Question Number 61327 by olalekan2 last updated on 01/Jun/19 Answered by mr W last updated on 01/Jun/19 $$\mathrm{1}. \\ $$$$\mathrm{5}^{\mathrm{5}} =\mathrm{3125} \\ $$$$ \\ $$$$\mathrm{2}.…
Question Number 192396 by moh777 last updated on 16/May/23 Answered by mehdee42 last updated on 16/May/23 $$\mathrm{4}{x}−{x}^{\mathrm{2}} =\mathrm{3}\Rightarrow{x}=\mathrm{1},\mathrm{3} \\ $$$${v}_{\mathrm{1}} =\pi\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{4}{x}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} {dx}=\pi\int_{\mathrm{0}}…
Question Number 61326 by maxmathsup by imad last updated on 01/Jun/19 $${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{sinx}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ Commented by perlman last updated on 01/Jun/19 $${exacte}\:{value}\:!…
Question Number 192399 by mnjuly1970 last updated on 17/May/23 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Algebra}\:\left(\mathrm{1}\:\right) \\ $$$$\:\:{G},\:{is}\:{a}\:{group}\:\:{and}\:\:\:{o}\left({G}\:\right)\:=\:{p}^{\:\mathrm{2}} \:. \\ $$$$\:\:\:{prove}\:{that}\:{G}\:{is}\:{an}\:{abelian}\:{group}. \\ $$$$\:\:\:{hint}:\:\:\left(\:{p}\:{is}\:{prime}\:{number}\:\:\right) \\ $$$$\:\:\:\:\:−−−−−−−−−−−−− \\ $$ Commented by aleks041103 last…
Question Number 192398 by Mastermind last updated on 16/May/23 $$\mathrm{Let}\:\mathrm{f}:\mathbb{R}^{\mathrm{3}} \rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{define}\:\mathrm{by}\: \\ $$$$\mathrm{f}\left(\mathrm{x},\:\mathrm{y},\:\mathrm{z}\right)\:=\:\mathrm{2x}^{\mathrm{2}} −\mathrm{y}+\mathrm{6xy}−\mathrm{z}^{\mathrm{3}} +\mathrm{3z}. \\ $$$$\mathrm{calculate}\:\mathrm{the}\:\mathrm{directional}\:\mathrm{deriva}− \\ $$$$\mathrm{tive}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\:\mathrm{u}=\left(\mathrm{2},\:\mathrm{1},\:−\mathrm{3}\right) \\ $$$$ \\ $$$$\mathrm{help}! \\ $$…
Question Number 126859 by sdfg last updated on 24/Dec/20 Answered by mr W last updated on 24/Dec/20 $${G}=\mathrm{10}×\mathrm{sin}\:\mathrm{60}°×\mathrm{cos}\:\mathrm{45}°\:{i}+\mathrm{10}×\mathrm{sin}\:\mathrm{60}°×\mathrm{cos}\:\mathrm{45}°\:{j}+\mathrm{10}×\mathrm{cos}\:\mathrm{60}°\:{k} \\ $$$$\Rightarrow\boldsymbol{{G}}=\frac{\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\boldsymbol{{i}}+\frac{\mathrm{5}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\boldsymbol{{j}}+\mathrm{5}\:\boldsymbol{{k}} \\ $$$$ \\ $$$${H}=−\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{sin}\:\mathrm{30}°\:{i}+\mathrm{15}×\mathrm{sin}\:\mathrm{45}°×\mathrm{cos}\:\mathrm{30}°\:{j}+\mathrm{15}×\mathrm{cos}\:\mathrm{45}°\:{k} \\…
Question Number 61322 by Tawa1 last updated on 31/May/19 Answered by MJS last updated on 02/Jun/19 $$\mathrm{the}\:\mathrm{cuboid}\:\mathrm{with}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{volume}\:\mathrm{at}\:\mathrm{a} \\ $$$$\mathrm{given}\:\mathrm{surface}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube} \\ $$$$ \\ $$$${S}\mathrm{urface}=\mathrm{2}{a}^{\mathrm{2}} +\mathrm{4}{ab}=\mathrm{216}\:\Rightarrow\:{b}=\frac{\mathrm{54}}{{a}}−\frac{{a}}{\mathrm{2}} \\…
Question Number 61320 by behi83417@gmail.com last updated on 31/May/19 $$\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{x}}: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\boldsymbol{\mathrm{x}}=\mathrm{12} \\ $$ Answered by MJS last updated on 31/May/19 $$\mathrm{no}\:“\mathrm{beautiful}''\:\mathrm{solution} \\ $$$${x}_{\mathrm{1}}…
Question Number 126854 by sdfg last updated on 24/Dec/20 Answered by mathmax by abdo last updated on 24/Dec/20 $$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{k}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\mathrm{r}^{\mathrm{2}} =−\mathrm{k}^{\mathrm{2}} \:\Rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{ik}\:\Rightarrow\mathrm{y}_{\mathrm{h}} =\mathrm{ae}^{\mathrm{ikx}}…
Question Number 192388 by sudipmoi last updated on 16/May/23 Answered by aleks041103 last updated on 22/May/23 $${p}>\mathrm{0} \\ $$$$\int_{\mathrm{1}} ^{\:\infty} \frac{{sin}\left({x}\right){dx}}{{x}^{{p}} }=\left[−\frac{{cos}\left({x}\right)}{{x}^{{p}} }\right]_{\mathrm{1}} ^{\infty} −\int_{\mathrm{1}}…