Physics, asked by mehtaindtrd, 9 months ago

pls answer with proper explanation​

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Answered by doausanse18
0

Answer:

Explanation:

  • x^{2} \sqrt{x} \sqrt[n]{x} \frac{x}{y} \leq \beta x_{123} \sqrt[n]{x}  \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \pi \alpha x_{123} x^{2} \leq \geq \neq \pi \alpha \beta x_{123} \\
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