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 Number 208105 by hardmath last updated on 05/Jun/24 $$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{8}}\:\:+\:\:\frac{\mathrm{3}}{\mathrm{64}}\:\:+\:\:\frac{\mathrm{4}}{\mathrm{256}}\:\:+\:\:…\:\:+\:\:=\:\:? \\ $$ Commented by Frix last updated on 05/Jun/24 $$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{2}}{\mathrm{16}}+\frac{\mathrm{3}}{\mathrm{64}}+\frac{\mathrm{4}}{\mathrm{256}}+…= \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{\infty}…
Question Number 208097 by efronzo1 last updated on 05/Jun/24 $$\:\:\:\:\downharpoonleft\underline{\:} \\ $$ Answered by MM42 last updated on 05/Jun/24 $$\mathrm{2}^{{x}−{m}+\mathrm{1}} −\mathrm{2}^{{x}−{m}} \:=\mathrm{8} \\ $$$$\Rightarrow\mathrm{2}^{{x}−{m}} =\mathrm{8}\Rightarrow{x}−{m}=\mathrm{3}\Rightarrow{x}={m}+\mathrm{3}…
Question Number 208076 by hardmath last updated on 04/Jun/24 $$\left(\mathrm{5}\:−\:\mid\mathrm{x}\mid\right)^{−\:\frac{\mathrm{1}}{\mathrm{3}}} \:\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{4}\right)\:<\:\mathrm{0} \\ $$ Answered by TonyCWX08 last updated on 04/Jun/24 $${This}\:{inequality}\:{are}\:{defined}\:{when}\:{x}\in\langle−\mathrm{5},\mathrm{5}\rangle \\ $$$${Two}\:{Possible}\:{Cases} \\…
Question Number 208090 by efronzo1 last updated on 04/Jun/24 $$\:\:\:\underline{\:} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208075 by hardmath last updated on 04/Jun/24 $$\mathrm{Find}:\:\:\:\int\:\mathrm{sin}\:\left(\mathrm{2x}\:−\:\frac{\pi}{\mathrm{4}}\right)\:=\:? \\ $$ Answered by TonyCWX08 last updated on 04/Jun/24 $${sin}\left(\mathrm{2}{x}−\frac{\pi}{\mathrm{4}}\right) \\ $$$$={sin}\left(\mathrm{2}{x}\right){cos}\left(\frac{\pi}{\mathrm{4}}\right)−{sin}\left(\frac{\pi}{\mathrm{4}}\right){cos}\left(\mathrm{2}{x}\right) \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left({sin}\left(\mathrm{2}{x}\right)−{cos}\left(\mathrm{2}{x}\right)\right) \\…
Question Number 208063 by hardmath last updated on 03/Jun/24 $$\begin{cases}{\mathrm{x}\:+\:\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{y}\:=\:\mathrm{5}}\\{\frac{\mathrm{3}}{\mathrm{5}}\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{11}}\end{cases}\:\:\:\:\:\Rightarrow\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:? \\ $$ Answered by A5T last updated on 03/Jun/24 $${x}+{y}+\frac{\mathrm{3}}{\mathrm{5}}\left({x}+{y}\right)=\mathrm{16}=\frac{\mathrm{8}\left({x}+{y}\right)}{\mathrm{5}}\Rightarrow{x}+{y}=\mathrm{10} \\ $$ Terms of Service…
Question Number 208071 by hardmath last updated on 03/Jun/24 $$\frac{\mathrm{4}\:\centerdot\:\mathrm{10}^{\boldsymbol{\mathrm{n}}} \:−\:\mathrm{52}}{\mathrm{3}} \\ $$$$\mathrm{The}\:\mathrm{numbers}\:\mathrm{are}\:\:\mathrm{38}\:\:\mathrm{in}\:\mathrm{total} \\ $$$$\mathrm{Find}:\:\:\boldsymbol{\mathrm{n}}\:=\:? \\ $$ Commented by mr W last updated on 04/Jun/24…
Question Number 208020 by Frix last updated on 02/Jun/24 $$\begin{cases}{\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{14}}{\mathrm{625}}}\\{\sqrt{{x}}+\sqrt{{y}}=\mathrm{8}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{all}\:\mathrm{solutions}. \\ $$ Answered by efronzo1 last updated on 02/Jun/24 $$\:\:\Rightarrow\mathrm{x}+\mathrm{y}\:+\mathrm{2}\sqrt{\mathrm{xy}}\:=\:\mathrm{64}\: \\ $$$$\:\Rightarrow\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{xy}}\:=\:\frac{\mathrm{14}}{\mathrm{625}}\:;\:\mathrm{x}+\mathrm{y}\:=\:\frac{\mathrm{14}}{\mathrm{625}}\mathrm{xy}\: \\…
Question Number 207984 by Thomaseinstein last updated on 02/Jun/24 Answered by Frix last updated on 02/Jun/24 $${p},\:{q}\:>\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{p}^{\mathrm{2}} }−\frac{\mathrm{1}}{{q}^{\mathrm{2}} }=\frac{\mathrm{4}}{\mathrm{3}}\:\Rightarrow\:{q}^{\mathrm{2}} =\frac{\mathrm{3}{p}^{\mathrm{2}} }{\:\mathrm{3}−\mathrm{4}{p}^{\mathrm{2}} } \\…