Answer:
"64" is the output of the above code.
Explanation:
- The function holds the expression with (**) operator which is used to calculate the power in python language.
- So if we write 2**4, then this will give 16 as output. It is because when we calculate the , then it will become 16.
- The above question passes the num1 and num2 value in the function as an argument which is 3 and 4.
- So 3 will be initialized for x variable and 4 is for y variable.
- So when we calculate then it will become which will become 16.
- So 64 is the output for the above question.
Answer:
Well, according to my parents... the correct answer is C. toward,away from
I hope this is correct!
Answer: provided in the explanation section
Explanation:
Given that:
Assume D(k) =║ true it is [1 : : : k] is valid sequence words or false otherwise
now the sub problem s[1 : : : k] is a valid sequence of words IFF s[1 : : : 1] is a valid sequence of words and s[ 1 + 1 : : : k] is valid word.
So, from here we have that D(k) is given by the following recorance relation:
D(k) = ║ false maximum (d[l]∧DICT(s[1 + 1 : : : k]) otherwise
Algorithm:
Valid sentence (s,k)
D [1 : : : k] ∦ array of boolean variable.
for a ← 1 to m
do ;
d(0) ← false
for b ← 0 to a - j
for b ← 0 to a - j
do;
if D[b] ∧ DICT s([b + 1 : : : a])
d (a) ← True
(b). Algorithm Output
if D[k] = = True
stack = temp stack ∦stack is used to print the strings in order
c = k
while C > 0
stack push (s [w(c)] : : : C] // w(p) is the position in s[1 : : : k] of the valid world at // position c
P = W (p) - 1
output stack
= 0 =
cheers i hope this helps !!!
The very first step of the lowest cost method is to find the cell with the lowest cost in the entire matrix representing the cost of transportation along with supply and demand.
C. Find the cell with the lowest cost from the remaining (not crossed out) cells.
<u>Explanation:</u>
The second step in the lowest cost method is to allocate as many units as possible to that cell (having the lowest cost) without exceeding the supply or demand.
Then cross out the row or column (or both) that is exhausted by the assignment made. These two steps are further repeated until all the assignments are made and the total cost of transportation is calculated at the end.