Adding Several Numbers

Introduction to Adding Several Numbers:

An integer is a set of whole numbers. Whole numbers above zero is said to be positive numbers denoted as ‘+’ sign and whole numbers below zero is said to be negative numbers denoted as ‘-‘. A number with zero is said to be neither negative nor positive and it does not have any sign in math. In addition there may be different digits numbers to add. The positive numbers can be written with or without the sign. Let us see about adding numbers in this article.

Rules for Adding Several Numbers

For adding several numbers, write the numbers one by one in column-wise.
Write the numbers of one’s digit in the right side of the column.
Write the numbers of tens place; hundred’s place one before the unit’s place and ten’s place in a row.
First add the one’s place and the sum of one’s place is more than one digit add the last digit to the columns top of the ten’s place, and vice versa.

Example Problems to Adding Several Numbers

Example 1:

Adding several numbers 6 + 7 + 3 + 5 + 9

Solution:

Adding all these numbers we get the carry of 3 and write before the number zero.

6

7

3

5

9

----------

30

----------



Example 2:

Adding several numbers 45 + 23 + 78 + 55+ 68+ 19

Solution:

Adding all these numbers we get the carry of 3 to the ten’s place column and add it to the ten’ place digit.

45

23

78

55

68

19

--------------

258

--------------

Example 3:

Adding several numbers 3564 + 4767 + 2433 + 7548

Solution:

Add all these numbers together to get 2 carry by adding one’s place and then carry of 2 by adding ten’s place and then the carry of 2 by adding hundred’s place and adding thousands place we get the carry of 1 it can be placed before the one’s digit by adding thousand place sum because there is no place digits before the thousands place.

3564

4767

2433

7548

--------------

18312

--------------



Problems to Practice for Adding Several Numbers

1. Adding 64 + 47 + 9 + 32 + 81

Key: 233

2. Adding 476 + 568 + 324 + 978

Key: 2346

3. Adding 7586 +9798 + 2434 + 9791

Key: 29609

4. Adding 94079 + 23247 + 86485 + 7607

Key: 211418

Probability Histograms

Introduction to probability histograms:

Histograms are used to plot density of an data, and often for density estimation: estimating the probability density function of the underlying variable. The total area of in a histogram used for probability density is always normalized to 1. If the length of the intervals on the x-axis are all 1, then a histogram is identical to a relative frequency plot. Now let us  see about the probability histogram.    

Solving Problem for Probability Histograms:

Example

When we toss a coin for three times. What is the probability of getting a heads? And also find the mean, standard deviation, expected value, standard error using probability histogram.

Solution:

Let us consider a coin to toss. If we toss a coin. We will get head or tail. Assume we calculate the number of heads. If we got head means probability is 1. Otherwise probability is 0.

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Calculate mean:

Mean = 0 + 1 / 2

= 1 / 2

probability histograms value of   Mean = 0.5.

Calculate Standard deviation:

Standard deviation = (1 – 0) v (1 / 2) (1 /2)

probability histograms value of standard deviation = 0.5.

Calculate Expected value:

Expected value = number of tosses × 0.5

If we toss a coin for three times means expected value is

= 3 × 0.5

probability histograms value of expected vaue  = 1

Calculate standard error:

Calculate standard error = vnumber of toss ×0.5

probability histograms value of  standard error  = v3 × 0.5

Practice Problems for Solving Probability Histograms:

Problem 1:

Compare the probability histogram and expect the sum with the actual results. What do conclude about that 4 dice result?

Answer :

Result of the 4 dice is not fair with the actual size.

Problem 2:

When we toss a coin for 13 times. What is the probability to get a tails? And also find the mean, standard deviation, expected value, standard error using probability histogram.

Answer:

Mean = 0.76 Standard deviation = 0.76 Expected value = 0.76 Standard error = 0.76

Trinomial Factoring Program

Introduction to trinomial factoring program:-
An algebraic expression containing three terms is called a trinomial.for example 2x+3y + 4z is a trinomial.
When we multiply two binomials, we get a trinomial. For ex (2x+5)(x+4) = 2x2 + 13x + 20 which is a trinomial.
So it follows that  we can find the factors of a trinomial.
Let us see some formulas that we use for trinomial factoring program.I like to share this What is a Trinomial with you all through my article.

Formulas Used in Trinomial Factoring Program:-

There are certain formulas that assist us when we need to factorise the trinomials.v
1. a2 + 2ab + b2  = (a + b)2
Let us do a problem bases on this formula.
# Factorize 25x2 + 30xy + 9y2
Using the above formula we find a2 = 25x2 => a = 5x;  2ab = 30xy = 2*5*3*x*y ; b2 = 9y2 = > b = 3y
Hence the solution is (a+b)2  =  (5x+3y)2= (5x+3y)(5x+3y)
Solution: The factors of 25x2+ 30xy + 9y2 = (5x+3y)(5x+3y)
2. Here is another formula to assist us to do the trinomial factoring program.
It is a2 - 2ab + b2 = (a - b)2
# Factorize 16x2 - 24xy + 9y2
This problem confirms to the second formula given here
Hence if a2 = 16x2 then a = 4x; b2 = 9y2 => b= 3y and  -2ab = -24xy => 2*4*3*x*y
Hence the factors of 16x2 - 24xy + 9y2  = (4x - 3y)2  = (4x-3y)(4x -3y)
Solution: The factors of 16x2 - 24xy + 9y2 = (4x - 3y)(4x - 3y)
Here is another formula to help trinomial factoring program
3. a2 + (a+b)x + ab =  (x+a)(x+b)
The second term is addition of two factors and the third term is the multiplication of two terms.
Hence our steps would be (1) to find the factors of a and b (2) select the factors that satisfy a+b
# Factorise x2 + 12x + 35
Solution:-
Here a+b= 12 and ab = 35
Step 1 : Let us find the factors of ab that is 35
Factors of 35 are (1,35), (5,7)
Step 2 : Let us add the factors and see which satisfies (a+b)
If we add 1+35 = 36 which is not what we want.
If we add 5+7= 12 which is what we want.
So the factors are (x+5)(x+7)
Hence  x2 + 12x + 35  can be factored as (x+5)(x+7)
Solution of the problem is (x+5)(x+7).

Probles Based on Trinomial Factoring Program:

Let us do one more problem based on the third formula .
# Factorize x2 +6x - 27
Solution:-
In this problem (a+b) = 6 and ab = -27
Step 1 find the factors of 27
(1,27), (3,9)
Let us add them 1+27=28 which is not the 2nd term 6 So we discard it.
Let us add 3+9 = 12 which is again not +6
But the third term has a negative sign.
So let us do 9-3 = +6
Now our formula changes slightly.  we need to put a negative number also
Hence we write it as (x+9)(x-3) and note that 9-3=6 which is the middle term and 9*-3 = -27 which is the last term             Hence the solution is x2 + 6x - 27 = (x+9)(x-3)
Answer : (x+9)(x-3)
Thus trinomial factoring program can be made easy if we learn the formulas that assit the factoring.

How to Solve Matrix Equality

Introduction for how to solve matrix equality:

Let us see how to solve the matrix equality in this article. The matrix equality normally represents the problems that involving the matrices for equating the right hand side with the left hand side using some identities in matrix like  `[[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]]` which is generally represented as I having the diagonal elements equal to one. Some of the problems using with some properties that exist with the matrices.

Consider the two matrices P and Q are said to be equal if

(i) both the matrices P and Q are of the same order or size.

(ii) the corresponding entries in both the matrices P and Q are equal.

i.e. the matrices P = [pij]m × n and Q = [qij]a × b are equal if m = a, n = b and pij = qij  for every i and j.

Examples to Explain "how to Solve Matrix Equality"

Let us see some of the example problems about how to solve the matrix equality.

If  `[[p,o],[i,u]]`  =`[[-115,215],[115,-415]]`  then find the values of p, o, i, u .
Solution:

Here we know that the two matrices are equal, their corresponding entries are also equal.

`[[p,o],[i,u]]`  =`[[-115,215],[115,-415]]` 

? The answer is p = -115, o = 215, i = 115, u = -415

Find the value of  a by solving if `[[p,3p - q],[2p + r,3q - r]]` = `[[0,-7],[3,2a]]`
Solution:

Given `[[p,3p - q],[2p + r,3q - r]]` = `[[0,-7],[3,2a]]`

On equating we have

p = 0              ---------------(1)

3p - q = -7     ---------------(2)

2p + r = 3      ---------------(3)

3q - r = 2a    ---------------(4)

Put (1) i.e p = 0  in (2)

3(0) - q = -7

q = 7   ---------------(4)

Put (1) i.e p = 0  in (3)

2p + r = 3

2(0) + r = 3

r = 3        ---------------(6)

Put r=3 and q=7 in (4)

3q - r = 2a

3(7) - (3) = 2a

21 - 3 = 2a

18 = 2a

2a = 18

a = `18/2`

a=9

Hence we find the a value using the matrix equality after solving the equations.

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Practice Problems to Explain "how to Solve Matrix Equality"

If  `[[p,o],[u,u+a]]`  =`[[-105,25],[15,-45]]`  then find the value of a .


The answer is that a = -60

Find the value of  a by solving if `[[p,3p - q],[2p + r,3q + r]]` = `[[0,-7],[3,2a]]`


The answer is that  a = 12.

Prime Number Algorithm

Introduction to prime number algorithm:

A prime number is a positive natural number with no other divisors except one and itself but the number 1 is excluded by definition. That is to say one is not a prime number. That leaves us with the first prime number as 2  and it is quite obvious that all other prime numbers after 2 have to be odd numbers only. The concept of prime numbers has fascinated mathematicians from the earliest times. Euclid proved infinitude of prime numbers as early as 300 B.C. There have been many failed attempts to find a formula that will generate prime numbers. A search for efficient algorithms to find prime numbers gained momentum with the advent of computers and today many efficient programs exist to calculate prime numbers limited only by the processor memory and speed.

Finding Prime Numbers by Division:

In order to check whether a given number n is a prime or not, it has to be divided one by one by all prime numbers less than `sqrt(n)`. The reasoning is simple; if the number is not a prime then there will be at least one prime factor which is less than  `sqrt(n)`. This can be used to create a simple algorithm for checkng numbers one by one and discarding compound numbers will leave a list of prime numbers.

In simple language we can write the algorithm as:

Take numbers 2 to N (one by one). Let the number by x

Divide x by all prime numbers less than or equal to `sqrt(x)`

If it is divisible by any of the prime numbers, discard it as non-prime and go to next number.

Simple! Isn't it? But it is not a very useful algorithm for use with computers, where too much processor is lost in division operation at every stage of this algorithm. Processors are more comfortable with read and write commands and simple addition and substraction operations. It is interesting to note that one of the oldest algorithms Sieve of Eratosthenes is one of the efficient algorithms for use with modern computers.

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Sieve of Eratosthenes of Prime Number Algorithm:

Concept behind this prime number algorithm is quite simple and uses tool of skip counting learnt by all of us in elementary school. It has only two steps. Write all the numbers up to a specified number. 2 is a prime, so after 2 strike out every second number. Next is 3 again a prime, so after 3 strike out every third number; and so on. Every time you complete the process for a prime p, you would be left with only primes up to the number p2.

Step 1:

2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  14,  15,  16,  17,  18,  19,  20,  21,  22,  23,  24,  25

Step 2: Striking out every second, then every third and then every fifth number, we are left with prime numbers less than 52 = 25 (shown in bold).

2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  14,  15,  16,  17,  18,  19,  20,  21,  22,  23,  24,  25

We are left with prime numbers up to 25 as: 2, 3, 5, 7, 11, 13, 17, 19, 23.

It is interesting to note that this algorithm is found quite suitable for listing prime numbers using present day computing devices.

Perimeter Semi Circle

Introduction to Semi circle perimeter:

A perimeter is a path that surrounds an area. The word comes from the Greek peri (around) and meter (measure). The term may be used either for the path or its length - it can be thought of as the length of the outline of a shape. The perimeter of a circular area is called circumference.

(Source: wikipedia)

Semi circle formula:

Fomula for finding semi circle is p =  1/2 p × d + d.

Semi Circle Perimeter Definition - Examples:

Semi circle perimeter definition - Example 1:

Find the perimeter of a circle known that its diameter is 6

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 6 + 6 = 15.4285714

Semi circle perimeter definition - Example 2:

Find the perimeter of a circle known that its diameter is 25

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 25 + 25 = 64.2857142

Semi circle perimeter definition - Example 3:

Find the perimeter of a circle known that its diameter is 54

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 54 + 54 = 138.857143

Semi circle perimeter definition - Example 4:

Find the perimeter of a circle known that its diameter is 45

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 45 + 45 = 115.714286

Semi circle perimeter definition - Example 5:

Find the perimeter of a circle known that its diameter is 63

Solution:

The perimeter is always multiplied by ½.`pi` into diameter then adds with diameter

Perimeter = 1.57142857 x 63 + 63 = 162

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Semi Circle Perimeter Definition - Practice Problems:

Practice Problem 1:

Find the perimeter of a circle known that its diameter is 5

Answer:

12.8571428

Practice Problem 2:

Find the perimeter of a circle known that its diameter is 8

Answer:

20.5714286

Line Segments with Numbers

Introduction for line segments with numbers:
Line segments are the important one in geometry chapters of mathematics subject. Line segments have more definitions; Line is an endless straight mark. Line segment is a part of one line, which are both directions having an end points with name. And line segments have so many names or types. Here, we use numbers for line segments names.

Line Segments:

A line segment is one part of line, which is having two end points. And it is defined as a straight line, which is joining with two points with coordinates and with out extending the line after that the point. It means all the given points between the given two numbers.

The above figure is having a line segment like 3,6                     

Generally we know a normal line is extending in both two directions, so this word “segment” is very important in this “line segment”.

In the above figure 3, 6 are the two end points, the length of the line segment has used with numbers for name as two end points 3, 6.

Line Segments distance formula:

In line segments we have to find the distance of line segments with end points name or numbers,

Line segment distance formula,

XY = √(x2-x1)2+ (y2-y1)2

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Examples of Line Segments with Numbers:

Example 1:

Using the distance formula Find line segment between two end point numbers, end points are

M = (3, 2), N = (5, 2)

Solution:

x1=3, y1= 2, x2=5, y2=2

Line segment distance formula  

XY = √(x2-x1)2+ (y2-y1)2

= √ (5-3)2 + (2-2)2

= √ (2)2 + (0)2

= √4+ 0

= √4

XY distance = 2 units

Example 2:

Using the distance formula Find line segment between two end point numbers, end points are

S = (3, 2), T = (6, 4)

Solution:

x1=3, y1= 2, x2=6, y2=4

Line segment distance formula  

ST = √(x2-x1)2+ (y2-y1)2

= √ (6-3)2 + (4-2)2

= √ (3)2 + (2)2

= √9+ 4

Answer is ST distance = √13 units.

Kinds of Angles in Trigonometry

Introduction to kinds of angles in trigonometry:

In mathematics, angles are nothing but the combining of two lines in trigonometry. In trigonometry, there are different kinds of angles. In trigonometry, the different kinds of angles are named as right angles, obtuse angle, acute angles, straight angles etc. The classification of different kinds of angles with diagrams is given below.

Kinds of Angles in Trigonometry – Types:

The diagrams and explanation for the different kinds of angles are given below:

Right angles:                                                              

The Diagram this kind of angles is given below, 

           
These are right angles where the value of degree of angle is 90 degree.

Acute angles:

The Diagram this kind of angles is given below, 

             
These are acute angles where the value of degree of angle is smaller than 90 degree.

Obtuse angles:

The Diagram this kind of angles is given below,   

             
These are obtuse angles where the value of degree of angle is larger than 90 degree and less than 180 degrees.

Reflex angles:

The Diagram this kind of angles is given below, 

                
These are reflex angles where the value of degree of angle is larger then 180 degree.

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Kinds of Angles in Trigonometry – Examples:

Example 1: Find what kind of angle it is given that the value of degree is 120?

Solution:

Let as assume x is value of degree.

Then, x = 120 degree.

The degree of angle is larger than 90 degree and also less than 180 degree.

Therefore, it is obtuse angles.

Example 2: Find what kind of angle it is given that the value of degree is 220?

Solution:

Let as assume x is value of degree.

Then, x = 220 degree.

The degree of angle is larger than 180 degree.

Therefore, it is reflex angles.

Kinds of angles in trigonometry – Practice problems:

Problem 1: Find what kind of angle it is given that the value of degree is 90?

Answer is given below:

These kinds of angles are right angles.

Problem 2: Find what kind of angle it is given that the value of degree is 30?

Answer is given below:

These kind of angles are acute angles.

Elementary Math Methods

Introduction for elementary math methods:

Elementary math methods cover all the basic operations and function in algebra topic. The elementary math is the main part covered arithmetic. All the basic operation is presented in this area. This is covering all basic operation of addition multiplication, subtraction, and division. The elementary math methods cover kindergarten level to middle school level. This methods also helps to solve real life problems.

Example: 3x+ 2 = 10

Elementary Math Methods Cover

Elementary algebra lessons are contain this basic methods

Arithmetic Operations:

The real numbers have the following properties:

a + b= b +  a    ab  = ba                            (Commutative Law)

(a+ b)+ c= a+ (b + c)      (ab)c = a(bc)        (Associative Law)

a+(b +c)= ab +ac                                       (Distributive law)

Fractions:

To add two fractions numbers with the same denominator, we use the Distributive Law property:

` a/b+c/b` = `1/(b*a)` +`1/(b*c)`   =`1/(b(a+c))`   =`(a+c)/b`

To add two fraction with different denominators, we use a frequent denominator:

`a/b+c/d` =  `(ad+bc)/(bd)`

Factoring

Here we can make use of Distributive Law to expand certain algebraic conditions. In rare case we need to repeal this method (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest condition occurs when the given expression has a common factor as given below,

3x(x-2)=3x2 – 6

Example Problems i Elementary Math Methods

Which is the larger number -13 or -16?

Solution:

The large number in the negative signed numbers we considered, which number is zero to negative side have been going that numbers small number. And near to zero numbers are called large numbers

-16,-15,,-14,-13,-12,-111,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0

so,-13 is larger than -16.

2. List all the integers between -2 and 4.

Solution:

-2,-1, 0, 1,2,3,4 these number are present in the -2 ,4

The -2, 4 between numbers are -1,0,1,2,3

3. x-4=8?

We add +4 on both sides

x=8+4

x=12

4. Simplify 8 `-:` ` 2/3` ?

Solution:

We divide the given equation this is simple method the division inverse of multiplication

8

4*3   = 12

5. Reduce `14/35` .?

Solution:

`14/35`

We divide 7 on both sides

`2/5`

6.12x=4?

Solution:

x=`4/12`

x=`1/3`

7. Factorize the given expression x2-9   ?\

Solution;

The general form of

(a2-b2)=(a-b)(a+b)

So, x2-9=  (x-3)(x+3)

8. Simplify `1/2` + `2/3` ?

Solution   :

`1/2` + `2/3`

We take L.C. M   on 2, 3   

`(1*3)/(2*3)` +`(2*2)/(3*2) `

=`3/6` +`4/6`  =`(3+4)/6`

=`7/6`

9. 3X+4y=5x-2y ?

Solution:

4Y+2y=5x-3x

6y=2x

y=`(2x)/6`

y=`x/3`