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mestny [16]
3 years ago
10

a technician has just finished setting a soho wireless router but the client does not want the pc on the wireless network for se

curity reasons. the technician connects an rj-45 cable to the computer and the router, does not receive network connectivity. which of the following is the next step to diagnose the problem?
Computers and Technology
1 answer:
neonofarm [45]3 years ago
4 0

Answer:

Check for the link light activities on the computer system and the router.

Explanation:

A technician has just completed installing the following wifi connection, and for privacy purposes, the user doesn't want the computer system on the wirelessly. The technician links the RJ-45 cable to the device as well as the router, but it is not linked to the network.  

So, another phase to diagnose the issue is to search on the computer system and router for the connection light activities.

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5 0
3 years ago
(70 points) This is a legit question that I have for a device FOR my homework.
guapka [62]

Answer:

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7 0
3 years ago
Read 2 more answers
Weak Induction
slega [8]

Answer:

Following are the answer to this question:

Explanation:

In option 1:

The value of n is= 7, which is (base case)

\to 3^7

when n=k for the true condition:

\to 3^k

when n=k+1 it tests the value:

\to 3^{(k+1)}= 3^k,3\\\to < (k!) 3 \ substituting \ equation \\\to

since k>6  hence the value is KH>3 hence proved.

In option 2:

when:

for n=1:(base case)

\log(1!)

0<=0 \\ condition is true

when the above statement holds value n=1

when n=k

\log(k!)

when n=k+1

\log(k+1)!=\log(k!)+\log(k+1)\\

             

\because k \log k      [\therefore KH>K \Rightarrow  \log(KH>\loK)]

In option 3:

when n=1:

A_1 \cup B=A_1 \cup B\\\\

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 \\\\

hence n=k+1 is true.

7 0
3 years ago
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