Here is the code for youimport java.util.; class PrintNumberWi.pdf

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Here is the code for you: import java.util.*; class PrintNumberWithCommas { public static void main(String[] args) { int number; Scanner sc = new Scanner(System.in); System.out.print(\"Enter an integer between 1000 and 999999: \"); number = sc.nextInt(); while(number < 1000 || number > 999999) { System.out.println(\"Invalid input. Retry...\"); System.out.print(\"Enter an integer between 1000 and 999999: \"); number = sc.nextInt(); } String output = Integer.toString(number); output = output.substring(0, output.length() - 3) + \",\" + output.substring(output.length() - 3); System.out.println(output); } } And the output screenshot is: Solution Here is the code for you: import java.util.*; class PrintNumberWithCommas { public static void main(String[] args) { int number; Scanner sc = new Scanner(System.in); System.out.print(\"Enter an integer between 1000 and 999999: \"); number = sc.nextInt(); while(number < 1000 || number > 999999) { System.out.println(\"Invalid input. Retry...\"); System.out.print(\"Enter an integer between 1000 and 999999: \"); number = sc.nextInt(); } String output = Integer.toString(number); output = output.substring(0, output.length() - 3) + \",\" + output.substring(output.length() - 3); System.out.println(output); } } And the output screenshot is:.

Here is the code for you:
import java.util.*;
class PrintNumberWithCommas
{
public static void main(String[] args)
{
int number;
Scanner sc = new Scanner(System.in);
System.out.print("Enter an integer between 1000 and 999999: ");
number = sc.nextInt();
while(number < 1000 || number > 999999)
{
System.out.println("Invalid input. Retry...");
System.out.print("Enter an integer between 1000 and 999999: ");
number = sc.nextInt();
}
String output = Integer.toString(number);
output = output.substring(0, output.length() - 3) + "," + output.substring(output.length() - 3);
System.out.println(output);
}
}
And the output screenshot is:
Solution
Here is the code for you:
import java.util.*;
class PrintNumberWithCommas
{
public static void main(String[] args)
{
int number;
Scanner sc = new Scanner(System.in);
System.out.print("Enter an integer between 1000 and 999999: ");
number = sc.nextInt();

while(number < 1000 || number > 999999)
{
System.out.println("Invalid input. Retry...");
System.out.print("Enter an integer between 1000 and 999999: ");
number = sc.nextInt();
}
String output = Integer.toString(number);
output = output.substring(0, output.length() - 3) + "," + output.substring(output.length() - 3);
System.out.println(output);
}
}
And the output screenshot is:

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import java.util.*; public class Program { public static void main(String[] args) { Scanner input = new Scanner(System.in); String x=\"X\"; int amount; System.out.println(\"Enter amount\"); amount=input.nextInt(); System.out.println(\"Amount is: \"+amount); for(int i=1; i<=amount; i++) { System.out.println(x); for(int j=1; j<=i; j++) { System.out.print(x); } } } } Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain using the most basic codes. import java.util.*; public class Program { public static void main(String[] args) { Scanner input = new Scanner(System.in); String x=\"X\"; int amount; System.out.println(\"Enter amount\"); amount=input.nextInt(); System.out.println(\"Amount is: \"+amount); for(int i=1; i<=amount; i++) { System.out.println(x); for(int j=1; j<=i; j++) { System.out.print(x); } } } } Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain using the most basic codes. import java.util.*; public class Program { public static void main(String[] args) { Scanner input = new Scanner(System.in); String x=\"X\"; int amount; System.out.println(\"Enter amount\"); amount=input.nextInt(); System.out.println(\"Amount is: \"+amount); for(int i=1; i<=amount; i++) { System.out.println(x); for(int j=1; j<=i; j++) { System.out.print(x); } } } } Why does It print out the last set of X\'s twice? I am doing basic Java so please try to explain using the most basic codes. Solution Hi, I have modified the code. it is working as expected now. Highlighted the code changes below. Issue here is with you have written this statement twice System.out.println(x); which causes the issue. Program.java import java.util.*; public class Program { public static void main(String[] args) { Scanner input = new Scanner(System.in); String x=\"X\"; int amount; System.out.println(\"Enter amount\"); amount=input.nextInt(); System.out.println(\"Amount is: \"+amount); for(int i=1; i<=amount; i++) { System.out.println(); for(int j=1; j<=i; j++) { System.out.print(x); } } } } Output: Enter amount 10 Amount is: 10 X XX XXX XXXX XXXXX XXXXXX XXXXXXX XXXXXXXX XXXXXXXXX XXXXXXXXXX.

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It is extreme important for a firm to able to meet its obligation as they become due. liquidity ratio measures the ability of the firm to meet its current obligations(liabilities).A firm should ensure that it does not suffer from lack of liquidity and also that it does not have excess liquidity. The higher the ratio, the more liquid the company is. Low values indicate that a firm may have difficulty meeting obligations. All other things being equal, In general point of view a high liquid ratio to be better than a low liquid ratio, because a high liquid ratio means that the company is more likely to meet its liabilities which are due . In general, low or decreasing ratios generally suggest that a company is over-leveraged, struggling to maintain or grow sales, paying bills too quickly or collecting receivables too slowly. On the other hand, a high or increasing ratio generally indicates that a company is experiencing solid top-line growth, quickly converting receivables into cash, and easily able to cover its financial obligations. Such companies often have faster inventory turnover and cash conversion cycles. Since the circumstances for every business and industry are different, making it difficult to create a universal benchmark for healthy liquidity ratios. The best practice for investors, lenders and managers is to consider liquidity ratios of successful competitors and historical trends. In general terms the current ratio of 2 to 1 is considered satisfactory and quick ratio of1 to 1 is considered to represent a satisfactory current financial condition. It is possible to have higher values of current and quick ratio but the higher value of these ratio indicates that that the business managers is unable to find investment opportunities .A higher degree of liquidity is also bad as idle asset earns nothing, the firms fund will be unnecessary tied up in current asset. Solution It is extreme important for a firm to able to meet its obligation as they become due. liquidity ratio measures the ability of the firm to meet its current obligations(liabilities).A firm should ensure that it does not suffer from lack of liquidity and also that it does not have excess liquidity. The higher the ratio, the more liquid the company is. Low values indicate that a firm may have difficulty meeting obligations. All other things being equal, In general point of view a high liquid ratio to be better than a low liquid ratio, because a high liquid ratio means that the company is more likely to meet its liabilities which are due . In general, low or decreasing ratios generally suggest that a company is over-leveraged, struggling to maintain or grow sales, paying bills too quickly or collecting receivables too slowly. On the other hand, a high or increasing ratio generally indicates that a company is experiencing solid top-line growth, quickly converting receivables into cash, and easily able to cover its financial obligations. Such companies often have faster invento.

It is extreme important for a firm to able to meet its obligation as.pdf

It is extreme important for a firm to able to meet its obligation as.pdf

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This is a decomposition reaction. Notes: (1)combustion: things react with oxygen, therefore they get oxidized. e.g. burning fuels, where hydrocarbons react with oxygen to form oxides (water and carbon dioxide). (2)decomposition: only one reactant to the left , but more than one species form on the right. Obviously this reaction is decomposition. (3)synthesis: the reverse of decomposition. Several reactants react to form one product. (4)water forming: water must forms on the right side. e.g. Burn hydrogen in oxygen to get water. Solution This is a decomposition reaction. Notes: (1)combustion: things react with oxygen, therefore they get oxidized. e.g. burning fuels, where hydrocarbons react with oxygen to form oxides (water and carbon dioxide). (2)decomposition: only one reactant to the left , but more than one species form on the right. Obviously this reaction is decomposition. (3)synthesis: the reverse of decomposition. Several reactants react to form one product. (4)water forming: water must forms on the right side. e.g. Burn hydrogen in oxygen to get water..

This is a decomposition reaction. Notes (1)combu.pdf

This is a decomposition reaction. Notes (1)combu.pdf

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sp sp2 .pdf

Given that [H+ ] = 2.2x10 -7 M We know that [H+ ] [ OH- ] = Kw ( ionic product of water ) [H+ ] [ OH- ] = 1.0 x10 -14 M 2 [ OH- ] = (1.0 x10 -14 ) / [H+ ] = (1.0 x10 -14 ) / (2.2x10 -7 ) = 4.54 x 10 -8 M Solution Given that [H+ ] = 2.2x10 -7 M We know that [H+ ] [ OH- ] = Kw ( ionic product of water ) [H+ ] [ OH- ] = 1.0 x10 -14 M 2 [ OH- ] = (1.0 x10 -14 ) / [H+ ] = (1.0 x10 -14 ) / (2.2x10 -7 ) = 4.54 x 10 -8 M.

Given that [H+ ] = 2.2x10 -7 MWe know that [H+ ] [ OH- ] = Kw ( io.pdf

Given that [H+ ] = 2.2x10 -7 MWe know that [H+ ] [ OH- ] = Kw ( io.pdf

Given that [H+ ] = 2.2x10 -7 MWe know that [H+ ] [ OH- ] = Kw ( io.pdf

Graphite is carbon in the form of hexagonal crystalline layers, in which covalent bonding exists between atoms in the layers, and the (parallel) layers are bonded by van der Waals forces, thus leading to highly anisotropic properties. Solution Graphite is carbon in the form of hexagonal crystalline layers, in which covalent bonding exists between atoms in the layers, and the (parallel) layers are bonded by van der Waals forces, thus leading to highly anisotropic properties..

Graphite is carbon in the form of hexagonal crystalline layers, in w.pdf

Graphite is carbon in the form of hexagonal crystalline layers, in w.pdf

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During fermentation, when O2 is not present, NAD+ is regenerated from NADH by transferring electrons to organic molecules. In this process, pyruvate is converted into organic molecules like lactate, ethanol. Solution During fermentation, when O2 is not present, NAD+ is regenerated from NADH by transferring electrons to organic molecules. In this process, pyruvate is converted into organic molecules like lactate, ethanol..

During fermentation, when O2 is not present, NAD+ is regenerated fro.pdf

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Ellipse : locus of points which moves such that sum of the distances from two fixed point is constant. Hyperbola : locus of points which moves such that differences of the distances from two fixed point is constant. Ellipse : two focci, two axis ,major and minor, closed and bounded region ,four vertices and centre Hyperbola : two focci, two axis ,transverse and conjugate , un bounded region ,two vertices and centre. Ellipse : eccentricty less than one. Hyperbola : eccentricity more than one. Solution Ellipse : locus of points which moves such that sum of the distances from two fixed point is constant. Hyperbola : locus of points which moves such that differences of the distances from two fixed point is constant. Ellipse : two focci, two axis ,major and minor, closed and bounded region ,four vertices and centre Hyperbola : two focci, two axis ,transverse and conjugate , un bounded region ,two vertices and centre. Ellipse : eccentricty less than one. Hyperbola : eccentricity more than one..

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1. Here is the code for you: import java.util.*; class PrintNumberWithCommas { public static void main(String[] args) { int number; Scanner sc = new Scanner(System.in); System.out.print("Enter an integer between 1000 and 999999: "); number = sc.nextInt(); while(number < 1000 || number > 999999) { System.out.println("Invalid input. Retry..."); System.out.print("Enter an integer between 1000 and 999999: "); number = sc.nextInt(); } String output = Integer.toString(number); output = output.substring(0, output.length() - 3) + "," + output.substring(output.length() - 3); System.out.println(output); } } And the output screenshot is: Solution Here is the code for you: import java.util.*; class PrintNumberWithCommas { public static void main(String[] args) { int number; Scanner sc = new Scanner(System.in); System.out.print("Enter an integer between 1000 and 999999: "); number = sc.nextInt();

2. while(number < 1000 || number > 999999) { System.out.println("Invalid input. Retry..."); System.out.print("Enter an integer between 1000 and 999999: "); number = sc.nextInt(); } String output = Integer.toString(number); output = output.substring(0, output.length() - 3) + "," + output.substring(output.length() - 3); System.out.println(output); } } And the output screenshot is: