Question Number 198419 by Nimnim111118 last updated on 19/Oct/23 $$\:\:{Please}\:{Help}… \\ $$$$\:\:\int\underset{{S}} {\int}{x}^{\mathrm{2}} {dydz}+{y}^{\mathrm{2}} {dzdx}+\mathrm{2}{z}\left({xy}−{x}−{y}\right){dxdy}\:{where} \\ $$$$\:\:\:{S}\:{is}\:{the}\:{surface}\:{of}\:{the}\:{cube}.\:\mathrm{0}\leqslant{x}\leqslant\mathrm{1},\:\mathrm{0}\leqslant{y}\leqslant\mathrm{1}, \\ $$$$\:\:\:\:\mathrm{0}\leqslant{z}\leqslant\mathrm{1} \\ $$$$ \\ $$ Terms of…
Question Number 198413 by mokys last updated on 19/Oct/23 Commented by mokys last updated on 19/Oct/23 $${find}\:{log}_{\mathrm{12}} \mathrm{60} \\ $$ Answered by cortano12 last updated…
Question Number 198414 by ajfour last updated on 19/Oct/23 $$\frac{\mathrm{3}}{\mathrm{8}}\left(\mathrm{3}{h}−{p}\right)^{\mathrm{2}} +\mathrm{3}{ph}=\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${and} \\ $$$$\frac{\left(\mathrm{3}{h}−{p}\right)^{\mathrm{3}} }{\mathrm{16}}+{p}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{1}\right)={h}^{\mathrm{3}} −{h}−{c} \\ $$$${Find}\:\:{p}\:{and}\:{h}\:\:{in}\:{terms}\:{of}\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\centerdot \\ $$ Terms of…
Question Number 198400 by cortano12 last updated on 19/Oct/23 $$\:\:\mathrm{20}^{\mathrm{11}} −\mathrm{1}\:=\:…\left(\mathrm{mod}\:\mathrm{1000}\right) \\ $$ Answered by MM42 last updated on 19/Oct/23 $$\mathrm{20}^{\mathrm{11}} −\mathrm{1}\overset{\mathrm{1000}} {\equiv}−\mathrm{1}\overset{\mathrm{1000}} {\equiv}\mathrm{999} \\…
Question Number 198403 by Mingma last updated on 19/Oct/23 Answered by MM42 last updated on 19/Oct/23 $$\sqrt{{x}+\mathrm{2}}={u}\Rightarrow{x}={u}^{\mathrm{2}} −\mathrm{2}\Rightarrow{dx}=\mathrm{2}{udu} \\ $$$$\Rightarrow\int\:\frac{\mathrm{2}{udu}}{{u}^{\mathrm{2}} −{u}−\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{3}}\int\left(\frac{\mathrm{2}}{{u}−\mathrm{2}}+\frac{\mathrm{1}}{{u}+\mathrm{1}}\right){du} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{ln}\left({u}−\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left({u}+\mathrm{1}\right)+{c} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}{ln}\left(\sqrt{{x}+\mathrm{2}}−\mathrm{2}\right)+\frac{\mathrm{2}}{\mathrm{3}}{ln}\left(\sqrt{{x}+\mathrm{2}}+\mathrm{1}\right)+{c}\:\:\checkmark…
Question Number 198367 by universe last updated on 18/Oct/23 $$\:\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{{x}^{\mathrm{2}} −\left({b}+\mathrm{1}\right){x}+{b}}{{x}^{\mathrm{2}} −\left({a}+\mathrm{1}\right){x}+{a}}\:\:\left({a}\neq{b}\:\&\:{a},{b}\:\in\:\mathbb{R}\:\sim\:\left\{\mathrm{1}\right\}\right) \\ $$$$\:\:\mathrm{can}\:\mathrm{take}\:\mathrm{all}\:\mathrm{values}\:\mathrm{except}\:\mathrm{two}\:\mathrm{values}\:\alpha\:\&\:\beta \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\alpha+\beta\:=\:\mathrm{0}\:\:\mathrm{then}\:\mid\frac{\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} −\mathrm{8}}{\mathrm{ab}}\mid\:\:=\:\:?? \\ $$ Commented by Frix last updated…
Question Number 198363 by BHOOPENDRA last updated on 18/Oct/23 Answered by mahdipoor last updated on 18/Oct/23 $${F}_{{A}} ={A}\left({cos}\mathrm{30}\boldsymbol{\mathrm{j}}+{sin}\mathrm{30}\boldsymbol{\mathrm{i}}\right) \\ $$$${F}_{{B}} ={B}_{{x}} \boldsymbol{\mathrm{i}}+{B}_{{y}} \boldsymbol{\mathrm{j}} \\ $$$$\Sigma{F}=\mathrm{0\begin{cases}{\boldsymbol{\mathrm{j}}\Rightarrow\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{A}−\frac{\mathrm{12}}{\mathrm{13}}\left(\mathrm{350}\right)+{B}_{{y}}…
Question Number 198357 by pticantor last updated on 18/Oct/23 $$\boldsymbol{{let}}\:\:\boldsymbol{{a}}\in\mathbb{R},\:\boldsymbol{{z}}\in\mathbb{C} \\ $$$$\boldsymbol{{resolve}}\: \\ $$$$\left(\boldsymbol{{z}}+\mathrm{1}\right)^{\boldsymbol{{n}}} =\boldsymbol{{e}}^{\boldsymbol{{i}}\pi\boldsymbol{{na}}} \\ $$$$\boldsymbol{{deduce}}\:\boldsymbol{{that}}\:\boldsymbol{{P}}_{\boldsymbol{{n}}} =\underset{{k}=\mathrm{0}} {\overset{\boldsymbol{{n}}−\mathrm{1}} {\prod}}\boldsymbol{{sin}}\left(\boldsymbol{{a}}+\frac{\boldsymbol{{k}}\pi}{\boldsymbol{{n}}}\right) \\ $$ Terms of Service…
Question Number 198352 by sonukgindia last updated on 18/Oct/23 Commented by Rasheed.Sindhi last updated on 18/Oct/23 $${Q}#\mathrm{198313} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 198385 by BHOOPENDRA last updated on 18/Oct/23 Terms of Service Privacy Policy Contact: info@tinkutara.com