Question Number 74582 by Maclaurin Stickker last updated on 26/Nov/19 $${Find}\:{all}\:{values}\:{of}\:{x}: \\ $$$$\left(\mathrm{2}^{{x}} \right)^{{x}^{\mathrm{2}} −\mathrm{8}} =\mathrm{32} \\ $$ Answered by MJS last updated on 26/Nov/19…
Question Number 74580 by chess1 last updated on 26/Nov/19 Answered by mind is power last updated on 26/Nov/19 $$\sqrt{\mathrm{1}+\mathrm{t}}=\mathrm{1}+\frac{\mathrm{t}}{\mathrm{2}}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{o}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{t}+\mathrm{1}\right)}=\mathrm{1}+\frac{\mathrm{t}}{\mathrm{3}}+\mathrm{o}\left(\mathrm{t}\right)\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }=\mathrm{1}+\frac{\mathrm{x}^{\mathrm{2}}…
Question Number 140113 by EnterUsername last updated on 04/May/21 $$\mathrm{Let}\:{z}_{\mathrm{1}} =\mathrm{1}+{i},\:{z}_{\mathrm{2}} =−\mathrm{1}−{i}\:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers} \\ $$$$\mathrm{such}\:\mathrm{that}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} \:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{form}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}. \\ $$$$\mathrm{Then}\:{z}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left(\mathrm{A}\right)\:\sqrt{\mathrm{3}}\left(\mathrm{1}+{i}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\sqrt{\mathrm{3}}\left(\mathrm{1}−{i}\right) \\ $$$$\left(\mathrm{C}\right)\:\sqrt{\mathrm{3}}\left({i}−\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\sqrt{\mathrm{3}}\left(−\mathrm{1}−{i}\right)…
Question Number 74579 by malikmasood3535@gmail.com last updated on 26/Nov/19 $${find}\:{the}\:{gradient}\:{of}\:{scalar}\:{point}\:{function}\:{being}\:{expressed}\:{in}\:{term}\:{of}\:{scalar}\:{triple}\:{product}\:{as}\:{u}=\left(\bar {{a}},\bar {{b}},\bar {{c}}\right)=\bar {{a}}.\bar {{b}}×\bar {{c}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9041 by tawakalitu last updated on 16/Nov/16 Commented by sou1618 last updated on 16/Nov/16 $$ \\ $$$$ \\ $$$$\begin{pmatrix}{}&{{X}}&{}\\{{A}}&{{B}}&{{C}}\end{pmatrix} \\ $$$${X}=\frac{{A}×{B}×{C}}{\mathrm{10}} \\ $$…
Question Number 140114 by EnterUsername last updated on 04/May/21 $$\mathrm{If}\:\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta+\mathrm{cos}\:\gamma=\mathrm{0}=\mathrm{sin}\:\alpha+\mathrm{sin}\:\beta+\mathrm{sin}\:\gamma,\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{cos}\left(\mathrm{2}\alpha\right)+\mathrm{cos}\left(\mathrm{2}\beta\right)+\mathrm{cos}\left(\mathrm{2}\gamma\right)=\mathrm{0} \\ $$$$\left(\mathrm{B}\right)\:\mathrm{sin}\left(\mathrm{3}\alpha\right)+\mathrm{sin}\left(\mathrm{3}\beta\right)+\mathrm{sin}\left(\mathrm{3}\gamma\right)=\mathrm{3sin}\left(\alpha+\beta+\gamma\right) \\ $$$$\left(\mathrm{C}\right)\:\mathrm{cos}\left(\mathrm{3}\alpha\right)+\mathrm{cos}\left(\mathrm{3}\beta\right)+\mathrm{cos}\left(\mathrm{3}\gamma\right)=\mathrm{3cos}\left(\alpha+\beta+\gamma\right) \\ $$$$\left(\mathrm{D}\right)\:\mathrm{sin}\left(\mathrm{2}\alpha\right)+\mathrm{sin}\left(\mathrm{2}\beta\right)+\mathrm{sin}\left(\mathrm{2}\gamma\right)=\mathrm{0} \\ $$ Terms of Service Privacy Policy…
Question Number 74573 by Raxreedoroid last updated on 26/Nov/19 $$\mathrm{Find}\:\left(\mathrm{turn}\:\mathrm{it}\:\mathrm{into}\:\mathrm{non}-\mathrm{segma}\:\mathrm{expression}\right) \\ $$$$\mathrm{1}+\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} +\mathrm{3}}{\mathrm{2}} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 74570 by Learner-123 last updated on 26/Nov/19 $${lim}_{{n}\rightarrow\infty} \frac{\mathrm{1}}{\left({n}\right)^{\frac{\mathrm{1}}{{n}}} }\:=\:? \\ $$ Answered by mind is power last updated on 26/Nov/19 $$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}n}^{−\frac{\mathrm{1}}{\mathrm{n}}}…
Question Number 140104 by EnterUsername last updated on 04/May/21 $$\mathrm{Let}\:{z}=\mathrm{1}+\mathrm{cos}\left(\mathrm{10}\pi/\mathrm{9}\right)+{i}\mathrm{sin}\left(\mathrm{10}\pi/\mathrm{9}\right).\:\mathrm{Then} \\ $$$$\left(\mathrm{A}\right)\:\mid{z}\mid=\mathrm{2cos}\left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{arg}\:{z}=\frac{\mathrm{8}\pi}{\mathrm{9}} \\ $$$$\left(\mathrm{C}\right)\:\mid{z}\mid=\mathrm{2cos}\left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{arg}\:{z}=\frac{\mathrm{5}\pi}{\mathrm{9}} \\ $$ Commented by EnterUsername last updated on 04/May/21 $$\mathrm{One}\:\mathrm{or}\:\mathrm{more}\:\mathrm{answers}\:\mathrm{may}\:\mathrm{be}\:\mathrm{correct}. \\…
Question Number 140107 by EnterUsername last updated on 04/May/21 $$\mathrm{If}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:\mathrm{and}\:{z}_{\mathrm{3}} \:\mathrm{are}\:\mathrm{the}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{counterclock}\:\mathrm{sense}\:\mathrm{and}\:{w}\neq\mathrm{1}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}\:\mathrm{root} \\ $$$$\mathrm{of}\:\mathrm{unity},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:{z}_{\mathrm{1}} −{z}_{\mathrm{3}} =\left({z}_{\mathrm{3}} −{z}_{\mathrm{2}} \right){w} \\ $$$$\left(\mathrm{B}\right)\:{z}_{\mathrm{1}}…