Answer:
10G Ethernet
Explanation:
These are the options for the question;
A) 10BASE Ethernet
B) Gigabit Ethernet
C) Fast Ethernet
D) 10G Ethernet
From the question, we are informed about instance of my company deciding to upgrade the older office Ethernet network and needs the fastest speed possible but has decided against fiber optic cable. In this case my solution for this problem is getting
10G Ethernet. Ethernet can be regarded as traditional technology that connects devices in LAN(wired local area network) as well as WAN(wide area network) which allows them to have communication with each other through a protocol, this protocol is reffered to as common network language, it also be regarded as rules. 10 Gigabit Ethernet which is a technology ofgroup of computer networking that enables transmission of Ethernet frames at high rate of 10 gigabits per second. Therefore, 10G Ethernet is the solution since we need
the fastest possible speed.
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Answer:
speed = float(input("Enter the speed: "))
hours = int(input("Enter the hours: "))
distance = 0
for i in range(hours):
distance += speed * 1
print("The distance after " + str(i+1) + ". hour(s): " + str(distance))
Explanation:
*The code is in Python.
Ask the user to enter the speed and the hours
Initialize the distance as 0
Create a for loop that iterates hours times. Inside the loop, calculate the cumulative distance traveled at the end of each hour and print it (Note that the distance = speed x hour)
Answer:
Try another browser
Explanation:
Some browsers do not support advanced features like flex box.
Answer:
/*
Find Largest and Smallest Number in an Array Example
This Java Example shows how to find largest and smallest number in an
array.
*/
public class FindLargestSmallestNumber {
public static void main(String[] args) {
//array of 10 numbers
int numbers[] = new int[]{32,43,53,54,32,65,63,98,43,23};
//assign first element of an array to largest and smallest
int smallest = numbers[0];
int largetst = numbers[0];
for(int i=1; i< numbers.length; i++)
{
if(numbers[i] > largetst)
largetst = numbers[i];
else if (numbers[i] < smallest)
smallest = numbers[i];
}
System.out.println("Largest Number is : " + largetst);
System.out.println("Smallest Number is : " + smallest);
}
}
/*
Output of this program would be
Largest Number is : 98
Smallest Number is : 23
*/
Explanation:
