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Author: Tinku Tara

S-1-n-1-2n-2-1-4n-4-1-8n-8-S-1-n-i-1-1-2-i-n-2-i-Solvable-

Question Number 7999 by FilupSmith last updated on 27/Sep/16 $${S}=\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{\mathrm{2}{n}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{4}} }+\frac{\mathrm{1}}{\mathrm{8}{n}^{\mathrm{8}} }+… \\ $$$${S}=\frac{\mathrm{1}}{{n}}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}^{{i}} {n}^{\mathrm{2}^{{i}} } } \\ $$$$\mathrm{Solvable}? \\ $$ Answered…

given-the-3-rd-degree-polynomial-P-x-2x-1-x-3-Q-x-12x-8-given-that-x-1-is-a-factor-of-P-x-and-P-0-10-find-Q-x-

Question Number 73530 by Rio Michael last updated on 13/Nov/19 $${given}\:{the}\:\mathrm{3}^{{rd}} \:{degree}\:\:{polynomial} \\ $$$${P}\left({x}\right)\:=\:\left(\mathrm{2}{x}\:−\mathrm{1}\right)\left({x}−\mathrm{3}\right){Q}\left({x}\right)\:+\:\mathrm{12}{x}−\mathrm{8} \\ $$$${given}\:{that}\:\left({x}−\mathrm{1}\right)\:{is}\:{a}\:{factor}\:{of}\:{P}\left({x}\right)\:{and}\:\:{P}\left(\mathrm{0}\right)\:=\:\mathrm{10} \\ $$$${find}\:{Q}\left({x}\right) \\ $$ Answered by MJS last updated…

Question-73525

Question Number 73525 by arkanmath7@gmail.com last updated on 13/Nov/19 Commented by mathmax by abdo last updated on 13/Nov/19 $${z}^{\mathrm{2}} +\left(\mathrm{1}−{i}\right){z}−\mathrm{3}{i}\:=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{1}−{i}\right)^{\mathrm{2}} −\mathrm{4}\left(−\mathrm{3}{i}\right)\:=\mathrm{1}−\mathrm{2}{i}−\mathrm{1}+\mathrm{12}{i}\:=\mathrm{10}{i} \\ $$$${z}_{\mathrm{1}}…

Feedback-request-Any-symbols-that-needs-to-be-added-We-are-aware-of-the-following-which-will-be-added-in-next-update-vertical-dots-contour-integral-is-congurant-to-Pleas

Question Number 7988 by Tinku Tara last updated on 26/Sep/16 $$\boldsymbol{\mathrm{Feedback}}\:\boldsymbol{\mathrm{request}} \\ $$$$\bullet\:\mathrm{Any}\:\mathrm{symbols}\:\mathrm{that}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{be} \\ $$$$\:\:\:\:\mathrm{added}.\:\mathrm{We}\:\mathrm{are}\:\mathrm{aware}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\:\:\:\:\mathrm{which}\:\mathrm{will}\:\mathrm{be}\:\mathrm{added}\:\mathrm{in}\:\mathrm{next}\:\mathrm{update}: \\ $$$$\:\:\:\:-\:\mathrm{vertical}\:\mathrm{dots}\: \\ $$$$\:\:\:\:-\:\mathrm{contour}\:\mathrm{integral} \\ $$$$\:\:\:\:-\:\mathrm{is}\:\mathrm{congurant}\:\mathrm{to} \\ $$$$\mathrm{Please}\:\mathrm{let}\:\mathrm{us}\:\mathrm{know}\:\mathrm{any}\:\mathrm{symbols}…

The-area-of-the-region-in-the-complex-plane-satisfying-the-inequality-log-cos-pi-6-z-2-5-4-z-2-4-lt-2-is-

Question Number 139057 by EnterUsername last updated on 21/Apr/21 $$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}\:\mathrm{satisfying} \\ $$$$\mathrm{the}\:\mathrm{inequality}\:\mathrm{log}_{\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)} \left[\frac{\mid\mathrm{z}−\mathrm{2}\mid+\mathrm{5}}{\mathrm{4}\mid\mathrm{z}−\mathrm{2}\mid−\mathrm{4}}\right]<\mathrm{2}\:\mathrm{is}\:? \\ $$ Answered by MJS_new last updated on 22/Apr/21 $$\mid{z}−\mathrm{2}\mid={x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\frac{{x}+\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{1}\right)}}{\mathrm{ln}\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}<\mathrm{2}\:\Leftrightarrow\:\mathrm{ln}\:\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\:>\mathrm{ln}\:\mathrm{3}…