Question Number 79883 by MJS last updated on 29/Jan/20 $$\mathrm{to}\:\mathrm{Sir}\:\mathrm{Jagoll}\:\left({and}\:{of}\:{course}\:{everybody}\:{else}\right) \\ $$$$ \\ $$$$\left(\mathrm{1}\right) \\ $$$${y}=\frac{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{4}}= \\ $$$$=\frac{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}{\left({x}−\mathrm{4}\right)\left({x}+\mathrm{1}\right)}\:\Rightarrow \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{zeros}\:\mathrm{at}\:{x}=−\mathrm{2};\:{x}=\mathrm{3}}\\{\mathrm{vertical}\:\mathrm{asymptotes}\:\mathrm{at}\:{x}=−\mathrm{1};\:{x}=\mathrm{4}}\end{cases} \\ $$$$\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\left\{−\mathrm{1};\:\mathrm{4}\right\} \\…
Question Number 79876 by mathocean1 last updated on 28/Jan/20 $${f}\:{is}\:{derivable}\:{in}\:\mathbb{R}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:\mathrm{is}\:\mathrm{pair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{odd}\left(\mathrm{unpair}\right). \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that}\:\mathrm{if}\:{f}\:{is}\:\mathrm{unpair}\:,\:{f}\:'\:\mathrm{is}\:\mathrm{pair}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sirs} \\ $$ Commented by MJS last updated…
Question Number 145411 by puissant last updated on 04/Jul/21 $$\mathrm{un}\:\mathrm{espace}\:\mathrm{vectoriel}\:\mathrm{n}'\mathrm{as}\:\mathrm{un}\:\mathrm{seul}\: \\ $$$$\mathrm{hyper}-\mathrm{plan}..\:\mathrm{quelle}\:\mathrm{est}\:\mathrm{sa}\:\mathrm{dimension}.? \\ $$ Answered by Olaf_Thorendsen last updated on 04/Jul/21 $$\mathrm{L}'\mathrm{une}\:\mathrm{des}\:\mathrm{definitions}\:\mathrm{des}\:\mathrm{hyperplans} \\ $$$$\mathrm{entraine}\:\mathrm{dim}\left(\mathrm{H}\right)\:=\:\mathrm{dim}\left(\mathrm{E}\right)−\mathrm{1}\:=\:{n}−\mathrm{1} \\…
Question Number 145406 by mathdanisur last updated on 04/Jul/21 $${if}\:\:\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}+\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{b}}=\mathrm{2}\:;\:\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}+\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{c}}=\mathrm{0} \\ $$$${find}\:\:\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{a}}} \boldsymbol{{b}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{b}}} \boldsymbol{{c}}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{log}}_{\boldsymbol{{c}}} \boldsymbol{{a}}}\:=\:? \\ $$ Answered by som(math1967) last…
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Question Number 79861 by M±th+et£s last updated on 28/Jan/20 Commented by mr W last updated on 29/Jan/20 $${proof}\:{see}\:{Q}\mathrm{79932}! \\ $$ Terms of Service Privacy Policy…
Question Number 145393 by physicstutes last updated on 04/Jul/21 $$\mathrm{The}\:\mathrm{number}\:\frac{\mathrm{2}}{\mathrm{13}}\:\mathrm{expressed}\:\mathrm{as}\:\mathrm{a}\:\mathrm{decimal}\:\mathrm{is}\:\mathrm{0}.\mathrm{153846153846}… \\ $$$$\mathrm{The}\:\mathrm{200th}\:\mathrm{and}\:\mathrm{300th}\:\mathrm{digits}\:\mathrm{are}? \\ $$ Answered by mr W last updated on 04/Jul/21 $${mod}\:\left(\mathrm{200},\mathrm{6}\right)=\mathrm{2} \\ $$$$\Rightarrow\mathrm{200}{th}\:{digit}\:{is}\:\mathrm{5}…
Question Number 79852 by Pratah last updated on 28/Jan/20 Commented by MJS last updated on 28/Jan/20 $$\mathrm{Sir}\:\mathrm{Pratah},\:\mathrm{how}\:\mathrm{old}\:\mathrm{are}\:\mathrm{you}?\:\mathrm{this}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{solved}\:\mathrm{by}\:\mathrm{any}\:\mathrm{13}\:\mathrm{years}\:\mathrm{old}\:\mathrm{student} \\ $$ Answered by key of…
Question Number 145383 by mathdanisur last updated on 04/Jul/21 $${if}\:\:{a};{b};{c}\in\mathbb{R}^{+} \\ $$$${find}\:\:\left(\sqrt[{\mathrm{3}}]{{abc}}\:+\:\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{\mathrm{2}{b}}\:+\:\frac{\mathrm{1}}{\mathrm{4}{c}}\right)_{\boldsymbol{{min}}} =\:? \\ $$ Answered by mnjuly1970 last updated on 04/Jul/21 $$\:\mathrm{A}\:\geqslant\:\sqrt[{\mathrm{3}}]{{abc}}\:+\mathrm{3}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{8}{abc}}} \\ $$$$\:\:\:\:\:\:=\sqrt[{\mathrm{3}}]{{abc}}\:+\frac{\mathrm{3}}{\mathrm{2}}\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{abc}}}\:\overset{{am}−{gm}}…
Question Number 79826 by Pratah last updated on 28/Jan/20 Commented by Pratah last updated on 28/Jan/20 $$\mathrm{please}\:\:\mathrm{solution}.\:\:\mathrm{help} \\ $$ Commented by mr W last updated…