Question Number 126835 by mohammad17 last updated on 24/Dec/20 Answered by liberty last updated on 25/Dec/20 $$\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:\:−\mathrm{sin}\:\frac{\pi}{\mathrm{6}}}\\{\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix}^{−\mathrm{1}} \begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{1}\begin{pmatrix}{\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{−\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\end{pmatrix}\:\begin{pmatrix}{{x}'}\\{{y}'}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\:{x}'+{y}'\right)}\\{\frac{\mathrm{1}}{\mathrm{2}}\left(−{x}'+\sqrt{\mathrm{3}}\:{y}'\right.}\end{pmatrix}…
Question Number 126827 by harckinwunmy last updated on 24/Dec/20 Answered by AST last updated on 24/Dec/20 $$\mathrm{Q33}. \\ $$$$\frac{\mathrm{4}+\mathrm{7}{i}}{\mathrm{1}−\mathrm{3}{i}} \\ $$$$\mathrm{Rationalising} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{4}+\mathrm{7}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}{\left(\mathrm{1}−\mathrm{3}{i}\right)\left(\mathrm{1}+\mathrm{3}{i}\right)}=\frac{\mathrm{4}+\mathrm{12}{i}+\mathrm{7}{i}+\left(\mathrm{21}{i}^{\mathrm{2}} \right)}{\mathrm{1}^{\mathrm{2}} −\left(\mathrm{3}{i}\right)^{\mathrm{2}}…
Question Number 61284 by naka3546 last updated on 31/May/19 $$\left(\mathrm{998}^{\mathrm{999}} \:×\:\mathrm{999}^{\mathrm{998}} \:×\:\mathrm{2019}^{\mathrm{2019}} \right)\:\:{mod}\:\left(\mathrm{1000}\right)\:\:=\:\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
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Question Number 192349 by 073 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${I}=\int\frac{{cos}^{\mathrm{4}} {x}\sqrt{\mathrm{1}+{sinx}}}{{cosx}}\:{dx}=\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\sqrt{\mathrm{1}+{sinx}}\:{cosxdx} \\ $$$${let}\::\:\mathrm{1}+{sinx}={u}^{\mathrm{2}} \:\Rightarrow{cosxdx}=\mathrm{2}{udu} \\ $$$$\Rightarrow{I}=\mathrm{2}\int\left(−{u}^{\mathrm{2}} +\mathrm{2}{u}\right){u}^{\mathrm{2}}…
Question Number 61283 by naka3546 last updated on 01/Jun/19 $$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \right)\:\:=\:\:\mathrm{2} \\ $$$$\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)\left({x}^{\mathrm{8}} \:+\:{y}^{\mathrm{8}} \right)\:\:=\:\:\mathrm{4} \\ $$$$\left({x}^{\mathrm{3}} +\:{y}^{\mathrm{3}}…
Question Number 192348 by leandrosriv02 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${M}\:\:{is}\:{midpoint}\:<{AB}>\Rightarrow{M}\left(\frac{{a}}{\mathrm{2}}\:,\:\frac{{b}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{0}{M}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow{OM}={BM}={AM}…
Question Number 61270 by necx1 last updated on 31/May/19 Answered by Askash last updated on 31/May/19 $$\mathrm{24} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 192338 by josemate19 last updated on 15/May/23 $${y}=\:\left(\frac{\left(\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}}\right)\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)}{\int_{\mathrm{0}} ^{\:{x}} {lnt}\:{dt}}\right) \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}? \\ $$ Answered…
Question Number 126758 by bounhome last updated on 24/Dec/20 $${solve}: \\ $$$$\mathrm{1}.{y}'^{\mathrm{3}} +{y}^{\mathrm{2}} −{y}'^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}.\mathrm{4}{y}={x}^{\mathrm{2}} +{y}'^{\mathrm{2}} \\ $$ Answered by liberty last updated…