Find and replace is the answer
Answer:
n := length(A)
repeat
swapped := false
for i := 1 to n-1 inclusive do
<em> /* if this pair is out of order */</em>
if A[i-1] > A[i] then
<em> /* swap them and remember something changed */</em>
swap(A[i-1], A[i])
swapped := true
end if ⇒
end for
until not swapped
end procedure
Explanation:
This is pseudocode
The code of five digits is of the form:
A B C D F
Where A (the first digit) may be any of 9 digits (because 0 is excluded)
B, C, and D, may be each any of 10 digits (0 - 9)
F (the last digit) may be 0, 2, 4, 6 or 8, i.e. 5 different digits.
So the number of different available codes is:
9 * 10 * 10 * 10 * 5 = 45,000.
Answer: 45,000
Answer:yo bro I don’t know but the answer is b
Explanation:
Answer:
vehicle super class 9.1.4
Explanation:
So you need to create a super class containig all the animals use the class above for referance