Question Number 192705 by Mingma last updated on 25/May/23 Commented by Frix last updated on 25/May/23 $$\mathrm{For}\:{n}=\mathrm{1}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{gives}\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$ Commented by Frix last updated on…
Question Number 192663 by Mingma last updated on 24/May/23 Answered by cherokeesay last updated on 24/May/23 Commented by Mingma last updated on 25/May/23 Perfect �� Terms…
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Question Number 192344 by Mingma last updated on 15/May/23 Commented by malwan last updated on 17/May/23 $${I}\:{think}\:{its}\:{greet}\:{question} \\ $$$${please}\:{any}\:{one}\:{try}\:{to}\:{solve}\:{it} \\ $$ Commented by Mingma last…
Question Number 191958 by Rupesh123 last updated on 04/May/23 Commented by Rupesh123 last updated on 04/May/23 x=? Commented by mr W last updated on 04/May/23…
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Question Number 191854 by Rupesh123 last updated on 01/May/23 Answered by a.lgnaoui last updated on 02/May/23 $$\mathrm{l}\:\mathrm{image}\:\mathrm{comporte}\:\mathrm{3}\:\mathrm{principale}\boldsymbol{\mathrm{S}}\mathrm{s}\:\mathrm{surfaces} \\ $$$$\bullet\boldsymbol{\mathrm{S}}\mathrm{1}:\:\:\boldsymbol{\mathrm{S}}\mathrm{urface}\:\mathrm{du}\:\mathrm{quart}\:\mathrm{ceecle}\:\mathrm{C1}\:\left(\mathrm{rayon}\boldsymbol{\mathrm{r}}_{\mathrm{1}} =\mathrm{2}\right) \\ $$$$\bullet\mathrm{S2}:\:\boldsymbol{\mathrm{S}}\mathrm{urface}\:\mathrm{petit}\:\mathrm{cercleC2}\left(\mathrm{ryon}\::\boldsymbol{\mathrm{r}}_{\mathrm{2}} \right) \\ $$$$\bullet\boldsymbol{\mathrm{S}}\mathrm{3}:\mathrm{Surface}\:\mathrm{portion}\:\mathrm{cercle}\:\left(\mathrm{rayon}\:\boldsymbol{\mathrm{r}}_{\mathrm{3}}…
Question Number 60717 by peter frank last updated on 24/May/19 Commented by Prithwish sen last updated on 25/May/19 $$=\begin{vmatrix}{\frac{\partial\mathrm{x}}{\partial\mathrm{r}}\:\:\:\:\:\frac{\partial\mathrm{x}}{\partial\theta}\:\:\:\:\:\frac{\partial\mathrm{x}}{\partial\emptyset}}\\{\frac{\partial\mathrm{y}}{\partial\mathrm{r}}\:\:\:\:\frac{\partial\mathrm{y}}{\partial\theta}\:\:\:\:\:\:\frac{\partial\mathrm{y}}{\partial\phi}}\end{vmatrix} \\ $$$$\:\:\:\:\begin{vmatrix}{\frac{\partial\mathrm{z}}{\partial\mathrm{r}}\:\:\:\:\:\frac{\partial\mathrm{z}}{\partial\theta}\:\:\:\:\:\:\frac{\partial\mathrm{z}}{\partial\phi}}\end{vmatrix} \\ $$$$=\mathrm{sin}\theta\mathrm{cos}\phi\left(\mathrm{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:\mathrm{cos}\phi\:\right)+\mathrm{sin}\theta\mathrm{sin}\phi\left(\mathrm{r}^{\mathrm{2}}…
Question Number 191756 by Shlock last updated on 30/Apr/23 Answered by aleks041103 last updated on 30/Apr/23 $${side}\:{of}\:{whole}\:{square}\:\rightarrow\:{a} \\ $$$${side}\:{blue}/{red}/{yellow}\:{square}\:\rightarrow\:{b}/{r}/{y} \\ $$$${b}^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({a}−\mathrm{2}{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +{y}^{\mathrm{2}}…
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