Answer:(A) usability
Explanation:
The term usability refers to the usage value of products, how much they can be used and their value after their use. So adding on these points usability helps to quantify the financial value of IT components.
Answer:
Blocking Mode
Explanation:
Spanning Tree Protocol is used to allow path redundancy in the network without creating cycles/circles also called loops.
When two parts of the switched network are connected via two or more Layer 2 switches this result in a loop.
This affects the performance of the network as the result of broadcast packets flooding.
STP puts one port of the switch to forwarding mode and the rest of the ports within the same part of the network to the blocking mode to avoid broadcast packet flooding. STP puts all the ports that are allowing redundant paths to blocking mode and the one port that is left after this is placed in forward mode.
Spanning Tree Algorithm is used by STP to determine the optimal path of switch to the network.
Bridge Protocol Data Units are used to share the information about the optimal path determined by the spanning tree algorithm with other switches.
This information helps STP to eliminate the redundant paths.
So this is how STP allows only one active path to the destination while blocking all other paths to avoid switching loop.
The simulation, player 2 will always play according to the same strategy.
Method getPlayer2Move below is completed by assigning the correct value to result to be returned.
Explanation:
- You will write method getPlayer2Move, which returns the number of coins that player 2 will spend in a given round of the game. In the first round of the game, the parameter round has the value 1, in the second round of the game, it has the value 2, and so on.
#include <bits/stdc++.h>
using namespace std;
bool getplayer2move(int x, int y, int n)
{
int dp[n + 1];
dp[0] = false;
dp[1] = true;
for (int i = 2; i <= n; i++) {
if (i - 1 >= 0 and !dp[i - 1])
dp[i] = true;
else if (i - x >= 0 and !dp[i - x])
dp[i] = true;
else if (i - y >= 0 and !dp[i - y])
dp[i] = true;
else
dp[i] = false;
}
return dp[n];
}
int main()
{
int x = 3, y = 4, n = 5;
if (findWinner(x, y, n))
cout << 'A';
else
cout << 'B';
return 0;
}
Answer: Isolation, Depression, Humiliation.
I hope this helps you out! ☺