Question Number 177812 by Spillover last updated on 09/Oct/22 Answered by JDamian last updated on 09/Oct/22 $$\mathrm{cosh}\:\left(\mathrm{ln}\:{x}\right)=\frac{{e}^{\mathrm{ln}\:{x}} +{e}^{−\mathrm{ln}\:{x}} }{\mathrm{2}}=\frac{{x}+\frac{\mathrm{1}}{{x}}}{\mathrm{2}}=\frac{{x}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{x}}\:={A} \\ $$$$\mathrm{sinh}\:\:\left(\mathrm{ln}\:{x}\right)=\frac{{e}^{\mathrm{ln}\:{x}} −{e}^{−\mathrm{ln}\:{x}} }{\mathrm{2}}=\frac{{x}−\frac{\mathrm{1}}{{x}}}{\mathrm{2}}=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{x}}={B}…
Question Number 177815 by infinityaction last updated on 09/Oct/22 Answered by mahdipoor last updated on 09/Oct/22 $$\Rightarrow\begin{cases}{{xy}+{yy}+{zy}=\mathrm{2}{y}}\\{{xx}+{yx}+{zx}=\mathrm{2}{x}}\\{{xy}+{yz}+{zx}=\mathrm{1}}\end{cases}\Rightarrow{i}+{ii}−{iii}\Rightarrow \\ $$$${xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}\left({y}+{x}\right)−\mathrm{1}\Rightarrow \\ $$$${g}\left({x},{y}\right)={xy}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}\left({x}+{y}\right)=−\mathrm{1}\Rightarrow…
Question Number 46738 by Umar last updated on 30/Oct/18 $${please}\:{help} \\ $$$$ \\ $$$${Write}\:{an}\:{algorithm}\:{that}\:{will}\:{find} \\ $$$${the}\:{solution}\:{of}\:{the}\:{equation} \\ $$$$\:\:\:\:\:\:\:\:\:\:{f}\left({x}\right)=\begin{cases}{−{x},\:{when}\:{x}<\mathrm{0}}\\{{x},\:\:\:\:\:{when}\:{x}\geqslant\mathrm{0}}\end{cases} \\ $$ Commented by hassentimol last updated…
Question Number 177811 by Spillover last updated on 09/Oct/22 Answered by CElcedricjunior last updated on 09/Oct/22 $$\boldsymbol{\mathrm{Tan}}\:\boldsymbol{\mathrm{x}}=\boldsymbol{\mathrm{sinh}\theta}=\frac{\boldsymbol{{e}}^{\boldsymbol{\theta}} −\boldsymbol{{e}}^{−\boldsymbol{\theta}} }{\mathrm{2}} \\ $$$$=>\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\theta}} −\mathrm{2}\boldsymbol{{e}}^{\boldsymbol{\theta}} \boldsymbol{\mathrm{tanx}}−\mathrm{1}=\mathrm{0} \\ $$$$\boldsymbol{\Delta}=\mathrm{4}\boldsymbol{\mathrm{tan}}^{\mathrm{2}}…
Question Number 177810 by Spillover last updated on 09/Oct/22 Answered by Beginner last updated on 09/Oct/22 Commented by Spillover last updated on 09/Oct/22 $$\mathrm{thanks} \\…
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Question Number 46713 by agniv last updated on 30/Oct/18 $$\frac{{x}−\mathrm{1}}{{x}−\mathrm{2}}−\frac{{x}−\mathrm{2}}{{x}−\mathrm{3}}=\frac{{x}−\mathrm{5}}{{x}−\mathrm{6}}−\frac{{x}−\mathrm{6}}{{x}−\mathrm{7}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{for}}\:\boldsymbol{{x}} \\ $$ Answered by Kunal12588 last updated on 30/Oct/18 $$\frac{\left({x}−\mathrm{1}\right)\left({x}−\mathrm{3}\right)−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)}=\frac{\left({x}−\mathrm{5}\right)\left({x}−\mathrm{7}\right)−\left({x}−\mathrm{6}\right)^{\mathrm{2}} }{\left({x}−\mathrm{6}\right)\left({x}−\mathrm{7}\right)} \\…
Question Number 46681 by Aditya789 last updated on 30/Oct/18 Commented by math1967 last updated on 30/Oct/18 $${is}\:{it}\:{solve}? \\ $$ Commented by Aditya789 last updated on…
Question Number 46680 by Sanjarbek last updated on 30/Oct/18 Answered by MJS last updated on 30/Oct/18 $$\mathrm{we}\:\mathrm{had}\:\mathrm{this}\:\mathrm{before}.\:\mathrm{infinite}\:\mathrm{continued}\:\mathrm{fractions} \\ $$$$\mathrm{without}\:\mathrm{cyclic}\:\mathrm{patterns}\:\mathrm{represent}\:\mathrm{transcendental} \\ $$$$\mathrm{numbers}.\:\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$ Terms of…
Question Number 177741 by infinityaction last updated on 08/Oct/22 Answered by Frix last updated on 08/Oct/22 $$\mathrm{let}\:{p}={a}_{{n}} \wedge{q}={a}_{{n}+\mathrm{1}} \\ $$$${q}^{\mathrm{2}} −\mathrm{2}{pq}−{p}=\mathrm{0}\:\Rightarrow\:{q}={p}+\sqrt{{p}\left({p}+\mathrm{1}\right)} \\ $$$$\Rightarrow\:{q}>{p}\:\Leftrightarrow\:{a}_{{n}+\mathrm{1}} >{a}_{{n}} \\…