Multiplying Negative and Positive Numbers

Multiplication is a short way of doing repeated addition. Just as 3 × 8 means 3 addends of 8: 8 + 8 + 8 = 24
3 × (-8) means 3 addends of -8.
-8 + (-8) + (-8) = -24; therefore, 3 × (-8) = -24
When we multiply (a positive number) × (a negative number), the product is a negative number. The Commutative Property of Multiplication is also used with signed numbers.
3 × (-8) = -8 × 3 = -24

Multiplying negative and positive numbers results in negative number. Let us see sample problems for multiplying negative and positive numbers.

Multiplying Negative and Positive Numbers:

MULTIPLICATION RULES:

From these patterns we can write the rules for multiplication of two numbers. (The rules in your text should emphasize that they apply to two numbers.)

1. Multiplication of two numbers with the same signs:

a. Multiply each  absolute values

b. The answer is positive

RULE 1:

4 × 6 = 24 -8 × (-7) = 56

2. Multiplication of two numbers with different signs:

a. Multiply each absolute values

b. The answer is negative

RULE 2:

             -8 × 5 = -40 3 × (-1) = -3

Now that you have used the rules for multiplication, you may find that the addition rules confuse you! It helps to remember how

we added on the number line to derive the addition rules. Then think about these patterns if you forget rules for multiplication.

There are several ways to write multiplication:

             6 × 4;                6 × 4 and 6(4)    or     (6)(4)

We will continue to put parentheses around negative numbers that follow the operations:

a. 8 × (-4) 8 × (−4) 8(-4)

b. -5 × (-2) − 5× (−2) -5(-2)

The first factor can be written with parentheses, but it usually is written as line b above.

Now let's see what happens when more than two factors are multiplied:

(REMEMBER no symbols between parentheses means to multiply).
          -3(-4)(2)                            -6(-2)(-1)(-4)                              -5(6)(-2)(3)

            12 × 2                          12 × (−1)(−4)                             -30(-2)(3)

                24                                     -12(-4)                                     60(3)

                                                           48                                          180

(2 negative factors)                   (4 negative factors)                     (2 negative factors)

When we had an even number of negative factors the products were positive!


            -3(4)(2)                               -6(-2)(-1)(4)                               -5(6)(-2)(-3)

            -12 × 2                                12 × (-1)(4)                               -30(-2)(-3)

               -24                                     -12(4)                                        60(-3)
                                      

                                                          -48                                          -180


           (one negative factor)        (3 negative factors)                   (3 negative factors)

When we had and odd number of negative factors, the products were negative.

Multiplying Negative and Positive Numbers:

RULE for multiplying any number of factors:

1. Multiply the absolute values

2. Count the negative factors

a. The product is positive for an even number of negative factors.


b. The product is negative only If the number of negative factors is odd..

When you divide, you think of the related multiplication problem:

              20 ÷ 4 = 5 because 4 × 5 = 20

dividend ÷ divisor = quotient, and divisor × quotient = dividend

a. 27 ÷ (-3) = ?
       

        -3 × ? = 27 The product 27, is positive; therefore,

the two factors must have the same sign. Since the first factor is negative, the second factor is also negative

Since -3 × (-9) = 27, we know

          27 ÷ (-3) = -9

b. -12 ÷ 2 = ?                 The product, -12, is negative; therefore, the two factors must have

2 × ? = -12                     different signs. Since the first factor is positive, the second factormust be negative.

c. -28 ÷ (-4) = ? The product, -28, is negative;

therefore, the two factors must have -4 × ? = -28 different signs. Since the first factor is negative the second factor must be positive.

Since             -4 × 7 = -28

we know -28 ÷ (-4) = 7

	PROBLEMS:	ANSWERS:
1	-6(-9)	54
2	8(-7)	-56
3	-9(4)	-36
4	-3(-4)(2)	24
5	(-5)(6)(3)	-90
6	-5(-2)(5)(-3)	-150
7	-2(-1)(7)(3)	42

Percentage in Line Graphs

In this article we discuss the percentage in line graphs. The graph can be defined as relationship between the varying things. The line graph is also called as line chart or drawing things. Graph is mostly used for bars and lines to display data. The use of graph is to comparing the numbers in easiest way. The straight line graph is used to connect the points in real data.

Explanations of Percentage in Line Graphs:

Let we are take some data and draw the line graph in below details,

Year	1970	1980	1990	2000	2010
Percentage	70.00%	80.00%	90.00%	100.00%	110.00%

Where,

P=percentage and Y=Year.

The above table describes the year and population percentage from the year 1970 to 2010. The following table details can be taken and make the percentage line graphs.

Step1:  The above percentage in line graph for the given table is shown below, in which the year is represented along x-axis and the population percentage is represented along y-axis.

Step2: For percentage of population interval is 20 and for year the interval is 10.

Step3: From the table at first for year 1970 the corresponding percentage 70 is marked it is placed in between 60 and 80.

Step 4: From the table at second for year 1980 the corresponding percentage 80 is marked in percentage 80.

Step5: From the table at third for year 1990 the corresponding percentage 90 is marked it is placed in between 80 and 100.

Step6: From the table at fourth for year 2000 the corresponding percentage 100 is marked in percentage 100.

Step7: From the table at first for year 2010 the corresponding percentage 110 is marked it is placed in between 100 and 120.

Step8: Now join all these points by using a line this gives the percentage in line graphs or line chart.

Examples for Percentage in Line Graphs:

Example 1:

The following table shows the fruits and people like the fruits percentage details. Then draw the percentage in line graphs for the below table,

Fruits	Apple	Orange	Mango	Graphs	Banana	Pineapple
Percentage	20.00%	55.00%	40.00%	30.00%	45.00%	35.00%

Where,

P represents percentage in y-axis and Fruits represents x-axis.

Exponential and Logarithmic Calculator

In math exponential function is a function of ex, where e is the digit such to the significance ex the same its contain consequent.

For example,

`e^x` is an exponential function

In math logarithmic function is an inverse function of the expoential function.

`log_a (x)`

Use exponential and logarithmic calculator easily get an answer.

Basic Solving Exponential and Logarithmic Calculator: -

Enter the exponential equations then click the answer button. Perform the exponential operation and get the answer.Check the answer in manual calculation `3^(2x-1)`

`3^(2x-1) = (3^3)^x`

`= 3^3x`

If the base value is same equal to one another

2x-1  = 3x

-1 = 3x-2x

-1= x  (or)  x = -1

Calculator answer as 0.698970004336.

Example Problems for Exponential and Logarithmic Calculator: -

Problem 1:-

Find x in the equation `(1+(.10/12))^12x= 2`

Solution:-

`(1+(.10/12))^12x - 2`

Multiply Ln on both sides

`Ln(1+(.10/12))^12x = Ln(2)`

Simplify the equation

`12x Ln(1+(.10/12)) = Ln(2)`

`12x Ln(1+(.10/12))`

`x = (Ln(2))/(12Ln(1+.10/12))`

x = 6.9603

Using calculator to verify the answer.

Problem 2:-

Find x in the equation 400 = 5000 `(1- (4/4+e^(-0.002x)))`

Solution:-

Divide 5000 on both sides

0.08 = `(1- (4/4+e^(-0.002x)))`

Subtract `e^(-0.002x)` on above equation

0.092 = `4/(4+e^(0.002))`

Multiply 4+`(e^-0.002x)` on both sides

0.092 `(4+e^(0.002x))` = 4

Divide 0.92 on both sides

`(4+e^(0.002x))` = 4.34782608696

Subtract 4 on both sides

`e^(0.002x)` = 0.34782608696

Multiply Ln on both sides

`Ln(e^(0.002x)) = Ln (0.34782608696)`

Simplify the equations

-0.002xLn(e) = Ln (0.34782608696)

= -1.0560526

We know that Ln(e) = 1

Divide -0.002 on both sides

`x = (Ln(0.34782608696))/-0.002`

= 528.02633

Using calculator to verify the answer.



Problem 3:-

Find log 1000 verify the answer by using a calculator.

Solution:-

Expressing the given expression in scientific notation:

log 1,000 = log 103

Since 1,000 is the 3rd power of 10,

log 1,000 = 3

Using calculator to verify the answer.

Problem 4:-

Find log 0.0001 verify the answer by using a calculator.

Solution:-

Rewriting the given expression in scientific notation:

log 0.0001 = log 10–4

Since 0.0001 is –4th power of 10,

log 0.0001 = –4

Using calculator to verify the answer.

Trigonometric Functions of Special Angles

Trigonometry  is branch of  mathematics which deals with triangles.Trigonometry means measure of triangles . Trigonometry deals with the relationship between sides and angles of triangle. There are six basic trigonometric functions and they are sine, cosine, tangent, secant, cosecant and cotangent. They are defined based on the sides of a right triangle.

Explanation of Trigonometric Functions of Special Angles

Basic functions of  a right triangle following are defined as follows:



i)Sine abbrevated as "sin"                and   sin(theta) = ` c/b`

ii)Cosine abbrevated as "cos"          and   cos(theta) = ` a/b`

iii)Tangent abbrevated as "Tan"         and   Tan(theta) = ` c/a`

iv)Secant abbrevated as  "sec"         and   sec(theta) = ` b/a`

v)Cosecant abbrevated as "cosec"    and   cosec(theta) = ` b/c`

vi)cotangent abbrevated as "cot"       and   cot(theta) = ` a/c`

There are 4 types trigonometric identities.

1:  Reciprocal Identiy :

i) cosecA = `(1)/(sin A)`

ii) sec A= `(1)/(cos A)`

iii) cotA = `(1)/(tanA)`

2:  Pythagorean Identites :

Sin2 A  + Cos2 `A =1`

Sec2 `A`   - Tan2 `A = 1<br>`

Cosec2 A  - Cot2 A`=1`

3:  Quotent Identites :-

tan`A` =  sin`A`` /` cos`A`

cotA =  cos`A`` /` sin`A`

4 : Even- odd Identites:

1)sin(-`A) =`sin`A`

2)cos(-`A) = cosA`

`3)`tan(-`A) =`tan`A`

`4)`cosec(-`A)= `-cosec`A`

`5)`sec(-`A) = `sec`A`

`6)cot(-A) = -cot(A)<br>`

Chart for trigonometric functions of special angles :-

here "U" stands for undefined or infinite.

Examples on Trigonometric Functions of Special Angles:

Given an right  triangle whose base is 3 cms and base angle with hypotenuse is 60o.find the following

i) height of triangle

ii) area of triangle

Solution:

Given an right  triangle and base side is 3 cms and base angle is 60o.

Tan of base angle is height /base

so tan 60o = height /3

=> `sqrt(3)`  = height /3

=>height of triangle is 3`sqrt(3)` cms.

ii)

Area of a right triangle A is 1/2(base * height)

=> A =(1/2)*3*3`sqrt(3)`

`=> A=4.5``sqrt(3)`

`A= 7.79`


Ex 2 :   In a right triangle given angle A is 60o and side adjacent to angle A is 5cms .

find the hypotenuse of triangle

solution:

Cos A = adjacent side/hypotenuse

Cos 60o = 5/hypotenuse

=>1/2=5/hypotenuse

=> hypotenuse =5/2 =

=> hypotenuse side= 2.5cms

Ex3:

find value of cos 60o + sin 30o

solution:

value of cos 60o=1/2

value of sin 30o=1/2

hence

cos 60o + sin 30o  =1/2 +1/2 =1

Easy Way to Multiply Big Numbers

Multiplication is one of the basic operation in mathematics and its symbolic representation is ‘’×’’. It is difficult while multiplying big numbers. So, we are going to use the easy for multiplying big numbers. Easy way is to split the numbers into smaller parts and then multiply with 5, 10 and then add the result to the multiplication of remaining parts, we get the answer. Let us see about easy way to multiply big numbers in this article.

Worked Examples for Easy Way to Multiply Big Numbers

Example 1 for Easy Way to Multiply Big Numbers:

Multiply the Number which Ends with 5:

How to square the number 85?

Step 1:

Take the number which is on the left of the given number. The left of the given number is 8.

Step 2:

Multiply the left of the given number 8 with its consecutive number 9.

Multiply 8 × (8 + 1) = 8 × 9 = 72

Step 3:

Then multiply the ending number with the same ending number 5 is 5×5 = 25.

Step 4:

Grouping the step 2 and step 3 results, we get 7225.

Step 5:

Hence, squaring the given number 85 is 7225.

Example 2 for Easy Way to Multiply Big Numbers:

Multiplying Big Numbers by 11:

How to multiply 986 by 11?

Step 1:

Multiply 986 × 11

Step 2:

The number 11 can be written as (10 + 1).

986 × (10 + 1)

Step 3:

Now multiply 986 by 10, we get 9860.

Step 4:

Multiply 986 by 1, we get 986.

Step 5:

Add the result of above two steps.

9860 + 986 = 10846

Step 6:

Hence, multiplying 986 by 11 is 10846.

Example 3 for Easy Way to Multiply Big Numbers:

Multiplying two Big Digit Numbers:

55 × 57

13 × 14

Solution:

a) 55 × 57

Step 1:

Split the multiplier into two parts.

(55) × (55 + 2)

Now, multiply 55 with 55 and 2 respectively and add both the results, we get the answer.

Step 2:

Multiply the first part 55 × 55, by using the multiplication concept of numbers end with 5.

Step 3:

Take the number which is on the left of the given number 55. The left of the given number is 5.

Step 4:

Multiply the left of the given number 5 with its consecutive number 6.

Multiply 5 × (5 + 1) = 5 × 6 = 30

Step 5:

Then multiply the ending number with the same ending number 5 is 5×5 = 25.

Step 6:

Grouping the step 4 and step 5 results, we get 3025.

Step 7:

Multiply the second part 55 ×2, by using the multiplication concept of numbers end with 5.

55 × 2 = 110

Step 8:

Hence, multiplying 55 × 57, we have to add the step 6 and step 7 results, we get the answer.

3025 + 110 = 3135

Step 9:

Hence, the solution for multiplying 55 × 57 is 3135.

b) 13 × 15

Step 1:

13 × (10 + 5)

Step 2:

13 × 10 = 130

Step 3:

13 × 5 = 65

Step 4:

130 + 65 = 195

Step 5:

Hence, the solution for multiplying 13 × 15 is 195.

Practice Problems for Easy Way to Multiply Big Numbers

1. Multiply the 788 by 11.

2. What is the value of square of 75?

Solutions:

1. 8668

2. 5625

Adding Sine and Cosine

Here we are going to see the article as adding sine and cosine, generally sine and cosine are trigonometric functions it is used for measuring the side length and angle of the right angle triangle .Sine is defined as the ration of the opposite side to the hypotenuse and cosine is defined as the ratio of the adjacent side to the hypotenuse. So here the right angle triangle will be given with the corresponding sides name like adjacent, opposite and hypotenuse.

Example 1-adding Sine and Cosine:

Sin 45 +cos 45

Solution:

We know that the value of sin 45 and cos 45 (both are equal) =1/`sqrt 2` , plug these values into above expression.

`1/sqrt 2` +`1/sqrt2`

If the denominators of both fractions are same so we add the numerator part only we get the answer.

`(1+1)/sqrt 2 =2/sqrt 2`

Example 2-Adding sine and cosine:

Sin260+cos260

Solution:

Already we know that the formula for sin2 A+cos 2A=1

Here the value of A=60 so we use the formula as Sin260+cos260=1.

It will be check with the another method,

We know that the value of sin 60 is sqrt `3/2 ` and cos 60 is `1/2` plug these values into above expression

`(sqrt(3)/2)^2 + (1/2)^2` =`3/4` +`1/4` =1

We get the same answer in both the method.

Example 3-adding Sine and Cosine:

Sin 30+cos 90

Solution:

We know that the value of sin 30 and cos 90 the value of sin 30 is `1/2` and cos 90 is 0plug these values into above expression.

`1/ 2` +0= `1/2`



Example 4-Adding sine and cosine:

Sin 0+cos (-30)

Solution:

We know that the value of sin 0 is 0 and cos (-30) =cos 30 value of cos 30 is sqrt 3/2

0+`(sqrt3)/2` =`(sqrt 3)/2.`

Practice problem -Adding sine and cosine:

1)    Add sin2 25+cos2 25

Answer =1

2)    Add sin 67+cos 13

Answer =1.8949

Statistical Techniques

Let fp( .| `theta` )g be a family of probability densities indexed by a parameter `theta` .Let y = (y1,....., yn) be a sample from p(.|`theta` ) for some unknown `theta` . Let T(y) be a statistic such that the joint distribution factors as follows

`prod` p(yi | `theta` ) = g(T(y); `theta` )h(y):

for some functions g and h. Then T is called a sufficient statistic for `Theta``theta` .

The  statistical techniques involve to study the two important parts namely mean deviation and standard deviation.

Mean Deviation for un grouped data:

Let x1,x2,x3,………xn be the given n observation. Let  `barx`   be the AM and M be the median .then

i.) MD(x¯) = `(sum_(i= 1)^n)/(n)` | xi –x|           where `barx`

Statistical Techniques in Mean, Mean Deviation

1.) Find the mean deviation about the mean for the given data.

15,17,10,13,7,18,9,6,14,11.

Sol: Let the mean of the given data be `barx` Then,

`barx` = `(sum_(i= 1)^n)/(n)` | xi –x|
= `(120)/(10)`  = 12    [n = 10]

so, the values of (xi – x) are:

3, 5, 2, 1,-5,6,-3,-6,2,-1

So, the values of | xi –x| are:

3 , 5 , 2 , 1 , 5 , 6 , 3 , 6 , 2 , 1

MD(`barx` ) =   `(sum_(i= 1)^n)/(n)` | xi –x|


=  `(34)/(10)`  = 3.4

Hence the MD (`barx` ) = 3.4

Statistical Techniques in Variance with Example:

Variance and Standard Deviation for statistical techniques:

Definition of variance :

Mean of the squares of the deviations from the mean is called the variance , to be denoted by σ2. The variance of n observations x1 , x2 ,.....xn is given by σ2 =   `(1)/(n)` `sum_(i=1)^n` (xi - x¯)2 .

Definition  of Standard deviation :

The  positive square root of variance is called standard deviation and it is denoted by σ  .

σ =`sqrt((sum_(i=1)^n( x i-x)^2)/(n))`

Ex: Find the mean, standard deviation and variance of first n natural numbers

Sol:

Mean,x = ( 1 + 2 + 3 +........+ n)    = 1  .1   n( n + 1 ) =  1  ( n + 1 )
n               n   2                      2

Variance, σ2 = `(sum)/(n)` x-2
=  x¯2



Variance, σ2 = `(sum)/(n)`  n2  =  { 1/2 (n + 1) }2



=          `(1)/(n)` `(n(n+1)(2n+1))/(6)`  -  `((n+1)^2)/(4)`

=     `(1)/(n)` `(n(n+1)(2n+1))/(6)`  -  `((n+1)^2)/(4)`

=  `((n+1))/(n)` ×   `(n(2n+1))/(6)`  -  `((n+1))/(4)`

= `((n+1))/(1)`   × `((2n+1))/(6)`  -  `((n+1))/(4)`

=`((n+1)(n-1))/(12)`

Variance,σ2  = `((n^2 -1))/(12)`

Standard deviation ,σ2 = `((sqrt(n^2 -1)))/(12)`

= `((1/2) (sqrt(n^2 -1)))/(3)`

Change in Percentage

Change is something which never dies and never changes. Change can be termed in the mathematical form as Variable, which has no constant value and it can take any value. It is usually denoted using some alphabet like the usual one ‘x’, the value of x is always unknown. There is a change in lot of things around us like every instant that is every second, and even that every second is a change in terms of time, distance and anything possible. The word possible is very challenging and there is nothing impossible in the world, impossible can be changed into possible through hard work and perseverance.

Coming to the concept of numbers, in which the topic concerned is about percentage  This concept when thought of, the general perception that every single has is that it is tough and not an easy concept to understand. The Percentage Change can always be an increase or a decrease in per-cent-age, it can be either way round. So it is necessary for us to learn both the method of calculations, actually both require the same Percentage Change Formula, the Calculating Percentage Change here will be only difference in the final answer, that too in the interpretation part, since obviously without understanding the number one can never say if it is an increase or a decrease in the change occurred.

Let us now try to answer the question of  How to Calculate Percentage Change. Answering this question is never a great task, it is all about the numbers we get for solving, and one can easily solve such type of questions, if one understands the problem and interprets the Percentage of Change. Let us consider an example to get a better understanding Calculate Percentage Change of the numbers whose initial value that is the original value is 50 and the final value is 100; the answer for this question is 50 – 100 / 50 * 100 = 100%, the answer shows us that there is an 100 % increase in the final value from the original value and that is the process of Calculating Percentage Change.

The fact is that anyone can solve such problems, but the interpretation part can only be done by the person who completely understood the concept. The final interpretation shows us the direction of change that is it shows whether the change is positive or negative and that makes the concept complete.

First Decimal Place

Binary numerals are usually read digit-by-digit, in order to distinguish them from decimal numbers. For example, the binary numeral 100 is pronounced one zero zero, rather than one hundred, to make its binary nature explicit, Since the binary numeral 100 is equal to the decimal value four, it would be confusing to refer to the numeral as one hundred. (Source: Wikipedia)

First Decimal Place:

Let us take a number, 19.7.

1 indicates place value ten.

9 indicates place value one.

7 indicate 7/10(tenths decimal place)

Example problems for first decimal place:

Example 1:

Calculate the following 15.9+15.7-0.15+0.1

Solution:

Given that, 15.9+15.7-0.15+0.1

Numbers of digits behind decimal point are equal.

Step 1: Adding positive numbers,

= 15.9+15.7+0.1

= 31.7.

Step 2: Negative integer add to the primary step

= 31.7 – 0.15

= 31.55

The solution is 31.55

Example 2:

Calculate the following 18.2+0.1-18.7+1.8

Solution:

= 18.2+0.1-18.7+1.8

Numbers of digits behind decimal points are equal.

Step 1: Adding positive numbers,

= 18.2+0.1+1.8

= 20.1

Step 2: Negative number add to the first step

= 20.1– 18.7

= 1.4

The solution is 1.4

Addition for first decimal place:

Problem 1:

Add the following 1.3+1.5+1.7+1.9

Add the positive value,

=1.3+1.5+1.7+1.9

= 6.4

The solution is 6.4

Problem 2:

Add the following 1.6+1.5+1.7+1.7

Add the positive value,

=1.6+1.5+1.7+1.7

=6.5

The solution is 6.5

Example problem for first decimal place:

Subtraction for first decimal place:

Problem 1:

Find 9.3-9.5-9.8-9.7

= 9.3-9.5-9.8-9.7

Subtract the negative value,

= -19.7

The solution is -19.7

Problem 2:

Find 8.6-8.5-8.7-8.9

Subtract the negative value,

= 8.6-8.5-8.7-8.9

= -17.5

The solution is -17.5

Multiplication for first decimal place:

Problem 1:

3.5 * 3.5 * 3.5

Multiply the positive value,

= 3.5 * 3.5 * 3.5

= 12.25* 3.5

= 42.8

The solution is 42.8

Problem 2:

1.8 * 1.8 * 1.8

Multiply the positive terms,

= 1.8 * 1.8 * 1.8

= 3.24 * 1.8

= 5.8

The solution is 5.8

Division for first decimal place:

Problem 1:

`7.8/0.6`

Divide the positive numerator value,

= `7.8/0.6`

= 13.0

The solution is 13.0

Problem 2:

`9.7/8.8`

Divide the numerator value,

= `9.7/8.8`

=1.1

The solution is 1.1

Practice Problem for first Decimal Place:

Problem1: 7.2+8.3+8.2

Answer: 23.7

Problem2: 9.2+9.2+9.2

Answer: 27.6

Problem3: 15.3+15.2+15.2

Answer: 45.7

Practice Model Definition

If any student wants to know about the definition for math practice model with examples, they can be referring the below examples for them. In this we can see some definitions for practice model in statistics with example problems. Let us see some of the definition in statistics such as mean, median and mode with example and practice problems with solve as follows.

Practice Model Definition - Definitions:

In this following we can see some definition for math practice model in statistics.

Practice Model Definition 1:

Mean:

It is a set of data which can find the average in dividing by sum of numbers with total numbers in the given data.

Practice Model Definition 2:

Median:

It is a set of numbers which can find the middle numbers and arrange the numbers in order to select the middle number.

Mode:

It is a set of numbers that can occur frequently in a set of data and no number can occur more than once.
Practice Model Definition - Examples and Practice Problems:

In this following we can see some definition for math practice model in statistics with example and practice problems.

Find the mean of weights for 6 peoples in kilograms are 15, 9, 10, 7, 16, and 23.

Solution:

                Sum of numbers
Mean = ------------------------
                  Total number

= `(15+9+10+7+16+23)/6`

= `80/6`

= 13.33

Find the median of 10, 24, 12, 21, 7 and 18.

Solution:

Arrange the data in ascending order as 7, 10, 12, 18, 21 and 24.

N = 6

Since n is even, median = `1 / 2` [`(nth)/2` item value + (`n / 2` + 1)th item value]

= `1 / 2` [6th item value + (`6/2` + 1)th item value]

= `1 / 2` [3rd item value + 4th item value]

= `1 / 2` [12 + 18]

= `1 / 2 xx 30`

= 15

Find the mode of 22, 17, 19, 15, and 22.

Solution:

22 are repeated twice.

Mode = 22

Practice Problems:

Find the mean of weights for 6 peoples in kilograms are 5, 9, 10, 7, 16, and 23.

Solution:

= 11.66

Find the median of 8, 25, 11, 21, 6 and 17.

Solution:

= 14

Find the mode of 20, 17, 11, 13, and 20.

Solution:

= 20

Integral of X Sinx

The term integral may also refer to the notion of anti-derivative, a function F whose derivative is the given function ƒ. In this case it is called an indefinite integral, while the integrals discussed in this article are termed definite integrals. Integrate the function f(x) can be written as`int ` f(x) dx. The integral is used to find the surface area, volume geometric solids by using stokes theorem, divergence theorem (like double integral, triple integral).
Source Wikipedia.

Integral Formulas:

1. `int` x n dx = `(x^n+1) / (n+1) ` + c

2. `int`sin x. dx = - cos x + c

3.` int ` cos x .dx = sin x. + c

4.  ` int` sec2x. dx = tan x + c

5. `int` cosec x.cot x. dx = -cosec x + c

6.`int dx / sqrt(a^2 - x^2) = sin^-1(x / a) + c`

Integral Problems:

Integral problem 1:

Find the integration of  x sin x with respect to x.

Solution:

Given term is x sin x

Integral of x sin x can be expressed as `int` x sin x dx

let u = x                         dv = sin x dx

`(du)/(dx)` = 1            and        v = - cos x

We know `int ` u dv  = uv - `int` v du

= x . (-cos x ) - ` int` (-cos x) 1.dx

= - x cos x +` int` cos x dx

= - x cos x + sin x + c

Answer:  `int` x sin x dx  = - x cos x + sin x + c

Integral problem 2:

Find the integration of  (2x + sin x) with respect to x.

Solution:

Given term is 2x + sin x

Integral of (2x + sin x )can be expressed as `int`(2x + sin x )  dx

`int`(2x + sin x )  dx =  `int` 2x dx +` int` (sin x).dx

= 2 `int` x dx +` int`sin x dx

= 2` (x^2/2)`+ (-cos x) + c

= x2 - cos x + c

Answer:  `int` (2x + sin x) dx  =   x2 - cos x + c

Integral problem 3:

Find the integration of  (sin x  + 3cosec2x)  with respect to x

Solution:

Given term is  (sin x  + 3cosec2x)

Integral of (sin x  + 3cosec2x) can be expressed as  `int ` (sin x + 3cosec2x) dx

`int ` (sin x + 3cosec2x) dx  =` int` (sin x dx) +` int` (3cosec2x) dx

= `int` (sin x dx) + 3` int ` (cosec2x) dx

=  (- cos x) + 3 (-cot x) + c

= - cos x - 3 cot x + c

Answer: `int `(sin x  + 3cosec2x) dx  = - (cos x + 3 cot x) + c

Integral problem 4:

Find the integration of  x2 sin x with respect to x.

Solution:

Given term is x2 sin x

Integral of x2 sin x can be expressed as `int`x2 sin x  dx

let u = x2                          dv = sin x dx

`(du)/(dx)` = 2x            and        v = - cos x

We know `int ` u dv  = uv - `int` v du

`int`x2 sin x  dx   = x2 . (-cos x ) - ` int` (-cos x) 2x.dx

= - x2 cos x +` int` 2x cos x dx

=  - x2 cos x + 2` int` x cos x dx + c

= - x2 cos x + 2 I1+c

Take I1 = ` int` x cos x dx

let u = x                       and     dv = cos x dx

`(du)/(dx)` = 1                                   v =  sin x

We know `int ` u dv  = uv - `int` v du

I1 =` int` x cos x dx = x sin x - `int` sin x dx

= x sin x - (-cos x) + c

I1 = x sin x + cos x + c

`int`x2 sin x  dx  =  - x2 cos x + 2 I1+ c

=  - x2 cos x + 2 (x sin x + cos x )+ c

= - x2 cos x + 2x sin x + 2cos x + c

= 2x sin x - (x2 - 2 )cos x + c

Answer:  `int`x2 sin x  dx   =  2x sin x - (x2 - 2 )cos x + c

Factor out Square Roots

The square root is the radical form of the method that are used in the calculation mathematical problems. The symbol for the square root is meant by root √. This could be in the form to describe their nature of working with the square roots. There are numerous methods are available in the rooting, they are square (second) root `sqrt(x)` , cube (Third) root `root(3)(x)` up to nth root `root(n)(x)` .  Here we are going to see about how to factor out square roots and solved example problems based on the factor out square roots.

Some Examples for Square Root Operations:

`sqrt(x) xx sqrt(y)` =`sqrt(xy)`   [By the first property]
`xsqrty` = `sqrt(x^2y)`              [By the second property]
`sqrt(x/y)` = `sqrtx/sqrty`          [By the third property]


Steps to factor out square roots:

Get the values to find the factor out square root values.
Now check for the factors in the given square root. The factor should be in the squared form for factoring the value out of the root symbol.
The squared values are factor out from the square root symbol and
Write the calculated value in the correct form.

Factor out Square Roots - Example Problems:

Factor out square roots -  Problem 1:

Solve for factor out square root of 448

Solution:

Factoring out the square root of 448

`sqrt 448`

Finding factors by L Division method

2  |448

2  |224

2  |112

2  |56

2  |28

2  |14

7

The factor for the `sqrt(448)` is `sqrt(2^6xx7)`

=  ` sqrt(2^6 xx 7)`

= `8 sqrt 7`

The factor out value for the square root value of 448 is `8sqrt7`

Factor out square roots -  Problem 2:

Solve for factor out square root of 648

Solution:

Factoring out the square root of 648

`sqrt 648`

Finding factors by L Division method

2 |648

2 |324

2 |162

3 |81

3 |27

3 |9

3

The factor for the `sqrt(648)` is `sqrt(2^3xx 3^4)`

=  ` 2xx9 sqrt(2 )`

= `18 sqrt 2`
The factor out value for the square root value of 448 is `18sqrt2`


Factor out square roots -  Problem 2:

Solve for factor out square root of 528

Solution:

Factoring out the square root of 528

`sqrt 528`

Finding factors by L Division method

2 |528

2 |264

2 |132

2 |66

3 |33

11

The factor for the `sqrt(528)` is `sqrt(2^4xx 3 xx 11)`

=  ` 2sqrt(33)`
The factor out value for the square root value of 528 is ` 2sqrt33`

Integers Prime Numbers

In math, the whole numbers are called as integers. Likewise, the prime numbers are always integers that are whole numbers. It does not contain any sign in front of the number. It is also called as natural numbers. Now we are going to see about prime numbers are integers.
Explanation for Integers Prime Numbers

Some notes about prime numbers in math:

A number which has only two distinct divisors is called as prime numbers that is indicated as integers. The property primality is included in prime numbers. In two distinct divisors, 1 is constant divisor of all prime numbers. The prime number’s factor series is n = p1, p2,……pt.

Determine the prime numbers in math:

Step 1: The multiple factors of number are listed.

Step 2: State the given number is prime number when it has two divisors. If there are more than two divisors, we can call that number is composite number.

If we take the number as greater than 100 means, how we find whether that number is prime or not. Do the following procedures.

Procedure 1: Find out the square root of number.

Procedure 2: List the prime numbers of rounded value.

Procedure 3: Find out whether the listed prime numbers are divide the rounded value or not.

Procedure 4: If the listed prime numbers are not divide the given number means that is called as prime number.

More about Integers Prime Numbers

Example problems for integers prime numbers:

Problem 1: State the given integer is prime or not.

134

Answer:

Given integer are 134.

Square root value of 134 is 11.57.

The square root value is rounded as 12.

List the below 12 prime numbers are 2, 3, 5, 7, 11.

In above prime numbers, the integer 2 is dividing the 134.

Finally, the given integer 134 is not a prime number.

Problem 2: Find out which is not a prime number from the below list.

31, 43, 53, 29, 23, 60

Answer:

The number is 60 is not a prime number. Because it has factors more than two.

Exercise problems for integers prime numbers:

1. Find out the prime numbers from the list.

73, 60, 25, 80, 30

Answer: The prime number is 73.

2. Find out the prime number from the list

120, 101, 140, 90

Answer: The prime number is 101.

Metric Conversion Equations

Let us see about metric conversion equations. The conversion is mainly done between the measurements. Generally, there are two types of measurements like standard measurement and metric measurements. There are more measurements in metric. The metric measurements used to measure the area, distance, energy, force, mass, power, pressure, velocity or speed, volume or capacity and volume of liquid.

Conversion Equations of Metric:

The metric conversion equations are following below. The equations are forming by the table format. The basic metric units are meter, gram and liter. Some prefixes are,

1000  = kilo.

100 = Hecto.

10 = deca.

0.1= deci.

0.01 = centi.

0.001 = milli.

Conversion of length:

The metric conversion of length equation is following below.

Length conversion
1 kilometer	0.62137 miles
1 meter	3.2808 feet
1 centimeter	0.3937 inch
1 mile	1.6093 kilometer
1 foot	0.3048 meter
1 inch	2.54 centimeter

Conversion of weight:

The metric conversion of weight equation is following below.

Weight conversion
1 kilogram	2.2046 ibs.
1 gram	0.0353 ounce
1 short ton	907.1847 kilos
1 pound	0.4536 kilos
1 ounces	28.3495 grams.

Conversion of volume:

The metric conversion equation of volume is described below.

Volume conversion
1 hectoliter	100 liter
1 liter	100 centiliter
1 gallon	3.7853 liters
1 cubic inch	16.3871 cubic cm
1 hectoliter	0.1 cubic meter
1 liter	1 cubic diameter
1.0567 quarts	1 cubic decimeter

Conversion of Area:

The metric conversion of area equation is described below.

	Area conversion
	1 square kilometer	1 million square meters	0.3861 square miles
	1 hectare	10,000 square meters	2.471 acres
	1 square meter	10.7639 square feet	1550.003 sq inch
	1 square mile	640 acres	258.9988 hectares
	1 square foot	144 square inches	929.0304 sq centimeters

These are the conversion equation for the standard to metric.

Examples of Conversion Equations:

Let us see some examples of metric conversions.

Example 1:

Convert the 2 millimeter scale into inches.

Solution:

1 millimeter = 0.03937 inches.

So, 2 millimeter = 2 * 0.03937.

= 0.07874.

Example 2:

Convert the 2 inch thread into centimeter.

Solution:

1 inch = 2.54 centimeter.

2 inch =  2.54 * 2.

= 5.08 centimeter.

Example 3:

Convert the 2 square foot into square inches.

Solution:

1 square foot = 144 square inches.

2 square foot = 144 * 2.

= 288 square inches.

Exponents

Introduction to exponents:

           The exponent is usually shown as a superscript to the right of the base. The exponentiation an can be read as: a raised to the n-th power, a raised to the power [of] n, or possibly a raised to the exponent [of] n, or more briefly as a to the n. Some exponents have their own pronunciation: for example, a2 is usually read as a squared and a3 as a cubed.

(Source – Wikipedia.)

Properties of Exponents:

           The following four properties are generally used in exponents. These properties are very helpful to crack problems in exponents.

• The first and foremost significant uniqueness fulfilled by integer exponentiation is

           p^x+y= p^x . p^y

• This distinctiveness has the result

           p^x-y = p^x /p^y

• for p ≠ 0, and

           (p^x)^y=p^x.y

• Another basic identity is (p. q)ⁿ = pⁿ . qⁿ

Examples Word Problems for Exponents:

Word problems exponents- Example 1:

         Solve: (s⁴) (s³)

Solution:

         Conditions of what those exponents indicate. "To the 4th" demonstrates multiplying four reproductions and "to the 3rd" demonstrates multiplying three reproductions. By the generalization method the factors are then multiplied.

          (d⁴)(d³) = (d d d d) (d d d) [multiply the terms]
                     = d d d d d d d [simplify the terms]
                     = d⁷

          Answer is: d⁷

Word problems exponents- Example 2:

          Solve:  (n³) (n⁶)

Solution:

         Conditions of what those exponents indicate. "To the 3rd" demonstrates multiplying three reproductions and "to the 6th" demonstrates multiplying six reproductions. By the generalization method the factors are then multiplied.

          (n⁴)(n⁵) = (n n n n) (n n n n n) [multiply the terms]

                        = n n n n n n n n n [simplify the terms]

                        = n⁹

         Answer is: n⁹

Word problems exponents- Example 3:

         Solve: (st)⁴/s³

Solution:

         Conditions of what those exponents indicate. "To the 4th" demonstrates multiplying four reproductions and "to the 3rd" demonstrates multiplying three reproductions. By the generalization method the factors are then multiplied.

          (st)⁴ / s³= t⁴ s^4-3 [simplify the terms]

                         =t⁴s¹

          Answer is: t⁴s¹

Word problems exponents- Example 4:

        Shorten the following Zero exponents using the quotient rule:

          7³/ 7³

Solution:

        Given 7³/ 7³

           = 7⁴×7^-4  [multiply the terms]

           = 7⁰ [add the exponents]

           = 1

           Answer is: 1

Practice word Problems for exponents:

• Solve : (s⁴) ( st³)            

             Answer: s⁵ t³

• Solve :(a³b³c⁴) (ab) (a² b c³) 

             Answer: a⁶ b⁵ c⁷         

• Simplify: (t³)⁴                      

             Answer: t⁷

• Solve: (a . b)⁶                     

             Answer: a⁶.b⁶