Menu Close

Author: Tinku Tara

Question-208149

Question Number 208149 by mnjuly1970 last updated on 06/Jun/24 Answered by A5T last updated on 06/Jun/24 $$\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor>\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\Rightarrow{x}−\lfloor\mathrm{2}{x}^{\mathrm{2}} \rfloor<\mathrm{1}−\mathrm{2}{x}^{\mathrm{2}} +{x} \\ $$$${Suppose}\:{D}_{{f}} ,{R}_{{f}} \subseteq\mathbb{R}…

Question-208139

Question Number 208139 by efronzo1 last updated on 06/Jun/24 $$\:\:\:\: \\ $$ Answered by Ghisom last updated on 06/Jun/24 $${x}>\mathrm{0} \\ $$$$\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:+\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{5}\:+\mathrm{ln}\:\mathrm{2}}=\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\frac{\mathrm{ln}\:\left({x}^{\mathrm{2}}…

Solve-for-x-x-2-x-2-1-x-2-x-2-1-x-2-2-x-2-1-x-2-3-x-2-1-x-2-100-1-

Question Number 208164 by Fridunatjan08 last updated on 06/Jun/24 $${Solve}\:{for}\:{x}: \\ $$$${x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} +…+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} =\mathrm{1} \\…

y-3-cos-2-2-cos-find-max-y-

Question Number 208167 by hardmath last updated on 06/Jun/24 $$\mathrm{y}\:=\:\mathrm{3}\:\mathrm{cos}^{\mathrm{2}} \:\alpha\:+\:\mathrm{2}\:\mathrm{cos}\:\alpha \\ $$$$\mathrm{find}:\:\:\:\mathrm{max}\left(\mathrm{y}\right)\:=\:? \\ $$ Answered by A5T last updated on 07/Jun/24 $${y}\leqslant\mathrm{3}×\mathrm{1}+\mathrm{2}×\mathrm{1}=\mathrm{5}\Rightarrow{max}\left({y}\right)=\mathrm{5}\left({Equality}\:{at}\:\alpha=\mathrm{0}\right] \\ $$…

Question-208135

Question Number 208135 by efronzo1 last updated on 06/Jun/24 $$\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by A5T last updated on 06/Jun/24 $${r}^{{n}+\mathrm{2}} ={r}^{{n}+\mathrm{1}} +\frac{{r}^{{n}} }{\mathrm{2}}\Rightarrow{r}^{\mathrm{2}} ={r}+\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{r}=\frac{\mathrm{1}\underset{−} {+}\sqrt{\mathrm{3}}}{\mathrm{2}}…