Question Number 45500 by Sanjarbek last updated on 13/Oct/18 Answered by MrW3 last updated on 13/Oct/18 $${x}=\mathrm{64}^{\frac{\mathrm{1}}{{x}}} ={e}^{\frac{\mathrm{ln}\:\mathrm{64}}{{x}}} \\ $$$$\mathrm{1}=\frac{\mathrm{1}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{64}}{{x}}} \\ $$$$\mathrm{ln}\:\mathrm{64}=\frac{\mathrm{ln}\:\mathrm{64}}{{x}}{e}^{\frac{\mathrm{ln}\:\mathrm{64}}{{x}}} \\ $$$$\frac{\mathrm{ln}\:\mathrm{64}}{{x}}=\mathbb{W}\left(\mathrm{ln}\:\mathrm{64}\right)\:\leftarrow{Lambert}\:{W}\:{function} \\…
Question Number 111039 by mathdave last updated on 01/Sep/20 $${solve}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{6}} }{dx} \\ $$$${mr}\:\:{abbo}\:{your}\:{question}\: \\ $$ Answered by mathdave last updated on…
Question Number 111035 by ZiYangLee last updated on 01/Sep/20 $$\mathrm{If}\:{b}\in\mathbb{Z}^{+} \:\forall\:\mathrm{both}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} −{bx}+\mathrm{132}=\mathrm{0}\:\mathrm{are}\:\mathrm{integers}. \\ $$$$\mathrm{Find} \\ $$$$\left(\mathrm{1}\right)\mathrm{the}\:\mathrm{largest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{b} \\ $$$$\left(\mathrm{2}\right)\mathrm{the}\:\mathrm{smallest}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{b}. \\ $$ Answered by 1549442205PVT…
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Question Number 176563 by MikeH last updated on 21/Sep/22 $$\mathrm{is}\:\mathrm{there}\:\mathrm{an}\:\mathrm{iOS}\:\mathrm{version}\:\mathrm{for}\:\mathrm{this}\:\mathrm{app}?\:\mathrm{please} \\ $$ Answered by Tinku Tara last updated on 23/Sep/22 $$\mathrm{There}\:\mathrm{is}\:\mathrm{no}\:\mathrm{ios}\:\mathrm{version} \\ $$ Terms of…
Question Number 111023 by mohammad17 last updated on 01/Sep/20 $$\int\frac{{sin}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$ Commented by mohammad17 last updated on 01/Sep/20 $${help}\:{me}\:{sir} \\ $$ Answered by…
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Question Number 111011 by mohammad17 last updated on 01/Sep/20 $${solve}:\:{y}^{''} +{y}^{'} ={tanx} \\ $$ Answered by mathmax by abdo last updated on 01/Sep/20 $$\left.\mathrm{h}\right)\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{r}\:=\mathrm{0}\:\Rightarrow\mathrm{r}\left(\mathrm{r}+\mathrm{1}\right)\:=\mathrm{0}\Rightarrow\mathrm{r}=\mathrm{0}\:\mathrm{or}\:\mathrm{r}=−\mathrm{1}\:\Rightarrow\mathrm{y}\left(\mathrm{x}\right)=\mathrm{ae}^{−\mathrm{x}}…
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