Question Number 204152 by universe last updated on 07/Feb/24 $$\:\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{xy}−\mathrm{x}^{\mathrm{3}} \mathrm{y}^{\mathrm{2}} \\ $$$$\:\mathrm{attained}\:\mathrm{over}\:\mathrm{the}\:\mathrm{square}\:\mathrm{0}\leqslant\mathrm{x}\leqslant\mathrm{1};\mathrm{0}\leqslant\mathrm{y}\leqslant\mathrm{1}\:\mathrm{is} \\ $$ Answered by ajfour last updated on 08/Feb/24 $${f}\left({x},{y}\right)={z}={t}−{t}^{\mathrm{3}} \\ $$$$\frac{{dz}}{{dt}}=\mathrm{1}−\mathrm{3}{t}^{\mathrm{2}}…
Question Number 204145 by cortano12 last updated on 07/Feb/24 Answered by deleteduser1 last updated on 07/Feb/24 Commented by cortano12 last updated on 08/Feb/24 $$\:\mathrm{i}\:\mathrm{got}\:\mathrm{x}=\mathrm{40}° \\…
Question Number 204168 by mr W last updated on 07/Feb/24 Commented by mr W last updated on 07/Feb/24 $${find}\:{the}\:{locus}\:{of}\:{point}\:{P}. \\ $$ Answered by mr W…
Question Number 204171 by cherokeesay last updated on 07/Feb/24 Commented by AST last updated on 07/Feb/24 $${Are}\:{those}\:{points}\:{on}\:{the}\:{circle}\:{the}\:{vertices}\:{of}\:{a} \\ $$$${regular}\:{hexagon}?\:{Is}\:{H}\:{the}\:{midpoint}\:{of}\:\Gamma\:{and}\:{A}? \\ $$ Answered by AST last…
Question Number 204123 by Ari last updated on 06/Feb/24 Commented by Rasheed.Sindhi last updated on 06/Feb/24 $$\mathrm{0}\:{is}\:{even}\:{because}\:{it}'{s}\:{divisible}\:{by}\:\mathrm{2} \\ $$ Answered by Frix last updated on…
Question Number 204117 by mokys last updated on 06/Feb/24 $${solve}\:{for}\:\theta\:{in}\:{terms}\:{of}\:{y}\:{when}\:{y}\:=\:{e}^{\theta} \left({cos}\theta+{sin}\theta\right)\:\:\: \\ $$ Commented by mr W last updated on 06/Feb/24 $${impossible}! \\ $$ Commented…
Question Number 204141 by Tawa11 last updated on 06/Feb/24 Answered by AST last updated on 07/Feb/24 $${A}_{\mathrm{3}} \cap{A}_{−\mathrm{3}} =\left(−\infty,−\mathrm{3}\right]=\cap_{{i}=−\mathrm{3}} ^{\mathrm{3}} \\ $$ Commented by Tawa11…
Question Number 204142 by Tawa11 last updated on 06/Feb/24 $$\mathrm{Solve}:\:\:\mathrm{x}^{\mathrm{x}} \:\:=\:\:\mathrm{27}^{\mathrm{x}\:\:−\:\:\mathrm{3}} \\ $$ Answered by Frix last updated on 07/Feb/24 $${x}^{{x}} =\mathrm{27}^{{x}−\mathrm{3}} \\ $$$${x}\mathrm{ln}\:{x}\:=\left({x}−\mathrm{3}\right)\mathrm{ln}\:\mathrm{27} \\…
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Question Number 204105 by CrispyXYZ last updated on 06/Feb/24 $$\mathrm{find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\left(\mathrm{1}\:+\:\mathrm{cos}{x}\right)\:\mathrm{sin}{x} \\ $$$$\mathrm{without}\:\mathrm{derivative}. \\ $$ Answered by AST last updated on 06/Feb/24 $$\left(\mathrm{1}+{cosx}\right)\frac{\sqrt{\mathrm{3}}\left({sinx}\right)}{\:\sqrt{\mathrm{3}}}\leqslant\left(\frac{\mathrm{1}+{cos}\left({x}\right)+\sqrt{\mathrm{3}}{sinx}}{\mathrm{2}}\right)^{\mathrm{2}} ×\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\leqslant\left(\frac{\mathrm{1}+\sqrt{\mathrm{1}^{\mathrm{2}}…