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Sophie [7]
3 years ago
13

Given the following program: public class MysteryNumbers { public static void main(String[] args) { String one = "two"; String t

wo = "three"; String three = "1"; int number = 20; ​ sentence(one, two, 3); sentence(two, three, 14); sentence(three, three, number + 1); sentence(three, two, 1); sentence("eight", three, number / 2); } ​ public static void sentence(String three, String one, int number) { System.out.println(one + " times " + three + " = " + (number * 2)); } } Write the output of each of the following calls. Sound F/X sentence(one, two, 3); sentence(two, three, 14); sentence(three, three, number + 1); sentence(three, two, 1); sentence("eight", three, number / 2);
Computers and Technology
1 answer:
enyata [817]3 years ago
7 0

Answer:

Check the explanation

Explanation:

/*Given the following program:

public class MysteryNumbers {

   public static void main(String[] args) {

       String one = "two";

       String two = "three";

       String three = "1";

       int number = 20;

​

       sentence(one, two, 3);

       sentence(two, three, 14);

       sentence(three, three, number + 1);

       sentence(three, two, 1);

       sentence("eight", three, number / 2);

   }

​

   public static void sentence(String three, String one, int number) {

       System.out.println(one + " times " + three + " = " + (number * 2));

   }

}*/

Write the output of each of the following calls.

sentence(one, two, 3);     three times two = 6

sentence(two, three, 14); 1 times three = 28

sentence(three, three, number + 1); 1 times 1 = 42

sentence(three, two, 1);   three times 1 = 2

sentence("eight", three, number / 2); 1 times eight = 20

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NAND is logically complete. Use only NAND gates to constructgate-level circuits that compute the
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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

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1         0            1

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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:

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

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