Remember for a moment a recent trip you have made to the grocery store to pick up a few items. What pieces of data did the Point

Answer

Continue on your current heading and maintain your speed unless the vessel required to give way does not act.

Explanation

When two vessels approach each other, the one positioned on the right side is designated as the stand-on vessel, which means the other vessel, the give-way vessel, must yield. In this circumstance, you should keep your course and speed unless the give-way vessel fails to do so. If that happens, you must take evasive maneuvers to avoid collision by steering clear, never turning toward or crossing in front of the other vessel.

Selecting the option to trim audio upon clicking would be your best answer

Response:

C++ code provided below with suitable annotations

Clarification:

pattern.cpp

#include<iostream>

using namespace std;

void printCross(int n)

{

int i,j,k;

if(n%2) //odd number of lines

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==i || j==(n-i+1))

cout<<j;

else

cout<<" ";

}

cout<<"\n";

}

}

else //even number of lines

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==i || j==(n-i+1))

{

cout<<" "<<j<<" ";

}

else

cout<<" ";

}

cout<<"\n";

}

}

void printForwardSlash(int n)

{

if(n%2)

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==n-i+1)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

else

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==(n-i+1))

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

void printBackwardSlash(int n)

{

if(n%2) // odd number of lines

{

for(int i=n;i>=1;i--)

{

for(int j=n;j>=1;j--)

{

if(j==i)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

else //even number of lines

{

for(int i=1;i<=n;i++)

{

for(int j=1;j<=n;j++)

{

if(j==i)

{

cout<<j;

}

else

cout<<" ";

}

cout<<"\n";

}

}

}

int main()

{

int num;

char ch;

cout<<"Create a numberes shape that can be sized."<<endl;

cout<<"Input an integer [1,50] and a character [x,b,f]."<<endl;

cin>>num>>ch;

if(ch=='x' || ch=='X')

printCross(num);

else if(ch=='f' || ch=='F')

printForwardSlash(num);

else if(ch=='b' || ch=='B')

printBackwardSlash(num);

else

cout<<"\nWrong input"<<endl;

return 0;

}

The code relevant to the problem in question:

import java.util.Scanner;

public class Test {

   public static void main(String[] args) {

       double number;

       double squareRootOfNumber;

       String userInput = null;

       Scanner scanner = new Scanner(System.in);

       System.out.println("Please enter a number: ");

       userInput = scanner.next();

       number = Double.parseDouble(userInput);

       squareRootOfNumber = Math.sqrt(number);

       if (number < 0) {

           throw new ArithmeticException("Cannot compute the square root of a negative number");

       }

       System.out.format("The square root of the entered number is %.2f %n", squareRootOfNumber);

   }

}

The program will output:

Please enter a number:

-90

Exception in thread "main" java.lang.ArithmeticException: Cannot compute the square root of a negative number at com..ans.Test.main(Test.java:18)

Explanation:

The standard Java library function java.lang.Math.sqrt does not throw an ArithmeticException for negative arguments, hence there’s no need to enclose your code in a try/catch block.

Instead, we manually trigger ArithmeticException using the throw keyword:

throw new ArithmeticException("Cannot compute the square root of a negative number");