Answer:
Tetris
Explanation:
What is the result of executing the following code? You can assume the code compiles and runs. #include using namespace std; void factorial(int n) { cout << n << '*'; factorial(n-1); } int main() { factorial(4); return 0; }
Answer:
The result of executing the code is 24.
Explanation:
Factorial of a number:
The factorial of a number is the multiplication of a number by all it's previous numbers until one. For example:
0! = 1
1! = 1
2! = 2*1 = 2
3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
In this question:
This is a C++ code, which is a recursive function to calculate a factorial of a number.
The input, given by factorial(4), is 4, so the result of executing the code is 24.
cout << n << '*'; factorial(n-1);
This means for each input, until n = 1, the output is the factorial of the number. This is the recursive function.
Write a function process_spec that takes a dictionary (cmap), a list (spec) and a Boolean variable (Button A) as arguments and returns False: if the spec has a color that is not defined in the cmap or if Button A was pressed to stop the animation Otherwise return True
Answer:
Explanation:
The following code is written in Python. The function takes in the three arguments and first goes through an if statement to check if Button A was pressed, if it was it returns False otherwise passes. Then it creates a for loop to cycle through the list spec and checks to see if each value is in the dictionary cmap. If any value is not in the dictionary then it returns False, otherwise, it returns True for the entire function.
def process_spec(cmap, spec, buttonA):
if buttonA == True:
return False
for color in spec:
if color not in cmap:
return False
return True
6. Which of the following games will most likely appeal to a casual player?
a) Wii Sport
b) Contra
c) Farmville
d) Both A and C
Answer:
A
Explanation:
Wii sports is a good game for a casual player. :)
Suppose the program counter is currently equal to 1700 and instr represents the instruction memory. What will be the access of the next instruction if:
instr[PC] = add $t0, $t1, $t2
Answer:
C
Explanation:
omputer chip
sorry btw
Susan needs to configure a WHERE condition in the code for a macro. Which option will she use?
Database Analyzer
Macro Builder
Expression Builder
Condition Formatter
Answer:condition formatter have a blessed day.
Explanation:
Answer:
D) Condition Formatter took the test
Explanation:
a user called to inform you that the laptop she purchased yesterday is malfunctioning and will not connect to her wireless network. You personally checked to verify that the wireless worked properly before the user left your office. the user states she rebooted the computer several times but the problems still persists. Which of the following is the first step you should take to assist the user in resolving this issue?
Write a script called pow.sh that is located in your workspace directory to calculate the power of two integer numbers (e.g. a ^ b); where a and b are passed as parameters from the command line (e.g. ./pow.sh 2 3). The result should be echoed to the screen as a single integer (e.g. 8).
Answer:
#! /usr/bin/bash
echo $[$1**$2]
Explanation:
(a) The first line of the code is very important as it tells the interpreter where the bash shell is located on the executing computer.
To get this on your computer, simply:
1. open your command line
2. type the following command: which bash
This will return the location where the bash shell is. In this case, it returned
/usr/bin/bash
(b) The second line is where the actual arithmetic takes place.
The variables $1 and $2 in the square bracket are the first and second arguments passed from the command line. The ** operator raises the left operand (in this case $1) to the power of the second operand (in this case $2).
Then, the echo statement prints out the result.
(c) Save this code in a file named as pow.sh then run it on the command line.
For example, if the following is run on the command line: ./pow.sh 2 3
Then,
the first argument which is 2 is stored in $1 and;
the second argument which is 3 is stored in $2.
The arithmetic operator (**) then performs the arithmetic 2^3 which gives 8.
-) Files are organized in:
i) RAM
iii) Directories
ii) Cache
iv) None of the above
Answer:
iii) Directories.
Explanation:
Files are organized in directories. These directories are paths that describes the location wherein a file is stored on a computer system. Thus, it is generally referred to as file folders.
For example, a directory would be C:/Users/Lanuel/Music/songs of joy.mp3.
Annapurno
Page
Date
corite
a programe to input length and the
breath of the rectangle and cliculate the
anecan perimeter .write a program to input length and breadth of the rectangle and calculate the area and perimeter
Answer:
The program in Python is as follows:
Length = float(input("Length: "))
Width = float(input("Width: "))
Area = Length * Width
print("Area: ",Area)
Explanation:
This gets input for length
Length = float(input("Length: "))
This gets input for width
Width = float(input("Width: "))
This calculates the area
Area = Length * Width
This prints the calculated area
print("Area: ",Area)
what is the output of this line of code?
print("hello"+"goodbye")
-"hello"+"goodbye"
-hello + goodbye
-hello goodbye
-hellogoodbye
Answer:
“hello” + “goodbye”
Explanation:
Define a class StatePair with two template types (T1 and T2), a constructor, mutators, accessors, and a PrintInfo() method. Three vectors have been pre-filled with StatePair data in main():vector> zipCodeState: ZIP code - state abbreviation pairsvector> abbrevState: state abbreviation - state name pairsvector> statePopulation: state name - population pairsComplete main() to use an input ZIP code to retrieve the correct state abbreviation from the vector zipCodeState. Then use the state abbreviation to retrieve the state name from the vector abbrevState. Lastly, use the state name to retrieve the correct state name/population pair from the vector statePopulation and output the pair.Ex: If the input is:21044the output is:Maryland: 6079602
Answer:
Here.Hope this helps and do read carefully ,you need to make 1 .cpp file and 1 .h file
Explanation:
#include<iostream>
#include <fstream>
#include <vector>
#include <string>
#include "StatePair.h"
using namespace std;
int main() {
ifstream inFS; // File input stream
int zip;
int population;
string abbrev;
string state;
unsigned int i;
// ZIP code - state abbrev. pairs
vector<StatePair <int, string>> zipCodeState;
// state abbrev. - state name pairs
vector<StatePair<string, string>> abbrevState;
// state name - population pairs
vector<StatePair<string, int>> statePopulation;
// Fill the three ArrayLists
// Try to open zip_code_state.txt file
inFS.open("zip_code_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file zip_code_state.txt." << endl;
return 1; // 1 indicates error
}
while (!inFS.eof()) {
StatePair <int, string> temp;
inFS >> zip;
if (!inFS.fail()) {
temp.SetKey(zip);
}
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetValue(abbrev);
}
zipCodeState.push_back(temp);
}
inFS.close();
// Try to open abbreviation_state.txt file
inFS.open("abbreviation_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file abbreviation_state.txt." << endl;
return 1; // 1 indicates error
}
while (!inFS.eof()) {
StatePair <string, string> temp;
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetKey(abbrev);
}
getline(inFS, state); //flushes endline
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetValue(state);
}
abbrevState.push_back(temp);
}
inFS.close();
// Try to open state_population.txt file
inFS.open("state_population.txt");
if (!inFS.is_open()) {
cout << "Could not open file state_population.txt." << endl;
return 1; // 1 indicates error
}
while (!inFS.eof()) {
StatePair <string, int> temp;
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetKey(state);
}
inFS >> population;
if (!inFS.fail()) {
temp.SetValue(population);
}
getline(inFS, state); //flushes endline
statePopulation.push_back(temp);
}
inFS.close();
cin >> zip;
for (i = 0; i < zipCodeState.size(); ++i) {
// TODO: Using ZIP code, find state abbreviation
}
for (i = 0; i < abbrevState.size(); ++i) {
// TODO: Using state abbreviation, find state name
}
for (i = 0; i < statePopulation.size(); ++i) {
// TODO: Using state name, find population. Print pair info.
}
}
Statepair.h
#ifndef STATEPAIR
#define STATEPAIR
#include <iostream>
using namespace std;
template<typename T1, typename T2>
class StatePair {
private:
T1 key;
T2 value;
public:
// Define a constructor, mutators, and accessors
// for StatePair
StatePair() //default constructor
{}
// set the key
void SetKey(T1 key)
{
this->key = key;
}
// set the value
void SetValue(T2 value)
{
this->value = value;
}
// return key
T1 GetKey() { return key;}
// return value
T2 GetValue() { return value;}
// Define PrintInfo() method
// display key and value in the format key : value
void PrintInfo()
{
cout<<key<<" : "<<value<<endl;
}
};
#endif
In this exercise we have to use the knowledge in computational language in C++ to write the following code:
We have the code can be found in the attached image.
So in an easier way we have that the code is
#include<iostream>
#include <fstream>
#include <vector>
#include <string>
#include "StatePair.h"
using namespace std;
int main() {
ifstream inFS;
int zip;
int population;
string abbrev;
string state;
unsigned int i;
vector<StatePair <int, string>> zipCodeState;
vector<StatePair<string, string>> abbrevState;
vector<StatePair<string, int>> statePopulation;
inFS.open("zip_code_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file zip_code_state.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <int, string> temp;
inFS >> zip;
if (!inFS.fail()) {
temp.SetKey(zip);
}
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetValue(abbrev);
}
zipCodeState.push_back(temp);
}
inFS.close();
inFS.open("abbreviation_state.txt");
if (!inFS.is_open()) {
cout << "Could not open file abbreviation_state.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <string, string> temp;
inFS >> abbrev;
if (!inFS.fail()) {
temp.SetKey(abbrev);
}
getline(inFS, state);
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetValue(state);
}
abbrevState.push_back(temp);
}
inFS.close();
inFS.open("state_population.txt");
if (!inFS.is_open()) {
cout << "Could not open file state_population.txt." << endl;
return 1;
}
while (!inFS.eof()) {
StatePair <string, int> temp;
getline(inFS, state);
state = state.substr(0, state.size()-1);
if (!inFS.fail()) {
temp.SetKey(state);
}
inFS >> population;
if (!inFS.fail()) {
temp.SetValue(population);
}
getline(inFS, state);
statePopulation.push_back(temp);
}
inFS.close();
cin >> zip;
for (i = 0; i < zipCodeState.size(); ++i) {
}
for (i = 0; i < abbrevState.size(); ++i) {
}
for (i = 0; i < statePopulation.size(); ++i) {
}
}
Statepair.h
#ifndef STATEPAIR
#define STATEPAIR
#include <iostream>
using namespace std;
template<typename T1, typename T2>
class StatePair {
private:
T1 key;
T2 value;
public:
StatePair()
{}
void SetKey(T1 key)
{
this->key = key;
}
void SetValue(T2 value)
{
this->value = value;
}
T1 GetKey() { return key;}
T2 GetValue() { return value;}
void PrintInfo()
{
cout<<key<<" : "<<value<<endl;
}
};
See more about C code at brainly.com/question/19705654
describe how sodium ammonium chloride can be separated from a solid mixture of ammonium chloride and anhydrous calcium chloride
Answer:
There are way in separating mixtures of chlorides salts such as that of sodium chloride and ammonium chloride. It can be done by crystallization, filtration or sublimation. If we want to separate the mixture, we have to heat up to 330-350 degrees Celsius and collect the gas that will be produced.
Need help with 9.2 Lesson Practice edhesive asap please
Answer:
Can you add an attachment please
In this exercise we have to use the computer language knowledge in python, so the code is described as:
This code can be found in the attached image.
So the code can be described more simply below, as:
height = []
height.append([16,17,14])
height.append([17,18,17])
height.append([15,17,14])
print(height)
See more about python at brainly.com/question/26104476
What is a principle of DevOps?
Answer:
the main principles of DevOps are automation, continuous delivery, and fast reaction to feedback. You can find a more detailed explanation of DevOps pillars in the CAMS acronym: Culture represented by human communication, technical processes, and tools.
Write a script called fact.sh that is located in your workspace directory to calculate the factorial of a number n; where n is a non-negative integer between 1 and 20 that is passed as a parameter from the command line (e.g. ./fact.sh 5). The result should be echoed to the screen as a single integer (e.g. 120).
NOTE: Do not include any other output, just output the single integer result.
Submit your code to the auto-grader by pressing Check-It!
NOTE: You can submit as many times as you want up until the due date.
Answer:
hope this helps
Explanation:
10.The front end side in cloud computing is
Answer:
The front end is the side the computer user, or client, sees. The back end is the "cloud" section of the system.
The front end includes the client's computer (or computer network) and the application required to access the cloud computing system.
Explanation:
can i have brainliest please
1. If you were on Earth and wanted to call someone via the radio on the moon, and they immediately sent a message back as soon as they received your message, how long from when you sent your message would you get a response
Answer:
About 2.5 seconds.
Explanation:
The speed of radio waves is 299,750 km/s whereas the distance between moon and earth is 384,400 km. By dividing speed of radio waves over the distance of moon and earth, we get 1.22 seconds of time from one side or in other words the person on the moon receives message in 1.22 seconds from the earth and the person on the earth receives a response from the moon in also 1.2 seconds so total time taken by the sender in receiving response from the moon is 2.5 to 3 seconds.
This project must target a Unix platform and execute properly on our CS1 server.
The project must be written in C, C++, or Java.
Problem Overview
This project will simulate a scheduler scheduling a set of jobs.
The project will allow the user to choose a scheduling algorithm from among the six presented in the textbook. It will output a representation of how the jobs are executed.
Design
You may design your own implementation approach, but here are a few constraints.
Your program should read in a list of jobs from a tab-delimited text file named jobs.txt. The format of the text file should have one line for each job, where each line has a job name, a start time and a duration. The job name must be a letter from A-Z. The first job should be named A, and the remaining jobs should be named sequentially following the alphabet, so if there are five jobs, they are named A-E. The arrival times of these jobs should be in order.
The scheduler choice should be a command-line parameter that is one of the following: FCFS, RR, SPN, SRT, HRRN, FB, ALL. If ALL is input, the program should produce output for all six scheduling algorithms. RR and FB should use a quantum of one. FB should use three queues.
Your output should be a graph as shown in the slides. The graph can be text or you can use a graphics package such as JavaFX to draw the graph. For text, you may draw the graph down the page rather than across.
Your program should be able to reproduce the sample shown in the book as well as any similar set of jobs.
Sample Output
Below is sample text-based output. For graphical output, you can make the graph look like the ones in the textbook and slides.
FCFS
A XXX
B XXXXXX
C XXXX
D XXXXX
E XX
FCFS (this is another way you may print the output instead of the one above)
A B C D E
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
X
Answer:
hope this helps ,if it did pls do consider giving brainliest
Explanation:
// Java program for implementation of FCFS scheduling
import java.text.ParseException;
class GFG {
// Function to find the waiting time for all
// processes
static void findWaitingTime(int processes[], int n,
int bt[], int wt[]) {
// waiting time for first process is 0
wt[0] = 0;
// calculating waiting time
for (int i = 1; i < n; i++) {
wt[i] = bt[i - 1] + wt[i - 1];
}
}
// Function to calculate turn around time
static void findTurnAroundTime(int processes[], int n,
int bt[], int wt[], int tat[]) {
// calculating turnaround time by adding
// bt[i] + wt[i]
for (int i = 0; i < n; i++) {
tat[i] = bt[i] + wt[i];
}
}
//Function to calculate average time
static void findavgTime(int processes[], int n, int bt[]) {
int wt[] = new int[n], tat[] = new int[n];
int total_wt = 0, total_tat = 0;
//Function to find waiting time of all processes
findWaitingTime(processes, n, bt, wt);
//Function to find turn around time for all processes
findTurnAroundTime(processes, n, bt, wt, tat);
//Display processes along with all details
System.out.printf("Processes Burst time Waiting"
+" time Turn around time\n");
// Calculate total waiting time and total turn
// around time
for (int i = 0; i < n; i++) {
total_wt = total_wt + wt[i];
total_tat = total_tat + tat[i];
System.out.printf(" %d ", (i + 1));
System.out.printf(" %d ", bt[i]);
System.out.printf(" %d", wt[i]);
System.out.printf(" %d\n", tat[i]);
}
float s = (float)total_wt /(float) n;
int t = total_tat / n;
System.out.printf("Average waiting time = %f", s);
System.out.printf("\n");
System.out.printf("Average turn around time = %d ", t);
}
// Driver code
public static void main(String[] args) throws ParseException {
//process id's
int processes[] = {1, 2, 3};
int n = processes.length;
//Burst time of all processes
int burst_time[] = {10, 5, 8};
findavgTime(processes, n, burst_time);
}
}
// Java program to implement Shortest Job first with Arrival Time
import java.util.*;
class GFG {
static int[][] mat = new int[10][6];
static void arrangeArrival(int num, int[][] mat) {
for (int i = 0; i < num; i++) {
for (int j = 0; j < num - i - 1; j++) {
if (mat[j][1] > mat[j + 1][1]) {
for (int k = 0; k < 5; k++) {
int temp = mat[j][k];
mat[j][k] = mat[j + 1][k];
mat[j + 1][k] = temp;
}
}
}
}
}
static void completionTime(int num, int[][] mat) {
int temp, val = -1;
mat[0][3] = mat[0][1] + mat[0][2];
mat[0][5] = mat[0][3] - mat[0][1];
mat[0][4] = mat[0][5] - mat[0][2];
for (int i = 1; i < num; i++) {
temp = mat[i - 1][3];
int low = mat[i][2];
for (int j = i; j < num; j++) {
if (temp >= mat[j][1] && low >= mat[j][2]) {
low = mat[j][2];
val = j;
}
}
mat[val][3] = temp + mat[val][2];
mat[val][5] = mat[val][3] - mat[val][1];
mat[val][4] = mat[val][5] - mat[val][2];
for (int k = 0; k < 6; k++) {
int tem = mat[val][k];
mat[val][k] = mat[i][k];
mat[i][k] = tem;
}
}
}
// Driver Code
public static void main(String[] args) {
int num;
Scanner sc = new Scanner(System.in);
System.out.println("Enter number of Process: ");
num = sc.nextInt();
System.out.println("...Enter the process ID...");
for (int i = 0; i < num; i++) {
System.out.println("...Process " + (i + 1) + "...");
System.out.println("Enter Process Id: ");
mat[i][0] = sc.nextInt();
System.out.println("Enter Arrival Time: ");
mat[i][1] = sc.nextInt();
System.out.println("Enter Burst Time: ");
mat[i][2] = sc.nextInt();
}
System.out.println("Before Arrange...");
System.out.println("Process ID\tArrival Time\tBurst Time");
for (int i = 0; i < num; i++) {
System.out.printf("%d\t\t%d\t\t%d\n",
mat[i][0], mat[i][1], mat[i][2]);
}
arrangeArrival(num, mat);
completionTime(num, mat);
System.out.println("Final Result...");
System.out.println("Process ID\tArrival Time\tBurst" +
" Time\tWaiting Time\tTurnaround Time");
for (int i = 0; i < num; i++) {
System.out.printf("%d\t\t%d\t\t%d\t\t%d\t\t%d\n",
mat[i][0], mat[i][1], mat[i][2], mat[i][4], mat[i][5]);
}
sc.close();
}
}
The Freeze Panes feature would be most helpful for which situation?
A.when the functions used in a worksheet are not working correctly
B.when a large worksheet is difficult to interpret because the headings disappear during scrolling
C.when the data in a worksheet keeps changing, but it needs to be locked so it is not changeable
D.when a large worksheet is printing on too many pages
Answer:
i think a but not so sure
Explanation:
Answer: Option B is the correct answer :)
Explanation:
Suppose a TCP segment is sent while encapsulated in an IPv4 datagram. When the destination host gets it, how does the host know that it should pass the segment (i.e. the payload of the datagram) to TCP rather than to UDP or some other upper-layer protocol
Answer and Explanation:
A host needs some datagram where TCP is encapsulated in the datagram and contains the field called a protocol. The size for the field is 8-bit.
The protocol identifies the IP of the destination where the TCP segment is encapsulated in a datagram.
The protocol field is 6 where the datagram sends through TCP and 17 where the datagram is sent through the UDP as well
Protocol with field 1 is sent through IGMP.
The packet is basic information that is transferred across the network and sends or receives the host address with the data to transfer. The packet TCP/IP adds or removes the field from the header as well.
TCP is called connection-oriented protocol as it delivers the data to the host and shows TCL receives the stream from the login command. TCP divides the data from the application layer and contains sender and recipient port for order information and known as a checksum. The TCP protocol uses the checksum data to determine that data transferred without error.
Which of the following is not an advantage of e-commerce for consumers?
a. You can choose goods from any vendor in the world.
b. You can shop no matter your location, time of day, or conditions.
c. You have little risk of having your credit card number intercepted.
d. You save time by visiting websites instead of stores.
Answer: c. You have little risk of having your credit card number intercepted.
====================================
Explanation:
Choice A is false because it is an advantage to be able to choose goods from any vendor from anywhere in the world. The more competition, the better the outcome for the consumer.
Choice B can also be ruled out since that's also an advantage for the consumer. You don't have to worry about shop closure and that kind of time pressure is non-existent with online shops.
Choice D is also a non-answer because shopping online is faster than shopping in a brick-and-mortar store.
The only thing left is choice C. There is a risk a person's credit card could be intercepted or stolen. This usually would occur if the online shop isn't using a secure method of transmitting the credit card info, or their servers are insecure when they store the data. If strong encryption is applied, then this risk can be significantly reduced if not eliminated entirely.
Answer:
c. You have little risk of having your credit card number intercepted.
Explanation:
Hope this helps
In an employee database, an employee's ID number, last name, first name, company position, address, city, state, and Zip Code make up a record. true false
Answer:
false
Explanation:
Commands from the Quick Access toolbar can be used by
Clicking on the command
Navigating to Backstage view
Selecting the appropriate options on the status bar
Selecting the function key and clicking the command
Answer:
Clicking on the command.
Explanation:
PowerPoint application can be defined as a software application or program designed and developed by Microsoft, to avail users the ability to create various slides containing textual and multimedia informations that can be used during a presentation.
Some of the features available on Microsoft PowerPoint are narrations, transition effects, custom slideshows, animation effects, formatting options etc.
A portion of the Outlook interface that contains commonly accessed commands that a user will require frequently and is able to be customized by the user based on their particular needs is referred to as the Quick Access toolbar.
Commands from the Quick Access toolbar can be used by clicking on the command.
What is Microsoft Excel? Why is it so popular
Answer:
Its a spreadsheet devolpment by Microsoft.
Explanation:
For this exercise, you are given the Picture class and the PictureTester class. The Picture class has two instance variables. You will need to finish the class by writing getter and setter methods for these two instance variables.
Then in the PictureTester class a Picture object has been created for you. You will need to update this object and then print using the getter methods you created.
public class Picture
{
private String name;
private String date;
public Picture(String theName, String theDate){
name = theName;
date = theDate;
}
// Add getter and setter methods here.
// method names should be:
// getName, setName, getDate, setDate
}
Answer:
hope this helps, do consider giving brainliest
Explanation:
public class Picture {
// two instance variables
private String name;
private String date;
// Parameterized constructor
public Picture(String name, String date) {
super();
this.name = name;
this.date = date;
}
/**
* return the name
*/
public String getName() {
return name;
}
/**
* param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* return the date
*/
public String getDate() {
return date;
}
/**
* param date the date to set
*/
public void setDate(String date) {
this.date = date;
}
}
----------------------------------------------------TESTER CLASS---------------------------------------------------------------------
package test;
public class PictureTester {
public static void main(String[] args) {
// TODO Auto-generated method stub
// creating Picture object (pic is the reference variable) and we are invoking the constructor by passing // values of name and date
Picture pic = new Picture("flower_pic", "28-04-2020");
// now we will first use get method
String name = pic.getName();
String date = pic.getDate();
// displaying the value fetched from get method
// value will be name - flower_pic and date - 28-04-2020
System.out.println("[ Name of the Picture - "+name+" and Date - "+date+" ]");
// now we will use set method to set the value of name and date and then display that value
// declaring two variable
String name1 = "tiger_pic";
String date1 = "28-03-2020";
//assigning those value using set method
pic.setName(name1);
pic.setDate(date1);
System.out.println("[ Name of the Picture - "+pic.getName()+" and Date - "+pic.getDate()+" ]");
}
}
Circle class
private members
double radius
double xPos
double yPos
public members
double diameter()
get the diameter of the Circle. It returns a value, diameter.
double area()
calculate the area of the Circle
double circumference()
calculate the circumference of the circle
double getRadius()
returns the radius
double getX()
returns the xPos value
double getY()
returns the yPos value
void setX(double x)
sets xPos, no requirements
void setY(double y)
sets yPos, no requirements
double distanceToOrigin()
returns the distance from the center of the circle to the origin
HINT: Find out how to calculate the distance between two points and recall the origin is at (0,0)
bool insersect(const Circle& otherCircle)
Take another Circle by const reference (see more notes below)
Returns true if the other Circle intersects with it, false otherwise
bool setRadius(double r)
sets the radius to r and returns true if r is greater than zero, otherwise sets the radius to zero and returns false
Remember, you will need a header file (.h) and an implementation file (.cpp)
You'll also need to update your Makefile
NOTE: The Circle class should not do any input or output.
CircleDriver class
private members
Circle circ1;
Circle circ2;
void obtainCircles()
Talk with the user to obtain the positions and radii for two Circles from the user. Repeats prompts until the user gives valid values for the radii
It does not validate the values, but rather checks the return value from a call to Circle's setRadius method
void printCircleInfo()
Prints the following information about each of the Circles to the screen:
The location of the Circle's center (xPos, yPos), the distance from the origin, each area, circumference, diameter
Lastly print whether or not the two circles intersect
Sample output from printCircleInfo().
Information for Circle 1:
location: (0, 50.0)
diameter: 200.0
area: 31415.9
circumference: 628.318
distance from the origin: 50.0
Information for Circle 2:
location: (50.0, 0)
diameter: 200.0
area: 31415.9
circumference: 628.318
distance from the origin: 50.0
The circles intersect.
public membersvoid run()
run merely calls all the other methods. Here's your definition for run()
//This will go in your CircleDriver.cpp
void CircleDriver::run()
{
obtainCircles();
printCircleInfo();
}
main
Main does very little. In fact, here is your main:
int main()
{
CircleDriver myDriver;
myDriver.run();
return(0);
}
Answer:
What was the actual the question? I am confused.
Explanation:
Explain the five generations of computer
Answer:
1 First Generation
The period of first generation: 1946-1959. Vacuum tube based.
2 Second Generation
The period of second generation: 1959-1965. Transistor based.
3 Third Generation
The period of third generation: 1965-1971. Integrated Circuit based.
4 Fourth Generation
The period of fourth generation: 1971-1980. VLSI microprocessor based.
5 Fifth Generation
The period of fifth generation: 1980-onwards. ULSI microprocessor based.
The phrase "generation" refers to a shift in the technology that a computer is/was using. The term "generation" was initially used to describe different hardware advancements. These days, a computer system's generation involves both hardware and software.
Who invented the 5th generation of computers?The Japan Ministry of International Trade and Industry (MITI) launched the Fifth Generation Computer Systems (FGCS) initiative in 1982 with the goal of developing computers that use massively parallel computing and logic programming.
The generation of computers is determined by when significant technological advancements, such as the use of vacuum tubes, transistors, and microprocessors, take place inside the computer. There will be five computer generations between now and 2020, with the first one starting around 1940.
Earlier models of computers (1940-1956)Computers of the Second Generation (1956-1963)Computers of the Third Generation (1964-1971)Computers of the fourth generation (1971-Present)Computers of the fifth generationLearn more about generations of computers here:
https://brainly.com/question/9354047
#SPJ2
The Quick Search capability is found on which control?
Database query
Backstage view
Record navigator
File management
Answer:
backstage view
Explanation:
In the code snippet below, pick which instructions will have pipeline bubbles between them due to hazards.
a. lw $t5, 0($t0)
b. lw $t4, 4($t0)
c. add $t3, $t5, $t4
d. sw $t3, 12($t0)
e. lw $t2, 8($t0)
f. add $t1, $t5, $t2
g. sw $t1, 16($t0)
Answer:
b. lw $t4, 4($t0)
c. add $t3, $t5, $t4
Explanation:
Pipeline hazard prevents other instruction from execution while one instruction is already in process. There is pipeline bubbles through which there is break in the structural hazard which preclude data. It helps to stop fetching any new instruction during clock cycle.
Write a piece of code that constructs a jagged two-dimensional array of integers named jagged with five rows and an increasing number of columns in each row, such that the first row has one column, the second row has two, the third has three, and so on. The array elements should have increasing values in top-to-bottom, left-to-right order (also called row-major order). In other words, the array's contents should be the following: 1 2, 3 4, 5, 6 7, 8, 9, 10 11, 12, 13, 14, 15
PROPER OUTPUT: jagged = [[1], [2, 3], [4, 5, 6], [7, 8, 9, 10], [11, 12, 13, 14, 15]]
INCORRECT CODE:
int jagged[][] = new int[5][];
for(int i=0; i<5; i++){
jagged[i] = new int[i+1]; // creating number of columns based on current value
for(int j=0, k=i+1; j jagged[i][j] = k;
}
}
System.out.println("Jagged Array elements are: ");
for(int i=0; i for(int j=0; j System.out.print(jagged[i][j]+" ");
}
System.out.println();
}
expected output:jagged = [[1], [2, 3], [4, 5, 6], [7, 8, 9, 10], [11, 12, 13, 14, 15]]
current output:
Jagged Array elements are:
1
2 3
3 4 5
....
Answer:
Explanation:
The following code is written in Java and it uses nested for loops to create the array elements and then the same for loops in order to print out the elements in a pyramid-like format as shown in the question. The output of the code can be seen in the attached image below.
class Brainly {
public static void main(String[] args) {
int jagged[][] = new int[5][];
int element = 1;
for(int i=0; i<5; i++){
jagged[i] = new int[i+1]; // creating number of columns based on current value
for(int x = 0; x < jagged[i].length; x++) {
jagged[i][x] = element;
element += 1;
}
}
System.out.println("Jagged Array elements are: ");
for ( int[] x : jagged) {
for (int y : x) {
System.out.print(y + " ");
}
System.out.println(' ');
}
}
}