Short Definition of Geometry

Geometry (Ancient Greek: γεωμετρία; geo- "earth", -metria "measurement") "Earth-measuring" is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Initially a body of practical knowledge concerning lengths, areas, and volumes, in the 3rd century BC geometry was put into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a standard for many centuries to follow.( Source- Wikipedia)

Short Basic Geometry Definitions:
Short Definition of Geometry for Square:

In mathematical geometry, a square consists of 4 equal angles every side which includes 90° and 4 equal sides. Square area is calculated by-product of two opposite sides.
Short Definition of Geometry for Rectangle:

In mathematical geometry, a rectangle is equal to the square, it consists of 4 equal angles with 90° and 4 equal sides. Rectangle area is calculated by using the product of length and width.
Short Definition of Geometry for Triangle:

In mathematical geometry, triangles consist of 3 straight lines that are equal length and it is called as equilateral triangle. Triangle area is calculated by half of the product of bottom and height.
Example for Short Definition of Geometry:
To solve the Rectangle problems:

Area of rectangle= Length x width.

Perimeter of rectangle= 2 (Length + Width)

Example 1:

Find the area of rectangle that sides measure 6 centimeters and 4 centimeters respectively.

Solution:

Given that

Length – 6 centimeter.

Breadth – 4 centimeter.

The formula used to find out the area of rectangle is = length * breadth.

Area = length * breadth.

Substitute the given value in the formula

Area   = 6 * 4

The result of 6 and 4 = 24

Hence the area of the rectangle given is 24 square centimeter.

Example 2:

Find out the volume of the cube that sides measure 6 centimeter.

Solution:

Given that

Measure of cube length = 6.

The formula used to find out the volume of the cube is = a3.

While we insert the value in the formula we get the result is

Volume = 63.

= 216

Hence the volume of the given cube is 216 cubic centimeters.

To solve Square problems:

Perimeter of square= 2 * side

Area of square= side * side

Problem 3:

To find out the area and perimeter of square while side length is 4cm?

Solution:

Area of square = (side*side)

=4 * 4

=>16 cm2

Perimeter of square = 4 * side

= 4 * 4

= 16 cm

The Multiplication Rules

In algebra fundamental arithmetic procedure (addition, subtraction, multiplication, and division) in general used in day-to-day life. The multiplication defined as product of numbers. Multiplication is usually expressed in a*b or ab form. The multiplying numbers are having two terms i) Multiplicand and ii) Multiplier. In this article, we are going to discuss about the multiplication rules

The Multiplication Rules – Rules and Example Problems:

The multiplication rules - Rules:

Rule 1: The multiplication of two integers by the connected signs resolve be positive sign

a) Positive `xx` positive = positive

b) Negative `xx` negative = positive

Rule 2:  The multiplication of two integers by the different signs will be negative

a) Positive `xx` negative = negative

b) Negative `xx` positive = negative.

The multiplication rules – Example problems:

Example 1:

Evaluate 6802 `xx` 322

Solution:

6802 `xx` 322

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

13604

13604

20406

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

2190244

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

Therefore, the final answer is 2190244.

Example 2:

Find the solution for given problem

(- 9667) `xx` (- 26)

Solution:

(- 9667) `xx` (- 26)

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

58002

19334

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

251342

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

Therefore, the final answer is 251342.

Example 3:

Find the solution for given problem

8655 `xx` (- 124)

Solution:

8654 `xx` (- 124)

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

34616

17308

8654

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

- 1073096

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

So, the final answer is (- 1073096)
The Multiplication Rules – Practice Problems:

Practice problem 1:

Find the solution for given problem

(- 7897) `xx` (- 53)

Answer: So, the final answer is 418541

Practice problem 2:

Find the solution for given problem

(- 7790) `xx` (- 48)

Answer: So, the final answer is 373920

Practice problem 3:

There are 280 students in every class at the performing arts school. If there are 62 complete classrooms, how many students attend this school?

Answer: 17360 students attend this school

Practice problem 4:

Jesse wants his giant chocolate chip cookies to have 28 chips each. Today he is make 355 cookies. How many chips does he need?

Answer: 9940 chips does he need.

Variance Covariance Formula

Variance: Variance is used in the statistical analysis to find the the extent to which a single variable is varying from its mean value given a set of values.

Variance of a random variable X is often denoted as VAR ( X). It is also denoted by the symbol `sigma` 2X

Covariance: Co-variance is used in statistical analysis to find the extent to which 2 variable are varying together given a set of values for both these variables.

Covariance of two random variable X and Y is denoted as COV ( X,Y). Unlike variance which has the mathematical symbol sigma ( `sigma` ), Covariance doesn't have any kind of symbol to depict it.
Formulae for Variance and Covariance.

Lets consider 2 sets of data namely X and Y such that the values in X are x1 , x2 , x3 , x4 , ......xn and similarly the values in Y are y1 , y2 , y3 , y4 , .......yn

Now lets denote the mean of the X set of variables to be    Xm

Mean Xm = X1 + X2 + X3 +........ + Xn / n

Similarly lets denote the mean of the Y set of variables to be Ym

Mean Ym = Y1 + Y2 + Y3 +.......+ Yn / n

Formulae for Variance of X = ( x1 - xm)2 + (x2 - xm)2 + ( x3 - xm)2 + .......+ (xn - xm)2 / n

VAR ( X ) = (1/n) `sum` (xi - xm)2

Formulae for Variance of Y = ( y1 - ym)2 + (y2 - ym)2 + ( y3 - ym)2 + .......+ (yn - ym)2 / n

VAR ( Y ) = (1/n) `sum` (yi - ym)2

Formulae for Covariance of (X,Y) = ( x1 - xm) ( y1 - ym) + (x2 - xm)(y2 - ym) + ( x3 - xm)( y3 - ym) + .......+ (xn - xm)(yn - ym) / n

COV ( X , Y ) = (1/n) `sum` (xi - xm)(yi - ym)
Example Problem 1 on Variance and co Variance

Lets assume the below set of marks received by Bill and Bob in Maths, Physics, Chemistry, English and Biology for the purpose of solving Variance and Co Variance

table

Based on the Formulae given in the previous paragraph, Lets calculate the Mean Marks for Bill and Bob

Mean Marks for Bill = 16+12+14+18+20 /5 = 16.00

Mean Marks for Bob = 14+18+15+18+20 /5 = 17.00

Now lets apply the formulae of variance and find out the Variance of Bill and Bob

Variance of Bill = (16-16)2 + (12-16)2 + (14-16)2 + (18-16)2 + (20 - 16)2 / 5 = 40 / 5 = 8

Variance of Bob = (14-17)2 + ( 18-17)2 + (15-17)2 + (18-17)2 + (20-17)2 / 5 = 24 / 5 = 4.8

Co variance of ( Bill , Bob ) = (16-16)*(14-17) + (12-16)*(18-17) + (14-16)*(15-17) + (18-16)(18-17) + (20-16)*(20-17) / 5

Co Variance of ( Bill , Bob ) = 21/5 = 4.2

I am planning to write more post on What are Line Segments and Perpendicular and Parallel Lines. Keep checking my blog.

Example Problem 2 on Variance and co Variance

Lets assume a class of 4 student who have been asked to rate their liking toward 2 musical instruments Guitar and Piano on a scale of 10

Based on the above data, lets calculate the Mean of the people liking Guitar and Mean of people liking Piano.

Mean(Guitar) = ( 5+3+7+9)/4 = 24/4 = 6

Mean(Piano) = (5+9+9+5)/4 = 28/4 = 7

Var(Guitar) = ((5-6)2 + (3-6)2 + (7-6)2 + (9-7)2 )/ 4 = (1+9+1+4)/4 = 15/4 = 3.75

Var(Piano) = ((5-7)2 + (9-7)2 + (9-7)2 + (5-7)2 ) / 4 = (4+4+4+4)/4 = 16/4 = 4

Cov(Guitar,Piano) = [(5-6)(5-7) + (3-6)(9-7) + (7-6)(9-7) + (9-7)(5-7) ] / 4 = (2+(-6)+2+(-4))/4 = -6/4 = -1.5

Binomial Probability Function

Binomial probability Problems represented by B(n,p,x). It gives the probability of exactly x successes in ‘n’ Bernoullian trials, p being the probability of success in a trial. The constants n and p are called the parameters of the distribution. A Binomial distribution can be used under the following condition.

(i) any trial, result in a success or a failure

(ii) There are a finite number of trials which are independent.

(iii) The probability of success is the same in each trial.

In a Binomial distribution function mean is always greater than the variance. The binomial probability function example problems and practice problems are given below.
Example Problems - Binomial Probability Function:

Ex 1:  By using the binomial distribution. If the sum of mean and variance is 4.8 for  5 trials find the distribution

Solution:  np + npq = 4.8 , np(1 + q) = 4.8

5 p [1 + (1 − p) = 4.8

p2 − 2p + 0.96 = 0 , p = 1.2 , 0.8

p = 0.8 ; q = 0.2 [p cannot be greater than 1]

The Binomial distribution is P[X = x] = 5Cx (0.8)x (0.2)5−x,  x = 0 to 5

Practice Problem - Binomial Probability Function:

Pro 1: Find the value for p by using the Binomial distribution if n = 5and P(X = 3) = 2P(X = 2).

Ans: p = `(2)/(3)`

Pro 2: Find the probability values by using the binomial distribution.  A pair of dice is thrown 10 times. If getting a doublet is considered a success (i) 4 success (ii) No success.

Ans: (a) (35/216)(5/6)6

(b) (5/6)10

Proportion Equation

The data is obtaining by the comparison of two ratios is called proportion data. Proportion data is represented as a:b = c:d. This proportion data can be written in the form of fraction as `a/b` = `c/d` . Where the pairs of data (a,b) and (c,d) are in proportion. When the proportions are equal, the cross product of the proportion will be also equal. That is, `a/b` = `c/d` can be written as ad=bc.

Examples for Proportion Equation:

Example 1 for proportion equation:

Martin read 63 pages of the book in 33 minutes. How many pages will he be able to read in 43 minutes?

Solution:

Martin takes 33 minutes to read 63 pages.

Martin will take 43 minutes to read x pages.

This can be written as,

`63/33` = `x/43`

Now we have to do the cross multiplication.

63 `xx ` 43 = x `xx` 33

2709 = 33x

This can be written as,

33x = 2709

Now we have to divide both sides by 33.

`(33x)/33` = `2709/33`

x = 82.09 now we have to round it to the unit place.

x = 82

Therefore, Martin will read 90 pages in 45 minutes.

Example 2 for proportion equation:

Paul bought 12 apples for dollar 48.  How many apples will he be able to buy in $ 93?

Solution:

Paul spends $48 for 12 apples.

Paul will spend $93 for x apples.

This can be written as,

`12/48` = `x/93`

Now we have to do the cross multiplication.

12 `xx` 93 = x `xx` 48

1116 = 48x

This can be written as,

48x = 1116

Now we have to divide both sides by 48.

`(48x)/48` = `1116/48`

x = 23

Therefore, Paul can buy 23 apples for $ 93.

Practice Problems for Proportion Equation:

Problem 1 for proportion equation:

Martin read 40 pages of the book in 28 minutes. How many pages will he be able to read in 52 minutes?

Solution: Martin will read 74 pages in 52 minutes.

Problem 2 for proportion equation:

Paul bought 8 apples for dollar 22. How many apples will he be able to buy in $ 66?

Solution: Paul can buy 24 apples for $ 66

Polynomials Calculator

An Algebraic expression is of the form axn is called a monomial. The variable a is called the coefficient of xn and n, the degree of monomial. For example, 7x3 is monomial in x of degree 3 and 7 is the coefficient of x3. The combination of two monomials is called a binomial and the combination of three monomials is called a trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is trinomial. The sum of n number of monomials, where n is finite and x is called a polynomial in x.
Illustration to Polynomials:

Polynomial Calculator Example 1:

The polynomial calculator of the equation x2 + ax + b gives the remainder 18, when divided by x – 2 and leaves the polynomial calculator of remainder –2 when that is been divided by (x + 3).

Find the values of a and b.

Solution to the polynomial calculator:

P(x) = x2 + ax + b.

In tyh is polynomial calculator,

When x – 2 divides P(x) then the remainder is P (2).

∴P (2) = 4 + 2a + b.

But remainder = 18 ⇒ 4 + 2a + b = 18;

2a + b = 14 (1)

When (x + 3) divides P(x)

, the remainder is P (–3).     ∴ P (–3) = (–3)2 + a (–3) + b

= 9 – 3a + b.

But remainder = –2;      ∴ 9 – 3a + b = –2;

⇒ –3a + b = –11 (2)

(1) ⇒   2a + b = 14

(2) ⇒ –3a + b = –11 (subtracting)

5a        = 25

(Or) a     = 5

Substituting a = 5 in equation (1) we get

10 + b = 14; b = 4, ∴ a = 5, b = 4
Subtraction of Polynomials Calculator:

Example for Polynomials calculator:

Subtract        2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Solution:

Using associative and distributive properties, we have

( x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the signs of the polynomial in Line (2), we get

Line (3): –2x3 + 3x2 + 1.

Adding the polynomials in Line (1) and Line (3), we get

–x3 + 8x2 – 4x – 5.

Integral Part Definition

In geometrical and other applications of integral calculus it becomes necessary to find the difference in the valves of an integral of a function f(x) for two assigned values of the independent variable x, a and b. this difference is called the definite integral of f(x) over the range (a,b) and is denoted by $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where F(x) is an integral of f(x), F(b) is the variable of F(x) at x=b, F(a) is the value of F(x) at x=a.

It is often written thus  : $\int_{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)$.

Note 1. The integral $\int_{a}^{b}f(x)dx $ is read as the integral of f(x) from a to b.The number a is called the lower limit and the number b, the upper limit of integration.

Note 2. It should be seen that the value of a definite integral is perfectly unique and is independent of the particular form of integral which may employ to calculate it. Considering F(x)+c instead of F(x), we get

$\int_{a}^{b}f(x)dx=[F(x)+c]_{a}^{b}=[F(b)+c]-[F(a)+c]=F(b)-F(a)$

so that the arbitrary constant disappears in the process and we get the same values as on considering F(x). This is why the name is given as definite integral.

Note 3. It is assumed that a and b are finite.
Definition of Integration by Parts

When the given function cannot be integrated directly by using standard formulae, we try other methods. The process of integration is largely of tentative nature and no systematic procedure can be given as in differentiation. However, the following are two important methods opf integration.

1. Integration by Substitution           2. Integration by Parts

Here, we will discuss about Integration by parts

An Integration is the inverse process of differentiation.By differentiaon we find the derivative of the given function, whereas by integration we find the function whose derivative is known.

If the derivative of F(x) is f(x) then we say that the antiderivative or integral of f(x) is F(x),such that

int f( x ) dx= F( x )

Thusd/dx F(x) = f(x) => intf(x) dx = F(x)

Integration by Parts:

Theorem: if u and v are two differentiable function of x then

int ( u v ) dx = [ u * int v dx ] - int { du/dx * int v dx } dx .

We can express this result as given below:

Integral of product of two function

= ( 1st function ) * ( integral of 2nd function ) - int { ( derivative of 1st function ) * ( integral of 2nd function ) } dx .

We should choose u and v in such a way that the second function v on the right hand side is easy to integrate. Sometimes, this rule has to be used repeatedly. This rule is also useful in integrating logarithmic and inverse t - functions of the type log x, log(ax2 + bs + c), sin-1 x, tan-1 x etc.The following guidelines will help you to see which of the two functions in the product should be taken as the first function.

Notes:

If the integrand is of the form f(x) * xn, we consider xn as the first function and f(x)as the second function.
If the integrand contains a logarithmic or an inverse trigonometric function, we take it as the first function. If the second function is not given in any case we take that as one.

Example Problems on Integration by Parts:

Pro 1: Evaluate, int xsin 2x dx

Sol : Given,

int   xsin 2x dx

=     int sin 2x dx - int { d/dx ( x ) * int sin 2x dx } dx

=     x * ( -(cos 2x)/2 ) - int 1 * ( -(cos 2x)/2 ) dx

=  (-x cos 2x)/2 + 1/2 int cos 2x dx

"-(x cos  + 1/2 * "(sin

= -(x cos 2x)/2 + 1/4 sin 2x + C , which is the Answer.

Pro 2: int logx/x^2 dx

Sol :   Integrating by parts, taking logx as the fi rst function and 1/x^2  as the second function, we get

int "log   dx

=  int (log x) * 1/x^2 dx

=   ( log x ) * int 1/x^2 dx - int { d/dx ( log x ) * int 1/x^2 dx } dx

=    ( log x )(-1/x )   - int 1/x  * ( -1/x ) dx

= - log x/x + int 1/x^2 dx

= - log x/x - 1/x + C, which  is the required Answer.

Practice problems on Integration by Parts:

Pro 1: Evaluate, int (  x cos x) dx    ( Answer:   x sinx +cosx+c )

Pro 2: Evaluate, int e2x sin x dx            ( Answer: 1/5 e2x ( 2 sin x - cos x ) + C )