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3 years ago

Plzzzzzzzzzzzzzzzz help its almost a missing assignment

Computers and Technology

2 answers:

kow [346]3 years ago

7 0

B) They were the first to fly a powered plane.

vampirchik [111]3 years ago

3 0

Answer:

They were the first to fly a powered airplane.

Explanation:

Waaa? You don't know the Wright Brothers!?! They built a plane!

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Digital security measures include _____. Select 2 options.

elixir [45]

Answer:

firewalls

passwords

8 0

3 years ago

Read 2 more answers

Write a C++ program that prompt the user to enter three points (x1, y1), (x2, y2), (x3,y3) of a triangle and

My name is Ann [436]

Answer:

#include<iostream>

using namespace std;

int main(){

int x1,x2,x3;

int y1,y2,y3;

cout<<"Enter the value of first point(x1,y1): ";

cin>>x1>>y1;

cout<<"\nEnter the value of second point(x2,y2): ";

cin>>x2>>y2;

cout<<"\nEnter the value of third point(x3,y3): ";

cin>>x3>>y3;

}

Explanation:

first include the library iostream for use the input/output commands

then, write the main function. within the main function declare the variable which store the value of points.

after that, use the 'cout' for output. It has the function which print the value or message on the screen.

and 'cin' is used to store the value in the variable.

So, in the above code cout ask for enter the value of point and after enter value by user, cin store in the variables.

Note: you have to enter two value with space or by enter.

5 0

3 years ago

Code a program that will go through all the numbers from 1 to 100 and print them. However, if the number is evenly divisible by

worty [1.4K]

Answer:

public class Main

{

// required method

public static void fizzBuzz(){

// looping through 1 to 100

for(int i = 1; i <= 100; i++){

//if number is evenly divisible by both 3 and 5

if(i%3 == 0 && i%5 == 0){

System.out.println("fiz buzz");

}

// if number is divisible by 3

else if (i%3 == 0){

System.out.println("fizz");

}

// if number is divisible by 5

else if (i%5 == 0){

System.out.println("buzz");

}

// if number is not divisible by both 3 and 5

else {

System.out.println(i);

}

// main method

public static void main(String[] args) {

//calling function

fizzBuzz();

}

Explanation:

8 0

3 years ago

When a program lets the user know that an invalid menu choice has been made, this is an example of?

kolezko [41]

The question has the multiple choices below

A) Input validation
B) output correction
C) compiler criticism
D) output validation

The answer is (A) Input validation

Also known as data validation, it is the correct and proper testing of any input supplied by an application or a user. It prevents wrongly formed data from entering the system. Applications should be able to check and validate every input entered into the system since it is hard to detect a malicious user trying to gain entry into software.

5 0

3 years ago

Data Structure in C++

agasfer [191]

The code .cpp is available bellow

#include<iostream>

using namespace std;

//declaring variables

void merge(int* ip, int sz, int* opt, bool opt_asc); //merging

int* mergesort(int* ip, int sz);

void mergesort(int *ip, int sz, int* opt, bool opt_asc);

void merge(int* ip, int sz, int* opt, bool opt_asc)

{

int s1 = 0;

int mid_sz = sz / 2;

int s2 = mid_sz;

int e2 = sz;

int s3 = 0;

int end3 = sz;

int i, j;

if (opt_asc==true)

{

i = s1;

j = e2 - 1;

while (i < mid_sz && j >= s2)

{

if (*(ip + i) > *(ip + j))

{

*(opt + s3) = *(ip + j);

s3++;

j--;

}

else if (*(ip + i) <= *(ip + j))

{

*(opt + s3) = *(ip + i);

s3++;

i++;

}

if (i != mid_sz)

{

while (i < mid_sz)

{

*(opt + s3) = *(ip + i);

s3++;

i++;

}

if (j >= s2)

{

while (j >= s2)

{

*(opt + s3) = *(ip + j);

s3++;

j--;

}

else

{

i = mid_sz - 1;

j = s2;

while (i >= s1 && j <e2)

{

if (*(ip + i) > *(ip + j))

{

*(opt + s3) = *(ip + i);

s3++;

i--;

}

else if (*(ip + i) <= *(ip + j))

{

*(opt + s3) = *(ip + j);

s3++;

j++;

}

if (i >= s1)

{

while (i >= s1)

{

*(opt + s3) = *(ip + i);

s3++;

i--;

}

if (j != e2)

{

while (j < e2)

{

*(opt + s3) = *(ip + j);

s3++;

j++;

}

for (i = 0; i < sz; i++)

*(ip + i) = *(opt + i);

}

int* mergesort(int* ip, int sz)

{

int* opt = new int[sz];

mergesort(ip, sz, opt, true);

return opt;

}

void mergesort(int *ip, int sz, int* opt, bool opt_asc)

{

if (sz > 1)

{

int q = sz / 2;

mergesort(ip, sz / 2, opt, true);

mergesort(ip + sz / 2, sz - sz / 2, opt + sz / 2, false);

merge(ip, sz, opt, opt_asc);

}

int main()

{

int arr1[12] = { 5, 6, 9, 8,25,36, 3, 2, 5, 16, 87, 12 };

int arr2[14] = { 2, 3, 4, 5, 1, 20,15,30, 2, 3, 4, 6, 9,12 };

int arr3[10] = { 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 };

int *opt;

cout << "Arays after sorting:\n";

cout << "Array 1 : ";

opt = mergesort(arr1, 12);

for (int i = 0; i < 12; i++)

cout << opt[i] << " ";

cout << endl;

cout << "Array 2 : ";

opt = mergesort(arr2, 14);

for (int i = 0; i < 14; i++)

cout << opt[i] << " ";

cout << endl;

cout << "Array 3 : ";

opt = mergesort(arr3, 10);

for (int i = 0; i < 10; i++)

cout << opt[i] << " ";

cout << endl;

return 0;

}

4 0

4 years ago