Question Number 84904 by john santu last updated on 17/Mar/20 $$\mathrm{3}^{\mathrm{2x}^{\mathrm{2}} } \:+\:\mathrm{3}^{\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{5}} \:\geqslant\:\mathrm{10}.\:\mathrm{3}^{\mathrm{4x}+\mathrm{6}} \\ $$$$ \\ $$ Commented by john santu last updated…
Question Number 84902 by john santu last updated on 17/Mar/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{5}^{\mathrm{x}} +\mathrm{5}^{\mathrm{2x}} \right)\:^{\frac{\mathrm{1}}{\mathrm{x}}} \:? \\ $$ Commented by john santu last updated on 17/Mar/20…
Question Number 19367 by Tinkutara last updated on 10/Aug/17 $$\mathrm{If}\:\alpha,\:\beta,\:\gamma\:\mathrm{and}\:\delta\:\mathrm{are}\:\mathrm{four}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{tan}\:\left(\theta\:+\:\frac{\mathrm{5}\pi}{\mathrm{4}}\right)\:=\:\mathrm{3}\:\mathrm{tan}\:\mathrm{3}\theta,\:\mathrm{then} \\ $$$$\left(\mathrm{1}\right)\:\Sigma\mathrm{tan}\:\alpha\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:=\:−\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\Sigma\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:=\:−\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}\:\alpha\:\mathrm{tan}\:\beta\:\mathrm{tan}\:\gamma\:\mathrm{tan}\:\delta\:=\:−\mathrm{3} \\ $$ Answered by ajfour…
Question Number 150432 by Jamshidbek last updated on 12/Aug/21 $$\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2n}+\mathrm{1}\right)!}{\mathrm{8}^{\mathrm{n}} \centerdot\left(\mathrm{n}!\right)^{\mathrm{2}} }=?\:\:\:\:\:\mathrm{Help}\:\mathrm{please} \\ $$ Answered by Olaf_Thorendsen last updated on 12/Aug/21 $${f}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\left(\mathrm{1}−{x}\right)^{\mathrm{3}/\mathrm{2}} }\:=\:\left(\mathrm{1}−{x}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}}…
Question Number 84899 by Power last updated on 17/Mar/20 Commented by Tony Lin last updated on 17/Mar/20 $${A}\left({ADE}\right)=\mathrm{9} \\ $$$$\Rightarrow{h}=\mathrm{6} \\ $$$${A}\left({EFC}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{EC}×\left({h}×\frac{\mathrm{6}}{\mathrm{7}}\right) \\…
Question Number 19362 by Tinkutara last updated on 10/Aug/17 $$\mathrm{The}\:\mathrm{block}\:{Q}\:\mathrm{moves}\:\mathrm{to}\:\mathrm{the}\:\mathrm{right}\:\mathrm{with}\:\mathrm{a} \\ $$$$\mathrm{constant}\:\mathrm{velocity}\:{v}_{\mathrm{0}} \:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure}. \\ $$$$\mathrm{The}\:\mathrm{relative}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{body}\:{P}\:\mathrm{with} \\ $$$$\mathrm{respect}\:\mathrm{to}\:{Q}\:\mathrm{is}\:\left(\mathrm{assume}\:\mathrm{all}\:\mathrm{pulleys}\:\mathrm{and}\right. \\ $$$$\left.\mathrm{strings}\:\mathrm{are}\:\mathrm{ideal}\right) \\ $$ Commented by Tinkutara last…
Question Number 150435 by mathdanisur last updated on 12/Aug/21 $$\boldsymbol{\mathrm{S}}\mathrm{olve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mid\mathrm{x}\:-\:\mathrm{3}\mid^{\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:-\:\mathrm{8x}\:+\:\mathrm{15}}{\boldsymbol{\mathrm{x}}\:-\:\mathrm{2}}} \:=\:\mathrm{1} \\ $$ Answered by amin96 last updated on 12/Aug/21 $${x}\neq\mathrm{2}\:\:\:\Rightarrow\:\:{x}^{\mathrm{2}} −\mathrm{8}{x}+\mathrm{15}=\mathrm{0}\:\:\left({x}−\mathrm{5}\right)\left({x}−\mathrm{3}\right)=\mathrm{0}…
Question Number 150429 by ZiYangLee last updated on 12/Aug/21 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{the}\:\mathrm{common} \\ $$$$\mathrm{tangents}\:\mathrm{to}\:\mathrm{the}\:\mathrm{parabola}\:{y}^{\mathrm{2}} =\mathrm{4}{x}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:{x}^{\mathrm{2}} =\mathrm{2}{y}−\mathrm{3}. \\ $$ Answered by Olaf_Thorendsen last updated on 12/Aug/21…
Question Number 84894 by Power last updated on 17/Mar/20 Commented by abdomathmax last updated on 18/Mar/20 $${I}=\int\sqrt{\frac{{ax}+{b}}{{cx}+{d}}}{dx}\:{we}\:{do}\:{the}\:{changement}\sqrt{\frac{{ax}+{b}}{{cx}+{d}}}={t}\:\Rightarrow \\ $$$$\frac{{ax}+{b}}{{cx}+{d}}={t}^{\mathrm{2}} \:\Rightarrow{ax}+{b}={ct}^{\mathrm{2}} {x}\:+{dt}^{\mathrm{2}} \:\Rightarrow\left({a}−{ct}^{\mathrm{2}} \right){x}={dt}^{\mathrm{2}} −{b}\:\Rightarrow \\…
Question Number 19358 by Tinkutara last updated on 10/Aug/17 $$\mathrm{In}\:\mathrm{the}\:\mathrm{arrangement}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{figure} \\ $$$$\mathrm{two}\:\mathrm{beads}\:\mathrm{slide}\:\mathrm{along}\:\mathrm{a}\:\mathrm{smooth} \\ $$$$\mathrm{horizontal}\:\mathrm{rod}.\:\mathrm{The}\:\mathrm{relation}\:\mathrm{between} \\ $$$${v}\:\mathrm{and}\:{v}_{\mathrm{0}} \:\mathrm{in}\:\mathrm{given}\:\mathrm{position}\:\mathrm{will}\:\mathrm{be} \\ $$ Commented by Tinkutara last updated on…