Answer:
752.71 kilobytes
Explanation:
Given ;
956-pixel x 1290-pixel
Shades of gray, = 32
Bit depth (n) = 2^n
32 = 2^5
Bit depth, n = 5 bits / pixel
The size of display memory :
Pixel dimension * n
(956 Pixel * 1290 pixel) * 5 bits /pixel
= 6166200 pixel * bits / pixel
= 6166200 bits
Converting bits to kilobytes :
1 bit = 1/8 bytes
1 byte = 1/1024 kilobytes
Hence,
1 bit = 1/(8 * 1024) kilobytes
1 bit = 1/8192 kilobytes
6166200 bits = (1/8192 * 6166200) kilobytes
= 752.7099609375 kilobytes
= 752.71 kilobytes
Usually back to their start point. Or a random place on the map.
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
Vi ‘filename’
For example file called main.py
vi main.py
