Question Number 85104 by jagoll last updated on 19/Mar/20 $$\mathrm{Given}\: \\ $$$$\begin{cases}{\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}−\mathrm{3x}\:=\:−\mathrm{1}}\\{\mathrm{4y}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{6y}\:=\:−\mathrm{1}}\end{cases} \\ $$$$\mathrm{find}\:\mathrm{2y}\:−\:\mathrm{x} \\ $$ Commented by john santu last updated on…
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Question Number 19557 by ajfour last updated on 12/Aug/17 Commented by ajfour last updated on 12/Aug/17 $$\mathrm{If}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{line}\:\mathrm{through}\:\:\mathrm{z}=\:\mathrm{z}_{\mathrm{0}} \\ $$$$\mathrm{and}\:\:\mathrm{z}=\mathrm{c}\:\:\:\mathrm{be}\:\:\bar {\mathrm{a}z}+\mathrm{a}\bar {\mathrm{z}}+\mathrm{k}=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{z}=\mathrm{a}\:\mathrm{and}\:\mathrm{scalar}\:\mathrm{k}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of} \\ $$$$\mathrm{z}_{\mathrm{0}}…
Question Number 150628 by aupo14 last updated on 14/Aug/21 Answered by som(math1967) last updated on 14/Aug/21 $$\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}.\boldsymbol{{x}}.\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{4}+\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{4}\boldsymbol{{y}} \\ $$$$\left(\boldsymbol{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{17}}{\mathrm{4}}−\mathrm{4}\boldsymbol{{y}}\right) \\ $$$$\left(\boldsymbol{{x}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =−\mathrm{4}\left(\boldsymbol{{y}}−\frac{\mathrm{17}}{\mathrm{16}}\right) \\…
Question Number 19555 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Find}\:\mathrm{center}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{having} \\ $$$$\mathrm{equation}\:{z}\bar {{z}}\:+\:\left(\mathrm{1}\:−\:{i}\right){z}\:+\:\left(\mathrm{1}\:+\:{i}\right)\bar {{z}}\:−\:\mathrm{1}\:=\:\mathrm{0}. \\ $$ Answered by ajfour last updated on 12/Aug/17 $$\mathrm{Equation}\:\mathrm{of}\:\mathrm{circle}: \\…
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Question Number 150601 by mathdanisur last updated on 13/Aug/21 Answered by Olaf_Thorendsen last updated on 13/Aug/21 $${u}_{\mathrm{1}} \:=\:\frac{\mathrm{sin}\:\mathrm{ln}{i}^{{i}} }{{i}}\:=\:\frac{\mathrm{sin}\left({i}\mathrm{ln}{i}\right)}{{i}}\:=\:\frac{\mathrm{sin}\left({i}\mathrm{ln}{e}^{{i}\frac{\pi}{\mathrm{2}}} \right)}{{i}} \\ $$$${u}_{\mathrm{1}} \:=\:\frac{\mathrm{sin}\left({i}\left({i}\frac{\pi}{\mathrm{2}}\right)\right)}{{i}}=\:\frac{−\mathrm{1}}{{i}}\:=\:{i} \\ $$$${u}_{\mathrm{2}}…
Question Number 150600 by mathdanisur last updated on 13/Aug/21 $$\mathrm{xyz}\:=\:\mathrm{10} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:-\:\mathrm{7} \\ $$$$\mathrm{xy}\:+\:\mathrm{xz}\:+\:\mathrm{yz}\:=\:\mathrm{2} \\ $$$$\mathrm{Find}\:\:\frac{\mathrm{xy}}{\mathrm{z}}\:+\:\frac{\mathrm{xz}}{\mathrm{y}}\:+\:\frac{\mathrm{yz}}{\mathrm{x}}\:=\:? \\ $$ Answered by amin96 last updated on 13/Aug/21…
Question Number 150597 by mathdanisur last updated on 13/Aug/21 Answered by Olaf_Thorendsen last updated on 13/Aug/21 $$\frac{\overline {{nnn}…{nn}}}{{n}+{n}+{n}…{n}}\:=\:\frac{{n}\underset{{p}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\sum}}\mathrm{10}^{{p}} }{{kn}}\:=\:\frac{\frac{\mathrm{1}−\mathrm{10}^{{k}} }{\mathrm{1}−\mathrm{10}}}{{k}} \\ $$$$=\:\frac{\mathrm{10}^{{k}} −\mathrm{1}}{\mathrm{9}{k}}…
Question Number 150603 by mathdanisur last updated on 13/Aug/21 $$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{x}}=\mathrm{sin}\left(\mathrm{165}°\right) \\ $$$$\boldsymbol{\mathrm{y}}=\mathrm{cos}\left(\mathrm{165}°\right) \\ $$$$\boldsymbol{\mathrm{z}}=\mathrm{tan}\left(\mathrm{165}°\right) \\ $$ Answered by ajfour last updated on 13/Aug/21…