CNG is a readily available alternative to ________

I WILL GIVE BRAINLIEST IF ANSWER FAST What is the measurement of this dial caliper?

Galina-37 [17]

The answer is B. 4.045 inches

6 0

3 years ago

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 rectangular channel 2 m wide carries 3 m3 /s of water at a depth of 1.2 m. If an obstruction 40 cm wide is placed in the middl

Marta_Voda [28]