Question Number 165461 by mathocean1 last updated on 01/Feb/22 $${show}\:{that}\:{argth}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\right) \\ $$$$\left.{for}\:{x}\:\in\:\right]−\mathrm{1};\mathrm{1}\left[\right. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 34365 by mondodotto@gmail.com last updated on 05/May/18 $$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{a}}>\boldsymbol{\mathrm{b}}>\mathrm{0}\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}} \\ $$$$\boldsymbol{\mathrm{b}}<\frac{\boldsymbol{\mathrm{ae}}^{\boldsymbol{{x}}} +\boldsymbol{\mathrm{be}}^{−\boldsymbol{{x}}} }{\boldsymbol{\mathrm{e}}^{\boldsymbol{{x}}} +\boldsymbol{\mathrm{e}}^{−\boldsymbol{{x}}} }<\boldsymbol{\mathrm{a}} \\ $$ Answered by MJS last updated on 05/May/18…
Question Number 34361 by mondodotto@gmail.com last updated on 05/May/18 $$\boldsymbol{\mathrm{A}}\:\boldsymbol{\mathrm{right}}\:\boldsymbol{\mathrm{prism}}\:\boldsymbol{\mathrm{has}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{regular}} \\ $$$$\boldsymbol{\mathrm{hexagonal}}\:\boldsymbol{\mathrm{base}}\:\boldsymbol{\mathrm{with}}\:\boldsymbol{\mathrm{sides}}\:\boldsymbol{\mathrm{of}} \\ $$$$\boldsymbol{\mathrm{length}}\:\mathrm{15}\boldsymbol{\mathrm{cm}},\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{height}}\: \\ $$$$\boldsymbol{\mathrm{of}}\:\mathrm{20}\boldsymbol{\mathrm{cm}}.\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{its}}\:\boldsymbol{\mathrm{volume}}\:\boldsymbol{\mathrm{and}} \\ $$$$\boldsymbol{\mathrm{total}}\:\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{area}}. \\ $$ Answered by MJS last updated…
Question Number 99894 by bachamohamed last updated on 23/Jun/20 $$\:\:\: \\ $$$$\underset{\boldsymbol{{n}}\geqslant\mathrm{0}} {\sum}\:\:\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} +\mathrm{1}}\:=\:? \\ $$ Answered by smridha last updated on 24/Jun/20 $$\boldsymbol{{s}}\left(\boldsymbol{{a}}\right)=\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 99895 by bachamohamed last updated on 23/Jun/20 $$\:\:\:\:\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\left(\mathrm{3}\boldsymbol{{n}}+\mathrm{1}\right)^{\mathrm{3}} }=\frac{\mathrm{13}}{\mathrm{27}}\:\boldsymbol{\zeta}\left(\mathrm{3}\right)\:+\frac{\mathrm{2}\boldsymbol{\pi}^{\mathrm{3}} }{\mathrm{81}\sqrt{\mathrm{3}}}\: \\ $$ Answered by maths mind last updated on 24/Jun/20 $$\pi{cot}\left(\pi{x}\right)=\frac{\mathrm{1}}{{x}}+\underset{{n}\geqslant\mathrm{1}}…
Question Number 165424 by SANOGO last updated on 01/Feb/22 Commented by alephzero last updated on 01/Feb/22 $$\mathrm{Please},\:\mathrm{translate}\:\mathrm{into}\:\mathrm{English} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 165422 by SANOGO last updated on 01/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 99887 by mr W last updated on 23/Jun/20 Commented by mr W last updated on 23/Jun/20 $${Additional}\:{question}: \\ $$$${Find}\:{the}\:{minimum}\:{initial}\:{velocity}\:{v} \\ $$$${such}\:{that}\:{the}\:{small}\:{block}\:{can}\:{reach} \\ $$$${the}\:{right}\:{end}\:{of}\:{the}\:{bigger}\:{block}.…
Question Number 165404 by MWSuSon last updated on 01/Feb/22 $$\mathrm{given}\:\mathrm{a}\:\mathrm{loan}\:\mathrm{from}\:\mathrm{a}\:\mathrm{bank}\:\mathrm{on}\:\mathrm{a}\:\mathrm{fixed}\:\mathrm{interest}\:\mathrm{rate}\:\mathrm{5\%} \\ $$$$\mathrm{suppose}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{make}\:\mathrm{monthly}\:\mathrm{payments}\:\mathrm{for}\:\mathrm{14months} \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{i}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{amount}\:\mathrm{i}\:\mathrm{should}\:\mathrm{pay}\:\mathrm{monthly}\:\mathrm{and} \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{i}\:\mathrm{know}\:\mathrm{how}\:\mathrm{much}\:\mathrm{i}\:\mathrm{will}\:\mathrm{pay}\:\mathrm{back}\:\mathrm{to}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{at}\:\mathrm{the} \\ $$$$\mathrm{end}\:\mathrm{of}\:\mathrm{the}\:\mathrm{14months}? \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{explanations}. \\ $$ Commented by mr…
Question Number 165403 by naka3546 last updated on 01/Feb/22 $$\mathrm{sec}^{\mathrm{2}} \mathrm{1}°\:+\:\mathrm{sec}^{\mathrm{2}} \:\mathrm{2}°\:+\:\mathrm{sec}^{\mathrm{2}} \:\mathrm{3}°\:+\:\ldots+\:\mathrm{sec}^{\mathrm{2}} \:\mathrm{89}°\:=\:? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com