Question Number 187780 by mnjuly1970 last updated on 21/Feb/23 $$ \\ $$$$\:\:{solve} \\ $$$$\:\:\:\:\lfloor\:\:\frac{\mathrm{1}}{\mathrm{4}\:}\:+\:\mathrm{2}^{\:{x}} \:\rfloor\:+\:\lfloor\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{2}^{\:{x}+\mathrm{1}} \:\rfloor=\mathrm{1} \\ $$$$ \\ $$ Answered by witcher3 last updated…
Question Number 56705 by gunawan last updated on 22/Mar/19 $$\mathrm{the}\:\:\mathrm{smallest}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{S}=\left\{\:^{\mathrm{3}} \sqrt{{n}}−^{\mathrm{3}} \sqrt{{m}}\:\mid\:{n},\:{m}\:\in\mathbb{N}\right\}\:\:\mathrm{is}… \\ $$ Answered by kaivan.ahmadi last updated on 25/Mar/19 $${S}\:{has}\:{no}\:{smallest}\:{value}\:{of}\:\:{the} \\…
Question Number 56696 by Tawa1 last updated on 21/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{distance}\:\mathrm{from}\:\:\left(\mathrm{1},\:\mathrm{7},\:\mathrm{1}\right)\:\:\mathrm{to}\:\:\mathrm{3x}\:−\:\mathrm{2y}\:+\:\mathrm{2z}\:\:=\:\:\mathrm{6} \\ $$ Commented by mr W last updated on 21/Mar/19 $${d}=\frac{\mid\mathrm{3}×\mathrm{1}−\mathrm{2}×\mathrm{7}+\mathrm{2}×\mathrm{1}−\mathrm{6}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{17}}} \\…
Question Number 56697 by Tawa1 last updated on 21/Mar/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\:\:\:\mathrm{L}\:\:=\:\:\left(\mathrm{1},\:\mathrm{4},\:\mathrm{2}\right)\:+\:\mathrm{N}\left(\mathrm{1},\:\mathrm{3},\:\mathrm{2}\right)\:\:\:\mathrm{and} \\ $$$$\:\:\:\mathrm{r}\:\:=\:\:\left(−\mathrm{1},\:\mathrm{1},\:−\mathrm{1}\right)\:+\:\lambda\left(\mathrm{1},\:\mathrm{2},\:−\mathrm{1}\right) \\ $$ Answered by mr W last updated on 21/Mar/19 $${many}\:{methods}\:{to}\:{solve}.…
Question Number 56685 by pieroo last updated on 21/Mar/19 $$\mathrm{show}\:\mathrm{that}\:\alpha^{\mathrm{4}} +\beta^{\mathrm{4}} \:=\:\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} \right)^{\mathrm{2}} \:−\mathrm{2}\alpha^{\mathrm{2}} \beta^{\mathrm{2}} \\ $$ Commented by pieroo last updated on 21/Mar/19…
Question Number 122214 by peter frank last updated on 14/Nov/20 Answered by floor(10²Eta[1]) last updated on 14/Nov/20 $$\mathrm{if}\:\mathrm{f}\left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{a}\:\mathrm{repeated}\:\mathrm{root}\:\mathrm{so}\:\mathrm{f}'\left(\mathrm{x}\right)\:\mathrm{has}\:\mathrm{that}\:\mathrm{root} \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{54x}^{\mathrm{2}} +\mathrm{6x}−\mathrm{88}=\mathrm{0}=\mathrm{27x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{44} \\ $$$$\mathrm{x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}.\mathrm{44}.\mathrm{3}}}{\mathrm{18}}=\frac{−\mathrm{1}\pm\mathrm{23}}{\mathrm{18}} \\…
Question Number 187731 by Rupesh123 last updated on 21/Feb/23 Answered by Frix last updated on 21/Feb/23 $$\mathrm{Very}\:\mathrm{obviously}\:{x}=\mathrm{0} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 56643 by Sr@2004 last updated on 20/Mar/19 Commented by Sr@2004 last updated on 20/Mar/19 $${please}\:{solve}\:\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9} \\ $$ Commented by malwaan last updated on…
Question Number 56641 by ajfour last updated on 20/Mar/19 $$\mathrm{x}^{\mathrm{5}} +\mathrm{ax}^{\mathrm{4}} +\mathrm{cx}^{\mathrm{2}} +\mathrm{dx}+\mathrm{e}=\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{x}=\frac{\mathrm{rt}+\mathrm{s}}{\mathrm{t}+\mathrm{p}}\:\:.\:\mathrm{Find}\:\mathrm{r},\mathrm{s},\mathrm{p}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{equation}\:\mathrm{gets}\:\mathrm{transformed} \\ $$$$\mathrm{to}\:\:\:\:\:\lambda\mathrm{t}^{\mathrm{5}} +\mathrm{Dt}+\mathrm{E}=\mathrm{0}. \\ $$ Commented by ajfour…
Question Number 122174 by AbdullahMohammadNurusSafa last updated on 14/Nov/20 $${If}\:\:\boldsymbol{{x}}\:=\:\sqrt{\mathrm{5}\:}\:+\:\sqrt{\mathrm{3}},{then}\:\boldsymbol{{x}}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{3}} }\:=\:? \\ $$$$\boldsymbol{{or}},\:\boldsymbol{{is}}\:\boldsymbol{{it}}\:\boldsymbol{{possible}}\:\boldsymbol{{at}}\:\boldsymbol{{all}}? \\ $$ Answered by behi83417@gmail.com last updated on 14/Nov/20 $$\mathrm{x}^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{3}}…