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Simora [160]
3 years ago
10

NAND"" and ""NOR"" circuits are commonly used as a basis for flash memory chips. A NAND B is defined to be the negation of ""A a

nd B."" A NOR B is defined to be the negation of ""A or B."" (i) Write truth tables for NAND and NOR connectives. (ii) Show that (A NAND B) (A NOR B) is equivalent to (A NAND B). (iii) Show that (A NAND B) (A NOR B) is equivalent to (A NOR B).

Mathematics
1 answer:
Anit [1.1K]3 years ago
6 0

Answer:

(i) A truth table shows how the truth or falsity of a compound statement depends on the truth or falsity of the simple statements from which it's constructed.

Since A ∧ B (the symbol ∧ means A and B) is true only when both A and B are true, its negation A NAND B is true as long as one of A or B is false.

Since A ∨ B (the symbol ∨ means A or B) is true when one of A or B is true, its negation A NOR B is only true when both A and B are false.

Below are the truth tables for NAND and NOR connectives.

(ii) To show that (A NAND B)∨(A NOR B) is equivalent to (A NAND B) we build the truth table.

Since the last column (A NAND B)∨(A NOR B) is equal to (A NAND B) it follows that the statements are equivalent.

(iii) To show that (A NAND B)∧(A NOR B) is equivalent to (A NOR B) we build the truth table.

Since the last column (A NAND B)∧(A NOR B) is equal to (A NOR B) it follows that the statements are equivalent.

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It would be easier to work the same units for volume

Since there are 1000 ml in a liter

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Taking the square root of both sides gives two possible cases,

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If \alpha=2x and \beta=\dfrac\pi4, we have

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Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

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and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

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\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

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