Answer:b)The item’s classification and category set
Explanation: Mandatory access control(MAC) is the security component in the computer system. It is regarding the controlling the access of the operating system by the administrator.The accessing is made limited by the MAC according to the sensitivity of the data .
The authorization for user to access the system is based on this sensitivity level known sensitivity label. The objects contain the information regarding the classification and categories or level of items. Thus, the correct option is option(b).
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Answer:
B. word processing
Explanation:
The appropriate software to use for creating research papers is word processing. This refers to any software that is designed with the main function of the production, storage, and manipulation of any kind of text document that is located on a computer or word processor. One of the most popular software example is Microsoft Studio's Word Processor from their Office Software Suite.
Answer:
We have proven that the property of the Fibonacci sequence
holds by Mathematical Induction.
Explanation:
For n≥2. We prove that it holds for n=2, Assume that it holds for n=k and prove that it holds for n=k+1
F(1)=1
F(2) = 1
F(3) = F(3 − 1) + F(3 − 2)= F(2) + F(1)=1+1=2
F(4) = F(4 − 1) + F(4 − 2)= F(3) + F(2)=2+1=3
In ![[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)](https://tex.z-dn.net/?f=%5BF%28n%2B1%29%5D%5E2%20%3D%5BF%28n%29%5D%5E2%2BF%28n-1%29F%28n%2B2%29)
We prove it is true for n=2.
When n=2
![L.H.S: [F(n+1)]^2 = [F(2+1)]^2 = [F(3)]^2 = 2^2 =4\\R.H.S: [F(n)]^2+F(n-1)F(n+2) \\=[F(2)]^2+F(2-1)F(2+2)\\= [F(2)]^2+F(1)F(4)\\=1^2+1*3=1+3=4](https://tex.z-dn.net/?f=L.H.S%3A%20%5BF%28n%2B1%29%5D%5E2%20%3D%20%5BF%282%2B1%29%5D%5E2%20%3D%20%5BF%283%29%5D%5E2%20%3D%202%5E2%20%3D4%5C%5CR.H.S%3A%20%5BF%28n%29%5D%5E2%2BF%28n-1%29F%28n%2B2%29%20%20%5C%5C%3D%5BF%282%29%5D%5E2%2BF%282-1%29F%282%2B2%29%5C%5C%3D%20%5BF%282%29%5D%5E2%2BF%281%29F%284%29%5C%5C%3D1%5E2%2B1%2A3%3D1%2B3%3D4)
We assume it is true for n=k and prove that it holds for n=k+1.
When n=k+1 in ![[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)](https://tex.z-dn.net/?f=%5BF%28n%2B1%29%5D%5E2%20%3D%5BF%28n%29%5D%5E2%2BF%28n-1%29F%28n%2B2%29)
Substituting n=k+1 in the LHS:
and applying: F(k+2)=F(k+1)+F(k)
![LHS: [F(k+2)]^2=[F(k+1)+F(k)]^2\\= [F(k+1)]^2+2F(k+1)F(k)+[F(k)]^2\\=[F(k+1)]^2+F(k)[2F(k+1)+F(k)]\\=[F(k+1)]^2+F(k)[F(k+1)+F(k+1)+F(k)]\\=[F(k+1)]^2+F(k)[F(k+1)+F(k+2)]\\=[F(k+1)]^2+F(k)F(k+3)](https://tex.z-dn.net/?f=LHS%3A%20%5BF%28k%2B2%29%5D%5E2%3D%5BF%28k%2B1%29%2BF%28k%29%5D%5E2%5C%5C%3D%20%5BF%28k%2B1%29%5D%5E2%2B2F%28k%2B1%29F%28k%29%2B%5BF%28k%29%5D%5E2%5C%5C%3D%5BF%28k%2B1%29%5D%5E2%2BF%28k%29%5B2F%28k%2B1%29%2BF%28k%29%5D%5C%5C%3D%5BF%28k%2B1%29%5D%5E2%2BF%28k%29%5BF%28k%2B1%29%2BF%28k%2B1%29%2BF%28k%29%5D%5C%5C%3D%5BF%28k%2B1%29%5D%5E2%2BF%28k%29%5BF%28k%2B1%29%2BF%28k%2B2%29%5D%5C%5C%3D%5BF%28k%2B1%29%5D%5E2%2BF%28k%29F%28k%2B3%29)
Substituting n=k+1 in the RHS :![[F(n)]^2+F(n-1)F(n+2)](https://tex.z-dn.net/?f=%5BF%28n%29%5D%5E2%2BF%28n-1%29F%28n%2B2%29)
RHS=![[F(K+1)]^2+F(K+1-1)F(K+1+2)=[F(k+1)]^2+F(k)F(k+3)](https://tex.z-dn.net/?f=%5BF%28K%2B1%29%5D%5E2%2BF%28K%2B1-1%29F%28K%2B1%2B2%29%3D%5BF%28k%2B1%29%5D%5E2%2BF%28k%29F%28k%2B3%29)
Since the LHS=RHS
Therefore, the property is true.
FOR REFERENCE
When n=k in ![[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)](https://tex.z-dn.net/?f=%5BF%28n%2B1%29%5D%5E2%20%3D%5BF%28n%29%5D%5E2%2BF%28n-1%29F%28n%2B2%29)
Substituting n=k in the LHS:
and applying: F(k+1)=F(k)+F(k-1)
![LHS: [F(k+1)]^2=[F(k)+F(k-1)]^2\\= [F(k)]^2+2F(k)F(k-1)+[F(k-1)]^2\\=[F(k)]^2+F(k-1)[2F(k)+F(k-1)]\\=[F(k)]^2+F(k-1)[F(k)+F(k)+F(k-1)]\\=[F(k)]^2+F(k-1)[F(k)+F(k+1)]\\=[F(k)]^2+F(k-1)F(k+2)\\Substituting \: n=k \:in\: th\:e RHS :[F(n)]^2+F(n-1)F(n+2)](https://tex.z-dn.net/?f=LHS%3A%20%5BF%28k%2B1%29%5D%5E2%3D%5BF%28k%29%2BF%28k-1%29%5D%5E2%5C%5C%3D%20%5BF%28k%29%5D%5E2%2B2F%28k%29F%28k-1%29%2B%5BF%28k-1%29%5D%5E2%5C%5C%3D%5BF%28k%29%5D%5E2%2BF%28k-1%29%5B2F%28k%29%2BF%28k-1%29%5D%5C%5C%3D%5BF%28k%29%5D%5E2%2BF%28k-1%29%5BF%28k%29%2BF%28k%29%2BF%28k-1%29%5D%5C%5C%3D%5BF%28k%29%5D%5E2%2BF%28k-1%29%5BF%28k%29%2BF%28k%2B1%29%5D%5C%5C%3D%5BF%28k%29%5D%5E2%2BF%28k-1%29F%28k%2B2%29%5C%5CSubstituting%20%5C%3A%20n%3Dk%20%5C%3Ain%5C%3A%20th%5C%3Ae%20RHS%20%3A%5BF%28n%29%5D%5E2%2BF%28n-1%29F%28n%2B2%29)
![RHS=[F(K)]^2+F(K-1)F(K+2)](https://tex.z-dn.net/?f=RHS%3D%5BF%28K%29%5D%5E2%2BF%28K-1%29F%28K%2B2%29)
LHS=RHS
Answer:
MsgBox("User name is missing", 1 Or 16, "User Name Error")
Explanation: