What is the complexity of the following algorithm in terms of Big-0? Algorithm func(x) sum =0 for i in 1:x for j in 1:10 sum= sum +j endFor endFor for i in 1:x/2 print "*****

sum = 0 for j in range(1,12):    sum = sum + (9*j + 3)    print(sum) What is printed as a result of executing this algorithm?

bool isprime(long n) /* fixed from to https://www.geeksforgeeks.org/euclid-euler-theorem/?ref=lbp */{    // check whether a number is prime or not    int i;    for (i = 2; i * i <= n; i++)        if (n % i == 0)            return false;    return true;}

The following code implements Newton's algorithm for finding the square root of a number using repetition: n = 17                # we want the square root of 17 for exampleg = 4                 # our guess is 4 initiallyerror = 0.0000000001  # we want to stop when we are this closewhile abs(n - (g**2)) > error:  g = g - ((g**2 - n)/(2 * g))# g holds the square root of n at this point   Implement this algorithm in a function sqRoot(n), that returns the square root using recursion. Assume that the function will be called with positive numbers only. Note: Your initial guess cannot be zero!