Answer:
data source
Explanation:
The main aim of a data source is for the gathering of all necessary information that is needed to access a data. Since he has used the information to create a pie chart, this means that some data were used for the creation of this pie chart. Hence the information used for the creation of the pie chart is the data source for the information illustrated on the pie chart.
You need some more criteria. Is this a webpage? If so.. What language(s) are applicable for writing your web application. My profession is systems development, do I do a lot of fronte-end/ back-end development. I don't mind helping, just need some more information. If not what resource does this need to be created in?
Answer:
The Firewall (Router) is the vulnerable hardware.
Explanation:
The entry into the network will be the firewall (If there is a firewall installed on the network), if this medium is compromised, every workstation on the network will be vulnerable.
In traditional programming, probably the most often used error-handling outcome was to terminate the program in which the offending statement occurred.
<h3>What is a traditional programming?</h3>
Traditional programming is known to be a form of manual way that one or a user makes a program.
Note that in the case above, In traditional programming, probably the most often used error-handling outcome was to terminate the program in which the offending statement occurred.
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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;
}
