Question Number 208218 by hardmath last updated on 07/Jun/24 $$\mathrm{a}_{\boldsymbol{\mathrm{n}}} \:\:\mathrm{numbers}\:\mathrm{series} \\ $$$$\mathrm{If}\:\:\mathrm{S}_{\mathrm{16}} \:−\:\mathrm{S}_{\mathrm{13}} \:\:=\:\:\mathrm{S}_{\mathrm{106}} \:−\:\mathrm{S}_{\mathrm{103}} \\ $$$$\mathrm{Find}:\:\:\:\:\frac{\mathrm{3a}_{\mathrm{3}} \:+\:\mathrm{4a}_{\mathrm{4}} \:+\:\mathrm{5a}_{\mathrm{5}} }{\mathrm{2a}_{\mathrm{12}} }\:\:=\:\:? \\ $$ Commented…
Question Number 208187 by hardmath last updated on 07/Jun/24 $$\mathrm{Find}:\:\:\:\mathrm{1},\mathrm{03}^{\mathrm{200}} \:=\:? \\ $$ Answered by Ghisom last updated on 07/Jun/24 $$=\mathrm{10}^{\mathrm{200log}\:\mathrm{1}.\mathrm{03}} \approx\mathrm{10}^{\mathrm{200}×.\mathrm{012837}} \approx\mathrm{10}^{\mathrm{2}.\mathrm{5675}} \approx\mathrm{369}.\mathrm{36} \\…
Question Number 208199 by efronzo1 last updated on 07/Jun/24 Answered by Frix last updated on 07/Jun/24 $${y}=\sqrt{\mathrm{18}+\mathrm{3}{x}−{x}^{\mathrm{2}} }\:\mathrm{is}\:\mathrm{a}\:\mathrm{semi}\:\mathrm{circle}\:\mathrm{with}\:{r}=\frac{\mathrm{9}}{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{3}}+\sqrt{\mathrm{6}−{x}}\:\mathrm{has}\:\mathrm{the}\:\mathrm{maximum}\:\begin{pmatrix}{\frac{\mathrm{3}}{\mathrm{2}}}\\{\mathrm{3}\sqrt{\mathrm{2}}}\end{pmatrix} \\ $$$$\mathrm{We}\:\mathrm{have}\:\mathrm{2}\:\mathrm{solutions}\:\mathrm{for}\:\mathrm{0}\leqslant{m}<\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}}\:\mathrm{and} \\ $$$$\mathrm{exactly}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{at}\:{m}=\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\…
Question Number 208149 by mnjuly1970 last updated on 06/Jun/24 Answered by A5T last updated on 06/Jun/24 $$\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} \\ $$$${Suppose}\:{D}_{{f}} ,{R}_{{f}} \subseteq\mathbb{R}…
Question Number 208164 by Fridunatjan08 last updated on 06/Jun/24 $${Solve}\:{for}\:{x}: \\ $$$${x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} +{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{3}} +…+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{100}} =\mathrm{1} \\…
Question Number 208135 by efronzo1 last updated on 06/Jun/24 $$\:\:\:\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}}…
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} \\…