Addition substraction and multiplication are examples of the set of _____ provided by the int data type
1 answer:
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Answer:
The code to calculate the area of a circle is:
from math import pi
def circleArea(radius):
if radius > 0:
return pi * (radius ** 2)
else:
return None
if __name__ == '__main__':
radius = 5
print("Radius: {} Area: {}".format(radius,circleArea(radius)))
Explanation:
A detailed explanation of each line of code is given below.
#Define the number pi used to calculate the area
from math import pi
#We define a function that calculates the area of a circle
def circleArea(radius):
#Check if the radius is valid ( radius > 0) since there aren´t negative radius
if radius > 0:
#Compute the area formula for a circle
return pi * (radius ** 2)
else:
#Return None if the radius is invalid
return None
#Run the function we´ve defined
if __name__ == '__main__':
#Define a radius
radius = 5
#Call the function and parse the radius through it, then print the result
print("Radius: {} Area: {}".format(radius,circleArea(radius)))
The recursive function would work like this: the n-th odd number is 2n-1. With each iteration, we return the sum of 2n-1 and the sum of the first n-1 odd numbers. The break case is when we have the sum of the first odd number, which is 1, and we return 1.
int recursiveOddSum(int n) {
if(2n-1==1) return 1;
return (2n-1) + recursiveOddSum(n-1);
}
To prove the correctness of this algorithm by induction, we start from the base case as usual:
by definition of the break case, and 1 is indeed the sum of the first odd number (it is a degenerate sum of only one term).
Now we can assume that returns indeed the sum of the first n-1 odd numbers, and we have to proof that
returns the sum of the first n odd numbers. By the recursive logic, we have
and by induction, is the sum of the first n-1 odd numbers, and 2n-1 is the n-th odd number. So,
is the sum of the first n odd numbers, as required: