Question Number 131974 by Salman_Abir last updated on 10/Feb/21 Answered by liberty last updated on 10/Feb/21 $$\mathrm{vol}\:=\frac{\mathrm{1}}{\mathrm{3}}\pi\mathrm{h}\left(\mathrm{R}^{\mathrm{2}} +\mathrm{Rr}+\mathrm{r}^{\mathrm{2}} \right) \\ $$$$ \\ $$ Answered by…
Question Number 131969 by mnjuly1970 last updated on 10/Feb/21 $$\:\:\:\:\:\:\:\:\:\:\:\:…\:{math}\:\:{analysis}… \\ $$$$\:\:\:\:\phi=\:\:\int_{−\infty} ^{\:+\infty} \frac{{xsin}\left({x}\right)}{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}{dx}=? \\ $$$$\:\:\:\:\:\phi=\int_{−\infty} ^{\:+\infty} \frac{{xsin}\left({x}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$$\:\:\:\:\:\:\:\overset{{x}+\mathrm{1}={t}} {=}\int_{−\infty} ^{\:+\infty} \frac{\left({t}−\mathrm{1}\right){sin}\left({t}−\mathrm{1}\right)}{{t}^{\mathrm{2}}…
Question Number 66434 by iklima_0412 last updated on 15/Aug/19 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{x}}{\mathrm{1}−\mathrm{2x}} \\ $$ Commented by mathmax by abdo last updated on 15/Aug/19 $${let}\:{f}\left({x}\right)=\frac{{cos}^{\mathrm{2}} {x}−{x}}{\mathrm{1}−\mathrm{2}{x}}\:\Rightarrow\:{for}\:{x}\:\neq\mathrm{0}\:\:{we}\:{have}\:{f}\left({x}\right)=\frac{{x}−{cos}^{\mathrm{2}}…
Question Number 66433 by yyuuuuuuuu last updated on 15/Aug/19 $$\mathrm{A}\:\mathrm{2kg}\:\mathrm{is}\:\mathrm{attached}\:\mathrm{to}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{a}\:\mathrm{vertical}\:\mathrm{wire}\:\mathrm{of}\:\mathrm{length}\:\mathrm{2m}\:\mathrm{with}\:\mathrm{a}\:\mathrm{diameter}\:\mathrm{of}\:\mathrm{0}.\mathrm{64mm}\:\mathrm{and}\:\mathrm{having}\:\mathrm{an}\:\mathrm{extension}\:\mathrm{of}\:\mathrm{0}.\mathrm{60m}.\mathrm{Calculate}\:\mathrm{the}\:\mathrm{tensile}\:\mathrm{strain}\:\mathrm{on}\:\mathrm{the}\:\mathrm{wire}\left(\mathrm{take}\:\mathrm{g}=\mathrm{9}.\mathrm{8}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 896 by 112358 last updated on 15/Apr/15 $${Given}\:{that}\:{the}\:{velocity}\:{v}\:{of}\:{a}\:{body} \\ $$$${t}\:{seconds}\:{after}\:{passing}\:{a}\:{point}\:{O} \\ $$$${is}\:{found}\:{by} \\ $$$$\:\:\:\:\:\:\:\:{v}^{\mathrm{2}} =\frac{\mathrm{1}}{{k}}\left[{P}−\left({P}−{kv}_{{o}} ^{\mathrm{2}} \right){e}^{−\frac{\mathrm{2}{kt}}{{m}}} \right] \\ $$$${determine}\:{the}\:{distance}\:{covered} \\ $$$${by}\:{the}\:{body}\:\:{one}\:{hour}\:{after}\: \\…
Question Number 131965 by liberty last updated on 10/Feb/21 Commented by liberty last updated on 10/Feb/21 Commented by TheSupreme last updated on 10/Feb/21 $${T}={t}_{\mathrm{1}} +{t}_{\mathrm{2}}…
Question Number 66431 by hmamarques1994@gmail.com last updated on 15/Aug/19 $$\: \\ $$$$\:\boldsymbol{\mathrm{Determine}}\:\:\boldsymbol{\mathrm{x}}\:\:\boldsymbol{\mathrm{e}}\:\:\boldsymbol{\mathrm{y}}: \\ $$$$\: \\ $$$$\:\begin{cases}{\boldsymbol{\mathrm{x}}^{\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{i}}}}} +\:\frac{\mathrm{1}}{\boldsymbol{\mathrm{y}}^{\boldsymbol{\mathrm{i}}\sqrt{\boldsymbol{\mathrm{i}}}} }\:=\:\mathrm{10}}\\{\frac{\mathrm{1}}{\left(\boldsymbol{\mathrm{xy}}\right)^{\boldsymbol{\mathrm{i}}\sqrt{\boldsymbol{\mathrm{i}}}} }\:=\:\mathrm{21}}\end{cases} \\ $$$$\: \\ $$ Answered by…
Question Number 893 by Karting7442 last updated on 14/Apr/15 $$\:\:\frac{−\mathrm{12}+{x}}{\mathrm{11}}=−\mathrm{3}\:\:\:\:\:\:\:\:{check} \\ $$$$ \\ $$$$\:\:\:{what}\:{is}\:{x}?\:{also}\:{check}\:{it}. \\ $$ Answered by prakash jain last updated on 14/Apr/15 $$−\mathrm{12}+{x}=−\mathrm{33}…
Question Number 892 by Karting7442 last updated on 14/Apr/15 $$\frac{−\mathrm{5}+{x}}{\mathrm{22}}=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:{check} \\ $$$$\:\: \\ $$$$\:\:\:{what}\:{is}\:{x}?{also}\:{check}\:{it}. \\ $$ Answered by prakash jain last updated on 14/Apr/15 $$\frac{−\mathrm{5}+{x}}{\mathrm{22}}×\mathrm{22}=−\mathrm{1}×\mathrm{22}…
Question Number 131961 by Zack_ last updated on 10/Feb/21 $$\int\left({sin}^{\mathrm{4}} {x}.{cos}^{\mathrm{4}} {x}\right){dx} \\ $$ Answered by liberty last updated on 10/Feb/21 $$\mathrm{I}=\int\:\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} \mathrm{dx} \\…