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Question Number 135112 by Ar Brandon last updated on 10/Mar/21 Commented by Ar Brandon last updated on 10/Mar/21 $$\mathrm{Trying}\:\mathrm{to}\:\mathrm{convert}\:\mathrm{text}\:\mathrm{into}\:\mathrm{latex}.\:\mathrm{What} \\ $$$$\mathrm{should}\:\mathrm{I}\:\mathrm{do}\:\mathrm{inorder}\:\mathrm{to}\:\mathrm{copy}\:\mathrm{the}\:\mathrm{text}\:? \\ $$ Commented by…
Question Number 69576 by Mr. K last updated on 25/Sep/19 $$\int_{−\mathrm{2}} ^{\:\mathrm{2}} \left({x}^{\mathrm{3}} {cos}\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)\sqrt{\mathrm{4}−{x}^{\mathrm{2}} }{dx} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 4040 by Filup last updated on 27/Dec/15 $${f}\left({x}\right)={x}^{\mathrm{sin}^{{x}} \left({x}\right)} \\ $$$${g}\left({x}\right)={x} \\ $$$$ \\ $$$$\mathrm{Why}\:\mathrm{is}\:{f}\:'\left({x}\right)=\mathrm{0}\:\mathrm{on}\:{g}\left({x}\right)? \\ $$$$ \\ $$$${Are}\:\boldsymbol{{all}}\:{extrema}\:\left(\mathrm{min},\:\mathrm{max},\:\mathrm{inflection}\right) \\ $$$$\mathrm{of}\:{f}\left({x}\right)\:\mathrm{on}\:{g}\left({x}\right)? \\ $$…
Question Number 69574 by Ajao yinka last updated on 25/Sep/19 Answered by MJS last updated on 25/Sep/19 $$\mathrm{36}_{\mathrm{7}} =\mathrm{27}_{\mathrm{10}} \\ $$$$\mathrm{27}_{\mathrm{10}} !=\mathrm{10}\:\mathrm{888}\:\mathrm{869}\:\mathrm{450}\:\mathrm{418}\:\mathrm{352}\:\mathrm{160}\:\mathrm{768}\:\mathrm{000}\:\mathrm{000}_{\mathrm{10}} = \\ $$$$\mathrm{36}_{\mathrm{7}}…
Question Number 135108 by bramlexs22 last updated on 10/Mar/21 $$\left(\sqrt{\frac{\mathrm{x}−\mathrm{1}}{\mathrm{x}}}\:\right)^{\mathrm{x}^{\mathrm{2}} } \:=\:\left(\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{x}+\mathrm{1}} \\ $$ Answered by MJS_new last updated on 10/Mar/21 $${x}^{\mathrm{2}} ={x}+\mathrm{1}\:\Leftrightarrow\:\frac{{x}−\mathrm{1}}{{x}}=\frac{\mathrm{1}}{{x}^{\mathrm{2}} } \\…
Question Number 135111 by faysal last updated on 10/Mar/21 Answered by Ñï= last updated on 10/Mar/21 $$\mathrm{2cos}\:\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}} \theta=\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{2}\left(\mathrm{2cos}\:^{\mathrm{2}} \theta−\mathrm{1}\right)=\mathrm{2cos}\:\mathrm{2}\theta=\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$$\mathrm{4cos}\:^{\mathrm{2}}…