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

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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.

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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?