

= 847.8

Result: Volume of paraboloid = 847.8

Thus, this is the require answer for solving the volume of paraboloid.

Multiple Coefficients of Determination

In this topic we will discuss about multiple coefficients determination. The multiple coefficients are the coefficients that have the many term in an expression. In a variable, a coefficient is a number in front of a variable. For example, in the expression `2x^3-7x+19` , the coefficient of the `2x^2` is 1 and the coefficient of the x is -7. The third term, 19, is known as a constant. Here we see about multiple coefficients and problems.

Additional Information about Multiple Coefficients of Determination:

In an expression the coefficients may be positive or negative or zero. When addition or subtract the polynomials, just we want to add or subtract the coefficients in the same terms. The Coefficient in an expression is the number which is multiplied by one or more variables or powers of variables in the term. The following some of the related terms in multiple coefficients determination.

1) Variable

2) Expression

3) Polynomial

4) Term
Example Problem for Multiple Coefficients of Determination:

Multiple coefficient of determination – Example 1

Identify the number of terms in the expression, 2x2 + 4x + 7y + 6xy + 5xy2 + 3x2y

Solution:

Step 1:

When an expression is writing as an addition, the parts that are added are the terms of the expression.

Step 2:

2x2 the variable has the coefficient of 2.

4x the variable have the coefficient of 4.

7y this variable has the coefficient of 7.

6xy this variable’s having the coefficient of 6.

5xy2 this variable’s having the coefficient of 5.

3x2y this variable’s having the coefficient of 3.

Therefore, there are multiple terms in the expression.

Multiple coefficient of determination – Example 2

2) Identify the number of terms in the expression, x3 + 4x2 + 3x + 6y3 + 2y2 + 7y

Solution:

Step 1:

When an expression is writing as an addition, the parts that are added are the terms of the expression.

Step 2:

x3 the variable having the coefficient of 1.

4x2 the variable having the coefficient of 4.

3x this variable have the coefficient of 3.

6y3 this variable’s having the coefficient of 6.

2y2 this variable’s having the coefficient of 2.

7y this variable’s having the coefficient of 7.

Therefore, there are multiple terms in the expression.

Symmetric Equation of a Line

There are several ways to denote a distinct line in R3. Symmetric equation is one such way of representing distinct lines. Initially to deal with symmetric equation familiarity with parametric equation is necessary. Obviously this notation is considered compact too. Getting into the concept of symmetric equation, it sets each component of the line equal to a common parameter. Following this all the components are set equal to one another. Knowledge in vector equation of line and parametric equation would help us in solving tutorials.

More about Symmetric Matrix Equation of a Line:

The parametric equation of the line is stated as follows:

`(x-a_x)/(m_1)=(y-a_y)/(m_2)=(z-a_z)/(m_3)`

To arrive at this equation, each expression in the parametric equation of the line should be treated equal. For example

`x: a_x + (t)(m_1)`

It should be taken as

`x=a_x + (t)(m_1)`

The expression should be solved for each value of t and ultimately the resultant expression would be equal to the constant t and therefore would be equal to each other. Interestingly this expression of line violates one of the common doctrines of mathematics which limits us to use only one equality sign per expression. It is this factor that helps us to use this mode of representation in fewer or more dimensions. In case we are to use it in R2 form then the z expression would be ignored thus containing only one equality sign per expression. Similarly if we are expected to express a line that is parallel to a co ordinate plane then the term of the axis to which it is parallel to is also ignored.
Example Problems – Symmetric Equation of a Line:

Example 1 – Symmetric equation of a line:

Find the symmetric equation of the line through point (6,-7,20) & perpendicular to the plane 2x+3y-6z-8=0.

Solution:

Let us consider a normal vector n

2x + 3y - 6z - 8 = 0 is n = <2 -6="-6" 3="3">

The vector n is also the directional vector of the line thru point P(6, -7, 20) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <6 -7="-7" 20="20"> + t<2 -6="-6" 3="3">

L = <6 -7="-7" -="-" 20="20" 2="2" 3t="3t" 6t="6t" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 6 + 2t

y = -7 + 3t

z = 20 - 6t

Symmetric form of equation.

`t` =` (x - 6)/2` = `(y + 7)/3` = `(z - 20)/-6`


Example 2 – Symmetric equation of a line:

Find the symmetric equation of the line through point (2, 3, 4) & perpendicular to the plane x+2y-3z-4=0.

Solution:

Let us consider a normal vector n

x+2y-3z-4=0 is n = <1 -3="-3" 2="2">

The vector n is also the directional vector of the line thru point P(2, 3, 4) and perpendicular to the given plane.

The equation of the line L is:

L = P + tn = <2 3="3" 4="4"> + t<1 -3="-3" 2="2">

L = <2 -="-" 2t="2t" 3="3" 3t="3t" 4="4" t="t">

By solving for t convert the equation of the line to symmetric form

L:

x = 2 + t

y = 3 + 2t

z = 4 - 3t

Symmetric form of equation.

`t` = `(x - 2)/1` = `(y - 3)/2` = `(z - 4)/-3`

Solving for x With Polynomials

Polynomial defined as the function p(x) of the form p(x) =a0 +a1x+a2x2+……….anxn. Where a0, a1…an real numbers and n is the non negative integer is called polynomial in x over reals.For example 4x2-7x+3 is a polynomial over integers. If one of the powers of x in p(x) is either a negative integer of a fraction (either positive or negative), then p(x) is not a polynomial. For example x+2/x  is not a polynomial. The highest exponent of the variable in polynomial is called the degree of the polynomial.. Here we are going to study about how to solving for x with polynomial and its example problems.

Solving for X with Polynomials - Example Problems

Example: 1

Solve for x in the following polynomial expression 3x+5+6x +7 = 3x+4

Solving steps:    In the left hand side combine the like term first

3x+ 6x+ 7 + 5 = 3x + 4

9x + 12 = 3x + 4

Add both sides -4 we get

9x + 12 - 3 = 3x + 4 - 4

In right hand side 4 - 4 will be cancelling

9x+ 9 = 3x

Add both sides -3x

9x -3x + 9 = 3x -3x

6x + 9 =0

Add both sides -9 we get

6x + 9 -9 = -9

6x = -9

Divide both sides 6

x = - 9/6

The simplest form is - 3/2

Therefore the value of x = - 3/2

Solving for X with Polynomials - Example: 3

Solve for x in the following polynomial x2 + 9x +18 =0

Solving steps:

First we have to find the factor for a given polynomial

We can write

x2 + 9x +18  = (x+3)(x+6)

These are the two factors the equation

Now we solve the both equation.

Both terms equating to zero we get

First x+3=0

Add both sides -3 we get

x = -3

Next term is x+6 = 0

Add both sides -6

x = -6

Therefore the value of x is -1,-6

Polynomials Algebra Tiles

In mathematics, a polynomial is an expression of finite length constructed from variables (also known as indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents. For example, x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number. In this article we shall discuss about polynomials algebra tiles. (Source: Wikipedia)


Sample Problem for Polynomials Algebra Tiles:

Solved polynomials algebra tiles problem:

Example 1:

Factorize the polynomial 2x3 – 10x2 – 24x+ 72

Solution

Sum of the coefficients of terms: 2–10–24 + 72 = 40 ≠ 0.

Therefore the value (x–1) is not a factor.

Sum of the coefficients of even degree terms = –10 + 72 = 62

Sum of the coefficients of odd degree terms = 2 – 24 = –22

Since they are not equivalent we estimate that (x + 1) is also not a factor. Let us check whether x – 2 is a factor. By synthetic division method

2 | 2       -10     -24        +72

|

|          +4      -12          -72
________________________

2        -6     -36     |    0

_________________________

Since the remainder is 0, (x – 2) is a factor. To find other factors

2x2 – 6x – 36 = 2x2 – 12 x + 6x –36

= 2x(x –6) + 6(x–6) = (2x + 6) (x – 6)

Therefore, x3 – 5x2 – 12x + 36 = (x–2) (x–6) (x+3)


Example 2:

Factorize 4x3 + 2x2 – 10x + 4

Solution:

Since the addition of the coefficients of all the terms: 4 + 2 – 10 + 4 = 10 – 10 = 0

We guess that (x – 1) is a factor.

By synthetic division,

1 | 4         +2        -10        +4

|

|              4        +6        -4

________________________

4         6         -4     |   0

_________________________

Remainder is 0. Quotient is 4x2 + 6x – 4

To find other factors, factorize the quotient,

4x2 + 6x – 4 = 4x2 + 8x – 2x – 4

= 4x (x + 2) – 2 (x + 2)

= (x + 2) (4x – 2)

∴ 2x3 + x2 – 5x + 2 = (x – 1) (x + 2) (2x – 1)
Practice Problem for Polynomials Algebra Tiles:

Factorize 2x2 + 3x -2

Answer: (x + 2) (2x – 1)

Factorize X2 – 3x – 18

Answer: (x + 3) (x – 6)

Add and Subtract Fractions Calculator

Add and subtract calculator which is used for finding the addition and subtraction for various value of a fraction. Calculator is a process of calculating fractions with different values.In general fraction contains two parts numerator and denominator. the upper part is the numerator and lower part is a denominator which depends upon the this two parts. there are types of fractions such as proper,improper fraction and mixed fraction.

Rules for Add and Subtract on Fractions Calculator:

fraction calculator

First choose the operations in the column shown(add,subtract,multiply and divide)
then,fraction #1 enter the fraction value and in the fraction#2 then press the calculate
We get final result in column below the calculate.
For the next operation press reset and follow the steps.

Fraction shown above will be the calculator for the fraction addition and subtraction steps in the calculator will be explained below.
Addition on fractions:

For adding same denominator fractions just add all the numerator and keep same denominator.


For addition with different denominators fraction:

Find the l.c.m. for all the denominators given .
Change into equivalent fraction with the same LCM denominator
By taking  the LCM common in  the denominator and add all the numerators.


Subtraction on fractions :

For  subtract same denominator fractions subtracting  all numerator and keep same denominator.


For subtraction with different  denominator:

Finding  l.c.m. for all the denominators given.
Change equivalent fraction with same l.c.m denominator and then subtract


Problems for Add and Subtract Calculator:
Example 1:
Add   1/4 and 2/4

Solution:
here we have the same denominator so, just add the numerator keep the denominator same.
1/4 +2/4 = (1+2) /4

=3/4

Example 2:
Add  4/5  and 1/3

Solution :
The LCM of 5 and 3 is 15.
Therefore,4/5+1/3 = (4xx3)/(5xx3)+(1xx5)/(3xx5)

=12/15+5/15

=17/15

Example 3:
Add 2 4/5 and 3 5/6

Solution:
The given fraction is a mixed fraction first change into proper fraction and add
 2 4/5 +3 5/6=(2xx5)+4/5+(3xx6)+5/6=14/5+23/6

so, the denominator is different LCM of 5,6 =30
=(14xx6)/(5xx6)+(23xx5)/(6xx5)

=84/30 +115/30

=119/30

Example 4:
Subtract  3/5 -1/5

Solution:
here, same denominator just subtract numerator and keep denominator is same.

 3/5-1/5 =(3-1)/5
=2/5
Example 5:
subtract 4 2/5-2 1/5

Solution :
The given fraction is a mixed fraction first change into proper fraction and subtract
(4xx5)+1/5 -(2xx5)+1/5

so, 21/5 -11/5

=(21-11)/5

=10/5

Example 6:
Subtract  3/4   from  5/6

Solution :
We need to find equivalent fractions of 3/4 and 5/6 , which have the same denominator.
This denominator is given by the LCM of 4 and 6. The required LCM is 12.

Therefore, 5/6-3/4 =(5xx2)/(6xx2)-(3xx3)/(4xx3)
=10/12-9/12

=1/12

Definition of Subset Learning

We come across set of different kinds in everyday life. A football team is set of players, a class is a set of students, a set of books in mathematics and so on. Basically, the set is an undefined term.  Anyhow, it can be a well defined  collection of objects. Here the word 'objects' has been taken in wider and broader sense. Mathematically, the numbers, words, letters, signs, symbols, thoughts, etc. all are the objects. Let us learn about a set and  definition of a subset in this article.

Learning Definition of a Subset

Definition

If each element of A is an element of B, then the set A is said to be a sub set of the set B.

Symbolically, we represent this by A `sub` B and read it as ' A is a subset of B'.

Thus A `sub` B `hArr` (  x `in` A   `rArr` x `in` B ).

If A `sub` B and A `!=` B then the set A is said to a proper subset of B.

learning examples of subset

1. {1, 2, 3, 4, 5 } `sub` { x : x is a natural number }

2. { a, b, c, d } `sub` { a, b, c, d, e}

3. Set of all vowels is a subset of set of all alphabets

4. A = { set of all integers}

B = { set of all positive integers}

B is a subset of A.

5. T = { x : x is a student of class 9 in your school }

W = { x : x is a student in your school}

T is a subset of W.



Definition of Subset Learning - Points to Remember

1. `phi` `sub` A ( since `phi` has no elements, it is a subset of every set)

2. A `sube` A ( every set is a subset of itself)

3. A `sube` B and B `sube` A `hArr`    A = B

4. A `sub` B and B `sub` D `=>`   A `sub` D

5. If A `sub` B, then x `!in` B `rArr` x `!in` A.

Adding Exponents Worksheet

Exponents are nothing but multiplying the number by itself.  For example, y2 is the same as y * y. Exponent denotes the number of times the number needed to be multiplied by itself. An exponent is nothing but a superscript, or small number that can be written at the top right edge of a number, variable, or set of parentheses. In this article we shall discuss the adding exponents worksheet.

Problems for Adding Exponents Worksheet:

Adding exponents worksheet:

Problem 1:

To find x3y4. x5y3

Solution:

Rearrange the factors and multiply the exponent terms, using the rule

Here,

am.an = am+n

= (x3. x5)( y3. y4)

= x8. Y7

X and y are different terms,

So the final result is x8. Y7

Problem 2 for Adding exponents worksheet:

To find  (32)3

Solution:

(32)3 = 93 =  729 using power rule,

The answer is 729

Problem 3:

To find  (22)3

Solution:

(22)3 =(2x2)3 =  43=64 using power rule,

The answer is 64

Mixed variables for multiply exponents worksheet:

Have a mix of variables:

Example 1:

=xy2z y3z

=x y2+3z1+1

=xy5z2

so the result is =x3y5z2

Example 2:

=x3y3z3 .  x2yz

=x3+2 y3+1 z3+1

=x5y4z4

so the result is =x5y4z4

With constant examples:

Example 1:

=5xyz .4xyz

=(5 4)x1+1 y1+1 z1+1

=20x2y2z2

The result is =20x2y2z2

Example 2:

=7xyz.3 x2yz

=(73 ) x1+2y1+1z1+1

=21x3 y2z2

The result is  =21x3 y2z2
More Examples for Multiply Exponents Worksheet:

Example 1:

=5xyz `xx` 4xyz

= (5`*` 5)x1+1 y1+1 z1+1

=25x2y2z2

The result is =25x2y2z2

Example 2:

=7xyz `xx` 4 x2yz

= (7`*` 4) x1+2y1+1z1+1

=28x3 y2z2

The result is  =28x3 y2z2

Example 3:

=5xyz `xx` 7xyz

=(7`*` 5)x1+1 y1+1 z1+1

=35x2y2z2

The result is =35x2y2z2

Example 4:

=7xyz  ` xx` 2 x2yz

= (7`*` 2 ) x1+2y1+1z1+1

=14x3 y2z2

The result is  =14x3 y2z2

Define Positive Integer

The definition of positive integer is one of the most important topic in mathematics. Positive integer is always present after the zero in the number line. The symbol used for representing the positive integer is called as the " + ". All the mathematical operations are done using the positive integer. In this article, we are going to see about the positive integer with brief explanation and some example problems.

Explanation to Define Positive Integer
The explanation for to define the positive integer is given below the following,

Addition problem for defining the positive integer
Subtraction problem for defining the positive integer
Multiplication problem for defining the positive integer
Division problem for defining the positive integer

Example Problems to Define Positive Integer

Example 1: Add the following positive integer, 36 and 24

Solution:

36 and 24

3 6

2 4

___

6 0

___

Result: Positive integer addition = 6 0

Thus, this is the result to define the addition problem problem of positive integer.

Example 2: Subtract the following to define positive integer, 36 and 24

Solution:

36 and 24

3 6

2 4

___

1 2

___

Result: Positive integer subtraction = 1 2

Thus, this is the result to define the subtraction problem of positive integer.

Example 3: Multiply the following to define positive integer, 36 and 24

Solution:

36 and 24

3 6

2 4

____

8 6 4

____

Result: Positive integer multiplication = 6 0

Thus, this is the result to define the multiplication problem of positive integer.


Example 4: Divide the following to define positive integer, 48 and 6

Solution:

48 and 6

6 ) 4 8 ( 8

4 8

____

0

____

Result: Positive integer division = 6 0

Thus, this is the result to define the division problem of positive integer.
Practice Problems to Define Positive Integer

Example 1: Add the following positive integer, 42 and 14.

Answer: 56

Example 2: Subtract the following positive integer, 42 and 14.

Answer: 28

Example 3: Multiply the following  positive integer, 42 and 14.

Answer: 588

Example 4: Divide the following positive integer, 56 and 7.

Answer: 6

Probability Distribution Definition

The word probability and chance are familiar to everyone. Many a time we come across statements like “It is possible that our school students may get state ranks in forthcoming public examination.

“Probably it may rain today”

Definition: The word chance, possible, probably, likely etc. convey some sense of uncertainty about the occurrence of some events. Our entire world is filled with uncertainty. We make decisions affected by uncertainty virtually every day. In order to think about and measure uncertainty, we turn to a branch of mathematics is called as probability.

Probability Distribution Definition-classical Definitions

Definition: If there are n exhaustive, likewise exclusive and in the same way likely outcomes of an experiment and m of them are favorable to an event A, and then the mathematical probability of A is defined as the ratio m/n.

Definition for random variable:

The outcomes of an experiment are represented by a random variable if these outcomes are numerical or if real numbers can be assigned to them. For example, in a die rolling experiment, the corresponding random variable is represented by the set of outcomes {1, 2, 3, 4, 5, 6} ; while in the coin tossing experiment the outcomes head (H) or tail (T) can be represented as a random variable by assuming 0 to T and 1 to H.

Types of Random variables:

(1) Discrete Random variable
(2) Continuous Random variable


Definition for Discrete Random Variable: If a random variable takes only a finite or a countable number of values, it is called a discrete random variable.

Example:
1. The number of heads obtained when two coins are tossed is a discrete random variable as X assumes the values 0, 1 or 2  which form a countable set.
2. Number of Aces when ten cards are drawn from a well shuffled pack of 52 cards.
Probability Distribution Definition-theoretical Distributions:

Theoretical probability distributions is classified into

1. Binomial Distribution

2. Poisson Distribution

3. Normal Distribution

4. Exponential Distribution

Multiplication Of Exponents

Exponents are significant in scientific notation, when large or small quantities are denoted as powers of 10. exponents are mentioned by superscripts, as in the examples above. But it is not always possible way to write them this method.  If x is the exponent to which is a minimum base quantity a is increased value, then a x can be written in ASCII as a power of x. In a scientific notation, the higher case letter E can be used to point out that a number is raised up to a positive or negative power of 10. For model take 125x3. Here 125 is coefficient of variable 'x’ Then 3 is the exponent value of x . Exponent value also known as power value.
Suitable Examples for Multiply Exponents


Exponents of 1 and 0
If the exponent is 1, then only have the variable itself (example a1 =a)

Generally need not to write the "1", but it sometimes helps to remember that y is also a1
Exponent of  0

If the exponent is 0, then the values are not multiplying by anything and the result is just "1" (example a0 = 1)
Multiplying Variables with Exponents

multiply this variable with exponents:

(z2)(z3)

So that z2 = zz, and z3 = zzz so that all the multiplies,

(z2)(z3)

= zzzzz

That is 5 "z"s multiplied mutually so the new exponent must be 5:

(z2)(z3)

= z5

The exponents say to that there are two "z"s multiplied by 3 "z"s for a total of 5 "z"s:

(z2)(z3)



= z2+3 =z5

So, the simplest method is to just add the components
Mixed Variables for Multiply Exponents:

Have a mix of variables:

Example 1:

=xy2z y3z

=x y2+3z1+1

=xy5z2

so the result is =x3y5z2    

Example 2:

=x3y3z3   x2yz

=x3+2 y3+1 z3+1

=x5y4z4

so the result is =x5y4z4

With constant examples:

Example 1:

=5xyz 4xyz

=(5 4)x1+1 y1+1 z1+1

=20x2y2z2

The result is =20x2y2z2

Example 2:

=7xyz 3 x2yz

=(73 ) x1+2y1+1z1+1

=21x3 y2z2

The result is  =21x3 y2z2    

Meaning Of Prime Numbers

In mathematics, a prime number (or a prime) is a natural number that has exactly two distinct natural number divisors: 1 and itself. The first twenty-five prime numbers are: Prime Numbers:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.Condition: - If x is the prime number then the next factors of the number x is 1 and X. Let us see meaning of prime numbers.(Source:Wikipidea)                                                                                                                                                                                                                                                                              

Meaning of Prime Numbers:

Prime number means that number should divide by one and the number itself only. Otherwise it is not a prime number. If any one number divisible by two means that is not a prime number.

            Let us take a number to find prime number or not. For that we will take 131.

Step 1:

            Let us find the square root for that given number 131 = 11.44
            Now we have to change the decimal number as a whole number. That should be the nearest big whole number. 11.44 = 12

Step 2:

            Find the prime numbers below 12. The prime numbers up to 12 are 2, 3, 5, 7 and 11.

Step 3:
            So from this factors no one number will not divide by 131. The number contains 1 and 131 only.  Hence it is considered as a prime number.
Example Problems - Meaning of Prime Numbers:

Problem 1:

            Find out the number 29 is prime number or not?

Solution:

            Here the number 29 is divisible by one and itself only. It has no more factors other than this. So 29 is considered as a prime number.

Problem 2:

            Find out the number 53 is prime number or not?

Solution:

            The getting number 53 is not divisible by two. 53 has only two factors. Those factors are one and itself only. So we can say the given number is prime number.

Problem 3:

            Find out the number 78 is prime number or not?

Solution:

            The obtain number 78 is not a prime number. 78 is divisible by two. The number 78 contains more than two factors. Those factors are 2 x 3 x 13. Therefore the given number 78 is not a prime number.

Problem 4:

            Find out the number 108 is prime number or not?

Solution:

            The obtain number 108 is not a prime number. Why means 108 is divisible by two. 108 includes more than two factors. Those prime factors are 2 x 2 x 3 x 3 x 3. Therefore we cannot say it is a prime number.
Practice problems - Meaning of prime numbers:

Problem 1:

            Find the number 13 is prime number or not?

Solution:

            The given number 13 is considered as a prime number.

Problem 2:

            Find out the number 54 is prime number or not?

Solution:

            The given number 54 is not a prime number.

Solving Polynomials

What are Polynomials?
The simple definition of Polynomial is that, it is an expression containing multiple terms that are combined together through addition, subtraction and multiplication. This polynomial definition is illustrated by this polynomial equation: 3x + 4y + 5. In this example, the polynomial equation is a combination of three terms.

Ways of Solving Polynomials
There are different ways of solving polynomials and the strategy differs based on the polynomial equation provided. In this article, let us discuss two different ways to solve polynomial through relevant examples.

Example 1: Solve polynomial (a+8) (2a-10)

In this equation, there are two sets of terms that are multiplied with each other. To solve this polynomial, multiply each term of the first set with all the terms in the second set. This is illustrated below:

When ‘a’ is multiplied with ‘2a’, it gives 2a2. Then ‘a’ is multiplied with -10, which results in ‘-10a’. As a next step, multiply 8 with the terms in the second set. When ‘+8’ is multiplied with ‘2a’, it gives ‘+16a’ and ‘+8’ multiplied with ‘-10’ gives ‘-80’. Now the equation is dissolved into:

(a+8) (2a-10) = 2a2 -10a + 16a -80

In the above equation, the second term (-10a) and the third term (+16a) contains the same term with the same degree, but the constant value alone varies. When there are several terms with same variable and same degree, they can be combined. In this case, -10a+16a can be rearranged as 16a-10a, which when solved results in 6a. Thus, the polynomial equation is further solved into the result 2a2 + 6a – 80.

Example 2: Solve polynomial equation a+2b = 5a+7b
In this equation, there are two polynomials, one on the left hand side and the other on the right hand side. In this case, these polynomials help each other to find the solution. The polynomial solver for this expression will be evolved through the steps below:
• Original Expression: a+2b = 5a+7b
• Moving right hand side expression to left: a +2b-5a-7b=0.
• Grouping terms of same variables and degree: a-5a+2b-7b=0
• Combining terms of same variables and degree: -4a-5b
• Take common factor out: -1(4a+5b)

Thus the result is -1(4a+5b).

When the polynomials are complex, the above strategies alone might not work. For instance, if you are solving quadratic polynomials then you have to perform series of steps such as finding zeros of the polynomial, finding roots and much more to solve them.

What is a Perfect Number?

A Perfect Number is nothing but a whole number, which is identical to the sum of its each and every proper divisor.

Perfect numbers

What are Divisors?
Divisors are same as the factors of a number.  A divisor or factor is a number which divides a number evenly and gives the remainder as zero. Finding factors of a number is simple. For example, consider the number 9.  The divisors or factors of 9 are 1, 3, 9, because only these three numbers can divide 9 without leaving any remainder.

Mathematical problems are solved by finding factors and multiples as we did now. In the above case, 3 is the factor of 9 as it divides 9 evenly. At the same time, 9 is a multiple of 3 because 3 X 3 = 9. This signifies that finding factors and multiples are necessary to find the perfect numbers and to solve many algebraic expressions.

What are Proper Divisors?
Proper divisors are the divisors of the number excluding the number itself. As already discussed, the divisors or factors of 9 are 1, 3 and 9. The proper divisors of number 9 are 1, 3, and 9. Number 9 is not a proper divisor of 9 because it is the same as the original number for which the divisors are identified.

What are Perfect Numbers?
Knowing what proper divisors are, now let us look at the question: What is a perfect numbers? Identify the proper divisors of a number and add them. If the result of addition is same as the actual number for which the divisors are identified, then that number is a perfect number.

For example, we identified 1 and 3 as proper divisors of9. Now add these proper divisors:
1 + 3 = 4

Here the sum is 4. Instead, if the sum was 9 (the actual number for which you found the divisors), then 9 will be called as perfect number. In this case, sum is 4 and not 9. This signifies that 9 is not a perfect number.

Let us take up another example. Are 6 a perfect number? The divisors or factors of 6 are 1, 2, 3, and 6. The proper divisors of number 6 will then be 1, 2 and 3. Now, add these proper divisors.  Addition of 1, 2 and 3 will result in 6. The result 6 is same as the number 6 for which we identified these proper divisors. Thus, 6 is a perfect number.

The other perfect numbers are:
• 28
• 496
• 8128

Applied Statistics learning

Let's learn about what is applied statistics in today's post.

Applied statistics is that section of statistics which can be applied in real life. There are different levels of measurements and these are:

• Nominal measurement
• Ordinal measurement
• Ratio measurement
• Interval measurement

Next time will help you learning on examples of analyzing data.

You can also avail to a math tutor for more help. On can also get help for other subjects such as biology tutoring and so on.

Do post your comments.