Question Number 156059 by cortano last updated on 07/Oct/21 Commented by john_santu last updated on 07/Oct/21 $${g}\left({x}\right)=\mathrm{3}{x}^{\mathrm{3}} +\left({h}\left({x}\right)\left({x}^{\mathrm{4}} −\mathrm{3}{x}+\mathrm{1}\right)+\mathrm{2}{x}^{\mathrm{3}} −\mathrm{7}{x}\right)^{\mathrm{2}} \\ $$$$\frac{{g}\left({x}\right)}{{x}^{\mathrm{4}} −\mathrm{3}{x}+\mathrm{1}}=\:{u}\left({x}\right)\left({x}^{\mathrm{4}} −\mathrm{3}{x}+\mathrm{1}\right)+{ax}^{\mathrm{3}} +{bx}^{\mathrm{2}}…
Question Number 156058 by VIDDD last updated on 07/Oct/21 $$\:\:{cos}\frac{\pi}{\mathrm{8}}=…?\:\:{with}\:{solution}\:{plz} \\ $$ Answered by cortano last updated on 07/Oct/21 $$\:\mathrm{cos}\:\frac{\pi}{\mathrm{4}}=\mathrm{2cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}−\mathrm{1} \\ $$$$\:\mathrm{cos}\:\frac{\pi}{\mathrm{8}}=\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{4}}}{\mathrm{2}}} \\ $$…
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Question Number 24945 by Tinkutara last updated on 29/Nov/17 $$\mathrm{A}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:{m}\:\mathrm{moving}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{speed}\:{v}\:\mathrm{hits}\:\mathrm{elastically}\:\mathrm{another} \\ $$$$\mathrm{stationary}\:\mathrm{particle}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{2}{m}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{smooth}\:\mathrm{horizontal}\:\mathrm{circular}\:\mathrm{tube}\:\mathrm{of} \\ $$$$\mathrm{radius}\:{r}.\:\mathrm{The}\:\mathrm{time}\:\mathrm{in}\:\mathrm{which}\:\mathrm{the}\:\mathrm{next} \\ $$$$\mathrm{collision}\:\mathrm{will}\:\mathrm{take}\:\mathrm{place}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by Tinkutara…
Question Number 24934 by Tinkutara last updated on 29/Nov/17 $$\mathrm{A}\:\mathrm{uniform}\:\mathrm{circular}\:\mathrm{disc}\:\mathrm{of}\:\mathrm{mass}\:\mathrm{1}.\mathrm{5}\:\mathrm{kg} \\ $$$$\mathrm{and}\:\mathrm{radius}\:\mathrm{0}.\mathrm{5}\:\mathrm{m}\:\mathrm{is}\:\mathrm{initially}\:\mathrm{at}\:\mathrm{rest}\:\mathrm{on}\:\mathrm{a} \\ $$$$\mathrm{horizontal}\:\mathrm{frictionless}\:\mathrm{surface}.\:\mathrm{Three} \\ $$$$\mathrm{forces}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitude}\:{F}\:=\:\mathrm{0}.\mathrm{5}\:\mathrm{N} \\ $$$$\mathrm{are}\:\mathrm{applied}\:\mathrm{simultaneously}\:\mathrm{along}\:\mathrm{the} \\ $$$$\mathrm{three}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle} \\ $$$${xyz}\:\mathrm{with}\:\mathrm{its}\:\mathrm{vertices}\:\mathrm{on}\:\mathrm{the}\:\mathrm{perimeter} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{disc}.\:\mathrm{One}\:\mathrm{second}\:\mathrm{after}\:\mathrm{applying} \\…
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Question Number 24893 by sushmitak last updated on 28/Nov/17 $$\mathrm{About}\:\mathrm{a}\:\mathrm{collision}\:\mathrm{which}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{are}\:\mathrm{not}\:\mathrm{correct} \\ $$$$\mathrm{a}.\:\mathrm{Physical}\:\mathrm{touch}\:\mathrm{is}\:\mathrm{a}\:\mathrm{must} \\ $$$$\mathrm{b}.\:\mathrm{Particles}\:\mathrm{cannot}\:\mathrm{change} \\ $$$$\mathrm{c}.\:\mathrm{Effect}\:\mathrm{of}\:\mathrm{external}\:\mathrm{force}\:\mathrm{is}\:\mathrm{not}\:\mathrm{considered} \\ $$$$\mathrm{d}.\:\mathrm{Momentum}\:\mathrm{may}\:\mathrm{or}\:\mathrm{may}\:\mathrm{not}\:\mathrm{change} \\ $$$$\mathrm{multi}−\mathrm{correct}\:\mathrm{question} \\ $$ Commented…