Answer:
word = input('Enter a single word: ', 's');
n = length(word);
nodupWord = [];
for i = 1:n
dup = false;
c = word(i);
for j = 1:i-1
if word(j) == c
dup = true;
break;
end
end
if ~dup
nodupWord = [nodupWord, c]; %add the non-duplicate char to end
end
end
disp(['Adjusted word: ', nodupWord])
Explanation:
The code is in Python.
Answer:
for (int h = k; h >= 0; h--)
Explanation:
From the list of given options, option C answers the question.
In the outer loop
Initially, k = 0
In the inner loop,
h = k = 0
The value of h will be printed once because h>=0 means 0>=0 and this implies once
To the outer loop
k = 1
The inner loop will always assume value of k;
So,
h = 1
This will be printed twice because of the condition h>=0 means 1>=0.
Since 1 and 0 are >=0; 1 will be printed twice
To the outer loop
k = 2
The inner loop
h = 2
This will be printed thrice because of the condition h>=0 means 2>=0.
Since 2, 1 and 0 are >=0; 2 will be printed thrice
To the outer loop
k = 3
The inner loop
h = 3
This will be printed four times because of the condition h>=0 means 3>=0.
Since 3, 2, 1 and 0 are >=0; 3 will be printed four times
1) A
2)D
3)C
4)A
5)D
Hope this will help u
The answer to your question is true.
Answer:
I am not sure on this one I am guessing it is True
Explanation:
