Answer:
Operating systems have some code called an 'interrupt handler', which prioritises the interrupts and saves them in a queue. Buffers are used in computers as a temporary memory area, and they are essential in modern computers because hardware devices operate at much slower speeds than the processor.
Answer:
Option C is the correct answer to the following question.
Explanation:
Because network externality is that in which a consequence of the industrial activities or the other commercial activities which are affects another parties without involving the reflected in cost of goods and services. The other folks used the website as compares to any other folks enjoys. So, that's why it is correct.
Answer:
// here is code in c++.
#include <bits/stdc++.h>
using namespace std;
// function that return greatest common divisor
int g_c_d(int num1, int num2)
{
if (num1 == 0)
return num2;
return g_c_d(num2 % num1, num1);
}
// main function
int main()
{
// variables
int num1,num2;
cout<<"enter two numbers:";
// read two numbers from user
cin>>num1>>num2;
// call the function and print the gcd of both
cout<<"greatest common divisor of "<<num1<<" and "<<num2<<" is "<<g_c_d(num1.num2)<<endl;
}
Explanation:
Read two numbers from user and assign them to variables "num1" and "num2".Call the function g_c_d() with parameter "num1" and "num2".According to Euclidean algorithm, if we subtract smaller number from the larger one the gcd will not change.Keep subtracting the smaller one then we find the gcd of both the numbers.So the function g_c_d() will return the gcd of both the numbers.
Output:
enter two numbers:5 9
greatest common divisor of 5 and 9 is 1
enter two numbers:-25 15
greatest common divisor of -25 and 15 is 5
I belive their were answer chocies to this question am i right
Answer:
Answer given below
Explanation:
a) 10 coins ( 25*8 + 1*2)
b) 14 coins (27*10 + 12*1 + 7*1 + 3*1 + 1*1)
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<u>Source Code in C++:
</u>
#include <iostream>
using namespace std;
//main function
int main()
{
int n,n1;
cout << "Enter the amount : "; //taking amount as input
cin >> n;
if(n<=0)
{
cout << "Fatal Error : Input Failure." << endl;
return 1;
}
cout << "Enter the number of denominations : "; //taking number of denominations as input
cin >> n1;
if(n1<=0)
{
cout << "Fatal Error : Input Failure." << endl;
return 1;
}
int a[n1],c[n1],sum=0; //array to keep count of each amount, and sum to store total coins
cout << "Enter the denominations in descending order : ";
for(int i=0;i<n1;i++)
{
cin >> a[i];
}
for(int i=0;i<n1;i++)
{
c[i]=n/a[i];
sum=sum+c[i];
n=n%a[i];
}
for(int i=0;i<5;i++)
{
cout << a[i] << " : " << c[i] << endl;
}
return 0;
}
