The correct answer is d. It will print i equal to 1 to 5 in increasing order.

Here's the step by step explanation:

1. The variable `i` is initialized to 1.
2. The `for` loop condition checks if `i` is less than or equal to 5. If it is, it enters the loop.
3. Inside the loop, it prints the value of `i` and then increments `i` by 1.
4. This process repeats until `i` is no longer less than or equal to 5.
5. Therefore, the output will be `i` equal to 1 to 5 in increasing order.

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The correct answer is d. It will print i equal to 1 to 5 in increasing order.

Here's the step by step explanation:

1. The variable `i` is initialized to 1.
2. The `for` loop condition checks if `i` is less than or equal to 5. If it is, it enters the loop.
3. Inside the loop, it prints the value of `i` and then increments `i` by 1.
4. This process repeats until `i` is no longer less than or equal to 5.
5. Therefore, the output will be `i` equal to 1 to 5 in increasing order.

Knowee AI · Accepted Answer

The correct answer is d. It will print i equal to 1 to 5 in increasing order.

Here's the step by step explanation:

1. The variable `i` is initialized to 1.
2. The `for` loop condition checks if `i` is less than or equal to 5. If it is, it enters the loop.
3. Inside the loop, it prints the value of `i` and then increments `i` by 1.
4. This process repeats until `i` is no longer less than or equal to 5.
5. Therefore, the output will be `i` equal to 1 to 5 in increasing order.

Select the most appropriate answer for the given code#include<stdio.h>void main(){ int i=1; for(;i<=5;) { printf("i=%d\n",i); i++; }}a.Errorb.It will print i equal to 1 for five timesc.It will print i equal to 1 to 5 in decreasing orderd.It will print i equal to 1 to 5 in increasing order

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