Question Number 59549 by hhghg last updated on 11/May/19 $$\frac{\mathrm{1}}{\mathrm{5}}×\mathrm{i}\:\:\mathrm{i}=\mathrm{7} \\ $$ Answered by ajfour last updated on 11/May/19 $$\mathrm{11i}=\mathrm{35} \\ $$$$\mathrm{i}=\frac{\mathrm{35}}{\mathrm{11}} \\ $$ Commented…
Question Number 59547 by hhghg last updated on 11/May/19 $$\mathrm{4}+\mathrm{t}×\mathrm{c}\:\mathrm{t}=\mathrm{3}\:\mathrm{c}=\mathrm{6} \\ $$ Answered by Forkum Michael Choungong last updated on 11/May/19 $$\mathrm{4}\:+\:{t}\:×\:\mathrm{6}{t}=\mathrm{3} \\ $$$$\mathrm{6}{t}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}…
Question Number 190611 by a.lgnaoui last updated on 07/Apr/23 $$\mathrm{Montrer}\:\mathrm{que}: \\ $$$$\mathrm{1}\bullet\boldsymbol{\mathrm{c}}^{\mathrm{2}} =\boldsymbol{\mathrm{a}}^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \\ $$$$\mathrm{2}\bullet\:\mathrm{rayon}\:\:\:\:\:\boldsymbol{\mathrm{r}}=\frac{\boldsymbol{\mathrm{c}}}{\mathrm{1}+\sqrt{\mathrm{2}}}−\frac{\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}}{\mathrm{2}+\sqrt{\mathrm{2}}} \\ $$$$ \\ $$ Commented by a.lgnaoui last updated…
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Question Number 59499 by MathGuru last updated on 11/May/19 Commented by maxmathsup by imad last updated on 11/May/19 $${we}\:{have}\:{x}^{\mathrm{2}} −\mathrm{3}{x}\:={x}\left({x}−\mathrm{3}\right)\:\:{and}\:\mathrm{2}{x}−\mathrm{6}\:=\mathrm{2}\left({x}−\mathrm{3}\right)\:\:{and} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}\:+\mathrm{6}\:={x}^{\mathrm{2}} −\mathrm{9}\:−\mathrm{5}{x}\:+\mathrm{15}\:=\left({x}−\mathrm{3}\right)\left({x}+\mathrm{3}\right)\:−\mathrm{5}\left({x}−\mathrm{3}\right)=\left({x}−\mathrm{3}\right)\left({x}−\mathrm{2}\right)\:{so} \\…
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Question Number 190536 by cortano12 last updated on 05/Apr/23 $$\:\mathrm{If}\:\mathrm{p},\mathrm{q}\:\mathrm{and}\:\mathrm{r}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$$\:\mathrm{x}^{\mathrm{3}} −\mathrm{3x}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\:\mathrm{of}\:\sqrt[{\mathrm{3}}]{\mathrm{3p}−\mathrm{2}}\:+\sqrt[{\mathrm{3}}]{\mathrm{3q}−\mathrm{2}}+\sqrt[{\mathrm{3}}]{\mathrm{3r}−\mathrm{2}}\: \\ $$ Answered by Frix last updated on 05/Apr/23…
Question Number 124988 by bemath last updated on 07/Dec/20 Commented by bemath last updated on 07/Dec/20 $${x}\:=\:\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}}\:=\:\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} \:=\:−\frac{\mathrm{1}}{\mathrm{2}}+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow{x}_{\mathrm{1}} =\:\mathrm{cos}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+\:{i}\mathrm{sin}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2007}} \:=\:\mathrm{cos}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right)+{i}\mathrm{sin}\:\left(\frac{\mathrm{4014}\pi}{\mathrm{3}}\right)…
Question Number 190520 by mathlove last updated on 04/Apr/23 $${if}\:{a},{b}\:{and}\:{c}\:{root}\:{of}\:{the} \\ $$$${x}^{\mathrm{3}} −\mathrm{16}{x}^{\mathrm{2}} −\mathrm{57}{x}+\mathrm{1}=\mathrm{0} \\ $$$${thi}\:{find}\:{thd}\:{volue}\:{of} \\ $$$${a}^{\frac{\mathrm{1}}{\mathrm{5}}} +{b}^{\frac{\mathrm{1}}{\mathrm{5}}} +{c}^{\frac{\mathrm{1}}{\mathrm{5}}} =? \\ $$ Answered by…
Question Number 190522 by Rupesh123 last updated on 04/Apr/23 Answered by gatocomcirrose last updated on 05/Apr/23 $$\mathrm{g}'\left(\mathrm{y}\right)=\mathrm{f}\left(\mathrm{y}\right)\Rightarrow\mathrm{g}''\left(\mathrm{y}\right)=\mathrm{f}'\left(\mathrm{y}\right) \\ $$$$\mathrm{f}'\left(\mathrm{x}\right)=\mathrm{cosx}\sqrt{\mathrm{1}+\mathrm{sen}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{g}''\left(\pi/\mathrm{6}\right)=\mathrm{f}'\left(\pi/\mathrm{6}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{15}}}{\mathrm{4}} \\ $$ Terms…