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Question-226936

Question Number 226936 by Estevao last updated on 19/Dec/25 Answered by Ghisom_ last updated on 19/Dec/25 $$\int\mathrm{2}^{\mathrm{ln}\:{x}} {dx}=\int{x}^{\mathrm{ln}\:\mathrm{2}} {dx}=\frac{{x}^{\mathrm{1}+\mathrm{ln}\:\mathrm{2}} }{\mathrm{1}+\mathrm{ln}\:\mathrm{2}}+{C} \\ $$ Commented by Estevao…

Question-226919

Question Number 226919 by Spillover last updated on 19/Dec/25 Answered by Ghisom_ last updated on 19/Dec/25 $$\underset{{n}} {\underbrace{\mathrm{6}…}}\mathrm{8}^{\mathrm{2}} =\underset{{n}} {\underbrace{\mathrm{4}…}}\mathrm{6}\underset{{n}} {\underbrace{\mathrm{2}…}}\mathrm{4} \\ $$$$\Rightarrow\:\underset{{digits}} {\sum}=\left(\mathrm{4}+\mathrm{2}\right){n}+\mathrm{6}+\mathrm{4}=\mathrm{6}{n}+\mathrm{10} \\…

two-small-balls-are-hung-from-a-point-same-mass-same-charge-and-rope-length-are-same-the-two-strings-make-an-angle-30-0-when-immersed-in-a-liquid-of-0-8g-cc-the-angle-remains-same-ball-1-6g

Question Number 226907 by fantastic2 last updated on 18/Dec/25 $${two}\:{small}\:{balls}\:{are}\:{hung}\:{from}\:{a}\:{point} \\ $$$$\left({same}\:{mass},\:{same}\:{charge}\:{and}\:{rope}\:{length}\:{are}\:{same}\right) \\ $$$${the}\:{two}\:{strings}\:{make}\:{an}\:{angle}\:\mathrm{30}^{\mathrm{0}} \\ $$$${when}\:{immersed}\:{in}\:{a}\:{liquid}\:{of}\:\rho=\mathrm{0}.\mathrm{8}{g}/{cc} \\ $$$${the}\:{angle}\:{remains}\:{same}.\rho_{{ball}} =\mathrm{1}.\mathrm{6}{g}/{cc} \\ $$$${what}\:{is}\:{the}\:{value}\:{of}\:\kappa\left({dielectric}\:{const}.\right){of} \\ $$$${the}\:{liquid} \\ $$…

Reduce-to-canonical-form-sin-2-x-2-u-x-2-sin-2-2x-2-u-x-y-cos-2-x-2-u-y-2-0-

Question Number 226898 by TonyCWX last updated on 18/Dec/25 $$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{canonical}\:\mathrm{form}: \\ $$$$\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}^{\mathrm{2}} }+\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{x}\partial\mathrm{y}}+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{x}\right)\frac{\partial^{\mathrm{2}} \mathrm{u}}{\partial\mathrm{y}^{\mathrm{2}} }=\mathrm{0} \\ $$ Terms of Service…