Question: def squareRootOf ( X , T ) : # Initialize the initial guess and error C = X / 2 . 0 E = X

def squareRootOf(X, T):
# Initialize the initial guess and error
C = X /2.0
E = X -(C * C)
while E > T:
# Update the current estimate
C =(C + X / C)/2.0
# Calculate the new error
E = X -(C * C)
return C, E

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