Answer:
abacus . Mesopotamia or China, possibly several thousand years BCE. ...
binary math . Pingala, India, 3rd century BCE. ...
punched card . Basile Bouchon, France, 1725. ...
Explanation:
:)
It depends on what digital pen you're using.
Ink pens are better because they don't run out of ink like digital and you dont have to re-charge every time.
Hope this helps!:)<span />
Answer:
#include <iostream>
#include <time.h>
#include <string>
using namespace std;
int main(){
srand(time(NULL));
cout<<"Throw dice"<<endl;
int b =0;
int a=0;
a=rand()%6;
b=rand()%6;
for (int i =0;i<1;i++)
{cout<<"dice one: "<<a<<endl;}
for (int i =0;i<1;i++)
{cout<<"dice two: "<<b<<endl;}
if(a>b)
{cout<<"first dice won"<<endl;}
if(b>a)
{cout<<"second dice won"<<endl;}
else{cout<<"they are same"<<endl;
return main();
}
return 0;
}
Explanation:
/*best dice roll game just for you change it as you want but all necessary things are there/*
Answer:
Following are the solution to the given choices:
Explanation:
Given:
double currentBalance[91];//defining a double array
In point a:
The name of the array is= currentBalance.
In point b:
91 double values could be saved in the array. It requires 8bytes to hold a double that if the array size is 91*8 = 728
In point c:
Each element's data type is double.
In point d:
The array index range of values is between 0 and 90 (every array index starts from 0 and ends in N-1, here N=91).
In point e:
To access first element use currentBalance[0], for middle currentBalance[91/2] , for last currentBalance[90]
