y=x23+0.9sin⁡(kx)⋅3−x2y = {x}^{\frac{2}{3}} + 0.9 \sin\left( k x \right) \cdot \sqrt{3} - {x}^{2}

asked by guest
on Apr 06, 2025 at 10:53 pm



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Investigate the equation: y=x23+0.9sin⁡(kx)⋅3−x2y = {x}^{\frac{2}{3}} + 0.9 \sin\left( k x \right) \cdot \sqrt{3} - {x}^{2}.

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