When you do a vehicle check, what do you NOT need to keep an eye on?

This question allows you to practice proving a language is non-regular via the Pumping Lemma. Using the Pumping Lemma (Theorem 1

Ulleksa [173]

Answer:

L is not a regular language with formal proofs

Explanation:

(a) To prove that L is not a regular language, we will use a proof by contradiction. the assumption entails that L is a regular language. Then by the Pumping Lemma for Regular Languages, 

there exists a pumping length p for L such that for any string s ∈ L where |s| ≥ p, 

s = xyz subject to the following conditions: 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L

(b) To determine that L is not a regular language, we mke use of proof by contradiction. lets assume, that L is regular. Then by the Pumping Lemma for Regular Languages, it states also,

The pumping length, p for L such that for any string s ∈ L where |s| ≥ p, s = xyz subject to the condtions as follows : 

(a) |y| > 0 

(b) |xy| ≤ p, and 

(c) ∀i > 0, xyi 

z ∈ L. 

Choose s = 0p10p 

. Clearly, |s| ≥ p and s ∈ L. By condition (b) above, it follows is shown. by the first condition x and y are zeros.

for some k > 0. Per (c), we can take i = 0 and the resulting string will still be in L. Thus, xy0 

z should be in L. xy0 

z = xz = 0(p−k)10p 

It is shown that is is not in L. This is a contraption with the pumping lemma. our assumption that L is regular is incorrect, and L is not a regular language

6 0

3 years ago

(a) Draw the Moore finite state machine (FSM) of an electronic combination lock with a RESET button, two number buttons (0 and 1

Dafna11 [192]

Answer:

Explanation:

There are a total of 6 states and 3 bits in this problem. Whenever the Reset button is pressed, RESET state is called otherwise the state according to the diagram is called. For the combination to be "01011", the input sequence has to be in the same order. If 0 is pressed instead of 1 in state "010", the last state of output ending with 0 will be called and likewise in all the states that follow.

7 0

3 years ago

Where can you find free air pods that look real

Tema [17]

You can find air pods that look real on letgo. or you can go to wish.com but if you want a good pair jus get the real ones

7 0

3 years ago

An air standard cycle with constant specific heats is executed in a closed system with 0.003 kg of air and consists of the follo

Vsevolod [243]

Answer:

a) Please see attached copy below

b) 0.39KJ

c) 20.9‰

Explanation:

The three process of an air-standard cycle are described.

Assumptions

1. The air-standard assumptions are applicable.

2. Kinetic and potential energy negligible.

3. Air in an ideal gas with a constant specific heats.

Properties:

The properties of air are gotten from the steam table.

b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.

P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K

T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K

Qin=m(u₂₋u₁)=mCv(T₂-T₁)

=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ

Qout=m(h₃₋h₁)=mCp(T₃₋T₁)

=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ

Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ

c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰

7 0

2 years ago

Water flows at low speed through a circular tube with inside diameter of 2 in. A smoothly contoured body of 1.5 in. diameter is

Art [367]

Answer:

Pressure = 11.38 psi

Force = 13.981 Ibf

Explanation:

Step by step solution is in the attached document.

5 0

3 years ago