Question Number 7988 by Tinku Tara last updated on 26/Sep/16 $$\boldsymbol{\mathrm{Feedback}}\:\boldsymbol{\mathrm{request}} \\ $$$$\bullet\:\mathrm{Any}\:\mathrm{symbols}\:\mathrm{that}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{be} \\ $$$$\:\:\:\:\mathrm{added}.\:\mathrm{We}\:\mathrm{are}\:\mathrm{aware}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following} \\ $$$$\:\:\:\:\mathrm{which}\:\mathrm{will}\:\mathrm{be}\:\mathrm{added}\:\mathrm{in}\:\mathrm{next}\:\mathrm{update}: \\ $$$$\:\:\:\:-\:\mathrm{vertical}\:\mathrm{dots}\: \\ $$$$\:\:\:\:-\:\mathrm{contour}\:\mathrm{integral} \\ $$$$\:\:\:\:-\:\mathrm{is}\:\mathrm{congurant}\:\mathrm{to} \\ $$$$\mathrm{Please}\:\mathrm{let}\:\mathrm{us}\:\mathrm{know}\:\mathrm{any}\:\mathrm{symbols}…
Question Number 139057 by EnterUsername last updated on 21/Apr/21 $$\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 $$\mid{z}−\mathrm{2}\mid={x}\geqslant\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\frac{{x}+\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{1}\right)}}{\mathrm{ln}\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}<\mathrm{2}\:\Leftrightarrow\:\mathrm{ln}\:\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\:>\mathrm{ln}\:\mathrm{3}…
Question Number 139056 by sahnaz last updated on 21/Apr/21 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{10}×+\mathrm{6}\right)^{\mathrm{x}−\mathrm{2}} >\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 73523 by 01 last updated on 13/Nov/19 $$\mathrm{prove}\:\mathrm{that}\::\: \\ $$$$\:\mathrm{2}^{\pi} >\mathrm{8} \\ $$ Answered by MJS last updated on 13/Nov/19 $$\mathrm{3}<\pi\:\Rightarrow\:\pi=\mathrm{3}+{p};\:{p}>\mathrm{0} \\ $$$$\mathrm{2}^{\mathrm{3}+{p}}…
Question Number 139059 by sahnaz last updated on 21/Apr/21 Answered by mr W last updated on 21/Apr/21 $$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{2}×\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{n}} +\mathrm{9}×\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} +\mathrm{64}}{\left(\frac{\mathrm{2}}{\mathrm{4}}\right)^{{n}} +\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{{n}} +\mathrm{1}} \\ $$$$=\frac{\mathrm{2}×\mathrm{0}+\mathrm{9}×\mathrm{0}+\mathrm{64}}{\mathrm{0}+\mathrm{0}+\mathrm{1}}…
Question Number 139053 by mathocean1 last updated on 21/Apr/21 Commented by mathocean1 last updated on 21/Apr/21 $${Determinate}\:{R} \\ $$ Answered by mr W last updated…
Question Number 73518 by aliesam last updated on 13/Nov/19 Commented by mathmax by abdo last updated on 13/Nov/19 $${let}\:{A}\left({x}\right)=\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)^{\mathrm{2}{x}−\mathrm{1}} \:\Rightarrow{A}\left({x}\right)={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\frac{{x}+\mathrm{1}}{{x}−\mathrm{1}}\right)} \\ $$$$={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\frac{{x}−\mathrm{1}+\mathrm{2}}{{x}−\mathrm{1}}\right)} \:={e}^{\left(\mathrm{2}{x}−\mathrm{1}\right){ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}}\right)} \:\:{we}\:{have}\:{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{x}−\mathrm{1}}\right)\sim\frac{\mathrm{2}}{{x}−\mathrm{1}}\left({x}\rightarrow+\infty\right) \\…
Question Number 139052 by EnterUsername last updated on 21/Apr/21 $$\left(\mathrm{1}+\mathrm{z}\right)^{\mathrm{n}} =\left(\mathrm{1}−\mathrm{z}\right)^{\mathrm{n}} \\ $$$${where}\:{z}\:{is}\:{a}\:{complex}\:{number} \\ $$ Answered by mathmax by abdo last updated on 21/Apr/21 $$\mathrm{z}=−\mathrm{1}\:\mathrm{is}\:\mathrm{not}\:\mathrm{solution}\:\:\mathrm{let}\:\mathrm{z}\neq−\mathrm{1}…
Question Number 139055 by EnterUsername last updated on 21/Apr/21 $$\mathrm{Let}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{be}\:\mathrm{complex}\:\mathrm{numbers}\:\mathrm{representing}\:\mathrm{the}\:\mathrm{points} \\ $$$$\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{respectively}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}. \\ $$$$\mathrm{If}\:\frac{\mathrm{a}}{\mathrm{b}}+\frac{\mathrm{b}}{\mathrm{a}}=\mathrm{1}\:\mathrm{and}\:\mathrm{O}\:\mathrm{is}\:\mathrm{the}\:\mathrm{origin}.\:\mathrm{Then}\:\Delta\mathrm{OAB}\:\mathrm{is}\:? \\ $$ Answered by MJS_new last updated on 22/Apr/21 $$\frac{{a}}{{b}}+\frac{{b}}{{a}}=\mathrm{1}\:\Rightarrow\:{b}={a}\left(\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right) \\…
Question Number 139051 by mathsuji last updated on 21/Apr/21 $${a};{b};{c}\in\mathbb{N} \\ $$$$\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)=\mathrm{105}^{\boldsymbol{{c}}} \\ $$ Commented by Rasheed.Sindhi last updated on 22/Apr/21 $$\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)=\mathrm{105}^{\boldsymbol{{c}}} \\ $$$$\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)\left({a}+\mathrm{3}{b}\right)=\mathrm{3}^{{c}} .\mathrm{5}^{{c}}…