Answer: Reusing software device has some economical challenges such as:
- Investment cost in reusing the device is considered as the extra cost.It might require some extra components for the working to become usable
- Requires skilled person who can develop and design the software to be used again.The creating and designing of new software design is comparatively easy but making the system reusable requires someone who has more designing skill who will be highly paid for the work
- Cost of writing and reading of the software can also be considered as the economical challenge as the reused system created is to be studied by some other organization members or sources not familiar with the functioning. .
Answer:
It involves a matter involving doubt, uncertainty, or difficulty that may be solved, problem ... getting into a frame of mind to be creative and solve problems.
Explanation:
Hope i am marked as brainliest answer
Answer:
Everything.It doesn't matter the information because there's always a way to plagiarize.No matter what the information, always, ALWAYS, cite the source.
Explanation:
Answer:
ISO 27002
Explanation:
The acronym ISO stand for The International Organization for Standardization. its a non governmental body that is responsible for setting standards guide internationally.
while NIST stand for National Institute of Standards and technology and it is also a body that is responsible for setting up standard guide in the US..
ISO 27002 is a standard code for implementing information security management systems. while ISO 27017 is for cloud security. NIST 800-12 is a general security and NIST 800-14 is for policy development. thus option A is correct.
Note :
All code preceded by ISO is from The International Organization for Standardization and are international standard. while those preceded by NIST is from National Institute of Standards and technology and are not international standard.
Answer:
#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string str)
{
char a,b;
int length = str.length();
for (int i = 0; i < length / 2; i++)
{
a=tolower(str[i]);//Converting both first characters to lowercase..
b=tolower(str[length-1-i]);
if (b != a )
return false;
}
return true;
}
int main() {
string t1;
cin>>t1;
if(isPalindrome(t1))
cout<<"The string is Palindrome"<<endl;
else
cout<<"The string is not Palindrome"<<endl;
return 0;
}
Output:-
Enter the string
madam
The string is Palindrome
Enter the string
abba
The string is Palindrome
Enter the string
22
The string is Palindrome
Enter the string
67876
The string is Palindrome
Enter the string
444244
The string is not Palindrome
Explanation:
To ignore the cases of uppercase and lower case i have converted every character to lowercase then checking each character.You can convert to uppercase also that will also work.
