Learning Sample Space

Percentage is a number expressed  as a fraction of  100. The word percent is short form of the Latin word "Percentum" meaning out of hundred.  It means ratio of some number to hundred. We use the sign" %" to denote percentage.  It is basically a fraction, with 100 as the denominator and the number as the numerator.

For example 25% :

25% = (25)/(100)

In everyday life, we relate percentage to profit, loss, discount,rate of interest in banks, sales tax, income tax.

(i) Convert a ratio into percent , we write it as a fraction and multiply it with 100.

Example:  To express 20 : 50 as percent , we write (20)/(50) x 100 = 40%

(ii) To convert decimal into percent  we also multiply with 100

Example:   0.175 to percent   0.175 x 100 = 17.5%

(iii) To convert percent to fraction or decimal.  We drop the sign % and divide the remaining number by 100

Example:   45% = (45)/(100)

(iv) To find the certain percent of a given quantity, we multiply the given percent with the given quantity

Example:  30 % of 120   = (30)/(100) x 120 = 36

Some examples on solve finding percentages

1) what is 25 % of 200 ?

Solution:   (25)/(100)  x 200   = 0.25 x 200 = 50

2) What is 45% of 20 ?

Solution:   (45)/(100) x 20 = 0.45 x 20 = 9

3) We have to pay sales tax of 2% on product for $6.25. What is the amount we pay in all for that product.

Solution:  2% = (2)/(100) = 0.02

0.02 x 6.25 = 0.125

Total amount paid is $ ( 6.75 +0.125) = $  6.875

4) A shoe shop marks 40% discount on each pair of shoes , what is the cost price of a pair of shoes that cost $ 67.

Solution:     40 % = (40)/(100) = 0.4

0.4 x 67 = 26.8,

Selling price = cost price - discount = 67 - 26.8 = $ 40.2

5) 43 % of a number is 53.75, find the number .

Solution:   Let the number be x

43% = (43)/(100) = 0.43

0.43 x  = 53.75   divide by 0.43

x = (53.75)/(0.43) = 125

Word problems on solve finding percentages

1) Mary had 24 pages to write .  By the evening, she completed 25% of her work .  How many pages did she have left ?

Solution:    Mary completed 25% means  (25)/(100) x 24  = 6 pages
Number of pages left is 24 - 6 = 18 pages

2) John's income is 20% more than that of Tom .  How much percent is the income of Tom less than that of Jim.

Solution:   Let Tom's income is  $ 100

John's income is 20% more means 20 + 100 = $ 120

Tom's income is 100

Jim's income is 120

Tom's income is $20

% is  (20)/(120) x 100 = (100)/(6) % = (50)/(3) % = 16 (2)/(3) %

3)A 90 kg solution has 10% salt .  How much water must be evaporated to leave the solution  with 20% salt.

Solution:    Let amount of water evaporated is x
90 kg solution has 10  % salt means   (10)/(100) x 90  = 9 kg
After evaporation total quantity left is ( 90 -x )
In this solution salt is 20% means   (20)/(100)  (90-x) 20/(100) = (1)/(5) (90-x)
Quantity of salt remains same in both of these , as only water evaporated
(1)/(5) (90 - x) = 9
90 - x = 9 . 5,   90 -x = 45 ,  90 -45 = x ,  x = 45
45 Kg water evaporated.

Logistic Growth

Logistic growth is always explained with the help of the logistic curve. Logistic curve can be shown in form of sigmoid curve. Logistic growth models the S-shaped curve of growth of  set P, where P might be referred as  population. At the initial stage of the growth the graph is in exponential then saturation begins and the growth slows and finally at maturity the growth stops.

I like to share this Logistic Regression Model with you all through my article.

Learn the definition of Logistic growth:

learn the Logistic Growth formula:

P (t) = 1/ (1+e –t)

where P is the population and t is considered as a time.

The S-curve is obtained if the range of the time over the real numbers from −∞ to +∞. In practice, due to the nature of the exponential function e−t, it is then enough to calculate time t over a small range of real numbers like [−7, +7].

Learn the derivative which is most commonly used:

d/dt P(t) = P(t).(1-P(t))

The computation of function P is

1-P(t) = P(-t)


Learn the logistic differential equation:

learn the logistic growth function, it can be used to calculate the first order nonlinear differential equation.

d/dt P(t) =P(t)(1- P(t))

Here P is a variable with respect to time t and by applying the condition P(0) we can get ½ .

One may readily find the solution to be

P(t) = et / (e t + e c )

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Decide the constant of integration e c = 1 gives the other well-known form of the definition of the logistic curve

P (t) = e t / (e t + 1) = 1 / (1 + e –t)

The logistic curve demonstrates the exponential increase for negative t, which can slow down the curve to linear growth. It is in the slope of 1/4 at t = 0, then it approach exponentially decaying gap at y=1

The relationship among the logistic sigmoid function and the hyperbolic tangent is given by,

2P(t) = 1 + tanh (t/2)

Laws of Exponents

The laws of exponents are used for combining exponents of numbers. Exponents is a number raised to another number, it is denoted as,   a n,  here, n is known as the exponent of the nth power of a.

Laws of exponent are as follows:

x1 = x

x0 = 1

Negative exponent

x - n = (1)/(x^n)

Multiplication law of exponent

x a x b =  (x) a+b

Division law of exponent

(x^a)/(x^b)   =  (x) a-b

Power of power law of exponent

(x a) b  =  x ab

(xy)a   =  xa y a(x/y)^a = (x^a)/(y^a)

Fractional law of exponent x^(a/b)     =    (x^a)^(1/b)  =   root(b)(x^a)


Examples on laws of exponents

1)  Solve  the exponent (32) 5 =  (3) 2x5 = (3) 10      ( using Multiplication law)

2) Solve the exponent  (5^4)/(5)  =  (5) 4-1 =  5 3 = 125    (Using division law)

3) Simplify the exponent 2 (- 5)  = (1)/(2^5)  = (1)/(32)

4) Simplify the exponent  (1/4)^(-3) = (1)/[(1/4)^3] = (1)/(1^3/4^3)  = (4^3)/(1^3) = 4 3 = 64   (using Division law)

5)Simplify using the law of exponents  (sqrt(4) ) -3 = (4^(1/2))^(-3)        (using fractional law)

=  (4)^[(1/2)*(-3)]   (using power of power  law)

= 4^(-3/2)   (using multiplication law)

=  (1)/(4^(3/2))           (using negative law)

= (1)/((4^3)^(1/2)) = (1)/((64)^(1/2))

= (1)/((8^2)^(1/2)) = (1)/(8^(2*(1/2))) = (1)/(8)

Solved examples

Below are the solved examples on laws of exponents:

1)Solve the exponents  3 7 * 3 2 = 3 (7+2) = 3 9       (using Multiplication law)

2) Solve the exponents 2 (-3) * (-7) (-3) =  (2 * (-7)) (-3) =  (-14) (-3)    (using Power of power law)

3) root(3)((343)^-2)  = (343^(-2))^(1/3)      (using Fractional law)

=  (343^(1/3))^(-2) =  (1)/((343^(1/3))^2)   ( using negativel law)

=  (1)/(7^3^(1/3))^2   = (1)/(7^(3*1/3))^2       (using power of power law)

=  (1)/(7^2)  =   (1)/(49)

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4) Simplify using the law of exponents [{(1/5)^(-2)}^2]^(-1) =  {(1/5)^(-2)}^(2*(-1))

=  {(1/5)^(-2)}^(-2)

= (1/5)^((-2)*(-2))

= (1/5)^4

= (1^4)/(5^4) = (1)/(625)

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 )


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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.