<span>When you examine a computer chip under a microscope, you will see </span>integrated circuits.
Answer:
O(n^2)
Explanation:
The number of elements in the array X is proportional to the algorithm E runs time:
For one element (i=1) -> O(1)
For two elements (i=2) -> O(2)
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.
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For n elements (i=n) -> O(n)
If the array has n elements the algorithm D will call the algorithm E n times, so we have a maximum time of n times n, therefore the worst-case running time of D is O(n^2)
Answer:
def find_max(num_1, num_2):
max_val = 0.0
if (num_1 > num_2): # if num1 is greater than num2,
max_val = num_1 # then num1 is the maxVal.
else: # Otherwise,
max_val = num_2 # num2 is the maxVal
return max_val
max_sum = 0.0
num_a = float(input())
num_b = float(input())
num_y = float(input())
num_z = float(input())
max_sum = find_max(num_a, num_b) + find_max(num_y, num_z)
print('max_sum is:', max_sum)
Explanation:
I added the missing part. Also, you forgot the put parentheses. I highlighted all.
To find the max_sum, you need to call the find_max twice and sum the result of these. In the first call, use the parameters num_a and num_b (This will give you greater among them). In the second call, use the parameters num_y and num_z (This will again give you greater among them)
Explanation:
1 make sure only you know the password
2 having weak security on one browser is basically a doorway for someone to get into your network
