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Question Number 225503    Answers: 2   Comments: 0

$${find}\:{x}\in\mathbb{C}\:{for}\:{r}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$$\left(−{r}\right)^{{x}} ={r} \\ $$

Question Number 225497    Answers: 9   Comments: 0

Question Number 225490    Answers: 0   Comments: 19

Question Number 225488    Answers: 0   Comments: 0

$$\mathrm{Diffenrantial}\:\mathrm{Geometry}.....=\wedge= \\ $$$$\mathrm{Christoffel}\:\mathrm{symbol}\: \\ $$$$\Gamma_{\sigma\mu\nu} \:\mathrm{first}\:\mathrm{kind}\:\mathrm{and}\:\Gamma_{\mu\nu} ^{\:\sigma} \:\mathrm{second}\:\mathrm{kind}... \\ $$$$\mathrm{and}\:\mathrm{Chritoffel}\:\mathrm{symbol}\:\mathrm{satisfy} \\ $$$$\Gamma_{\mu\nu} ^{\:\sigma} =\frac{\mathrm{1}}{\mathrm{2}}{g}^{\sigma{l}} \Gamma_{{l}\mu\nu} \\ $$$$\Gamma_{{ijk}} \:\:\left\{\theta,\rho\right\}\in\left\{{i},{j},{k}\right\} \\ $$$$\Gamma_{\theta\theta\theta} \:,\:\Gamma_{\rho\rho\rho} \:,\:\Gamma_{\theta\theta\rho} \\ $$$$\Gamma_{\theta\rho\theta} \:,\:\Gamma_{\rho\theta\rho} \:,\:\Gamma_{\rho\theta\theta} \\ $$$$\Gamma_{\theta\rho\rho} \:,\:\Gamma_{\rho\rho\theta} \\ $$$$\Gamma_{{jk}} ^{{i}} =\frac{\mathrm{1}}{\mathrm{2}}{g}^{{il}} \Gamma_{{ljk}} \\ $$$${g}_{\mu\nu} =\begin{pmatrix}{\:{r}^{\mathrm{2}} }&{\:\:\:\:\:\:\mathrm{0}}\\{\:\mathrm{0}}&{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\end{pmatrix} \\ $$$$\left({g}_{\mu\nu} \right)^{−\mathrm{1}} ={g}^{\mu\nu} \\ $$$${g}^{\mu\nu} =\begin{pmatrix}{\:\frac{\mathrm{1}}{{r}^{\mathrm{2}} }}&{\:\:\:\:\:\:\:\:\mathrm{0}}\\{\:\mathrm{0}}&{\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}}\end{pmatrix} \\ $$$${g}_{\theta\theta} ={r}^{\mathrm{2}} ,\:{g}_{\theta\rho} =\mathrm{0}\:,\:{g}_{\rho\theta} =\mathrm{0}\:,\:{g}_{\rho\rho} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\partial_{\theta} {g}_{\theta\theta} =\mathrm{0}\:,\:\partial_{\rho} {g}_{\theta\theta} =\mathrm{0}\:,\:\partial_{\theta} {g}_{\theta\rho} =\mathrm{0}\:,\:\partial_{\rho} {g}_{\theta\rho} =\mathrm{0} \\ $$$$\partial_{\theta} {g}_{\rho\theta} =\mathrm{0}\:,\:\partial_{\rho} {g}_{\rho\theta} =\mathrm{0}\:,\:\partial_{\theta} {g}_{\rho\rho} =\mathrm{2}{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)\:,\:\partial_{\rho} {g}_{\rho\rho} =\mathrm{0} \\ $$$$\Gamma_{{abc}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{{b}} ^{\:} {g}_{{ac}} +\partial_{{c}} {g}_{{ab}} −\partial_{{a}} {g}_{{bc}} \right) \\ $$$$\Gamma_{\theta\theta\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\theta\theta} +\partial_{\theta} {g}_{\theta\theta} −\partial_{\theta} {g}_{\theta\theta} \right)=\mathrm{0} \\ $$$$\Gamma_{\rho\rho\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\rho\rho} +\partial_{\rho} {g}_{\rho\rho} −\partial_{\rho} {g}_{\rho\rho} \right)=\mathrm{0} \\ $$$$\Gamma_{\theta\rho\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\theta\rho} +\partial_{\rho} {g}_{\theta\rho} −\partial_{\theta} {g}_{\rho\rho} \right)=−{r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\Gamma_{\rho\theta\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\rho\rho} +\partial_{\rho} {g}_{\rho\theta} −\partial_{\rho} {g}_{\theta\rho} \right)={r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\Gamma_{\rho\theta\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\rho\theta} +\partial_{\theta} {g}_{\rho\theta} −\partial_{\rho} {g}_{\theta\theta} \right)=\mathrm{0} \\ $$$$\Gamma_{\theta\rho\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\theta\theta} +\partial_{\theta} {g}_{\theta\rho} −\partial_{\theta} {g}_{\rho\theta} \right)=\mathrm{0} \\ $$$$\Gamma_{\theta\theta\rho} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\theta} {g}_{\theta\rho} +\partial_{\rho} {g}_{\theta\theta} −\partial_{\theta} {g}_{\theta\rho} \right)=\mathrm{0} \\ $$$$\Gamma_{\rho\rho\theta} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\rho} {g}_{\rho\theta} +\partial_{\theta} {g}_{\rho\rho} −\partial_{\rho} {g}_{\rho\theta} \right)={r}^{\mathrm{2}} \mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\Gamma_{{jk}} ^{{i}} ={g}^{{il}} \Gamma_{{ljk}} \\ $$$${g}^{\theta\theta} =\frac{\mathrm{1}}{{r}^{\mathrm{2}} }\:,\:{g}^{\theta\rho} =\mathrm{0}\:,\:{g}^{\rho\theta} =\mathrm{0}\:,\:{g}^{\rho\rho} =\frac{\mathrm{1}}{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)} \\ $$$$\Gamma_{\theta\theta} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\theta\theta} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\theta\theta} =\mathrm{0}}\\{{g}^{\theta\rho} \Gamma_{\rho\theta\theta} =\mathrm{0}}\end{cases},\:\Gamma_{\rho\theta} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\rho\theta} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\rho\theta} =\mathrm{0}}\\{{g}^{\theta\rho} \Gamma_{\rho\rho\theta} =\mathrm{0}}\end{cases}\:,\:\Gamma_{\theta\rho} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\theta\rho} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\theta\rho} =\mathrm{0}}\\{{g}^{\theta\rho} \Gamma_{\rho\theta\rho} =\mathrm{0}}\end{cases}\:,\:\Gamma_{\rho\rho} ^{\theta} ={g}^{\theta{l}} \Gamma_{{l}\rho\rho} =\begin{cases}{{g}^{\theta\theta} \Gamma_{\theta\rho\rho} =−\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)}\\{{g}^{\theta\rho} \Gamma_{\rho\rho\rho} =\mathrm{0}}\end{cases} \\ $$$$\Gamma_{\theta\theta} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\theta\theta} =\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\theta\theta} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\theta\theta} =\mathrm{0}}\end{cases}\:,\:\Gamma_{\rho\theta} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\rho\theta} \:=\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\rho\theta} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\rho\theta} =\mathrm{cot}\left(\theta\right)}\end{cases},\:\Gamma_{\theta\rho} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\theta\rho} =\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\theta\rho} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\theta\rho} =\mathrm{cot}\left(\theta\right)}\end{cases}\:,\:\Gamma_{\rho\rho} ^{\rho} ={g}^{\rho{l}} \Gamma_{{l}\rho\rho} =\begin{cases}{{g}^{\rho\theta} \Gamma_{\theta\rho\rho} =\mathrm{0}}\\{{g}^{\rho\rho} \Gamma_{\rho\rho\rho} =\mathrm{0}}\end{cases} \\ $$$$\therefore\Gamma_{\rho\rho} ^{\theta} =−\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right) \\ $$$$\:\:\:\Gamma_{\mu\nu} ^{\:\rho} =\Gamma_{\nu\mu} ^{\rho} \\ $$$$\:\:\:\Gamma_{\rho\theta} ^{\rho} =\Gamma_{\theta\rho} ^{\rho} =\mathrm{cot}\left(\theta\right) \\ $$$$\:{R}_{{jkl}} ^{{i}} =\partial_{{k}} \Gamma_{{jl}} ^{{k}} −\partial_{{l}} \Gamma_{{jk}} ^{{i}} +\Gamma_{{km}} ^{{i}} \Gamma_{{jl}} ^{{m}} −\Gamma_{{lm}} ^{{i}} \Gamma_{{jk}} ^{{m}} \\ $$$${R}_{\theta\theta\theta} ^{\theta} \:,\:{R}_{\theta\rho\theta} ^{\theta} \:,\:{R}_{\rho\theta\theta} ^{\theta} \:,\:{R}_{\rho\rho\theta} ^{\theta} \:,\:{R}_{\theta\rho\rho} ^{\theta} \:,\:{R}_{\rho\rho\rho} ^{\theta} \:,\:{R}_{\rho\theta\rho} ^{\theta} \:,\:{R}_{\theta\rho\rho} ^{\theta} \\ $$$${R}_{\theta\theta\theta} ^{\rho} \:,\:{R}_{\theta\rho\theta} ^{\rho} \:,\:{R}_{\rho\theta\theta} ^{\rho} \:,\:{R}_{\rho\rho\theta} ^{\rho} \:,\:{R}_{\theta\rho\rho} ^{\rho} \:,\:{R}_{\rho\rho\rho} ^{\rho} \:,\:{R}_{\rho\theta\rho} ^{\rho} \:,\:{R}_{\theta\rho\rho} ^{\rho} \: \\ $$$$\mathrm{and}\:\mathrm{Riemann}\:\mathrm{metric}\:\mathrm{tensor}\:\mathrm{have}\:\mathrm{symmetries} \\ $$$${R}_{{abcd}} =−{R}_{{bacd}} \\ $$$${R}_{{abcd}} =−{R}_{{abdc}} \\ $$$${R}_{{abcd}} ={R}_{{cdab}} \\ $$$$\mathrm{Non}-\mathrm{Zero}\:\Gamma_{\mu\nu} ^{\:\rho} \: \\ $$$$\Gamma_{\rho\rho} ^{\:\theta} =−\mathrm{sin}\left(\theta\right)\mathrm{cos}\left(\theta\right)\:,\:\Gamma_{\rho\theta} ^{\:\rho} \:,\:\Gamma_{\theta\rho} ^{\:\rho} =\mathrm{cot}\left(\theta\right) \\ $$$$\left\{{i},{j},{k},\ell\right\}\in\left\{\theta,\rho\right\} \\ $$$${R}_{\rho\theta\rho} ^{\theta} =\partial_{\theta} \Gamma_{\rho\rho} ^{\theta} −\partial_{\rho} \Gamma_{\rho\theta} ^{\theta} +\Gamma_{\theta{m}} ^{\:\theta} \Gamma_{\rho\rho} ^{{m}} −\Gamma_{\rho{m}} ^{\theta} \Gamma_{\rho\theta} ^{{m}} \\ $$$$=\mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$${R}_{\alpha\rho\theta\rho} =\mathrm{g}_{\alpha\mu} {R}_{\rho\theta\rho} ^{\mu} \\ $$$${R}_{\theta\rho\theta\rho} =\mathrm{g}_{\theta\mu} {R}_{\rho\theta\rho} ^{\mu} =\begin{cases}{{g}_{\theta\theta} {R}_{\rho\theta\rho} ^{\theta} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right)}\\{{g}_{\theta\rho} {R}_{\rho\theta\rho} ^{\rho} =\mathrm{0}}\end{cases} \\ $$$${R}_{\theta\rho\theta\rho} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\therefore\:{R}_{\rho\theta\rho} ^{\theta} =\mathrm{sin}^{\mathrm{2}} \left(\theta\right)\:,\:{R}_{\theta\rho\theta\rho} ={r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \left(\theta\right) \\ $$$$\sim\mathrm{2}-\mathrm{dimensional}\:\mathrm{Riemann}\:\mathrm{Manifold}\sim \\ $$$${R}_{{abcd}} ={K}\left({g}_{{ac}} {g}_{{bd}} −{g}_{{ad}} {g}_{{bc}} \right) \\ $$$${R}_{\mu\nu} ^{\mathrm{Ricci}} ={Kg}_{\mu\nu} \\ $$$${K}\:\mathrm{is}\:\mathrm{gaussian}\:\mathrm{curvature}\:\:\mathrm{aka}\:{K}=\frac{\mathrm{det}\mathbb{II}_{{p}} }{\mathrm{det}\:\mathbb{I}_{{p}} }=\frac{{LM}−{N}^{\mathrm{2}} }{{EG}−{F}^{\mathrm{2}} } \\ $$$${Q}\mathrm{225479} \\ $$

Question Number 225479    Answers: 1   Comments: 0

$$\mathrm{Let}\:\mathcal{S};\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}^{\mathrm{3}} \:\mathrm{Sphere}\: \\ $$$${Q}\mathrm{1}.\:\mathrm{Find}\:\mathrm{metric}\:\mathrm{tensor}\:\mathrm{g}_{\mu\nu} \: \\ $$$${Q}\mathrm{2}.\:\mathrm{Find}\:\mathrm{Riemann}\:\mathrm{metric}\:\mathrm{tensor}\:{R}_{{jkl}} ^{{i}} \\ $$$${Q}.\mathrm{3}\:\:\mathrm{Find}\:\mathrm{Ricci}\:\mathrm{tensor}\:\mathrm{R}_{\alpha\beta} \\ $$$${Q}.\mathrm{4}\:\:\mathrm{Find}\:\mathrm{Ricci}\:\mathrm{Scalar}\:\mathcal{R} \\ $$$$\mathrm{Christoffel}\:\mathrm{symbol}\:\mathrm{first}\:\mathrm{kind}\: \\ $$$$\Gamma_{\mu\nu\sigma} =\frac{\mathrm{1}}{\mathrm{2}}\left(\partial_{\nu} ^{\:} \mathrm{g}_{\mu\sigma} +\partial_{\sigma} \mathrm{g}_{\mu\nu} −\partial_{\mu} \mathrm{g}_{\nu\sigma} \right) \\ $$$$\mathrm{Christoffel}\:\mathrm{symbol}\:\mathrm{second}\:\mathrm{kind} \\ $$$$\Gamma_{{jk}} ^{{i}} =\mathrm{g}^{{il}} \Gamma_{{ljk}} =\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}^{{i}\sigma} \left(\partial_{{j}} ^{\:} \mathrm{g}_{{lk}} +\partial_{{k}} \mathrm{g}_{{lj}} −\partial_{{l}} \mathrm{g}_{{jk}} \right) \\ $$$$\mathrm{Riemann}\:\mathrm{metric}\:\mathrm{tensor} \\ $$$${R}_{{jkl}} ^{{i}} =\partial_{{k}} \Gamma_{{jl}} ^{{i}} −\partial_{{l}} \Gamma_{{jk}} ^{{i}} +\Gamma_{{km}} ^{{i}} \Gamma_{{jl}} ^{{m}} −\Gamma_{{lm}} ^{{i}} \Gamma_{{jk}} ^{{m}} \\ $$

Question Number 225474    Answers: 1   Comments: 2

Question Number 225461    Answers: 1   Comments: 0

$${Which}\:{one}\:{is}\:{the}\:{oddest}\:{prime}\:{number}? \\ $$

Question Number 225449    Answers: 2   Comments: 0

Question Number 225416    Answers: 1   Comments: 2

Question Number 225407    Answers: 4   Comments: 3

Question Number 225391    Answers: 0   Comments: 0

$$\:\:\:\:\:\:\:\underset{\mathrm{n}\:\:\:=\:\:\:\mathrm{1}} {\overset{\infty} {\sum}}\:\left(−\:\:\:\:\mathrm{1}\right)^{\mathrm{n}\:\:\:\:−\:\:\:\mathrm{1}} \:\frac{\mathrm{H}_{\mathrm{n}} \:\mathrm{H}_{\mathrm{2n}} ^{\left(\mathrm{2}\right)} }{\mathrm{n}}\:\:\:\:\:\:\:=\:\:\:\:\:? \\ $$

Question Number 225392    Answers: 2   Comments: 1

Question Number 225382    Answers: 1   Comments: 0

$$\mathrm{Been}\:\mathrm{a}\:\mathrm{while}\:\mathrm{guys} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{xln}\left(\mathrm{1}+\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx} \\ $$

Question Number 225365    Answers: 0   Comments: 1

$$\mathrm{who}'\mathrm{s}\:\mathrm{trying}\:\mathrm{this}\:\mathrm{scholars}? \\ $$

Question Number 225397    Answers: 1   Comments: 1

Question Number 225356    Answers: 4   Comments: 1

Question Number 225354    Answers: 1   Comments: 1

Question Number 225338    Answers: 1   Comments: 1

Question Number 225330    Answers: 1   Comments: 0

$${if}\:\:\left({fogoh}\right)\left({x}\right)={cos}^{\mathrm{2}} \left({x}+\mathrm{9}\right) \\ $$$${then}\:\:\:{f}\left({x}\right)=?\:,\:\:{g}\left({x}\right)=?\:\:,\:{h}\left({x}\right)=? \\ $$

Question Number 225323    Answers: 1   Comments: 0

$$\mathrm{Find}:\:\:\:\left(\frac{\mathrm{3}\:+\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}{\mathrm{3}\:-\:\mathrm{2}\:\sqrt[{\mathrm{4}}]{\mathrm{5}}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} .\:\:\:\frac{\sqrt[{\mathrm{4}}]{\mathrm{5}}\:-\:\mathrm{1}}{\:\sqrt[{\mathrm{4}}]{\mathrm{5}}\:+\:\mathrm{1}}\:\:=\:? \\ $$

Question Number 225394    Answers: 1   Comments: 2

Question Number 225307    Answers: 1   Comments: 1

Question Number 225299    Answers: 0   Comments: 1

$$\mathrm{We}\:\mathrm{have}\:\mathrm{integrated}\:\mathrm{artificial} \\ $$$$\mathrm{intelligence}: \\ $$$$-\:\mathrm{images}\:\mathrm{are}\:\mathrm{checked}\:\mathrm{during}\:\mathrm{upload} \\ $$$$\:\:\:\mathrm{and}\:\mathrm{will}\:\mathrm{be}\:\mathrm{rejected}\:\mathrm{if}\:\mathrm{not}\:\mathrm{related} \\ $$$$\:\:\:\mathrm{to}\:\mathrm{science} \\ $$$$-\:\mathrm{daily}\:\mathrm{processing}\:\mathrm{of}\:\mathrm{other}\:\mathrm{posts} \\ $$$$\mathrm{A}\:\mathrm{few}\:\mathrm{senior}\:\mathrm{users}\:\mathrm{are}\:\mathrm{exempted}\:\mathrm{from} \\ $$$$\mathrm{these}\:\mathrm{checks}. \\ $$

Question Number 225303    Answers: 0   Comments: 0

$${Let} \\ $$$${S}_{{n}} \left({x}\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{sin}\:\left(\left(\mathrm{2}{r}−\mathrm{3}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{10}{r}+\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)−\mathrm{sin}\left(\:\left(\mathrm{6}{r}−\mathrm{1}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)\mathrm{cos}\:\left(\left(\mathrm{2}{r}+\mathrm{5}\right)\mathrm{2}^{−\left({r}+\mathrm{1}\right)} {x}\right)}{\mathrm{2}^{{r}−\mathrm{1}} \left(\mathrm{sin}\:\left({r}\mathrm{2}^{\mathrm{3}−{r}} −{x}\right)\mathrm{sin}\:\left(\mathrm{2}^{\mathrm{2}−{r}} −{x}\right)\right)} \\ $$$${then}\:{find}\:{the}\:{value}\:{of} \\ $$$$\underset{{m}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{cos}\:\left({x}\right)}{\mathrm{2}^{{n}} \left(\mathrm{1}+\mathrm{2}{S}_{{n}} \left({x}\right)\mathrm{cos}\:\left({x}\right)\right)}\right){dx}}{\left[\frac{{d}}{{dx}}\left(\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\mathrm{cos}^{−\mathrm{1}} \left(\frac{\mathrm{cos}\:\left({x}\right)}{\mathrm{2}^{{n}} \left(\mathrm{1}+\mathrm{2}{S}_{{n}} \left({x}\right)\mathrm{cos}\:\left({x}\right)\right)}\right)\right]_{{x}=\mathrm{0}} \right.}\right) \\ $$

Question Number 225282    Answers: 2   Comments: 4

$${Q}\mathrm{225146} \\ $$$${here}\:{if}\:{the}\:{wedge}\:{and}\:{the} \\ $$$${ground}\:\:{had}\:{no}\:{friction} \\ $$$${it}\:{would}\:{start}\:{going}\:\rightarrow \\ $$$${what}\:{is}\:{acc}.\:{at}\:{time}\:{t}? \\ $$$$\mu={kx} \\ $$

Question Number 225230    Answers: 1   Comments: 3

$$\:\:\:{tg}^{\mathrm{4}} \mathrm{10}°+{tg}^{\mathrm{4}} \mathrm{50}°+{tg}^{\mathrm{4}} \mathrm{70}°=? \\ $$

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