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qwelly [4]
3 years ago
8

As time goes on, technology is likely to _______. A. Increase in complexity b. Decrease in complexity c. Stay at the same comple

xity d. No one really has any idea Please select the best answer from the choices provided A B C D
Computers and Technology
1 answer:
amid [387]3 years ago
5 0

Answer:

The answer to this question is the option "A".

Explanation:

In this question, the answer is Increase in complexity because, In computer science, the computerized or simply complexity is an algorithm. In this algorithm, the number of the resource is required for moving it (a quality separate to “complexity” in a conventional reason). So in this question the answer is option A that is  Increase in complexity.

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What type of task can be defined to allow you fine-grained control over the management tasks a user can perform in an OU
erma4kov [3.2K]

The type of task can be defined to allow you fine-grained control over the management tasks a user can perform in an OU is custom.

<h3>What is custom priority in Outlook tasks?</h3>

Outlook is known to have a  New Tasks window that often shows a created-in Priority field that has  Low, Normal, and High setting.

This setting above is a custom one and it is one where a person can sort tasks by using its priority level instead of due date. This help to have  more perspective and one can reorganize their tasks when needed.

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3 0
2 years ago
Can you answer this question?
patriot [66]

Answer:

To do this you'll need to use malloc to assign memory to the pointers used.  You'll also need to use free to unassign that memory at the end of the program using the free.  Both of these are in stdlib.h.

#include <stdlib.h>

#include <stdio.h>

#define SIZE_X 3

#define SIZE_Y 4

int main(void){

       int **matrix, i, j;

       // allocate the memory

       matrix = (int**)malloc(SIZE_X * sizeof(int*));

       for(i = 0; i < SIZE_X; i++){

               matrix[i] = (int *)malloc(SIZE_Y * sizeof(int));

       }

       // assign the values

       for(i = 0; i < SIZE_X; i++){

               for(j = 0; j < SIZE_Y; j++){

                       matrix[i][j] = SIZE_Y * i + j + 1;

               }

       }

       // print it out

       for(i = 0; i < SIZE_X; i++){

               for(j = 0; j < SIZE_X; j++){

                       printf("%d, %d:  %d\n", i, j, matrix[i][j]);

               }

       }

       // free the memory

       for(i = 0; i < SIZE_X; i++){

               free(matrix[i]);

       }

       free(matrix);

       return 0;

}

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