Question Number 61605 by Mr X pcx last updated on 05/Jun/19 $${solve}\:\:{at}\:{Z}^{\mathrm{2}} \:\:\:\:\mathrm{2}{x}\:+\mathrm{5}{y}\:=\mathrm{4} \\ $$ Commented by mr W last updated on 05/Jun/19 $${x}=\mathrm{2}−\mathrm{5}{n} \\…
Question Number 127139 by MathSh last updated on 27/Dec/20 $$\mathrm{2}^{{y}} \:=\:{y}^{\mathrm{2}} \:\Rightarrow\:{y}\:=\:? \\ $$ Answered by Ar Brandon last updated on 27/Dec/20 $$\mathrm{2} \\ $$…
Question Number 192675 by ajfour last updated on 24/May/23 Answered by a.lgnaoui last updated on 26/May/23 Commented by ajfour last updated on 26/May/23 $${not}\:{legible} \\…
Question Number 61601 by Mr X pcx last updated on 05/Jun/19 $${calvulate}\:\int\int_{{w}} \left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right){e}^{−{x}−{y}} {dxdy} \\ $$$${with}\:{W}=\left\{\left({x},{y}\right)\in{R}^{\mathrm{2}} /\mathrm{0}\leqslant{x}\leqslant\mathrm{1}\:{and}\right. \\ $$$$\left.\mathrm{1}\leqslant{y}\leqslant\mathrm{3}\right\} \\ $$ Commented by…
Question Number 192668 by mathocean1 last updated on 24/May/23 $$\int_{\mathrm{0}} ^{\mathrm{4}} \frac{\mathrm{1}}{{x}}{e}^{{x}} {dx}\:−\:\int_{\mathrm{0}} ^{\mathrm{4}} \frac{\mathrm{1}}{{x}}{e}^{\frac{\mathrm{1}}{\mathrm{2}}{x}} {dx}=\:? \\ $$ Answered by witcher3 last updated on 24/May/23…
Question Number 127132 by mathocean1 last updated on 27/Dec/20 $${knowing}\:{that}\:\mathrm{7}^{\mathrm{3}{k}+\mathrm{1}} \equiv\mathrm{7}\left[{mod}\:\mathrm{9}\right]\:;{k}\:\in\:\mathbb{N}. \\ $$$${show}\:{that}\:\mathrm{2005}^{\mathrm{2005}} \equiv\mathrm{7}\left[{mod}\:\mathrm{9}\right] \\ $$ Answered by JDamian last updated on 27/Dec/20 $$\mathrm{2005}\equiv\mathrm{7}\:{mod}\:\mathrm{9} \\…
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Question Number 192661 by Mingma last updated on 24/May/23 Commented by Mingma last updated on 24/May/23 Prove that Commented by Mingma last updated on 25/May/23 Perfect �� …
Question Number 61591 by MJS last updated on 05/Jun/19 $$\mathrm{solve}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$$$\sqrt[{\mathrm{2}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{3}}]{{z}}=−\mathrm{1} \\ $$$$\sqrt[{\mathrm{4}}]{{z}}=−\mathrm{1} \\ $$ Commented by Smail last updated on 05/Jun/19…