Question Number 221359 by Engr_Jidda last updated on 31/May/25 Answered by Rasheed.Sindhi last updated on 31/May/25 $${x}\ast{y}=\mathrm{2}{x}+\mathrm{2}{y}−\frac{{xy}}{\mathrm{5}} \\ $$$${let}\:{e}\:{is}\:{an}\:{identity}\:{element}\:{of}\:\ast \\ $$$${x}\ast{e}={e}\ast{x}={x} \\ $$$$\Rightarrow\mathrm{2}{x}+\mathrm{2}{e}−\frac{{xe}}{\mathrm{5}}={x} \\ $$$${e}\left(\frac{\mathrm{10}−{x}}{\mathrm{5}}\right)=−{x}…
Question Number 221352 by Nicholas666 last updated on 31/May/25 $$ \\ $$$$\:\:\:\:\mathrm{Given}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c}\:>\:\mathrm{0}\:, \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:{a}\:+\:{b}\:+\:{c}\:=\:{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:, \\ $$$$\:\mathrm{Prove}\:;\:\frac{{a}^{\mathrm{3}} }{{a}^{\mathrm{4}} \:+\:{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{3}} }{{b}^{\mathrm{4}} \:+\:{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{3}} }{{c}^{\mathrm{4}} \:+\:\:{a}\:+\:{b}}\:\leqslant\:\mathrm{1}…
Question Number 221354 by Nicholas666 last updated on 31/May/25 $$ \\ $$$$\:\:\mathrm{Let}\:{a},{b},{c}\:\mathrm{be}\:\mathrm{there}\:\mathrm{real}\:\mathrm{numbers}, \\ $$$$\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}; \\ $$$$\:\mathrm{sin}\:{a}\:+\:\mathrm{sin}\:{b}\:+\:\mathrm{sin}\:{c}\:\geqslant\:\mathrm{2}\:\:\Rightarrow\:\mathrm{cos}\:{a}\:+\:\mathrm{cos}\:{b}\:+\:\mathrm{cos}\:{c}\:\leqslant\:\sqrt{\mathrm{5}}\:\:\:\:\:\:\:\:\:\: \\ $$$$\mathrm{and}, \\ $$$$\:\mathrm{sin}\:{a}\:+\:\mathrm{sin}\:{b}\:+\:\mathrm{sin}\:{c}\:\geqslant\:\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow\:\mathrm{cos}\left({a}−\pi/\mathrm{6}\right)\:+\:\mathrm{cos}\left({b}−\pi/\mathrm{6}\right)\:+\:\mathrm{cos}\left({c}−\pi/\mathrm{6}\right)\:\geqslant\:\mathrm{0}\:.\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$ Commented…
Question Number 221348 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−\mathrm{2}^{{x}} }{{x}−\mathrm{2}} \\ $$ Answered by mr W last updated on 31/May/25 $$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{4}\left(\mathrm{1}−\mathrm{2}^{{x}} \right)}{{x}}…
Question Number 221350 by MrGaster last updated on 31/May/25 $$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{cos}\:\pi{x}}{\Gamma\left(\mathrm{2}+{x}\right)\Gamma\left(\mathrm{2}−{x}\right)}{dx} \\ $$$$ \\ $$$$ \\ $$ Commented by Ghisom last updated on 31/May/25…
Question Number 221347 by RoseAli last updated on 31/May/25 $$\underset{{x}\rightarrow\mathrm{2}} {\mathrm{lim}}\:\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$ Answered by AntonCWX8 last updated on 31/May/25 $$\frac{\mathrm{4}−{x}^{\mathrm{2}} }{{x}−\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}+{x}\right)\left(\mathrm{2}−{x}\right)}{{x}−\mathrm{2}}…
Question Number 221321 by dr1001sa last updated on 30/May/25 $${if}\:\mathrm{0}<{x}<{y}<{e}^{\mathrm{2}} \:{then} \\ $$$${y}^{\sqrt{{x}}} +{x}^{\mathrm{2}} +\mathrm{6}{xy}+\mathrm{18}{y}^{\mathrm{2}} +\frac{\mathrm{8}}{{x}}+\frac{\mathrm{16}}{\mathrm{9}{x}^{\mathrm{2}} {y}^{\mathrm{2}} }>\mathrm{2}+{x}^{\sqrt{{y}}} \\ $$ Terms of Service Privacy Policy…
Question Number 221322 by Frix last updated on 30/May/25 $$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}} }={x}^{\frac{\mathrm{1}}{{b}}} \\ $$ Answered by fantastic last updated on 30/May/25 $${or}\:{x}^{\frac{\mathrm{1}}{{a}}} +\sqrt{{x}^{\frac{{a}+{b}}{{ab}}}…
Question Number 221306 by Gbenga last updated on 30/May/25 $$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{csch}^{\mathrm{2}} \left(\pi{n}\right)}{{n}^{\mathrm{2}} } \\ $$ Answered by SdC355 last updated on 30/May/25 $$\underset{{l}=\mathrm{1}} {\overset{\infty}…
Question Number 221332 by fantastic last updated on 30/May/25 Answered by mr W last updated on 30/May/25 $${AB}={OB}=\sqrt{\mathrm{3}}\:{OD} \\ $$$$\frac{\Delta{DOE}}{\Delta{ABC}}=\left(\frac{{OD}}{{AB}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\Rightarrow\Delta{DOE}=\frac{\mathrm{12}}{\mathrm{3}}=\mathrm{4}\:{sq}.\:{units} \\…