You should avoid those types of writing because it doesn't make you look like you are a professional.
hope this helps!
4 Types Of Dimmers :
1. Incandescent/Halogen
2. Magnetic Low Voltage (MLV)
3. Fluorescent
4. Light Emitting Diode (LED)
These are the different types of switches :
1. A single-pole switch controls lights from a single location. ...
2. A 3-way switch provides two separate control locations and is best used with recessed lights. ...
3. A 4-way switch provides for three or more dimming locations.
4. Switches are wired to the "hot" conductor in a wall box.
<em>I hope that my answer helps!</em>
William had to manually edit the data if he wants to remove the null values and incorrect column headers.
<h3>What is data editing?</h3>
Data editing is known to be a term that connote the act of making changes, reviewing or adjustment some survey data.
Note that by editing one can remove want one do not want from a group of data and as such, William had to manually edit the data if he wants to remove the null values and incorrect column headers.
Learn more about data from
brainly.com/question/26711803
#SPJ1
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:
Following are the program in python language the name of the program is factors.py
num= int(input("Please enter a positive integer: "))#Read the number by user
print("The factors of ",num,"are:")
for k in range(2,num): #iterating over the loop
if(num%k==0): #checking the condition
print(k)#display the factor
Output:
Please enter a positive integer: 12
The factors of 12 are:
2
3
4
6
Explanation:
Following are the description of the program
- Read the number by user in the "num" variable
- Iterating the for loop from k=2 to less then "num".
- In the for loop checking the factor of "num" variable by using % operator.
- Finally display the factor by using print function
