1. The system admin staff would like you to create a cron job that runs only on weekdays (Monday - Friday) at 1am, continuously throughout the year. The job should write a list of all processes that are currently running on the host to a file called processes.txt 2. The Helpdesk would like to keep metrics on disk usage for a short period of time. Write a cron job that writes the output of the 'df -h' command to a local file titled metric.txt. The job should run on Wednesday from 10am till 2pm and update every 2 minutes. Make sure that you don't OVERWRITE the data that's already in the file. You'll need to structure your command so that it APPENDS to the file, saving the data that's already been written.

Answer:

#include<iostream>

using namespace std;

//main function

int main(){

   //initialization

   float a1,a2;

//display the message

cout<<"Enter the first number: ";

   cin>>a1;  //store the value

   cout<<"Enter the second number: ";

   cin>>a2;   //store the value

   //display the calculation result

   cout<<"The sum is: "<<a1+a2<<endl;

   cout<<"The Subtraction is: "<<a1-a2<<endl;

   cout<<"The product is: "<<a1*a2<<endl;

   cout<<"The Quotient is: "<<a1/a2<<endl;

}

Explanation:

Create the main function and declare the two variables of float but we can enter the integer as well.

display the message on the screen and then store the input enter by the user into the variable using the cin instruction.

the same process for second input as well, display the message and store the input.

after that, print the output of the calculation. the operator '+' is used to add the two numbers like a1+a2, it adds the number stored in the variable.

similarly, the subtraction operator is '-', product '*' and quotient operator '/'.

and finally, we get the desired output.

Answer:

D. 22 % 2 != 1

Explanation:

Since 22 is even than it would have to stay 0.

Therefore, your only answer choice would be D.

Answer:

D is the answer

Explanation:

because arteriales don't requare valves because they only go to one direction

Answer:

10

Explanation:

An enqueue operation is a function that adds an element(value) to a queue array. A dequeue operations removes an element from a queue array. Queue arrays follow a first-in-first-out approach, so elements that are first stored in the queue are removed/accessed first(enqueue operations add elements at the rear of the queue array).

The following operations leave 10 elements in the queue of array size 12 after its done:

10 enqueue operations= adds 10 elements

5 dequeue operations= removes 5 elements( 5 elements left in queue)

6 enqueue operations= adds 6 elements(11 elements in queue)

10 dequeue operations= removes 10 elements(1 element left in queue)

8 enqueue operations= adds 8 elements(9 elements in queue)

2 dequeue operations= removes 2 elements(7 elements left in queue)

3 enqueue operations= adds 3 elements(10 elements in queue)

Therefore there are 10 elements in the queue after enqueue and dequeue operations.