Question Number 131094 by Chhing last updated on 01/Feb/21 $$\:\: \\ $$$$\:\:\:\mathrm{Calculate} \\ $$$$\:\:\mathrm{1}/\:\mathrm{I}\:=\:\oint_{\mathrm{c}^{+} } \frac{\mathrm{zdz}}{\left(\mathrm{z}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\mathrm{2z}+\mathrm{1}−\mathrm{2i}\right)}\:\:,\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}\mid=\mathrm{2}\right\}\: \\ $$$$\:\:\mathrm{2}/\:\mathrm{J}\:=\oint_{\mathrm{c}^{+} } \frac{\mathrm{ch}\left(\mathrm{z}\right)\mathrm{dz}}{\mathrm{z}\left(\mathrm{e}^{\mathrm{z}} −\mathrm{1}\right)}\:\:,\:\:\mathrm{C}=\left\{\mathrm{z}/\mid\mathrm{z}−\mathrm{3i}\mid=\mathrm{4}\right\} \\ $$$$\:\:\mathrm{3}/\:\mathrm{K}=\oint_{\mathrm{c}^{+}…
Question Number 19 by user1 last updated on 25/Jan/15 $$\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{the}\:\mathrm{regression} \\ $$$$\mathrm{line}\:\mathrm{of}\:{y}\:\mathrm{on}\:{x}\:\mathrm{and}\:\mathrm{the}\:\mathrm{regression}\:\mathrm{line}\:\mathrm{of} \\ $$$${x}\:\mathrm{on}\:{y}.\:\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{tan}\:\theta=\left\{\frac{\left(\mathrm{1}−{r}^{\mathrm{2}} \right)}{{r}}×\frac{\sigma_{{x}} ×\sigma_{{y}} }{\left(\sigma_{{x}} ^{\mathrm{2}} +\sigma_{{y}} ^{\mathrm{2}} \right)}\right\} \\ $$…
Question Number 17 by user1 last updated on 25/Jan/15 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{regression}\:\mathrm{coefficient}\:{b}_{{xy}} \:\mathrm{between} \\ $$$${x}\:\mathrm{and}\:{y}\:\mathrm{for}\:\mathrm{the}\:\mathrm{following}\:\mathrm{data}: \\ $$$$\Sigma{x}=\mathrm{24},\:\Sigma{y}=\mathrm{44},\:\Sigma{xy}=\mathrm{306},\:\Sigma{x}^{\mathrm{2}} =\mathrm{164}, \\ $$$$\Sigma{y}^{\mathrm{2}} =\mathrm{574}\:\mathrm{and}\:{n}=\mathrm{4}. \\ $$ Answered by user1 last…
Question Number 15 by user1 last updated on 25/Jan/15 $$\mathrm{If}\:{A}=\begin{bmatrix}{\:\:\mathrm{3}}&{−\mathrm{5}}\\{−\mathrm{4}}&{\:\:\:\mathrm{2}}\end{bmatrix}, \\ $$$$\mathrm{show}\:\mathrm{that}\:{A}^{\mathrm{2}} −\mathrm{5}{A}−\mathrm{14}{I}=\mathrm{0} \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{We}\:\mathrm{have}: \\ $$$${A}^{\mathrm{2}}…
Question Number 131084 by greg_ed last updated on 01/Feb/21 $$\mathrm{let}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{in}\:\mathbb{R}\::\: \\ $$$${x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0}. \\ $$ Answered by liberty last updated on 01/Feb/21 $$\mathrm{x}_{\mathrm{1}}…
Question Number 13 by user1 last updated on 25/Jan/15 $$\mathrm{Expand}\:\mathrm{the}\:\mathrm{determnent}\: \\ $$$$\:\:\bigtriangleup=\begin{vmatrix}{{a}}&{{h}}&{{g}}\\{{h}}&{{b}}&{{f}}\\{{g}}&{{f}}&{{c}}\end{vmatrix} \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{Expanding}\:\mathrm{by}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{row},\:\mathrm{we}\:\mathrm{have}: \\ $$$$\bigtriangleup={a}\centerdot\begin{vmatrix}{{b}}&{{f}}\\{{f}}&{{c}}\end{vmatrix}−{h}\centerdot\begin{vmatrix}{{h}}&{{f}}\\{{g}}&{{c}}\end{vmatrix}+{g}\centerdot\begin{vmatrix}{{h}}&{{b}}\\{{g}}&{{f}}\end{vmatrix}…
Question Number 131086 by EDWIN88 last updated on 01/Feb/21 $$\:{y}''−{y}\:=\:{e}^{−\mathrm{2}{x}} \:\mathrm{sin}\:\left({e}^{−{x}} \right)\: \\ $$ Answered by liberty last updated on 01/Feb/21 $$\left[\:\mathrm{D}^{\mathrm{2}} −\mathrm{1}\:\right]\mathrm{y}\:=\:\mathrm{e}^{−\mathrm{2x}} \mathrm{sin}\:\left(\mathrm{e}^{−\mathrm{x}} \right)…
Question Number 11 by user1 last updated on 25/Jan/15 $$\mathrm{Evaluate}:\:\: \\ $$$$\:\begin{vmatrix}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}&{{x}−\mathrm{1}}\\{\:\:\:\:\:{x}+\mathrm{1}}&{{x}+\mathrm{1}}\end{vmatrix} \\ $$ Answered by user1 last updated on 30/Oct/14 $$=\left[\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\left({x}+\mathrm{1}\right)−\left({x}+\mathrm{1}\right)\left({x}−\mathrm{1}\right)\right] \\…
Question Number 131083 by Dwaipayan Shikari last updated on 01/Feb/21 $$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}−\frac{\mathrm{16}}{\mathrm{13}−\frac{\mathrm{81}}{\mathrm{25}−\frac{\mathrm{256}}{\mathrm{41}−\frac{\mathrm{625}}{\mathrm{61}−\frac{\mathrm{1296}}{\mathrm{85}−\frac{\mathrm{2401}}{\mathrm{113}−…}}}}}}}=\frac{\mathrm{6}}{\boldsymbol{\pi}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 9 by user1 last updated on 25/Jan/15 $$\mathrm{Let}\:{A}=\begin{bmatrix}{\:\:\:\mathrm{0}}&{−\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}\\{\mathrm{tan}\frac{\mathrm{x}}{\mathrm{2}}}&{\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{bmatrix}\:\mathrm{and}\:{I}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{identity}\:\mathrm{matrix}\:\mathrm{of}\:\mathrm{order}\:\mathrm{2}.\:\mathrm{Show}\:\mathrm{that}\: \\ $$$$\left({I}+{A}\right)=\left({I}−{A}\right)\centerdot\begin{bmatrix}{\mathrm{cos}\:{x}}&{−\mathrm{sin}\:{x}}\\{\mathrm{sin}\:{x}}&{\:\:\:\:\mathrm{cos}\:{x}}\end{bmatrix}. \\ $$ Answered by user1 last updated on 30/Oct/14 $$\mathrm{Let}\:\:\:\:\mathrm{tan}\:\frac{{x}}{\mathrm{2}}={t} \\…