Menu Close

Author: Tinku Tara

lim-n-2n-1-2n-3-4n-1-2n-2n-2-4n-

Question Number 205716 by universe last updated on 28/Mar/24 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)}{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}\:\:=\:\:? \\ $$ Answered by MM42 last updated on 28/Mar/24 $$\frac{\left(\mathrm{2}{n}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)…\left(\mathrm{4}{n}\right)}{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)…\left(\mathrm{4}{n}−\mathrm{1}\right)}<{A}<\frac{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{4}\right)…\left(\mathrm{4}{n}+\mathrm{2}\right)}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)…\left(\mathrm{4}{n}+\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{\mathrm{4}{n}+\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right){A}}<{A}<\frac{\mathrm{4}{n}+\mathrm{2}}{\left(\mathrm{2}{n}\right){A}} \\ $$$$\Rightarrow\frac{\mathrm{4}{n}+\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}<{A}^{\mathrm{2}}…

If-a-b-c-R-and-a-b-c-6-Prove-that-a-2-4-4a-2-9a-6-b-2-4-4b-2-9b-6-c-2-4-4c-2-9c-6-0-

Question Number 205733 by hardmath last updated on 28/Mar/24 $$\mathrm{If}\:\:\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{R}^{+} \:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{6} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4a}^{\mathrm{2}} −\mathrm{9a}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4b}^{\mathrm{2}} −\mathrm{9b}\:+\:\mathrm{6}}\:+\:\frac{\mathrm{c}^{\mathrm{2}} −\mathrm{4}}{\mathrm{4c}^{\mathrm{2}} −\mathrm{9c}\:+\:\mathrm{6}}\:\leqslant\:\mathrm{0} \\ $$ Answered by…

Figure-Shows-that-Object-A-is-connected-to-Object-B-by-thread-along-the-Slope-it-shows-a-costant-acceleration-motion-the-mass-of-A-of-B-are-3m-2m-respectively-and-when-A-communicates-from-point-P-to

Question Number 205684 by MathedUp last updated on 27/Mar/24 $$\mathrm{Figure}\:\mathrm{Shows}\:\mathrm{that}\:\mathrm{Object}\:\boldsymbol{\mathrm{A}}\:\mathrm{is}\:\mathrm{connected}\:\mathrm{to} \\ $$$$\mathrm{Object}\:\boldsymbol{\mathrm{B}}\:\mathrm{by}\:\mathrm{thread}\:\mathrm{along}\:\mathrm{the}\:\mathrm{Slope} \\ $$$$\mathrm{it}\:\mathrm{shows}\:\mathrm{a}\:\mathrm{costant}\:\mathrm{acceleration}\:\mathrm{motion} \\ $$$$\mathrm{the}\:\mathrm{mass}\:\mathrm{of}\:\boldsymbol{\mathrm{A}}\:\mathrm{of}\:\boldsymbol{\mathrm{B}}\:\mathrm{are}\:\mathrm{3m}\:,\:\mathrm{2m} \\ $$$$\mathrm{respectively}\:\mathrm{and}\:\mathrm{when}\:\boldsymbol{\mathrm{A}}\:\mathrm{communicates}\:\mathrm{from}\:\mathrm{point}\:\boldsymbol{\mathrm{P}}\:\mathrm{to}\:\boldsymbol{\mathrm{Q}}\: \\ $$$$\boldsymbol{\mathrm{B}}'{s}\:\mathrm{the}\:\mathrm{decrease}\:\mathrm{in}\:\mathrm{potential}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{10}\:\:\mathrm{times} \\ $$$$\mathrm{the}\:\mathrm{decrease}\:\mathrm{in}\:\mathrm{Kinetic}\:\mathrm{energy}\:\mathrm{of}\:\boldsymbol{\mathrm{B}}\: \\ $$$$\mathrm{find}\:\mathrm{accerate}\:\mathrm{of}\:\boldsymbol{\mathrm{A}}\:\left(\mathrm{3}\:\mathrm{point}\right) \\…

Question-205681

Question Number 205681 by naka3546 last updated on 27/Mar/24 Answered by mr W last updated on 27/Mar/24 $${a}=\mathrm{1}\:{cm} \\ $$$${R}=\frac{\mathrm{1}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{6}} \\ $$$${r}=\frac{\sqrt{\mathrm{3}}{b}}{\mathrm{6}} \\ $$$$\frac{{b}}{{a}}=\frac{\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}−\mathrm{2}{R}}{\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{3}} \\…