♥ <span>Boot your Chromebook into recovery Mode
(Escape refresh and power keys all need to be help down)
It </span><span>will then reboot into recovery mode
♥ </span><span>Press Ctrl+D at the recovery screen
♥ </span><span>To turn the Verification off you will need to press the enter button.
♥ and then you have it :D
</span>
It is true that in the array implementation of a queue, the pop operation is most efficient if the front of the queue is fixed at index position. The correct option is a.
<h3>What is pop operation? </h3>
The removal of an element is referred to as a pop operation. Again, because we only have access to the element at the top of the stack, we can only remove one element. We simply take the top of the stack off.
A push operation decrements the pointer before copying data to the stack; a pop operation copies data from the stack before incrementing the pointer.
The pop operation in an array implementation of a queue is most efficient if the queue's front is fixed at index position.
Thus, the correct option is a.
For more details regarding pop operation, visit:
brainly.com/question/15172555
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The type of physical drives does windows disable defragmenting, but provides another method of optimization is known as windows.
<h3>What is windows?</h3>
It should be noted that windows is a separate viewing area on a computer display screen in a system.
In this case, the type of physical drives does windows disable defragmenting, but provides another method of optimization is known as windows.
Learn more about window on:
brainly.com/question/25243683
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Here is a somewhat cryptic solution that works:
#include <algorithm>
#include <cstdlib>
using namespace std;
void q(char c, int count)
{
for (int i = 0; i < count; i++) {
putchar(c);
}
}
void p(int b1, int plusses)
{
q(' ', b1);
q('+', plusses);
}
int main()
{
for (int i = -3; i <= 3; i++)
{
int pl = min(6, (3 - abs(i)) * 2 + 1);
p(6-pl, pl);
i == 0 ? p(0, 6) : p(6, 0);
p(0, pl);
putchar('\n');
}
getchar();
}
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