Question Number 207385 by hardmath last updated on 13/May/24 $$\mathrm{Find}: \\ $$$$\sqrt{\left(\mathrm{2},\mathrm{5}−\sqrt{\mathrm{5}}\right)^{\mathrm{2}} }\:−\:\sqrt[{\mathrm{3}}]{\left.\left(\mathrm{1},\mathrm{5}−\sqrt{\mathrm{5}}\right)^{\mathrm{3}} \right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:−\:\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{4}} \\ $$ Commented by Frix last updated on 13/May/24 $$=\frac{\mathrm{7}}{\mathrm{2}}−\sqrt{\mathrm{5}}−\frac{\sqrt{\mathrm{6}\left(−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right)}}{\mathrm{4}}−\frac{\sqrt{\mathrm{2}\left(−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{5}}\right.}}{\mathrm{4}}\mathrm{i}…
Question Number 207408 by hardmath last updated on 13/May/24 $$\boldsymbol{\mathrm{z}}\:\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{z}}^{\mathrm{11}} \:=\:? \\ $$ Answered by Frix last updated on 13/May/24 $${z}=\mathrm{e}^{−\frac{\pi}{\mathrm{6}}\mathrm{i}} \\ $$$${z}^{\mathrm{11}} =\mathrm{e}^{−\frac{\mathrm{11}\pi}{\mathrm{6}}\mathrm{i}} =\mathrm{e}^{\left(\mathrm{2}\pi−\frac{\mathrm{11}\pi}{\mathrm{6}}\right)\mathrm{i}}…
Question Number 207407 by hardmath last updated on 13/May/24 Commented by Frix last updated on 13/May/24 $$\mathrm{2}.\:…=\mathrm{2}\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{10}° \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 207394 by hardmath last updated on 13/May/24 $$\frac{\mathrm{a}}{\mathrm{b}}\:\:=\:\:\frac{\mathrm{c}}{\mathrm{d}} \\ $$$$\mathrm{a}^{\mathrm{3}} \:−\:\mathrm{b}^{\mathrm{3}} \:=\:\mathrm{625} \\ $$$$\mathrm{c}^{\mathrm{3}} \:−\:\mathrm{d}^{\mathrm{3}} \:=\:\mathrm{1} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:=\:? \\ $$ Answered by mr…
Question Number 207395 by hardmath last updated on 13/May/24 $$\mathrm{Geometric}\:\mathrm{series}: \\ $$$$\frac{\mathrm{b}_{\mathrm{4}} \:\centerdot\:\mathrm{b}_{\mathrm{7}} \:\centerdot\:\mathrm{b}_{\mathrm{10}} }{\mathrm{b}_{\mathrm{1}} \:\centerdot\:\mathrm{b}_{\mathrm{3}} \:\centerdot\:\mathrm{b}_{\mathrm{5}} }\:\:=\:\:\mathrm{2}^{\mathrm{12}} \:\:\:\:\:\mathrm{find}:\:\:\:\frac{\mathrm{b}_{\mathrm{5}} }{\mathrm{b}_{\mathrm{2}} }\:\:=\:\:? \\ $$ Answered by…
Question Number 207332 by mustafazaheen last updated on 12/May/24 $$\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)×\left(\overset{\rightarrow} {\mathrm{a}}\right)=? \\ $$$$\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$ Commented by Ghisom last updated on 12/May/24 $$\left(\begin{pmatrix}{{a}_{\mathrm{1}}…
Question Number 207361 by hardmath last updated on 12/May/24 $$\mathrm{y}\:=\:\frac{\mathrm{tg}\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{ctg}\boldsymbol{\mathrm{x}}}{\mathrm{8}}\:\:\:\:\:,\:\:\:\:\:\left(\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{y}\right)\:=\:? \\ $$ Answered by Berbere last updated on 12/May/24 $${ctg}\left({x}\right)=\frac{\mathrm{1}}{{y}};{y}={tan}\left({x}\right) \\ $$$$\Leftrightarrow{Min}\left(\frac{\mathrm{1}}{\mathrm{8}}\left({y}+\frac{\mathrm{1}}{{y}}\right);{y}\in\right]\mathrm{0},\infty\left[\right) \\…
Question Number 207326 by hardmath last updated on 11/May/24 $$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\:? \\ $$$$\mathrm{2}.\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°\:+\:\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207327 by hardmath last updated on 11/May/24 $$\left(\mathrm{x}−\mathrm{3}\right)\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}}\:\:=\:\:\mathrm{0} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by A5T last updated on 11/May/24 $${x}=\mathrm{3}\:{or}\:\mathrm{2}\:{or}\:−\mathrm{1} \\ $$…
Question Number 207320 by BaliramKumar last updated on 11/May/24 Commented by A5T last updated on 11/May/24 $$\mathrm{19}.\:{The}\:{remainder}\:{when}\:{a}\:{number}\:{is}\:{divided}\:{by}\:\mathrm{16} \\ $$$${is}\:{the}\:{same}\:{as}\:{the}\:{remainder}\:{when}\:{its}\:{last} \\ $$$${four}\:{digits}\:{are}\:{divided}\:{by}\:\mathrm{16} \\ $$$$\mathrm{9100}\equiv\mathrm{12}\left({mod}\:\mathrm{16}\right)\:\Rightarrow\left({b}\right) \\ $$…