Question Number 111611 by mathdave last updated on 04/Sep/20 $${solve}\:{for}\:{x}\:{ad}\:{y}\:{in} \\ $$$$\mathrm{5}{x}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{12}………\left(\mathrm{1}\right) \\ $$$$\mathrm{5}{y}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }\right)=\mathrm{4}……….\left(\mathrm{2}\right) \\ $$ Answered by Her_Majesty last updated…
Question Number 177133 by Matica last updated on 01/Oct/22 $$\:\:{f}\left({x}\right)\:=\:{x}^{\mathrm{2022}} \:.\:{Find}\:\:{derivative}\:{f}^{\left({n}\right)} . \\ $$ Answered by JDamian last updated on 01/Oct/22 $${f}^{\left({n}\right)} =\begin{cases}{\frac{\mathrm{2022}!}{\left(\mathrm{2022}−{n}\right)!}\:{x}^{\mathrm{2022}−{n}} \:\:\:\:\:\forall{n}\leqslant\mathrm{2022}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\forall{n}>\mathrm{2022}}\end{cases} \\…
Question Number 177134 by DAVONG last updated on 01/Oct/22 $$\mathrm{I}=\int\mathrm{ln}\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{2x}} \right)\mathrm{dx}=? \\ $$ Answered by Frix last updated on 02/Oct/22 $$\int\mathrm{ln}\:\left(\mathrm{1}+\mathrm{2e}^{−\mathrm{2}{x}} \right)\:{dx}\overset{{t}=−\mathrm{2e}^{−\mathrm{2}{x}} } {=} \\…
Question Number 177132 by mokys last updated on 01/Oct/22 Commented by mr W last updated on 01/Oct/22 $${instead}\:{of}\:{posting}\:{these}\:?????,\:{you} \\ $$$${can}\:{easily}\:{try}\:{by}\:{yourself}: \\ $$$${A}=\left(−\frac{\mathrm{1}}{\cancel{\mathrm{3}}}\right)\left(−\frac{\cancel{\mathrm{3}}}{\cancel{\mathrm{5}}}\right)\left(−\frac{\cancel{\mathrm{5}}}{\cancel{\mathrm{7}}}\right)…\left(−\frac{\cancel{\mathrm{4}{k}−\mathrm{1}}}{\mathrm{4}{k}+\mathrm{1}}\right) \\ $$$$=\left(−\mathrm{1}\right)^{\mathrm{2}{k}} \frac{\mathrm{1}}{\mathrm{4}{k}+\mathrm{1}}…
Question Number 177131 by mokys last updated on 01/Oct/22 Commented by Frix last updated on 02/Oct/22 $$\frac{\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}+…+\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)×\mathrm{2}{n}}}{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)×\mathrm{2}{n}}+\frac{\mathrm{1}}{\left({n}+\mathrm{2}\right)×\left(\mathrm{2}{n}−\mathrm{1}\right)}+…+\frac{\mathrm{1}}{\left({n}+{n}\right)×\left(\mathrm{2}{n}−\left({n}−\mathrm{1}\right)\right)}}=\frac{\mathrm{3}{n}+\mathrm{1}}{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 111581 by mathdave last updated on 04/Sep/20 $${the}\:{age}\:{of}\:{a}\:{mother}\:{and}\:{daughter}\:{arein} \\ $$$${the}\:{ratio}\:{of}\:\mathrm{8}:\mathrm{5}.{if}\:{the}\:{age}\:{now}\:{is} \\ $$$$\mathrm{15},{what}\:{was}\:{the}\:{age}\:{of}\:{the}\:{mother} \\ $$$${six}\:{years}\:{ago}. \\ $$ Commented by Her_Majesty last updated on 04/Sep/20…
Question Number 46042 by naka3546 last updated on 20/Oct/18 Commented by Tawa1 last updated on 20/Oct/18 $$\mathrm{Sir},\:\mathrm{can}\:\mathrm{you}\:\mathrm{share}\:\mathrm{me}\:\mathrm{the}\:\mathrm{link}\:\mathrm{to}\:\mathrm{download}\:\mathrm{this}\:\mathrm{pdf}\:\mathrm{or}\:\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{get}\:\mathrm{it}\:\mathrm{sir} \\ $$ Commented by Kunal12588 last updated on…
Question Number 177099 by mokys last updated on 30/Sep/22 Answered by TheHoneyCat last updated on 04/Oct/22 $$\mathrm{taking}\:\mathrm{the}\:\mathrm{log}\:\mathrm{of}\:\mathrm{your}\:\mathrm{product} \\ $$$$ \\ $$$$\mathrm{L}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{{n}} \left(\mathrm{ln}\left(\mathrm{1}+{x}^{{n}+\mathrm{1}} \right)−\mathrm{ln}\left(\mathrm{1}+{x}^{{n}}…
Question Number 46025 by Sanjarbek last updated on 20/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18 $${ln}_{{sinx}} {cosx}=\mathrm{4}−{x} \\ $$$$\left({sinx}\right)^{\mathrm{4}−{x}} ={cosx} \\ $$ Terms of…
Question Number 46020 by mondodotto@gmail.com last updated on 20/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 20/Oct/18 $$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{{x}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}} \\…