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} \\ $$$${side}\:{blue}/{red}/{yellow}\:{square}\:\rightarrow\:{b}/{r}/{y} \\ $$$${b}^{\mathrm{2}} ={r}^{\mathrm{2}} +\left({a}−\mathrm{2}{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} +{y}^{\mathrm{2}}…
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Question Number 191714 by Mingma last updated on 29/Apr/23 Commented by AST last updated on 29/Apr/23 Answered by AST last updated on 29/Apr/23 $${Let}\:{AC}\:{intersect}\:{the}\:{circle}\:{from}\:{A} \\…
Question Number 191715 by Shlock last updated on 29/Apr/23 Answered by mr W last updated on 29/Apr/23 Commented by Shlock last updated on 29/Apr/23 Sir,canyon use the similarity of the right triangles? Can you give more explanation in details?…
Question Number 191676 by ajfour last updated on 28/Apr/23 Commented by ajfour last updated on 28/Apr/23 $${If}\:{first}\:{time}\:{dipped}\:{area}\:{of} \\ $$$${biscuit}\:{is}\:{A}_{\mathrm{1}} . \\ $$$${Find}\:\:{max}.\:{of}\:\frac{{A}_{\mathrm{2}} }{{A}_{\mathrm{1}} }.\:\: \\…
Question Number 191457 by Mingma last updated on 24/Apr/23 Commented by Tinku Tara last updated on 24/Apr/23 $$\mathrm{The}\:\mathrm{user}\:\mathrm{is}\:\mathrm{blocked}\:\mathrm{from}\:\mathrm{making}\:\mathrm{further} \\ $$$$\mathrm{posts}. \\ $$ Commented by mr…
Question Number 191458 by Mingma last updated on 24/Apr/23 Answered by a.lgnaoui last updated on 26/Apr/23 $$\bigtriangleup\mathrm{ACD}\:\:\mathrm{et}\:\bigtriangleup\mathrm{BCD}\:\:\mathrm{semblablables} \\ $$$$\mathrm{AB}\:\mathrm{tangdnte}\:\mathrm{au}\:\mathrm{quart}\:\mathrm{cercle}\:\mathrm{Rouge}\:\mathrm{en}\:\mathrm{D} \\ $$$$\mathrm{CD}=\mathrm{R2}\:\:\:\mathrm{CD}\bot\mathrm{AB}\:\:;\mathrm{BC}=\mathrm{R1}+\mathrm{R2} \\ $$$$\left(\mathrm{R1}\:\mathrm{rayon}\:\mathrm{du}\:\:\mathrm{quart}\:\:\mathrm{cercle}\:\mathrm{vert}.\right. \\ $$$$\mathrm{ABC}\:\mathrm{triangle}\:\:\mathrm{recrangle}\:\mathrm{en}\:\mathrm{C}…
Question Number 191335 by Mingma last updated on 23/Apr/23 Answered by mr W last updated on 23/Apr/23 $${radius}\:{of}\:{big}\:{quatercircle}\:={a} \\ $$$${radius}\:{of}\:{big}\:{semicircle}\:={b} \\ $$$${radius}\:{of}\:{small}\:{quatercircle}\:={c} \\ $$$${b}^{\mathrm{2}} +\left(\mathrm{2}{b}\right)^{\mathrm{2}}…
Question Number 191325 by Mingma last updated on 23/Apr/23 Answered by HeferH last updated on 23/Apr/23 $$\mathrm{16}\:=\:\mathrm{2}×\mathrm{a}\left(\mathrm{a}+\mathrm{b}\right) \\ $$$$\mathrm{Yellow}\:=\:\mathrm{8}\: \\ $$ Commented by Mingma last…