Answer:
The function in Python is as follows:
def d2x(d, x):
if d > 1 and x>1 and x<=9:
output = ""
while (d > 0):
output+= str(d % x)
d = int(d / x)
output = output[::-1]
return output
else:
return "Number/Base is out of range"
Explanation:
This defines the function
def d2x(d, x):
This checks if base and range are within range i.e. d must be positive and x between 2 and 9 (inclusive)
if d >= 1 and x>1 and x<=9:
This initializes the output string
output = ""
This loop is repeated until d is 0
while (d > 0):
This gets the remainder of d/x and saves the result in output
output+= str(d % x)
This gets the quotient of d/x
d = int(d / x) ----- The loop ends here
This reverses the output string
output = output[::-1]
This returns the output string
return output
The else statement if d or x is out of range
<em> else:</em>
<em> return "Number/Base is out of range"</em>
<em />
Answer:
Transition section helps us to move from one shot to the next.
Explanation:
Synopsis: This tells actually what is the story is all about. We can call that as a “short description about the story”.
Sketch: It is the drawing window, where we pictorially represent the story.
Transition: This actually tells us about the next move.
Shot description: We can consider a “shot” as one of the scene in the story. So, it shot contain image and its description.
Shot Sequence: This is for “Pre-visualizing” video.
Among all the choice, Transition option takes the write definition.
seach up this on goggle many great videos will pop up
Explanation:
<3
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 !!!
Answer:
You go to the option that says "snap to solve" an that's it
