Question Number 180449 by Mastermind last updated on 12/Nov/22 $$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{0} \\ $$$$\mathrm{2x}\:+\:\mathrm{4y}\:−\:\mathrm{z}\:=\:\mathrm{0} \\ $$$$\mathrm{3x}\:+\:\mathrm{2y}\:+\:\mathrm{2z}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x},\mathrm{y},\mathrm{and}\:\mathrm{z} \\ $$ Commented by mr W last…
Question Number 180446 by Mastermind last updated on 12/Nov/22 $$\mathrm{Solve}\::\: \\ $$$$\mathrm{x}_{\mathrm{1}} \:+\:\mathrm{2x}_{\mathrm{2}} \:−\:\mathrm{3x}_{\mathrm{3}} \:=\:\mathrm{0} \\ $$$$\mathrm{2x}_{\mathrm{1}} \:+\:\mathrm{4x}_{\mathrm{2}} \:−\:\mathrm{2x}_{\mathrm{3}} \:=\:\mathrm{2} \\ $$$$\mathrm{3x}_{\mathrm{1}} \:+\:\mathrm{6x}_{\mathrm{2}} \:−\:\mathrm{4x}_{\mathrm{3}} \:=\:\mathrm{3}…
Question Number 180447 by Mastermind last updated on 12/Nov/22 $$\mathrm{x}\:+\:\mathrm{y}\:−\:\mathrm{z}\:=\:\mathrm{0} \\ $$$$\mathrm{2x}\:−\:\mathrm{3y}\:+\:\mathrm{z}\:=\:\mathrm{0} \\ $$$$\mathrm{x}\:−\mathrm{4y}\:+\:\mathrm{2z}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x},\mathrm{y},\:\mathrm{and}\:\mathrm{z} \\ $$ Commented by mr W last…
Question Number 180438 by Mastermind last updated on 12/Nov/22 $$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{0}} {\mathrm{m}}\frac{\sqrt{\mathrm{4}+\mathrm{x}}−\sqrt{\mathrm{4}−\mathrm{x}}}{\mathrm{x}} \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{above} \\ $$ Answered by Frix last updated on 12/Nov/22 $$\mathrm{l}'\mathrm{H}\hat…
Question Number 180413 by Mastermind last updated on 12/Nov/22 $$\mathrm{li}\underset{\mathrm{h}\rightarrow\mathrm{0}} {\mathrm{m}}\left(\frac{\mathrm{e}^{\mathrm{h}} −\mathrm{1}}{\mathrm{h}}\right) \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{above} \\ $$ Answered by Frix last updated on 12/Nov/22…
Question Number 180414 by Mastermind last updated on 12/Nov/22 $$\mathrm{li}\underset{\mathrm{x}\rightarrow\mathrm{3}} {\mathrm{m}}\left(\frac{\mathrm{3}^{\mathrm{x}} −\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{x}} −\mathrm{3}^{\mathrm{3}} }\right) \\ $$$$ \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{limit}\:\mathrm{above} \\ $$ Answered by Frix last…
Question Number 49283 by Pk1167156@gmail.com last updated on 05/Dec/18 Answered by afachri last updated on 05/Dec/18 $$\:\:\mathrm{1}.\:\:\:{a}^{\mathrm{2}} +\:{b}^{\mathrm{2}} +\:{c}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\:\:\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{2}} \:−\:\mathrm{2}\left({ab}\:+\:{bc}\:+\:{ac}\right) \\ $$$$\:\:\:\:\:\:\:\mathrm{2}\left({ab}\:+\:{bc}\:+\:{ac}\right)\:\:\:\:\:\:=\:\:\:\mathrm{9}\:−\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\left(\boldsymbol{{ab}}\:+\:\boldsymbol{{bc}}\:+\:\boldsymbol{{ac}}\right)\:\:\:\:\:\:=\:\:\:\mathrm{0}…
Question Number 180301 by Noorzai last updated on 10/Nov/22 Commented by MJS_new last updated on 10/Nov/22 $$\mathrm{this}\:\mathrm{has}\:\mathrm{the}\:\mathrm{shape} \\ $$$${a}^{{x}} +\frac{\mathrm{1}}{{a}^{{x}} }={b}^{{x}} \\ $$$$\mathrm{which}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{for}\:{b}: \\ $$$${b}=\sqrt[{{x}}]{{a}^{{x}}…
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Question Number 180295 by Mastermind last updated on 10/Nov/22 $$\mathrm{Express}\:\mathrm{these}\:\mathrm{both}\:\mathrm{Cartesian}\:\mathrm{and}\: \\ $$$$\mathrm{polar}\:\mathrm{form} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{3z}^{\mathrm{2}} −\mathrm{2z}+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{f}\left(\mathrm{z}\right)=\mathrm{z}+\frac{\mathrm{1}}{\mathrm{z}} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thanks} \\ $$…