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Question-8338

Question Number 8338 by tawakalitu last updated on 09/Oct/16 Commented by tawakalitu last updated on 09/Oct/16 $$\mathrm{find}\:\mathrm{the}\:\mathrm{magnitude}\:\mathrm{and}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{resultant}\:\mathrm{of}\:\mathrm{the}\:\mathrm{forces}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{diagram} \\ $$ Answered by ridwan balatif…

Question-139393

Question Number 139393 by BHOOPENDRA last updated on 26/Apr/21 Answered by Ar Brandon last updated on 26/Apr/21 $$\mathrm{I}=\int\frac{\mathrm{cos}{x}}{\mathrm{2cos}{x}+\mathrm{sin}{x}+\mathrm{3}}{dx},\:{t}=\mathrm{tan}\frac{{x}}{\mathrm{2}}\Rightarrow{dt}=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}}{dx} \\ $$$$\:\:=\int\frac{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{2}\left(\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)+\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}}…

a-uniform-pole-30m-and-weight-40kg-is-carried-at-P-by-John-and-Ama-8m-away-from-Q-Find-the-distance-from-P-where-a-mass-of-20kg-should-be-attached-so-that-Amas-support-is-twice-that-of-John-if-tbe-s

Question Number 139390 by abenarhodym last updated on 26/Apr/21 $${a}\:{uniform}\:{pole}\:\mathrm{30}{m}\:{and}\:{weight}\:\mathrm{40}{kg}\:{is}\:{carried}\:{at}\:{P}\:\:{by}\:{John}\:{and}\:{Ama}\:\mathrm{8}{m}\:{away}\:{from}\:{Q}.\:{Find}\:{the}\:{distance}\:{from}\:{P}\:{where}\:{a}\:{mass}\:{of}\:\mathrm{20}{kg}\:{should}\:{be}\:{attached}\:{so}\:{that}\:{Amas}\:{support}\:{is}\:{twice}\:{that}\:{of}\:{John}\:{if}\:{tbe}\:{system}\:{is}\:{in}\:{equilibrim} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

Question-8300

Question Number 8300 by tawakalitu last updated on 06/Oct/16 Commented by ridwan balatif last updated on 07/Oct/16 $$\mathrm{solution} \\ $$$$\mathrm{1}.{v}=\sqrt{\mathrm{2gh}},\:\mathrm{where}\:\mathrm{h}\:\mathrm{is}\:\mathrm{height}\:\mathrm{and}\:\mathrm{g}\:\mathrm{is}\:\mathrm{gravitational}\:\mathrm{acceleration} \\ $$$$\:\:\:\:{v}=\sqrt{\mathrm{2}×\mathrm{9}.\mathrm{8}×\mathrm{5}} \\ $$$$\:\:\:\:{v}=\mathrm{9}.\mathrm{899}\:{m}/{s} \\…

n-0-1-6n-

Question Number 139359 by Dwaipayan Shikari last updated on 26/Apr/21 $$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{6}{n}\right)!} \\ $$ Commented by Dwaipayan Shikari last updated on 28/Apr/21 $$\frac{{cosh}\left(\mathrm{1}\right)+\mathrm{2}{cosh}\left(\frac{\mathrm{1}}{\mathrm{2}}\right){cos}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)}{\mathrm{3}} \\…

sin-50-sin-40-without-tables-or-calculators-

Question Number 73787 by Rio Michael last updated on 15/Nov/19 $${sin}\:\mathrm{50}\:+\:{sin}\:\mathrm{40}=\:?\:{without}\:{tables}\:{or}\:{calculators} \\ $$ Commented by mind is power last updated on 15/Nov/19 $$\mathrm{i}\:\mathrm{see}\:\mathrm{just}\:\mathrm{cardan}\:\mathrm{Methode} \\ $$$$\mathrm{sin}\left(\mathrm{50}\right)=\mathrm{sin}\left(\mathrm{90}−\mathrm{40}\right)=\mathrm{cos}\left(\mathrm{40}\right)…