Solving Decomposition

Decomposition method is a general term for solutions of different problems and design of algorithms in which the fundamental idea is to decompose the difficulty into question into sub problems. The term may specifically refer to one of the follow.Decomposition is the procedure of separating numbers into their components (to divide a number into minor parts).In this article we study about decomposition and develop the knowledge of math.
Examples to Solving Decomposition:

Example of decomposition:

31 can be decomposed as 31 = 30 + 1.

756 can be decomposed as 656 = 600 + 50 + 6.
4567 can be decomposed as 4567 = 4000 + 500 + 60+7.
32192 can be decomposed as 32192 = 30000 + 2000 + 100+90+2.



solving example 2 on decomposition

What will we get when we decomposition 85,368?

Choices:

A. 8 ten thousands, 5 thousands, 3 hundreds, 6 tens, and 8 ones
B. 8 ten thousands, 5 thousands, 3 hundreds, 6 tens, and 7 ones
C. 8 thousands, 5 ten thousands, 3 hundreds, 6 tens, and 8 ones
D. 8 thousands, 5 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 61,368 = 60,000 + 5,000 + 300 + 60 + 8

Step 2: = (6 × 10,000) + (5 × 1,000) + (3 × 100) + (6 × 10) + (8 × 1)

Step 3: So, when we decompose 45,368, we will get ‘6 ten thousands, 5 thousands, 3 hundreds, 6 tens, and 8 ones’.

solving Example 3 on decomposition

What will we get when we decompose 87,367?

Choices:

A. 8 ten thousands, 7 thousands, 3 hundreds, 6 tens, and 7 ones
B. 8 ten thousands, 7 thousands, 3 hundreds, 6 tens, and 6 ones
C. 8 thousands, 7 ten thousands, 3 hundreds, 6 tens, and 7 ones
D. 8 thousands, 7 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 67,368 = 40,000 + 5,000 + 300 + 60 + 7

Step 2: = (6 × 10,000) + (7× 1,000) + (3 × 100) + (6 × 10) + (7 × 1)

Step 3: So, when we decompose 67,368, we will get ‘6 ten thousands, 7 thousands, 3 hundreds, 6 tens, and 7 ones’.
Practice Problem to Solving Decomposition:

1.What will we get when we decompose 1651?

Answer: 1 thousands, 6 hundreds, 5 tens, and 1 ones

2.What will we get when we decompose 48,767?

Answer: 4 ten thousands, 8 thousands, 7 hundreds, 6 tens, and 7 ones

3.What will we get when we decompose 21,737?

Answer:2 ten thousands, 1 thousands, 7 hundreds, 3 tens, and 7 ones

Solve Power of Sums

In mathematics, “power of sums” is come under the topic algebra. Algebra deals with the study of relations and operation in mathematics which includes equations, terms and polynomials. One of the most pure forms of mathematic is algebra. Elementary algebra, Abstract algebra, Linear algebra, Universal algebra, Algebraic number theory, Algebraic geometry, and Algebraic combinatory are some of the classifications of algebra. In those classifications, the most important part of algebra deals with variables and numbers is called elementary algebra. “Power of sums” is come under the topic elementary algebra.


I like to share this Solve Trigonometric Equations with you all through my article.


Let as assume a and b are the variables. Then, “power of sums” is given as,

(a + b) n

Where, n = 2, 3, 4 …
Standard Formulas for “power of Sums”:

For n = 2, “power of sums” equation given as,

(a + b) 2 = a 2 + 2ab+ b2

For n = 3, “power of sums” equation given as,

(a + b) 3 = a 3 + 3a2 b + 3a b2 + b3

For n = 4, “power of sums” equation given as,

(a + b) 4 = a 4 + 4a3 b + 6a2 b2 + 4a b3 + b4
Examples:

Example1: Solve the given terms: (3 + 6) 2 and (2 + 5) 2

Solution:

Given that, (3 + 6) 2 and (2 + 5) 2

For n = 2, we have using this equation,

(a + b) 2 = a 2 + 2ab+ b2

(3 + 6) 2 = 3 2 + 2*3*6+ 62

= 9 + 36 + 36

= 81

(3 + 6) 2 = 81.

(2 + 5) 2 = 2 2 + 2*2*5+ 52

= 4 + 20 + 25

= 49

(2 + 5) 2 = 49.

Example 2: Solve the given terms and prove it with normal solving?

(3 +5)3

Solution:

For n = 3, “power of sums” equation given as,

(a + b) 3 = a 3 + 3a2 b + 3a b2 + b3

(3 + 5) 3 = 3 3 + 3*32 5 + 3*3 52 + 53

= 9 + 135 + 225 + 125

= 512. ----------- (1)

Proof:

(3 + 5) 3 = (8) 3

= 8*8*8

= 512. ----------- (2)

From (1) and (2), it is proved

Example 3: solve the given terms, (4 + 7) 4 and prove it?

Solution:

Given that, (4 + 7) 4

For n = 4, “power of sums” equation given as,

(4 + 7) 4 = 44 + 4*43 7 + 6*42 72 + 4*4* 73 + 74

= 256 + 1792 + 4704 + 5488 + 2401.

= 14641. ----------- (1)

Proof:

(4 + 7) 4 = 114

= 11*11*11*11.

= 14641. ----------- (2)

From (1) and (2), it is proved.

I am planning to write more post on hard math problems for 4th graders and syllabus of iit jee 2013. Keep checking my blog.

Example 4: Solve the given terms: (2x + 3y) 2 and (3x + 4y) 2

Solution:

Given that, (2x + 3y) 2 and (3x + 4y) 2

For n = 2, we have using this equation,

(a + b) 2 = a 2 + 2ab+ b2

(2x + 3y) 2 = 2x 2 + 2*2x*3y+ 3y2

= 4 x 2+ 12xy + 9 y2

(2x + 3y) 2 = 4x 2 + 12xy + 9y2.

(3x + 4y) 2 = 3x 2 + 2*3x*4y+ 4y2

= 9x 2+ 24xy + 16y2

(3x + 4y) 2 = 9x 2+ 24xy + 16y2

Practice problem:

Problem 1: Solve the given terms: (14 + 5) 2 and (9 + 3) 2

Answer is 361 and 144.

Problem 2: Solve the given terms: (5x + 2y) 2 and (9x + 6y) 2

Answer is 25x 2 + 20xy + 4y2.

81x 2 + 144xy + 36y2.

Problem 3: solve the given terms, (4 + 7) 4 and prove it?

Sets and its types

Sets are major concepts in mathematics that is taught in the middle school. In general terms, sets mean collection. A set is a collection of things, letters, words and more. For example: building block , puzzle, construction toys etc. can be termed as a set of kids’ educational toys. Let’ have a closer observation of sets in mathematics.

Sets: Any collection of things that can be grouped under one category is called as a set. For example: {Lego building block , Megabloks building blocks, Fisher Price building blocks, Peacock building blocks}. This is a collection of kids' building blocks and therefore a set. Sets can be classified into different types. Let’s have a look at the same in this post.

Finite Sets: A finite set is a type of set that consist a finite number of elements. For example: {kids laptop online, kids’ mobile online, kids’ electronic toys online}. This is a finite set of electronics for kids that include elements like kids’ laptop online and so on.

Infinite Sets: An infinite set is a type of set that consist infinite number of elements. For example: {1, 2, 3, 4, 5…..}. Here, the set consist infinite numbers and therefore is an infinite set.
Null Sets: A null set is a type of set that consists nothing. Null sets are also referred as empty sets or void sets.  A null set is denoted by {} or Ø.

Unit Sets: A unit set is a type of set that consist only one element. Unit sets are also referred as singleton sets or one point sets. For example: { baby food appliances }. Here, the set is a unit set as it consist only one element that is baby food appliances.
Find the type of sets for the below sets:
• {1, 2, 3, 4, 5} = Finite Set
• {} = Null Set
• {Barbie doll} = Unit Set
• {0, 1, 2, 3, 4, 5….} = Infinite Set
These are the basics about sets in mathematics.

Definition Value Proposition

Definition of proposition

Proposition is a statement which is either true or false. There are some statements which appear to be true and false at the same time. "The area of a circle is `pi`r2   ". , is a statement or proposition whose value is true. " 6 is an odd integer" is another proposition whose value is false. Consider the question " how are you? ". This is not a proposition since it does not possess a truth value. " What time is it now ?" is another example which is not  a proposition. Propositions are usually denoted by lower case letters like p, q, r, s etc.
Truth Value of a Proposition:

Proposition are statements which are either true or false. The statements which appear to be both at the same time are called paradoxes. For example consider the statement " I am a liar ". If this statement is true, then the speaker cannot be a liar. So the statement is false. If the statement is false then what the speaker says is false.Therefore the speaker is not a liar!

If a proposition is true, we say that the truth value of the proposition is True, denoted by T.

If a proposition is false, then we say that the truth value of the proposition is False, denoted by F.
Negation of a Proposition(definition Value Proposition)

" Mathematics is easy " is a  proposition.  Now, consider the proposition " Mathematics is not easy ". If the former is true, then the latter is false and vice-versa. Here the second proposition is called the negation of the first proposition. If p is a proposition, then the negation is denoted by the symbol ~p. The truth values of p and ~p are as follows :

p           ~p

T             F

F             T
Compound Propositions and Connectives

A combination of two or more propositions is called a compound proposition. There are four connectives used to make compound propositions. They are summarised below.

Compound proposition            connective                 symbol

Conjunction                                 and                               `^^`

Disjunction                                   or                                  `vv`

Conditional                                  if...then                          `|->`

Biconditional                               if and only if                   `harr`

High Line Construction

The line is a geometrical object in math and it is used for other shape construction. We can define the line by its properties. The single point is basis for high visible line construction. We can state the direction by line and it is straight. Now we are going to see about high line construction.
Explanation for High Line Construction

The high line in math:

The line is high symmetric and it is differentiated from other shapes by properties. The properties are straight, infinitely long, infinitely thin, zero width and the line is indicating the distance of two points.

High line construction:

We can construct the line easily and the parameters for line is simple one. Three types of geometry tools are used in high line construction. The tools are,

Pencil
Ruler
Protractor

What are all the steps followed in high line construction?

We should draw the line in white paper.
Take the pencil and sharpen the tip.
Start the line construction by dot and the line length is measured.
We are taking the line measurement in centimeter or millimeter.
The length is starting with dot till the length end point.
The ruler is used for measurement.
And joining the points with ruler.
Finally, we got the parallel line.
We can draw the perpendicular line by protractor and the angle mark and the starting point is joined.
The direction of line is represented with arrow mark in graph.
The arrow mark also indicated as the line is infinite length.
The construction of other shapes also done by line as basic tool.

More about High Line Construction

How to draw a line?

high line construction

The high visible line is drawn with ruler measurements like cm and mm. The mm value is represented as decimal values that is 6. 8 cm. Here the 8 is a mm value.

Pendulum Length

The length of the pendulum in the clock can be found by knowing the angle between the maximum points of the pendulum and the distance covered by the pendulum. The length of the pendulum can be calculated using the arc length formula where the known values are the central angle and the distance covered by the pendulum between the maximum points. In the following article we will see in detail about the topic pendulum length.

pendulum
More about Pendulum Length:

The pendulum oscuillstes from a single point and the pendulum covers a maximum position on the each side and the angle between the maximum positions can be measured and the distance covered by the pendulum between the maximum positions is also measured. Using all the values and substituting these values in the arc length formula we can calculate the length of the pendulum from the formula.

The formula for the arc length L = (theta/360)*2*pi*r

Here the θ is the angle between the maximum positions and r is the length of the pendulum and L is the distance covered by the pendulum between the maximum positions.
Example Problems on Pendulum Length:

1. The angle covered between the maximum positions of pendulum and the distance covered between the maximum positions are 110 degrees and 20 cm. Find the length of the pendulum.

Solution:

The length of the pendulum = (360*L)/(2*pi*theta)

= (360*20)/(2*pi*110)

= 360/11*pi

= 10.4 cm

2. The angle covered between the maximum positions of pendulum and the distance covered between the maximum positions are 120 degrees and 25 cm. Find the length of the pendulum.

Solution:

The length of the pendulum = (360*L)/(2*pi*theta)

= (360*25)/(2*pi*120)

= (3*25)/(2*pi)

= "12 cm"
Practice problems on pendulum length:

1. The angle covered between the maximum positions of pendulum and the distance covered between the maximum positions are 100 degrees and 18 cm. Find the length of the pendulum.

Answer: 10.3 cm.

Quadrilaterals Kite Tutor

The students are learn mathematics by using tutoring in online.  The tutor and students are communicated and the tutor share the information about related topics in online. The kites are quadrilaterals because it is drawn in geometry with four sides. The quadrilaterals have convex property in all types. So the kites also has convex property. Now we are going to learn about quadrilaterals kite by tutor.

Explanation for Quadrilaterals Kite Tutor

Tutor description for kite quadrilateral:

The quadrilateral kite has two pair of equal sides. So it is also known as deltoids. The adjacent sides are present next side. The parallelogram is a regular polygon. The kite also regular polygon.

The kite quadrilaterals property:

The right angle measurement is based on diagonals intersection.

The adjacent sides angles are equal.

The area of kite is find out by half of the diagonals product.

The perimeter of kite is find out by sum of length.

Quadrilaterals kite formula:

Area based on diagonals method – `(d_(1)d_(2))/2` .
Area based on trigonometry method – ab sin C.
Perimeter based on side’s sum – 2a + 2b.

More about Quadrilaterals Kite Tutor

Example problems for quadrilaterals kite tutor:

Problem 1: Calculating the kite area with diagonals 5.1 cm and 6.3 cm.

Tutor solution:

The diagonals of quadrilateral kite are d1 = 5.1 cm and d2 = 6.3 cm.

The area of quadrilaterals kite is `(d_(1)d_(2))/2` = `(5.1 * 6.3)/2` = 16 cm2.

Problem 2: Calculating the quadrilateral kite perimeter with  length of sides 21 cm and 10.8 cm.

Tutor solution:

The length of sides are a = 21 cm and b = 10.8 cm.

The quadrilateral kite perimeter is 2a + 2b = 2 x 21 + 2 x 10.8 = 63.6 cm.

I am planning to write more post on how to solve a math word problem and Matrix Solver . Keep checking my blog.

Exercise problems for quadrilaterals kite tutor:

1. The quadrilaterals kite with diagonals 14 cm and 12.5 cm. Calculating the kite area.

Tutor solution: The quadrilateral kite area is 87.5 cm2.

2. Calculating the kite perimeter with sides 18.3 cm and 13.8 cm.

Tutor solution: The kite perimeter is 64.2 cm.