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Lesechka [4]
3 years ago
14

Disconnecting means shall be capable of being locked in the open position. The provisions for locking do not have to reamin in p

lace. A. True B. False
Engineering
1 answer:
Lorico [155]3 years ago
7 0
The answer is a.true
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Answer: Idk but try downloading more apps it’s a lot easier to have more than one

Explanation:

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3 years ago
Prove the following languages are nonregular, once using the pumping lemma and once using the Myhill-Nerode theorem. When using
VashaNatasha [74]

Answer:

For any string, we use s = xyz

Explanation:

The pumping lemma says that for any string s in the language, with length greater than the pumping length p, we can write s = xyz with |xy| ≤ p, such that xyi z is also in the language for every i ≥ 0. For the given language, we can take p = 2.

Here are the cases:

  • Consider any string a i b j c k in the language. If i = 1 or i > 2, we take x = \epsilon   and y = a. If i = 1, we must have j = k and adding any number of a’s still preserves the membership in the language. For i > 2, all strings obtained by pumping y as defined above, have two or more a’s and hence are always in the language.
  • For i = 2, we can take    and y = aa. Since the strings obtained by pumping in this case always have an even number of a’s, they are all in the language.
  • Finally, for the case i = 0, we take x = \epsilon  , and y = b if j > 0 and y = c otherwise. Since strings of the form b j c k are always in the language, we satisfy the conditions of the pumping lemma in this case as well.
8 0
3 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

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