Odd Square Numbers

Introduction to odd square numbers:

Odd square numbers are one of the basis for mathematics. The odd numbers are 1,3,5,7 etc. The formula for representing the odd square numbers are 2M+1, where m value is used to represent the any type of variable function. Simply the odd number can be defined as the number which are not divisible the number two. These numbers are called as the odd number.

Explanation for Odd Square Number

Odd numbers are the alternatives of the even numbers. The number that are not divisible by the number 2, 4, 6 ,8 etc are called as the odd number. For example, the numbers 3, 5, 7, 9, 11 etc are represented as the odd numbers. The diagrammatic representation of odd numbers are shown below,

The odd square numbers are represented by using the formula, (2M+1)2 . The formula can be simplified as 2(m2 + m) +1. By using the this formula the odd square number problems are solved.

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Example Problem for Odd Square Numbers

Pro 1: Find the odd square number for the given value. M = 4.

Sol :  Step 1: The formula for finding the odd square numbers is given below,

Odd square numbers = (2M+1)2 = `4M^2 + 8M + 1` = 4M(M+2) + 1

Step 2: The value of m is given. Therefore substitute the value in the formula.

Step 3: By substituting the formula we get,

= (2(4)+1)2

= 2(42 +4)+1

= 2(20)+1

= 40+1

= 41.

This is the required odd square number.

Pro 2: Find the odd square number for the given value. M = 6.

Sol :  Step 1: The formula for finding the odd square numbers is given below,

Odd square numbers = (2M+1)2 = 4(m2 + 2m) +1.

Step 2: The value of m is given. Therefore substitute the value in the formula.

Step 3: By substituting the formula we get,

= (2(6)+1)2 

= (12+1)2 = 13x13 = 169

This is the required odd square number.

Practice Problem for Odd Square Numbers

Pro 1: Find the odd square number for the given value. M = 8.

Ans : 196

Pro 2: Find the odd square number for the given value. M = 10.

Ans : 221

Slant Height Square Pyramid

Introduction to slant height of a square pyramid:

Pyramid is one of the shapes in geometry. A square pyramid is a general pyramid consist of square base. It is octahedron type.The lateral edge length and slant height, s of a right square pyramid of side length and height are In pyramid, the outer surfaces are triangular and converge at a point. The base of pyramid can be any shape like triangular, square, rectangular or of any polygon shape. Pyramid is classified into Volume of a Pyramid, square pyramid, and rectangular pyramid. All types of pyramid have three triangular faces and a base. Let’s see about basic three shapes of pyramid.

Square Pyramid Figure

Types of pyramid: 

Square pyramid
Triangular pyramid and
Rectangular pyramid.


Formula for Square Pyramid Slant Height:

Formula for finding the slant height of square pyramid:

s² = h² + (b/2)²
here s, slant height
h, height of pyramid
b, base length

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Example Problems in Square Pyramid Slant Height:

Ex 1:

The  square pyramid of height 24 cm. Find the slant height if the base edges are given as 14 cm.

Sol:
Formula for finding slant height will be  
h=24 and b=14 `b/2` =7
s² = h² +` (b/2)^2 `
s= `sqrt(24^2 + 7^2) `

= `sqrt(576 + 49)`

= `sqrt(625)`
= 25 cm

Ex 2:
The height of square pyramid 350 ft. and each side of  base is 646 ft. calculate the slant height length.

Sol:
given h=350 b=646 b/2=323
s² = h² + (b/2)²
s² = 350² + 323²
s² = 226829
s = 426.27 ft.

Ex 3:
The  square pyramid of height 32 cm. Find the slant height if the base edges are given as 12 cm.

Sol:
Formula for finding slant height will be  
h=32 and b=12 `b/2` =6
s² = h² +` (b/2)^2 `
s= `sqrt(32^2 + 6^2) `

= `sqrt(1024 + 36)`

= `sqrt(1060)`
= 32.55 cm
Ex 4:
The  square pyramid of height 16 cm. Find the slant height if the base edges are given as 4 cm.

Sol:
Formula for finding slant height will be  
h=16 and b=4 `b/2` =2
s² = h² +` (b/2)^2 `
s= `sqrt(16^2 + 4^2) `

= `sqrt(256 + 16)`

= `sqrt(272)`
= 16.5 cm

Solve Implicit Function or Relation

Introduction :

The function of implicit function is related to the variables. These two variables are given by an equation. This function has not been solved explicitly. A relation between the variables of function is said to be implicit function. For example   x^2 + y^2 = 100, y is an implicit function of x and x is an implicit function of y. In this article, we shall discuss about solve implicit function or relation.

Solve Implicit Function or Relation - Problems:

Solve implicit function or relation - problem 1:

Calculate the implicit function of x and implicit function of y in the given equation    -x^2 = - 5y  .

Solution:

Given equation is        -x^2 = - 5y.                      --------------(1)

Adding by  x^2 + 5y on both side, So we get

- x^2 + 5y +  x^2 = -5y.+ x^2 + 5y

+ 5y = + x^2                     --------------(2)

Now  divided by 5 on both sides,

`(5y)/(5)` = `( x^2) /(5)` .

Implicit function of x is                    y =  `( x^2) /(5)`.

Take equation (2)    ,         5y =  x^2 

Take square root on both sides,       `sqrt(5y)` = `sqrt(x^2)` .

` sqrt(5y)`   = x .

x = `sqrt(5y)` .

Answer:     y =  `( x^2) /(5)`.  is Implicit function of x .                       

x = `sqrt(5y)` . is Implicit function of y .                               

Solve implicit function or relation - problem 2:

The implicit function function is 10xy^2 - 5y^2  = 5. Evaluate  `(dy/dx)` .

Solution:

Given implicit function is    10xy^2 - 5y^2  = 5.

Now Find the derivative of  xy^2

`d/dx`(10xy^2)    = x 20y `(dy/dx)` + 10y^2 (1).

Find the derivative of 5y^2

`d/dx`(5y^2)    =  10y `(dy/dx)` .

Find the derivative of 5 (constant)

`d/dx`(5)    = 0.

So,                     10xy^2 - 5y^2  = 5.

20xy `(dy/dx)` + 10y^2 - 10y `(dy/dx)` . = 0

Subtract by 10 y^2 on both sides,

20xy `(dy/dx)` + 10y^2 - 10y `(dy/dx)` - 10 y^2 . = 0 - 10y^2

20xy `(dy/dx)` - 10y `(dy/dx)` .=  - 10y^2

Take `dy/dx` in common

` (dy/dx)` (20xy - 10y) = - 10y^2

Divided by (20xy - 10y) on both side so we get,

` (dy/dx)`` ((20xy - 10y)/(20xy-10y))` = `((- 10y^2)/(20xy-10y))`.

` (dy/dx)`  =  `((- 10y^2)/(20xy-10y))`.

Answer:  ` (dy/dx)`  =  `((- 10y^2)/(20xy-10y))`.

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Solve Implicit Function or Relation - Practice Problems:

Solve implicit function or relation - practice problem 1:

Find the implicit function of y in the given equation    x  = 2 - 3xy .

Answer:     Implicit function of y is               x  = `((2)/(1 + 3y))`         ..

Solve implicit function or relation - practice problem 2:

Find the implicit function of x in the given equation    3y  = 9 - 3xy .

Answer:     Implicit function of x is               y  = `((3)/(1 + x))`         ..

Basic Math Ratios

Introduction for basic math ratios:
A relation is getting form the comparison of two quantities like in some wisdom is called a ratio. Ratio will be show by the terms of fraction, that is, a:b is equal to a/b For example 25 is `1/4` of the hundred, Therefore, the ratio of 25 to 100 is `1/4` We can write it in ratio as 1:4.

Example Problems for Basic Math Ratios:
Example 1 by basic math ratios:

Find the ratio of 70 centimeters to 5 meters in its simplest form.

Solution:

5 m = 5 `xx` 100 cm

Therefore, 5 m = 500 cm

Therefore, the ratio of 70 cm to 4  m = 50cm: 500 cm

Now we have to cancel out the units. So we get

70: 500

Now we have to divide both terms by 10. So we get

7:50

Example 2 by basic math ratios:

Find the ratio of 500 m to 1.4 km in its simplest form.

Solution:

1 km = 1000 m

1.4 km = 1.4 `xx` 1000

Therefore, 1.4 km = 1400 m

Therefore, the ratio of  500 meter to 1.4 km = 500 m : 1400 m

Now we have to cancel out the units. So we get,

500:1400

Now we have to divide both terms by100. So we get

5:14

Example 3 by basic math ratios:

Find the ratio of 40 minutes to 5 hours in its simplest form.

Solution:

1 hour = 60 minutes

5 hours = 5 `xx` 60 minutes

Therefore, 5 hours = 300 minutes

Therefore, the ratio of 40 minutes to 5 hours = 40 min : 300 min

Now we have to cancel out the units. So we get

40: 300

Now we have to divide both terms by 10. So we get

4:30

Now we have to divide both terms by 2. So we get

2:15

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For Problems for Basic Math Ratios:

Problem 1 by basic math ratios:

Find the ratio of 600 milliliters to 1.8 liters in its simplest form.

Solution: 1:3

Problem 2 by basic math ratios:

Find the ratio of 700 milligrams to 5 kilograms in its simplest form.

Solution: 7:50

Data Table an Introduction

Data in mathematics is the collection of facts which can be values or measurements. They consist of numbers or observations or values or just description of the things. Data is mainly of two types, quantitative data which consists of numbers and qualitative data which consists of descriptive information that which describes something. Quantitative data can be further categorized into discrete and continuous data. A discrete data can be only certain values or whole numbers and continuous data can be any value within the given range. These values can be tabulated for better understanding of the given data.  Data Table as the name suggests is the data tabulated in rows and columns. Data Table Definition is the tabulated display of information or data in named rows and columns.
Example of a data table : The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

Day Mon Tue Wed  Thu  Fri   Sat  Sun
Rainfall 0.0 12.4  3.2  0.0  20.4  6.8 1.2
(in mm)

Two way Table is a table in which the data is categorized in a two way that is a cross classification table.
In a two way table we have a column of data for one input and a row of data for the second input. At the intersection of the row and the column the
answer is written.

The following is a two way table showing the information about the number of girls and boys with their ages, in year 9, in year 10, in year 11 in a school
               Year 9     Year 10 Year 11
Boys   70    80   140
Girls   75   110   170
Total  145    190   310

In the above two way data table we have two parameters boys and girls at once in the rows and in the columns the ages with number of girls and boys.  This two way table is useful to estimate the probability of an outcome.
Creating a Data Table
• First the table is to be named. The title should be related to the data put in the table.
• Next we need to figure out how many rows and columns are required according to the data given
• We need to now draw the table with the necessary rows and columns. The top row and the first column are used for labeling. The leftmost column is for the independent variables for instance, if the data is about the rainfall in the previous year. Here, the independent variable will be the ‘months of the year’. So, the leftmost column is labeled ‘Month’ and the next column is labeled as ‘rainfall’
• The experiment outcomes are recorded in the appropriate columns. The information displayed in the table should be clear and obvious. All the spaces should be filled with a number, no space should be left.  If any derived result from the data or an average in the given data should be written in the right most column.
• Finally the table has to be checked thoroughly making sure all the data is clear and correct for further use of the information of the data table.

Sum and Difference Formulas in Trigonometry

In this article, we will discuss about the sum and difference formulas in various parts of mathematics. First we see the sum and difference formulas in trigonometry for various trig functions. The sum & difference formulas include two angles which will be defined and the angles are applied to the various fundamental trig functions. The formulas show the relationship between the two angles and trig functions. These formulas are very useful to solve the problems in trigonometry.

First we discuss about sum and difference formulas for sine function. Suppose we have two angles named as (a) and (b), then for the two angles we write the relationship as sin (a+b) =sin (a) cos (b) +cos (a) sin (b). the  difference formula is expressed as sin(a-b)=sin(a)cos(b)-cos(a)sin(b). To prove these formulas we have to use geometry calculus. Now we take cosine function, suppose we have same angles, then for the two angles we expressed sum formula as cos(a+b)=cos(a)cos(b)-sin(a)sin(b) and difference formula expressed as cos(a-b)=cos(a)cos(b)+sin(a)sin(b).

The sum & difference formulas for third trigonometric function mean Sum and difference formulas for tangent function. This formula is valid for all values where tan a, tan b and tan (a+b) are used. Where (a) and (b) are the two angles. The formula can be expressed as tan (a+b) = (tan a+tan b/1-tan a*tan b) and difference formula is tan (a-b) = (tan a-tan b/1+tan a*tan b). The formulas for tangent function also used for finding the angle between two lines. But the question is how to find the angle, so for finding angle we have to calculate the slope of both lines. The equations can be written as tan θ= (m1-m2/1+m1*m2), where m1 is slope of first line and m2 is slope of second line.

Now sum and difference formulas examples. First we take example for sine and cosine function then tangent function. First problem is, suppose we have to calculate exact value of sin (75°). For this we use sum angle formula such as sin(75°)=sin(30°+45°)=sin(30°)*cos(45°)+cos(30°)*sin(45°) and we know the value of sin(30) and sin(45). Second problem is, suppose we have cos x=1/2 and cos y=1/3 then we calculate the value of cos(x+y) and cos(x-y). So using sum and difference formula cos(x+y) = (1/2*1/3-1/3*1/2) =cos 0=1.

Now problem based on tangent formulas. First problem is, find the exact value of tan (105). Using sum formula tan (105°) =tan (60°) +tan (45°)/1-tan (60°)*tan (45°) and we well know the value of tan (60°) and tan (45°). Second problem is, suppose given data is y=3x-5 and y=-2x+2. From these two data we write slope of both lines m1=3 and m2=-2. Then we use angle formula tanθ= [3-(-2)/1+ (3)*(-2)]. After simplifying we calculate the angle.

Table of Integrals

Integration is a fundamental operation in integral calculus. Integration means calculating the area made by curve. While doing integration, table of known integral are very useful. There are various types of table of integral. First type is definite integral table, definite integral means integration of function with definite limit. Such as integration of fx dx where limit are x=a and x=b, which made an area. Suppose limits are indefinite then it is known as improper integral. For this type of problem we use suitable limiting procedure to convert indefinite limit to definite limit.Definite integral table contain many expressions. In contains elliptic integral, square root, arc tangent means inverse tangent functions, exotic function and some special functions.

Gaussian integral table contains expressions of erf function or error function, Gaussian integral table also known as probability integral.  In this type of integration  we have to integrate one dimensional Gaussian function over the limit from negative infinite to positive infinite(-ve8, +ve 8) it can be solve by using a technique like combining to, one dimensional function in Gaussian function. Here we integrate first dummy variable present in the integral and carryout the term in the end. The n it becomes function of one variable. Now we move to polar coordinate. It is not necessary to use polar coordinates, also we can proved in simple way. Now we take continued function whit erf, or error function.  To solve this function we use Laplas method.

Exponential integral table contain integration of exponential functions. Exponential functions means e^x functions, where e is a number whose value is 2.718. Integration of exponential functions means constant change in the independent variable gives same change that is proportional to input. Exponential functions some time written as exp(x), but it is unpractical to write for any independent variable. Other than mathematics exponential functions are also used in physics and chemistry. Some integral table exponential function are-

Exponential function = e^x
Inverse exponential function = ln x
Derivative of exponential function = e^x
Indefinite integral of exponential function = e^x +p

Increasing exponential function always above from x axis and closed to negative value of x. some time it can be expressed as cbx in this form base b is real number, Variable x can be real or complex number and c is a constant term. Decreasing exponential functions always below from x axis and closed to positive value of x.