Question Number 9298 by tawakalitu last updated on 29/Nov/16 Answered by mrW last updated on 29/Nov/16 $$\mathrm{radius}\:\mathrm{r}=\left(\mathrm{7}+\mathrm{3}\right)/\mathrm{2}=\mathrm{5}\:\mathrm{cm} \\ $$$$\mathrm{height}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{h}=\mathrm{3}\:\mathrm{cm} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{segment}\:=\:\mathrm{r}^{\mathrm{2}} \mathrm{cos}\:^{−\mathrm{1}} \left(\frac{\mathrm{r}−\mathrm{h}}{\mathrm{r}}\right)−\left(\mathrm{r}−\mathrm{h}\right)\sqrt{\mathrm{2rh}−\mathrm{h}^{\mathrm{2}} } \\…
Question Number 140364 by fadialfadifadi@gmail.com last updated on 06/May/21 Commented by fadialfadifadi@gmail.com last updated on 06/May/21 $$\boldsymbol{\mathrm{Real}}\:\boldsymbol{\mathrm{analysis}} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 140367 by aliibrahim1 last updated on 06/May/21 Answered by meetbhavsar25 last updated on 06/May/21 $${Answer}\:{is}\:\frac{\mathrm{9}}{\mathrm{25}}. \\ $$$${x}^{\mathrm{2}} \:{has}\:{minimum}\:{value}\:\mathrm{0}. \\ $$$${y}^{\mathrm{2}} \:{has}\:{minimum}\:{value}\:\mathrm{0}. \\ $$$$\left({z}−\mathrm{1}\right)^{\mathrm{2}}…
Question Number 140363 by cesarL last updated on 06/May/21 $${My}\:{friends}\:{can}\:{you}\:{help}\:{me}? \\ $$$${f}\left({x},{y}\right)={x}^{\mathrm{3}} {y}^{\mathrm{7}} −\mathrm{4}{e}^{{x}−{y}} \\ $$$${Find}: \\ $$$$\left.{a}\right)\:\frac{\partial^{\mathrm{2}} {f}}{\partial{x}\partial{y}}=? \\ $$$$\left.{b}\right)\:\frac{\partial^{\mathrm{2}} {f}^{\mathrm{2}} }{\partial{y}\partial{x}}=? \\ $$$${Please}…
Question Number 74825 by liki last updated on 01/Dec/19 Commented by liki last updated on 01/Dec/19 $$…{Anyone}\:{to}\:{help}\:{me}\:! \\ $$ Commented by $@ty@m123 last updated on…
Question Number 9287 by suci last updated on 28/Nov/16 $${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$ Answered by mrW last updated on 28/Nov/16 $$=−\mathrm{4}×\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$$$=−\mathrm{4}×\mathrm{2}×\begin{vmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\…
Question Number 9286 by suci last updated on 28/Nov/16 $${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{5}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$ Answered by mrW last updated on 28/Nov/16 $$=\mathrm{5}×\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{4}}&{\mathrm{0}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\ $$$$=\mathrm{5}×\left(−\mathrm{4}\right)×\begin{vmatrix}{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\\{\mathrm{0}}&{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{3}}&{\mathrm{0}}&{\mathrm{0}}\end{vmatrix} \\…
Question Number 9285 by suci last updated on 28/Nov/16 $${find}\:{the}\:{determinant}\:{of}\:{the}\:{matrix}\:{below} \\ $$$$\begin{bmatrix}{\mathrm{1}}&{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{0}}&{\mathrm{6}}&{\mathrm{3}}\\{\mathrm{4}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{4}}\end{bmatrix} \\ $$ Answered by mrW last updated on 28/Nov/16 $$\mid\mathrm{A}\mid=\begin{vmatrix}{\mathrm{1}}&{\mathrm{4}}&{−\mathrm{3}}&{\mathrm{1}}\\{\mathrm{2}}&{\mathrm{0}}&{\mathrm{6}}&{\mathrm{3}}\\{\mathrm{4}}&{−\mathrm{1}}&{\mathrm{2}}&{\mathrm{5}}\\{\mathrm{1}}&{\mathrm{0}}&{−\mathrm{2}}&{\mathrm{4}}\end{vmatrix} \\ $$$$\mathrm{R4}−\mathrm{R1}\:\mathrm{and}\:\mathrm{R3}−\mathrm{R2}×\mathrm{2}: \\…
Question Number 74821 by sridhar nayak last updated on 01/Dec/19 Answered by mind is power last updated on 01/Dec/19 $$\Leftrightarrow\left(\mathrm{6e}^{\mathrm{y}} −\mathrm{2x}\right)\mathrm{dy}−\mathrm{dx}=\mathrm{0}…\mathrm{E} \\ $$$$\mathrm{try}\:\mathrm{too}\:\mathrm{find}\:\mathrm{k}\left(\mathrm{y}\right)\:\mathrm{To}\:\mathrm{mak}\:\mathrm{it}\:\mathrm{exacte} \\ $$$$\Leftrightarrow\mathrm{k}\left(\mathrm{y}\right)\left(\mathrm{6e}^{\mathrm{y}}…
Question Number 140353 by Engr_Jidda last updated on 06/May/21 Commented by Engr_Jidda last updated on 06/May/21 $${complex}\:{analysis} \\ $$ Terms of Service Privacy Policy Contact:…