Answer:
The Answer would be C, i think
Step-by-step explanation:
Answer:
21
Step-by-step explanation:
I dunno, I used a calculator B)
Assuming the order required is as n-> inf.
As n->inf, o(log(n+1)) -> o(log(n)) since the 1 is insignificant compared with n.
We can similarly drop the "1" as n-> inf, the expression becomes log(n^2+1) ->
log(n^2)=2log(n) which is still o(log(n)).
So yes, both are o(log(n)).
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Answer:
You can choose which equation below suits your taste but I recommend the 2n + 1 = -33.
Step-by-step explanation:

Answer:
<u>Translate K to N and reflect across the line containing JK. </u>
Step-by-step explanation:
The rest of the question is the attached figure.
From the figure, we can deduce the following:
∠K = ∠N
JK = MN
HK = LN
So, N will be the image of K
By translating K to N, The segment JK will over-lap the segment MN,
Then, we need to reflect the point H across the the line containing JK to get the point L
So, the translation and a reflection that will be used to map ΔHJK to ΔLMN:
<u>Translate K to N and reflect across the line containing JK. </u>