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Mamont248 [21]
3 years ago
13

True of False - use T or F An interface is compiled into a separate bytecode file (.class).

Computers and Technology
1 answer:
olga55 [171]3 years ago
5 0

Answer:

T

Explanation:

An interface is compiled to a separate bytecode class file.

For example of out application has a class myClass.java and an interface myInterface.java:

class myClass{

   String name;

   int age;

}

interface myInterface{

  public String getName();

  public int getAge();

}

Both of these are compiled to their own respective class files by the compiler,namely, myClass.class and myInterface.class.

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A section-lined area is always completely bounded or outlined by an?
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Visible outline.
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Rank these three account types in order of decreasing liquidity. Start by picking the most liquid account type first
Ghella [55]
As there are no options given, I will simply mention the three most liquid accounts in order:

<span>1. Checking Account
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When creating any digital work, what is the MOST important thing to remember?
kvasek [131]

Answer:

ALWAYS  double check your work

Explanation:

3 0
3 years ago
14. In cell B14, create a formula without using a function that adds 1 to the value in cell B12 and then multiplies the result b
vagabundo [1.1K]

Answer:

The formula for the given problem is given below:

= (1+B$12)×B13

Explanation:

Immediately you do one, then you can autofill the formula to the mentioned range B15:B17 and then to C14 to H17

When been done correctly, this is how the formula will look in those cells if you do it correctly.

Check the file attached below to see it.

7 0
2 years ago
Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, give an O(n) time algorithm t
kari74 [83]

Answer:

See explaination

Explanation:

Keep two iterators, i (for nuts array) and j (for bolts array).

while(i < n and j < n) {

if nuts[i] == bolts[j] {

We have a case where sizes match, output/return

}

else if nuts[i] < bolts[j] {

this means that size of nut is smaller than that of bolt and we should go to the next bigger nut, i.e., i+=1

}

else {

this means that size of bolt is smaller than that of nut and we should go to the next bigger bolt, i.e., j+=1

}

}

Since we go to each index in both the array only once, the algorithm take O(n) time.

8 0
3 years ago
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