Question Number 207312 by Wuji last updated on 11/May/24 $$\mathrm{obtain}\:\mathrm{the}\:\mathrm{state}\:\mathrm{model}\:\mathrm{of}\:\mathrm{the}\:\mathrm{system} \\ $$$$\mathrm{whose}\:\mathrm{transfer}\:\mathrm{function}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by} \\ $$$$\frac{\mathrm{Y}\left(\mathrm{s}\right)}{\mathrm{U}\left(\mathrm{s}\right)}=\frac{\mathrm{s}}{\mathrm{15s}^{\mathrm{3}} +\mathrm{26s}+\mathrm{36}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207313 by Wuji last updated on 11/May/24 $$\mathrm{find}\:\mathrm{the}\:\mathrm{transfer}\:\mathrm{function}\:\mathrm{of}\:\mathrm{the}\:\mathrm{state} \\ $$$$\mathrm{model}\:\mathrm{of}\:\mathrm{the}\:\mathrm{system}\:\mathrm{given}\:\mathrm{by} \\ $$$$\overset{\bullet} {\mathrm{x}}=\begin{vmatrix}{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\mathrm{1}\:\:−\mathrm{2}\:\:\:−\mathrm{3}}\end{vmatrix}\mathrm{x}+\begin{vmatrix}{\mathrm{0}\:\:\:\mathrm{0}}\\{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{vmatrix} \\ $$$$\mathrm{and}\:\begin{vmatrix}{\mathrm{y}_{\mathrm{1}} }\\{\mathrm{y}_{\mathrm{2}} }\end{vmatrix}=\begin{vmatrix}{\mathrm{1}\:\:\mathrm{0}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{0}\:\:\mathrm{1}}\end{vmatrix}\mathrm{x} \\ $$ Terms of Service Privacy…
Question Number 207314 by universe last updated on 11/May/24 $$\:\mathrm{let}\:\mathrm{f}:\mathbb{R}\rightarrow\mathbb{R}\:\mathrm{be}\:\mathrm{a}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{then} \\ $$$$\:\mathrm{show}\:\mathrm{that} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{x}^{\mathrm{2}} \right)\:\forall\:\mathrm{x}\:\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\:\:\mathrm{function} \\ $$$$\:\left(\mathrm{2}\right)\:\mathrm{if}\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{f}\left(\mathrm{2x}+\mathrm{1}\right)\:\forall\mathrm{x}\in\mathbb{R}\:\mathrm{then}\:\mathrm{f}\:\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\:\:\mathrm{constant}\:\mathrm{function}\: \\ $$ Answered by…
Question Number 207315 by galva2000 last updated on 11/May/24 $${if}\:{ab}+{ac}+{bc}=\mathrm{2}\: \\ $$$${calculate}\:{minimum}\:{of}\:\mathrm{10}{a}^{\mathrm{2}} +\mathrm{10}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$ Answered by Berbere last updated on 11/May/24 $${S}=\mathrm{10}{a}^{\mathrm{2}} +\mathrm{10}{b}^{\mathrm{2}}…
Question Number 207328 by hardmath last updated on 11/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{sin}\:\mathrm{50}°\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:\:=\:\:? \\ $$ Answered by som(math1967) last updated on 12/May/24 $$\:\frac{\mathrm{4}{sin}\mathrm{50}{cos}\mathrm{20}−\mathrm{1}}{{cos}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left({sin}\mathrm{70}+{sin}\mathrm{30}\right)−\mathrm{1}}{{sin}\left(\mathrm{90}−\mathrm{20}\right)} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{70}+\mathrm{1}−\mathrm{1}}{{sin}\mathrm{70}}=\mathrm{2} \\…
Question Number 207330 by hardmath last updated on 11/May/24 $$\mathrm{Find}:\:\:\:\mathrm{4}\:\mathrm{cos}\:\mathrm{50}°\:\:+\:\:\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{20}°}\:\:=\:\:? \\ $$ Answered by som(math1967) last updated on 12/May/24 $$\:\frac{\mathrm{4}{sin}\mathrm{20}{cos}\mathrm{50}+\mathrm{1}}{{sin}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}\left({sin}\mathrm{70}−{sin}\mathrm{30}\right)+\mathrm{1}}{{sin}\mathrm{20}} \\ $$$$=\frac{\mathrm{2}{sin}\mathrm{70}−\mathrm{1}+\mathrm{1}}{{cos}\left(\mathrm{90}−\mathrm{20}\right)} \\…
Question Number 207292 by mr W last updated on 10/May/24 Commented by mr W last updated on 10/May/24 $${find}\:{the}\:{side}\:{length}\:{of}\:{the}\:{regular} \\ $$$${hexagon}. \\ $$ Answered by…
Question Number 207230 by mr W last updated on 10/May/24 Answered by mr W last updated on 10/May/24 Commented by mr W last updated on…
Question Number 207231 by efronzo1 last updated on 10/May/24 Answered by Sutrisno last updated on 11/May/24 $$\bullet{f}\left({x}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}{x}+\mathrm{1}\:…\left(\mathrm{1}\right) \\ $$$$\bullet{x}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$$\:\:{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)+{f}\left(\frac{−\mathrm{1}}{{x}−\mathrm{1}}\right)=\mathrm{2}−\frac{\mathrm{1}}{{x}}\:…\left(\mathrm{2}\right) \\ $$$$\bullet{x}=\frac{−\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${f}\left(\frac{−\mathrm{1}}{{x}−\mathrm{1}}\right)+{f}\left({x}\right)=\frac{−\mathrm{1}}{{x}−\mathrm{1}}+\mathrm{1}\:…\left(\mathrm{3}\right)…
Question Number 207225 by mdrashid last updated on 10/May/24 $$\mathrm{6}×\mathrm{5}=\mathrm{30} \\ $$ Commented by Frix last updated on 10/May/24 $$\mathrm{Aiming}\:\mathrm{at}\:\mathrm{the}\:\mathrm{Nobel}\:\mathrm{Prize}? \\ $$ Terms of Service…