Question Number 196915 by sonukgindia last updated on 03/Sep/23 Commented by mr W last updated on 23/Sep/23 $${see}\:{Q}\mathrm{197578} \\ $$ Terms of Service Privacy Policy…
Question Number 196893 by hardmath last updated on 02/Sep/23 Answered by Rasheed.Sindhi last updated on 02/Sep/23 $${x}=\sqrt{\left(−\mathrm{1}+{i}\right)+\sqrt{\left(−\mathrm{1}+{i}\right)+\sqrt{…}}}\: \\ $$$${x}=\sqrt{\left(−\mathrm{1}+{i}\right)+{x}}\: \\ $$$${x}^{\mathrm{2}} =\:\left(−\mathrm{1}+{i}\right)+{x} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}−{i}=\mathrm{0}…
Question Number 196885 by hardmath last updated on 02/Sep/23 Answered by MM42 last updated on 02/Sep/23 $$\mathrm{2}{cosx}=\sqrt{\mathrm{2}+\mathrm{2}{cos}\mathrm{2}{x}}=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{cos}\mathrm{4}{x}}} \\ $$$$=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+{cos}\mathrm{8}{x}}}}=….=\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}} \\ $$$${let}\:\:{x}=\mathrm{0}\Rightarrow\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+…}}}=\mathrm{2}\:\checkmark \\ $$$$ \\ $$…
Question Number 196886 by Amidip last updated on 02/Sep/23 Answered by MM42 last updated on 02/Sep/23 $${tan}\left(\alpha+\beta\right)=\frac{{tan}\alpha+{tan}\beta}{\mathrm{1}−{tan}\alpha{tan}\beta}=\frac{{p}}{{q}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{p}}{{q}−\mathrm{1}}=\frac{{sin}^{\mathrm{2}} \left(\alpha+\beta\right)}{{sin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)} \\ $$$$\left.\Rightarrow{psin}\left(\alpha+\beta\right){cos}\left(\alpha+\beta\right)=\left({q}−\mathrm{1}\right){sin}^{\mathrm{2}} \left(\alpha+\beta\right)\right) \\ $$$$\Rightarrow{sin}^{\mathrm{2}}…
Question Number 196880 by subheenoy last updated on 02/Sep/23 $$\mathrm{1}.\:{A}\:{container}\:{of}\:{milk}\:{is}\:\frac{\mathrm{4}}{\mathrm{5}}\:{full}.\:{When}\:\mathrm{10}{L}\:{of}\:{milk}\:{is}\:{poured}\:{into}\:{it}\:{the}\:{container}\:{becomes}\:\:\frac{\mathrm{9}}{\mathrm{10}}\:{full}.\:{What}\:{is}\:{the}\:{capacity}\:{of}\:{the}\:{container}? \\ $$ Answered by Frix last updated on 02/Sep/23 $$\frac{\mathrm{4}}{\mathrm{5}}{x}+\mathrm{10}=\frac{\mathrm{9}}{\mathrm{10}}{x} \\ $$$$\mathrm{10}=\left(\frac{\mathrm{9}}{\mathrm{10}}−\frac{\mathrm{4}}{\mathrm{5}}\right){x}=\left(\frac{\mathrm{9}−\mathrm{8}}{\mathrm{10}}\right){x}=\frac{\mathrm{1}}{\mathrm{10}}{x} \\ $$$${x}=\mathrm{100} \\…
Question Number 196913 by ERLY last updated on 02/Sep/23 $${soit}\:\left\{_{{r}_{{n}+\mathrm{1}} ={r}_{{n}} /\left(\mathrm{2}+{r}_{{n}} ^{\mathrm{2}} \right)} ^{{r}_{\mathrm{0}} =\mathrm{1}} \right. \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}} >\mathrm{0} \\ $$$${demontrer}\:{par}\:{recurrence}\:{que}\:{r}_{{n}+\mathrm{1}} \leq\frac{\mathrm{1}}{\mathrm{2}}{r}_{{n}} \\ $$$${demontrer}\:{sans}\:{recurrence}\:{que}\:{r}_{{n}}…
Question Number 196881 by HeferH last updated on 02/Sep/23 Commented by MathematicalUser2357 last updated on 10/Sep/23 $$\ast\mathrm{Tears}\:\mathrm{paper}\ast \\ $$$$\mathrm{Now}\:\mathrm{there}\:\mathrm{is}\:\mathrm{no}\:\mathrm{question} \\ $$ Answered by HeferH last…
Question Number 196883 by Amidip last updated on 02/Sep/23 Answered by som(math1967) last updated on 02/Sep/23 $$\:\:\mathrm{sin}^{−\mathrm{1}} {x}+\mathrm{sin}^{−\mathrm{1}} {y}=\pi−\mathrm{sin}^{−\mathrm{1}} {z} \\ $$$$\Rightarrow{sin}^{−\mathrm{1}} \left({x}\sqrt{\mathrm{1}−{y}^{\mathrm{2}} }+{y}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)=\mathrm{sin}^{−\mathrm{1}}…
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Question Number 196872 by York12 last updated on 02/Sep/23 $${Let}\:\xi\:{be}\:{a}\:{positive}\:{Root}\:{of}\:{x}^{\mathrm{2}} −\mathrm{2023}{x}−\mathrm{1} \\ $$$${Define}\:{a}\:{sequence}\:\varphi_{{i}} \:{such}\:{That}\:\varphi_{\mathrm{0}} =\mathrm{1} \\ $$$$\varphi_{{n}+\mathrm{1}} =\lfloor\varphi_{{n}} \xi\rfloor,\:{find}\:{The}\:{Remainder}\:{When}\:\varphi_{\mathrm{2023}\:} {is}\:{divided}\:{by}\:\sqrt{\varphi_{\mathrm{2}} } \\ $$ Answered by…