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Question-76167

Question Number 76167 by Master last updated on 24/Dec/19 Answered by behi83417@gmail.com last updated on 24/Dec/19 $$\Rightarrow\begin{cases}{\mathrm{3x}−\mathrm{x}^{\mathrm{2}} −\mathrm{3y}+\mathrm{xy}=\mathrm{1}−\mathrm{3x}−\mathrm{xy}+\mathrm{3x}^{\mathrm{2}} \mathrm{y}}\\{\mathrm{2x}−\mathrm{x}^{\mathrm{2}} +\mathrm{2y}−\mathrm{xy}=\mathrm{1}−\mathrm{2x}+\mathrm{xy}−\mathrm{2x}^{\mathrm{2}} \mathrm{y}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}−\mathrm{5y}+\mathrm{2xy}=−\mathrm{x}−\mathrm{2xy}+\mathrm{5x}^{\mathrm{2}} \mathrm{y}}\\{\mathrm{5x}−\mathrm{2x}^{\mathrm{2}} −\mathrm{y}=\mathrm{2}−\mathrm{5x}+\mathrm{x}^{\mathrm{2}}…

If-lim-x-1-x-1-k-1-x-1-L-0-Find-lim-x-0-x-1-1-x-1-1-k-1-

Question Number 141699 by cesarL last updated on 22/May/21 $${If}\:{lim}_{{x}\rightarrow\mathrm{1}} =\frac{\sqrt[{{k}}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}={L}\neq\mathrm{0}\:\:{Find}\:{lim}_{{x}\rightarrow\mathrm{0}} \frac{\sqrt{{x}+\mathrm{1}}−\mathrm{1}}{\:\sqrt[{{k}}]{{x}+\mathrm{1}}−\mathrm{1}} \\ $$ Answered by iloveisrael last updated on 22/May/21 $${Given}\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\:\frac{\sqrt[{{k}\:}]{{x}}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{L}\:{equivalent} \\ $$$${to}\:\underset{{t}\rightarrow\mathrm{1}}…

Daniel-and-Bruno-are-playing-with-perfect-cube-Daniel-is-the-first-player-if-he-obtains-1-or-2-he-wins-the-game-and-the-party-stopping-or-else-Bruno-plays-and-if-he-have-3-4-6-Bruno-won-and-the-game

Question Number 141692 by ArielVyny last updated on 22/May/21 $${Daniel}\:{and}\:{Bruno}\:{are}\:{playing}\:{with}\:{perfect}\:{cube} \\ $$$${Daniel}\:{is}\:{the}\:{first}\:{player}\:{if}\:{he}\:{obtains}\:\mathrm{1}\:{or}\:\mathrm{2} \\ $$$${he}\:{wins}\:{the}\:{game}\:{and}\:{the}\:{party}\:{stopping} \\ $$$${or}\:{else}\:{Bruno}\:{plays}\:{and}\:{if}\:{he}\:{have}\:\left\{\mathrm{3}.\mathrm{4}.\mathrm{6}\right\}\:{Bruno}\:{won}\:{and}\:{the}\:{game}\:{stopping} \\ $$$${Determine}\:{the}\:{probability}\:{that}\:{Daniel}\:{winand}\:{the}\:{probability}\:{that}\:{Bruno}\:{win} \\ $$$$ \\ $$ Answered by MJS_new…

Question-10621

Question Number 10621 by ketto last updated on 20/Feb/17 Answered by mrW1 last updated on 20/Feb/17 $${x}\:{boys}\:{and}\:{y}\:{girls} \\ $$$${x}−{y}=\mathrm{10}\:\:\:\:\:\left({i}\right) \\ $$$${x}=\mathrm{2}\left({y}+\mathrm{1}\right)\:{or} \\ $$$${x}−\mathrm{2}{y}=\mathrm{2}\:\:\:\:\:\left({ii}\right) \\ $$$$…

log-5-1-10-9e-2-2-1-2-9-2-10-2-3-3-1-3-9-3-10-3-4-4-1-4-9-4-10-4-Euler-Mascheroni-Constant-

Question Number 141694 by Dwaipayan Shikari last updated on 22/May/21 $${log}\left(\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{10}}\mathrm{9}{e}^{\gamma} \right)=\frac{\zeta\left(\mathrm{2}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} }\right)−\frac{\zeta\left(\mathrm{3}\right)}{\mathrm{3}}\:\left(\frac{\mathrm{1}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} }{\mathrm{10}^{\mathrm{3}} }\:\right)+\frac{\zeta\left(\mathrm{4}\right)}{\mathrm{4}}\left(\frac{\mathrm{1}^{\mathrm{4}} +\mathrm{9}^{\mathrm{4}} }{\mathrm{10}^{\mathrm{4}} }\right)−… \\ $$$$\gamma={Euler}\:{Mascheroni}\:{Constant} \\ $$…

Question-10620

Question Number 10620 by Saham last updated on 20/Feb/17 Commented by mrW1 last updated on 20/Feb/17 $${please}\:{check}: \\ $$$${after}\:{collision}\:\overset{\rightarrow} {{V}}_{{A}} =−\mathrm{5}.\mathrm{0}{i}+\mathrm{20}{j}\:\:\:\left({not}\:−\mathrm{5}.\mathrm{0}{j}+\mathrm{20}{j}\right) \\ $$$${answer}\:\left({b}\right)\:\mathrm{500}\:{J}\:\:\:\left({not}\:\mathrm{525}\:{J}\right) \\ $$…

Calculus-I-1-2-1-1-x-2-1-x-4-3-4-dx-

Question Number 141691 by mnjuly1970 last updated on 22/May/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:….{Calculus}\left({I}\right)…. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}:=\int_{\frac{\mathrm{1}}{\mathrm{2}\:}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{4}} \right)^{\frac{\mathrm{3}}{\mathrm{4}}} }{dx}=??? \\ $$ Answered by Dwaipayan Shikari last updated…