Your answer would be: P<span>redatory Pricing .</span>
Answer:
The correct answer to the following question will be "Data-in-use".
Explanation:
- Data-in-use is an IT term referring to active information that is usually preserved in a semi-persistent physical state in RAM of computer, CPU registers or caches.
- It might be created, modified or changed, deleted or accessed via different endpoints of the interface. This is indeed a useful term for IT departments to pursue institutional defense.
Therefore, it's the right answer.
A) IS MOSTLY JUST ENGINE Oil.
Answer:
C code for half()
#include<stdio.h>
void half(float *pv);
int main()
{
float value=5.0; //value is initialized
printf ("Value before half: %4.1f\n", value); // Prints 5.0
half(&value); // the function call takes the address of the variable.
printf("Value after half: %4.1f\n", value); // Prints 2.5
}
void half(float *pv) //In function definition pointer pv will hold the address of variable passed.
{
*pv=*pv/2; //pointer value is accessed through * operator.
}
- This method is called call-by-reference method.
- Here when we call a function, we pass the address of the variable instead of passing the value of the variable.
- The address of “value” is passed from the “half” function within main(), then in called “half” function we store the address in float pointer ‘pv.’ Now inside the half(), we can manipulate the value pointed by pointer ‘pv’. That will reflect in the main().
- Inside half() we write *pv=*pv/2, which means the value of variable pointed by ‘pv’ will be the half of its value, so after returning from half function value of variable “value” inside main will be 2.5.
Output:
Output is given as image.
Answer:
PROGRAM QuadraticEquation
Solver
IMPLICIT NONE
REAL :: a, b, c
;
REA :: d
;
REAL :: root1, root2
;
//read in the coefficients a, b and c
READ(*,*) a, b, c
WRITE(*,*) 'a = ', a
WRITE(*,*) 'b = ', b
WRITE(*,*) 'c = ', c
WRITE(*,*)
// computing the square root of discriminant d
d = b*b - 4.0*a*c
IF (d >= 0.0) THEN //checking if it is solvable?
d = SQRT(d)
root1 = (-b + d)/(2.0*a) // first root
root2 = (-b - d)/(2.0*a) // second root
WRITE(*,*) 'Roots are ', root1, ' and ', root2
ELSE //complex roots
WRITE(*,*) 'There is no real roots!'
WRITE(*,*) 'Discriminant = ', d
END IF
END PROGRAM QuadraticEquationSolver