3 p = 4t – 6tõ – – ≈³. - 2
Posted: Tue Jul 05, 2022 7:28 am
- 3 P 4t 6to 2 1 (4.72 KiB) Viewed 12 times
In this equation, why this equation states that the firstcritical exponent is 𝛿 = 3?
3 p = 4t – 6tõ – – ≈³. - 2
Hi, I have a question.
In this equation, why this equation states that the firstcritical exponent is 𝛿 = 3?
3 p = 4t – 6tõ – – ≈³. - 2