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