The shape of a bst approaches that of a perfectly balanced binary tree, (log2n) is the time complexity for a balanced binary search tree in case of insertions and search.
In computing, binary bushes are mainly used for looking and sorting as they offer a way to save statistics hierarchically. a few common operations that may be conducted on binary trees encompass insertion, deletion, and traversal.
A binary tree has a special situation that each node could have a most of two youngsters. A binary tree has the benefits of each an ordered array and a linked listing as search is as brief as in a taken care of array and insertion or deletion operation are as fast as in related listing.
In pc science, a binary tree is a tree information shape in which every node has at maximum two youngsters, that are known as the left baby and the proper toddler.
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Answer: 195.5 or 12 7/32
Step-by-step explanation:
There is no letter tetha in the table so I use α instead. However it is not sence to final result.
The expression is:
(sinα+cosα)/(cosα*(1-cosα))
Lets divide the nominator and denominator by cosα
(sinα/cosα+cosα/cosα)/(cosα*(1-cosα)/cosα)= (tanα+1)/(1-cosα)=
=(8/15+1)/(1-cosα)= 23/(15*(1-cosα)) (1)
As known cos²α=1-sin²α (divide by cos²α both sides of equation)
cos²a/cos²α=1/cos²α-sin²α/cos²α
1=1/cos²α-tg²α
1/cos²α=1+tg²α
cos²α=1/(1+tg²α)
cosα=sqrt(1/(1+tg²α))= +-sqrt(1/(1+64/225))=+-sqrt(225/(225+64))=
=+-sqrt(225/289)=+-15/17 (2)
Substitute in (1) cosα by (2):
1st use cosα=15/17
1) 23/(15*(1-cosα)) =23/(15*(1-15/17))= 23*17/2=195.5
2-nd use cosα=-15/17
2)23/(15*(1-cosα)) =23/(15*(1+15/17))= 23*17/32=12 7/32
Answer:
pretty sure its -2,3 :)
Step-by-step explanation:
So B is halfway between 0 and 1. What's half of 1? 1/2.
A is halfway between 0 and B. What's half of 1/2? 1/4.
So A is 1/4 of 1.
Alternatively -
0 - A - B - - - 1
If each - is an eighth, A is at 2/8, or 1/4
