Question Number 207339 by Ghisom last updated on 12/May/24 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left({x}+{a}\right)^{\mathrm{1}/{x}} −{x}^{\mathrm{1}/{x}} }{\left({x}+{b}\right)^{\mathrm{1}/{x}} −{x}^{\mathrm{1}/{x}} }\:=? \\ $$ Answered by sniper237 last updated on 12/May/24 $$\:\frac{{a}}{{b}}\:\:\:{cause}\:\:\overset{{X}=\mathrm{1}/{x}}…
Question Number 207332 by mustafazaheen last updated on 12/May/24 $$\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)×\left(\overset{\rightarrow} {\mathrm{a}}\right)=? \\ $$$$\mathrm{how}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution} \\ $$ Commented by Ghisom last updated on 12/May/24 $$\left(\begin{pmatrix}{{a}_{\mathrm{1}}…
Question Number 207361 by hardmath last updated on 12/May/24 $$\mathrm{y}\:=\:\frac{\mathrm{tg}\boldsymbol{\mathrm{x}}\:\:+\:\:\mathrm{ctg}\boldsymbol{\mathrm{x}}}{\mathrm{8}}\:\:\:\:\:,\:\:\:\:\:\left(\mathrm{0}\:;\:\frac{\pi}{\mathrm{2}}\right) \\ $$$$\mathrm{Find}:\:\:\:\mathrm{min}\left(\mathrm{y}\right)\:=\:? \\ $$ Answered by Berbere last updated on 12/May/24 $${ctg}\left({x}\right)=\frac{\mathrm{1}}{{y}};{y}={tan}\left({x}\right) \\ $$$$\Leftrightarrow{Min}\left(\frac{\mathrm{1}}{\mathrm{8}}\left({y}+\frac{\mathrm{1}}{{y}}\right);{y}\in\right]\mathrm{0},\infty\left[\right) \\…
Question Number 207362 by mr W last updated on 12/May/24 Answered by A5T last updated on 12/May/24 $${Ptolemy}'{s}\:{theorem}:\:{AC}×{BP}={AP}×{BC}+{AB}×{PC} \\ $$$${AB}={AC}={BC}\Rightarrow{PB}={PA}+{PC} \\ $$ Answered by sniper237…
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…