Question Number 86260 by ajfour last updated on 27/Mar/20 Commented by ajfour last updated on 27/Mar/20 $${Find}\:{the}\:{maximum}\:{force}\:{applied} \\ $$$${in}\:{upturning}\:{the}\:{eql}.\:\bigtriangleup,\:{expending} \\ $$$${minimum}\:{energy}. \\ $$ Answered by…
Question Number 20724 by Tinkutara last updated on 01/Sep/17 $$\mathrm{A}\:\mathrm{sphere}\:\mathrm{is}\:\mathrm{rolling}\:\mathrm{without}\:\mathrm{slipping}\:\mathrm{on} \\ $$$$\mathrm{a}\:\mathrm{fixed}\:\mathrm{horizontal}\:\mathrm{plane}\:\mathrm{surface}.\:\mathrm{In}\:\mathrm{the} \\ $$$$\mathrm{Figure},\:\mathrm{A}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{of}\:\mathrm{contact},\:{B}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sphere}\:\mathrm{and}\:{C}\:\mathrm{is}\:\mathrm{its}\:\mathrm{topmost} \\ $$$$\mathrm{point}.\:\mathrm{Then} \\ $$$$\left({a}\right)\:\overset{\rightarrow} {{v}}_{\mathrm{C}} \:−\:\overset{\rightarrow} {{v}}_{\mathrm{A}} \:=\:\mathrm{2}\left(\overset{\rightarrow} {{v}}_{\mathrm{B}}…
Question Number 86258 by lémùst last updated on 27/Mar/20 $${calculate}\:{I}=\int_{\mathrm{1}} ^{+\infty} \frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{4}} −{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$ Commented by mathmax by abdo last updated on…
Question Number 20723 by rajpurohithakshay999@gmail.com last updated on 01/Sep/17 $$\int\sqrt{{x}}\:.{sinx}.{dx} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 86256 by TawaTawa1 last updated on 27/Mar/20 Answered by ajfour last updated on 27/Mar/20 $$\pi{r}^{\mathrm{2}} =\pi{a}^{\mathrm{2}} \left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$ Answered by ajfour last…
Question Number 20721 by ajfour last updated on 01/Sep/17 $$\mathrm{tan}\:{x}\mathrm{tan}\:{z}=\mathrm{3} \\ $$$$\mathrm{tan}\:{y}\mathrm{tan}\:{z}=\mathrm{6} \\ $$$${x}+{y}+{z}\:=\:\pi \\ $$$${Solve}\:{for}\:{x},\:{y},\:{and}\:{z}. \\ $$ Answered by sma3l2996 last updated on 02/Sep/17…
Question Number 151789 by mathdanisur last updated on 23/Aug/21 Answered by ArielVyny last updated on 23/Aug/21 $${we}\:{have}\:\mathrm{0}\leqslant\lambda\leqslant\mathrm{1}\rightarrow\mathrm{0}\leqslant\lambda+{a}+{b}\leqslant\mathrm{1}+{a}+{b} \\ $$$${and}\:{we}\:{admit}\:{that}\:{a}+{b}\geqslant\lambda+\mathrm{1};{b}+{c}\geqslant\lambda+\mathrm{1} \\ $$$${c}+{a}\geqslant\lambda+\mathrm{1}\:\:\left({note}\:{that}\:{abc}=\mathrm{1}\right) \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{a}+{b}}\leqslant\frac{\mathrm{1}}{\lambda+{a}+{b}}\:\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}} \\ $$$$\mathrm{0}\leqslant\frac{\mathrm{1}}{\mathrm{1}+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+{b}+{c}}\leqslant\frac{\mathrm{1}}{\lambda+\mathrm{2}}…
Question Number 86254 by sakeefhasan05@gmail.com last updated on 27/Mar/20 $$\int\:\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}\sqrt{\mathrm{x}…….}}}}\:\:\:\mathrm{dx} \\ $$ Answered by TANMAY PANACEA. last updated on 27/Mar/20 $${y}=\sqrt{{x}\sqrt{{x}\sqrt{{x}\sqrt{{x}…\infty}\:\:\:}}} \\ $$$${y}=\sqrt{{xy}}\: \\ $$$${y}^{\mathrm{2}}…
Question Number 151791 by mathdanisur last updated on 23/Aug/21 Answered by ghimisi last updated on 23/Aug/21 $${x}_{{i}} ^{\mathrm{4}} +\mathrm{1}\geqslant\mathrm{2}{x}_{{i}} ^{\mathrm{2}} \Rightarrow{x}_{{i}} ^{\mathrm{4}} −{x}_{{i}} ^{\mathrm{2}} +\mathrm{1}\geqslant{x}_{{i}}…
Question Number 151790 by mathdanisur last updated on 23/Aug/21 Answered by Olaf_Thorendsen last updated on 24/Aug/21 $${u}_{\mathrm{1}} \:=\:\sqrt{\mathrm{99}} \\ $$$${u}_{{n}} \:=\:\sqrt{\mathrm{102}−\mathrm{3}{n}+{u}_{{n}−\mathrm{1}} } \\ $$$${u}_{\mathrm{33}} \:=\:\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+…+\sqrt{\mathrm{96}+\sqrt{\mathrm{99}}}}}}…