Square root of a number is that number when multiplied by itself yields the original number. For example 7 multiplied by 7 yields 49. We say square root of 49 is 7 and this mathematically written as`sqrt(49)` = 7.
Similarly we have higher roots like cube root, fourth root etc. Cube root of a number is that number when multiplied by itself thrice yields the original number. Example 2 x 2 x2 =8 so we say cube root of 8 is 3 and this is represented as `root(3)(8)` = 2.
In general form we have the nth root of n umber represented as `root(n)(x)` . If n=2 , then it is called square root and is normally not written. This means that square root of 10 is written as `sqrt(10)` and not as `root(2)(10)` . If the value of n is not written, then it is assumed to be 2 , the square root.

In this lesson we will see the different methods to find the square root of a number.

Steps to find the square root using averaging and continuous divison method

 

There are many methods to find the square root of a number. The most commonly used are

  • Averaging and continuous division method
  • Long Divison method 

Step by Step prcodeure to find the square root using averaging and continuous method:

Step 1: Divide the given value by the nearest square value, in which the square value should be lesser than the given value

Step 2: Take the average for the calculated result and the divisor which is used.

Step 3: Now divide using the given value and the calculated average value.

Step 4: once again take the average for the calculated above result and the divisor.

Step 5: Now square the above calculated value. The squared value should be equal to given value of the problem, if not repeat the steps 3 and 4.

Example

Find the square root of 2265

Solution:

The square root value for 2265 is nearly equal value to the square values between  47 and 48 , because 472 = 2209 and 48 2 = 2304

 

Step 1: Divide 2265  by 47.

` 2265 / 47` = 48.1914894

Step 2: Take average for   48.1914894 and 47.

`(48.1914894 + 47 )/2` =  47.5957447

Step 3: Divide 2265 by  47.5957447

`2265 / 47.5957447 ` =    47.5882879

 

Step 4: Take average for the     47.5882879  and 47.5957447

`( 47.5882879 + 47.5957447 )/2`  =    47.5920163

Step 5: check the result by taking square for the calculated value.

47.59201632 =  2265.00002

 

This value that is approximately equal to the given value of square root  `sqrt 2265`

If the value is not equal to the given value then repeat the steps 4 and 5.

Step 5: One value should be added to the above calculated result it is a divisor, that value should be less than the remainder value.

Step 6: If the requested result is not obtained repeat the steps 2, 3, 4 and 5

Steps to find the square root using long divison method

Steps for finding the square root using long division method:

Step 1: Find the square value that are related to the first value or first two values of the given problem. Approximately equal or exactly equal of the given value. That is dividend value for that divisor

Step 2: Multiply the value by the value itself and then subtract the selected value and the multiplied value.

Step 3: Now here add two values to the resulted value . If not add two zeros place a decimal place in quotient

Step 4: Now multiply the resulted value with 2. This is the dividend value for the divisor value

Example :  Find the  square root of 2265

Solution:

Long division method.

                                47.59201613
                           4 |  2265
                              |  16
                       87  |    665
                              |    609
                     945  |      5600
                              |      4725
                    9509 |        87500
|        85581
                  95182 |          191900
                              |          190364
             9518401 |              15360000
                              |                9518401
           95184026 |                 584159900
                              |                 571104156
         951840321 |                     1305574400
                              |                     951840321
      9518403223 |                       35373407900
                              |                       28555209669

|                         6818198231

The above resulted value is approximately equal to the given square root value, So the answer is  47.59201613

Steps to find square root – Practice Problems:

 

Problem 1:

Solve the square root of 3434

Answer: 58.6003413

Problem 2:

Solve the square root of 5654

Answer:   75.1930848

Let A be a set. Then the relation IA = {(aa) : a belong to A} on A is called the identity relation on A.

In other words, a relation IA on A is called the identity relation if every element of A is related to itself only.

Example:- the relation IA= {(1, 1), (2, 2), (3, 3)} is the identity relation on set A = {1, 2, 3}. But relations R1 = {(1, 1), (2, 2)} and R2 = {(1, 1), (2, 2), (3, 3), (1, 3)} are not identity relations on A, because (3, 3) does not belong to R1 and in R2 element 1 is related to elements 1 and 3.

 

Definition Of Identity Relations

 

Let A be a set. The relation IA defined by IA = {xy) : x belong to Ay belong to Ax = y} is called the identity relation in A.

Thus the identity relation in a set A is the set of the ordered pairs (xy) of A × A for which x = y. if A = {1, 2, 3, 4, 5}, then

IA = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}

The identity relation on a non-void set A is always reflexive relation on A.

The identity relation on a set A is an antisymmetric relation.

The identity relation on a finite set A is the smallest equivalence relation on A.

Identity Relations- Problem

 

Show that the identity relation on a set S is an equivalence relation.

Solution. Let be the identity relation on a set S

Then R = {(aa) : a belong to S}. to prove the that Ris an equivalence relation in S.

R is reflexive. Let a be any element of S. then by definition of R, we have (aa) belong to R → a R a

Thus a R a ”or” a belong to S. therefore R is reflexive

R is symmetric. Let ab belong to S be such that a R b. then (ab) belong to R.

Therefore we must have a = b and thus a R b → b R a. So R is symmetric

R is transitive. Let a, b, c belong toS be such that a R b and b R c. Then (ab) belong to R and (bc) belong to R. Therefore  we must have a = b and b = c. From this we get a = c and so we have (ac) belong to R i.e., we have a R c. thus a R b ad b R c → aR c. Therefore R is transitive.

Hence is an equivalence relation on S.

Scientific notation, also known as standard form or as exponential notation, is a way of writing numbers that accommodates values too large or small to be conveniently written in standard decimal notation. Scientific notation has a number of useful properties and is often favored by scientists, and mathematicians,

In scientific notation all numbers are written like this:   a × 10b                                                         (source: Wikipedia)

Rules in scientific notation:

Step 1: The coefficient should be bigger than or equivalent to 1 and less than ten,

Step 2: The base should be ten

Step 3: The exponent must explain the number of decimal place that the decimal need to be moved to modify the number to standard notation.

Step 4:  A negative exponent means so as to the decimal is moved to the left when varying to standard notation.

To write a number in rules in scientific notation: 

Standard scientific notation places the decimal following the first digit and drops the zeroes.

9.3300000000000

In the number 93,300,000,000,000 the coefficient will be 933

To find the exponent scientific notation calculate the number of places from the decimal to the last part of the number.

In 93,300,000,000,000 there are 11 places.

Therefore we write 93,300,000,000,000 as:

9.33×1013

Example:   58800000

5.88×107

Multiplication problems – rules in scientific notation: 

    Example 1: (2 x 104) (7x 105) = 14 x 109 = 1.4 `xx 10^10`

    Example 2: (8 x 10 3) (9x 105) = 72 x 108  = 7.2`xx10^9`

    Example 3: (2.3 x 10 5) (9.2x 10 6)

= 21.16 x 10 11

= 2.116 x 10 12

Division problems - rules in scientific notation:

    Example 1: (25x 106) / (5 x 103)

                  = 5 x 103

    Example 2: (3 x 10 8) / (7 x 103)

= 0.42 x 105

Shift the decimal point in excess of the right until the coefficient lies between 1 and 10. For each place shift the decimal over the exponent will be lowered 1 power of ten.

0.42×10 5= 4.2 x 10in scientific notation.

     Example 3: `(1xx 10^6) / (15 xx 10^3)`

     = 0.066 x 10-1

Shift the decimal point in excess of the right until the coefficient lies between 1 and 10. For each place shift the decimal over the exponent will be lowered 1 power of ten.

Practice problems – rules in scientific notation:

Problem 1: (2 x 104) (8x 106) = 16 x 1010

Problem 2: (2 x 104) (9x 105) = 18 x 109 

Problem 3: (2 x 103) (11x 105) = 22 x 108 

Problem 4: (2 x 102) (23x 106) = 46 x 108

 Prime factor is that a factor of number are prime and the factors are the divisors of number. The prime factors will be positive integers. The process of finding a prime factor is known as prime factorization. Now let us study what is prime factor.

Prime factor: The prime number of integers that of prime numbers are prime factors. The prime numbers are always combined to a original integer. The prime factor by exponent P is a large exponent of  “P” is “a” and it divides the original number.

Prime factorization: finding the prime numbers of divisors is called prime factorization. For example, take a number 8 that is it becomes 2 x 4, here 2 is smallest prime and there is 4 that is composite number so find the factors of 4 that is it becomes 2 x 2, now we can say that factors of 8 are 2 x 2 x 2. The prime factors are also represented as powers that is 23.

Prime factor problems:

Example problems:

Problem 1: Find out the prime factors of number 8.

Solution:

The given number is 8.

Factors of number 8 are 2 x 4.

The number 4 is prime and the prime factors of 4 are 2 x 2.

Therefore, the prime factors of 8 are 2 x 2 x 2 (or) 23.

Problem 2: Find the prime factors of 100.

Solution :

Here lowest prime is 2

100 / 2 = 50

50 is a complex number so it is further divided by 2

50 / 2 = 25

25 is a complex number so it is further divided by 5

25 / 5 = 5 and here 5 is a prime number.

Hence the Prime factors of 100 = 22 * 52.

Practice problems:

1. Find out the prime factors of number 36.

Solution: The prime factors are 2 x 2 x 3 x 3.

2. Find out the prime factors of number 81.

Solution: The prime factors are 34

In differentiation natural is represented by “ln”. ln = log e. Logarithmic differentiation is possible only when the base of the log is e. Otherwise we need to apply the change of base rule before going for differentiation.

Example log 10 x = log e (x )/ log e 10 = 1 / log e 10 `xx` lnx.

Now it can be differentitated.

The formula we need to remember are as follows:

(i) `d / dx` (log x) = `1 / x`

(ii) log (ab) = log a + log b.

(iii) log `(a / b)` = log a – log b.

(iv) log (a m) = m log a.

Here log means, we consider it as ln. That is log with base e.

Now let us see few problems on this topic differentiate natural log.

Example problems on differentitate natural log.

 

Ex 1: Differentiate: y = log x 2

Soln: Given: y = log x 2 = 2 log x

Therefore `dy / dx` = `d / dx` (2 log x) = 2 `d / dx` (log x) = `2 / x` .

Ex 2: Differentiate: y = log [(x + 2) (x + 3)]

Soln: Given: y = log [(x + 2) (x + 3)]

Therefore y = log (x + 2) + log (x + 3)

Therefore `dy / dx ` =` d / dx` [log (x + 2) + log (x + 3)]

`dy / dx ` = `d / dx` (log (x + 2)) + `d / dx` (log (x + 3))

= `1 /( x + 2)` + `1 /( x + 3)` =` (x + 3 + x + 2) /[ (x + 2) (x + 3)]`

=` (2x + 5) / (x ^2 + 5x + 6)`

Ex 3: Differentiate y =` log [(x + 3) /( x + 4)]`

Soln: Given: y = `log [(x + 3) / (x + 4)]`

`=>` y = log (x + 3) – log (x + 4)

Therefore `dy / dx ` = `d / dx ` (log (x + 3)) – `d / dx` log (x + 4)

= `1 / (x + 3) ** 1 /( x + 4)` = `[x + 4 ** (x + 3)] /[ (x + 3) (x + 4)]`

=`[ x + 4 ** x ** 3] /[ x ^2 + 7x + 12 ]` = `[2x + 1] / [x ^2 + 7x + 12]`

 

More example problems on differentitate natural log.

 

Ex 4: Differentiate y = log (e x)

Soln: Given: y = log (e x)

y = x . log e [Here log e ^e = 1]

`=>` y = x

Therefore `dy / dx ` = 1  [Since` d / dx` (x) = 1]

Ex 5: Differentiate y = x x

Soln: Since the variable in the power, we can apply log to the base e before going for differentiation.

Soln: Given: y = x x

Therefore log y = log x x = x . log x

Therefore on diff, `1 / y` . `dy / dx ` = x . `d / dx` (log x) + log x . `d / dx` (x)

`dy / dx` = y [x . `1 / x` + log x (1)] = y [1 + log x]

Therefore `dy / dx` = x x [ 1 + log x]

Different hypothesis tests create dissimilar assumptions regarding the distribution of the random variable being tested within the information. Hypothesis testing is the use of statistics to found the probability that a specified hypothesis is correct. Hypothesis is specified as declaration which may or may not be accurate. In statistics two hypothesis testing are used. They are null hypothesis and alternative hypothesis. These two hypothesis testing are opposed to every other. In statistics the significance level is symbolized through alpha. Let us see about the probability of hypothesis testing normal distribution.

Hypothesis testing normal distribution

Hypothesis testing consists of following steps:

  • Null hypothesis
  • Alternative hypothesis
  • Test statistic
  • Level of significance
  • Inference

Hypothesis testing normal distribution

  • Assume the z-test and the t-test.  These tests consider the data are separately tested from the normal distribution.
  • These tests are comparatively robust with respect to departure from this declaration, as a result long as the sample size n is large enough.
  • Both tests estimate a sample mean , which has been approximately normal sampling distribution among mean equivalent toward the population mean μ, in any case of the population distribution being tested.
  • σ is recognized for the z-test. The t-test must compute an approximation s of the standard deviation as of the test.

t- test:

`t = (barx-mu)/(s/sqrt(n))` ~ t(n-1)

z- test:

`z = (barx-mu)/(sigma/sqrt(n))` ~ N(0,1)

  • The null hypothesis to the population is distributed with mean μ, the z-statistic contain a standard normal distribution, N(0,1).

Examples for hypothesis testing normal distribution

A sample of 600 items is taken from a population whose mean is 25 and standard deviation is 15. The mean of sample come from the population with mean 26.8.

Solution

Null hypothesis:

H0: The given sample is taken from the population with mean 26.8

Alternative hypothesis:

H1: The given sample is not taken from the population with mean 26.8

Test statistic:

`z = (barx-mu)/(sigma/sqrt(n))` ~ N(0,1)

`barx`- mean of the sample =25

μ – population mean 26.8

σ is unknown we can use the unbiased estimator.

s = 15, n is the number of items.

`z = |(25-26.8)/(15/sqrt(600))|` ~ N(0, 1)

`z = |(-1.8)/(15/24.49)|`

z = 2.938

Level of significance:

At 5% level for the 25 degrees of freedom ‘z’ table value of 1.96

Therefore Zα = 1.96

Hence Z > Zα that is 2.938 >1.96

Therefore the null hypothesis is  rejected.

Inference:

These samples are not taken from the population with mean 26.8.

SSS:  The Side Side Side Triangle Congruence  known as the SSS  also called as the Edge Edge Edge Triangle Congruence If Any two triangles are having three sides as congruent then they are called as the Side side side Triangle congruencies

 

SAS:   The Side Angle Side Triangle Congruence  are known as the SAS  also called as the Edge Angle Edge Triangle Congruence If Any two triangles are having two sides and one angle as congruent then they are called as the Side angl -side Triangle congruencies

 

AAS : The Side Angle Side Triangle Congruence  are known as the AAS  also called as the Angle Angle Edge Triangle Congruence If Any two triangles are having two Angles and one Side as congruent then they are called as the Angle angle side Triangle congruencies.

ASA

ASA

 

   ASA

Angle-angle-side (AAS)

The Angle Angle Side Triangle Congruence  are known as the AAS  also called as the Angle Angle Edge Triangle Congruence. If Any two triangles are having two Angles and one Side as congruent then they are called as the Angle angle side Triangle congruencies.

AAS

AAS

 

Angle-side-angle (ASA)

The Angle Side Angle Triangle Congruence  are known as the ASA  also called as the Angle Edge Angle Triangle Congruence .If Any two triangles are having two Angles and one Side as congruent then they are called as the Angle Side angle Triangle congruencies

Included angle 
Polygon is formed by a side in between two adjacent Sides is a included Angle

Included side 
Polygon is formed by a side in between two adjacent angles  is a included side.

Postulate 
An initial statment wich is to be proved.

Proof 
A step by step procedure to prove the given postulate is known as proof.For example to prove the Side angle side te prof include step by step procedures.
Side-angle-side (SAS)

The Side Angle Side Triangle Congruence  are known as the SAS  also called as the Edge Angle Edge Triangle Congruence If Any two triangles are having two sides and one angle as congruent then they are called as the Side angle side Triangle congruencies

 

The side-side-side (SSS)

The Side-Side-Side Triangle Congruence  known as the SSS  also called as the Edge-Edge-Edge Triangle Congruence If Any two triangles are having three sides as congruent then they are called as the Side side side Triangle congruencies

The intersection of two lines are create the vertex and this vertex is produce an angle with line. These angles are measured by two units. They are degree and radians. Here we see about measure of angle with degree. The one 360th part of a circle is denoted as one degree.

 

Explanation for degree angle measure:

 

Measure of an angle:

The measure of an angle is degree in geometry. The circle has 360 degrees. When the subtended angles are used, the small measure is available and astronomical measurements and measure of latitude example of small measurement. The smaller measurements are divided into 60 seconds when the small angles are measured.

The degree is denoted as ° that is after a number the small circle is added.The minutes also used in degree measures that is if any seconds are measured that is represented as  ‘ after the degree value. For example, the measure of angle is 40° 12′. We can measure the angle using protractor. It is in semi-circle shape and the values from 0° to 180°.

If the reference meridian is used in angle means it is considered as sphere like earth, mars.The edges are incident to vertex is called degree of vertex in graph theory.

Examples for degree angle measure:

 

Problem 1: If the straight line is PQR and angle of RQA is 35° means measure the angle of PQA.

degree angle measure

degree angle measure

Answer:

PQA = straight line + RQA

= 180° + 35°

= 215°.

Problem 2: Measure the angle using sum of angle formula from given values.

A = 50°, C = 60° and B = ?

Answer:

        Using sum of angle formula,

A + B + C = 180°

50° + B + 60° = 180°

B = 180° – 110°

B = 70°.

Excercise problem for angle measurement:

1. Measure the AQR angle if PQA is 280°.

        Answer: The angle of AQR is 80°.

2. Find the unknown angle from B = 80°, C = 60° A = ?.

        Answer: A = 40°.

Let us see how to subtract the fractions in this topic. A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, etc… A fraction consist of a numerator and a denominator, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole.(Source: Wikipedia)

 

Example problems of subtracting fractions:

 

Subtracting fractions problem 1:

Subtract `1/2 -1/4 `

Solution:

We can subtract the given fractions by using the following methods.

Before we can go to subtracting, we can see the denominator part.

If denominator values are equal we need not to change the numerator.

But if denominator values are different we can find the L.C.M of denominators and then change the numerator value depends on L.C.M value.

In the above problem, denominators are different.

So we can take the L.C.M of 2, 4.

L.C.M (2, 4) = 4

Therefore, `(1*2)/(2*2)-(1*1)/(4*1)`

`= 2/4-1/4`

Here denominators are equal.

That is, `(2-1)/4`

`=` 1/4

Answer:` 1/4`

Subtracting fractions problem 2:

Subtract` 3/5-2/5`

Solution:

We can subtract the given fractions by using the following methods.

Before we can go to subtracting, we can see the denominator part.

If denominator values are equal we need not to change the numerator.

But if denominator values are different we can find the L.C.M of denominators and then change the numerator value depends on L.C.M value.

In the above problem, denominator values are equal.

Therefore, `(3-2)/5`

`= 1/5`

Answer: `1/5`

Subtracting fractions problem 3:

Subtract `4/8-5/12`

Solution:

In a given problem, denominator values are different.

So we can take the L.C.M of 8, 12.

L.C.M (8, 12) = 24

Therefore, `(4*3)/(8*3)- (5*2)/(12*2)`

` = 12/24-10/24`

Here denominator values are equal.

That is, `(12-10)/24`

`= 2/24`

We can simplify the above part.

That is ` (2-:2)/(24-:2)=1/12 `

Answer:` 1/12`

 

Practice problems of subtracting fractions:

 

  1. Subtract `2/5-1/5`
  2. Subtract `3/2-1/4`
  3. Subtract `1/3-2/4`

Answer:

  1. `1/5`
  2. `5/4`
  3. `-1/6`

Let us see about the solving probability of an event. The solving probability of simple event is ratio of the count of favorable outcomes to the total number of possible outcomes of event.

P(E) = Number of positive outcomes ⁄ Total number of possible outcomes

We can get the probability of an event, dividing the number of outcomes by total number of possible outcomes.

 

Solving probability of an event : Types of Events :

Types of solving events are,

  • Compound event
  • Impossible event
  • Sure event
  • Compound event:

It consists of two or more than two simple events.

Types:

1. Independent events :

Here, the outcome of one event does not affect the affect the outcome of another.

P (A and B) = P (A). P (B)      [A and B are two events]

2. Dependent Event:

Both events are dependent. That is, outcome of one event affects the outcome of another event.

P(U or V) = P(U) +P(V)

  • Impossible events

Impossible event do not occur and the probability is always zero.

  • Sure event

An event which  certainly occurs is a sure event.

These are the types of events.

Solving probability of an event :Examples:

Examples for solving probability of an event:

Ex 1:

A bag  contains 7  black  and 5 yellow  balls. Find the corresponding probabilities of drawing a black ball and yellow ball if the balls are replaced after each draw.

Sol:

Total number of possible outcome = 7+5= 12

P (Drawing black ball)      = 7/12

P (Drawing yellow ball) = 5/12

This is an independent event ( multiply the probabilities of two events)

P(Drawing black and yellow ball) = 7/12 * 5/12

=  35/144

Ex 2:

An urn contains 6 red and 4 white  balls. Find the corresponding probability of drawing a red ball and a white ball, if the balls are not replaced after each draw.

Sol:

Total number of possible outcome = 6+4 = 10

P (Drawing red ball)      = `6/10`

Now we have not replaced a ball taken at the first draw

Total number of possible outcome = 5+4 =9

P (Drawing white ball) = `4/9`

P(Drawing red and white ball) = `6/10` *` 4/9`

= ` 24/ 90`

Ex 3:

Find the probability of a person’s birthday falling on April 31st.

Sol:

April 31st never occurs in any year.(Impossible event)

Probability of birthday falling on April 31st is 0.

Next Page »