Question Number 204550 by Rasheed.Sindhi last updated on 21/Feb/24 $${Where}'{s}\:{sir}\:{AST}? \\ $$$${Why}\:{his}\:{posts}\:{are}\:{now}\:{by} \\ $$$$'{deleteduser}\mathrm{1}'\:! \\ $$$${Did}\:{he}\:{delete}\:{his}\:{account}? \\ $$$$\mathcal{T}{here}'{s}\:{no}\:{post}\:{by}\:{name}\:\:'{AST}\:\:' \\ $$ Commented by A5T last updated…
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Question Number 204318 by MASANJAJJ last updated on 12/Feb/24 Answered by Rasheed.Sindhi last updated on 12/Feb/24 $$\bullet{T}\left({U}+{V}\right)={T}\left(\:\left(−\mathrm{12},\mathrm{12}\right)+\left(\mathrm{6},−\mathrm{16}\right)\:\right) \\ $$$$\:\:\:\:\:={T}\left(−\mathrm{6},−\mathrm{4}\right)=\left(−\mathrm{6}+\mathrm{8},−\mathrm{4}+\mathrm{7}\right)=\left(\mathrm{2},\mathrm{3}\right) \\ $$$$\: \\ $$$$\bullet{T}\left({U}\right)+{T}\left({V}\right)={T}\left(−\mathrm{12},\mathrm{12}\right)+{T}\left(\mathrm{6},−\mathrm{16}\right) \\ $$$$\:\:\:=\left(−\mathrm{12}+\mathrm{8},\mathrm{12}+\mathrm{7}\right)+\left(\mathrm{6}+\mathrm{8},−\mathrm{16}+\mathrm{7}\right)…
Question Number 204313 by serenity last updated on 12/Feb/24 $${lim}\frac{\mathrm{3}×^{\mathrm{2}} −\mathrm{8}×−\mathrm{16}}{\mathrm{2}×^{\mathrm{2}} \mathrm{9}×+\mathrm{4}} \\ $$$$ \\ $$ Answered by Faetmaaa last updated on 27/Feb/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}{x}^{\mathrm{2}}…
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Question Number 204250 by Noorzai last updated on 10/Feb/24 Answered by AST last updated on 10/Feb/24 $${tan}\left(\mathrm{9}\right)+{tan}\left(\mathrm{81}\right)−\left({tan}\mathrm{27}+{tan}\mathrm{63}\right) \\ $$$$=\frac{{sin}\mathrm{9}}{{cos}\mathrm{9}}+\frac{{sin}\mathrm{81}}{{cos}\mathrm{81}}−\left(\frac{{sin}\mathrm{27}}{{cos}\mathrm{27}}+\frac{{sin}\mathrm{63}}{{cos}\mathrm{63}}\right) \\ $$$$=\frac{\mathrm{2}{sin}\left(\mathrm{9}+\mathrm{81}\right)=\mathrm{2}}{\mathrm{2}{cos}\mathrm{9}{cos}\mathrm{81}}−\frac{\mathrm{2}{sin}\left(\mathrm{27}+\mathrm{63}\right)=\mathrm{2}}{\mathrm{2}{cos}\left(\mathrm{27}\right){cos}\left(\mathrm{63}\right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{9}\right){cos}\left(\mathrm{9}\right)}−\frac{\mathrm{2}}{\mathrm{2}{sin}\left(\mathrm{27}\right){cos}\left(\mathrm{27}\right)}=\frac{\mathrm{2}}{{sin}\left(\mathrm{18}\right)}−\frac{\mathrm{2}}{{sin}\left(\mathrm{54}\right)} \\ $$$${Let}\:\theta=\mathrm{18}°\Rightarrow{sin}\left(\mathrm{5}\theta\right)=\mathrm{16}{sin}^{\mathrm{5}}…
Question Number 204142 by Tawa11 last updated on 06/Feb/24 $$\mathrm{Solve}:\:\:\mathrm{x}^{\mathrm{x}} \:\:=\:\:\mathrm{27}^{\mathrm{x}\:\:−\:\:\mathrm{3}} \\ $$ Answered by Frix last updated on 07/Feb/24 $${x}^{{x}} =\mathrm{27}^{{x}−\mathrm{3}} \\ $$$${x}\mathrm{ln}\:{x}\:=\left({x}−\mathrm{3}\right)\mathrm{ln}\:\mathrm{27} \\…
Question Number 204024 by Tawa11 last updated on 04/Feb/24 Find the Cartesian equation of x(t) = 2 cos t And y(t) = 3 cos t…