A. should be the answer hope this helps :)
Answer:
b they have high sensitivity to surrounding sound
Explanation:
tell me it's correct or not
if it's correct so I hope it is helpful for you
but no so I am sorry
Answer:
The program in C++ is as follows:
#include <iostream>
using namespace std;
int perimeter(int side1, int side2, int side3){
return side1+side2+side3;
}
struct Triangle {
int side1; int side2; int side3;
};
int main(void) {
int side1, side2, side3;
cout<<"Sides of the triangle: ";
cin>>side1>>side2>>side3;
struct Triangle T;
T.side1 = side1;
T.side2 = side2;
T.side3 = side3;
cout << "Perimeter: " << perimeter(T.side1,T.side2,T.side3) << endl;
return 0;
}
Explanation:
See attachment for complete code where comments are as explanation
Answer:
b) iteration.
Explanation:
There are various algorithms which are used in computer programming. Iteration is also a type of algorithm which is used to create loops. The instructions are inserted once and then these instructions are repeated to create a loop function.
JAVA programming was employed...
What we have so far:
* Two 2x3 (2 rows and 3 columns) arrays. x1[i][j] (first 2x3 array) and x2[i][j] (second 2x3 array) .
* Let i = row and j = coulumn.
* A boolean vaiable, x1rules
Solution:
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x1[i][j] = num.nextInt();
}
}// End of Array 1, x1.
for(int i=0; i<2; i++)
{
for(int j=0; j<3; j++)
{
x2[i][j] = num.nextInt();
}
}//End of Array 2, x2
This should check if all the elements in x1 is greater than x2:
x1rules = false;
if(x1[0][0]>x2[0][0] && x1[0][1]>x2[0][1] && x1[0][2]>x2[0][2] && x1[1][0]>x2[1][0] && x1[1][1]>x2[1][1] && x1[1][2]>x2[1][2])
{
x1rules = true;
system.out.print(x1rules);
}
else
{
system.out.print(x1rules);
}//Conditional Statement
