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Author: Tinku Tara

sdvanced-cslculus-if-x-R-and-x-0-x-e-t-1-t-ln-x-t-dt-then-prove-that-0-e-x-x-dx-2-

Question Number 136381 by mnjuly1970 last updated on 21/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:……{sdvanced}\:\:\:{cslculus}…… \\ $$$$\:{if}\:\:{x}\in\mathbb{R}^{+} \:{and}::\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\left({x}\right)=\int_{\mathrm{0}} ^{\:{x}} \frac{{e}^{{t}} −\mathrm{1}}{{t}}{ln}\left(\frac{{x}}{{t}}\right){dt} \\ $$$$\:\:{then}\:{prove}\:\:{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Psi=\int_{\mathrm{0}} ^{\:\infty} {e}^{−{x}} \boldsymbol{\phi}\left({x}\right){dx}=\zeta\left(\mathrm{2}\right)…

Question-70847

Question Number 70847 by A8;15: last updated on 08/Oct/19 Answered by mind is power last updated on 08/Oct/19 $$\begin{cases}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{2}}\\{\mathrm{2}{x}^{\mathrm{4}} +{xy}^{\mathrm{3}} +{x}^{\mathrm{2}} {y}^{\mathrm{2}} −{x}^{\mathrm{3}}…

Question-136382

Question Number 136382 by ajfour last updated on 21/Mar/21 Commented by ajfour last updated on 21/Mar/21 $${On}\:{a}\:{rough}\:{horizontal},\:{is}\:{placed} \\ $$$${a}\:{fixed}\:{V}\:{rail}\:{with}\:\angle=\alpha. \\ $$$${A}\:{solid}\:{double}\:{cone}\:{object}\:{of} \\ $$$${radius}\:{a}\:{and}\:{length}\:\mathrm{2}{b}\:{is} \\ $$$${now}\:{placed}\:{symmetricaly}\:{over}…

3-points-with-the-restriction-that-they-should-be-non-collinear-determine-circle-What-number-of-points-with-what-restriction-determine-sphere-

Question Number 5308 by Rasheed Soomro last updated on 06/May/16 $$\mathrm{3}\:\mathrm{points}\:\mathrm{with}\:\mathrm{the}\:\mathrm{restriction}\:\mathrm{that} \\ $$$$\mathrm{they}\:\mathrm{should}\:\mathrm{be}\:\mathrm{non}-\mathrm{collinear}\:\mathrm{determine} \\ $$$$\mathrm{circle}. \\ $$$$\mathrm{What}\:\mathrm{number}\:\mathrm{of}\:\mathrm{points}\:\mathrm{with}\:\mathrm{what} \\ $$$$\mathrm{restriction}\:\mathrm{determine}\:\mathrm{sphere}? \\ $$ Terms of Service Privacy…

Question-136377

Question Number 136377 by MJS_new last updated on 21/Mar/21 Commented by MJS_new last updated on 21/Mar/21 $$\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:“\mathrm{square}''\:\mathrm{is}\:\left(\frac{\pi}{\mathrm{3}}+\mathrm{1}−\sqrt{\mathrm{3}}\right){a}^{\mathrm{2}} \:\mathrm{but} \\ $$$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{this}\:{without}\:\mathrm{integration}? \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{afraid}\:\mathrm{I}'\mathrm{m}\:\mathrm{a}\:\mathrm{blockhead}… \\ $$ Answered…

Question-70838

Question Number 70838 by A8;15: last updated on 08/Oct/19 Answered by mind is power last updated on 08/Oct/19 $${X}^{\mathrm{3}} −\mathrm{1}=\left({X}−\mathrm{1}\right)\left({X}^{\mathrm{2}} +{X}+\mathrm{1}\right) \\ $$$${X}^{\mathrm{3}} +\mathrm{1}=\left({X}+\mathrm{1}\right)\left({X}^{\mathrm{2}} −{X}+\mathrm{1}\right)…

A-bowl-contains-carefully-shredded-confetti-6-of-which-are-blue-and-the-remaining-12-are-red-Each-piece-of-confetti-is-either-circular-triangular-or-rectangular-in-shape-and-are-in-the-ratio-1-3-2-f

Question Number 136375 by nadovic last updated on 22/Mar/21 $$\mathrm{A}\:\mathrm{bowl}\:\mathrm{contains}\:\mathrm{carefully}\:\mathrm{shredded} \\ $$$$\mathrm{confetti},\:\mathrm{6}\:\mathrm{of}\:\mathrm{which}\:\mathrm{are}\:\mathrm{blue}\:\mathrm{and}\:\mathrm{the} \\ $$$$\mathrm{remaining}\:\mathrm{12}\:\mathrm{are}\:\mathrm{red}. \\ $$$$\mathrm{Each}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{confetti}\:\mathrm{is}\:\mathrm{either}\:\mathrm{circular} \\ $$$$\mathrm{triangular}\:\mathrm{or}\:\mathrm{rectangular}\:\mathrm{in}\:\mathrm{shape}\:\mathrm{and} \\ $$$$\mathrm{are}\:\mathrm{in}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{1}:\mathrm{3}:\mathrm{2}\:\mathrm{for}\:\mathrm{each}\:\mathrm{colour} \\ $$$$\mathrm{respectively}.\:\mathrm{If}\:\mathrm{they}\:\mathrm{are}\:\mathrm{thrown}\:\mathrm{up}\: \\ $$$$\boldsymbol{\mathrm{thrice}}\:\mathrm{in}\:\mathrm{a}\:\mathrm{certain}\:\mathrm{competition}\:\mathrm{and}\:\mathrm{a} \\…

Question-70834

Question Number 70834 by Maclaurin Stickker last updated on 08/Oct/19 Answered by mind is power last updated on 08/Oct/19 $$\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{log}_{{x}} \left(\mathrm{2}^{{k}} \right){log}_{{x}} \left(\mathrm{2}^{\left.{k}+\mathrm{1}\right)}…