Question Number 201707 by ajfour last updated on 10/Dec/23 Answered by mr W last updated on 11/Dec/23 Commented by mr W last updated on 11/Dec/23…
Question Number 201702 by hardmath last updated on 10/Dec/23 $$\mathrm{Find}: \\ $$$$\int_{\mathrm{1}} ^{\:\mathrm{3}} \:\mathrm{dx}\:\int_{\boldsymbol{\mathrm{x}}} ^{\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} } \:\left(\mathrm{x}\:−\:\mathrm{y}\right)\:\mathrm{dy} \\ $$ Commented by mr W last updated…
Question Number 201693 by naka3546 last updated on 10/Dec/23 Answered by mr W last updated on 10/Dec/23 Commented by mr W last updated on 10/Dec/23…
Question Number 201689 by cherokeesay last updated on 10/Dec/23 Answered by witcher3 last updated on 10/Dec/23 $$\mathrm{x}^{\mathrm{3}} −\mathrm{6x}^{\mathrm{2}} +\mathrm{12x}−\mathrm{32}=\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{24} \\ $$$$\mathrm{x}−\mathrm{2}=\mathrm{y} \\ $$$$\Leftrightarrow\sqrt[{\mathrm{3}}]{\mathrm{y}+\mathrm{24}}=\mathrm{y}^{\mathrm{3}} −\mathrm{24}…
Question Number 201680 by cortano12 last updated on 10/Dec/23 $$\:\:\:\Subset \\ $$ Answered by Calculusboy last updated on 11/Dec/23 $$\boldsymbol{{Solution}}:\:\boldsymbol{{substitute}}\:\boldsymbol{{ditectly}},\:\boldsymbol{{we}}\:\boldsymbol{{get}}\:\frac{\mathrm{0}}{\mathrm{0}}\left(\boldsymbol{{indeterminant}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{p}}=\mathrm{2}\boldsymbol{{sinx}}−\boldsymbol{{sim}}\mathrm{2}\boldsymbol{{x}}\:\:\:\:\:\frac{\boldsymbol{{dp}}}{\boldsymbol{{dx}}}=\mathrm{2}\boldsymbol{{cosx}}−\mathrm{2}\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}} \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{q}}=\boldsymbol{{sinx}}−\boldsymbol{{xcosx}}\:\:\:\frac{\boldsymbol{{dq}}}{\boldsymbol{{dx}}}=\boldsymbol{{cosx}}−\left(\boldsymbol{{cosx}}−\boldsymbol{{xsinx}}\right) \\…
Question Number 201681 by jabarsing last updated on 10/Dec/23 $${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=\mathrm{3}{f}\left({x}\right)×{f}\left({x}+\mathrm{1}\right) \\ $$$${D}_{{f}} ={N} \\ $$$$\mathrm{2023}×{f}\left(\mathrm{1402}\right)=\mathrm{1} \\ $$$${have}\:{equation}\:{f}\left({x}\right)=\mathrm{1}\:{solution}? \\ $$ Answered by Frix last updated on…
Question Number 201683 by aurpeyz last updated on 10/Dec/23 $${Starting}\:{from}\:{substituting}\:{z}={x}+{iy}.\:{Identify} \\ $$$${the}\:{maximal}\:{region}\:{within}\:{which}\:{f}\left({z}\right)\:{is}\:{analytic} \\ $$$${f}\left({z}\right)=\frac{\mathrm{1}}{{z}\left({z}+\mathrm{1}\right)}.\: \\ $$$$ \\ $$$${Note}.\:{Do}\:{not}\:{start}\:{by}\:{just}\:{differentiating}\:{f}\left({z}\right).\: \\ $$$${Start}\:{by}\:\:{doing}\:{a}\:{substitution}\:{of}\:{x}\:{and}\:{iy}\:{and}\: \\ $$$${then}\:{verify}\:{Cauchy}\:{Rieman}\:{theorem}. \\ $$$$ \\…
Question Number 201679 by cherokeesay last updated on 10/Dec/23 Answered by Rasheed.Sindhi last updated on 10/Dec/23 $$\sqrt[{\mathrm{6}}]{\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\:+\sqrt[{\mathrm{6}}]{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:−\mathrm{1}}\:=\mathrm{1} \\ $$$${a}+{b}=\mathrm{1}\Rightarrow{b}=\mathrm{1}−{a} \\ $$$${a}^{\mathrm{6}} +{b}^{\mathrm{6}} =\mathrm{1}−\sqrt{{x}^{\mathrm{2}}…
Question Number 201629 by ali009 last updated on 09/Dec/23 Commented by ali009 last updated on 09/Dec/23 $${how}\:{is}\:{that}\:{cslculated}? \\ $$ Commented by aleks041103 last updated on…
Question Number 201631 by professorleiciano last updated on 09/Dec/23 Answered by mr W last updated on 10/Dec/23 $${totally}\:{number}\:{of}\:{words}:\:\mathrm{6}!=\mathrm{720} \\ $$$$ \\ $$$${number}\:{of}\:{words}\:{in}\:{which}\:\mathrm{3}\:{vowels} \\ $$$${are}\:{together}:\:\mathrm{4}!\mathrm{3}!=\mathrm{144} \\…