Question Number 161102 by cortano last updated on 12/Dec/21 Commented by Abdulazizov last updated on 12/Dec/21 $$ \\ $$ Answered by mr W last updated…
Question Number 95560 by I want to learn more last updated on 26/May/20 Commented by i jagooll last updated on 26/May/20 $$\mathrm{y}=\left(\mathrm{x}^{\mathrm{2}} +\mathrm{5}\right)^{−\mathrm{1}} \: \\…
Question Number 161091 by HongKing last updated on 11/Dec/21 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\mathrm{1}\:-\:\mathrm{x}}\:=\:\mathrm{2x}^{\mathrm{2}} \:-\:\mathrm{1}\:-\:\mathrm{2x}\:\sqrt{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} } \\ $$$$ \\ $$ Answered by MJS_new last updated on 11/Dec/21…
Question Number 161075 by mnjuly1970 last updated on 11/Dec/21 $$ \\ $$$$\:\:\:\:{simplify} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{n}}{\left(\:{n}^{\:\mathrm{2}} −\frac{\:\mathrm{1}}{\mathrm{4}}\:\right)^{\:\mathrm{3}} }\:=\:? \\ $$$$ \\ $$ Terms…
Question Number 161068 by blackmamba last updated on 11/Dec/21 $$\:\:\:\:\:\int\:\frac{\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\sqrt{{x}^{\mathrm{4}} −\mathrm{1}}}\:{dx}\:=? \\ $$ Answered by chhaythean last updated on 11/Dec/21 $$=−\int\frac{\mathrm{2x}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{x}^{\mathrm{4}} −\mathrm{1}}}\mathrm{dx} \\…
Question Number 161065 by cortano last updated on 11/Dec/21 $$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}}\end{cases} \\ $$$$\:{find}\:\sqrt{{x}+{abc}}\:+\sqrt{{x}−{abc}} \\ $$ Answered by Rasheed.Sindhi last updated on 12/Dec/21 $$\:\begin{cases}{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:+\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{a}….\left({i}\right)}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{8}}]{{x}−{abc}}\:=\:{b}…..\left({ii}\right)}\\{\sqrt[{\mathrm{4}}]{{x}+{abc}}\:−\sqrt[{\mathrm{4}}]{{x}−{abc}}\:=\:{c}…..\left({iii}\right)}\end{cases} \\ $$$$\mathrm{2}\sqrt[{\mathrm{4}}]{{x}+{abc}}\:={a}+{b} \\…
Question Number 161059 by HongKing last updated on 11/Dec/21 $$\mathrm{Find}: \\ $$$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{x}\right)\mathrm{ln}\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{y}}\right)\mathrm{dxdy} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 161058 by HongKing last updated on 11/Dec/21 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\mathrm{x}\left(\mathrm{y}-\mathrm{1}\right)\mathrm{dx}\:+\:\left(\mathrm{x}+\mathrm{1}\right)\mathrm{dy}\:=\:\mathrm{0} \\ $$$$ \\ $$ Answered by mr W last updated on 11/Dec/21 $$\frac{{dy}}{{dx}}+\frac{{x}}{{x}+\mathrm{1}}\left({y}−\mathrm{1}\right)=\mathrm{0}…
Question Number 95502 by bobhans last updated on 25/May/20 $$\mathrm{6}\:\mathrm{man}\:+\:\mathrm{8}\:\mathrm{woman}\:\Rightarrow\mathrm{working}\:\mathrm{a}\:\mathrm{job}\:\mathrm{in}\:\mathrm{10}\:\mathrm{days} \\ $$$$\mathrm{26}\:\mathrm{man}\:+\:\mathrm{48}\:\mathrm{woman}\:\Rightarrow\:\mathrm{in}\:\mathrm{2}\:\mathrm{days} \\ $$$$\mathrm{if}\:\mathrm{15}\:\mathrm{man}\:+\:\mathrm{20}\:\mathrm{woman}\:\Rightarrow\:??\:\mathrm{days} \\ $$ Commented by bobhans last updated on 25/May/20 $$\mathrm{my}\:\mathrm{answer}\:\mathrm{4}\:\mathrm{days}.\:\mathrm{it}\:\mathrm{correct}? \\…
Question Number 161039 by HongKing last updated on 11/Dec/21 $$\mathrm{let}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}: \\ $$$$\left(\mathrm{1}\:+\:\mathrm{x}\right)\:\mathrm{y}^{''} \left(\mathrm{x}\right)\:+\:\left(\mathrm{1}\:-\:\mathrm{x}\right)\:\mathrm{y}^{'} \left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}-\mathrm{x}}{\mathrm{1}+\mathrm{x}}\:\mathrm{y}\left(\mathrm{x}\right) \\ $$$$\mathrm{y}\left(\mathrm{0}\right)\:=\:\mathrm{1}\:,\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\left(\mathrm{y}^{''} \left(\mathrm{x}\right)\:+\:\mathrm{y}^{'} \left(\mathrm{x}\right)\:+\:\mathrm{y}\left(\mathrm{x}\right)\right)\:\mathrm{e}^{-\boldsymbol{\mathrm{x}}}…