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Question Number 196327 by pticantor last updated on 22/Aug/23 $$\boldsymbol{{calcul}}\:\boldsymbol{{la}}\:\boldsymbol{{somme}}\:\boldsymbol{{suivante}}: \\ $$$$\:\:\boldsymbol{{li}}\underset{\boldsymbol{{n}}\rightarrow+\infty} {\boldsymbol{{m}}}\:\underset{\boldsymbol{{k}}=\boldsymbol{{n}}} {\overset{\mathrm{2}\boldsymbol{{n}}} {\sum}}\boldsymbol{{sin}}\left(\frac{\boldsymbol{\pi}}{\boldsymbol{{k}}}\right) \\ $$$$\:\:\boldsymbol{{elrochi}} \\ $$ Answered by sniper237 last updated on…

there-is-a-cylinder-on-the-horizontal-plane-and-there-is-a-little-ball-with-mass-m-at-the-center-of-a-circle-on-the-vertical-plane-and-there-is-a-little-hole-in-the-wall-of-the-cylinder-on-the-lift-of

Question Number 196221 by liuxinnan last updated on 20/Aug/23 $${there}\:{is}\:{a}\:{cylinder}\:{on}\:{the}\:{horizontal}\:{plane} \\ $$$${and}\:{there}\:{is}\:{a}\:{little}\:{ball}\:{with}\:{mass}\:{m} \\ $$$${at}\:{the}\:{center}\:{of}\:{a}\:{circle}\:{on}\:{the}\:{vertical}\:{plane} \\ $$$${and}\:{there}\:{is}\:{a}\:{little}\:{hole}\:{in}\:{the}\:{wall}\:{of} \\ $$$${the}\:{cylinder}\:{on}\:{the}\:{lift}\:{of}\:{the}\:{ball}\:{which} \\ $$$${just}\:{enough}\:{for}\:{the}\:{ball}\:{to}\:{pass}\:{through} \\ $$$${the}\:{gravitational}\:{acceleration}\:{is}\:{g} \\ $$$${the}\:{collision}\:{between}\:{the}\:{ball}\:{and} \\…

Question-196251

Question Number 196251 by sonukgindia last updated on 20/Aug/23 Answered by MrGHK last updated on 20/Aug/23 $${x}+{y}=\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)} \\ $$$${x}−{y}=\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$$\left({x}+{y}\right)\left({x}−{y}\right)=\frac{{ln}\left(\mathrm{3}\right)}{{ln}\left(\mathrm{2}\right)}×\frac{\mathrm{2}{ln}\left(\mathrm{2}\right)}{{ln}\left(\mathrm{3}\right)} \\ $$$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{2}…

Question-196154

Question Number 196154 by sonukgindia last updated on 19/Aug/23 Answered by mr W last updated on 19/Aug/23 $${x}^{\mathrm{2}} =−\mathrm{2}\left({x}+\mathrm{2}\right) \\ $$$${x}^{\mathrm{4}} =\mathrm{4}\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}\right)=\mathrm{4}\left(\mathrm{2}{x}\right)=\mathrm{8}{x} \\ $$$${x}^{\mathrm{6}}…