Irene created a LAN , which generally uses Ethernet cables for communication.
<h3>What is LAN about?</h3>
The LAN ports are known to be employed to link computers that do not have Wi-Fi access and it is one that make use of an Ethernet cable.
Note that Irene created a LAN , which generally uses Ethernet cables for communication to access the internet.
Learn more about wireless connection from
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#SPJ1
<span>Dispute who should be the leader of the Muslims after the death of Mohammed.<span> </span></span>
def print_feet_inch_short(num_feet, num_inches):
#function named print_feet_inch_short takes in two arguments.
print(str(num_feet)+"'",str(num_inches)+"\"")
#output the feet and inch value supplied with the shorthand sign of feet and inch using string concatenation.
num_feet = int(input())
#allows user to specify a feet value and stores it in num_feet
num_inches = int(input())
#allows user to specify an inch value and stores it in num_inch
print_feet_inch_short(num_feet, num_inches)
#calling the function with the number of feets and inches given by user.
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Since both arrays are already sorted, that means that the first int of one of the arrays will be smaller than all the ints that come after it in the same array. We also know that if the first int of arr1 is smaller than the first int of arr2, then by the same logic, the first int of arr1 is smaller than all the ints in arr2 since arr2 is also sorted.
public static int[] merge(int[] arr1, int[] arr2) {
int i = 0; //current index of arr1
int j = 0; //current index of arr2
int[] result = new int[arr1.length+arr2.length]
while(i < arr1.length && j < arr2.length) {
result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
}
boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
result[i+j+index] = isArr1 ? arr1[index] : arr2[index]
}
return result;
}
So this implementation is kind of confusing, but it's the first way I thought to do it so I ran with it. There is probably an easier way, but that's the beauty of programming.
A quick explanation:
We first loop through the arrays comparing the first elements of each array, adding whichever is the smallest to the result array. Each time we do so, we increment the index value (i or j) for the array that had the smaller number. Now the next time we are comparing the NEXT element in that array to the PREVIOUS element of the other array. We do this until we reach the end of either arr1 or arr2 so that we don't get an out of bounds exception.
The second step in our method is to tack on the remaining integers to the resulting array. We need to do this because when we reach the end of one array, there will still be at least one more integer in the other array. The boolean isArr1 is telling us whether arr1 is the array with leftovers. If so, we loop through the remaining indices of arr1 and add them to the result. Otherwise, we do the same for arr2. All of this is done using ternary operations to determine which array to use, but if we wanted to we could split the code into two for loops using an if statement.
Answer:
one inch=25.4
12 meters=13.12
10 quarts=9.46
12 milliliters=0.0004
400 pounds= 181 kg
25 meters/second=82
68F=20C
Explanation:
heres the full ones
one inch=25.4
12 meters=13.1234 yards
10 quarts=9.46353
12 milliliters=0.000405768
400 pounds=181.437
25m/s=82.021f/s
68f=20C
