Question Number 207326 by hardmath last updated on 11/May/24 $$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°\:−\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\:? \\ $$$$\mathrm{2}.\:\:\:\mathrm{4}\:\mathrm{cos}^{\mathrm{2}} \:\mathrm{40}°\:+\:\:\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{20}°}\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 207327 by hardmath last updated on 11/May/24 $$\left(\mathrm{x}−\mathrm{3}\right)\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}−\mathrm{2}}\:\:=\:\:\mathrm{0} \\ $$$$\mathrm{Find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$ Commented by A5T last updated on 11/May/24 $${x}=\mathrm{3}\:{or}\:\mathrm{2}\:{or}\:−\mathrm{1} \\ $$…
Question Number 207320 by BaliramKumar last updated on 11/May/24 Commented by A5T last updated on 11/May/24 $$\mathrm{19}.\:{The}\:{remainder}\:{when}\:{a}\:{number}\:{is}\:{divided}\:{by}\:\mathrm{16} \\ $$$${is}\:{the}\:{same}\:{as}\:{the}\:{remainder}\:{when}\:{its}\:{last} \\ $$$${four}\:{digits}\:{are}\:{divided}\:{by}\:\mathrm{16} \\ $$$$\mathrm{9100}\equiv\mathrm{12}\left({mod}\:\mathrm{16}\right)\:\Rightarrow\left({b}\right) \\ $$…
Question Number 207317 by mr W last updated on 11/May/24 $${solve}\:{for}\:{y} \\ $$$$\frac{\mathrm{1}}{{y}'}+\frac{\mathrm{1}}{{y}''}=\mathrm{1} \\ $$ Answered by Berbere last updated on 11/May/24 $${y}'={z} \\ $$$$\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}'}=\mathrm{1}\Rightarrow{z}'=\frac{{z}}{{z}−\mathrm{1}}\Rightarrow\left(\frac{{z}−\mathrm{1}}{{z}}\right){dz}={dx}…
Question Number 207312 by Wuji last updated on 11/May/24 $$\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)} \\…