Option A because the environment should be selected for which the changes are to be applied. In a time Force IDE has many project environments for example maybe a java project and C++ project would be there on sandbox, so the environment selection is important.
Option B because the related changed sets should be specified so that other developers that have access to the project can see the changes being made.
Option D The data fields that are needed to be deployed should also be provided so that the updated version can be seen by other developers.
Rejected Options :
Option C user name and password has nothing to do with the production environment because if the user has it only then it can come and make changes.
Answer: setJMenubar
Explanation:Menus that are component of the Menu bar that display the options and tools for any function that can be performed in the system have the connection with the window in the field of java through the command of the setJMenubar. It is the command which is given to execute to bring the menu to the particular frame and get it attached according to the JFrame. Therefore, the correct option is setJMenubar.
<span>D. Page Layout i hope this helps </span>
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;
}
