Answer:
Maintenance Phase
Explanation:
One of the concepts employed in project management for describing stages involved when carrying out an information system development project is the systems development life cycle (SLDC). The cycle which starts from carrying out a feasibility study and ends in maintenance is a highly used conceptual model. There are 5 major stages or phase and they are the; Requirement Phase, Design Phase; Implementation Phase, Test Phase, and the Maintenance phase. The maintenance phase comes when testing has been complete and all enhancement and modifications have already been developed, and the system is operating.
Answer:
The following steps will help you design a safe and effective stretching program.
Explanation:
1. You will have to follow the ACSM's guidelines used for flexibility training.
2. evaluate your flexibility rate with the "sit-and-reach" test.
3. you have to also apply the basic principles of FITT in designing your own program.
4. have a "range-of-motion" tests performance.
5. make use of SMART guidelines when setting explicit flexibility goals.
Answer:
Description: Write a MASM 32bit program with a loop and indexed addressing that calculates the sum of all thegaps between successive array elements. The array elements are doublewords, sequenced in nondecreasing order.
;Include Irvine32.inc file used with link library for 32 bit applications
.386
.model flat,stdcall
.stack 4096
ExitProcess proto,dwExitCode:dword
INCLUDE Irvine32.inc
.data
myArray DWORD 0,2,5,9,10
arrSize = ($-myArray)/TYPE myArray
gapArr DWORD arrSize-1 DUP(?)
sum DWORD ?
.code
main PROC
;Call the procedure
call Clrscr
;Initialize ESI pointer
mov esi, 0
mov ecx, arrSize
dec ecx
L1:
mov eax, myArray[esi+4]
sub eax, myArray[esi]
mov gapArr[esi], eax
inc esi
loop L1
;Calculate the sum of gaps between array elements
mov sum, 0
mov esi, 0
mov ecx, arrSize
dec ecx
; move gapArr[esi] to a temporary register and then add that register value to sum
L2:
mov edx, gapArr[esi]
add sum, edx
inc esi
loop L2
INVOKE ExitProcess,0
main ENDP
END main
Explanation:
Answer:
a=4 , b=1
Explanation:
I'm not a computer science major at all but I think I can help you with this code.
Our program wants us to add 2 to a get new a value while also subtracting 1 from b value to obtain new b value. We we want to for for as long b is not 0 and a/b is nonnegative.
One round we get:
New a=0+2=2
New b=3-1=2
Let's see if we can go another round:
New a=2+2=4
New b=2-1=1
We can't go another round because b would be negative while a is positive which would make a/b negative. So our loop stops at this 2nd round.
a=4 , b=1
Other notes:
2nd choice makes no sense because a is always going to increase because of the addition on a and b was going to decrease because of the subtraction on it.
Third choice makes no sense because a/b doesn't even exist.
Fourth choice a/b is negative not nonnegative.
