Question Number 175829 by pete last updated on 07/Sep/22 $$\mathrm{The}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{Geometric}\:\mathrm{Progresion} \\ $$$$\left(\mathrm{G}.\mathrm{P}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\:\mathrm{8}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{Arithmetic} \\ $$$$\mathrm{Progresion}\:\left(\mathrm{A}.\mathrm{P}\right).\:\mathrm{The}\:\mathrm{first}\:\mathrm{terms},\:\mathrm{common} \\ $$$$\mathrm{difference}\:\mathrm{and}\:\mathrm{common}\:\mathrm{ratio}\:\mathrm{are}\:\mathrm{all}\:\mathrm{equal} \\ $$$$\mathrm{and}\:\mathrm{non}−\mathrm{zero}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{five} \\ $$$$\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{Geometric}\:\mathrm{Progresion}\left(\mathrm{G}.\mathrm{P}\right) \\ $$ Answered…
Question Number 175815 by Shrinava last updated on 07/Sep/22 $$\mathrm{If}\:\:\:\mathrm{n}\geqslant\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\mathrm{Then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\boldsymbol{\mathrm{n}}} \:\:+\:\:\mathrm{b}^{\boldsymbol{\mathrm{n}}} }{\mathrm{a}\:\:+\:\:\mathrm{b}}\:\:\geqslant\:\:\frac{\mathrm{a}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \:\:+\:\:\mathrm{b}^{\boldsymbol{\mathrm{n}}−\mathrm{1}} }{\mathrm{2}} \\ $$ Answered by mahdipoor…
Question Number 110268 by Khanacademy last updated on 28/Aug/20 Commented by som(math1967) last updated on 28/Aug/20 $$−\mathrm{2} \\ $$ Commented by Khanacademy last updated on…
Question Number 110269 by Khanacademy last updated on 28/Aug/20 Commented by bemath last updated on 28/Aug/20 $$−\frac{\mathrm{1}}{\mathrm{2}} \\ $$ Commented by Khanacademy last updated on…
Question Number 110265 by Khanacademy last updated on 28/Aug/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 44729 by behi83417@gmail.com last updated on 03/Oct/18 Answered by MJS last updated on 04/Oct/18 $$\mathrm{trivial}\:\mathrm{solutions}\:{a}={b}={c}=\mathrm{0}\:\vee\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${c}^{\mathrm{2}} +{bc}+{b}^{\mathrm{2}} −{a}=\mathrm{0} \\ $$$${c}^{\mathrm{2}}…
Question Number 175793 by infinityaction last updated on 07/Sep/22 $$\:\:\mathrm{if}\:\mathrm{xy}+\mathrm{y}^{\mathrm{2}} +\mathrm{zx}\:=\:\mathrm{48};\:\mathrm{where}\:\mathrm{x},\mathrm{y},\mathrm{z} \\ $$$$\:\:\mathrm{are}\:\mathrm{three}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers} \\ $$$$\:\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{possible} \\ $$$$\:\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{product}\:\left(\mathrm{xyz}\right) \\ $$ Commented by LordKazuma last updated on…
Question Number 110246 by Her_Majesty last updated on 28/Aug/20 $${solve}\:{for}\:{z}\in\mathbb{C}:\:\left({a}+{bi}\right)^{{z}} ={b}+{ai} \\ $$ Answered by mr W last updated on 28/Aug/20 $$\theta=\mathrm{tan}^{−\mathrm{1}} \frac{{b}}{{a}} \\ $$$${z}\mathrm{ln}\:\left({a}+{bi}\right)=\mathrm{ln}\:\left({b}+{ai}\right)…
Question Number 175770 by mnjuly1970 last updated on 06/Sep/22 $$ \\ $$$$\:\:\:\mathrm{P}_{{n}} \:=\:{e}^{\:\left(\frac{\mathrm{1}}{\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{2}}\right)\:+\left(\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{4}}\right)\:+…+\left(\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{{n}}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\:{e}^{\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{4}}\:\:+…+\frac{\mathrm{1}}{{n}−\mathrm{1}}\:−\frac{\mathrm{1}}{{n}}\right)} \\ $$$$\:\:\:\:\:\:\:\:=\:{e}^{\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\:\left(−\mathrm{1}\:\right)^{\:{k}+\mathrm{1}} }{{k}}} \\ $$$$\:\:\:\:\:\:\:\:\:\therefore\:\:\mathrm{P}\:=\:{lim}_{\:{n}\rightarrow\infty} \left({e}^{\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{\:{k}+\mathrm{1}}…
Question Number 44691 by ajfour last updated on 03/Oct/18 Commented by ajfour last updated on 03/Oct/18 $${Find}\:{equation}\:{of}\:{the}\:{biquadratic} \\ $$$${curve}. \\ $$ Answered by ajfour last…