Question Number 220263 by Tawa11 last updated on 10/May/25 Answered by mr W last updated on 10/May/25 $${blue}=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} ×{red} \\ $$$${pink}={green} \\ $$$${blue}+{pink}+{red}+{green}=\mathrm{24}.\mathrm{5} \\ $$$${blue}+{pink}=\frac{\mathrm{3}}{\mathrm{4}}\left({red}+{green}\right)…
Question Number 220257 by abbb last updated on 10/May/25 $${proof}\:{that}\:{volume}\:{of}\:{frustum}\:{of} \\ $$$$\:{circular}\:{cone}\:{is}\:\frac{\mathrm{1}}{\mathrm{3}}{h}\left[{A}\mathrm{1}+{A}\mathrm{2}+\sqrt{{A}\mathrm{1}{A}\mathrm{2}}\right. \\ $$$${A}_{\mathrm{1}} {and}\:{A}_{\mathrm{2}} \:{are}\:\:{areas}\:{of}\:{base} \\ $$ Answered by MrGaster last updated on 10/May/25…
Question Number 220232 by mnjuly1970 last updated on 09/May/25 $$ \\ $$$$\:\:\:\:{prove}\:{that} \\ $$$$ \\ $$$$\:\:\frac{\pi}{\mathrm{16}}\:<\:\int_{\mathrm{0}} ^{\:\mathrm{1}\:} \sqrt{\frac{{x}\left(\mathrm{1}−{x}\right)}{{sin}\left(\pi{x}\right)+{cos}\left(\pi{x}\right)+\mathrm{2}}}\:{dx}<\frac{\pi}{\mathrm{8}} \\ $$$$\:\:\:\:\: \\ $$ Answered by SdC355…
Question Number 220228 by SdC355 last updated on 09/May/25 $$\mathrm{1}-\mathrm{form} \\ $$$$\alpha={A}_{{x}} \mathrm{d}{x}+{A}_{{y}} \mathrm{d}{y}+{A}_{{z}} \mathrm{d}{z} \\ $$$$\mathrm{d}\alpha=\mathrm{d}{A}_{{x}} \wedge\mathrm{d}{x}+\mathrm{d}{A}_{{y}} \wedge\mathrm{d}{y}+\mathrm{d}{A}_{{z}} \wedge\mathrm{d}{z} \\ $$$$\mathrm{d}\alpha=\left(\frac{\partial{A}_{{z}} }{\partial{y}}−\frac{\partial{A}_{{x}} }{\partial{z}}\right)\mathrm{d}{y}\wedge\mathrm{d}{z}+\left(\frac{\partial{A}_{{x}} }{\partial{z}}−\frac{\partial{A}_{{z}}…
Question Number 220231 by fantastic last updated on 09/May/25 Answered by mr W last updated on 09/May/25 $$\mathrm{tan}\:\alpha=\frac{\mathrm{1}.\mathrm{5}}{\mathrm{1}+\mathrm{1}.\mathrm{5}+\sqrt{\left(\mathrm{1}+\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$?=\mathrm{2}{R}=\frac{\mathrm{2}×\mathrm{1}.\mathrm{5}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{\mathrm{3}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{5}\:\checkmark \\ $$…
Question Number 220224 by Mamadi last updated on 09/May/25 $${calcul}\:{together}\:{of}\:{definition}\:{of}\:{and} \\ $$$${calcul}\:{the}\:{derive}\:{f}^{\:} \:^{'} \\ $$$${f}\left({x}\right)=\:\:{x}\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}} \\ $$ Answered by hlwrc last updated on 09/May/25 $$\frac{{df}\left({x}\right)}{{dx}}=\sqrt{\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}}+\frac{{x}}{\mathrm{2}}\sqrt{\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}}×\frac{\left({x}−\mathrm{1}\right)−\left({x}+\mathrm{1}\right)}{\left({x}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 220226 by SdC355 last updated on 09/May/25 Commented by SdC355 last updated on 09/May/25 $$\mathrm{From}\:\mathrm{Generalized}\:\mathrm{Stoke}'\mathrm{s}\:\mathrm{theorem} \\ $$$$\oint_{\partial\boldsymbol{\Sigma}} \:\alpha=\int\int_{\:\boldsymbol{\Sigma}} \:\mathrm{d}\alpha \\ $$$$\oint_{\partial\boldsymbol{\Sigma}} \overset{\rightarrow} {\boldsymbol{\mathrm{F}}}\centerdot\mathrm{d}\boldsymbol{\mathrm{l}}=\int\int_{\:\boldsymbol{\Sigma}}…
Question Number 220221 by fantastic last updated on 08/May/25 $${Q}.{The}\:{density}\:{of}\:{an}\:{object}\:{of}\:{mass}\:{M}\:{is}\:\delta\:{and}\:{the}\:{density}\:{of}\:{the}\:{air}\:{is}\:\rho. \\ $$$${the}\:{mass}\:{of}\:{of}\:{the}\:{object}\:{is}\:{measured}\:{with}\:\:{the}\:{help}\:{of}\:{a}\:{metal}\:{weight}\:{of}\:{mass}\:{m}\:. \\ $$$${the}\:{density}\:{of}\:{the}\:{metal}\:{weight}\:{is}\:{d}. \\ $$$${if}\:\rho\ll\delta\:{them}\:{show}\:{that}\:{the}\:{real}\:{mass}\:{M}\:{will}\:{be} \\ $$$${m}\left(\mathrm{1}−\frac{\rho}{{d}}\:\right)\left(\mathrm{1}+\frac{\rho}{\delta}\right) \\ $$$${I}\:{have}\:{managed}\:{to}\:{M}=\frac{{m}\left(\mathrm{1}−\frac{\rho}{{d}}\right)}{\left(\mathrm{1}−\frac{\rho}{\delta}\right)} \\ $$$${but}\:{I}\:{can}\:{not}\:{figure}\:{it}\:{to}\:{the}\:{end} \\ $$$${please}\:{help} \\…
Question Number 220208 by Tawa11 last updated on 08/May/25 Commented by Tawa11 last updated on 08/May/25 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{and}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{shape}. \\ $$$$\mathrm{volume}\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\pi\mathrm{r}^{\mathrm{2}} \mathrm{h}\:\:\:\:\:\mathrm{but}\:\mathrm{I}\:\mathrm{am}\:\mathrm{not}\:\mathrm{getting}\:\mathrm{the}\:\mathrm{area}. \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{please}? \\ $$ Commented…
Question Number 220190 by SdC355 last updated on 07/May/25 $${i}^{{i}} ={e}^{−\frac{\pi}{\mathrm{2}}} \: \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:\mathrm{complex}\:\mathrm{number}\:\boldsymbol{{i}}\:\mathrm{as}\:\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix} \\ $$$$\boldsymbol{{i}}^{\boldsymbol{{i}}} =\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}^{\begin{pmatrix}{\mathrm{0}}&{−\mathrm{1}}\\{\mathrm{1}}&{\:\:\:\:\mathrm{0}}\end{pmatrix}} \: \\ $$$$\:\mathrm{But}\:\mathrm{why}\:\mathrm{Matrix}\:\mathrm{Exponent}\:\mathrm{Calculate}\:\mathrm{Dosen}'\mathrm{t}\:\mathrm{defined}?? \\ $$$$\:\mathrm{I}\:\mathrm{mean}\:{A},{B}\in\mathrm{mat}\left({m},{m}\right) \\ $$$$\mathrm{why}\:\mathrm{A}^{\mathrm{B}} \:\mathrm{dosen}'\mathrm{t}\:\mathrm{defined}??…