Question Number 155244 by mathdanisur last updated on 27/Sep/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$ \\ $$$$\left(\mathrm{sin}\boldsymbol{\mathrm{x}}\right)^{\mathrm{3}} \:+\:\mathrm{sin}\boldsymbol{\mathrm{x}}\:=\:\mathrm{cos}\boldsymbol{\mathrm{x}} \\ $$ Answered by ajfour last updated on 27/Sep/21 $$\mathrm{tan}\:{x}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{tan}\:^{\mathrm{2}}…
Question Number 24172 by ajfour last updated on 13/Nov/17 Commented by ajfour last updated on 13/Nov/17 $${Q}.\mathrm{24149}\:\left({solution}\right) \\ $$ Answered by ajfour last updated on…
Question Number 155243 by aliyn last updated on 27/Sep/21 $$\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\boldsymbol{{the}}\:\boldsymbol{{integral}}\:\int_{−\mathrm{1}} ^{\:\mathrm{1}} \:\boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{{p}}_{\mathrm{2}} \left(\boldsymbol{{x}}\right)\:\boldsymbol{{dx}}\:\:\:\boldsymbol{{is}}\:?\: \\ $$ Commented by aliyn last updated on 28/Sep/21 $$???\: \\…
Question Number 155242 by aliyn last updated on 27/Sep/21 $$\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\:\:\boldsymbol{\beta}\:\left(\mathrm{2},\:{n}\:\right)\:\boldsymbol{{is}}\:? \\ $$ Commented by tabata last updated on 27/Sep/21 $$=\:\frac{\Gamma\mathrm{2}\:\Gamma\left({n}\right)}{\Gamma\left({n}+\mathrm{2}\right)}\:=\:\frac{\left({n}−\mathrm{1}\right)!}{\left({n}+\mathrm{1}\right)!}\:=\:\frac{\left({n}−\mathrm{1}\right)!}{{n}\:\left({n}+\mathrm{1}\right)\left({n}−\mathrm{1}\right)!}\:=\:\frac{\mathrm{1}}{{n}\:\left({n}+\mathrm{1}\right)} \\ $$ Answered by puissant…
Question Number 24168 by Tinkutara last updated on 13/Nov/17 $$\mathrm{A}\:\mathrm{plank}\:\mathrm{of}\:\mathrm{mass}\:{M}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{sliding}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{surface}\:\mathrm{with}\:\mathrm{constant} \\ $$$$\mathrm{velocity}\:\mathrm{of}\:\mathrm{10}\:\mathrm{ms}^{−\mathrm{1}} .\:\mathrm{A}\:\mathrm{another}\:\mathrm{block}\:\mathrm{of} \\ $$$$\mathrm{mass}\:{M}\:\mathrm{kg}\:\mathrm{is}\:\mathrm{gently}\:\mathrm{placed}\:\mathrm{on}\:\mathrm{it}.\:\mathrm{The} \\ $$$$\mathrm{coefficient}\:\mathrm{of}\:\mathrm{friction}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{block}\:\mathrm{and}\:\mathrm{the}\:\mathrm{upper}\:\mathrm{surface}\:\mathrm{of}\:\mathrm{the}\:\mathrm{plank} \\ $$$$\mathrm{is}\:\mathrm{0}.\mathrm{2}.\:\mathrm{Assuming}\:\mathrm{that}\:\mathrm{plank}\:\mathrm{is}\:\mathrm{long} \\ $$$$\mathrm{enough}\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{block}\:\mathrm{does}\:\mathrm{not}\:\mathrm{fall}…
Question Number 155237 by mnjuly1970 last updated on 27/Sep/21 $$ \\ $$$$\:\:\mathrm{I}{f}\:\:\:\Omega\:=\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}^{\:\mathrm{2}} \left({x}\:\right).{sin}\left(\sqrt{{x}\:}\:\right)}{{x}}\:{dx} \\ $$$$\:\:\:\:\:{prove}\:{that}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Omega\:=\:\mathrm{4}\:\gamma^{\:\mathrm{2}} \:+\:\frac{\pi^{\:\mathrm{3}} }{\mathrm{3}}\:\:\:\:\:\blacksquare\:{m}.{n} \\ $$ Answered by…
Question Number 89702 by 974342176 last updated on 18/Apr/20 $$\mathrm{3}^{\mathrm{2}{t}+\mathrm{1}} +\mathrm{3}^{{t}+\mathrm{2}} =\frac{\mathrm{10}}{\mathrm{3}} \\ $$ Answered by Kunal12588 last updated on 18/Apr/20 $$\mathrm{3}^{{t}} ={p} \\ $$$$\mathrm{3}{p}^{\mathrm{2}}…
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Question Number 24164 by Tinkutara last updated on 13/Nov/17 $$\mathrm{Five}\:\mathrm{moles}\:\mathrm{of}\:\mathrm{an}\:\mathrm{ideal}\:\mathrm{gas}\:\mathrm{expand} \\ $$$$\mathrm{isothermally}\:\mathrm{and}\:\mathrm{reversibly}\:\mathrm{from}\:\mathrm{a} \\ $$$$\mathrm{pressure}\:\mathrm{of}\:\mathrm{10}\:\mathrm{atm}\:\mathrm{to}\:\mathrm{2}\:\mathrm{atm}\:\mathrm{at}\:\mathrm{300}\:\mathrm{K}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{largest}\:\mathrm{mass}\:\left(\mathrm{approx}\right) \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{lifted}\:\mathrm{through}\:\mathrm{a}\:\mathrm{height}\:\mathrm{of} \\ $$$$\mathrm{1}\:\mathrm{m}\:\mathrm{in}\:\mathrm{this}\:\mathrm{expansion}? \\ $$ Commented by Tinkutara…