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Q 5.Give outputs

a)

def reverse(s):

if len(s)==1:

return s[0]

else:
By Everleigh

Q 5.Give outputs

a)

def reverse(s):

if len(s)==1:

return s[0]

else:

return(s[-1]+reverse(s[0:len(s)-1)) print(reverse(“Recursion”))

About the author
Everleigh

EXAM QUESTIONS
QUESTION 1 Nestions taken from various sources)
Vanous options are provided as possible answers to the fo

find the roots of the following question
[tex]5x^{2}-29x+20[/tex]

1 thought on “Q 5.Give outputs<br /><br /> a)<br /><br /> def reverse(s):<br /><br /> if len(s)==1:<br /><br /> return s[0]<br /><br /> else:<br”

  1. Answer:

    Q 5.Give outputs

    a)

    def reverse(s):

    if len(s)==1:

    return s[0]

    else:

    return(s[-1]+reverse(s[0:len(s)-1)) print(reverse(“Recursion”)

    Q 5.Give outputs

    a)

    def reverse(s):

    if len(s)==1:

    return s[0]

    else:

    return(s[-1]+reverse(s[0:len(s)-1)) print(reverse(“Recursion”)

    Q 5.Give outputs

    a)

    def reverse(s):

    if len(s)==1:

    return s[0]

    else:

    return(s[-1]+reverse(s[0:len(s)-1)) print(reverse(“Recursion”)

    Q 5.Give outputs

    a)

    def reverse(s):

    if len(s)==1:

    return s[0]

    else:

    return(s[-1]+reverse(s[0:len(s)-1)) print(reverse(“Recursion”)

    Q 5.Give outputs

    a)

    def reverse(s):

    if len(s)==1:

    return s[0]

    else:

    return(s[-1]+reverse(s[0:len(s)-1)) print(reverse(“Recursion”)

    Reply

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