Question Number 82644 by Power last updated on 23/Feb/20 Commented by Power last updated on 23/Feb/20 $$\mathrm{sorry}\:\:\mathrm{it}\:\mathrm{worked} \\ $$ Commented by john santu last updated…
Question Number 148154 by mathdanisur last updated on 25/Jul/21 $$\sqrt{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!\:+\:…\:+\:\boldsymbol{{n}}!}\:\:\in\:\mathbb{N} \\ $$$$\boldsymbol{{n}}\:=\:? \\ $$ Answered by puissant last updated on 25/Jul/21 $$\mathrm{n}=\mathrm{1} \\ $$ Commented…
Question Number 17065 by mrW1 last updated on 30/Jun/17 $$\mathrm{if}\:\mathrm{y}=\mathrm{x}^{\mathrm{x}} \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:\mathrm{this}\:\mathrm{function}? \\ $$ Commented by ajfour last updated on 30/Jun/17 $$\mathrm{y}\in\left[\mathrm{e}^{−\mathrm{1}/\mathrm{e}} ,\infty\right). \\ $$…
Question Number 82596 by peter frank last updated on 22/Feb/20 Commented by mr W last updated on 23/Feb/20 $$\left({a}\right) \\ $$$${i}\:{think}\:{the}\:{beam}\:{will}\:{stay}\:{in}\:{water}. \\ $$$$\left({b}\right) \\ $$$${the}\:{beam}\:{will}\:{be}\:{partially}\:{reflected}…
Question Number 82593 by Power last updated on 22/Feb/20 Commented by mr W last updated on 23/Feb/20 $${t}=\sqrt{{x}}>\mathrm{0} \\ $$$${t}^{\mathrm{4}} −\mathrm{16}{t}−\mathrm{12}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{6}\right)=\mathrm{0}…
Question Number 82582 by M±th+et£s last updated on 22/Feb/20 $${solve} \\ $$$${x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$ Commented by mr W last updated on 23/Feb/20 $${x}^{\mathrm{3}} +\mathrm{3}\left(−\mathrm{1}\right){x}+\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{0}…
Question Number 82589 by Power last updated on 22/Feb/20 Commented by Power last updated on 23/Feb/20 $$\mathrm{sir}\:\mathrm{mathmax}\:\mathrm{pls} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 148118 by mathdanisur last updated on 25/Jul/21 Answered by liberty last updated on 25/Jul/21 Commented by liberty last updated on 25/Jul/21 $${AM}=\sqrt{\mathrm{12}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}}…
Question Number 148113 by mathdanisur last updated on 25/Jul/21 Answered by liberty last updated on 25/Jul/21 $$\:{Hatched}\:{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\mathrm{4}\right)^{\mathrm{2}} \\ $$$$=\mathrm{16}\sqrt{\mathrm{3}}−\mathrm{8}\pi \\ $$ Commented by mathdanisur…
Question Number 148109 by Opredador last updated on 25/Jul/21 $$\:\boldsymbol{\mathrm{x}}^{\sqrt{\mathrm{2}}} \:+\:\boldsymbol{\mathrm{x}}\:−\:\mathrm{3}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\:\boldsymbol{\mathrm{x}}\:\in\:\mathbb{R},\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com