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proof-that-the-equation-of-a-ellipse-at-a-center-0-0-is-x-2-a-2-y-2-b-2-1-

Question Number 148270 by abdurehime last updated on 26/Jul/21 $$\mathrm{proof}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{ellipse}\:\mathrm{at} \\ $$$$\mathrm{a}\:\mathrm{center}\left(\mathrm{0}.\mathrm{0}\right)\mathrm{is}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1} \\ $$ Commented by abdurehime last updated on 26/Jul/21…

f-x-x-2-x-1-x-1-where-x-1-find-the-range-of-the-function-

Question Number 148241 by 7770 last updated on 26/Jul/21 $$\:{f}:{x}\rightarrow\frac{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}−\mathrm{1}}\:{where}\:{x}\neq\mathrm{1} \\ $$$$\:{find}\:{the}\:{range}\:{of}\:{the}\:{function} \\ $$ Answered by Olaf_Thorendsen last updated on 26/Jul/21 $${f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{x}+\mathrm{2}+\frac{\mathrm{1}}{{x}−\mathrm{1}} \\…

Question-148221

Question Number 148221 by 0731619 last updated on 26/Jul/21 Answered by Olaf_Thorendsen last updated on 26/Jul/21 $$\mathrm{N}\left({x}\right)\:=\:\mathrm{tan}^{\mathrm{2}} \left(\mathrm{tan}{x}\right)−\mathrm{tan}^{\mathrm{2}} {x} \\ $$$$\mathrm{N}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:\mathrm{tan}^{\mathrm{2}} \left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{15}}\right)−\left({x}+\frac{{x}^{\mathrm{3}}…

find-the-residue-of-f-z-sin-z-cos-z-3-1-

Question Number 148174 by tabata last updated on 25/Jul/21 $${find}\:{the}\:{residue}\:{of}\:\:{f}\left({z}\right)=\frac{{sin}\left({z}\right)}{{cos}\left({z}^{\mathrm{3}} \right)−\mathrm{1}} \\ $$ Answered by mathmax by abdo last updated on 25/Jul/21 $$\mathrm{cosu}\sim\mathrm{1}−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{cos}\left(\mathrm{z}^{\mathrm{3}} \right)\sim\mathrm{1}−\frac{\mathrm{z}^{\mathrm{6}}…

Question-148157

Question Number 148157 by aliibrahim1 last updated on 25/Jul/21 Answered by puissant last updated on 26/Jul/21 $$\mathrm{pgcd}\left(\mathrm{a};\mathrm{b}\right)=\mathrm{pgcd}\left(\mathrm{a}−\mathrm{b};\mathrm{b}\right) \\ $$$$\Rightarrow\:\mathrm{pgcd}\left(\mathrm{2}^{\mathrm{a}} −\mathrm{1};\mathrm{2}^{\mathrm{b}} −\mathrm{1}\right)=\mathrm{pgcd}\left(\mathrm{2}^{\mathrm{a}} −\mathrm{2}^{\mathrm{b}} ;\mathrm{2}^{\mathrm{b}} −\mathrm{1}\right) \\…