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.
The author uses the text structure of compare and contrast to show multiple signals.
<h3>What is Text structures?</h3>
This is a term that connote the method used by authors to put together information in text.
Note that in the case above, The author uses the text structure of compare and contrast to show multiple signals.
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Answer:
Hot site.
Explanation:
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