Question Number 126363 by ajfour last updated on 19/Dec/20 $${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0}\:\:\:\:\:\:\:;\:\:\:\:\left[{c}<\mathrm{2}/\left(\mathrm{3}\sqrt{\mathrm{3}}\right)\right] \\ $$$$\left({Solve}\:{by}\:{a}\:\:{method}\:{other}\:{than}\right. \\ $$$$\left.{trigonometric}\:{solution}\right). \\ $$ Answered by mathmax by abdo last updated on…
Question Number 60816 by ANTARES VY last updated on 26/May/19 $$\boldsymbol{\mathrm{S}}=\mathrm{4}\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$ Answered by tanmay last updated on 26/May/19 $${ds}={Rd}\theta×{Rsin}\theta{d}\phi \\ $$$${S}={R}^{\mathrm{2}} \int_{\mathrm{0}}…
Question Number 60817 by ANTARES VY last updated on 26/May/19 $$\boldsymbol{\mathrm{V}}=\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{R}}^{\mathrm{3}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$ Commented by Prithwish sen last updated on 26/May/19 http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html Commented by…
Question Number 191867 by Spillover last updated on 02/May/23 $${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$ Answered by qaz last updated on 02/May/23…
Question Number 126329 by mnjuly1970 last updated on 19/Dec/20 Answered by Olaf last updated on 19/Dec/20 $${x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left({x}^{\mathrm{5}} −\mathrm{1}\right)−\mathrm{5}\left({x}^{\mathrm{4}} −{x}\right)−\mathrm{10}\left({x}^{\mathrm{3}}…
Question Number 191839 by MATHEMATICSAM last updated on 01/May/23 $$\mathrm{2}^{{a}} \:=\:\mathrm{3}^{{b}} \:=\:\mathrm{36}^{{c}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ab}\:=\:\mathrm{2}{c}\left({a}\:+\:{b}\right). \\ $$ Answered by AST last updated on 01/May/23 $${a}={blog}_{\mathrm{2}}…
Question Number 191833 by Mingma last updated on 01/May/23 Answered by AST last updated on 01/May/23 $${a}={log}_{\mathrm{2}} \mathrm{900};{b}={log}_{\mathrm{3}} \mathrm{900};{c}={log}_{\mathrm{5}} \mathrm{900} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}={log}_{\mathrm{900}} \mathrm{2}+{log}_{\mathrm{900}} \mathrm{3}+{log}_{\mathrm{900}} \mathrm{5}…
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Question Number 60745 by Tawa1 last updated on 25/May/19 Commented by Prithwish sen last updated on 25/May/19 $$\frac{\left[\left(\mathrm{n}+\mathrm{1}\right)!\right]^{\mathrm{n}} }{\left(\mathrm{n}!\right)^{\mathrm{n}+\mathrm{1}} }\:=\left[\frac{\left(\mathrm{n}+\mathrm{1}\right)!}{\mathrm{n}!}\right]^{\mathrm{n}} .\frac{\mathrm{1}}{\mathrm{n}!}\:=\frac{\left(\mathrm{n}+\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{1}}.\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{2}}…………\frac{\left(\mathrm{n}+\mathrm{1}\right)}{\mathrm{n}}\:>\mathrm{1} \\…
Question Number 60734 by Tawa1 last updated on 25/May/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\left(\mathrm{x}\:+\:\mathrm{2}\:+\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:+\:\mathrm{3}}\right)^{\mathrm{5}} \:−\:\:\mathrm{32}\left(\mathrm{x}\:+\:\mathrm{2}\:−\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{4x}\:+\:\mathrm{3}}\right)^{\mathrm{5}} \:\:=\:\:\mathrm{31} \\ $$ Commented by Prithwish sen last updated on…