Question Number 192288 by universe last updated on 14/May/23 Answered by mahdipoor last updated on 14/May/23 $${get}\:\angle{BCG}={a}\:\:\:\:\:,\:{BC}={b} \\ $$$${GD}={GB}+{BD}={BC}.{tana}+{BC}.{tan}\left(\mathrm{90}−{a}\right) \\ $$$$={b}\left({tana}+{cota}\right) \\ $$$${FD}={GD}.{tana}={b}\left({tan}^{\mathrm{2}} {a}+\mathrm{1}\right)={b}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {a}}\right)…
Question Number 126753 by bemath last updated on 24/Dec/20 $$\:\:\int\:\frac{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}+\mathrm{1}}\:{dx}\: \\ $$ Answered by Ar Brandon last updated on 24/Dec/20 $$\mathrm{sinx}+\mathrm{cosx}=\lambda\left(\mathrm{3sinx}+\mathrm{4cosx}+\mathrm{1}\right)+\mu\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{3sinx}+\mathrm{4cosx}+\mathrm{1}\right)\right\}+\gamma \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\lambda\left(\mathrm{3sinx}+\mathrm{4cosx}+\mathrm{1}\right)+\mu\left(\mathrm{3cosx}−\mathrm{4sinx}\right)+\gamma \\ $$$$\begin{cases}{\mathrm{3}\lambda−\mathrm{4}\mu=\mathrm{1}}\\{\mathrm{4}\lambda+\mathrm{3}\mu=\mathrm{1}}\\{\lambda+\gamma=\mathrm{0}}\end{cases}\Rightarrow\begin{cases}{\mathrm{25}\mu=−\mathrm{1}}\\{\mathrm{25}\lambda=\mathrm{7}}\\{\gamma=−\lambda}\end{cases}…
Question Number 61215 by naka3546 last updated on 30/May/19 Answered by ajfour last updated on 30/May/19 $${let}\:{bottom}\:{left}\:{corner}\:{of}\:{square} \\ $$$$\left({i}\:{assume}\:{so},\:{should}\:{be}\:{so}\right)\:{be} \\ $$$${origin}.\:\:{square}\:{side}\:\boldsymbol{{a}}\:,\:{radius}\:\boldsymbol{{r}} \\ $$$${centre}\:{of}\:{circle}\:{C}\left(\boldsymbol{{h}},\boldsymbol{{k}}\right) \\ $$$$\:\:\:{Then}\:{we}\:{have}…
Question Number 192286 by pete last updated on 14/May/23 $$\mathrm{Determine}\:\mathrm{whether}\:\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{x}}\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{is}: \\ $$$$\mathrm{1}.\mathrm{A}\:\mathrm{function} \\ $$$$\mathrm{2}.\:\mathrm{injective} \\ $$$$\mathrm{3}.\:\mathrm{surjective} \\ $$$$\mathrm{4}.\:\mathrm{bijective} \\ $$ Answered by mehdee42 last…
Question Number 126749 by bemath last updated on 24/Dec/20 Commented by MJS_new last updated on 24/Dec/20 $${t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{2}\int\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{10}{t}^{\mathrm{2}} +\mathrm{1}}{dt} \\ $$$$\mathrm{now}\:\mathrm{find}\:\mathrm{the}\:\mathrm{2}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{of}\:\mathrm{the}\:\mathrm{denominator} \\…
Question Number 192281 by uchihayahia last updated on 14/May/23 $$ \\ $$$$ \\ $$$$\:{help}\:{me}\:{proving}\:{this} \\ $$$$\:{f}:\mathbb{R}\rightarrow\mathbb{R}\:{dan}\:{c}\in\mathbb{R} \\ $$$$\:\underset{{x}\rightarrow{c}} {\mathrm{lim}}\:{f}\left({x}\right)={L}\:\Leftrightarrow\:\underset{{c}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{f}\left({x}+{c}\right)={L} \\ $$$$\: \\ $$$$ \\…
Question Number 126746 by bemath last updated on 24/Dec/20 $$\:\:\int\:\frac{{dx}}{\mathrm{tan}\:^{\mathrm{5}} {x}+\mathrm{1}}\:? \\ $$ Commented by MJS_new last updated on 24/Dec/20 $${t}=\mathrm{tan}\:{x}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{5}} +\mathrm{1}\right)}…
Question Number 61211 by alphaprime last updated on 30/May/19 $${Solve}\:{for}\:{x}\:{in}\:{terms}\:{of}\:{a}\: \\ $$$$\sqrt{{a}−\sqrt{{a}+{x}\:}}\:+\:\:\sqrt{{a}+\sqrt{{a}−{x}}}\:=\:\mathrm{2}{x} \\ $$$${Please}\:{sir}\:{i}\:{request}\:{you}\:{to}\:{solve}\:{this}\: \\ $$$${question}\:=\_= \\ $$ Commented by maxmathsup by imad last updated…
Question Number 192280 by Mingma last updated on 14/May/23 Answered by witcher3 last updated on 14/May/23 $$\Omega=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty}…
Question Number 61210 by pooja24 last updated on 30/May/19 $${for}\:{what}\:{value}\:{of}\:\theta,\:\:{e}^{{i}\theta} =\mathrm{0}\:\: \\ $$ Commented by Tony Lin last updated on 30/May/19 $${e}^{{i}\theta} ={cos}\theta+{isin}\theta \\ $$$$\:\:\:\:\:{cos}\theta\in\left[−\mathrm{1},\mathrm{1}\right]\in{R}…