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@Python_Dv Option C

@Python_Dv The answer is C [:3] takes Dhoni from the beginning but *not* including the 3 index, That's "D" (index 0), "h" (index 1) and "o" (index 2). Name 2[1:] goes from the index 1 of "Kohli" to the end. Index 0 is "K" and index 1 is "o". Hence "Dhoohli".

@Python_Dv C. "Dhoohli" ➡️ name1[:3] from the start (0) to 3 position "Dho" ➡️ name2[1:] from 1 position to the end "ohli"

@Python_Dv Dhoohli is the answer. Option ‘C’.

@Python_Dv The code combines the first three letters of "Dhoni" with all letters from the second letter onward in "Kohli," resulting in "Dhohli."

@Python_Dv C name1[:3] = Dho name2[1:] = ohli Concatenating the two would give you Dhoohli which isC

@Python_Dv Correct answer is option (A)

@Python_Dv @Python_Dv option 'C'

@Python_Dv D

@Python_Dv C. Because: The variables where sliced and concatenated. I.e. name1[:3] + name2[1:] => `Dho` +` ohli` => Dhoohli

@Python_Dv C

@Python_Dv == f"{name1[:3]}{name2[1:]}"

@Python_Dv C. Dhoohli This is a slicing operation Dho+ohli, therefore, Dhoohli

@Python_Dv C.

@Python_Dv D

@Python_Dv Kohni

@Python_Dv C

@Python_Dv C

@Python_Dv name1 = "Dhoni" name2 = "Kohli" result = name1[:3] + name2[1:] print(result) #Dhoohli☺️😚

@Python_Dv C

@Python_Dv C

@Python_Dv Number C

@Python_Dv C?

@Python_Dv C

@Python_Dv ٤

@Python_Dv c

@Python_Dv Answer : C

@Python_Dv C

@Python_Dv C

@Python_Dv c

@Python_Dv D

@Python_Dv C

@Python_Dv c

@Python_Dv Dhoohli

@Python_Dv C

@Python_Dv ᴅ