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Options :
A.) s1 < s2
B.) s1 <= s2
C.) s1.compareTo(s2) == −1
D.) s2.compareTo(s1) < 0
E.) s1.compareTo(s2) < 0
Answer: E.) s1.compareTo(s2) < 0
Explanation: Lexicographical ordering simply means the arrangement of strings based on the how the alphabets or letters of the strings appear. It could also be explained as the dictionary ordering principle of words based on the arrangement of the alphabets. In making lexicographical comparison between strings, the compareTo () method may be employed using the format below.
If first string = s1 ; second string = s2
To compare s1 with s2, the format is ;
s1.compareTo(s2) ;
If s1 comes first, that is, before s2, the method returns a negative value, that is a value less than 0 '< 0', which is the case in the question above.
If s2 comes first, that is, before s1, the method returns a positive value, that is a value greater than 0 '> 0'.
If both are s1 and s2 are the same, the output will be 0.
Answer:
I am pretty sure the answer is A
Explanation:
Answer:
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Answer:
t = [0:0.01:4];
x = 5*t - 10;
y = 25442 - 120*t + 144;
d = x.^2 + y.^2;
min = 1e+14;
for k = 1:length(t)
if d(k) < min
min = da(k);
tmin = t(k);
end
end
disp('The minimum distance is: ')
disp(sqrt(min))
disp('and it occurs at t = ')
disp(tmin)Output:
>> withLoop
The minimum distance is:
2.5106e+04
and it occurs at t =
4
Explanation:
