Learn Divisibility Tests

Divisibility tests means that to discover a number is divisible by a particular number (divisor) without perform the division operation. The rules shown below are used to translate a given number into a normally smaller number while preserve divisibility by the divisor. There are divisibility test for numbers in any radix, and they are all dissimilar, we represent rules simply for decimal numbers below. In this article we shall learn about divisibility tests rules with examples.

Divisibility tests rules

To learn a number is divisible by 2:

The number is end with even number (that is 0,2,4,6) the number are divisible by 2.

To learn a number is divisible by 3:

The sum of the digit is multiple of 3. The given number is divisible by 3.

To learn a number is divisible by 4:

The numbers are formed by the final pairs of digits is divisible by 4.

Example:

4672 => 72 ÷ 4 = 18

So 4672 ÷ 4 = 1168

A number is divisible by 8:

The numbers are formed by the final three digits are equally divisible by 8.

Example:

53,104 => 104 ÷ 8 = 13

So 53,104 ÷ 8 = 6638

To learn a number is divisible by 9:

The sum of the digits is a multiple of 9.

Example: 3,726

3,726 => 3 + 7 + 2 + 6 = 18

Since 18 = 9 × 2,

Then 3,726 ÷ 9 = 414

To learn a number is divisible by 6:

The numbers satisfy the rule for 2 and 3; that is, first it must be an even number, then a digit sum is a multiple of 3.

To learn a number is divisible by 12:

The numbers satisfy the rules for 3 and 4; that is, the digit sum is a multiple of 3, and its final digit pair is a multiple of 4.

Divisible tests practice problem

Problem1:

Check whether the given number 4652 is divisible by 4

Answer:

4652 => 52 ÷ 4 = 18

So 4672 ÷ 4 = 1163 therefore the given number is divisible by 4.

Problem2:

Check whether the given number 53112 is divisible by 8

Answer:

53,112 => 112 ÷ 8 = 14

So 53,112 ÷ 8 = 6639 therefore the given number is divisible by 8.

Vertical Stretch and Compression

Vertical stretch and compression mean transforming the graph based on the scale factors. Here we will see how we are performing the stretching and compression of graph using the scale factor. Stretching the graph is nothing but we are transforming the graph away from axis. Compression of the graph means squeezing the given graph towards the axis. We will see some example problems foe vertical stretch and compression of the graphs.
Vertical Stretch:

Vertical stretch is nothing but the stretching the graph away from x – axis. If the given function is f(x) then the vertical stretch of the given function is y = a f(x). Where 0 < a < 1

Example for vertical stretch:

Graph the following function and its vertical stretch. Where f(x) = x2 – 1.6x and the vertical stretch scale factor of the function f (x) is 0.4.

Solution:

Given function f(x) = x2 – 1.6x

We can write the given functions like y = x2 – 1.6x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the given function

We have to plug y = 0

So 0 = x2 – 1.6x

x (x – 1.6) = 0

x = 0 and x = 1.6

So the x intercept point is (0, 0) and (1.6, 0)

For y – intercept of the given function

We have to plug x = 0

So y = (x)2 – 1.6(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical stretch function.

The vertical stretch function is

f(x) = a f (x)

So y = 0.4 (x2 – 1.6x)

If we graph both the function we will get the following graph

vertical stretch - stretch graph
Vertical Compression:

Vertical compression is nothing but the squeezing the graph towards the x – axis. If the given function is f(x) then the vertical compression of the given function is y = af(x). Where a > 1

Example for vertical compression:

Graph the following function and its vertical compression. Where f(x) = x2 – 3.5x and the vertical compression scale factor of the function f (x) is 1.25

Solution:

The given function f(x) = x2 – 3.5x

We can write the given functions like y = x2 – 3.5x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the original function

We have to plug y = 0

So 0 = x2 – 3.5x

x (x – 3.5) = 0

x = 0 and x = 3.5

So the x intercept point is (0, 0) and (3.5, 0)

For y – intercept of the original function

We have to plug x = 0

So y = (x)2 – 3.5(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical compression function.

The vertical compression function is

f(x) = a f (x)

So y = 1.25 (x2 – 3.5x)

Solve Vector Spherical Coordinates

This article is to solve vector spherical coordinates. Solve vector spherical coordinates is nothing but solving problems under spherical coordinate with the help of the tutors. Tutor vista tutor have many highly qualified tutors in online. A quantity which has magnitude and direction is a vector. There are three methods  to describe a vector.  Specific lengths, direction, angles, projections or components are the methods. Below we can see about solve vector spherical coordinates.

Solve Vector Spherical Coordinates

From this system any point as the point of intersection of the spherical surface (radius r = constant) conical surface (theta phitheta and phi by dr, rd theta and dphi .The element sides are dr,rd theta and rsin theta dphi . = constant). A differential volume element is obtained in spherical coordinate. This happens by increasing r value,

The differential length dl = sqrt(dr^2+(rd theta)^2+(rsin theta dphi)^2)

The differential areas ds = dr.rd theta = rdrd theta

= dr.rsin theta dphi = r sin theta dphi dr

= rd theta. rsin theta dphi = r2sin2 theta d theta dphi

The differential volume dv = dr.rd theta.rsin theta dphi

dv = r2sin theta dphi dr.

So spherical co ordinates = (r, theta, phi)
Solve Vector Spherical Coordinates

A vector in cartesian co-ordinate system can be converted to spherical coordinate.

Conversion of  cartesian to spherical system

The Cartesian co-ordinates is (x,y,z). This cartesian coordinate is converted into spherical co-ordinates ( r, theta, phi )

Given                                  Transform

x                                     r = sqrt (x2+y2+z2)

y                                     theta = cos-1(z/(sqrt(x2+y2+z2))) = cos-1(z/r)

z                                    phi = tan-1(y/x)

Problem 1: Convert the cartesian coordinates x = 2, y = 1, z = 3 into spherical co-ordinates.

Given                        Transform

x = 2                           r = sqrt (x2+y2+z2)

= sqrt (4+1+9)

= sqrt (14) = 3.74

y = 1                            cos-1(z/r) =  cos-1(3/sqrt(14)) =  36.7°

z = 3                           phi = tan-1(y/x) tan-1(1/2)26.56°

Spherical co- ordinates are (3.74, 36.7°, 26.56°)

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.