Ur answer is: a-half-inch
<em>r = 15 cm</em>
<em>formula</em>
<em>V = πr²×h/3</em>
<em>replace</em>
<em>4950 = 22/7×r²×21/3</em>
<em>4950 = 22/7×r²×7</em>
<em>4950 = 22×r²</em>
<em>r² = 4950/22</em>
<em>r² = 225</em>
<em>r = √225</em>
<em>r = √15²</em>
<em>r = 15 cm</em>
If an event occurs, the agent logs details regarding the event. what is this event called GET.
The information in the agent log file is known to be the beginning of the log file, which is stated to show the agent's launch and handling of the services and configuration settings.
Keep in mind that the agent log also contains a history of the activities performed by the agent during runtime, along with any errors, and that it is utilised to investigate deployment issues.
As a result, if an event happens, the agent logs information about it. What is this GET event, exactly?
The agent monitoring services' startup and configuration settings are displayed at the log file's beginning. The sequence of agent runtime activity and any observed exceptions are also included in the agent log.
Learn more about agent logs:
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Answer:
a. gpupdate /force
Explanation:
Based on the information provided within the question it can be said that if the administrator does not want to wait she can use the command gpupdate /force. This command allows the individual to update both the local Group Policy settings and Active Directory-based settings. This the force tag makes it so that the policy is immediately update.
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;
}