Question Number 5013 by Rasheed Soomro last updated on 01/Apr/16 $${The}\:{sign}\:{for}\:“\:{is}\:{not}\:{congruent}\:{to}'' \\ $$$${is}\:{present}\:{on}\:{the}\:{keyboard}\:\left(\ncong\right). \\ $$$${But}\:{its}\:{opposite}\:“\:{is}\:{congruent}\:{to}'' \\ $$$${doesn}'{t}\:{exist}. \\ $$$${Although}\:{by}\:{adding}\:{two}\:{signs}\:'='\:{and}\:'\sim' \\ $$$${we}\:{can}\:{make}\:{the}\:{required}\:{sign}\:\left(\overset{\sim} {=}\right)\:{but} \\ $$$${I}\:{think}\:{it}\:{should}\:{be}\:{directly}\:{present}. \\…
Question Number 5011 by Yozzii last updated on 01/Apr/16 Commented by Yozzii last updated on 01/Apr/16 $${S}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{ln}\left({n}\right)}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{tan}^{−\mathrm{1}} {i}}{{n}−{i}+\mathrm{1}}\right) \\ $$$${S}=?\: \\ $$…
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Question Number 5008 by Rasheed Soomro last updated on 31/Mar/16 $$\mathrm{A}\:\mathrm{piece}\:\mathrm{of}\:\mathrm{plastic}\:\mathrm{strip}\:\mathrm{1}\:\mathrm{metre}\:\mathrm{long} \\ $$$$\mathrm{is}\:\mathrm{bent}\:\mathrm{to}\:\mathrm{form}\:\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle} \\ $$$$\mathrm{with}\:\mathrm{95}°\:\mathrm{as}\:\mathrm{measure}\:\mathrm{of}\:\mathrm{its}\:\mathrm{largest}\:\mathrm{angle}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}. \\ $$ Commented by prakash jain last updated…
Question Number 136076 by physicstutes last updated on 18/Mar/21 $$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the}\:\mathrm{curve}\:{y}\:=\:\mathrm{3}\:+\:\mathrm{2}{x}\:−{x}^{\mathrm{2}} ,\:\mathrm{the}\:{x}−\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:{y}\:=\:\mathrm{3}. \\ $$$$\:\mathrm{find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid}\:\mathrm{generated}\:\mathrm{when}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{is} \\ $$$$\mathrm{rotated}\:\mathrm{completely}\:\mathrm{about}\:\mathrm{the}\:\mathrm{line}\:{y}\:=\:\mathrm{3} \\ $$ Answered by mr W last updated…
Question Number 70543 by mr W last updated on 05/Oct/19 Commented by mr W last updated on 05/Oct/19 $${a}\:{semi}−{cylinder}\:{with}\:{radius}\:{R}\:{and} \\ $$$${mass}\:{M}\:{rests}\:{on}\:{a}\:{rough}\:{table}\:{as}\:{shown}. \\ $$$${when}\:{a}\:{small}\:{impulse}\:{is}\:{given}\:{to}\:{the} \\ $$$${semi}−{cylinder},\:{it}\:{begins}\:{to}\:{oscillate}.…
Question Number 5003 by 314159 last updated on 30/Mar/16 $${Find}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{17}} \:{in}\:{the}\:{expansion}\:{of} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)…\left({x}+\mathrm{19}\right). \\ $$ Commented by 123456 last updated on 30/Mar/16 $$\mathrm{1}+\mathrm{1}+…+\mathrm{1}\:\left(\mathrm{17}\right) \\ $$$$\mathrm{1}+\mathrm{1}+…+\mathrm{1}\:\left(\mathrm{17}\right)+\mathrm{0}…
Question Number 136070 by Dwaipayan Shikari last updated on 18/Mar/21 $$\frac{{sin}\left(\sqrt{\mathrm{2}}\right)}{\mathrm{1}^{\mathrm{3}} }+\frac{{sin}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)}{\mathrm{2}^{\mathrm{3}} }+\frac{{sin}\left(\mathrm{3}\sqrt{\mathrm{2}}\right)}{\mathrm{3}^{\mathrm{3}} }+…=\frac{\pi^{{b}} +\mathrm{1}}{\:{a}\sqrt{{b}}}−\frac{\pi}{{b}} \\ $$$${Find}\:{a}−{b} \\ $$ Answered by mnjuly1970 last updated on…
Question Number 136064 by bramlexs22 last updated on 18/Mar/21 $$\:\:\:\:\:\mathrm{3}^{\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{3}^{{x}} −\mathrm{1}\right)} \:=\:\mathrm{2}^{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{2}^{{x}} +\mathrm{1}\right)} +\:\mathrm{1} \\ $$ Answered by Kintinu last updated on 18/Mar/21…
Question Number 136067 by bramlexs22 last updated on 18/Mar/21 $$\Lambda\:=\:\int\:{x}^{\mathrm{3}} \:\left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{10}} \:{dx}\: \\ $$ Answered by Olaf last updated on 18/Mar/21 $$\Lambda\:=\:\int{x}^{\mathrm{3}} \left({x}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{10}}…