What is the output of the code below assuming that global variable x has value 2 and global y has value 3? def f1(): return "ab" def f2(): return f1() * x def f3(): return f2() + f1() * y print(f3())

Following are the output to the given method:

Program Explanation:

Program:

#declaring the global variable

x=2#defining x variable  that hold integer value  

y=3#defining y variable that hold integer value

def f1():#defining a method f1

   return "ab"#using return keyword that return string value

def f2():#defining a method f2

   return f1() * x#using return keyword that call f1 method

def f3():#defining a method f3

   return f2() + f1() * y#using return keyword that calls f2 method and call f1 method and multiple value by 3  

print(f3())#calling method f3

Output:

Please find the attached file.

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ababababab

Explanation:

The code above is written in python and python uses indentation .So let me rephrase the code accordingly and explain what the code really do.

Note x and y is a global variable which can be used by any of the function declared.  According to the question x and y are 2 and 3 respectively

The first block of code describes a function f1 without any argument but the code should return the string "ab"

def f1():

      return "ab"

The second block of code defines a function f2 and returns the value of f1 multiply by x. This means you are multiplying the string "ab" by 2 which will be equals to abab

def f2():

           return f1() * x

The third block of code declared a function f3 and returns the sum of  f2 and product of f1 and y. using PEMDAS principle the multiplication aspect will be solved first so, ab × 3 = ababab, then we add it to f2  . ababab + abab = ababababab.

def f3():

        return f2() + f1() * y

Finally, we print the function f3 value to get ababababab

print(f3())

If you run the code on your IDE like below you will get  ababababab

x = 2

y = 3

def f1():

      return "ab"  

def f2():

      return f1() * x  

def f3():  

      return f2() + f1() * y  

print(f3())

a.s= 230t– 110; 1500 lbc.t= 230s– 110; 1500 lbb.s= 230t+ 110; 1720 lbd.s= 199t– 110; 1280 lb

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