Answer:
void printC()
{
int i, j;
for (i = 0; i < 4; i++) //i indicate row number. Here we have 5 rows
{
printf("C"); //print C for every row
for (j = 0; j < 6; j++) //j indicate column number. Here we have 7 Rows
{
if (i == 0 || i == 4) //For first and last row
printf("C"); //print 'CCCCCCC'
else if (i = 1|| i= 3) //for Second forth row
printf("C + +"); //print 'C + +'
else if (i = 2) For second row
printf("C +++++"); //print 'C +++++'
else
continue; //to jump to next iteration
}
printf("\n"); // print in next line
}
}
Answer:
HAHHAHAHA HAAHA g abbabanjaja abunjing <em><u>abunjin</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em><em><u>✌</u></em>
Answer:
t= 8.7*10⁻⁴ sec.
Explanation:
If the signal were able to traverse this distance at an infinite speed, the propagation delay would be zero.
As this is not possible, (the maximum speed of interactions in the universe is equal to the speed of light), there will be a finite propagation delay.
Assuming that the signal propagates at a constant speed, which is equal to 2.3*10⁸ m/s (due to the characteristics of the cable, it is not the same as if it were propagating in vaccum, at 3.0*10⁸ m/s), the time taken to the signal to traverse the 200 km, which is equal to the propagation delay, can be found applying the average velocity definition:
If we choose x₀ = 0 and t₀ =0, and replace v= 2.3*10⁸ m/s, and xf=2*10⁵ m, we can solve for t:
⇒ t = 8.7*10⁻⁴ sec.
Answer:
providing real-time data feeds on millions of people with wearable devices
Explanation:
Answer:
D. It does not reflect any changes made in the document
Explanation:
A limited access is usually done by middle level and top level managers in an organisation to prevent other staff member to edit or make changes to confidential documents when they are out of the office, though the staff can read it, they can not make changes to it.
