Learn Online Inequalities

Definition of learn online inequalities:

Inequality is described as two real numbers or two algebraic expressions are communicated with functioning a symbol as ‘<’ (less than), ‘>’ (greater than), ‘≤’ (less than or equal) and ≥ (greater than or equal). Learn online inequalities are the method of learning the inequalities by online. The various inequalities are learned online as follows,

Numerical inequalities
Literal inequalities
Double inequalities
Strict inequalities
Slack inequalities
Linear inequalities

Types :


From learning online inequalities, they can be explained as follows,

1) Numerical inequalities:

Inequalities which enclose arithmetical lone without any variables is known as numerical inequalities

Eg: 5 < 8; 5 > 4

2) Literal inequalities:

Inequalities which have one or more variables are labeled as literal inequalities.

Eg: p< 5; q > 2; m ≥ 4; n ≤ 6

3) Double inequalities:

An inequality which contains two sign (< or > or ≤ or ≥) is named as double inequality.

Eg: 2 < b < 8; 3 ≥ t ≥ 8

4) Strict inequalities:

If an inequality holds a symbol < or >, then it is learned as strict inequalities.

Eg: Ax + B < 0; Ax2 + Bx + C > 0

5) Slack inequalities:

If an inequality involves a sign ≤ or ≥, then it is called slack inequalities.

Eg: Ax + By ≤ C; Ax + By ≥ C

6) Linear inequalities:

An inequality may have one variable with linear is called linear inequality through one variable; If it contains two variables, then it is called linear inequality through two variables.

Eg: Ax + By < C; Mx + C > Y


Rules & Example for learn online inequalities :


RULES :    Inequality contains the following rules for learning online,

Rule 1: All sides of an inequality can be added or subtracted by means of equal numbers without affecting the symbol.

Rule 2: Same numbers may be multiplied or divided as of both sides of an inequality.

Ex :  Solve online 40 u < 200 when

(i) ‘u’ is a natural number,

(ii) ‘u’ is an integer.

Sol :      Given 40 u < 200

From online,

40u / 40 < 200 / 40 (Rule 2)

u < 5.

(i) When ‘u’ is a natural number, then the statement gives,

1, 2, 3, 4

The solution set is {1, 2, 3, and 4}.

(ii) When ‘u’ is an integer, then the solution is given as,

..., – 3, –2, –1, 0, 1, 2, 3, 4

The solution set is {...,–3, –2,–1, 0, 1, 2, 3, 4}

Sat Math Practice Problems

Introduction:

Scholastic Assessment Test or Sat reasoning test is a test which is used to get admissions in US universities and colleges. Sat test deals with quantitative questions and English skills. Sat test is conducted for 3 hrs and 45 minutes. It consists of three parts; they are i) Critical reasoning ii) Math aptitude and iii) Writing.  Math section of Sat questions deal with quantitative questions and logical reasoning questions. Math questions are given with multiple choices. Students should Practice math problems to get good scores in Sat test. I like to share this Define Geometry with you all through my article.


Sat Math Problems:


Example 1:

There are 240 balls, 4 times as many are brown, and the rest is black. How many are green and how many are black colors?

Solution:

There are 4 times as many brown balls as there is black color,

Hence, there must be 4 brown for each 1 of black color.

Given that 4 + 1 = 5,

Therefore,

240/5 = 48

On solving this we get, 48 sets of balls with 4 greens and one of black color in each set.

The total is then,

48 * 4 = 192 green and 48 of black color

Answer Check:

Number of green balls = 192 balls

Remaining black balls = 48

Total balls = 192 + 48

= 240 balls.



Example 2:

Mani drives from his house to hills 150 miles away, and at the end of the day drives home. If Mani drives at a standard speed of 50 miles per hour, how long does Mani takes to drive the round trip?

Solution:

Here Mani takes 150 miles to reach the destination,

The total distance covered by Mani during the round trip = 150 + 150

= 300 miles

Mani drives at an average of 50 miles per hour.

Using the formula

Distance = speed x time

300 = 50 x X

Divide 50 on both sides,

300/50 = 50X/50

6 = X

Mani takes 6 hrs to complete the round trip.

Answer is 6 hrs.


Sat Math Practice Problems:

Practice Problems1:

In a class of 78 students 41 are taking Tamil, 22 are taking English and 9 students are taking both Tamil and English. How many students are not enrolled in any of the course?

A) 10

B) 15

C) 24

D) 34

Answer: C


Practice Problems 2:

Six years ago rani was X times as old as raji was. If rani is now 17 years old, how old is raji now in terms of X?

A) 12/X + 6

B) 11/X + 6

C) 17X

D) 18/X

Answer: B

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Practice Problems 3:

Which of the following numbers can be used to demonstrate that all prime numbers are not odd?

A) 2

B) 5

C) 11

D) 13.

Answer: A

Direction Vectors

Introduction:

In mathematics a direction vector that describes a line segment D is any vector of the direction vector,

AB?

Where, A and B are two distinct points on the line D. If v is a direction vector for the D, so is kv for any nonzero scalar k; and these are in fact all of the direction vectors for the line D. Under the some definitions, the direction vector is required to be a unit vector, in which case each line has exactly two direction vectors, which are negatives of each other equal in magnitude, opposite in direction. (Source:Wikipedia)

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Direction vector for a line in two-dimensional:


Any other line in two-dimensional Euclidean space can be described as the set of solutions to an equation of the form

ax + by + c = 0

Where a, b, c are real numbers. Then one direction vector of (D) is (- b,a). Any multiple of (- b, a) is also a direction vector.


Basic properties:


The following section uses the Cartesian coordinate system with basis vectors

a=a1e1+a2e2+a3e3 and
b=b1e1+b2e2+b3e3
Are equal if

a1=b1, a2=b2, a3=b3.


Magnitude and Direction of a Vector:


Let v can be a vectors given in component form by

                        v = < a,b >

The magnitude of the || v || of vector v is given by

                 || v || = sort (a 2 + b 2)

and the direction of the vector v is angle t in standard position and in counterclockwise direction is such that

                  tan(t) = v / u

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Vector methods:


Vectors and vector addition
Unit vectors
Base vectors and vector components
Rectangular components in 2-D
Rectangular coordinates in 3-D
Direction cosines
A vector connecting two points
Dot product
Rectangular coordinates
Projection of a vector onto a line
The cross product
The triple product
Rectangular coordinates
Triple vector product

Example:

The equation of a line is 2x - 3y + 15 = 0.

2x-3y=-15

2(-3)-3(3)=-15

-6-9=-15

So (-3, 3) is direction vectors for this line

Geometry in Our Daily Life

Introduction to Geometry in our daily life:

Geometry is a part of mathematics which is concerned with questions of shape, size, the relative position of figures and the properties of space. In earlier, geometry was a collection of empirically discovered principles which concerns about lengths, areas, angles and volumes which were developed to meet some practical need in surveying, construction, and various crafts. There are various fields of life where geometry is considered as an important field of study. Please express your views of this topic Similar Polygon by commenting on blog.


Geometry in Daily Life

The general applications of geometry in our daily life are as follows:

Geometry is the very useful to define relative physical locations.
Geometry can be used to analyze or design physical structures and objects and to determine the various location and elevation of physical features on the Earth's surface.
Major examples of the geometry in real life:  The shape and volume of food containers, the design of buildings, and the layout of streets and roadways. I have recently faced lot of problem while learning Finding Surface Area of a Cylinder, But thank to online resources of math which helped me to learn myself easily on net.

Major Applications of Geometry in Daily Life

Major Applications of Geometry in Daily Life are as follows:

In real life, Geometry is particularly useful in home building and improvement projects. If we want to find the area of the floor of a house, we can use geometry.
In turn, this information is very useful for laying carpet or tiles and also for telling an estate agent regarding the information that how big our house is when we want to put it on the market. Geometry involved in many our daily life problem.
If we need to reupholster a piece of furniture, it is necessary to estimate the amount of fabric that we need by calculating the furniture’s surface area.
Geometry also finds its application in hobbies. In order for the fish to thrive, the water in a goldfish tank needs to have a certain volume as well as surface area. We can calculate it easily using geometry.
Pastimes like quilting and some other design projects will use geometry extensively. How the shapes of a quilt block fit together can be understand which is basically dependent on geometry; so that is determining the amount of fabric we need. Geometry is used for find the angle of the object.

Math Division Problems

INTRODUCTION FOR DIVISION PROBLEMS

The operation of finding how many times one number, the divisor, is contained in a second, the dividend. The result is called the quotient, and if the divisor is not contained an integral number of times in the dividend, and number left over is called the remainder. Indicated either by the division sign, or by a stroke or bar, in which case the repression as a whole is called a fraction and the dividend and the divisor the numerator and denominator respectively. Understanding Matrix Division is always challenging for me but thanks to all math help websites to help me out.


DIVISION PROBLEMS EXAMPLE:


DIVISION PROBLEM 1:-
Solve 413 ÷ 7.
Set the divisor (7) before the bracket and place the dividend (3654) under it.
7)413(

Examine the first digit of the dividend (4). It is lesser than 7 so it can't be divided by 7 to produce a whole number. Next take the initial two digits of the dividend (41) and determine how many 7's it contains. In this case 41 has five sevens (5x7=35) but not six (6x7=42). Place the 5 after the division bracket.

7)413(5

Multiply 5 by 7 and place the result (35) below the 41 of the dividend.

7)413(5
35

Draw a line under 35 and subtract it from 41 (41-35=6). Bring down 3 from the 413 and place it to the right of the 6.

7)413(5
35
63

Divide 63 by 7 and place that answer after the division bracket to the right of the five.

7)413(59
35
63

Multiply 9 of the quotient by the divisor (7) to get 63 and place this below the 63 under the dividend. Subtract 63 from 63 to give an answer of 0. This indicates that there is not anything left over and 7 can be evenly divided into 413 to produce a quotient of 59.

7)413(59
35
63
63
0

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Practice division problems:


Division problem -1: solve 658 by 2
Division problem -2: solve 265 by 5.
Answers for Practice division problems.
Answer for division problem -1 = 329
Answer for division problem -2 = 53

Ballpark Estimate in Math

Introduction to ballpark estimate in math:

Ballpark estimate in math is defined as the method which is used to simply the calculation in easy way. In ballpark estimate in math the values taken for adding and subtracting process are changed to nearest and easier numbers for simplification. In considerations of small values in calculations they are changed to their nearest 10’s and for largest values of calculations they are changed to their nearest 50’s and 100’s. Understanding Answers for Math Problems is always challenging for me but thanks to all math help websites to help me out.


Examples for ballpark estimate in math:


1)      Evaluate 73 + 88.

Solution:

73 is changed to the nearest value 70 ( i.e. nearest 10’s since it is the smallest values )

88 is changed to the nearest value 90 ( i.e. nearest 10’s since it is the smallest values )

70

+90

-------

160

-------

2)      Evaluate 67 - 51.

Solution:

67  is changed to the nearest value 70 ( i.e. nearest 10’s since it is the smallest values )

51  is changed to the nearest value 50 ( i.e. nearest 10’s since it is the smallest values )

70

- 50

-------

20

-------

3)      Evaluate 163 + 188.

Solution:

163 is changed to the nearest value 150 ( i.e. nearest 50’s since it is the largest values )

188 is changed to the nearest value 200 ( i.e. nearest 100’s since it is the largest values )

150

+200

-------

350

-------

4)      Evaluate 107 - 65.

Solution:

107  is changed to the nearest value 100 ( i.e. nearest 100’s since it is the largest values )

65  is changed to the nearest value 50 ( i.e. nearest 50’s since it is the largest values )

100

-  50

-------

50

-------


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Some excercise about the ballpark estimate in maths.

1)      Evaluate 64 + 47. ( answer: 110 )

2)      Evaluate 194 - 173. ( answer: 50 )

Subset Learning

Subset :

A and B are sets such that every element of A belongs to B then we state that A is a subset of B and write A subset B.
w is said to be subset of set X if and only if each element of set w belong to set X as well.
w = { a ,d } and X = { a , b , c , d } .
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Main resource of subset learning:


The subset learning will execute all exceptional repetition in a subset .
Review will complete all exceptional repetition and well as force mid-interval repetition on all basics in a subset .evaluate of non-outstanding element is equivalent to Learning : Execute repetition existing from the element menu.
Review topics works like review all excluding it do not include items, i.e. it forces a review of all topics in a subset .
The such case of subset learning, the attach learn at the bottom of the inside gap can be used to execute outstanding repetition on a selected branch of the information.
The parameter subset learning in the statistics window indicate the progress of repetition in subset learning. This field display the number of objects, the number of topic, and the number of until elements in subset learning. I have recently faced lot of problem while learning Define Decimal, But thank to online resources of math which helped me to learn myself easily on net.

Feature of Subset learning:


The element subset choice problem is well known in data and pattern
Recognition. However, many of the technique deal exclusively with features that are continuous, or, make assumption that do not hold for many sensible machine learning algorithms.
For example, one regular assumption says that increasing the digit of features can never decrease performance.
Although assumptions such as monotonicity are often null for machine learning,one approach to quality subset selection in machine learn has on loan search and valuation technique from figures and pattern recognition.
This advance, dubbedthe covering estimate the accuracy of feature subsets via anumerical re-sampling method using the real machinelearn algorithm.
The wrapping has prove useful but is very slow to carry out as theinduction algorithm is called constantly.