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 question a, it simplifies. If you re-express it in boolean algebra, you get:
(a + b) + (!a + b)
= a + !a + b
= b
So you can simplify that circuit to just:
x = 1 if b = 1
(edit: or rather, x = b)
For question b, let's try it:
(!a!b)(!b + c)
= !a!b + !a!bc
= !a!b(1 + c)
= !a!b
So that one can be simplified to
a = 0 and b = 0
I have no good means of drawing them here, but hopefully the simplification helped!
Answer:
Hence the answer is False.
Explanation:
While there are some exceptions such as an if-statement, the scheme has one primary syntax (<thing I want to do> <things I want to do it to>).
The given statement is False.
Answer:
A. They are typically played in doors.
Explanation:
Most logical answer
