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Category: Algebra

Question-150628

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) \\…

Find-center-and-radius-of-circle-having-equation-zz-1-i-z-1-i-z-1-0-

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}: \\…

Question-150601

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}}…

xyz-10-x-y-z-7-xy-xz-yz-2-Find-xy-z-xz-y-yz-x-

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-150597

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}}…

lim-x-0-x-x-

Question Number 150599 by mathdanisur last updated on 13/Aug/21 $$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{x}}} \:=\:? \\ $$ Commented by n0y0n last updated on 13/Aug/21 $$\mathrm{limit}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$$$ \\…

Prove-that-three-points-z-1-z-2-z-3-are-collinear-if-determinant-z-1-z-1-1-z-2-z-2-1-z-3-z-3-1-0-

Question Number 19507 by Tinkutara last updated on 12/Aug/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{three}\:\mathrm{points}\:{z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} ,\:{z}_{\mathrm{3}} \:\mathrm{are} \\ $$$$\mathrm{collinear}\:\mathrm{if}\:\begin{vmatrix}{{z}_{\mathrm{1}} }&{\bar {{z}}_{\mathrm{1}} }&{\mathrm{1}}\\{{z}_{\mathrm{2}} }&{\bar {{z}}_{\mathrm{2}} }&{\mathrm{1}}\\{{z}_{\mathrm{3}} }&{\bar {{z}}_{\mathrm{3}} }&{\mathrm{1}}\end{vmatrix}=\:\mathrm{0} \\…