Answer:
b. The Hello timer as configured on the root switch.
Explanation:
There are differrent timers in a switch. The root switch is the only forwarding switch in a network, while non root switches blocks traffic to prevent looping of BPDUs in the network. Since the root switch is the only forwarding switch, all timing configuration comes from or is based on the configuration in the root.
The hello timer is no exception as the nonroot switch only sends 802.1D DTP hello BPDU messages forwarded to it by the root switch and its frequency depends on the root switch hello timer.
<span>An associate's degree requires two years of academic study and is the highest degree available at a community college</span>
Answer: Which of these is a compound morphology? (A. bookkeeper = book + keeper)
The software first uses (NLU) to map and analyze the input. It then creates the output using (NLG).
(It works by mapping input to representation and analyzing them.)
What type of morphology does this follow? (D. derivational morphology)
What is such an error called? (C. lexical ambiguity)
Explanation: Plato
Answer:
def leap_year_check(year):
return if int(year) % 4 == 0 and (int(year) % 100 != 0 or int(year) % 400 == 0)
Explanation:
The function is named leap_year_check and takes in an argument which is the year which we wish to determine if it's a new year or not.
int ensures the argument is read as an integer and not a float.
The % obtains the value of the remainder after a division exercise. A remainder of 0 means number is divisible by the quotient and a remainder other wise means it is not divisible by the quotient.
If the conditions is met, that is, (the first condition is true and either the second or Third condition is true)
the function leap_year_check returns a boolean ; true and false if otherwise.
The below code will help you to solve the given problem and you can execute and cross verify with sample input and output.
#include<stdio.h>
#include<string.h>
int* uniqueValue(int input1,int input2[])
{
int left, current;
static int arr[4] = {0};
int i = 0;
for(i=0;i<input1;i++)
{
current = input2[i];
left = 0;
if(current > 0)
left = arr[(current-1)];
if(left == 0 && arr[current] == 0)
{
arr[current] = input1-current;
}
else
{
for(int j=(i+1);j<input1;j++)
{
if(arr[j] == 0)
{
left = arr[(j-1)];
arr[j] = left - 1;
}
}
}
}
return arr;
}
