Answer:
Check explanation
Explanation:
Two stacks can make use of one array by utilizing various stack pointers that begins from different ends of an array. Looking at the array A[1...
], the first stack will drive elements that starts from position 1 as well as to move its' pointer to .
The Second stack will begin at the position and motion its' pointer to 1. The best likely divide is to offer each stack a half of an array. whenever any of two stacks transverse the half-point, an overflow can happen but for that overall number of elements, it must be
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
The answer is C.) PostScript
The answer is bevel, glow, reflection, shadow, and soft edges
