Question Number 121097 by sdfg last updated on 05/Nov/20 Answered by TITA last updated on 05/Nov/20 $${is}\:{x}'=\frac{{dx}}{{dt}}\:? \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 186473 by nadovic last updated on 04/Feb/23 $$\mathrm{A}\:\mathrm{metallic}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{subjected}\:\mathrm{to} \\ $$$$\mathrm{heating}\:\mathrm{such}\:\mathrm{that}\:\mathrm{as}\:\mathrm{the}\:\mathrm{metal} \\ $$$$\mathrm{expands},\:\mathrm{the}\:\mathrm{total}\:\mathrm{surface}\:\mathrm{area} \\ $$$$\mathrm{increases}\:\mathrm{at}\:\mathrm{rate}\:\mathrm{of}\:\mathrm{6}.\mathrm{25}\:\mathrm{cm}^{\mathrm{2}} \mathrm{s}^{−\mathrm{1}} . \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{rate}\:\mathrm{at}\:\mathrm{which}\:\mathrm{each} \\ $$$$\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{is}\:\mathrm{increasing}\:\mathrm{when} \\ $$$$\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{51}.\mathrm{2}\:\mathrm{cm}^{\mathrm{3}} .…
Question Number 120912 by bemath last updated on 04/Nov/20 $$\:\mathrm{find}\:\frac{\mathrm{dy}}{\mathrm{dx}}\:\mathrm{from}\:\mathrm{equation}\: \\ $$$$\:\frac{\mathrm{xy}^{\mathrm{3}} }{\mathrm{sec}\:\mathrm{y}−\mathrm{3}}\:=\:\mathrm{1}+\mathrm{y}^{\mathrm{4}} \\ $$ Answered by liberty last updated on 04/Nov/20 $$\mathrm{by}\:\mathrm{implicit}\:\mathrm{differential}\: \\ $$$$\Rightarrow\:\frac{\mathrm{xy}^{\mathrm{3}}…
Question Number 186445 by myint last updated on 04/Feb/23 $$\mathrm{Show}\:\:\mathrm{that}\:\:\mathrm{the}\:\:\mathrm{function}\:\:\mathrm{y}\:=\:\:\mid\:\mathrm{x}\:−\mathrm{5}\:\mid\:\:\mathrm{has}\:\:\mathrm{no}\:\:\mathrm{derivative}\:\:\mathrm{at}\:\:\mathrm{x}\:\:=\:\mathrm{5}. \\ $$ Answered by ARUNG_Brandon_MBU last updated on 04/Feb/23 $${f}\:'\left(\mathrm{5}\right)=\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left(\mathrm{5}\right)}{{x}−\mathrm{5}}=\underset{{x}\rightarrow\mathrm{5}} {\mathrm{lim}}\frac{\mid{x}−\mathrm{5}\mid}{{x}−\mathrm{5}} \\ $$$$\underset{{x}\rightarrow\mathrm{5}^{>} }…
Question Number 120593 by snipers237 last updated on 01/Nov/20 $$ \\ $$$$\:\underset{{z}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\Gamma\left({z}\right)+\Gamma\left(−{z}\right)}{\mathrm{2}}\:\overset{?} {=}\:−\gamma \\ $$ Answered by mnjuly1970 last updated on 02/Nov/20 $${solution}: \\…
Question Number 120553 by snipers237 last updated on 01/Nov/20 $$\:\: \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \mid\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\:−{ix}\right)\mid^{\mathrm{2}} {dx}\:=\:\frac{\pi}{\mathrm{2}}\: \\ $$ Answered by mnjuly1970 last updated on 01/Nov/20 $${solution}:\:…
Question Number 120528 by snipers237 last updated on 01/Nov/20 $${Let}\:\:{a}>\mathrm{0}\:,\:{g}\left({x}\right)=\underset{{n}\in\mathbb{Z}} {\sum}{exp}\left(−\frac{\left({x}−{n}\right)^{\mathrm{2}} }{{a}}\right) \\ $$$${Find}\:{the}\:{Fourier}\:{coefficients}\:{of}\:\:\:{g} \\ $$$${and}\:{deduce}\:{that}\:{for}\:{x}\in\mathbb{R}^{\ast} \\ $$$$\:\underset{{n}\in\mathbb{Z}} {\sum}{exp}\left(−\frac{\pi{n}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)=\:{x}\:.\underset{{n}\in\mathbb{Z}} {\sum}{exp}\left(−\pi{n}^{\mathrm{2}} {x}^{\mathrm{2}} \right) \\…
Question Number 186027 by Michaelfaraday last updated on 31/Jan/23 Answered by qaz last updated on 31/Jan/23 $$\frac{{d}\mid{y}\mid}{{dx}}=\begin{vmatrix}{\mathrm{sin}\:{x}}&{\mathrm{sin}\:{x}}\\{−\mathrm{sin}\:{x}}&{−\mathrm{sin}\:{x}}\end{vmatrix}+\begin{vmatrix}{\mathrm{cos}\:{x}}&{\mathrm{cos}\:{x}}\\{\mathrm{cos}\:{x}}&{\mathrm{cos}\:{x}}\end{vmatrix}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 120463 by bramlexs22 last updated on 31/Oct/20 $${Find}\:{the}\:{slope}\:{of}\:{the}\:{line}\:{tangent} \\ $$$${to}\:{the}\:{graph}\:{r}=\mathrm{3cos}\:^{\mathrm{2}} \left(\mathrm{2}\theta\right)\:{at} \\ $$$$\theta=\frac{\pi}{\mathrm{6}}.\: \\ $$ Answered by mr W last updated on 31/Oct/20…
Question Number 120452 by john santu last updated on 31/Oct/20 $${Find}\:{the}\:{equation}\:{of}\:{the}\:{line}\:{tangent}\:{to}\:{the}\:{parametric} \\ $$$${curve}\:{given}\:{by}\:{the}\:{equations}\:\begin{cases}{{x}=\left(\mathrm{1}+{t}^{\mathrm{3}} \right)^{\mathrm{4}} +{t}^{\mathrm{2}} }\\{{y}={t}^{\mathrm{5}} +{t}^{\mathrm{2}} +\mathrm{2}}\end{cases}\:{at}\:{t}=\mathrm{1} \\ $$ Answered by physicstutes last updated…