Learning Line Segment

A Line segment can be defined as the line joining two end points. Each and every point of the line lies between the end points. For example for line segment is triangle sides and square sides. In a polygon, the end points are the vertices, then the line joining the vertices are said to be an edge or adjacent vertices or diagonal. If both the end points lie on a curve, then the line segment is said to be chord.

Definition to line segment:

Let us see the learning of line segment,

If S is a space of vector lies on A or B, and H is an element of V, then H is a line segment if H can be given by,

H = {i + tj| t `in`|0,1|}

for vectors i, j`in`S having vectors are i and i+j which are known as the end points of H.

Often one wants to differentiate "open line segments" and "closed line segments". Then he explains a closed line segment ,and open line segment as an element of L it can be given by

H = {i+ tj| t`in` |0,1|}

for vectors i, j`in`S,

This is the definition to learning line segment.


Properties of line segment:

Some properties are there to learning line segment,

A line segment is a non zero set,connected together.
If S is a space of vector , then a closed line segment is a closed element in S. thus, an open line segment is an           open element in E if and only if S is one-dimensional.
The above are the features of the line segment.

In proofs:

In geometry to learning line segment, it is defined that a point D is between two other points C and D, if the distance CD added to the distance DE is equal to the distance CE.

Line Segments learning plays a key role in other fields. Such as, the group is convex, if the line joins two end points of the group is lie in the group.

Mathematics Form 3 Exercise

Mathematics form is structure of solution formula.It is used to solve the problem in nice manner.mathematics has different types.They are algebra,geometry,differentiation,integration and so on.We can easily evaluate the problems by using this forms and each has standard formulas.Mathematics fully based on parameters.In problem analysis,forms are very important.Forms of mathematics are use the notations and language is some hard to understand.

Example exercise for mathematics form

In maths, theorems are very important one because it is basis for entire application process.Each theorems should have proof.These proof derived by mathematicians.

Axioms are basic in structure of form and it is string of symbols.

Algebra:

Algebra is rules of operations and relations.It is one branch of pure mathematics.In this, variables representing numbers.It include terms,polynomials and equations.It allows arithmetic law and references to unknown number and functional relationship.

Example: x=y+5, x=y2

These forms are fully based on parameters when we given.

Sets:

Set is collection of unique objects.It is fundamental concept in mathematics.Venn diagram is used to evaluate the set problem.Set is based on members.One is subset and another one is power set.It is use some basic operations. They are union, intersection, complement and cartesian of product.

Example: 1). A={1,3,4,5}

2). C={blue, red, yellow}

Trignometry:

Trignometry is study the triangle.Pure mathematics and applied mathematics use the trignometry.Internal angle of triangle is based on sum operation.Commonly use the basic laws. Sine,cosine and tangent are important basic law.

Example: 1). cosA+sinB

2). SinA+sinB



Exercise for mathematics form

Exercise for algebra form:

1). Evaluate the equation x=2y using y=1,2,3.

Answer: x=2 x 1=2

x=2 x 2=4

x=2 x 3=6.

Exercise for set:

1). Find the set A combined with set B.

A={2,3,4,5} B={2,3,6,7}

Answer: AÙB={2,3,4,5,2,3,6,7}.

Exercise for trignometry:

1). Find the third angle of triangle from given angle.

A= 40° and B=40°

Answer: Using sum of internal angle rule,

A+B+C=180°

40°+40°+C=180°

C=180°- 80°

C=100°

Evaluating Definite Integrals

The definite integral f(x) between the limits x=a and x=b is defined by

int_a^bf(x)dx and its value is F(b) - F(a).

Here a is called the lower limit and b is the the upper limit of the integral, and F(x) is integral of f(x).The value fo the definite integral is obtained by finding out the indefinite integral first and then substituting the upper limit and lower limit for the variable in the indefinite integral.

Please express your views of this topic Evaluate the Definite Integral by commenting on blog.

Properties of Definite Integral for evaluating definite integrals

Let  int f(x)dx =F(x) + c.

Then int_a^bf(x)dx = F(b) - F(a) = [F(x)]a to b

Property1.

int_a^bf(x)dx =int_a^bf(t)dt

Proof:

int_a^bf(x)dx = [F(x)]a to b = F(b) - F(a).

int_a^bf(t)dt = [F(t)]a to b = F(b) - F(a).

Therefore

int_a^bf(x)dx = int_a^bf(t)dt

Property:2

int_a^bf(x)dx = - int_b^af(x)dx

Proof:

= - int_b^af(x)dx = - [F(x)] b to a

=-[F(a) - F(b)]

=F[b]-F[a].

=int_a^bf(x)dx

Property 3:

int_a^bf(x)dx = int_a^cf(x)dx + int_c^bf(x)dx

Proof:

= int_a^cf(x)dx +int_a^bf(x)dx

=[f()x]a to c + [F(x)]c to b

=F(c) - F(a) + F(b) - F(c).

=F(b) - F(a).

Property 4:

int_a^0f(x)dx = int_0^af(a - x)dx

put a-x=t

dx=-dt

When x=0, t=a, when x=a, t=0.

=- int_a^0f(x)dx

= int_0^af(t)dt

int_0^af(t)dt

int_0^af(x)dx

=int_0^af(a-x)dx.

Using Trignonmentry Problem

Evaluate: int_0^(pi/2)sin2xdx

Solution:

Let I = int_0^(pi/2)sin2xdx

= int_0^(pi/2)sin2[(pi/2)-x]dx

int_0^(pi/2)cos2xdx

Here First I, and Second I

Adding (1) and  (2)

we get 2I=int_0^(pi/2)(sint2x+cos2x)dx

=int_0^(pi/2)dx

=[x] 0 to pi The value for x will be assing as 0 and pi

=(pi /2). The value of pi assume  as 180.

2I = (pi/2 )

I=(pi /4). Answer

I am planning to write more post on factoring degree 3 polynomials and cbse sample papers for class 9 sa2. Keep checking my blog.

Evaluate:

int_0^(pi/2)log (tan x)dx

int_0^(pi/2)log (tan x)dx

int_0^(pi/2)log (tan x)dx

int_0^(pi/2)log (tan(pi/2)- x)dx

= int_0^(pi/2)log (cot x)dx

= Adding both First and Second Equations We get

int_0^(pi/2)[log (tan x +log (cot x))]dx

= int_0^(pi/2)log(tan x cot x)

= int_0^(pi/2)log 1dx

=0

I=0.

Reduction formulae:

A formula which expresses the integral of the nth indexed function interms of that of (n-1) th indexed (or lower). the function is called reduction formulae

Differentiation of Exponential Functions

Derivatives are most propably used to solve an equation by the application of some of the simple properties.

So, let me explain this statement by a simple illustration

If y = sin x which is a trignometric funtion then it's derivative can be taken as y' =cos x .                                                                                                             In this way we can solve the various homogeneous functions with a simple illustration which ought  to implement  various formulae.the few among them are as follows

d(xn)/dx =nxn-1
d(ex)/dx =ex
d(log x)/dx =1? x



The uv theorem of differentiation is applicable only when the given two functions are of different functions like one is of logarthmic and one is of algebraic function.The application of differentiation is mainly used in calculus specially to find out the rate of change

Differential and derivatives of exponential functions.

PARTIAL DIFFERENTIATION

Partial differentiation arise in variety of problems in science and engineering usually the independent variables are scalars for example,pressure, temperature, density, velocity, force ect.To formulate the partial differential equation from the given physical problem and to solve the mathematical problem.

DERIVATIVES OF TRIGNOMETRIC FUNCTIONS

The various trignometric functions like sin x ,cos x, tan x, cot x, sec x, cosec x can all be solved easily with application of derivatives as shown in the first illustration.

The derivative of an odd function is always even.
IF y=f(x) is a homogenous function of degree n in x then the relative error in y is n times the relative error in x.
The change in y is represented by ?y and the change in is represented by ?x.

DERIVATIVES OF HYPERBOLIC FUNTIONS

The hyperbolic funtions of trignomety as sin hx, cos hx, tan hx, cot hx, sec hx,cosec hx can all be implemented with the application of derivatives to solve the problem in few steps and in a simple way.

SOLVED PROBLEMS

x sin x

sol:         u(x)=x   v(x)= sin x

d ? dx  uv( x ) = u(x) d ? dx v(x) + v(x) d ? dx u(x)

= x d ? dx sin x + sin x dx ? dx

=x cos x +sin x

2. log (sin-1 (ex ))

sol:  u =ex    v = sin-1 u    y = log v

=d(u) ?dx ×d (v) ? dx ×dy ? dx

=ex × 1 ? ?1-u 2 × 1? v

=ex ? sin-1 (ex )?1-e2x

3. tan (ex )

sol:       d ? dx tan ( ex ) =sec2 (ex )  d ? dx (ex )

=ex sec2 (ex )


Algebra is widely used in day to day activities watch out for my forthcoming posts on answers for algebra 2 problems and cbse syllabus for class 9. I am sure they will be helpful.

Differentiation of Exponential Functions-Problems .

4) y  =  e2x log x

derivative is done in the following ways.

y'  =     e2x log x  d(2x log x)

y' =  e2x log x   [log x  d(2x)  + 2x d(log x)]

y' = e2x log x   [2 log x  + 2x/x]

y' =e2x log x  [ 2log x + 2]  Answer.

Heights and Distances

The height of a tower or the width of a river can be measured without climbing or crossing it. In this chapter we will show how it is made possible. Some suitable distances and angles will be measured to achieve the above results.

In this chapter often the terms, "Angle of Elevation", "Angle of Depression" are used.

some definitions of heights and distances

Angle of Elevation:

The angle of elevation of the point viewed is defined as the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level.

Let P be the position  of an object above the horizontal line OX where O is the eye of the observer looking at the object. Join OP. Then, angle XOP is called Angle of Elevation.

Angle of Elevation


Angle of Depression:

The angle of depression of a point  viewed is defined as the angle formed by the line of sight with the horizontal when the point is below the horizontal level.

Let P be the position  of an object below the horizontal line OX where O is the eye of the observer looking at the object. Join OP. Then, angle XOP is called Angle of Depression.

Angle of Depression

Solved examples of heights and distances

1. The angle of elevation of the top of  tower from a point 60m from it's foot is 300. What is the height of a tower?

Solution:   heights and distances problem(1)Let AB be the tower with it's foot at A.
Let C be the point of observation.
Given angle ACB=300  and AC = 60m
From right ? BAC :    AB/ AC = tan30
=> AB=60*tan30 = 20?3 m


2. From a ship mast head 100m high, the angle of depression of a boat is tan-1(5/12) . Find it's  distance from a ship?

Solution:   AB = ship mast = 100m, with head at B:boat problem
BD is the horizontal line.C is the boat.
Given angle DBC=`theta` =Tan-1(5/12)=angle of
depression of the boat from B.=> tan`theta` = 5/12.
AC/AB =cot `theta ` = 12/5 =>AC = 100(12/5) = 240m.

Summary of heights and distances :

The distance between two distant objects can be determined with the help of trigonometric ratios.

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.

Fixed Rate Calculator

A loan in which the interest rate do not alter through the whole expression of the loan reverse of adjustable rate A loan in which the interest rate does not modify through the whole term of the loan for an entity taking out a loan when rates are low, the fixed rate loan would permit him or her to "lock in" the short rates and not be worried with fluctuations. On the another hand, if interest rates were in the past elevated at the time of the loan, he or she would profit from a floating rate loan, since as the prime rate cut down to historically normal levels, the rate on the loan would reduce. Reverse of adjustable rate. A calculator is a small electronic device; it is used to achieve the fundamental operations of arithmetic. Modern calculators are more convenient than computers, while most PDAs are similar in size to handheld calculators.

Fixed rate calculator
Examples for the Fixed Rate Calculator:
Fixed rate calculator – Example 1:

Determine the payments and interest for a fixed rate loan, using monthly interest compounding and monthly payments. The purchase price $150, no of monthly payments is 2 months, and interest rate 5.000%, and the payment calculator computes the payment amount for you.
Solution:

Fixed rate calculator

The down payment amount is $11.00

The Loan amount is $ 139.00

Payment amount is $ 69.93

Interest rate is      5.000 %

Interest compounding: Monthly

Total amount financed: $139.00

Total payments: $139.87

Total finance charge: $0.87

Payment schedule:

Date           Payment InterestPrincipal Balance

Loan     08-26-2010                            139.00

1           09-26-2010   69.93       0.58    69.35    69.65

2           10-26-2010   69.94       0.29    69.65

2010 Total                139.87      0.87

Grand Total              139.87      0.87
More Examples for the Fixed Rate Calculator:
Fixed rate calculator – Example 1:

Determine the payments and interest for a fixed rate loan, using monthly interest compounding and monthly payments. The purchase price $100, no of monthly payments is 5 months, and interest rate 5.000%, and the payment calculator computes the payment amount for you.
Solution:

Fixed rate calculator

The down payment amount is $18.02

The Loan amount is $ 89.00

Payment amount is $ 18.02

Interest rate is      5.000 %

Interest compounding: Monthly

Total amount financed: $89.00

Total payments: $90.11

Total finance charge: $1.11

Payment schedule:

Event      Date             Payment Int    PrincipalBalance

Loan       08-26-2010                          89.00

1             09-26-2010  18.02    0.37    17.65  71.35

2             10-26-2010  18.02    0.30    17.72  53.63

3             11-26-2010  18.02    0.22    17.80  35.83

4             12-26-2010  18.02    0.15    17.87  17.96

2010 Total                  72.08    1.04    71.04

5             01-26-2011  18.03    0.07    17.96

2011 Total             18.03    0.07    17.96

Grand Total            90.11    1.11

Score as in Mathematical Terms

“Score” means 20! .The term "score" was came from the old English word “scoru”. It is the Middle age English word. Also, in turn, it came from the Norse skor. In mathematical terms score carried a value 20. Abraham Lincoln's celebrated Gettysburg consists of the phrase, "Four score and 7 years ago". In bible the term mentioned as “three score years and ten”.

Orgin of score:

In mathematical terms "Scores" way "groups of 20", just as "dozens" way "groups of 12". "Score" is eventually from the Proto-Indo-European root (s)ker-, sense "to cut" . Another English offspring is "shear". English still uses the word "score" to submit to an indentation or line made by a sharp instrument. Scores cut in wooden tally firewood were used in including, a fact we still make the unwitting suggestion to today when we talk regarding "keeping score". For a while in the history of the Germanic languages, the word for the cut used to record the number 20 became a word for the number itself.
Examples

4 score

Score mean 20 in mathematical terms

`4xx20=80`

Therefore four score is 80

3 score and 10

Score mean 20 in mathematical terms

`3 x 20` and 10

60+10=70.

Therefore 3 score and 10 is 70.

In Abraham Lincoln's celebrated Gettysburg consists of the phrase, "Four score and 7 years ago".

The phrase denotes, in mathematical terms

`4xx20` and 7

80+7

87

The phrase means 87 years ago.

Three score difference 10

Three score

`3xx20`

60 difference 10

60-10

50

Some mistakes arises that in some cases score is taken as 10. It wii lead to wrong answer

We cant assume any value to score. It means to 20 alone.

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Related examples like score:

Like score there are a lot of examples are in mathematical terms.

Foursquare

It means symmetrical in all dimensions or equal in all dimensions like cubical.

Like the words numbers also holds some meaning, Let take an example

223

It bibbiblically mean things involving to an order of survival beyond the visible, demonstrate-able space; that which is above and away from the laws of nature; and/or matters concerning the stroke or influence of the ghostly or psychic powers and the top-secret knowledge of them.

General Probability

General probability is the part of mathematics that learns the feasible outcomes of given events together with the outcomes' relative possibilities and distributions. In general usage, the word "probability" is used to mean the opportunity that a particular event will occur expressed on a linear scale from 0 to 1, also expressed as a percentage between 0 and 100.Now we will see the examples of probability.

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Examples- General Probability

Example 1

In a class there are 8 students got top eight marks in English. The marks are 78,83,86,88,92,93,95,98.  What is the probability for the following outcomes?

i) Select the marks are below 85.

ii) Select the marks between 80 and 90.

Solution:

i) Take P(A) is the probability for the marks are below 85.

Given marks are 78,83,86,88,92,93,95,98.

Total numbers n(S)=8

Here the following marks are below 85 n(A)={78,83}=2

So P(A)=(n(A))/(n(S))

=2/8

=1/4 .

ii) Take P(B) is the probability for the marks between 80 and 90.

The marks 83,86,88  are available between the 80 and 90.So n(B)=3

Total outcomes n(S)=8

So P(B)=(n(B))/(n(S))

= 3/8 .

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Example 2

What is the probability for select the letter ‘G’ from the word ‘GENERAL KNOWLEDGE’?

Solution:

Given word is GENERAL KNOWLEDGE.

Total letters n(S)=16

Number of ‘G’ letter n(A)=2

So the probability=2/16

=1/8 .

Example 3

Paul has 20 caps and Alex has 15 caps. What is the probability for select the Alex’s caps?

Solution:

Paul’s caps n(A)=20

Alex’s caps n(B)=15

Total number of caps n(S)=20+15

=35

Probability for select the Paul’s caps= 15/35

= 3/7 .
Practice Problems- General Probability

1) What is the probability for select the letter ‘E’ from the word ‘ELEMENT’?

Answer:

Probability=3/7 .

2) Tom has 5 white balls, Joseph has 7 yellow balls. What is the probability for getting the yellow balls?

Answer:

Probability=7/12 .

These examples and practice problems are used to study the general probability.

Solving Summation Notation

A summation notation (Σ) is used for sum the numbers or quantities indicated. We can sum all types of numbers such as real, integer, complex, rational numbers using sigma notation. Summation notation is also called as sigma notation. Using summation notation we can solve the problems easily. Summation a concise and convenient way of writing long sums.


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Solving Summation Notation – Solving Example Problems

Example 1: Find the sum for the given expression and explain the terms of summation notation;

7

∑ (n + 1)3 =?

n=1

Solution:

7

∑ (n + 1)3 =?

n=1

Lower Bound: ‘n’ is the lower bound of the summation notation.

Upper Bound:  ‘7’ is the upper bound of the summation notation.

Formula: (n + 1)3 is the formula to describe the each term of the summation notation.

Start: n = 1, here 1 is indicating the first number of the terms of summation notation. Usually value of start is zero (0) or one (1).

End: 7 is indicating the last umber of the terms of the summation notation.

Index Variable: Index variable ‘n’ is used to labeled each term of summation notation.

7

∑ (n + 1)3 = 8 + 27 + 64 + 125 + 216 + 343 + 512 = 1295

n=1

7

Therefore, ∑ i3 = 1295

i=1

Example 2: Solve and find the sum for the given expression;

6

∑ 3(n + n2) =?

n=1

Solution:

6

∑ 3(n + n2) =?

n=1

Formula: 3(n + n2), starting term: 1, end term: 6, and index variable: n

6

∑ 3(n + n2) / n = 3(1 + 12) + 3(2 + 22) + 3(3 + 32) + 3(4 + 42) + 3(5 + 52) + 3(6 + 62)

i=1


= 3(1 + 1) + 3(2 + 4) + 3(3 + 9) + 3(4 + 16) + 3(5 + 25) + 3(6 + 36)

= 3(2) + 3(6) + 3(12) + 3(20) + 3(30) + 3(42)

= 6 + 18 + 36 + 60 + 90 + 126

= 336

6

Therefore, ∑ 3(n + n2) = 336

i=1

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Example 3: Solve and find the sum for the given expression;

6

∑ n(2n + 1) / 2 =?

n=1

Solution:

6

∑ n(2n + 1) / 2 =?

n=1

Formula: n(2n + 1) / 2, starting term: 1, end term: 6, and index variable: n

6

∑ n(2n + 1) / 2 = (1(2(1) + 1) / 2) + (2(2(2) + 1) / 2) + (3(2(3) + 1) / 2) + (4(2(4) + 1) / 2)

n=1  + (5(2(5) + 1) / 2) + (6(2(6) + 1) / 2)


= (1(3) / 2) + (2(5) / 2) + (3(7) / 2) + (4(9) / 2) + (5(11) / 2) + (6(13) / 2)

= (3 / 2) + (10 / 2) + (21 / 2) + (36 / 2) + (55 / 2) + (78 / 2)

= 1.5 + 5 + 10.5 + 18 + 27.5 + 39

= 101.5

6

Therefore, ∑ n(2n + 1) / 2 = 101.5

n=1
Solving Summation Notation – Solving Practice Problems

Problem 1: Solve and calculate the sum for the given expression;

5

∑ n2(n) =?

n=1

Answer: 225

Problem 2: Solve and calculate the sum for the given expression;

5

∑ (2n + 2) / n =?

n=1

Answer: 14.57

Define Expanded Form

Expanded Form is a method to break up a number to illustrate how much each digit in the number represents and in other words, expanded form is the method of pulling a number separately and expressing it as a sum of the values of every digit. In this article we study about expand form define of and develop the knowldge of the math .

Example on Define Expanded Form

Choose the expanded form of the number 15,793.

Solution:

The expanded form of the number 15,793 can be found using a place value chart.

Now, let’s aim to write the number in words and define how it helps us identify the expanded form.

The number 15,793 is read as: 15 thousands 7 hundreds 93 which is the same as 1 ten thousand  5 thousands 7 hundreds 9 tens and 3 ones.

So, the expanded form of 15,793 is 10,000 + 5,000 + 700 + 90 + 3.

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Example to Define Expand Form Math:

Example 1:

57  can be expand as 57 = 50 + 7.

226 can be expand as 296 = 200 + 20 + 6.

5467 can be expand as 5467 = 5000 + 400 + 60+7.

89192 can be expand as 89192 = 80000 + 9000 + 100+90+2

Example 2:

What will we get when we expand form 23,368?

Choices:
A. 2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 8 ones
B. 2 ten thousands, 3 thousands, 6 hundreds, 6 tens, and 7 ones
C. 5 thousands, 5 ten thousands, 6 hundreds, 6 tens, and 8 ones
D. 5 thousands, 8 hundreds, 3 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 23,368 = 20,000 + 3,000 + 300 + 60 + 8
Step 2: = (2 × 10,000) + (3 × 1,000) + (3 × 100) + (6 × 10) + (8 × 1)
Step 3: So, when we expand 23,368, we will get ‘2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 8 ones’.

Example 3:

What will we get when we expand form 13,369?

Choices:
A. 1 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 9 ones
B. 2 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 5 ones
C. 3 thousands, 7 ten thousands, 9 hundreds, 6 tens, and 9 ones
D. 4 thousands, 7 hundreds, 9 tens, and 68 ones

Correct Answer: A

Solution:

Step 1: 13,369 = 10,000 + 3,000 + 300 + 60 + 9
Step 2: = (1 × 10,000) + (3× 1,000) + (3 × 100) + (6 × 10) + (9 × 1)
Step 3: So, when we expand 13,368, we will get ‘1 ten thousands, 3 thousands, 3 hundreds, 6 tens, and 9 ones’.

Systematic Sampling Method

In this article, let us study what is sampling and the types of sampling.

Preliminary concepts of sampling:

In testing the quality of bulbs produced, by a company it is impossible to test every bub manufactured by a company; it is quite sufficient if a sample is taken for testing and based on this test it is possible to draw conclusion about the population. If each and every item is tested, in some cases it may not be possible to send them to the market. thus, process of sampling is to get information about the population from the sample. To the extent to which we can do this with any accuracy depends on the choice of our sample or samples.

Samples are classified as follows:

non-probability samples

probability samples

Some non -probability samples  are 1. deliberate sampling , 2. Quota sampling 3. Block sampling.

Some probability samples  are1. Random sampling , 2. Stratified sampling , 3. Systematic sampling , 4. Multi stage sampling
What is Systematic Sampling

Systematic sampling with a random beginning is a form of restricted random sampling which is highly useful in surveys concerning enumerable population. In this method, every member of the population is numbered in a serial order and every ith element, starting from any of the first i items is chosen.

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Example for Systematic Sampling:

Suppose we require 5% sample of students from a college where there are 2000 students and where everyone is given a departmental number from 1 to 2000, we should first select one number at random from 1 to 20. Suppose the number chosen is 12; then our sample consists of students with departmental numbers 12, 32, 52, 72,..1992

If, however, the students are arranged in groups of 20 such that the first student belongs to the first of 20 specified communities, the second to a second community and so on, then our sample may consist of students of all belonging to the same community and consequently we cannot consider the sample to be random

Integral of ln u

Integral calculus is a branch of calculus that deals with integration. If f is a function of real variable x, then the definite integral is given by

int_a^bf(x)dx

Where a and b are intervals of a real line.

Integral of ln u means integration of natural logarithmic (ln) function with respect to the variable ' u '. In this article, we are going to see the list of natural logarithmic integral (ln) rules with few example problems.
List of Natural Logarithmic (ln) Integral Rules:

int 1/u dx = ln |u| + C

int  ln u du = uln u  - u + C

int uln u du = (u^2)/2 ln u  - 1/4 u2 + C

int u2 ln u du = (u^3)/3 ln u - 1/9 u3 + C

int (ln u)2 du = u(ln u)2 - 2uln u + 2u + C

int ((ln u)^n)/u dx = 1/(n + 1) (ln u)n+1 + C

int  (du)/(uln u) = ln (ln u) + C

Learning Natural Logarithmic (ln) Integral Rules with Example Problems:

Example problem 1:

Integrate the function f(u) = 5/u

Solution:

Step 1: Given function

f(u) = 5/u

int f(u) du = int5/u du

Step 2: Integrate the given function f(u) = 5/u with respect to ' u',

int5/u du = 5ln (u)+ C

Example problem 2:

Integrate the function f(u) = log (9u)

Solution:

Step 1: Given function

f(u) = log (9u)

int f(u) du = intlog (9u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let x = log (9u)    dv = du

dx = 1/u  du         v = u

int x dv = xv - int v dx

int log (9u) du = ulog (9u) - int u 1/u du

= ulog (9u) - int du

= ulog(9u) - u + C

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Example problem 3:

Integrate the function f(u) = 10log (3u)

Solution:

Step 1: Given function

f(u) = 10log (3u)

int f(u) du = int10log (3u) du

The above function can be written as

int f(u) du = 10intlog (3u) du

Step 2: Integrate the given function  with respect to ' u' using integration by parts,

Let u = log (3u)    dv = du

du = 1/u  du         v = u

int x dv = xv - int v dx

10 int log (3u) du = 10[ulog (3u) - int u 1/u du]

= 10ulog (3u) - 10int du

= 10ulog(3u) - 10u + C

Equilateral Pentagon Angles

In geometry an equilateral pentagon is a polygon with five sides of equal length. Its five internal angles, in turn, can take several values, thus permitting to form a family of pentagons. In contrast, the regular pentagon is unique, because is equilateral but at the same time its five angles are equal. Four intersecting equal circles disposed in a closed chain, are sufficient to describe an equilateral pentagon. ( Source Wikipedia )


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Examples

Problems 1

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 100, 58 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=100

Angle 3 a3=58

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+100+58+67+x=540

250+125+x=540

375+x=540

Subtract 375 on both sides

375-375+x=540-375

x=165

The missing angle is 165

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Problems 2

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 150, 70, 158 and 67

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=150

Angle 2 a2=70

Angle 3 a3=158

Angle 4 a4=67

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

150+70+158+67+x=540

220+225+x=540

445+x=540

Subtract 375 on both sides

455-455+x=540-455

x=95

The missing angle is 95

Problems 3

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 100, 150, 90 and 40

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=100

Angle 2 a2=150

Angle 3 a3=90

Angle 4 a4=40

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

100+150+90+40+x=540

250+130+x=540

380+x=540

Subtract 375 on both sides

380-380+x=540-380

x=160

The missing angle is 160

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Problems 4

Find the missing angle on the pentagon

Is the four angle on the equilateral pentagon are 155, 120, 60 and 70

Solution

The sum of the 5 angles of pentagon is 540

The given angles are

Angle 1 a1=155

Angle 2 a2=120

Angle 3 a3=60

Angle 4 a4 =70

Let the missing angle be x

Angle 5 a5=x

a1+a2+a3+a4+a5=540

155+120+60+70+x=540

275+130+x=540

405+x=540

Subtract 375 on both sides

405-405+x=540-405

x=135

The missing angle is 135

Distance Formula Proof

For distance formula proof statistics will be deals with the co-ordinates parameters.Formula for distance calculation by the co ordinates of the points given. For calculating distance statistics between the points we have the distance formula proof and we substitute the points in that formula we find the distance between the points.In this article we have the distance formula proof and the problems using the distance formula.
Distance Formula Proof:
Distance between two points co ordinates is a basic concept in geometry.Now, we give an algebraic expression for the same.

Let P1  (x1, y1) and P2 (x2, y2) be two points in a Cartesian plane and denotes the distance between P1 and P2 by d(P1, P2) or  by  P1P2. Draw the line segment ` bar(P_1P_2)`

distance formula proof

The segment ` bar(P_1P_2)` is parallel to the x axis  Then y1 = y2. Draw P1 L and P2 M, perpendicular to the x-axis. Then d(P1,P2) is equal to the distance between L and M. But L is (x1, 0) and M is (x2, 0).

So the length LM = |x1-x2| Hence d (P1, P2) = |x1-x2|.

therefore, [d(P1,P2)]2= |x1-x2|2+ |y1-y2|2

=(x1-x2)2+(y1-y2)2

=(x2-x1)2+(y2-y1)2
d(P1,P2) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2)`

Problems Using Distance Formula Proof:
Example 1:
Find the distance between the points A(-2,3) and B(3,7).

Solution:
Assume d be the distance between A and B.             (x1,y1) = (-2,3)

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                   (x2,y2) =  (3,7).

=`sqrt((3+2)^2 +(7-3)^2)`

=`sqrt(5^2+4^2)`

=`sqrt(25+16)`

=`sqrt41`
Example 2:
Find the distance between the points A(-1,1) and B (3,3).

Solution:
Assume d be the distance between A and B.                (x1,y1)= (-1,1)

Then d (A, B) =`sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                                (x2-y2)= (3,3).

=`sqrt((3+1)^2 +(3-1)^2)`

=`sqrt(4^2+(2)^2)`

=`sqrt(16+4)`

=`sqrt20`
=2`sqrt 5`

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Example 3:
Find the co ordinates distance between the points A(-1,5) and B (1, 4).

Solution:
Assume d be the distance between A and B.                       (x1,y1)= (-1,5)

Then d (A, B) = `sqrt((x_2-x_1)^2+(y_2-y_1)^2) `                          (x2-y2)= (1,4).

=`sqrt((1+1)^2 +(4-5)^2)`

= `sqrt(2^2+(-1)^2)`

=`sqrt(4+1)`
=`sqrt5`