Programimport java.util.Scanner; class Arithmetic Here we .pdf

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Program: import java.util.Scanner; class Arithmetic // Here we are taking class name as Arithmetic { public static void main(String args[]) { int x, y, a,b,c,e; double d; String s; Scanner in = new Scanner(System.in); // here we are scanning input from key board System.out.println(\"Please enter your name >>\"); s = in.nextLine(); System.out.println(\"Please enter an integer >>\"); x = in.nextInt(); System.out.println(\"Please enter another integer >>\"); y = in.nextInt(); System.out.println(\"hey \"+s+\" here are your answers : \"); a = x + y; System.out.println(x+\" + \"+y+\" is \"+a); b= x - y; System.out.println(x+\" - \"+y+\" is \"+b); c = x * y; System.out.println(x+\" * \"+y+\" is \"+c); d = (double)x / (double)y; // Here we are converting integer value in to double System.out.println(x+\" / \"+y+\" is \"+d); e = x%y; System.out.println(x+\" % \"+y+\" is \"+e); } } Output : Please enter your name >> Budda Please enter an integer >> 5 Please enter another integer >> 4 hey Budda here are your answers : 5 + 4 is 9 5 - 4 is 1 5 * 4 is 20 5 / 4 is 1.25 5 % 4 is 1 Solution Program: import java.util.Scanner; class Arithmetic // Here we are taking class name as Arithmetic { public static void main(String args[]) { int x, y, a,b,c,e; double d; String s; Scanner in = new Scanner(System.in); // here we are scanning input from key board System.out.println(\"Please enter your name >>\"); s = in.nextLine(); System.out.println(\"Please enter an integer >>\"); x = in.nextInt(); System.out.println(\"Please enter another integer >>\"); y = in.nextInt(); System.out.println(\"hey \"+s+\" here are your answers : \"); a = x + y; System.out.println(x+\" + \"+y+\" is \"+a); b= x - y; System.out.println(x+\" - \"+y+\" is \"+b); c = x * y; System.out.println(x+\" * \"+y+\" is \"+c); d = (double)x / (double)y; // Here we are converting integer value in to double System.out.println(x+\" / \"+y+\" is \"+d); e = x%y; System.out.println(x+\" % \"+y+\" is \"+e); } } Output : Please enter your name >> Budda Please enter an integer >> 5 Please enter another integer >> 4 hey Budda here are your answers : 5 + 4 is 9 5 - 4 is 1 5 * 4 is 20 5 / 4 is 1.25 5 % 4 is 1.

Program:
import java.util.Scanner;
class Arithmetic // Here we are taking class name as Arithmetic
{
public static void main(String args[])
{
int x, y, a,b,c,e;
double d;
String s;
Scanner in = new Scanner(System.in); // here we are scanning input from key board
System.out.println("Please enter your name >>");
s = in.nextLine();
System.out.println("Please enter an integer >>");
x = in.nextInt();
System.out.println("Please enter another integer >>");
y = in.nextInt();
System.out.println("hey "+s+" here are your answers : ");
a = x + y;
System.out.println(x+" + "+y+" is "+a);
b= x - y;
System.out.println(x+" - "+y+" is "+b);
c = x * y;
System.out.println(x+" * "+y+" is "+c);
d = (double)x / (double)y; // Here we are converting integer value in to double
System.out.println(x+" / "+y+" is "+d);
e = x%y;
System.out.println(x+" % "+y+" is "+e);
}
}
Output :
Please enter your name >>
Budda
Please enter an integer >>
5

Please enter another integer >>
4
hey Budda here are your answers :
5 + 4 is 9
5 - 4 is 1
5 * 4 is 20
5 / 4 is 1.25
5 % 4 is 1
Solution
Program:
import java.util.Scanner;
class Arithmetic // Here we are taking class name as Arithmetic
{
public static void main(String args[])
{
int x, y, a,b,c,e;
double d;
String s;
Scanner in = new Scanner(System.in); // here we are scanning input from key board
System.out.println("Please enter your name >>");
s = in.nextLine();
System.out.println("Please enter an integer >>");
x = in.nextInt();
System.out.println("Please enter another integer >>");
y = in.nextInt();
System.out.println("hey "+s+" here are your answers : ");
a = x + y;
System.out.println(x+" + "+y+" is "+a);
b= x - y;
System.out.println(x+" - "+y+" is "+b);
c = x * y;
System.out.println(x+" * "+y+" is "+c);
d = (double)x / (double)y; // Here we are converting integer value in to double

System.out.println(x+" / "+y+" is "+d);
e = x%y;
System.out.println(x+" % "+y+" is "+e);
}
}
Output :
Please enter your name >>
Budda
Please enter an integer >>
5
Please enter another integer >>
4
hey Budda here are your answers :
5 + 4 is 9
5 - 4 is 1
5 * 4 is 20
5 / 4 is 1.25
5 % 4 is 1

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there are several theory on defining acid and base, Lewis acids and bases is the broadest. The species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid. For the first one, it would be easier when you think their product in water solution. SO2 will give H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and Ca(OH)2 is a brownsted base(also lewis base). Solution there are several theory on defining acid and base, Lewis acids and bases is the broadest. The species donate electron pair is lewis base and the species accept electron pair is lewis acid. so for the second one. the oxygen in OH- has 3 pair of electrons. It donate one pair to the Al which has empty p-orbits thus Al accept one pair of electons. So OH- is lewis base, Al(OH)3 is lewis acid. For the first one, it would be easier when you think their product in water solution. SO2 will give H2SO3 and CaO will give Ca(OH)2, H2SO3 is clearly a brownsted acid(also lewis acid) and Ca(OH)2 is a brownsted base(also lewis base)..

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The Statement is False. As there are many applications of hyperbo;ic geometry.The hyperbolic geometry is used in 1. Cosmology and geometry 2.The theory of general relativity. 3. Application of spherical geometry 4. Celestial mechanics Solution The Statement is False. As there are many applications of hyperbo;ic geometry.The hyperbolic geometry is used in 1. Cosmology and geometry 2.The theory of general relativity. 3. Application of spherical geometry 4. Celestial mechanics.

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The Sarbanes-Oxley , 2002 contains following provisions which determines more adequacy of control over financial statements. One of the core elements of Sarbanes-Oxley was to clearly define and place responsibility for a company’s financial statements with its cchief executive officer (CEO) and chief financial officer (CFO), SOX mandated that these executives certify the following items (among others) for each annual and quarterly report: • They have reviewed the report • Based on their knowledge, the financial information included in the report is fairly presented • Based on their knowledge, the report does not contain any untrue statement of material fact or omit a material fact that would make the financial statements misleading • They acknowledge their responsibility for establishing and maintaining internal controls over financial reporting and other disclosures • They have evaluated the effectiveness of these controls, presented their conclusion as to effectiveness and disclosed any material changes in the company’s controls. SOX required companies to provide audit committees with the resources and authority to engage independent counsel and advisers to help them carry out their duties. SOX also required audit committees to establish procedures for receiving whistleblower complaints regarding accounting, auditing and internal control irregularities and to provide for the confidential and anonymous treatment of employee concerns regarding such matters. In addition, SOX enhanced the external auditor’s required communications with the audit committee to include the following: • A discussion of all critical accounting policies and practices used by the company • All alternative accounting treatments that have been discussed with management, the ramifications of the use of alternative disclosures and accounting treatments and the accounting treatment preferred by the audit firm • Other material written communications between the auditor and management. Conclusion: I think the provisions of Sarbanes oxely Act are adequate, We should also focus on the practicality of ever changing corporote world . A recent report showed that chances of reinstatement of financial statements are 46% higher for the companies who do follow sarbanes- oxley act. See this provision of Sarbanes oxely act. Officers and directors are prohibited from trading during pension “blackout” periods: The Act prohibits corporate officers and directors from trading company securities during a pension fund “blackout” period . It was never there . It was aptly introduced by Sarbanes-oxely Act. Solution The Sarbanes-Oxley , 2002 contains following provisions which determines more adequacy of control over financial statements. One of the core elements of Sarbanes-Oxley was to clearly define and place responsibility for a company’s financial statements with its cchief executive officer (CEO) and chief financial officer (CFO), SOX mandated that these executives certify the following it.

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sharnapiyush773

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Note Modified codecode#includeiostream #include stdio.h.pdf

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Note Modified codecode#includeiostream #include stdio.h.pdf

sharnapiyush773

Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about. Solution Oligotrophic area Usually, an olidgotropic organism is a one that can survive in an environment which has low levels of nutrients. So, an oligotropic area would be an environment which has less nutrient to support life. These environments are usually of air, ice, soil or water with low levels of nutrients. Eutropicarea In contrast to oligotropic, eutropic areas are nutrient rich areas. Here, due to excess nutrients/organic matter, a large number of algae and other plants would grow excessively, depleting the oxygen content of the water body. Due to depletion of oxygen, the other aquatic organisms might be affected. Upwelling area This occurs in oceanic ecosystems. Usually, the top or the upper surface of the ocean will be warmer and nutrient-depleted. Due to the wind, the nutrient-rich water will be moved to the upper surface, replacing the water present earlier (nutrient-depleted). This area is nothing but a nutrient-rich area, that was once nutrient-depleted. So, the difference between the three is in terms of the nutrient availability. Oligotropic is a one which has low amounts of nutrients. Eutropic is a one which has high amounts of nutrients and Upwelling area is a one which is nutrient-rich but was once nutrient-depleted. The second question cannot be answered because, stratification is a very common term used to refer to layers. Here, I do not know which stratification you are talking about..

Oligotrophic area Usually, an olidgotropic organism is a one that .pdf

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sharnapiyush773

n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf

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n = N 2^(rt)n = no of people at a given timeN = initial populat.pdf

sharnapiyush773

Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if he felt that the application was wrongly rejected. The situation should have been handled before Vitek hacking the system. The reason for this is because he affected the larger populace and environment by his act. The sewage spill was significant. It polluted over 500 metres of open drain in a residential area and flowed into a tidal canal. Cleaning up the spill and its effects took days and required the deployment of considerable resources. \"Marine life died, the creek water turned black and the stench was unbearable for residents,\" said Janelle Bryant, investigations manager for the Australian Environmental Protection Agency. Solution Maroochy Shire Sewage Spill Case Study Student’s Name Institution Affiliation The Maroochy Shire Sewage Spill case study is as story of an Australian man who was sent to prison for two years. He was found guilty of hacking into the Maroochy Shire, Queensland computerized waste management system and caused millions of liters of raw sewage to spill out into local parks, rivers and even the grounds of a Hyatt Regency hotel. The Maroochy dore District Court heard that 49-year-old Vitek Boden had conducted a series of electronic attacks on the Maroochy Shire sewage control system after a job application he had made was rejected by the area\'s Council. At the time he was employed by the company that had installed the system. Boden made at least 46 attempts to take control of the sewage system during March and April 2000. On 23 April, the date of Boden\'s last hacking attempt, police who pulled over his car found radio and computer equipment. Later investigations found Boden\'s laptop had been used at the time of the attacks and his hard drive contained software for accessing and controlling the sewage management system. Vitek should have appealed to the board if.

Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf

Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf

Maroochy Shire Sewage Spill Case Study Student’s Name Institution Af.pdf

sharnapiyush773

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sharnapiyush773

In developmental biology, an embryo is divided into two hemispheres: the animal pole and the vegetal pole within a blastula. The animal pole consists of small cells that divide rapidly, in contrast with the vegetal pole below it. In some cases, the animal pole is thought to differentiate into the later embryo itself, forming the three primary germ layers and participating in gastrulation. The vegetal pole contains large yolky cells that divide very slowly, in contrast with the animal pole above it. In some cases, the vegetal pole is thought to differentiate into the extraembryonic membranes that protect and nourish the developing embryo, such as the placentain mammals and the chorion in birds. The development of the animal-vegetal axis occurs prior to fertilisation. Sperm entry can occur anywhere in the animal hemisphere. The point of sperm entry defines the dorso-ventral axis - cells opposite the region of sperm entry will eventually form the dorsal portion of the body a. Males release so many sperm that the egg is covered by them. b. The egg has a plasma membrane, a vitelline envelope, and a jelly coat. c. Acrosome enzymes digest away the zona pellucida around the egg as it extrudes a filament that attaches to a receptor on the vitelline jelly layer envelope. d. This interaction between filament and receptor is a lock-and-key reaction that is species- specific. e. The egg plasma membrane and the sperm nuclear membrane fuse, allowing the nucleus to enter. f. Fusion takes place and the zygote begins development. g. As soon as the plasma membranes of sperm and egg fuse, the plasma membrane and the vitelline envelope undergo changes that prevent entrance of any other sperm. h. The vitelline envelope now becomes the fertilization envelope. Early Developmental Stages 1. Development includes events and processes that occur as a single cell becomes a complex organism. 2. All chordate embryos go through same early developmental stages: zygote, morula, blastula, early and late gastrula. 3. The presence of yolk, dense nutrient material, affects how the embryonic cells complete the first three stages. 4. Following fertilization, a zygote undergoes cleavage, cell division without growth. a. DNA replication and mitosis occur repeatedly; the cells get smaller each division. b. As deuterostomes, lancelets have a radial and indeterminate pattern of cleavage. 1) In radial cleavage, any plane passing through will divide the embryo into symmetrical halves. 2) In indeterminate cleavage, cells have not differentiated; if separated, each one develops a complete organism. 5. Because the lancelet has little yolk, the cell divisions are equal in the resulting morula. 6. A cavity called the blastocoel develops forming the hollow ball called the blastula. 7. Gastrulation is invagination of some cells of the blastocyst into blastocoel to form three primary germ layers. a. The outer layer of cells becomes the ectoderm. b. The inner layer of cells becomes the endoderm. c.

In developmental biology, an embryo is divided into two hemispheres.pdf

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In developmental biology, an embryo is divided into two hemispheres.pdf

sharnapiyush773

Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water. Solution Bryophytes- Development of primitive vasculature for water transport, though water absorption mechanism not present. Pteridophytes- Specialised vascular tissue for water transport, ability to collect water from moist soil. Gymnosperms- development of roots and vascular tissue for water absorption and conduction. Development of naked seeds that are wind pollinated. Sporophyte is dominant over the gametophyte. Angiosperms- Well developed vascular system, seed enclosed in seed coat. The earlier groups faced challenges in their transition from an aquatic to a terrestrial mode of life as they faced the problem of desciccation. Hence, they dried out quickly. Most earlier land plants were restricted to moist environments as a result. Minimizing the gametophyte and magnifying the sporophyte has led to the development of seed habit. Further, with the development of heterospory, the megaspore remains in the parent sporophyte and the microgametophyte travels to reach the ovule. Seed plants can reproduce independently of water..

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Bryophytes- Development of primitive vasculature for water transport.pdf

Bryophytes- Development of primitive vasculature for water transport.pdf

sharnapiyush773

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Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf

Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf

Post Exam Fun(da) Intra UEM General Quiz - Finals.pdf

Neurulation and the formation of the neural tube

Neurulation and the formation of the neural tube

Neurulation and the formation of the neural tube

Matatag-Curriculum and the 21st Century Skills Presentation.pptx

Matatag-Curriculum and the 21st Century Skills Presentation.pptx

Matatag-Curriculum and the 21st Century Skills Presentation.pptx

Application of Matrices in real life. Presentation on application of matrices

Application of Matrices in real life. Presentation on application of matrices

Application of Matrices in real life. Presentation on application of matrices

[GDSC YCCE] Build with AI Online Presentation

[GDSC YCCE] Build with AI Online Presentation

[GDSC YCCE] Build with AI Online Presentation

Championnat de France de Tennis de table/

Championnat de France de Tennis de table/

Championnat de France de Tennis de table/

Programimport java.util.Scanner; class Arithmetic Here we .pdf

1. Program: import java.util.Scanner; class Arithmetic // Here we are taking class name as Arithmetic { public static void main(String args[]) { int x, y, a,b,c,e; double d; String s; Scanner in = new Scanner(System.in); // here we are scanning input from key board System.out.println("Please enter your name >>"); s = in.nextLine(); System.out.println("Please enter an integer >>"); x = in.nextInt(); System.out.println("Please enter another integer >>"); y = in.nextInt(); System.out.println("hey "+s+" here are your answers : "); a = x + y; System.out.println(x+" + "+y+" is "+a); b= x - y; System.out.println(x+" - "+y+" is "+b); c = x * y; System.out.println(x+" * "+y+" is "+c); d = (double)x / (double)y; // Here we are converting integer value in to double System.out.println(x+" / "+y+" is "+d); e = x%y; System.out.println(x+" % "+y+" is "+e); } } Output : Please enter your name >> Budda Please enter an integer >> 5

2. Please enter another integer >> 4 hey Budda here are your answers : 5 + 4 is 9 5 - 4 is 1 5 * 4 is 20 5 / 4 is 1.25 5 % 4 is 1 Solution Program: import java.util.Scanner; class Arithmetic // Here we are taking class name as Arithmetic { public static void main(String args[]) { int x, y, a,b,c,e; double d; String s; Scanner in = new Scanner(System.in); // here we are scanning input from key board System.out.println("Please enter your name >>"); s = in.nextLine(); System.out.println("Please enter an integer >>"); x = in.nextInt(); System.out.println("Please enter another integer >>"); y = in.nextInt(); System.out.println("hey "+s+" here are your answers : "); a = x + y; System.out.println(x+" + "+y+" is "+a); b= x - y; System.out.println(x+" - "+y+" is "+b); c = x * y; System.out.println(x+" * "+y+" is "+c); d = (double)x / (double)y; // Here we are converting integer value in to double

3. System.out.println(x+" / "+y+" is "+d); e = x%y; System.out.println(x+" % "+y+" is "+e); } } Output : Please enter your name >> Budda Please enter an integer >> 5 Please enter another integer >> 4 hey Budda here are your answers : 5 + 4 is 9 5 - 4 is 1 5 * 4 is 20 5 / 4 is 1.25 5 % 4 is 1