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Evgesh-ka [11]
3 years ago
11

What is 4x+2x(-3-3) thanku

Computers and Technology
2 answers:
vichka [17]3 years ago
6 0

Answer:

-8x

Explanation:

Let's simplify step-by-step.

4x+2x(−3−3)

=4x+−12x

Combine Like Terms:

=4x+−12x

=(4x+−12x)

=−8x

have a wonderful day

Alex_Xolod [135]3 years ago
6 0

Answer:

-8x

Explanation:

4x + 2x( -3 - 3 )

Let's simplify.

4x + 2x( <u>-3 - 3</u> )

Step 1:- Calculate the difference.

4x + <u>2</u><u>x</u><u> ( -6 )</u>

Step 2 :- Distribute 2

4x + <u>2 × -6</u>

Step 3 :- Multiply 2 and -6

4x - 12x

Step 4 :- Combine like terms

-8x

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<h3>Why should a filename reflect its contents?</h3>

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Hence, In the case above, file name do not reflect best practices and as such it is No. A file name must include the date, the author, and the version number.

See options below

Does this file name reflect best practices?1 point

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No. A file name must include the date, the author, and the version number.

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Answer:

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Execution

Completion

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The purpose of the planning implementation stage is to plan and agree on specific goals by considering all stakeholders. There is also a need to set deadlines, time-to-time evaluation, and effective coordination of the stakeholders.

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The completion phase includes all actions necessary to bring out the results within the organizzation. Initially, there is a need to agree on the dimension of the IT strategy, the time dimension, and the detailed dimension.

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The small organization majorly face the problem of addressing and it also reduce the stability of the company and organization.

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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
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int[] result = new int[arr1.length+arr2.length]
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result[i+j] = Math.min(arr1[i], arr2[j]);
if(arr1[i] < arr2[j]) {
i++;
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boolean isArr1 = i+1 < arr1.length;
for(int index = isArr1 ? i : j; index < isArr1 ? arr1.length : arr2.length; index++) {
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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.


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In Blocks, Functions is known to be the element that one can find under the Advanced section.

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Hence, in make code arcade, The Advanced section is the part of the interface can be used to answer questions about how a block functions.

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