Whole Numbers and properties

Whole Number:-

In this lesson.. let me help you through on whole number and properties.

The set of whole numbers include all positive numbers from {0,1, 2,3,4----}.It does not include any fractions and any negative numbers.

Properties of whole number:-

Properties of Addition

1. The sum of two whole numbers is a whole number.

For eg; 2 and 5 are whole number 2 + 5 = 7 ,where 7 is also a whole number. 5 and 4 are whole numbers 5 +4 =9 9m is also a whole number.

2. The sum of two whole numbers is the same irrespective of the order in which they are added.

For example: 12 + 8 = 20 and 8 + 12 =20 So 12 + 8 = 8 + 12 11 + 0 = 0 + 11 = 11

3. The sum of the whole numbers is the same though they may be added in any of the following ways.

Sample Problems on Algebra

Here we begin the practice of solving algebra problems. Problems include the questions involving algebraic equations and basic operations had to be performed to arrive at the solution. Set of solved problems are as shown below.

Question - 1

Question: Solve the following equation:
3(x-1)=8

Answer: 3(x-1)=8
3x-3=8
3x=8+3
3x=11

Question - 2

Question: Solve the following equation:

Answer:


2(3x-19)=10

3x-19 = 5
3x=5+19
3x=24

x=8

Elementary Number Theory Problems

Elementary Number Theory Problems have proofs,word problems and solutions for the equations as follows.

Question - 1

What is the value of M and N respectively? If M39048458N is divisible by 8 and 11; Where M and N are single digit integers?

(1) 7, 8
(2) 8, 6
(3) 6, 4
(4) 5, 4

Answer -

If the last three digits of a number is divisible by 8, then the number is divisible by 8 (test of divisibility by 8).

Here, last three digits 58N is divisible by 8 if N = 4. (Since 584 is divisible by 8.)

For divisibility by 11. If the digits at odd and even places of a given number are equal or differ by a number divisible by 11, then the given number is divisible by 11.

Therefore, (M+9+4+4+8)-(3+0+8+5+N)=(M+5) should be divisible by 11 => when M = 6.

Question - 2

Question:

Answer:

Introduction for pre calculus home work

Introduction for pre calculus home work:

Pre calculus, (or Algebra 3 in some areas) advanced form of secondary school algebra, is a foundational mathematical discipline. It is also called Introduction to Analysis. In many schools, pre calculus is the actually two separate courses: Algebra and Trigonometry.

Homework, or homework assignment, refers to the tasks assigned to students by their teachers to be completed mostly outside of class, and derives its name from the fact that most students do the majority of such work at home.

Examples on Pre Calculus home work -

Following is the example on Pre Calculus home work

Example :

If a triangle ABC, we have that sinA = 7/9, what are the values of cosA and tanA? Solution: We have the basic trigonometric identity:

sin2 A + cos2 A = 1

In our case, we know that sin A = 7 / 9, so we replace in the previous equation to get: cos2 A = 1-(7 / 9)2

We want to solve for cosA so we get

cos2 A = 1 − (49/81) = 32 / 81 so

we apply square root to get

cos A =sqrt (32 / 81) = 4 √2/9.

To find tan A we recall that the tan A = sin A / cos A, so using the information we already have we find that

Tan A = ( 7/9 ) / ( 4√2/9 ) = 7/ 4√2 = 7√2/8

Note on Process of Air Demand Combustion

Introduction:

We will discuss about the process of combustion. Combustion should be an exothermic chemical response, which is go with by growth of heat and light at a quick rate. Combustion is an exothermic reaction, in which a fuel burns in the presense of oxygen with evolution of heat and light. The chief elements present in most of the fuels are C and H. In addition, a trace amount of S, N are also present.

The Process of Combustion

These are the process of combustion

• The formula should be based on the following process assumptions.
• 1- Most of the fuels have C, H, S and O.
• 2- Calorific value is the sum of the calorific value of each element.

Therefore, the Dulong's formula can be written as follows:

GCV = 1/100 [8080 C + 34500 (H-O/8) + 2240 S] k cal/kg

C, H, O & S - % of C, H, O & S in the fuel.

LCV = (HCV-0.09H×587) k cal/kg

Process of Air Demand Combustion:-

To find the amount of oxygen and therefore the amount of air requisite process for the combustion of a component of a fuel, it is essential to be relevant the subsequent basic principles:

• Substances always combine in definite proportions and these proportions are determined by the combustion molecular masses of the substances process involved and the products formed. For example, when carbon combines with oxygen to form carbon dioxide, the equation will be,

C + O₂ ---> CO₂ + 97k cal

indicates that mass extent of hydrogen, oxygen and carbon dioxide produced are 12:32:44 correspondingly.

• Air contains 21% of oxygen by volume, and mass percent of oxygen is 23. This means that 1 kg of oxygen process is supplied by,

1×100/23 = 4.35kg of air.

• Molecular mass of air is in use as 28.94 g/mol.

• Minimum oxygen required = Theoretical oxygen required - Oxygen present in the fuel.

• Minimum oxygen required should be calculated on the basis of complete combustion. If the combustion products contain CO and O₂ is found by subtracting the amount of O₂ required to burn CO to CO₂.

Type of Quadrant

Intordction to Quadrant

A quadrant is an instrument,if used to measure angles up to 90°. It was originally propose by Ptolemy as a better kind of astrolabe Several different variations of the instrument were later produced by medieval Muslim astronomers.

Types of Quadrant

Following are the types of Quadrant

1) . The sine quadrant also known as the Sine cal Quadrant was used for solving trigonometric problems and taking astronomical observations. It was developed by all Khwarizmi in 9th century Baghdad and prevalent until the nineteenth century. Its defining feature is a graph paper like grade if one side that is divided into sixty equal intervals on each axis and is also bounded by a 90 degree graduated arc. A cord was attached to the apex of the quadrant with in bead at the end of it to act as a plumb bob. They were also sometimes drawn on the back of astrolabes.

2). The universal quadrant used for solving astronomical problems for any latitude. These quadrants had either one or two sets of Shakespeare grids and were developed in the fourteenth century in Syria. Some astrolabes are all so printed on the back with the universal quadrant like an astrolabe created by IbnalSarrāj.

3). The hoary quadrant used for finding the time with the sun. The hoary quadrant to be find the time is either equaled or unequal a hours. Different sets of markings were created for either equal or unequal hours. For measuring the time in equal hours, the hoary quadrant could only be used for one specific latitude while a quadrant for unequal hours could be used anywhere based on an approximate formula. One edge of the quadrant had to be aligned with the sun and one aligned, a bead on the end of a plumb line attached to the center of the quadrant showed the time of the day.

4). The astrolabe alimentary quadrant a quadrant developed to the astrolabe. This quadrant make with a one half of a typical astrolabe plate. if astrolabe plates are symmetrical. A cord attached from the center of the quadrant with a bead at the other end was moved to represent the position of a celestial body . The ecliptic and star positions were make on the quadrant for the above. It is not known where and when the astrolabe quadrant was invented,if existent astrolabe quadrants are either of Ottoman or Mameluke origin, while there have been discovered twelfth century Egyptian and fourteenth century Syrian treatises on the astrolabe quadrant. These quadrants proved to be very popular alternative of astrolabes.