- 3 P 4t 6to 2 1 (4.72 KiB) Viewed 11 times
In this equation, why this equation states that the firstcritical exponent is 𝛿 = 3?
3 p = 4t – 6tõ – – ≈³. - 2
3 p = 4t – 6tõ – – ≈³. - 2
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answerhappygod
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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
In this equation, why this equation states that the firstcritical exponent is 𝛿 = 3?
3 p = 4t – 6tõ – – ≈³. - 2