Pronouns and its types

Pronoun is one of the parts of speech in English grammar. Pronoun is a part of speech that is used instead of noun. It is used to reduce repeated use of noun in a conversation. Examples of pronouns are he, she, they, them, him, and it and so on. For example: Mary got a HCL kids laptop yesterday. She bought it for her niece. (Here, ‘she’ is used instead of using ‘Mary’ again.) There are six types of pronouns. Let’s have a look at each of them.

Personal Pronouns:
Personal pronouns are types of pronouns that are used as substitute for proper or common nouns. Examples of personal nouns are: I, we, our, he, she, mine, his, her, and they and their. For example: I bought a HCL kids’ laptop from online kids store few days back.

Demonstrative Pronouns:
Demonstrative pronouns are types of pronouns that are used to point out objects. Examples of demonstrative pronouns are: this, that, these, those and more. For example: That baby bottle is of Nuby bottles brand.

Indefinite Pronouns:
Indefinite pronouns are types of pronouns that are used to refer general things or people in general way without specifying anyone. Examples of indefinite pronouns are: nobody, somebody, everybody and more. For example: One can buy Nuby bottles for kids’ online stores as well.

Distributive Pronouns:
Distributive pronouns are types of pronouns that are used to refer two or more persons or things at the same time. Examples of distributive pronouns are: each, either, neither and so on. For example: She can buy either Himalaya baby soap or Johnson & Johnson’s baby soap for her baby.

Relative Pronouns:
Relative pronouns are types of pronouns that are used to relate a subordinate clause to the rest of the sentence. Examples of relative pronouns are: who, whose, whom, which and so on. For example: For whom are you buying this Himalaya baby soap ?

Interrogative Pronouns:
Interrogative pronouns are used for asking questions. Examples of interrogative pronouns are what, which, who, whose, whom and so on. For example: What Christmas gift do you want?
These are the types of pronouns.

Verb and its types

Verb is one of the eight parts of speech in English grammar. Verb is a part of speech that describes an action or indicates a state of being. For example: Mary bought BSA stroller for her baby from online baby shop. There are different types of verbs. Let’s identify the various types of verbs and its definitions in this post.
Transitive and Intransitive Verb:
Transitive verb is a type of verb that involves a direct object while on the other hand; intransitive verb is a type of verb that doesn’t involve a direct object.

Examples of Transitive Verb:
• Mary bought BSA stroller from online baby shop.
• She eats fish.
• Kids love games.
Here, all the sentences have direct object such as BSA stroller, fish and games respectively.

Examples of Intransitive Verb:
• The boy throws.
• He walks very fast.
• She reads well.
Here, all the three sentences do not have a direct object.

Main and Auxiliary Verb:
A sentence can have two inter-related verb that are termed as main and auxiliary verb. The main verb is the type of verb that describes the primary action and auxiliary verb is the type of verb that adds detail to it. Some of the most popularly used auxiliary verbs are to be, to have, to do and more. Auxiliary verbs are also referred as helping verbs. For example:
• She will buy gravity destroyers toys for her son’s third birthday. (Here, ‘will’ is the auxiliary verb that is helping the main verb ‘buy’ to convey the meaning.)
• Gravity destroyers toy can be bought from online shopping centers. (Here, ‘can be’ is the auxiliary verb that is helping the main verb ‘bought’.)

Modal Verbs:
Modal verbs are type of auxiliary verb that adds mood to a sentence, most commonly imperative and probability. For example:
• She should buy the gift before Christmas. (Here, ‘should’ is the modal verb.)
• I might visit my cousin’s place this weekend. (Here, ‘might’ is the modal verb.)
• I wouldn’t have done that if I was you. (Here, ‘wouldn’t’ is the modal verb.)
These are some of the basic types of verb and its examples.

Adverbs and its types

Adverb is one of the eight parts of speech in English grammar. Adverb is a part of speech that modifies the meaning of a verb, adjective or another adverb in a sentence. For example: Mary instantly bought the cutlery stand after hearing about the discount. Adverbs can be classified into seven classes, namely time, frequency, place, manner, degree, affirmation and negation and reason. Let’s have a closer look at the same in this post.

Adverbs of Time:
Adverbs of time are the type of adverbs that are used to answer to the question “when”. Most commonly used adverbs of time are now, yesterday, today, once. For example: Mary bought the cutlery stand yesterday. Here, the answer is to the question: when Mary bought the cutlery stand?

Adverbs of Frequency:
Adverbs of frequency are the type of adverbs that are used to answer to the question “how often” i.e. answering about the frequency of occurrence. Popularly used adverbs of frequency are: seldom, rarely, frequently, often and more. For example: She often buys branded girls’tops  from online stores. Here, the answer is to the question: how often she buys branded girls’ tops from online stores?

Adverbs of Place:
Adverbs of place are used to answer to the question “where”. Commonly used adverbs of place are: forward, everywhere, out, in and more. For example: She has gone out of the house. Here, the answer is to the question: where has she gone?

Adverbs of Manner:
Adverbs of manner are used to answer to the question “how”. Some of the most popularly used adverbs of manner are: honestly, bravely, happily and more. For example: She happily accepted the gift from him. Here, the answer is to the question: How did she accept the gift from him?

Adverbs of Degree:
Adverbs of degree are used to answer to the question “how much”. Popularly used adverbs of degree are: fully, partly, altogether, almost etc. For example: She bought almost every baby essential from Chicco Talcum India collection.  Here, the answer is to the question: How much she bought from Chicco Talcum India collection.

Adverbs of Affirmation and Negation:
Adverbs of affirmation and negation are used either to confirm or deny a statement. Certainly, absolutely are some of the commonly used adverbs of affirmation and negation. For example: I trust my friend absolutely.

Adverbs of Reason:
As the term suggests, adverbs of reason are used to give the reason. Therefore and hence are the most commonly used adverbs of reason. For example: Chicco is one of the trusted baby brands hence, I buy products from this brand. These are the seven classes of adverbs.

Basic Understanding of Sets and Its Operations

There are various operations on sets which must be mastered to completely understand them. Some of them are intersection, union and so on. These operations have to be learnt. The sets and set operations form an integral part of mathematics. The terms Venn diagrams and set operations are closely related to each other. Venn diagrams can be used to explain various operations. The operations with sets can be fun to work with.  The union will give all the elements present in both the sets. The intersection will give the common elements. This is the basic difference between the two operations and can be easily understood.

Venn diagrams can be very helpful in this. They are graphical representation of the sets. It is always to better to explain a concept using a diagram. It improves the clarity of the concept and also makes the concept easy to understand. Venn diagrams do a great deal in achieving this. The set operations math can be understood using them.  Cardinality is another important concept that has to be learnt. It basically denotes the number of elements. It is not tough to find the number of elements but a term has been assigned for the same. So, there is nothing much to worry about it. It is quite a easy concept and can be easily learnt. There can be various types of sets like empty, unit and sometimes even special ones. Empty ones are those which have no elements in them. Unit ones are those which will be having only a single element in them.

There are several others like ones for prime numbers which is denoted by the letter ‘P’. There is also one for the natural numbers which is usually denoted by the letter ‘N’. The integer one contains both the positive and negative integers. Also there is one which contains all the real numbers in it. There can also be sets for rational numbers and complex. Rational numbers are contained in the former and the complex numbers are contained in the latter. There is similarity between all of them. The only difference being that the natures of elements vary in each of them. This is the basic difference and has to be understood. Once this is understood the concept becomes easy to understand and apply. It is not a complex subject and can be understood with a little bit of practice.

Past Tense and its Types

In English, there are three types of tenses – Present tense, Past tense and Future tense. Past tense is a type of tense that expresses something in past or states about some action that takes place in the past. For example: I got a baby bather India online yesterday. Here, the action of getting baby bather India online is done in past and therefore, the sentence is in past tense.  Past tense further can be classified into sub categories. Let’s have a look at the types of past tense along with examples.
Types of Past Tense:
•         Past Perfect Tense
•         Past Continuous Tense
•         Past Perfect Continuous Tense

Past Perfect Tense:
Past perfect tense refers to an action that is completed in the past. For example: My cousin bought branded baby foods online for her baby. Here, the action of buying branded baby foods online is completed in the past and therefore, the sentence is in past perfect tense.

Past Continuous Tense:
Past continuous tense refers to an action that was completed over a period of time in the past. For example: I was exploring a lot of baby online shops few days back and finally I got few good online baby stores. Here, the action of finding baby online shops was completed over a period of time and therefore the sentence is in past continuous tense.

Past Perfect Continuous Tense:
Past perfect continuous tense refers to an action that started in the past and continued up until another time in the past. Unlike present perfect continuous tense, in past perfect continuous tense the action doesn’t continue till the present. For example: I had been working on this problem from five days. Here, the action of working out the problem started and ended in two different times in the past and therefore, the sentence is in past perfect continuous tense.
These are the basics about past tense and its types.

Perpendicular Postulate

The number refers to the ordinal position is, thus the number is the sequence of the result set. And the ordinal numbers does not show the quantity.

When we learn about numbers and its applications, we will study the three types of numbers, there are,

Ordinal numbers
Cardinal numbers and
Nominal numbers.

Explanatory:

Cardinal numbers:

Cardinal numbers shows that “how many”in the group or list. This cardinal numbers are also called counting numbers. Because it shows the quantity. Also it does not have any decimal numbers or fractions, only used for counting.

Example: how many months in a year? Answer is 12 months.

Nominal numbers:

Nominal numbers used to only identify the things. It does not show the quantity or rank.

Example: A bike number on the race.

But here we have to go for only ordinal number and its positions, so let us see what are ordinal numbers?

Ordinal Numbers:

An Ordinal Number is a number that shows the position of something in a group or Ordinal numbers shows the Position (order) of the things in a group from - first; second, third, fourth, etc…and the ordinal numbers does not shown the quantity.

Example: in a bag contains “apple, mango, papaya, jack fruit, pineapple and chappotta” the word “pineapple” is in fifth from order.

Easy to remember: "ordinal" shows what "order" things are in.

Example Problems in Ordinal Numbers:

Example 1:

There are six children in a running race means find the ordinal positions in this?

Solution:

Ordinals locate a place in a sequence, when we describe the child who came in sixth; we are using an ordinal number

A number that tells the position of something in a race in 1st, 2nd, 3rd, 4th,  5th, 6th.

Example 2:

Indian cricket team players are ordered from Sachin, Shewag, Raina, Yuvraj, Yousuf, Dhoni, Virat,  Irfan, Harbhajan, Praveen, Zaheer,  who is 1st and who is 8th place from the list?

Answer is:       Sachin is 1st from the order of list and Irfan is 8th from the order of given list.

What is a Integer?

What is a Integer- Integers includes all whole numbers along with all negative numbers. That is all negative numbers and positive numbers along with zero are called integers. Definition of Integer says all positive and negative whole numbers are termed as integers. For example: - -3, 15, -8, 0, 78, 91 they all are integers. We can plot integers on the number line. Numbers towards right side on the number line are greater than number on its left. That is if we plot the number line with integers we see that all negative numbers lie on left side of number line and all positive lies on right hand side. They follow a trend of increasing towards right side. For example:- -3  falls on left side of 0 as -3 is smaller than zero, similarly 6 falls on right of 0 as 6 is greater than 0. Integer Definition says all the positive and negative whole numbers along with zero so the list of integers goes on.

To add, subtract, multiply and divide integers, we follow certain Integers Rules. For adding two integers, we should remember that on adding two negative integers, we get a negative integer and we get a positive integer if we add two positive integers. For example -3 + (-2) = -5 and 4 + 6 = 10. And if we are adding a positive and negative integer, we find their absolute values and then subtract the smaller number from the larger one and put the sign of the integer which has a larger absolute value. For example, if we have to add -3 and 5, then we find their absolute values which will be 3 and 5. We subtract 3 from 5 which gives 2 and then we add sign of 5 which is plus. Hence the solution will be +2.

Subtracting Integers– To subtract two integers, we need to change the sign of the integer to be subtracted that is if we have to subtract -3 from -5 then the equation becomes -5 + 3 because -3 was the integer to be subtracted. Now we can find their absolute values and can perform the action of addition. Subtracting two positive integers gives a positive integer. Subtracting two negative integers will give a negative integer. There are many Integer Word Problems that can be solved by using the integer rules of addition and subtraction.

Octal Representation of Numbers

Numbers is a vast subject but not a very tough topic in mathematics to understand. The vastness is because of the deductive skills of scholars and different representation of the same number to increase the security of the same. Octal Numbers is one such type of representation of number like the other representations like decimal, binary, etc. The word Octal or octa number system refers to the number eight and in Octal Number System the number is supported with the base 8. Octals are very similar to the decimal number other than in decimal we have 0 – 9 and in this we have 0 – 7.

It is not a very big task to convert decimal to octa number system or from this to decimal number. The Decimal to Octal Conversion can be understood through an example. But before going to an example let us get to know the steps to be followed while converting the decimal to octa number system.

The following are the steps used for the conversion: One should know the division and multiplication of numbers well so that he / she can solve the problem in ease. The decimal number should be written. The number should be divided by 8.

The quotient of the number should be written below. The remainder of the number after dividing has to be written on the right end. The number written below has to be further divided by 8. This division process has to be done until the number that is the remainder is less than the octa number system digits.

The process is completed once the above steps are followed. The answer has to be written now. The answer can be found by writing the numbers that is the remainder starting from the final remainder bottom to top. The answer should be written with subscript 8 to avoid confusion. Before going to an example for further explanation we should know the Octal Table which represents the octa number system in a table form with the corresponding decimal number for them.

This representation helps in the conversion of octa number system to decimal and the conversion of decimal to octa number system.

Let us consider the following example for understanding the conversion of decimal to octa number system.
Convert 9910 to octa number system: 99 / 8 = 12 with a remainder 3, 12 / 8 = 1 with a remainder 4, the answer for the question which is given will be: 9910 = 1438.

linear algebra sample final

Linear algebra is mainly use to find the solution of systems of linear equations in some unknowns. Linear algebra has a demonstration in analytic geometry and is generalize in operator theory. Linear algebra associates with the families of vectors called vector spaces, and with functions contain one input vector and output another vector, according to rules. Linear algebra has the application with the theory of systems of linear equations, determinants, and matrices. The sample problems are discussed below for final review.

Linear Algebra Sample Problems for Final:

The sample problems are solved in detail for final review as shown below.

Problem 1:

Evaluate the linear equation   -3(x - 4) / 6 - (x - 2) / 3 = -2 x

Sol:

The given equation has rational expressions. To eliminate the denominators by multiplying by the LCD
-3(x - 4) / 6 - (x - 2) / 3 = -2 x

The LCD is equal to 6*3 = 18. Multiply both sides by LCD.
18 * [-3 (x - 4) / 6 - (x - 2) / 3] = 18* [-2 x]

Simplify to eliminate the denominator.
-9(x - 4) - 6(x - 2) = -36x

Multiply factors and grouping the terms

-15x + 48 = -36x

Subtract 48 to both sides
-15x + 48 - 48 = -36x -48

Group like terms
-15x = -36x - 48

Add 36x to both sides
-13x + 36x = -36x - 48 +36x

Grouping the terms
23x = -48

Multiply both sides by `1/23 `
x = - `48/23`

Conclusion:
x = - `48/23` is the solution for the above given equation.

Sample Linear Algebra Practice Problems for Final:

1) Evaluate the linear equation   -3(x - 4) / 7 - (x - 2) / 2 = -2 x

2) Solve the linear equation -5(-x - 7) = 5x – 32.

3) Evaluate the linear equation   -3(-x +3) = x - 7

Answers:

1)   X = -38/15 is the solution for the above given equation.

2) The above equation has no solution.

3)  X=One is the solution for the above equation.

Order of Operations

Order or operations is one of the most important topics in mathematics. Order of operations is also known as PEMDAS or BODMAS. As the term suggests, order of operations refers to the steps to perform a mathematical operation. When it is a simple addition or subtraction, you can add or subtract instantly. For example: Sunny bought Nerf guns India from Fisher Price India collection. He already has Angry Birds toys India and Dartboard games.

How many variety of toys and games collection he has in total: Nerf guns + Angry Birds toys India + Dartboard = 3 variety of toys and games. This is addition operation in mathematics. John has three Nerf guns of Fisher Price India brand. He gave one to his cousin Andrew. How many Nerf guns does he have? 3 Nerf guns – 1 Nerf gun = 2 Nerf gun. This is subtraction operation in mathematics. But what do you do when a mathematical expression has more than one operation. For example: Sunny had 3 guns, his father gave him 2 new guns and he gave one to his friend John. How many he has now? Mathematically: (3+2) – 1 = 4. This is a mathematical expression.

While calculating a mathematical expression, it is essential to follow some rules or order of operation. This order of operation is referred to as BODMAS or PEMDAS as abbreviation for the functions in mathematics. The order of operations is expressed as below.
1. [B] Brackets or Parenthesis
2. [O] Orders (Exponents, roots etc)
3. [DM] Division and Multiplication
4. [AS] Addition and Subtraction

The above order of operation states that in a mathematical expression, at first the operation inside brackets needs to be calculated, thereafter the powers, roots and more, then division and multiplication and finally addition and subtraction. This should be the order of operation while simplifying or calculating a mathematical expression.
For example: There are six kids. Father bought one Nerf Guns India and two building block toys for each kid. How many toys father bought in total?
Mathematically, (1+2) x 6
Applying order of operations (1+2) = 3 x 6 = 18
Therefore, father bought 18 toys in total.

This is referred as order of operations or BODMAS in mathematics.

Estimating with Confidence

Concept of ‘estimation with confidence’ is normally used by everybody in their daily lives. For example, a person may say ‘I believe there are 90 percent chances that Congress will win in the next elections with more than 350 seats’. Here, the person’s confidence level is 90 percent and he is estimating by how much seats congress will win. Let’s move forward to the formal concept of ‘estimation with confidence’.

There are two types of tests one-tailed and two-tailed, we will discuss ‘estimation with confidence’ in each of the case separately.

Estimating with Confidence in a One Tailed Test

In the single tailed test we estimate with the given level of confidence whether the estimated value is same as the population value or either statistically greater than or less than the population value/expected value (for simplicity we are assuming normal distribution). In this we can have right or left tailed test.

Left Tailed : Let us assume the given level of confidence is 85%, from Fig-I we see yellow area is the rejection region. So, with 85% confidence we can say that our estimated value won’t b statistically smaller from the population value/expected value and there is only 15 % probability that it will be statistically smaller.

FIG-I

Right Tailed: In this case the calculation remains the same, the only difference we don’t want statistically greater value than population value/expected value rather than less than as in the previous case. We represent right tailed estimation as in Fig-II. Here yellow shaded region is the rejection area. Confidence level is 85%. So, with 85% confidence we can say that our estimated value won’t b statistically greater from the population value/expected value and there is only 15 % probability that it will be statistically greater.

FIG-II

Estimating with Confidence in Two Tailed Test

Under a two tailed estimation with confidence we do not want either too big or too small value in comparison to the population value/expected value. In other words, we have rejection area on the both tails of the curve (as shown in Fig-III). Here, the yellow shaded area is the rejection area. Here, the confidence level is of 70%. So, with 70% confidence we can say that our estimated value won’t b statistically different from the population value/expected value and there is only 30% probability that it will be statistically different.

FIG-III

Compare and Order Integers

A number line is good place to start when describing integers. As integers are all the counting numbers so starting from 0, the numbers to the right side of zero are called positive integers, and from 0 to the left side, all the numbers are called the negative integers.

Comparing Integers - We can Compare Integers with their Opposites which means the numbers on the opposite sides with equal distance from zero on the number line. For example, on a number line, 4 and -4 are the opposites. The number zero is the opposite of itself, and it is considered neither positive nor negative. Let us graph the integers and their opposites. Let’s graph 4 and its opposite on the number line. Firstly graph 4 by placing a dot at positive 4 on the number line. We know the opposite of positive 4 is negative 4. So place a dot at a negative 4 on the number line to complete the problem. Next example lets graph -2 and its opposite on the number line. First, graph negative 1 by placing a dot on the negative on the number line. Next step we know the opposite of negative 1 is positive is 1. So we place a dot on number line to complete our graph.

Ordering Integers – Ordering integers means Ordering Integers from least to Greatest and for this we write integers in the ascending order. The numbers on left of number line is smaller than the number on the right. Comparing and Ordering Integers can be done by plotting them on the number line. For example: - If we have numbers -3, 0, 5,-5,-1 and 4 and we need Compare and Order Integers, we first plot them on the number line and see the integers to the left on the number line are smaller to those on the right. So reading the dots on the number line from left to right gives us the order from least to the greatest. Here we have to do is read the dots from left to right and we have our order. The smallest number is here on the extreme left will be -5 and then it’s -3, then -1, then 0, then 4 and then 5 which is on extreme right. So we can write the order as -5 < -3 < -1 < 0 < 4 < 5.

Numerator and Denominator

Numerator and Denominator Definition - In mathematics, when we study fractions, the numerator and denominator are very important words. In a fraction, the top part is known as the numerator and the bottom part is termed as the denominator.  To understand the concept better, let us draw a circle that is divided into four equal parts, we call them fourths where one of the parts has been shaded. Then we say that one fourth of the circle has been shaded and it is written as ¼, here 4 is the denominator and 1 is the numerator. In Fractions Numerator and Denominator are represented like this Numerator/Denominator.

How do we Define Numerator and Denominator- The denominator describes the number of parts, and the numerator describes the number of parts that are shaded.

How to Remember Numerator and Denominator - Where do these names come from and how do they make sense? Denominator means the name. It is the name for the fraction. Well there is a history to the word denominator, which helps to understand the connections. When we go to the church we might say it has a certain denominations, Luther is one denomination, catholic is another denomination. There is another use of word denomination, when people have bills, a dollar bill or five dollar bill, those are different denominations, or the name that they have.

In school election, we might have noticed, that we nominate somebody to be the president if require. It is a similar word nominate. Nominate means to name somebody to be the president. Or in Spanish, the word nombre that means the name, so nombre, nominate, denomination they all relate to the word denominator. They come from the same land root of denomino.

So denominator is just a name. Whatdoes numerator mean? The number of parts that we talking about. So we have drawn four equal parts, how many shaded parts is the number that represents a numerator. Let us understand it by using an example that is identifying the numerator and denominator in the fraction 3/4.

The numerator is the number on the top in fraction. Here we have ¾ where 3 is on the top so 3 is the numerator and the denominator is the number in the bottom of the fractions. It is 4 here. This fraction represents 3 out of 4 pieces of a pie. The denominator shows total number of pieces of pie and 3 is represented as the part of those 4 pieces.

Definition of Mean

It is also called as arithmetic average, or simply average. I is defined as the arithmetic average of the set of data value and is calculated by the sum of all the values from the data and divide by size of the data. Especially the term arithmetic average is used in mathematics and statics, since it help us to differentiate it from other methods such as harmonic and geometric average. It is most frequently used in mathematics and statistics fields such as economics, sociology and history.

It is generally sample average or population average. If the set of observation is a statistical sample, then that average is called as sample average. If the set of observation is a statistical population, then that mean is called as population average.  The sample average is denoted by the symbol ( ) which also called as x bar and the population average is denoted by the symbol of Greek letter (µ). Both the sample and population average is calculated by adding all the observations and divided by the number of observations.

Formula for Mean
Formula for Sample average
= (∑¦x)/n
Where,
- Sample average
Sx - Sum of all the sample observation
n - Number of sample observation
Formula for Population average formula
µ =  (∑¦x)/N
Where,
µ - Population average
SX - Sum of all population observation
N - Number of population observation


Define Mean
It is defined as it the average of numbers from given set of data and is calculated by just addition of all numbers from set of data and divides them by the size of the data. Simply it is a just average of the given numbers. Average is represented by the symbol .
  = (∑¦x)/n
Where,
- Mean
S - Symbol of summation
x - Values present in given set of data
n - Number of data or size of data or how many numbers present in set of data

Definition of Mean
It is an arithmetic average of set of a data values and is not to be a middle value or any one value from the set of data values. It is a central value of a set of numbers or data. Calculation of average is very simple. Averages is calculated by adding all the numbers and divide them by how many numbers there are, in other words, it is the sum of the numbers divided by the count of numbers.
For example to find the average for given set of data such as 6, 11, 7, 4. To calculate average first add all the numbers,
6 + 11 + 7 + 4 = 28
Then dived the resulting number by how many numbers are their in data set, here there are 4 numbers are their, so
28/4 = 7
The average value is 7.

Sample Mean Formula
Where Sn = x1, x2, x3,………xn random sample of size n, then the formula for sample average is,
= (∑¦x)/n
Where,
- Sample average
S - Symbol of summation
x - Values present in the sample set of data
n - Number of sample data

How to find Directional Derivatives?

In calculus, the directional derivative is a differentiable function which multivariate along a vector V with a point P. It represents nothing but the rate of change of a function which moves in the direction of V, through the point P. The directional derivatives are the generalization of the partial derivatives; where in one of the coordinate axis is always parallel in direction.

Mathematical Definition of Directional Derivatives
The directional derivative is defined as the rate of change of the scalar function f(x) = f(x1, x2, x3... xn) along the unit vector u = (u1, u2, u3….. un), in which the unit vector is defined by a limit in terms of h. Therefore the mathematical definition for the directional derivative can be written as,

 Du f(x) = lim h->0+ f(x+hu) – f(x) / h.

Derivation of Directional derivative

It is simple method to find the directional derivative formula which will be used to simplify the problems. Let us define a function that is made up of a single variable, such as g(z) which equals f(sum of x0 + az and y0 + bz). Here x0, y0, a, b are some numbers which are fixed. Now, z is the letter which does not represent a fixed number.  Then, by the actual definition, for a single variable function we know that, g’(z) equals the limit which tends to h->0 of g (sum of z and h) – g(z) whole divided by h.  Now applying the value of the derivative z as zero, we will get, g’(0) which equals limit which tends to h->0 of subtraction of g (h) and g(0) whole divided by h. Substituting the equation of g(z) mentioned earlier we will get g’(0) as lim h->0 of f(x0 + az, y0 + bz) – f(x0,y0) whole divided by h. Then this equation will be obviously equal to the Du f(x0, y0). Therefore we will finally come with a relationship of g’(0) which equals Du f(x0, y0). Now we can rewrite g(z) as g(z) = f(x, y) where x will be equal to x0 + az and y will be equal to y0 + bz. Then, applying the chain rule we will get the equation as,  g’(z) = dg/dz = Sum of (multiplication of df/dx and dx/dz ) and (multiplication of df/dy and dy/dz), which will be equal to sum of fx (x,y) a and  fy (x,y).After taking z=0, we will get x as x0 and y as y0 and applying these, we will get, g’(0) as sum of            fx (x0,y0) a and fy (x0,y0) b. Now equating this equation with the g’(z) equation of very first we derived, we will arrive with a formula as shown below:

Du f(x, y) = fx (x, y) a + fy (x, y) b.

This equation will provide a simple way of calculation than the previous limit based definition.
Directional derivatives are used in solving the matrix symmetrical valued functions; especially the second directional derivative and it plays a vital role in solving non linear type problems.

Algebra Help

Algebra is the first branch of math which students come across, that is a complete change from arithmetic and basic geometry that students are used to studying. Algebra introduces the concept of the unknown variable and teaches students how the unknown can be derived from known parameters. Making use of both numbers and alphabets, algebra is a whole new ball game and needs complete attention right from the beginning .

How do keep up with the subject from the day you start learning it? To begin with, algebra is not as difficult as many students claim it is. Generally, students who require a little more time and explanation to understand algebraic concepts, seldom get it and as a result, come to the conclusion that it is hopelessly difficult. Needless to say, this leaves them with precious little motivation to continue learning the subject.

One of the first steps to learning algebra is to ensure that you understand each concept or equation when it is taught. Too many students tend to make light of their lack of clarity, telling themselves they'll take the time to clear it later on. Instead, take a proactive approach and ask a lot of questions in class. You don't have to worry about sounding dumb since it's a new subject and everyone is just as clueless as you are. Moreover when students ask questions, it helps teachers gauge the level of understanding among the students and proceed accordingly.

Memorize any equations or formulas immediately. You will be using them a lot so it's good to familiarize yourself with them and learn when and how to use these formulas. The best way to commit them to memory is to write them down. Practice sessions are very important so try to make time for them on a daily basis. Include different types of questions and don't just stick with the ones your teacher provides you with. As you get more comfortable with algebra and learn the formulas, vary the difficulty level to challenge yourself and learn new methods. It will also ensure that you are prepared for any question during the exams.

If you find that things are going above your head, then get help with algebra straightaway. Irrespective of whom you go to for help, they will find it easier to start from the beginning rather than having to finish the whole lot two weeks prior to the exam. For minor doubts you can approach your teacher or a classmate to point you in the right direction. For more extensive help, use a tutoring service which provides good quality, experienced tutors who will explain the theory, clear any doubts, and provide practice material. Algebra help online can be found on a number of websites which feature detailed tutorials as well as one on one tutoring, as per students' convenience.

Surface area of an ellipse cross section

Area of ellipse:
Like a circle or any other closed figure, an ellipse is also a closed figure. Therefore it is possible to find its area. Area of an ellipse can be found using the values of a and b. But what are these a and b? That can be understood using the general equation of an ellipse.

General equation of an ellipse:
[(x-h)/a]^2 + [(y-k)/b]^2, where a is the semi major axis and b is the semi minor axis. (h,k) is the centre of the ellipse.

Area of ellipse formula:
From the above general equation we know that the semi major axis is a and the semi minor axis is b. The formula for the area of an ellipse can now be written as follows:
Area = A =  pi * ab. The semi major axis can also be termed as the distance between the centre and the vertex, which is ‘a’ here. The semi minor axis can also be terms as the distance between the centre and the co-vertex, which is ‘b’ here.  See the picture below:

Implicit equation of an ellipse:

Another method for finding the area of an ellipse is using its implicit equation. The general form of the implicit equation of an ellipse is:

Px^2 + Qxy + Ry^2 = 1

To calculate area of an ellipse if the equation of the ellipse is given in this form, we use the following formula:

Area = A = 2 pi/√(4PR – Q^2)

Let us now try to understand how to calculate the area of an ellipse using an example.

Example 1: 
Find the area of an ellipse with centre at (2,3), length of major axis being 6 cm and length of minor axis being 4 cm.

Solution:
 In this problem, major axis = 6, therefore semi major axis = a = 6/2 = 3 cm
Minor axis  = 4, therefore semi minor axis = b = 4/2 = 2 cm
Using the formula for area of ellipse,
A =pi* ab =  pi * 3 * 2 = 6 pi cm^2

Statistics Mode

Mode definition (math):
In statistics, a branch of mathematics, mode of a set of observations (also called data) is the particular observation that occurs most number of times. In other words, the observation with the maximum frequency is the mode of the data.

Finding mode:
The following steps are to be followed when calculating the mode of a set of observations of ungrouped data.
Step 1: Arrange the data in ascending order
Step 2: Look for the entry that occurs the most number of times.
Step 3: The number found in the step 2 above is your mode.

Mode Math examples:
To properly understand what is a mode in math or statistics, let  us try the following examples:

Example 1: The height of 10 students in centimeters is as follows, find the mode of the data.
145, 142, 143, 146, 144, 146, 143, 141, 140, 143
Solution:
Step 1: Arrange the data in ascending order. So now the numbers would look like this:
140, 141, 142, 143, 143, 143, 144, 145, 146, 146
Step 2: Look for the entry that occurs the most number of times. In the above data, the number 143 occurs thrice, whereas all the other numbers occur either once or twice.
Step 3: The number found in step 2 is the mode. Therefore for this problem, mode = 143 since it occurs most number of times.

Example 2: The lengths of screws manufactured by a firm in inches are as follows:
1.55, 1.50, 1.52, 1.53, 1.50, 1.52, 1.51, 1.54, 1.56
Find the mode of the data.
Solution:
Step 1: Arranging the data in ascending order we have:
1.50, 1.50, 1.51, 1.52, 1.52, 1.53, 1.54, 1.55, 1.56
Step 2: From the above arrangement we see that two numbers 1.50 and 1.52, both occur twice.
Step 3: Therefore this data has two modes: 1.50 and 1.52. Such a data is called bimodal.

Example 3: 
The ages of students in college algebra class is as follows:
25, 23, 24, 22, 23, 25, 22, 21, 27, 28, 20, 29
Find the modes.
Solution:
Step 1: Arrange the data in ascending order
20, 21, 22, 22, 23, 23, 24, 25, 25, 27, 28, 29
Step 2: There are three numbers, each occurring twice. They are 22, 23 and 25.
Step 3: These three numbers are therefore the modes. Such a data is called trimodal.
If there are more than three modes in a data set, it is called multimodal data.

Identifying conic sections

See the picture above. Consider a double cone as shown in the picture. Also consider a plane. Let this plane intersect the double cone. The way in which the plane intersects the double cone, would give rise to conic sections. As we can see in the picture the axis of the cone is vertical.

1. Conic sections circles:  If the intersecting plane is perfectly horizontal, that means that if the intersecting plane is perpendicular to the vertical axis, then the conic section thus produced would be a circle.

2. Conic sections ellipse: If the intersecting plane is inclined to the vertical at an angle that is between α to 90 degrees, where α is the angle between the vertical axis and the slant length of the cone, then the conic section thus produced would be an ellipse.

3. Conic sections parabola: If the intersecting plane is inclined to the vertical axis at an angle that is between 0 to α (α is same as described above), then the conic section produced would be a parabola.

4. Conic sections hyperbola: If the intersecting plane is parallel to the vertical axis, then the conic section thus produced would be a hyperbola.

Identifying conic sections:
Conic sections can be identified by two methods. They are, graphical and algebraic.
Graphical method of identifying conic sections:
The graphs of various conic sections are as shown below:

1. Graph of a circle:

The graph of a circle would look like above, it would have a centre and the radius on all sides would be equal.

2.Graph of an ellipse:

The graph of an ellipse can look like any of the two above figures. One is a horizontal ellipse and other is a vertical ellipse. An ellipse would have a major and a minor axis, two vertices and two co-vertices.

3.Graph of a parabola:

A parabolic curve would look like above. It would essentially have a vertex, a focus, a directrix and an axis of symmetry.

A hyperbola can be horizontal or vertical. Algebraic method for identifying conic sections:
In a conic sections practice problem, the equation of the curve should be similar to any of the following general equations:

Equation of the curve	Type of conic section
(x-h)^2 + (y-k)^2 = r^2	Circle
((x-h)/a)^2 + ((y-k)/b)^2 = 1	Ellipse
y-k = 4a(x-h)^2	Parabola
((x-h)/a)^2 - ((y-k)/b)^2 = 1	Hyperbola

Introduction to discrete math applications

Discrete mathematics is defined as the study of mathematical structures that are logically discrete rather than continuous. Discrete mathematics had a concept in topics in "continuous mathematics" such as calculus and analysis

The set of conepts studied in discrete mathematics can be finite or infinite. Some of the mathematical relations often considered i part of discrete mathematics are Boolean algebra, the mathematics of social choice, linear programming, and number theory

Discrete mathematics includes sets, functions and relations, matrix algebra, combinatory and finite probability, graph theory, finite differences and recurrence relations, logic, mathematical induction, and algorithmic thinking.

Applications of Discrete Mathemetics:

It can be applicable in various fields such as combinational analysis, functional system theory, codings, crptology,etc

It can also be used in graph theory, probabilistic problems of discrete mathemetics, and it has the applications of algorithms and their complexity and computational problems of number theory and of algebra

It can be also used in vital, exciting and useful information about mathemetics that will be very useful to taught the lower and higher grade

Problems by Using Discrete Math Applications:

1. By using discrete mathematics find the value of n.

(1). (n+ 1)! = 10 * 9(n!)

Solution:

(n+1)! = 10 * 9(n!)

here expanding the factorization that gives

(n + 1) n! = 10 * 9n!

now cancel n! on both the sides

we get n+1 = 90

n = 90 – 1 = 89

therefore n = 90 .

(2). 1/4! + 1/5! = n/6!

Solution:

1/4 + 1/5(4!) = n/6.5(4!)

now cancel 4! On both sides

1 + 1/5 = n/6.5

6/5 = n/30

n = (30 * 6)/5 = 180 /5 = 36

Therfore we get n = 36.

2. Find the following factorial question by using discrete mathemetic

Where n = 4 and r = 2 find the values of

(1). n! / r!

Solution:

n! / r! = 4! / 2! = 4.3.2.1 / 2.1 = 12.

Therfore n! / r! =12

(2). n! /r!(n- r )!

Solution:

n! / r!(n-r)! = 4! / 2!(4-2 )!

= 4.3.2.1 / 2!

= 4.3.2.1 / 2.1

therefore n! / r!(n-r)! =12.

Divergence theorem

Divergence theorem commonly known as divergence theorem of gauss. This theorem also help to integrate the functions, divergence theorem is a combination of vector and integral. To understand the theorem we have to know about divergence of vector field.
Vector calculus is very important in engineering and physics to the gradient, divergence and curl. Lets take vector V(x,y,z), be a differential vector function where x y z are Cartesian coordinate and if v1,v2,v3 are component of vector V then the function div.V is called divergence of vector field. The physical meaning of divergence is value of a function that characterize a physical or geometrical property must be independent of the particular choice of coordinate that means those values must be invariant with respect to coordinate transformation.
Now divergence-theorem in integrals, this theorem is used for triple integration. Divergence-theorem, which transforms surface integral to triple integral, it also involves the divergence of vector function. The triple integration is a generalization the double integration. triple integral can be transform in to surface integral, over the boundary surface of a region in space and conversely. This is a practical interest because one of the two kinds of integral is often simpler than other. It also helps in establishing fundamental equation in fluid flow, heat conduction etc. the transformation is done by gauss divergence-theorem which include the divergence of a vector field function.
Divergence-theorem of gauss means also transformation between volume integral and surface integral. Let constant T be a closed bounded region in space whose boundary is a piece wise smooth Orientale surface and let F(x,y,z) be a vector function that is continuous and continuous first partial derivatives in some domain containing T. then we proof the divergence-theorem of gauss. In this theorem a unit normal vector of surface which is pointing to the out side of surface can also used.
Let’s clear this theorem with some Divergence Theorem Examples. Suppose we have to evaluate the surface integral by this divergence-theorem. In our problem surface S is a closed surface consisting cylindrical coordinates and also coordinates of circular disc are given.  To solve this problem we need to know triple integration. Then we can solve this type of problem, when we solve this types of problem we mostly use polar coordinates, which is defined by X=rcosθ and y = rsin θ
Let’s take one more Divergence Theorem Example that is transformation of volume integral in to surface integral. Suppose function which is in vector form is given and function also involve unit normal vector. Here in place of surface volume of sphere is given, now by using gauss divergence-theorem we integrate the function from volume integral to surface integral.

Introduction to algebra test questions

Algebra is one of a part of arithmetic that uses letters in place of some unknown numbers. It is concerning about learn the rules of operations and dealings, and the constructions. It also was including the conditions, polynomial expressions, equations and algebraic arrangement. In other words, we use numerals like 1, 2, 3, etc to represent numbers. In algebra we use numbers as well as letters of the alphabet such as a, b, c, etc for any numerical values we choose. In this article we shall discuss about algebra test questions.First we will solve some sample questions for algebra test and then do some practice questions.

Sample Questions for Algebra Test :

Ex 1: Solve the linear questions

19x - 7 = 21x – 9

Solution: Subtract 21 x from both sides of the above equation, we get the below term
            -2x – 7 = -9
Add 7 to both sides of the above equation, we get the below term

-2x – 7 + 7 = - 9 + 7

-2x = - 2

Divide both sides by -2 of the equation, we get  

        X = 1

The answer is x = 1

Test the answer for the given equation. Replace with in the above given equation for x. If the left side of the equation sum value is equal to the right side of the equation sum value after the replacement of x value, you have got the exact answer.

Left side of the equation:

 19(1) - 7 = 12

Right side of the equation:

 21(1) - 9 = 12

Ex 2: Solve the given Questions

12− (3x + 7) = 2x

Solution: In the given equation is expanding and the terms are out the bracket.

12 − (3x + 7) = 2x

12 −3 x − 7 = 2x

(12 – 7) − 3x = 2x

5 – 3x = 2x

Now we are identify that it is easier to get all the x value on the right side of the given equation, by adding x to both sides of the above equation.

5 = 5x

Now we divide both sides by 5 and swap the sides:

We get the x value is 1

Test the answer for the given equation. Replace with in the above given equation for x.

We get

12 – 3(1) + 7 = 2(1)

12 – 3 - 7 = 2

Practice Questions for Algebra Test :

Solve 17 x - 6 = 23 x – 18
                    Ans: x = 2

Solve 13− (5x + 9) = -3x
                    Ans: x = 2

Intermediate algebra test

Intermediate algebra is mainly used for solving the equations. Intermediate algebra mainly uses the simplification technique for solving the equations. Intermediate algebra uses the letters and variables. Intermediate algebra also deals with the expressions and exponents of polynomials. Algebra test problems are very simple and easy. Intermediate algebra test covers the topics,

• linear Equations
• Equation of a line
• Quadratic Equations
• Polynomial Functions
• Slope formula

Intermediate Algebra Test Problems

Algebra test Problem: 1

Solve the equation x + y = 6, x - 3y = 2.

A) x = 2, y = 4 B) x = 5, y = 2 C) x = 4, y = 2 D) x = 5, y = 1.

Answer: D

Algebra test Problem: 2

Find the slope of the equation 4x + 3y = 21

A) M = - 4 / 5 B) M = - 4 / 3 C) M = - 3 / 4 D) M = - 3 / 5.

Answer: B

Algebra test Problem: 3

Factorize the equation 2x2 - 5x + 3.

A) X = 2, X = 4 B) X = - 2, X = - 3 C) X = 1, X = 3 / 2 D) X = 1, X = 2 / 3.

Answer: C

Algebra test Problem: 4

Find the vertex of the equation Y = x2 + 4x + 7

A) (2, 4) B) ( -2, 3) C) (-3, 2) D) (1,-1).

Answer: B

Algebra test Problem: 5

Find the distance between the two points (1, 5) and (3, 9).

A) d = 6 B) d = 3 C) d = 7 D) d = 5.

Answer: D

Problems for Intermediate Algebra Test

Algebra test Problem: 6

 Find the value of x for the expression: 4x + (4 - 7x) = (2x + 3) - 4.

A) 3 B) 2 C) 8 D) 1

Answer: D

Algebra test Problem: 7

Add the two expressions (3x + 2y - 8) and (5x - 3y -4)

A) 5x + 4y - 12 B) 8x + y + 12 C) 8x - y - 12 D) 2x - y - 4.

Answer: C

Algebra test Problem: 8

The sum of the number and 32 is 67. Find the unknown number.

A) 32 B) 33 C) 31 D) 35

Answer: D

Algebra test Problem: 9

Multiply (3a2 + 4b) (2a + 6b)

A) 6a3 + 8ab + 18a2b + 24b2

B) 3a3 + 8ab + 18a2b + 24b2

C) 4a3 + 6ab + 16a2b + 24b2

D) 12a3 + 9ab +18a2b + 24b2

Answer: A

Solving Trigonometric Equations made easy

The six trigonometric functions are the sine, cos, tan, csc, sec and cot. An equation consisting of one or more trigonometric functions is called a trigonometric equation, for instance cot^2(x) + 1 = csc^2(x) is a trigonometric equation.  Some of Trigonometric Equations Examples are 2 cos^2(x) –sin(x) – 1 =0, 2sin(theta) – 1 = 0, cos(theta/2) = 1 + cos(theta), 6 sin^2(theta) – sin(theta) = 1, cot(2 theta) – tan(2 theta) = 0

Quadratic Trigonometric Equations
Trigonometric equations that involve trigonometric functions as products are called the quadratic trigonometric functions. In a quadratic equation with degree 2 written in the form of  ax^2 + bx + c. So, a quadratic trigonometric equation will consist of trigonometric functions with degree 2. For instance, 2 sin^2(x) – sin(x) +1 = 0 is an example for a quadratic trigonometric equation.

Let us now learn the steps involved in simplifying trigonometric equations. We solve trigonometric equations as we solve any other algebraic equation except that we use the trigonometric identities in solving. They are also solved by the sign of the trigonometric value by determining the quadrant(s) for that particular trigonometric value. Let us simplify a trigonometric equation, 2 sin(pi/2 -alpha) =1. We shall use the identity sin (pi/2 – alpha) = cos(alpha) in solving the given equation. So, the trigonometric equation reduces to, cos(alpha)=1/2 . Now we need to check for what angle of cosine alpha gives the value ½. We know that cos(pi/3) equals ½. So, we write it as cos(+/-) pi/3 =1/2. The positive and negative sign depends  on which quadrant the angle alpha lies and hence we can generalize the solution as alpha = (+/-)pi/3 + 2k pi where k is any integer. There is a significance for adding 2k pi to the solution, as per the periodic behavior of the function cosine. So, the solution set is written as {alpha = (+/-)pi/3 + 2k pi: k belongs to set of integers}

Sin(x) + cos(x) = 0
Simplifying the above equation we get, sin(x) = -cos(x). Here, the value of sine and cosine have to be equal with opposite sign. Let us write sine in terms of cosine using trigonometric identity, which gives us
  cos(pi/2 – x+2 k pi) = cos(pi + x)
      pi/2 – x + 2 k pi = pi + x

The general solution would be, x = -pi/4 + k pi where k belongs to set of integers
Now, in the interval [0, 2pi) let us consider k = 1 and k = 2, we get, x = (3/4) pi and x = (7/4) pi as the solution that is at 135 degrees and 315 degrees the value of sine and cosine are equal and opposite.

Steps for solving standard deviation problems

A standard deviation example would typically have a set of observations x_1, x_2, x_3,  …… x_n. The number of observations being n. In general it can be written like this: x_i, where i = 1, 2, 3, … n. To calculate the standard deviation of such a problem the following steps are followed:

Step 1: List all the observations in the first column of a table. The title of column would be x_i and the observations would be x_1, x_2, x_3,  …… x_n.

Step 2: Find the mean of the observations. If we denote the mean by x ¯, then the formula for calculating the sample mean would be:  x ¯ = (x_1+ x_2 + x_3+  ……+ x_n)/n = [∑x_i]/n

Step 3: We now come to the second column. In this column we find the deviation of each of the observation from the mean. For that we calculate the value of x_i - x ¯ for each of the observations.
Note that in this step it is possible that we get negative values. We shall take care of the negatives in the next step.

Step 4: Now we construct the third column. In this column we square the results of column (3) for each of the observations. That means we find the value of (x_i - x ¯)^2 for each of the n observations. Note here that since we squared the difference, the negatives are got rid of so we need not worry about the negative numbers canceling out the positive numbers and affecting our end result.

Step 5: Now we find the standard deviation for each of the observation. For that we compute the following value: (x_i - x ¯)^2 / n. That would be placed in our fourth column.

Step 6: Adding standard deviations for each of the observations we get the value of the expression : ∑((x_i - x ¯)^2 / n). That will give us our sample standard deviation formula. The symbol for standard deviation is the Greek small alphabet ‘s’ pronounced as ‘sigma’. Therefore we say that:
s = sqrt[∑((x_i - x ¯)^2 / n)] or s = sqrt [∑(x_i - x ¯)^2/n]

Inverse Functions and their derivatives in a nutshell

In Math, Inverse Function is a function whose relation to a given function is such that their composite is the identity function. Inverse Function can also be defined as the function obtained by expressing the dependent variable of one function as the independent variable of another; for instance f and g are inverse-functions if f(x)=y and f(y)=x. It is denoted by (-1) in the power, inverse of f(x) is f-1(x).

One important point we need to remember is that the inverse of a function may not always be a function. While finding Inverse of Function of a function y in terms of x, we just switch the x and y and then solve for y. The new function [y-1] we get is the inverse-function of the given original function. For instance, finding Inverse of a Function f(x)= (x/3) -1 involves a list of steps. First let us take f(x)=y=(x/3) -1. In the next step we switch the x and y, that gives us, x = (y/3)-1. The last step would be to solve for y, finally we get y= y-1= 3x+3 which is the inverse-function of f(x) = (x/3) -1

There are four main steps involved in Solving Inverse Functions. Given a function f(x) in terms of x,
1. First step, we write the given function f(x) equal to y
2. Second step involves interchanging the x and y
3. In the third step, we solve for y
4. Fourth step, we write y as f-1(x) which is the inverse-function of f(x)

Considering an example, let us now learn how to solve Inverse functions using the above steps. Given the function f(x) = 2x-3, let us solve inverse-function [f-1(x)]. The first step would be to write y = 2x-3. Next we interchange x and y which gives, x = 2y-3. Now, we solve for y; y = (x+3)/2. The last step is to put f-1(x) in place of y which gives us f-1(x)= (x+3)/2  which is the inverse-function of the given function.

Derivative of an Inverse Function, if for a function f an inverse-function f-1 exists and if f is differentiable at f-1(x) and f’[f-1(x)] is not equal to zero, then f-1 is differentiable at x and the formula is given as:

Considering an example, let us now learn how to solve Inverse functions using the above steps. Given the function f(x) = 2x-3, let us solve inverse-function [f-1(x)]. The first step would be to write y = 2x-3. Next we interchange x and y which gives, x = 2y-3. Now, we solve for y; y = (x+3)/2. The last step is to put f-1(x) in place of y which gives us f-1(x)= (x+3)/2  which is the inverse-function of the given function.

Derivative of an Inverse Function, if for a function f an inverse-function f-1 exists and if f is differentiable at f-1(x) and f’[f-1(x)] is not equal to zero, then f-1 is differentiable at x and the formula is given as:

A square root function is the inverse-function of a square function f(x) = x2. So, the derivative of an Inverse Function of square function f(x) = x2 can be solved using the power rule

An Introduction to Quartiles in Statistics

Quartile is a term derived from the word ‘quarter’ which is one fourth of something. So, we can say  Quartiles Statistics is a certain fourth of a data set. When a given data is arranged in the ascending order that is from the lowest value to the highest value and this data is divided into groups of four, we get what we call Statistics Quartiles.

In Quartiles Statistics there are three quartiles which are, the first quartile also called as lower quartile is denoted as Q1 which lies in the twenty five percent of the bottom data. The second quartile is the median of the data that divides the data in the middle and has fifty percent of the data below it and the other fifty percent of the data above it. It is denoted as Q2.The third quartile also called the upper quartile denoted as Q3 has seventy five percent of the data below it and twenty five percent of the data above it.

We can state the Quartiles Definition as, Quartiles are the three values that divide an ordered data set into four approximately equal parts. Definition of Quartiles can also be given as, A statistical term describing a division of a data set into four defined intervals based upon the values of the data and how they compare the entire data set.

Now that we are a bit familiar with Quartiles Statistics, let us learn how to calculate Quartiles of a given data set. For a better understanding of the steps involved let us consider a data set. Calculate the three
 quartiles of the data set of scores,  27 18 20 20 23 29 24.

The steps involved are:
• Count the total number of scores in the given data; n= 7
• Arrange the data set of scores in the ascending order, that is, from the lowest score to the highest score, we get, 18 20, 20, 23, 24, 27, 29
• Find the median (Q2) which is the middle value of the data set
Median = ½(n+1) = ½(7+1)= 4
So, the median (Q2)=23 [the fourth term of the ordered data set]
• Next we shall calculate the lower quartile (Q1) which is the median of the lower half of the data set.    Q1 = ¼(n+1) = ¼(7+1) = 2
So, the lower quartile (Q1) = 20 [the second term of the ordered data set]
• Finally we shall calculate the upper quartile (Q3) which is the median of the upper half of the data set. Q3=3/4(n+1) = ¾(7+1) = 6
So, the upper quartile (Q3)= 27 [the sixth term of the ordered data set]

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Derivatives of Cot functions for some special values

We can find the derivative of the cot function using the quotient rule of differentiation. The quotient rule can be states as follows: For a function f such that f(x) = u/v, where u and v are both function of x, the derivative of f(x) is given by the formula:
f’(x) = (v*u’ – u*v’)/v^2, where u’ is derivative of function u and v’ is derivative of function v.
We know that, cot x = cos x / sin x,
Therefore, (d/dx) cot x = (d/dx) (cos x / sin x),
= [sin x (d/dx) (cos x) – cos x (d/dx) (sin x)]/sin^2 (x)  … using the quotient rule stated above.
=[ sin x (-sin x) – cos x * cos x]/sin^2(x)
= [-sin^2(x) – cos^2(x)]/sin^2(x)
= -[sin^2(x) + cos^2(x)]/sin^2(x)
= -1/sin^2(x)
= -csc^2(x)
Therefore the derivative of cot x is –csc^2(x)

The above derivative can also be proved using the definition of derivative which is like this:
F’(x) = lim(t->x) [(f(t) – f(x)]/(t-x), for the function y = f(x) = cot (x) the definition would be like this:
(d/dx) cot x = lim(t->x) [cot t – cot x]/(t-x)
= lim(t->x) [1/tan t – 1/tan x]/(t-x)
  = lim(t->x) [tan x – tan t]/[(tan t * tan x)(t-x)]
= lim(t->x) {[tan x – tan t]/(t-x)} * 1/[tan t * tan x]
= lim(t->x) {-[tan t – tan x]/(t-x)} * lim(t->x) 1/[tan t * tan x]
= lim(t->x) {-[ tan(t-x) (1+tan t tan x)]/(t-x)} * 1/tan^2(x)
We know that lim(t->x) of tan(t-x)/(t-x) = 1. Therefore,
= -{1+tan^2(x)} *1/tan^2(x)
= - sec^2(x)/tan^2(x)
= [-1/cos^2(x)]/[sin^2(x)/cos^2(x)] … here the cos^2(x) cancel each other out.
= -1/sin^2(x)
= -csc^2(x)

Derivative of cot-x:
For finding the derivative of cot(-x) we use the chain rule.
Let –x = u, then du/dx = -1.
Cot (-x) = cot u
(d/du) cot u = - csc^2(u) …  from what we proved above.
Using the chain rule we have:
(d/dx) cot (-x) = (d/dx) cot u * du/dx = -csc^2(u) * (-1)
= -csc^2(-x) * (-1)
= csc^2(-x)
Thus we see that the derivative of cot(-x) is csc^2(-x)

Derivative of cot^-x (k):
Cot^(-x) (k) where k is some constant can be written like this also
= (cot(k))^(-x)
= 1/(cot k)^x
= (1/cot k)^x
Since k is a constant, cot k is also a constant and in turn, 1/cot k is also constant. So let 1/cot k = some constant say p. Therefore,
= p^x is the function.
(d/dx) p^x = p^x ln p
= (1/cot k)^x ln(1/cot k)
That is the derivative of the above function.

Triple integrals spherical co-ordinates

The concept of definite integration can be extended to two or three or multi dimensional system. For a bounded function f(x,y,z) defined on a rectangular box B (x0 = x = x1, y0 = y = y1, z0 = z = z1), the triple integral of f over B,
Integral (B) [f(x,y,z)] dV or  Integral  (B) [f(x,y,z)] dx dy dz,
can be defined as a suitable limit of Riemann sums corresponding to partitions of B into sub boxes by planes parallel to each of the co-ordinate planes. Triple integrals over more general domains are defined by extending the functions to be zero outside the domain and integrating over a rectangular box containing the domain.
All properties of double integrals hold good for triple integrals as well. In particular, a continuous function is integrable over a closed, bounded domain. If f(x,y,z) = 1 on the domain D, then the triple integral gives the volume of D.
Volume of D =  Integral (B)  dV
Triple integrals spherical coordinates: The spherical co-ordinates related to Cartesian co-ordinates x, y and z are given by the equations:

The triple integrals in spherical coordinates relationships are illustrated in the picture below

For the co-ordinate surfaces in the spherical co-ordinates, see picture below:

The volume in spherical co ordinates would be like this:

Spherical co ordinates are suited to problems involving  spherical symmetry and, in particular, to regions bounded by spheres centered at origin, circular cones with axes along the z axis, and vertical plains containing the z axis.

Infinite and removable discontinuity

Introduction To Discontinuity : Let us first define discontinuity  , a function f(x) is said to be continuous in an interval if it is continuous at every point in the domain of f(x). If the interval is closed, say [a, b], the function f(x) is continuous at every point of the interval including the end points a and b. If the interval is open, say ]a, b[, we exclude the end points of the interval. If the function is not continuous at a point, say c, we say that the function is discontinuous at c and c is called a point of discontinuity of f(x) .The graph of the function y = f(x)= 5 for x > 0 and 1 for x = 0, when drawn, we see that the function is discontinuous as there is a jump in the graph at x = 0. If we carefully observe that the function is continuous to the left of O and also to the right of O.

In other words, we say that left hand limit is existing and also the right hand limit is existing but they are not equal. ).Now let us more understand what is discontinuity? The function f is said to be discontinuous at a point a of its domain D if it is not continuous thereat. The point a is then called the point of discontinuity of the function. The discontinuity may arise due to any of the following situations: (i) lim x -> a+ f(x) or lim x -> a- f(x) of both may not exist. (ii) lim x -> a+ f(x) as well as lim x -> a- f(x) may exist, but are unequal. (iii) lim x -> a+ f(x) as well as lim x -> a- f(x) both may exist, but either of the two or both may not be equal to f(a).

We classify the points of discontinuity according to the various situations discussed above:
1. Removable discontinuity: Now let us understand What is a Removable Discontinuity ? A function f is said to have removable discontinuity at x = a if lim x -> a- f(x) = lim x -> a+ f(x) but their common value is not equal to f(a). Such a discontinuity can be removed by assigning a suitable value to the function f at x = a.

2. Discontinuity of the first kind: What is infinite discontinuity ? A function f is said to have a discontinuity of the first kind at x = a if lim x -> a- f(x) and lim x -> a+ f(x) both exist but are not equal. f is said to have a discontinuity of the first kind from the left at x = a if lim x -> a- f(x) exists but not equal to f(a). Discontinuity of the first kind from the right is similarly defined.

3. Discontinuity of the second kind: A function f is said to have a discontinuity of the second kind at x = a if neither lim x -> a- f(x) nor lim x-> a+ f(x) exists. F is said to have discontinuity of the second kind from the left at x = a if lim x -> a- f(x) does not exist. Similarly, if lim x -> a+ f(x) does not exist, then f is said to have discontinuity of the second kind from the right at x = a.

Know more about Free calculus help.
Let I be an open interval let c belongs to I, and let f be a function defined on I except possibly at the point c. The function f has a jump discontinuity at c if the one sided limits of f exists at c but are not equal. The function f is said to have a removable discontinuity at c if lim x -> c f(x) exists, but f(x) is neither not defined or has a value different from the limit. As a specific example, the function g defined by g(x) = x/|x| has a jump discontinuity at 0 since the left hand limit is -1 and the right hand limit is 1. A removable discontinuity can be removed by giving the function the value of the limit at the given point. The function h defined by h(x) = x sin (1/x) has a removable discontinuity at 0; assigning h(0) = 0 makes h a continuous function at 0.