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Author: Tinku Tara

Prove-i-i-e-pi-2-i-2-1-

Question Number 4556 by FilupSmith last updated on 07/Feb/16 $$\mathrm{Prove}\:{i}^{{i}} ={e}^{−\pi/\mathrm{2}} ,\:\:{i}^{\mathrm{2}} =−\mathrm{1} \\ $$ Answered by Yozzii last updated on 07/Feb/16 $${i}=\mathrm{0}+{i}×\mathrm{1}={cos}\mathrm{0}.\mathrm{5}\pi+{isin}\mathrm{0}.\mathrm{5}={e}^{\mathrm{0}.\mathrm{5}\pi{i}} \\ $$$$\therefore{i}^{{i}}…

nice-calculus-evaluation-0-pi-2-sin-x-ln-sin-x-dx-solution-cos-x-y-1-2-0-1-ln-1-y-2-dy-

Question Number 135627 by mnjuly1970 last updated on 14/Mar/21 $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\:{nice}\:……………..\:{calculus}\:… \\ $$$$\:\:\:\:\:\:\:{evaluation}:::::\:\:\:\boldsymbol{\phi}\overset{???} {=}\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} {sin}\left({x}\right){ln}\left({sin}\left({x}\right)\right){dx} \\ $$$$\:\:\:\:\:\:\:{solution}::::: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\overset{\langle{cos}\left({x}\right)={y}\rangle} {=}\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left(\mathrm{1}−{y}^{\mathrm{2}} \right){dy} \\ $$$$\:\:\:\:\:\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}}…

Question-4548

Question Number 4548 by Yozzii last updated on 07/Feb/16 Commented by Yozzii last updated on 07/Feb/16 $${In}\:{the}\:{diagram}\:{is}\:{a}\:{parallelogram}\:{ABCD} \\ $$$${with}\:{diagonal}\:{CB}. \\ $$$${E}\:{and}\:{F}\:{are}\:{the}\:{midpoints}\:{of}\:{CD}\:{and} \\ $$$${BD}\:{respectively}.\:{Using}\:{vectors},\:{prove} \\ $$$${that}\:{AE}\:{and}\:{AF}\:{trisect}\:{CB}.…

At-what-speed-should-a-basket-ball-player-project-a-ball-from-a-height-of-2-1-m-above-the-ground-at-an-angle-of-38-to-the-horizontal-so-that-it-just-passes-through-a-basket-ball-net-which-is-at-a-ho

Question Number 135618 by physicstutes last updated on 14/Mar/21 $$\:\mathrm{At}\:\mathrm{what}\:\mathrm{speed}\:\mathrm{should}\:\mathrm{a}\:\mathrm{basket}\:\mathrm{ball}\:\mathrm{player}\:\mathrm{project}\:\mathrm{a}\:\mathrm{ball}\:\mathrm{from}\:\mathrm{a}\:\mathrm{height} \\ $$$$\mathrm{of}\:\mathrm{2}.\mathrm{1}\:\mathrm{m} \\ $$$$\mathrm{above}\:\mathrm{the}\:\mathrm{ground}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{of}\:\mathrm{38}° \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{horizontal}\:\mathrm{so}\:\mathrm{that}\:\mathrm{it}\:\mathrm{just}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{a}\:\mathrm{basket}\:\mathrm{ball}\:\mathrm{net}\:\mathrm{which} \\ $$$$\mathrm{is}\:\mathrm{at}\:\mathrm{a}\:\mathrm{horizontal}\:\mathrm{distance}\:\mathrm{of}\:\mathrm{5}\:\mathrm{m}\:\mathrm{from}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{position}\:\mathrm{of}\:\mathrm{the}\:\mathrm{ball}\:\mathrm{and}\:\mathrm{3}\:\mathrm{m}\:\mathrm{above}\:\mathrm{the}\:\mathrm{playground}.\:\mathrm{at}\:\mathrm{determine}\:\mathrm{the}\:\mathrm{balls} \\ $$$$\mathrm{maximum}\:\mathrm{height}\:\mathrm{above}\:\mathrm{the}\:\mathrm{ground}. \\ $$ Answered…

Triangle-ABC-has-midpoints-D-E-and-F-By-connecting-each-verticie-with-the-opposite-midpoint-we-create-a-cress-section-called-G-Prove-that-all-three-lines-cross-at-point-G-regardless-of-the-type

Question Number 4543 by FilupSmith last updated on 06/Feb/16 $$\mathrm{Triangle}\:{ABC}\:\mathrm{has}\:\mathrm{midpoints} \\ $$$${D},\:{E}\:\mathrm{and}\:{F}. \\ $$$$ \\ $$$$\mathrm{By}\:\mathrm{connecting}\:\mathrm{each}\:\mathrm{verticie}\:\mathrm{with}\:\mathrm{the} \\ $$$$\mathrm{opposite}\:\mathrm{midpoint},\:\mathrm{we}\:\mathrm{create}\:\mathrm{a}\:\mathrm{cress}−\mathrm{section} \\ $$$$\mathrm{called}\:{G}. \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{all}\:\mathrm{three}\:\mathrm{lines}\:\mathrm{cross}\:\mathrm{at}\: \\…