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egoroff_w [7]
3 years ago
7

Linux distributions overview? information​

Computers and Technology
1 answer:
disa [49]3 years ago
6 0

Answer:

Debian

Arch Linux

Ubuntu

Fedora

Manjaro

Linux Mint

Elementary OS

openSUSE

Explanation:

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What is GIGO ?<br>plz answer me​
DanielleElmas [232]

Answer:

Garbage in, garbage out

Explanation:

In computer science, garbage in, garbage out is the concept that flawed, or nonsense input data produces nonsense output or garbage

4 0
2 years ago
When you heat a pot on a stove, the handle gets warm. which type of heat transfer is possible
valina [46]

Answer:

conduction

Explanation:

The handle will warm up until it's total heat losses equal the total heat coming in. Heat comes in mostly by conduction from the body of the pot.

5 0
3 years ago
Read 2 more answers
Using truth table, prove that:<br><br> (A + B). C = (A . C)+ (B .C) ?
Genrish500 [490]

Answer:

The image of truth table is attached.

Explanation:

In the truth table there is a separate table for the expression (A+B).C and for the expression (A.C)+(B.C) you can see in the truth table that the columns of (A+B).C is having same values as the (A.C)+(B.C).Hence we can conclude that (A+B).C is equal to (A.C)+(B.C).

4 0
3 years ago
Suppose you have two arrays of ints, arr1 and arr2, each containing ints that are sorted in ascending order. Write a static meth
telo118 [61]
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.


4 0
3 years ago
Full from of tmc computer
Butoxors [25]

Answer:

Traffic Message Channel,

7 0
3 years ago
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