Answer:
Following are the answer to this question:
Explanation:
In option 1:
The value of n is= 7, which is (base case)
when n=k for the true condition:
when n=k+1 it tests the value:
since k>6 hence the value is KH>3 hence proved.
In option 2:
when:
for n=1:(base case)
0<=0 \\ condition is true
when the above statement holds value n=1
when n=k
when n=k+1
![[\therefore KH>K \Rightarrow \log(KH>\loK)]](https://tex.z-dn.net/?f=%5B%5Ctherefore%20KH%3EK%20%5CRightarrow%20%20%5Clog%28KH%3E%5CloK%29%5D)
In option 3:
when n=1:
when n=k
![\to (A_1\cap A_2 \cap.....A_k) \cup B\\=(A_1\cup B) \cap(A_2\cup B_2)....(A_k \capB).....(a)\\\to n= k+1\\ \to (A_1\cap A_2 \cap.....A_{kH}) \cup B= (A_1\cup B)\\\\\to [(A_1\cap A_2 \cap.....A_{k}) \cup B]\cap (A_{KH}\cup B)\\\\\to [(A_1\cup B) \cap (A_2 \cup B) \cap (A_3\cup B).....(A_k\cup B)\cap (A_{k+1} \cup B)\\\\ \ \ \ \ \ \ substituting \ equation \ a \\\\](https://tex.z-dn.net/?f=%5Cto%20%28A_1%5Ccap%20A_2%20%5Ccap.....A_k%29%20%5Ccup%20B%5C%5C%3D%28A_1%5Ccup%20B%29%20%5Ccap%28A_2%5Ccup%20B_2%29....%28A_k%20%5CcapB%29.....%28a%29%5C%5C%5Cto%20n%3D%20k%2B1%5C%5C%20%5Cto%20%28A_1%5Ccap%20A_2%20%5Ccap.....A_%7BkH%7D%29%20%5Ccup%20B%3D%20%28A_1%5Ccup%20B%29%5C%5C%5C%5C%5Cto%20%20%5B%28A_1%5Ccap%20A_2%20%5Ccap.....A_%7Bk%7D%29%20%5Ccup%20B%5D%5Ccap%20%28A_%7BKH%7D%5Ccup%20B%29%5C%5C%5C%5C%5Cto%20%20%5B%28A_1%5Ccup%20B%29%20%5Ccap%20%28A_2%20%5Ccup%20B%29%20%5Ccap%20%28A_3%5Ccup%20B%29.....%28A_k%5Ccup%20B%29%5Ccap%20%28A_%7Bk%2B1%7D%20%5Ccup%20B%29%5C%5C%5C%5C%20%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20substituting%20%5C%20equation%20%5C%20a%20%5C%5C%5C%5C)
hence n=k+1 is true.
Answer:
Given that:
A= 40n^2
B = 2n^3
By given scenario:
40n^2=2n^3
dividing both sides by 2
20n^2=n^3
dividing both sides by n^2 we get
20 = n
Now putting n=20 in algorithms A and B:
A=40n^2
= 40 (20)^2
= 40 * (400)
A= 16000
B= 2n^3
= 2 (20)^3
= 2(8000)
B= 16000
Now as A and B got same on n = 20, then as given:
n0 <20 for n =20
Let us take n0 = 19, it will prove A is better than B.
We can also match the respective graphs of algorithms of A and B to see which one leads and which one lags, before they cross at n= 20.
Answer:
A. the BY statement.
Explanation:
The BY statement aids the procedure MEANS to develop the tree for the current BY group only, thereby analyze the stats, and clean the tree prior to the start and development of the next BY group.
However, without the BY statement, procedure MEANS develops its AVL tree for both the whole file and all sector estimates crossings established in CLASS.
Hence, in this case, the correct answer is the BY Statement.
Fed jig. defect bhds. if. bugs. bffs. hex.
