Which body of water is most likely to be a marine ecosystem?

Answer:

  hardware-initiated reset

Explanation:

Once in protected mode, the 80286 is designed to remain there until it is reset by hardware.

__

External hardware can be designed so that software can cause such a reset. In the 1984 PC/AT, such hardware combined with code in the BIOS allowed real mode re-entry and returned execution control to the program that caused the reset.

Answer:

The program is written using PYTHON SCRIPT below;

N=int(input(" Enter number of Rows you want:"))

M=[] # this for storing the matrix

for i in range(N):

l=list(map(int,input("Enter the "+str(i+1)+" Row :").split()))

M.append(l)

print("The 2D Matrix is:\n")

for i in range(N):

print(end="\t")

print(M[i])

W=[] # to store the first non zero elemnt index

T=[] # to store that value is positive or negative

L=len(M[0])

for i in range(N):

for j in range(L):

if (M[i][j]==0):

continue

else:

W.append(j) # If the value is non zero append that postion to position list(W)

if(M[i][j]>0): #For checking it is positive or negative

T.append(+1)

else:

T.append(-1)

break

print()

print("The first Non Zero element List [W] : ",end="")

print(W)

print("Positive or Negative List [T] : ",end="")

print(T)

Explanation:

In order for the program to determine a set of test cases it takes in input of 2D matrix in an N numbet of rows.

It goes ahead to program and find the column index of the first non-zero value for each row in the matrix A, and also determines if that non-zero value is positive or negative. The If - Else conditions are met accordingly in running the program.

Answer:

The program goes as follows  

Comments are used to explain difficult lines

#include<iostream>

#include<fstream>

#include<sstream>

#include<string>

using namespace std;

int main()

{

//Declare String to accept file name

string nm;

//Prompt user for file name

cout<<"Enter File Name: ";

getline(cin,nm);

//Create File

ofstream Myfile(nm.c_str());

//Prompt user for 4 names of movies

string movie;

for(int i = 1;i<5;i++)

{

 cout<<"Please enter a movie title #"<<i<<": ";

 cin>>movie;

 //Write to file, the names of each movies

 Myfile<<movie<<endl;

}

Myfile.close(); // Close file

//Create an Array for four elements

string myArr[4];

//Prepare to read from file to array

ifstream file(nm.c_str());

//Open File

if(file.is_open())

{

 for(int i =0;i<4;i++)

 {

  //Read each line of the file to array (line by line)

  file>>myArr[i];

 }

}

file.close(); // Close file

//Create a reverseOrder.txt file

nm = "reverseOrder.txt";

//Prepare to read into file

ofstream Myfile2(nm.c_str());

for(int j = 3;j>=0;j--)

{

 //Read each line of the file to reverseOrder.txt (line by line)

 Myfile2<<myArr[j]<<endl;

}

Myfile2.close(); //Close File

return 0;

}

See attached for .cpp program file

Answer:

The final tables in 3NF are as follows.

Project( ProjectNbr, ProjectName)

Employee( EmployeeNbr, EmployeeName, JobClass)

Job( JobClass, HourlyRate)

ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)

Explanation:

The given table is given below.

ProjectBilling( ProjectNbr, ProjectName, EmployeeNbr, EmployeeName, JobClass, HourlyRate, HoursBilled)  

ProjectNbr -> ProjectName

EmployeeNbr -> EmployeeName  

JobClass -> HourlyRate

ProjectNbr, EmployeeNbr -> HoursBilled

1NF

1. All the fields in the given table contain only a single value. The table is in 1NF.

2NF

2. New tables are formed based on the given functional dependencies.

Project( ProjectNbr, ProjectName)

Employee( EmployeeNbr, EmployeeName)

Job( JobClass, HourlyRate)

ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)

3. Every table is assigned a primary key which are as follows.

ProjectNbr is the primary key for Project table.

EmployeeNbr is the primary key for Project table.

JobClass is the primary key for Project table.

(ProjectNbr, EmployeeNbr) is the composite primary key for the ProjectBilling table.

4. The tables which are related to each other are linked via primary key and foreign key.

5. In the ProjectBilling table, the composite primary key, ProjectNbr, EmployeeNbr is composed of the primary keys of the Project and Employee tables, i.e., ProjectNbr and EmployeeNbr respectively.

6. Job table is related to Employee table. Hence, primary key of Job table, JobClass, is introduced as foreign key in Employee table.

Employee( EmployeeNbr, EmployeeName, JobClass)

7. Partial dependency arises when composite primary key exists and non-prime attributes (columns other than the primary key) depend on a part of the primary key, i.e., partial primary key.

In the ProjectBilling table, no partial dependency exists.

8. All the tables are in 2NF as given below.

Project( ProjectNbr, ProjectName)

Employee( EmployeeNbr, EmployeeName, JobClass)

Job( JobClass, HourlyRate)

ProjectBilling( ProjectNbr, EmployeeNbr, HoursBilled)

3NF

9. All the tables are in 2NF.

10. In every table, all non-prime attribute depend only on the primary key.

11. No transitive dependency exists, i.e., all non-prime attributes do not depend on other non-prime attributes.

12. Hence, all the conditions are satisfied and the tables are in 3NF.

Answer:

?

Explanation:

?

#!/bin/bash

usage() {

       echo "Usage: $0 [ -f outfile ] name password" 1>&2

       exit 1

}

while getopts "f:" o; do

       case "${o}" in

       f)

               filename=${OPTARG}

           ;;

       *)

               usage

           ;;

   esac

done

shift $((OPTIND-1))

name=$1

password=$2

if [ -z "$password" ] ; then

       usage

fi

set -x

if [ -z "$filename" ] ; then

       useradd -p `crypt $password` $name

else

       useradd -p `crypt $password` $name > $filename

fi

Answer:

it is two by five credit limit

Answer:

Your answer is A Tamera’s credit limit is most likely higher than Rupert’s, and her interest rate is most likely lower.

Explanation:

I got it right on the test and I hope this answer helps

Answer:

C. Maximal Flow Method

Explanation:

When it comes to optimization theory, Maximal Flow Method or problems have to do with finding a possible or viable flow through a flow network that obtains the maximum potential flow rate. The maximum flow problem can be viewed as an exceptional case of more complex and complicated network flow problems, such as that of the circulation problem.

Answer:

Print Area

Explanation:

First, I'll assume Donte is making use of a spreadsheet application (Microsoft Office Excel, to be precise).

From the given options, only "Print Area" fits the given description in the question.

The print area feature of the Excel software allows users to print all or selected workbook.

To print a selection section, Donte needs to follow the steps below.

Go to File -> Print.

At this stage, Donte will have the option to set the print features he needs.

Under settings, he needs to select "Print Selection"; this will enable him Print sections of the entire workbook.

Refer to attachment for further explanation

Answer:

C. print area

Explanation:

Answer:

We made use of the dynamic programming method to solve a linear-time algorithm which is given below in the explanation section.

Explanation:

Solution

(1) By making use of the  dynamic programming, we can solve the problem in linear time.

Now,

We consider a linear number of sub-problems, each of which can be solved using previously solved sub-problems in constant time, this giving a running time of O(n)

Let G[t] represent the sum of a maximum sum contiguous sub-sequence ending exactly at index t

Thus, given that:

G[t+1] = max{G[t] + A[t+1] ,A[t+1] } (for all   1<=t<= n-1)

Then,

G[0] = A[0].

Using the above recurrence relation, we can compute the sum of the optimal sub sequence for array A, which would just be the maximum over G[i] for 0 <= i<= n-1.

However, we are required to output the starting and ending indices of an optimal sub-sequence, we would use another array V where V[i] would store the starting index for a maximum sum contiguous sub sequence ending at index i.

Now the algorithm would be:

Create arrays G and V each of size n.

G[0] = A[0];

V[0] = 0;

max = G[0];

max_start = 0, max_end = 0;

For i going from 1 to n-1:

// We know that G[i] = max { G[i-1] + A[i], A[i] .

If ( G[i-1] > 0)

G[i] = G[i-1] + A[i];

V[i] = V[i-1];

Else

G[i] = A[i];

V[i] = i;

If ( G[i] > max)

max_start = V[i];

max_end = i;

max = G[i];

EndFor.

Output max_start and max_end.

The above algorithm takes O(n) time .

Answer:

False.

Explanation:

The correct option is false that is to say the arrangement is not true. So, let us check one or two things about the RSA System.

The RSA system simply refer to as a cryptosystem that was first established in the year 1973 but was later developed in the year 1977 by Ron Rivest, Adi Shamir and Leonard Adleman.

The main use of the RSA system is for encryption of messages. It is known as a public key cryptography algorithm.

The way the algorithm should have been arranged is;

(1). Randomly select two large prime numbers p and q, which always must bekept secret.

(2). Determine the product n = pq, known as public modulus.

(3). Select an integer number E, known as the public exponent, such that (p 1)and E have no common divisors, and (q 1) and E have no commondivisors.

(4). Determine the private exponent, D, such that (ED 1) is exactly divisible by both (p 1) and (q 1). In other words, given E, we choose D such that the integer remainder when ED is divided by (p1)(q 1) is 1.

(5). Release publicly the public key, which is the pair of numbers n and E, K =(n, E). Keep secret the private key, K = (n, D).

Therefore, The way it should have been arranged in the question should have been;

(1) , (4), (2), (3 ) and ( 5).

Answer:

Explanation:

public class ItemToPurchase {

private String itemName ;

private int itemPrice;

private int itemQuantity;

public ItemToPurchase(){

itemName = "none";

itemPrice = 0;

itemQuantity = 0;

}

public String getName() {

return itemName;

}

public void setName(String itemName) {

this.itemName = itemName;

}

public int getPrice() {

return itemPrice;

}

public void setPrice(int itemPrice) {

this.itemPrice = itemPrice;

}

public int getQuantity() {

return itemQuantity;

}

public void setQuantity(int itemQuantity) {

this.itemQuantity = itemQuantity;

}

}

ShoppingCartPrinter.java

import java.util.Scanner;

public class ShoppingCartPrinter {

public static void main(String[] args) {

Scanner scan = new Scanner(System.in);

ItemToPurchase item1 = new ItemToPurchase();

ItemToPurchase item2 = new ItemToPurchase();

System.out.println("Item 1");

System.out.print("Enter the item name: ");

String name1 = scan.nextLine();

System.out.print("Enter the item price: ");

int price1 = scan.nextInt();

System.out.print("Enter the item quantity: ");

int quantity1 = scan.nextInt();

item1.setName(name1);

item1.setPrice(price1);

item1.setQuantity(quantity1);

scan.nextLine();

System.out.println("Item 2");

System.out.print("Enter the item name: ");

String name2 = scan.nextLine();

System.out.print("Enter the item price: ");

int price2 = scan.nextInt();

System.out.print("Enter the item quantity: ");

int quantity2 = scan.nextInt();

item2.setName(name2);

item2.setPrice(price2);

item2.setQuantity(quantity2);

System.out.println("TOTAL COST");

int item1Total = item1.getPrice() * item1.getQuantity();

int item2Total = item2.getPrice() * item2.getQuantity();

System.out.println(item1.getName()+" "+item1.getQuantity()+" for $"+item1.getPrice()+" = $"+item1Total);

System.out.println(item2.getName()+" "+item2.getQuantity()+" for $"+item2.getPrice()+" = $"+item2Total);

System.out.println();

System.out.println("Total: $"+(item1Total + item2Total));

ItemToPurchase.java

======================================================

package brainly;

//Class header definition for class ItemToPurchase

public class ItemToPurchase {

   //Write the private fields

   private String itemName;

   private int itemPrice;

   private int itemQuantity;

   //Create the default constructor

   //And initialize all the fields to their respective initial values

   public ItemToPurchase(){

       this.itemName = "none";

       this.itemPrice = 0;

       this.itemQuantity = 0;

   }

   //Create the accessor methods as follows:

   public String getItemName() {

       return itemName;

   }

   public void setItemName(String itemName) {

       this.itemName = itemName;

   }

   public int getItemPrice() {

       return itemPrice;

   }

   public void setItemPrice(int itemPrice) {

       this.itemPrice = itemPrice;

   }

   public int getItemQuantity() {

       return itemQuantity;

   }

   public void setItemQuantity(int itemQuantity) {

       this.itemQuantity = itemQuantity;

   }

}     // End of class definition

=======================================================

ShoppingCartPrinter.java

=======================================================

package brainly;

//import the Scanner class

import java.util.Scanner;

//Class header definition for class ShoppingCartPrinter

public class ShoppingCartPrinter {

   public static void main(String[] args) {

       //Create an object of the Scanner class

       Scanner scnr = new Scanner(System.in);

       //prompt the user to enter the name of first item

       System.out.println("Item 1: Enter the item name");

       //Store the input in a String variable

       String firstItemName = scnr.nextLine();

       //Prompt the user to enter the price of the first item

       System.out.println("Enter the item price");

       //Store the input in an int variable

       int firstItemPrice = scnr.nextInt();

       //Prompt the user to enter the quantity of the first item

       System.out.println("Enter the item quantity");

       //Store the input in an int variable

       int firstItemQuantity = scnr.nextInt();

       //Call the scnr.nextLine() method to allow user enter the

       // second item

       scnr.nextLine();

       //Prompt the user to enter the name of the second item

       System.out.println("Item 2: Enter the item name");

       //Store the input in a String variable

       String secondItemName = scnr.nextLine();

       //Prompt the user to enter the price of the second item

       System.out.println("Enter the item price");

       //Store the input in an int variable

       int secondItemPrice = scnr.nextInt();

       //Prompt the user to enter the quantity of the second item

       System.out.println("Enter the item quantity");

       //Store the input in an int variable

       int secondItemQuantity = scnr.nextInt();

       //Create the first object using the first item

       ItemToPurchase item1 = new ItemToPurchase();

       //Create the second object using the second item

       ItemToPurchase item2 = new ItemToPurchase();

       //Get the total cost of the first item

       int totalcostofitem1 = firstItemPrice * firstItemQuantity;

       //Get the total cost of the second item

       int totalcostofitem2 = secondItemPrice * secondItemQuantity;

       //Calculate total cost of the two items

       int total = totalcostofitem1 + totalcostofitem2;        

       //Print out the result.

       System.out.println("Total: $" + total);

   }

}

=======================================================

=======================================================

>> Item 1: Enter the item name

Chocolate Chips

>> Enter the item price

3

>> Enter the item quantity

1

>> Item 2: Enter the item name

Bottled Water

>> Enter the item price

1

>>Enter the item quantity

10

>>Total: $13

======================================================

The above code has been written in Java and it contains comments explaining every line of the code. Please go through the comments carefully.

The actual lines of code are written in bold face to distinguish them from comments.

Also, in the sample output, the user inputs are written in bold face.

Answer:

Check the explanation

Explanation:

1.

void func(void){

int a,b,c; /*definition 1*/

/////* a,b,c for definition 1 are visible */

//// d, e are not visible

while(...){

int b, c, d; /* definition 2*/

////*

a from definition 1 is visible

b, c, d from definition 2 are visible

*/ ///

while(...){

int c, d, e;/* definition 3*/

////*

a from definition 1 is visible

b from definition 2 is visible

c, d, e from definition 3 are visible

*/ ///

}

////*

a from definition 1 is visible

b, c, d from definition 2 are visible

e not visible

*////

}

/////* a,b,c for definition 1 are visible */

///// d, e are not visible

}

Answer:

see explaination

Explanation:

#include <fstream>

#include <iostream>

#include <string>

//to set precision

#include <iomanip>

#include <bits/stdc++.h>

//structure of student is created.

struct Student

{

std::string name;

double grade1;

double grade2;

double grade3;

double Average;

// Constructor initializes semester scores to 0.

Student()

{

grade1 = 0;

grade2 = 0;

grade3 = 0;

Average = 0;

}

//Read from the file and calculate the average, min and max grade with name.

void read(std::fstream& fil)

{

std::string temp;

fil >> name;

fil >> temp;

grade1 = std::stod(temp);

fil >> temp;

grade2 = std::stod(temp);

fil >> temp;

grade3 = std::stod(temp);

Average = ((grade1 + grade2 + grade3) /3.0);

}

//write to file- name and average with precision 2.

void write(std::fstream& fil)

{

fil << name << " " << std::setprecision(2) << std::fixed << Average << std::endl;

}

//Logic to calculate the minimum grade of the class.

double min()

{

double min = grade1;

if(min > grade2)

{

min = grade2;

}

if(min > grade3)

{

min = grade3;

}

return min;

}

//Logic to calculate the maximum grade of the class.

double max()

{

double max = grade1;

if(max < grade2)

{

max = grade2;

}

if(max < grade3)

{

max = grade3;

}

return max;

}

//Print the name and average with precision 2 in the console.

void print()

{

std::cout << name << " " ;

std::cout << std::setprecision(2) << std::fixed << Average << std::endl;

}

};

//As per this question,students are set to 3.

Student students[3];

//main method.

int main()

{

// Read from the file question2.txt

std::fstream fil;

fil.open("question2.txt", std::ios::in);

for(int i = 0; i < 3; i++)

students[i].read(fil);

fil.close();

// finding the max grade of the class.

int max = students[0].max();

int temp = students[1].max();

if(max <temp)

max = temp;

temp = students[2].max();

if(max < temp)

max = temp;

// finding the min grade of the class.

int min = students[0].min();

temp = students[1].min();

if(min > temp)

min = temp;

temp = students[2].min();

if(min > temp)

min = temp;

// print the output in console.

std::cout << "Max Grade " << max << std::endl;

std::cout << "Min Grade " << min << std::endl;

for(int i = 0; i < 3; i++)

students[i].print();

// write to output file output2.txt .

fil.open("output2.txt", std::ios::out);

fil << "Max Grade " << max << std::endl;

fil << "Min Grade " << min << std::endl;

for(int i = 0; i < 3; i++)

students[i].write(fil);

fil.close();

return 0;

}

Answer:

Big data is a field that treats ways to analyze, systematically extract information from, or otherwise deal with data sets that are too large or complex to be dealt with by traditional data-processing application software. ... Big data was originally associated with three key concepts: volume, variety, and velocity.

SPANISH TRANSLATION

Big data es un campo que trata formas de analizar, extraer sistemáticamente información de, o de otra manera tratar con conjuntos de datos que son demasiado grandes o complejos para ser manejados por el software tradicional de aplicación de procesamiento de datos. ... Big data se asoció originalmente con tres conceptos clave: volumen, variedad y velocidad.

Answer:

A  C++ program was used in writing a function in merging ( vector a, vector b)   or that will merge the two vectors into one new one and return the merged vector.

Explanation:

Solution

THE CODE:

#include <iostream>

#include <vector>

using namespace std;

vector<int> merge(vector<int> a, vector<int> b) {

   vector<int> vec;

   int i = 0, j = 0;

   while(i < a.size() || j < b.size()) {

       if(i >= a.size() || (j < b.size() && b[j] < a[i])) {

           if(vec.empty() || vec[vec.size()-1] != b[j])

               vec.push_back(b[j]);

           j++;

       } else {

           if(vec.empty() || vec[vec.size()-1] != a[i])

               vec.push_back(a[i]);

           i++;

       }

   }

   return vec;

}

int main() {

   vector<int> v1 = {2, 4, 6, 8};

   vector<int> v2 = {1, 3, 7, 10, 13};

   vector<int> vec = merge(v1, v2);

   for(int i = 0; i < vec.size(); ++i) {

       cout << vec[i] << " ";

   }

   cout << endl;

   return 0;

}

Answer:

see explaination

Explanation:

The data is collected in a comma-separated values (CSV) format that always includes the following fields:

? date: string

? time: string

? client_ip: string

? server_ip: string

? url_stem: string

? url_query: string

? client_bytes: integer

? server_bytes: integer

What should you do?

a. Load the data as lines of text into an RDD, then split the text based on a comma-delimiter and load the RDD into DataFrame.

# import the module csv

import csv

import pandas as pd

# open the csv file

with open(r"C:\Users\uname\Downloads\abc.csv") as csv_file:

# read the csv file

csv_reader = csv.reader(csv_file, delimiter=',')

# now we can use this csv files into the pandas

df = pd.DataFrame([csv_reader], index=None)

df.head()

b. Define a schema for the data, then read the data from the CSV file into a DataFrame using the schema.

from pyspark.sql.types import *

from pyspark.sql import SparkSession

newschema = StructType([

StructField("date", DateType(),true),

StructField("time", DateType(),true),

StructField("client_ip", StringType(),true),

StructField("server_ip", StringType(),true),

StructField("url_stem", StringType(),true),

StructField("url_query", StringType(),true),

StructField("client_bytes", IntegerType(),true),

StructField("server_bytes", IntegerType(),true])

c. Read the data from the CSV file into a DataFrame, infering the schema.

abc_DF = spark.read.load('C:\Users\uname\Downloads\new_abc.csv', format="csv", header="true", sep=' ', schema=newSchema)

d. Convert the data to tab-delimited format, then read the data from the text file into a DataFrame, infering the schema.

Import pandas as pd

Df2 = pd.read_csv(‘new_abc.csv’,delimiter="\t")

print('Contents of Dataframe : ')

print(Df2)

Answer:

Explanation:

The objective here is to write a collection of MySetOperations in Python.

During the process of  submitting this answer after computing the program ; I keep getting a response message that goes thus "It appears that your answer contains either a link or inappropriate words. Please correct and submit again!"  . In the bid to curb such dilemma , i have created a word document for the collection of MySetOperations in Python.

The document can be seen in the attached file below.

Answer:

import java.util.Scanner;

public class StudentScores

{

public static void main(String[] args) {

 Scanner scnr = new Scanner(System.in);

 final int NUM_POINTS = 4;

 int[] dataPoints = new int[NUM_POINTS];

 int controlValue;

 int i;

 controlValue = scnr.nextInt();

 for (i = 0; i < dataPoints.length; ++i) {

     dataPoints[i] = scnr.nextInt();

 }

 for (i = 0; i < dataPoints.length; ++i) {

     System.out.print(dataPoints[i] + " ");

 }

 System.out.println();

 for (i = 0; i < dataPoints.length; ++i) {

     if(dataPoints[i] < controlValue){

         dataPoints[i] = dataPoints[i] * 2;          

     }

     System.out.print(dataPoints[i] + " ");

 }

}

}

Explanation:

*Added parts highligted.

After getting the control value and values for the array, you printed them.

Create a for loop that iterates through the dataPoints. Inside the loop, check if a value in dataPoints is smaller than the contorolValue. If it is, multiply that value with 2 and assign it to the dataPoints array. Print the elements of the dataPoints

Answer:

import java.util.Scanner;

public class numm3 {

   public static void main (String [] args) {

       Scanner scnr = new Scanner(System.in);

       final int NUM_POINTS = 4;

       int[] dataPoints = new int[NUM_POINTS];

       int controlValue;

       int i;

       System.out.println("Enter the control Variable");

       controlValue = scnr.nextInt();

       System.out.println("enter elements for the array");

       for (i = 0; i < dataPoints.length; ++i) {

           dataPoints[i] = scnr.nextInt();

       }

       for (i = 0; i < dataPoints.length; ++i) {

           System.out.print(dataPoints[i] + " ");

       }

       System.out.println();

       //Doubling elements Less than Control Variable

       for (i = 0; i < dataPoints.length; ++i) {

           if (dataPoints[i]<controlValue){

               dataPoints[i] = dataPoints[i]*2;

           }

           System.out.print(dataPoints[i] + " ");

       }

       System.out.println(); } }

Explanation:

See the additional code to accomplish the task in bold

The trick is using an if statement inside of a for loop that checks the condition (dataPoints[i]<controlValue) If true, it multiplies the element by 2

Answer:

A program in C++ was written to prompts the user to enter the area of the flat cardboard.

Explanation:

Solution:

The C++ code:

#include <iostream>

#include <iomanip>

#include <cmath>

using namespace std;

double min(double,double);

void max(double,double,double&,double&);

int main()

{double area,length,width=.001,vol,height,maxLen,mWidth,maxHeight,maxVolume=-1;

cout<<setprecision(3)<<fixed<<showpoint;

cout<<"Enter the area of the flat cardboard: ";

cin>>area;

while(width<=area)

{length=area/width;

max(length,width,vol,height);

if(vol>maxVolume)

{maxLen=length;

mWidth=width;

maxHeight=height;

maxVolume=vol;

}

width+=.001;

}

cout<<"dimensions of card to maximize the cardboard box which has a volume "

<<maxVolume<<endl;

cout<<"Length: "<<maxLen<<"\nWidth: "<<maxLen<<endl;

cout<<"dimensions of the cardboard box\n";

cout<<"Length: "<<maxLen-2*maxHeight<<"\nWidth: "

<<mWidth-2*maxHeight<<"\nHeight: "<<maxHeight<<endl;

return 0;

}

void max(double l,double w,double& max, double& maxside)

{double vol,ht;

maxside=min(l,w);

ht=.001;

max=-1;

while(maxside>ht*2)

{vol=(l-ht*2)*(w-ht*2)*ht;

if(vol>max)

{max=vol;

maxside=ht;

}

ht+=.001;

}

}

double min(double l,double w)

{if(l<w)

return l;

return w;

}

Note:  Kindly find the output code below

/*

Output for the code:

Enter the area of the flat cardboard: 23

dimensions of card to maximize the cardboard box which has a volume 0.023

Length: 4.796

Width: 4.796

dimensions of the cardboard box

Length: 4.794

Width: 4.794

Height: 0.001

*/

Answer:

a.

Explanation:

The option that correctly declares an array is int array.

This is a term that connote a regular system or arrangement. Here,  numbers or letters are said to be arranged in rows and columns.

In computing, An array is a term that describe memory locations box having the same name. All data in an array is said to  be of the same data type.

Learn more about array from

https://brainly.com/question/26104158

Answer:

The reason is due to proprietary design of the Operating System (OS) which require a virtualization software to blanket or "disguise" the hardware (processor) borderlines of the computer onto which it is to be installed.

Explanation:

An Apple system that has the RISC processor and system architecture which has an operating system made entirely for the Apple system architecture

If the above Apple OS is to be installed on a windows computer, then the procedure to setup up the OS has to be the same as that used when on an Apple system, hence, due to the different processors and components of both systems, a virtualization will be be needed to be provided by a Virtual box, Parallels desktop or other virtualization software.

Answer:

Check the explanation

Explanation:

INCLUDE Irvine32.inc

TRUE = 1

FALSE = 0

.data

gradeAverage WORD ?

credits WORD ?

oKToRegister BYTE ?

str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0

main PROC

call CheckRegs

exit

main ENDP

CheckRegs PROC

push edx

mov OkToRegister,FALSE

; Check credits for valid range 1-30

cmp credits,1 ; credits < 1?

jb E1

cmp credits,30 ; credits > 30?

ja E1

jmp L1 ; credits are ok

; Display error message: credits out of range

E1:

mov edx,OFFSET str1

call WriteString

jmp L4

L1:

cmp gradeAverage,350 ; if gradeAverage > 350

jna L2

mov OkToRegister,TRUE ; OkToRegister = TRUE

jmp L4

L2:

cmp gradeAverage,250 ; elseif gradeAverage > 250

jna L3

cmp credits,16 ; && credits <= 16

jnbe L3

mov OkToRegister,TRUE ; OKToRegister = TRUE

jmp L4

L3:

cmp credits,12 ; elseif credits <= 12

ja L4

mov OkToRegister,TRUE ; OKToRegister = TRUE

L4:

pop edx ; endif

ret

CheckRegs ENDP

END main

Answer:

The code implements a recursive function by decrementing number  by 1 from 10 to 4 recursively from the stack and add an extra space 2 using  the macro function

Explanation:

Solution

The  code implements a recursive function by decrementing number  by 1 from 10 to 4 recursively from the stack and add an extra space 2 using  the macro function mwritespace

The first number is printed after the code executes is 9

Code to be used:

Include Irvine32.inc

INCLUDE Macros.inc               ;Include macros for space

.code

main proc

  push 4                   ;Push 4 to the stack

  push 10                   ;Push 10 to the stack

  call rcrsn                   ;call the function  

  exit

  ;invoke ExitProcess,0           ;After return from the function

                      ;call exit function

  main ENDP               ;end the main

  rcrsn PROC               ;define the function

      push ebp               ;push the ebp into stack

      mov ebp,esp           ;set ebp as frame pointer

      mov eax,[ebp + 12]       ;This is for second parameter

      mov ebx,[ebp + 8]       ;This is for first parameter

      cmp eax,ebx           ;compare the registers

      jl recurse           ;jump to another function  

      jmp quit               ;call quit

      recurse:               ;Implement another function  

      inc eax               ;increment count value

      push eax               ;push the eax value into stack

      push ebx               ;push ebx into stack

      call rcrsn           ;call above function

      mov eax,[ebp + 12]       ;This is for second parameter

      call WriteDec           ;write the value on the screen

      mWritespace 2           ;Space macro print 2 spaces

                      ;pause the screen

      quit:               ;Implement quit function

      pop ebp               ;pop the pointer value from the stack

      ret 8               ;clean up the stack  

 rcrsn ENDP          

end main

Answer:Backspace

Explanation:

If you want to delete words you use the backspace button.

Answer:

Backspace

Explanation:

Answer:

Flaws and limitations identified in this program includes;

1.There was a not necessary usage of variable retrieval. Would have made use of canConvert.

2. Looking at the program, one will notice numerous typos. One of which is the fact that in JAVA we make use of Boolean instead of bool.

3.We rather use Integer.parseInt in JAVA and not Int16, cant make use of Int16.

4. The exception cant be printed

5. JAVA makes use of checkConversion instead of convertNumber as used in the program.

6. It cant work for decimal numbers, 0 and big integers.

Explanation:

See Answer for the detailed explaination of the flaws and limitations identified in the program.

Answer:

The program code is written in the explanation.

Explanation:

def turn_clockwise(d):

if d=="N":

return "E"

elif d=="E":

return "S"

elif d=="S":

return "W"

elif d=="W":

return "N"

else:

return "None"

Output:

Answer:

t

Explanation:

Answer:

dude what r u doin

Explanation:

Answer:

ierror(VLOOKUP(Q2,CBFStaff[[Staff ID]:[Name]],2,FALSE), "Invalid Staff ID")

Explanation:

Let me try as much as I can to explain the concept or idea of iferror in vlookup.

iferror have a typically function and result like an if else statement, its syntax is IFERROR(value,value _ if _ error), this simply means that if the the error is equal to value, value is returned if not, the next argument is returned.

Having said that, feom the question we are given,

let's substitute the value with vlookup function and add an else argument, it will look exactly this way;

IFERROR(VLOOKUP(),"Invalid Staff ID")// now this will set the message if vlookup cannot find the.

On the other hand using the values given, we will have;

ierror(VLOOKUP(Q2,CBFStaff[[Staff ID]:[Name]],2,FALSE), "Invalid Staff ID")