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Question Number 206227 by bett last updated on 09/Apr/24 $${OA}=\left(\overset{{x}} {\mathrm{4}}\right)\:{OB}=_{\mathrm{7}} ^{\mathrm{5}} \:{and}\:{AB}=\mathrm{5}\:{units} \\ $$ Answered by A5T last updated on 09/Apr/24 $$\sqrt{\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{5}\Rightarrow\mathrm{25}−\mathrm{16}=\mathrm{3}^{\mathrm{2}}…
Question Number 205820 by mustafazaheen last updated on 31/Mar/24 $$\begin{pmatrix}{\sqrt{\mathrm{3}}}\\{\mathrm{1}}\end{pmatrix}\:\:\mathrm{and}\:\:\begin{pmatrix}{\mathrm{1}}\\{\sqrt{\mathrm{3}}}\end{pmatrix}\:\:\:\mathrm{vector}\:\mathrm{find}\:\theta=? \\ $$ Answered by mr W last updated on 31/Mar/24 $$\mathrm{cos}\:\theta=\frac{\sqrt{\mathrm{3}}×\mathrm{1}+\mathrm{1}×\sqrt{\mathrm{3}}}{\:\mathrm{2}×\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\theta=\mathrm{30}° \\ $$…
Question Number 205321 by gopikrishnan last updated on 16/Mar/24 $$\overset{\rightarrow} {{a}}=\hat {{i}}+\mathrm{3}\hat {{j}}+\mathrm{4}\hat {{k}}\:\overset{\rightarrow} {{b}}=\mathrm{2}\hat {{i}}−\mathrm{3}\hat {{j}}+\mathrm{4}\hat {{k}}\:\overset{\rightarrow} {{c}}=\mathrm{5}\hat {{i}}−\mathrm{2}\hat {{j}}+\mathrm{4}\hat {{k}}\:{given}\:{that}\:\overset{\rightarrow} {{p}}×\overset{\rightarrow} {{b}}=\overset{\rightarrow} {{b}}×\overset{\rightarrow}…
Question Number 205164 by depressiveshrek last updated on 11/Mar/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{determinant}: \\ $$$$\begin{vmatrix}{\mathrm{1}−{x}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}−{x}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}−{x}}&{\ldots}&{{n}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}−{x}}\end{vmatrix} \\ $$ Answered by aleks041103 last updated on 12/Mar/24 $${By}\:{subtracting}\:{the}\:{first}\:{row}\:{from}\:{all}\:{other} \\ $$$$\begin{vmatrix}{\mathrm{1}−{x}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}−{x}}&{\mathrm{3}}&{\ldots}&{{n}}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}−{x}}&{\ldots}&{{n}}\\{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\ldots}&{{n}−{x}}\end{vmatrix}= \\…
Question Number 205156 by depressiveshrek last updated on 11/Mar/24 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{determinant}: \\ $$$$\begin{vmatrix}{\mathrm{5}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}&{\ldots}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{2}}&{\mathrm{5}}&{\mathrm{3}}&{\mathrm{0}}&{\ldots}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{5}}&{\mathrm{3}}&{\ldots}&{\mathrm{0}}&{\mathrm{0}}\\{\vdots}&{\vdots}&{\vdots}&{\vdots}&{\ddots}&{\vdots}&{\vdots}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\ldots}&{\mathrm{5}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\ldots}&{\mathrm{2}}&{\mathrm{5}}\end{vmatrix} \\ $$ Answered by pi314 last updated on 11/Mar/24 $$\Delta_{{n}} =\begin{vmatrix}{\mathrm{5}\:\mathrm{3}\:\:\:\mathrm{0}\:\mathrm{0}……\mathrm{0}\:\mathrm{0}}\\{\mathrm{2}\:\:\mathrm{5}\:\:\mathrm{3}\:\mathrm{0}……\mathrm{0}\:\mathrm{0}}\\{\mathrm{0}\:\:\mathrm{2}\:\:\mathrm{5}\:\mathrm{3}……\mathrm{0}\:\:\mathrm{0}}\\{………………\mathrm{5}\:\mathrm{3}}\\{\mathrm{0}\:\mathrm{0}\:\mathrm{0}\:\:\mathrm{0}…….\:\mathrm{2}\:\mathrm{5}}\end{vmatrix} \\ $$$$\Delta_{{n}}…
Question Number 204509 by JohnSmith last updated on 19/Feb/24 Commented by Rasheed.Sindhi last updated on 20/Feb/24 $$\mathcal{N}{ot}\:{clear}! \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 203419 by essaad last updated on 18/Jan/24 Commented by mr W last updated on 18/Jan/24 $${question}\:{is}\:{wrong}!\:{you}\:{can}\:{not} \\ $$$${determine}\:\alpha\:{uniquely}. \\ $$ Commented by essaad…
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Question Number 201526 by 281981 last updated on 08/Dec/23 Answered by AST last updated on 08/Dec/23 $${WLOG},{let}\:{O}\:{be}\:{the}\:{origin};\:{g}=\frac{{a}+{b}+{c}}{\mathrm{3}} \\ $$$$\mid{o}−{a}\mid=\mid{o}−{b}\mid=\mid{o}−{c}\mid={R} \\ $$$$\Rightarrow\left({o}−{a}\right)\left(\overset{−} {{o}}−\overset{−} {{a}}\right)={R}^{\mathrm{2}} \Rightarrow{a}\overset{−} {{a}}={R}^{\mathrm{2}}…