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1-1-z-1-z-2-

Question Number 215639 by Wuji last updated on 12/Jan/25 $$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{z}}\right)^{\left(\mathrm{1}+\mathrm{z}\right)} =\mathrm{2} \\ $$ Answered by mr W last updated on 12/Jan/25 $$\left(\mathrm{1}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{1}+{z}} =\mathrm{2}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{−\mathrm{1}} =\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{1}−\mathrm{2}} \\…

Question-215601

Question Number 215601 by SaidYousaf last updated on 11/Jan/25 Commented by mr W last updated on 11/Jan/25 $${please}\:{post}\:{only}\:{questions}, \\ $$$${comments}\:{or}\:{answers},\:{no}\:{photos} \\ $$$${which}\:{have}\:{nothing}\:{to}\:{do}\:{with}\:{the} \\ $$$${questions}!\:{thanks}! \\…

given-x-y-y-x-630-y-y-x-x-604-find-x-and-y-

Question Number 215564 by Wuji last updated on 10/Jan/25 $$\boldsymbol{\mathrm{given}};\:\:\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{y}}}+\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{x}}}=\mathrm{630} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{y}}\sqrt{\boldsymbol{\mathrm{y}}}+\boldsymbol{\mathrm{x}}\sqrt{\boldsymbol{\mathrm{x}}}=\mathrm{604} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}} \\ $$ Answered by Frix last updated on 10/Jan/25 $$\sqrt{{x}}={a}+{b}\mathrm{i}\wedge\sqrt{{y}}={a}−{b}\mathrm{i};\:{b}\geqslant\mathrm{0} \\…

Find-y-x-log-x-y-y-x-below-2x-2-xy-3y-2-0-

Question Number 215519 by walterpieuler last updated on 09/Jan/25 $$ \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{Find}}\:\left(\frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}\right)^{\boldsymbol{\mathrm{log}}_{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}} \frac{\boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{x}}}\:} \boldsymbol{\mathrm{below}}: \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{xy}}\:−\:\mathrm{3}\boldsymbol{\mathrm{y}}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$ \\…