In this lab, you complete a partially written C++ program that includes a function that returns a value. The program is a simple calculator that prompts the user for two numbers and an operator ( +, -, *, or / ). The two numbers and the operator are passed to the function where the appropriate arithmetic operation is performed. The result is then returned to the main() function where the arithmetic operation and result are displayed. For example, if the user enters 3, 4, and *, the following is displayed: 3 * 4 = 12


The source code file provided for this lab includes the necessary variable declarations and input and output statements. Comments are included in the file to help you write the remainder of the program.


Instructions

1. Write the C++ statements as indicated by the comments.

2. Execute the program by clicking the Run button at the bottom of the screen.


The output works but I get 5 errors. I get a declared performOperation() error, and then I'm getting a Division, Multiplication, Addition and Subtraction error.

Code:

// Calculator.cpp - This program performs arithmetic, ( +. -, *. / ) on two numbers.

// Input: Interactive

// Output: Result of arithmetic operation


#include

#include


using namespace std;


// Write performOperation() function declaration here


int performOperation(double numberOne, double numberTwo, string op);

int main()

{

double numberOne, numberTwo;

string operation;

double result;

cout << "Enter the first number: ";

cin >> numberOne;

cout << "Enter the second number: ";

cin >> numberTwo;

cout << "Enter an operator (+.-.*,/): ";

cin >> operation;


// Call performOperation method here


result = performOperation(numberOne,numberTwo,operation);

cout << numberOne;

cout << " " << operation << " ";

cout << numberTwo;

cout << " = ";

cout << result << endl;

return 0;

} // End of main() function


// Write performOperation function here

int performOperation(double numberOne, double numberTwo,std::string(op)){

double result;

if (op == "+")

result = numberOne + numberTwo;

else if (op == "-")

result = numberOne - numberTwo;

else if (op == "*")

result = numberOne*numberTwo;

else if (op == "/")

result = numberOne/numberTwo;

else

cout<<"Wrong input. Please try again.";

return result;

}

Answer:

I think you should off it and let it cool down by itself.

Explanation:

Hope it helps and have a nice day! :)

BRAINIEST is appreciated it would really help

Answer:

Scanner

Explanation:

The correct answer is Scanner. Input devices are used to enter data or instructions into a computer.

Option c is correct. Node.insertBefore(new, child) is the method that includes or embeds a new node directly ahead of a child node.

The Node interface's insertBefore() method places a node as a child of a given parent node ahead of a reference node.

InsertBefore() transfers the specified node to the new place if it already exists in the document. (That is, before adding it to the given new parent, it will automatically remove it from its current parent.)

As a result, a node cannot be present in two places in the document at once. The full contents of the DocumentFragment are moved into the child list of the chosen parent node if the given child is a DocumentFragment. the node that newNode is added before. In the event that this is null, newNode is appended to the end of the node's child nodes.

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Answer:

I'm generally puzzled

Explanation:

The statements that catch the exception in the program are:

      age = int(parts[1]) + 1

      print('{} {}'.format(name, age))

      # Get next line

      parts = input ( ) . split ( )

      name = parts[0]

  except ValueError:

      print ( ' Invalid value ! ' )

      # Get next line

      parts = input().split()

      name = parts [ 0 ]

# Split input into 2 parts: name and age

parts = input().split()

name = parts[0]

while name != '-1':

  # FIXME: The following line will throw ValueError exception.

  # Insert try/except blocks to catch the exception.

  try:

      age = int ( parts [ 1 ] )  + 1

      print( ' { } { } ' . format ( name, age ) )

      # Get next line

     parts = input( ) . split ( )

      name = parts[0]

  except ValueError:

      print ( ' Invalid value ! ' )

      # Get next line

      parts = input ( ) . split ( )

      name = parts[0]

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A tool which Kyla can use to store the document with a different name and makes changes in the new document is: B. Save As.

A document simply refers to a computer resource that is designed and developed to enable end users to easily store data as a single unit on a computer storage device.

Generally speaking, all computer documents can be identified by a title (name), date modified, size, and type such as the following;

A Save as command can be defined as a type of command associated with the file menu of a software application or document and it causes a copy of the current file to be created and stored to a different location, file name, and/or file type.

In this context, we can reasonably infer and logically deduce that Kyla  should save the document by using the "Save as" command.

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Since Tom is interested in becoming a dental hygienist and he asks the hygienist there if she would be willing to talk to him about what her job is like, this talk is called  D. counseling appointment.

In order to make decisions about your job, education, and overall well-being, career counseling is a process that will help you get to know and understand yourself as well as the working environment.

Note that Choosing a major and determining what profession you want to acquire after graduation are only two aspects of career development.

A career counselor will assist a client in exploring abilities and strengths, taking into account education levels and others.

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A script that(1) gets from a user: (i) a paragraph of plain text and (ii) a distance value.distance value. next, encrypts the text using Caeser cipher, then prints it as an encrypted text given below:

The below script also reads the file and then writes it to a file named ‘copyfile.txt’ and print the text file.

mapping = {}

with open ( " copyfile . txt " , " r " ) as keyFile:

   for line in copyFile:

      l1, l2 = line . split ( )

       mapping [ upper ( l1 ) ] = upper ( l2 )

       decrypt = " "

       with open ( " encrypted.txt " , " r " ) as encryptedFile:

           for line in encryptedFile:

               for char in line:

                   char = upper ( char )

                   if char in mapping:

                       decrypt + = mapping [ char ]

                   else:

                       decrypt += char print ( decrypt )

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Bidirectional microphones have a very constricted pickup angle, making them useful for separating a single voice or mechanism in the presence of other sound sources, as lengthy as nothing is directly behind the mic. They are useful for picking up two sources that are next to each other.

The pressure gradient principle underpins this type of microphone. The sound can travel through the membrane on both sides.

When the sound on the front and backside of the diaphragm differs, the microphone sends a signal.

Bidirectional microphones have a particularly narrow pickup angle, which makes them advantageous for distinguishing a single voice or framework in the involvement of other sound sources, as long as nothing is directly behind the mic.

They are useful for detecting two sources that are close together.

Thus, in this setting a bidirectional microphone can be used.

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Answer: search on web it’s faster than brainly

Explanation:

Answer: Its B. 1,000,000 per second

Explanation: One hertz is one tick per second. A computer with a one million hertz processer would have 1,000,000 ticks per second.

A C program segment for prompting a user to enter an integer and save it in an option variable is given below:

#include <stdio.h>

#include <ctype.h>

int main(void)

{

   char ch;

   float num1,num2,answer;

   printf("Enter the operation of your choice:\n");

   printf("a. add     s. subtract\n");

   printf("m. multiply q. divide\n");

   printf("q. quit\n");

   while ((ch = getchar())!='q')

   {

       ch=tolower(ch);

       //Ignore whitespace

       if (ch=='\n')

           continue;

       else

       {

           switch(ch)

           {

               //Addition part

               case 'a':

                   //First number

                   printf("Enter first number: ");

                   //Check to see if input is a number

               while (scanf("%f",&num1)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               //Second number

              printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               //Do math for respective operation

              answer = num1 + num2;

               //Print out result

               printf("%.3f + %.3f = %.3f\n", num1,num2,answer);

               break;

           //Subtraction part

           case 's':

              printf("Enter first number: ");

               while (scanf("%f",&num1)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               answer = num1 - num2;

               printf("%.3f - %.3f = %.3f\n", num1,num2,answer);

               break;

           //Multiplication part

           case 'm':

               printf("Enter first number: ");

               while (scanf("%f",&num1)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               answer = num1 * num2;

               printf("%.3f * %.3f = %.3f\n", num1,num2,answer);

               break;

           //Division part

           case 'd':

               printf("Enter first number: ");

               while (scanf("%f",&num1)==0)

               {

                  printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

               printf("Enter second number: ");

               while (scanf("%f",&num2)==0)

               {

                   printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                   scanf("%*s");

               }

              //Check for if number is a zero

               while (num2==0)

               {

                   printf("Please enter a non-zero number, such as 2.5, -1.78E8, or 3: ");

                   while (scanf("%f",&num2)==0)

                   {

                       printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");

                       scanf("%*s");

                   }

               }

               answer = num1 / num2;

               printf("%.3f / %.3f = %.3f\n", num1,num2,answer);

               break;

           //For if a non-valid operation is entered

           default:

               printf("That is not a valid operation.\n");

               break;

       }

   }

  printf("Enter the operation of your choice:\n");

   printf("a. add      s. subtract\n");

   printf("m. multiply q. divide\n");

   printf("q. quit\n");

}

printf("Bye.\n");

return 0;

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Answer:

c = 'z'

while c >= 'a':

 print(c)

 c = c h r(ord(c)-5)

Explanation:

Keep in mind, you could have also chosen to do:

print("z\nu\np\nk\nf\na\n")

but it seems the character arithmetic is what is asked for...

The edited and complete program is as follows

theSum = 0.0

count = 0

while True:

   number= input("Enter a number or press Enter to quit:")

   if not number:

       print("The sum is", theSum)

       if count>0:

           print("The average is ", theSum/count)

       exit()

   theSum+=float(number)

   count+=1

The complete program that calculates the average and the sum of numbers entered by the user is as follows:

Note that the program ends when the user presses the enter key and comments are used to explain each line

#This initializes the sum

theSum = 0.0

#This initializes the counter

count = 0

#This ensures the user enters numbers repeatedly

while True:

   #This gets the input

   number= input("Enter a number or press Enter to quit:")

   #When the user presses enter

   if not number:

       #This prints the sum and the average

       print("The sum is", theSum)

       if count>0:

           print("The average is ", theSum/count)

       #This exits the program

       exit()

   #This calculates the sum

   theSum+=float(number)

   count+=1

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Answer:

b(ii) prices[1]= 20.50;  // do not declare with double again

b (iii) scanf("%lf", &prices[4]);  //prices[4] is last element

d) See below

Explanation:

/* This program uses the string.h header file

The strncat() function in this file concatenates two strings. It takes three arguments, dest and source strings and number of characters to concatenate

*/

#include <stdio.h>
#include <string.h>

int main() {
char s1[10] = "Happy";
char s2[10] = " Birthday";
char s3[10]="";

strncat(s3, s1, 5);  //concatenate 5 characters of s1 to end of s3 and                            
                                    //store result in s3; s = "Happy"
strncat(s3, s2, 10); //concatenate s2 with current contents of s3
                                      //s3 becomes "Happy Birthday
printf("%s", s3);

return 0;

}

Answer:

Storage space

Explanation:

To partition and format a hard drive when no operating system has been installed, you may use either GParted or Windows installation media to boot the system and run disk setup utilities.

A hard drive simply refers to an electro-mechanical, non-volatile data storage device that is made up of magnetic disks (platters) that rotates at high speed.

Partition can be defined as a terminology which is used to describe the division of a hard-disk drive into two or more fragments.

In Computer technology, GParted or Windows installation media should be used to boot a computer system and run disk setup utilities, especially when you wish to partition and format a hard drive when there is no operating system (OS) installed.

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Complete Question:

To partition and format a hard drive when no operating system has been installed, you may use either__ or __ to boot the system and run disk setup utilities.

Please choose from a list of words:

basic disk

chkdsk

defragmentation

Disk Cleanup

Disk Management

dynamic disk

Error-checking

format

GParted

partition

volumes

Windows installation media

In this code block, the monkey sprite is an example of both an input and a option d: function.

It is an Another illustration is a function that directs a rat or a monkey to an object. "goto" is the name of this function. You can utilize the same code with multiple values by using arguments in a function. When using the function, the main code sends these values to it.

Note that In Game Lab, we refer to a group of values that depicts a character from a narrative, animation, or game as a sprite. Sprites can be saved in variables with labels, however unlike other types of values that you might have previously placed in variables, such numbers, sprites allow you to group together several related values under a single label.

Therefore, one can say that CoffeeScript and Python are two text-based programming languages covered by codeMonkey. The CoffeeScript is the name of the programming language employed in Coding Adventure. This computer language, which is related to JavaScript, is mostly used for web applications in the sector.

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Answer: parameter

Explanation:

Creating more accurate maps of a remote mountain range would be a good news for people in this line of work: C. surveying.

A topographic map can be defined as a type of map which contains an accurate and detailed graphical (two-dimensional) description, representation or records of both natural and man-made (artificial) features on the surface of planet Earth such as the following:

Generally speaking, surveying typically involves an examination and record of various information about the geographical area, landforms, and features of a particular region, especially through the use of maps.

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The time that is required for the packet to move over the the distance d at the speed s is  0.028 seconds.

The speed that It would take to move it in transmission is given as 0.008 seconds.

The details that we would have to solve the problem that we have here are given below

The distance = 5000 km, we convert to meters = 5000 * 1000

The speed of the propagation is given as (s) 2.5 x 108 m/s

The propagation delay is given as d / s

= 5000000 / 2.5 x 10^8

= 0.02 this is the value of the propagation

Next we have to find the length of the packet. L = 1000 bytes

this is equal to 8000 bits

The transmission rate is 1 Mbps = 1000000 bits

The transmission delay would be = L/R

= 8000 / 1000000

= 8 / 1000

= 0.008 seconds in transmission

The total required time that is to be gotten would be

Propagation + transmission

= 0.02 + 0.008

= 0.028

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One disadvantage each for list, array, and queue data structures are:

It is difficult to quickly search the queue. You must begin at the very end and might need to keep another queue. Don't even consider utilizing a queue if you have data that you want to make searchable in the future. It is difficult to add or remove items from the queue's middle.

Note that An array doesn't verify its borders: In the C programming language, we are unable to determine whether the values placed in an array exceed its size or not.

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Answer:

As a student I answered for students benefits only

Explanation:

I am explaining into list form so that everyone can understand it easily.

Benefits:

By sticking to this time plan, you have a better chance of tackling the task than getting to it with no predefined time allocated to it.

Proper time management enables you to prioritize tasks and tackle them first.

Instead of getting caught up in multitasking, you focus on one activity at a time for a specified duration.

Managing your time well allows you to get your work done on time. This elicits a sense of confidence and accomplishment in your capabilities. Getting through a long to-do list can also evoke these feelings.

A program that, given a square matrix, identifies the largest number and what position it has, and indicates how many even and odd numbers it has, is given below:

// C++ implementation to arrange

// odd and even numbers

#include <bits/stdc++.h>

using namespace std;

// function to arrange odd and even numbers

void arrangeOddAndEven(int arr[], int n)

{

  int oddInd = 1;

   int evenInd = 0;

   while (true)

   {

       while (evenInd < n && arr[evenInd] % 2 == 0)

           evenInd += 2;

       while (oddInd < n && arr[oddInd] % 2 == 1)

           oddInd += 2;

      if (evenInd < n && oddInd < n)

           swap (arr[evenInd], arr[oddInd]);

       else

           break;

   }

}

// function to print the array

void printArray(int arr[], int n)

{

   for (int i = 0; i < n; i++)

       cout << arr[i] << " ";

}

// Driver program to test above

int main()

{

   int arr[] = { 3, 6, 12, 1, 5, 8 };

   int n = sizeof(arr) / sizeof(arr[0]);

   cout << "Original Array: ";

  printArray(arr, n);

   arrangeOddAndEven(arr, n);

   cout << "\nModified Array: ";

   printArray(arr, n);

   return 0;

}

Original Array: 3 6 12 1 5 8

Modified Array: 6 3 12 1 8 5

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In this exercise we have to use the knowledge of C++ to write a program that is possible Define a function with no parameters that reads integers from input until integer 0 is read.

#include <iostream>

using namespace std;

bool

{

   int num;

   bool =true;

   do

   {

       cin>>num;

       if(num>=0 && num!=1)

           =false;

   }while(num!=1);

   return;

}

int main()

{

   bool allNegative;

   allNegative=

   if(allNegative==true)

       cout<<"All match";

   else

       cout<<"All not match";

   return 0;

}

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Answer:

A. freeze panes

Explanation:

num = int(input("User Input: "))

mylist = []

while(num !=0):

   mylist.append(num)

   num = int(input("User Input: "))

mylist.reverse()

for i in mylist:

   print(i,end = ' ')

What is a program?

A program is a set of instructions that a computer follows to complete a specific task.

This question is answered in Python, and it employs 0 as a sentinel value, which means that the program stops prompting the user for input when the user enters 0.

This line requests input from the user

num = int(input("User Input: "))

This line declares a list that is empty

mylist = []

This loop is repeated as long as user enters input other than 0

while(num !=0):

This appends user input to the list

   mylist.append(num)

This prompts user for another input

   num = int(input("User Input: "))

This reverses the list

mylist.reverse()

The following iteration the list in reverse order

for i in mylist:

   print(i,end = ' ')

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Answer:

Explanation: WE can do it justin university

Since Dana is using a cross-functional team to improve collaboration. She is using the Agile method of project development.

Agile is one that lets teams offer value to their clients more quickly and with fewer difficulties through an iterative approach to project management and software development.

Note that an agile team produces work in manageable, small-scale increments rather than staking all on a "big bang" launch.

Since the Agile technique divides a project into many parts and uses these phases to manage the project. Continuous improvement at every stage and ongoing collaboration with stakeholders are required. Teams cycle through a process of planning, carrying out, as well as assessing once the job starts.

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In this exercise we have to use the knowledge of C++ to write a program that is possible that asks the user to enter a number of seconds.

#include <iostream>

#include <string>

using namespace std;

int main() {

  int n;

  cout<<"Time Calculator"<<endl;

  cout<<"Enter the seconds you would like to calculate into days, hours, minutes: ";

  cin>>n;

  cout<<n<<" seconds have:"<<endl;

  cout<<"\t"<<(n/(24*60*60))<<" day(s)"<<endl;

  n = n % (24*60*60);

  cout<<"\t"<<(n/(60*60))<<" hours(s)"<<endl;

  n = n % (60*60);

  cout<<"\t"<<(n/(60))<<" minutes(s)"<<endl;

  n = n % (60);

  cout<<"\t"<<(n)<<" seconds(s)"<<endl;

  return 0;

}

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Answer:

See explanation below and try out this code to see what exactly is happening

# Program

j = 3

z = { ]  # empty dictionary

print(z)

z[2] = 1 # z

print(z)

z[7] = 3

print(z)

z[5] = 2

print(z)   # prints out {2: 1, 7: 3, 5: 2}

for l in z:

   print("Initial value of j = ", j)

   print("Value of l(key) = ", l)

   j += l

  print("Updated value of j = ", j)

print("Final value of j=", j)

Explanation:

What you have is a dictionary object. We use { } to represent a dictionary. A dictionary consists of keys and values related to the keys. For example, you can have a  dictionary which contains names of students and their GPAs

This is declared as
student_grades = {"Amy": 4.3, "Zach":4.0", "Bill": 3.2}

The names are referred to as keys and the GPA as values
You can also have both key and value as integers. Fir example you could store the cubes of numbers from 1 to 5 in a dictionary

cubes = {1: 1, 2:8, 3: 27, 4: 64, 5:125}

If you wanted to add the cube for 6 then you would use cubes[6] = 216 This creates a new entry in the dictionary 6:216 so that the dictionary is now

cubes = {1: 1, 2:8, 3: 27, 4: 64, 5:125, 6:216}

In this short program here is what is happening

j = 3   # variable j created and value set to 3

z = { }  # this creates an empty dictionary named z

z[2] = 1  #adds an entry with key = 2 and value = 1; z = {2 : 1}

z[7} = 3  #key  7 and value 2 added => z = {2 : 1, 7 : 3}

z[5] = 2  #key 5, value 2 => z = {2: 1, 7 : 3, 5 : 2}

So now there are 3 entries in the dictionary i.e. 3 pairs of key-values

When you iterate through the dictionary as in

for l in z:
 - you are actually iterating over the keys so l = 2, 7 and 5 on each iteration

So you are accumulating the key values

Here is what happens

First execution of  loop => j = 3 , l = 2 (first key =2), j = 3 + 2 = 5

Second time: l = 7 (second key value) , j = 5 + 7 = 12

Third time: l = 5, j = 12 + 5 = 17

There is a lot more to dictionaries and sometimes it can get complicated. Suggest you search the internet for the same

Putting print() statements also helps you visualize what is happening. Code is provided in the answer section to help you