Question Number 22463 by Tinkutara last updated on 18/Oct/17 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:{x}: \\ $$$$\frac{\mathrm{1}}{\left[{x}\right]}\:+\:\frac{\mathrm{1}}{\left[\mathrm{2}{x}\right]}\:=\:\left({x}\right)\:+\:\frac{\mathrm{1}}{\mathrm{3}}, \\ $$$$\mathrm{where}\:\left[{x}\right]\:\mathrm{is}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{integer}\:\mathrm{less} \\ $$$$\mathrm{than}\:\mathrm{or}\:\mathrm{equal}\:\mathrm{to}\:{x}\:\mathrm{and}\:\left({x}\right)\:=\:{x}\:−\:\left[{x}\right], \\ $$$$\left[\mathrm{e}.\mathrm{g}.\:\left[\mathrm{3}.\mathrm{4}\right]\:=\:\mathrm{3}\:\mathrm{and}\:\left(\mathrm{3}.\mathrm{4}\right)\:=\:\mathrm{0}.\mathrm{4}\right]. \\ $$ Answered by ajfour last updated…
Question Number 87988 by M±th+et£s last updated on 07/Apr/20 $${solve}\:{the}\:{PDE} \\ $$$$\mathrm{1}−{Z}={px}+{py}−{q}\sqrt{{pq}} \\ $$$$\mathrm{2}−{Z}={px}+{qy}+{sin}\left({p}+{q}\right) \\ $$$$\mathrm{3}−{p}\left(\mathrm{1}+{q}^{\mathrm{2}} \right)={q}\left({Z}−{a}\right) \\ $$$$\mathrm{4}−{Z}={xyp}^{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 153513 by yeti123 last updated on 08/Sep/21 $$\left(\frac{{x}}{\mathrm{5}}\:+\:\frac{{y}}{\mathrm{3}}\right)\left(\frac{\mathrm{5}}{{x}}\:+\:\frac{\mathrm{3}}{{y}}\right)\:=\:\mathrm{139},\:\forall{x},{y}\:\in\:\mathbb{R}_{>\mathrm{0}} \\ $$$$\mathrm{find}\:\mathrm{maximum}\:\mathrm{and}\:\mathrm{minimum}\:\mathrm{of}\:\:\frac{{x}\:+\:{y}}{\:\sqrt{{xy}}} \\ $$ Answered by liberty last updated on 08/Sep/21 $$\Rightarrow\frac{{x}+{y}}{\:\sqrt{{xy}}}\:=\:\sqrt{\frac{{x}}{{y}}}+\sqrt{\frac{{y}}{{x}}}=\sqrt{\frac{\mathrm{5}\lambda}{\mathrm{3}}}+\sqrt{\frac{\mathrm{3}}{\mathrm{5}\lambda}} \\ $$$${from}\:{condition} \\…
Question Number 153508 by mathdanisur last updated on 07/Sep/21 Answered by maged last updated on 08/Sep/21 $${x}=\mathrm{2},{y}=\mathrm{2},{z}=\mathrm{2}\Rightarrow\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} =\mathrm{12} \\ $$$${S}_{{min}} =\mathrm{2}+\mathrm{2}+\mathrm{2}+\mathrm{8}+\frac{\mathrm{1}}{\mathrm{4}+\mathrm{4}+\mathrm{4}} \\ $$$${S}_{{min}}…
Question Number 153505 by mathdanisur last updated on 07/Sep/21 Commented by MJS_new last updated on 08/Sep/21 $$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{5}}−\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{5}}=\mathrm{5}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\in\mathbb{C} \\ $$ Answered by…
Question Number 153500 by Engr_Jidda last updated on 07/Sep/21 $${Evaluate}\:{the}\:{line}\:{integral}\:{space} \\ $$$${if}\:{f}\left({r}\right)={Zi}+{Xj}+{Yk}\:{and}\:{C}\:{is}\:{a}\:{helix} \\ $$$${given}\:{by}\:{C}:\:{r}\left({t}\right)=\left({cost},{sint},−\mathrm{3}{t}\right)\:\:\mathrm{0}\leqslant{t}\leqslant\mathrm{2}\pi \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22423 by Tinkutara last updated on 17/Oct/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{no}\:\mathrm{three}\:\mathrm{consecutive} \\ $$$$\mathrm{binomial}\:\mathrm{coefficient}\:\mathrm{can}\:\mathrm{be}\:\mathrm{in}\:\mathrm{G}.\mathrm{P}.\:\mathrm{or}\:\mathrm{H}.\mathrm{P}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22392 by Tinkutara last updated on 17/Oct/17 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{coefficients} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}\:+\:\mathrm{2}{x}\:−\:\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{1025}} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{integer}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22394 by Tinkutara last updated on 17/Oct/17 $$\mathrm{Prove}\:\mathrm{that}\::\:\frac{\:^{{n}} {C}_{\mathrm{0}} }{{n}}−\frac{\:^{{n}} {C}_{\mathrm{1}} }{{n}+\mathrm{1}}+\frac{\:^{{n}} {C}_{\mathrm{2}} }{{n}+\mathrm{2}}−…+\left(−\mathrm{1}\right)^{{n}} .\frac{\:^{{n}} {C}_{{n}} }{\mathrm{2}{n}}=\frac{\mathrm{1}}{{n}.^{\mathrm{2}{n}} {C}_{{n}} } \\ $$ Terms of…
Question Number 22389 by Tinkutara last updated on 17/Oct/17 $${Simplify}: \\ $$$$\:^{{n}−\mathrm{1}} {C}_{\mathrm{2}} +\mathrm{2}\:^{{n}−\mathrm{2}} {C}_{\mathrm{2}} +\mathrm{3}\:^{{n}−\mathrm{3}} {C}_{\mathrm{2}} +…+\left({n}−\mathrm{2}\right)\:^{\mathrm{2}} {C}_{\mathrm{2}} \\ $$ Terms of Service Privacy…