Answer:
1.)
- C.) Optional
2.)
- D.) Short-term Notes Payable
3.)
- A.) Payroll Sinking Funds
4.)
- A.) A Formal Timekeeping System Is Used
(I'm possibly wrong on the last question, if so then my apologies and I wish you the best of luck.)
Answer:
<em><u>5G Will Quickly Become The New Standard For Cellular Networks. The Internet of Things (IoT) is rapidly developing and expanding. ... 5G will increase cellular bandwidth by huge amounts, making it much easier for the Internet of Things to network large numbers of devices together.</u></em>
Explanation:
Hope its help
A series of instructions written in a programming language for a computer to follow is referred to as a computer program.
<h3>
What is program?</h3>
Software, which also contains documentation and other intangible components, comprises computer programs as one of its components.
The source code of a computer program is the version that can be read by humans. Since computers can only run their native machine instructions, source code needs to be run by another software.
Using the language's compiler, source code may be converted to machine instructions. (An assembler is used to translate programs written in machine language.) An executable is the name of the generated file.
Therefore, A series of instructions written in a programming language for a computer to follow is referred to as a computer program.
To learn more about program, refer to the link:
brainly.com/question/11023419
#SPJ1
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;
}
