Question Number 18264 by Tinkutara last updated on 17/Jul/17 $$\mathrm{A}\:\mathrm{sky}\:\mathrm{diver}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{drops}\:\mathrm{out}\:\mathrm{with} \\ $$$$\mathrm{an}\:\mathrm{initial}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:=\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{law} \\ $$$$\mathrm{by}\:\mathrm{which}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{speed}\:\mathrm{varies} \\ $$$$\mathrm{before}\:\mathrm{the}\:\mathrm{parachute}\:\mathrm{is}\:\mathrm{opened}\:\mathrm{if}\:\mathrm{the} \\ $$$$\mathrm{drag}\:\mathrm{is}\:\mathrm{proportional}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sky}\:\mathrm{diver}'\mathrm{s} \\ $$$$\mathrm{speed}.\:\mathrm{Also}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{problem}\:\mathrm{when}\:\mathrm{the} \\ $$$$\mathrm{sky}\:\mathrm{diver}'\mathrm{s}\:\mathrm{initial}\:\mathrm{velocity}\:\mathrm{has}\:\mathrm{horizontal} \\ $$$$\mathrm{component}\:{v}_{\mathrm{0}}…
Question Number 18262 by Tinkutara last updated on 17/Jul/17 $$\mathrm{A}\:\mathrm{point}\:{P}\:\mathrm{is}\:\mathrm{located}\:\mathrm{above}\:\mathrm{an}\:\mathrm{inclined} \\ $$$$\mathrm{plane}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{by}\:\mathrm{sliding}\:\mathrm{under}\:\mathrm{gravity}\:\mathrm{down}\:\mathrm{a}\:\mathrm{straight} \\ $$$$\mathrm{frictionless}\:\mathrm{wire}\:\mathrm{joining}\:\mathrm{to}\:\mathrm{some}\:\mathrm{point} \\ $$$${P}\:'\:\mathrm{on}\:\mathrm{the}\:\mathrm{plane}.\:\mathrm{How}\:\mathrm{should}\:{P}\:'\:\mathrm{be} \\ $$$$\mathrm{chosen}\:\mathrm{so}\:\mathrm{as}\:\mathrm{to}\:\mathrm{minimize}\:\mathrm{the}\:\mathrm{time} \\ $$$$\mathrm{taken}? \\ $$ Commented…
Question Number 18261 by Tinkutara last updated on 17/Jul/17 $$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{angle}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{horizontal}\:\mathrm{at}\:\mathrm{which}\:\mathrm{a}\:\mathrm{stone}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{thrown}\:\mathrm{and}\:\mathrm{always}\:\mathrm{be}\:\mathrm{moving}\:\mathrm{away} \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{thrower}? \\ $$ Answered by ajfour last updated on 26/Jul/17…
Question Number 149329 by mnjuly1970 last updated on 04/Aug/21 Commented by Kamel last updated on 04/Aug/21 $${I}\:{think}\:{I}^{\mathrm{100}} +{k}\equiv\mathrm{0}\left[\mathrm{14}\right] \\ $$ Answered by Kamel last updated…
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Question Number 149331 by Samimsultani last updated on 04/Aug/21 Commented by john_santu last updated on 04/Aug/21 $$\mathrm{may}\:\mathrm{be}\:\mathrm{x}=\sqrt[{\mathrm{3}}]{\mathrm{7}+\sqrt{\mathrm{5}}}\:−\sqrt[{\mathrm{3}}]{\mathrm{7}−\sqrt{\mathrm{5}}}\: \\ $$ Commented by amin96 last updated on…
Question Number 18257 by aplus last updated on 17/Jul/17 Commented by mondodotto@gmail.com last updated on 17/Jul/17 $$\mathrm{Already}\:\mathrm{solved} \\ $$ Commented by prakash jain last updated…
Question Number 83791 by jagoll last updated on 06/Mar/20 $$\mathrm{Let}\:\mathrm{x},\:\mathrm{y}\:\mathrm{are}\:\mathrm{two}\:\mathrm{different}\:\mathrm{real} \\ $$$$\mathrm{numbers}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{equation}\: \\ $$$$\sqrt{\mathrm{y}+\mathrm{4}}\:=\:\mathrm{x}−\mathrm{4}\:\mathrm{and}\:\sqrt{\mathrm{x}+\mathrm{4}}\:=\:\mathrm{y}−\mathrm{4}. \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} \:\mathrm{mod}\left(\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{3}} \right)\:\mathrm{is} \\ $$ Commented by mr…
Question Number 83786 by jagoll last updated on 06/Mar/20 $$\frac{{x}^{\mathrm{2}} }{\mathrm{log}_{\left(\mathrm{5}−{x}\right)} \:\left({x}\right)}\:\leqslant\:\left(\mathrm{5}{x}−\mathrm{4}\right)\:\mathrm{log}_{{x}} \:\left(\mathrm{5}−{x}\right)\: \\ $$ Answered by john santu last updated on 06/Mar/20 $${x}^{\mathrm{2}} \:\mathrm{log}_{{x}}…
Question Number 83787 by john santu last updated on 06/Mar/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{abc}\:\mathrm{if}\: \\ $$$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}^{\mathrm{2}} +\sqrt{\mathrm{2}^{\mathrm{3}} +\mathrm{2}^{\mathrm{4}} +\sqrt{…}}}}\:=\:\frac{\sqrt{\mathrm{a}}+\sqrt{\mathrm{b}}}{\mathrm{c}} \\ $$ Answered by jagoll last updated on 06/Mar/20…