Answer:
Algorithm 1 uses:
n*log(n) operations.
While algorithm 2 uses:
n^(3/2) operations.
We want to see, as n grows, which algorithm uses fewer operations.
So we would want to first solve:
n*log(n) = n^(3/2)
This will give us the exact value of n such that the number of operations is the same in both algorithms.
dividing both sides by n we get:
log(n) = n^(3/2)/n = n^(3/2 - 1) = n^(1/2)
where we can use:
log(n) = ln(n)/ln(10)
ln(n) = ln(10)*n^(1/2)
This equation actually has no solutions.
This happens because the right side is always larger than the left side.
Then, the same thing happens for our two initial equations:
n^(3/2) is always larger than n*log(n), as you can see in the graph below, where n^(3/2) is represented with the orange graph:
So we can conclude that the fist algorithm uses less operations as n grows.
Answer:
5 miles
Step-by-step explanation:
2 as a baseline then add 2.40 for first mile which then comes up to 4.40. 4.40 plus 1.40 = 5.80 plus 1.40 = 7.20 plus 1.40 = 8.60 plus 1.40 = 10. then we count the amount of 1.40's we added which was 4. and then we account for the first mile so there we have 5 miles
Answer: 28
Step-by-step explanation:
If I understand the question correctly, are you implying that 15 more dogs are added to the shelter
Answer:
Mine's going good. Just doing school at home and playing volleyball
For this case we must simplify the following expression:
So, if we apply distributive property to the terms within parentheses we have:
We simplify taking into account that:
- Equal signs are added and the same sign is placed.
- Different signs are subtracted and the major sign is placed.
Answer: