Question Number 47768 by Cesar Larez last updated on 14/Nov/18 $$\mathrm{Somebody}\:\mathrm{know}\:\mathrm{spanish}? \\ $$ Commented by Cesar Larez last updated on 14/Nov/18 $$\mathrm{have}\:\mathrm{u}\:\mathrm{whatsapp}? \\ $$ Commented…
Question Number 113278 by mathdave last updated on 12/Sep/20 $${solve} \\ $$$$\int\frac{\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{2}−\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{4}}}}{{x}\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} }}{dx} \\ $$ Answered by 1549442205PVT last updated on…
Question Number 113271 by mathdave last updated on 12/Sep/20 $${solve}\:{the}\:{initial}\:{boundary}\:{value} \\ $$$${problem}\:{of}\:{wave}\:{equation} \\ $$$$\frac{\partial^{\mathrm{2}} {u}\left({x},{t}\right)}{\partial{t}^{\mathrm{2}} }=\mathrm{9}\frac{\partial^{\mathrm{2}} {u}\left({x},{t}\right)}{\partial{x}^{\mathrm{2}} },\mathrm{0}<{x}<\mathrm{2},{t}>\mathrm{0} \\ $$$${u}\left(\mathrm{0},{t}\right)=\mathrm{1},{u}\left(\mathrm{2},{t}\right)=\mathrm{3},{t}>\mathrm{0} \\ $$$${u}\left({x},\mathrm{0}\right)=\mathrm{2},\mathrm{0}<{x}<\mathrm{2} \\ $$$$\frac{\partial{u}}{\partial{t}}\left({x},\mathrm{0}\right)=\mathrm{sin2}{x},\mathrm{0}<{x}<\mathrm{2} \\…
Question Number 113246 by mohammad17 last updated on 11/Sep/20 Commented by mohammad17 last updated on 11/Sep/20 $${please}\:{sir}\:{help}\:{me} \\ $$ Answered by Aziztisffola last updated on…
Question Number 113237 by A8;15: last updated on 11/Sep/20 Answered by MJS_new last updated on 11/Sep/20 $$\mathrm{I}\:\mathrm{solved}\:\mathrm{it}.\:\mathrm{question}\:\mathrm{113109} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 113219 by naka3546 last updated on 11/Sep/20 Answered by 1549442205PVT last updated on 11/Sep/20 Commented by Aina Samuel Temidayo last updated on 11/Sep/20…
Question Number 178725 by ComplexPrime last updated on 20/Oct/22 $$\mathrm{Let}\:{a}\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{number}.\:\mathrm{Then} \\ $$$$\sqrt{{a}}=\sqrt{\left(−{a}\right)\left(−\mathrm{1}\right)}=\sqrt{−{a}}\centerdot{i}=\sqrt{{a}\left(−\mathrm{1}\right)}\centerdot{i}=\sqrt{{a}}\centerdot{i}^{\mathrm{2}} =−\sqrt{{a}} \\ $$$$\mathrm{Where}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mistake}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 113191 by ZiYangLee last updated on 11/Sep/20 $$\mathrm{Solve}\:\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}+\mathrm{3cos}\frac{{x}}{\mathrm{2}}+\mathrm{1}=\mathrm{0} \\ $$ Answered by bobhans last updated on 11/Sep/20 $$\mathrm{2q}^{\mathrm{2}} +\mathrm{3q}+\mathrm{1}\:=\:\mathrm{0}\:,\:\mathrm{where}\:\mathrm{q}\:=\:\mathrm{cos}\:\left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$$$\left(\mathrm{2q}+\mathrm{1}\right)\left(\mathrm{q}+\mathrm{1}\right)=\mathrm{0} \\…
Question Number 47627 by hassentimol last updated on 12/Nov/18 $$\mathrm{Could}\:\mathrm{someone}\:\mathrm{explain}\:\mathrm{me}\:\mathrm{how}\:\mathrm{cellular} \\ $$$$\mathrm{automata}\:\mathrm{theory}\:\mathrm{can}\:\mathrm{be}\:\mathrm{used}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{bioligical}\:\mathrm{field}\:\mathrm{in}\:\mathrm{relation}\:\mathrm{to}\:\mathrm{the}\:\mathrm{spread} \\ $$$$\mathrm{of}\:\mathrm{disease}\:\mathrm{and}/\mathrm{or}\:\mathrm{cancer}\:\mathrm{cells}\:? \\ $$$${Thank}\:{you}\:{very}\:{much}\:! \\ $$ Terms of Service Privacy Policy…
Question Number 113160 by mohammad17 last updated on 11/Sep/20 Commented by abdomsup last updated on 11/Sep/20 $$\left.{not}\:{correct}\:\Gamma\:{is}\:{defined}\:{on}\right]\mathrm{0},\infty\left[\right. \\ $$ Commented by mohammad17 last updated on…