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Phantasy [73]
2 years ago
9

Define an interface GUIComponent consisting of four method declaration: open: void-returning, accepts no parameters close: both

accepts no parameters, return boolean indicating whether or not the component is willing to be closed at that point (for example, a text window might have unsaved data and should not be closed) resize: void-returning accepts new width and height (integers) in that order move: void returning, accepts new x and y positions (in that order) respectively
Computers and Technology
1 answer:
Lesechka [4]2 years ago
3 0

Answer:

interface GUIComponent{

  public void open();

  public boolean close();

  public void resize(int width, int height);

  public void move(int x, int y);

}

Explanation:

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These tools give presenters more freedom to move about the room and interact with their audience.
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Which statement describes borders and shading ?
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3 years ago
NAND is logically complete. Use only NAND gates to constructgate-level circuits that compute the
Drupady [299]

Answer:

Hi, for this exercise we have two laws to bear in mind:

Morgan's laws

NOT(А).NOT(В) = NOT(A) + NOT (B)

NOT(A) + NOT (B) = NOT(А).NOT(В)

And the table of the Nand

INPUT OUTPUT

A B A NAND B

0 0         1

0 1         1

1 0         1

1 1         0

Let's start!

a.

Input            OUTPUT

A       A     A NAND A

1         1             0

0        0            1

b.

Input            OUTPUT

A       B     (A NAND B ) NAND (A NAND B )

0         0            0

0         1            0

1          0             0

1          1             1

C.

Input            OUTPUT

A       B     (A NAND A ) NAND (B NAND B )

0        0           0

0        1             1

1         0            1

1          1            1

Explanation:

In the first one, we only need one input in this case A and comparing with the truth table we have the not gate

In the second case, we have to negate the AND an as we know how to build a not, we only have to make a nand in the two inputs (A, B) and the make another nand with that output.

In the third case we have that the OR is A + B and we know in base of the morgan's law that:

A + B = NOT(NOT(А).NOT(В))

So, we have to negate the two inputs and after make nand with the two inputs negated.

I hope it's help you.

3 0
3 years ago
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