Give a big-O estimate for the number of operations, where an operation is a comparison or a multiplication, used in this segment of an algorithm (ignoring comparisons used to test the conditions in the for loops, where a1, a2, ..., an are positive real numbers). m :

O(n2)

Step-by-step explanation:

the first iteration algorithm of the i-for loop (the outer loop), the j-for loop (the inner loop) will run 2 to

n times which is represented as

(n − 1 times).

the second iteration algorithm of the i-for loop, the j-for loop will run 3 to n times represented as

(n − 2 times).

the third to the last iteration algorithm of the i-for loop, the j-for loop will run n − 1 to n times (2 times).

And the second to the last iteration of the i-for loop, the j-for loop will run from n to n times (1 time)

For the last iteration of the i-for loop, the j-for loop will run 0 times because i + 1 > n.

Now we know that the number of times the loops are run is

1 + 2 + 3 + . . . + (n − 2) + (n − 1) = n(n − 1)/2

So we can express the number of total iterations as n(n − 1)/2.

Since we have two operations per loop (one comparison and one multiplication), we have

2 ·n(n−1)/2 = n

2 − n operations.

So f(n) = n2 − n

f(n) ≤ n2

for n > 1.

Therefore, the algorithm is O(n2) with

C = 1 and k = 1.

start with a line segment pq that we will copy. mark a point r that will be one endpoint of the new line segment. set the compasses' point on the point p of the line segment to be copied. adjust the compasses' width to the point q. the compasses' width is now equal to the length of the line segment pq.

step-by-step explanation:

49%

step-by-step explanation:

when you realize that you dont know just realize i just took a test and im so smart