Question Number 206224 by mathzup last updated on 09/Apr/24 $${find}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left({x}^{\mathrm{5}} \right){dx} \\ $$ Answered by Frix last updated on 10/Apr/24 $$\int\mathrm{tan}^{−\mathrm{1}} \:{x}^{{n}} \:{dx}\:\underset{{u}'=\mathrm{1}\wedge{v}=\mathrm{tan}^{−\mathrm{1}}…
Question Number 206227 by bett last updated on 09/Apr/24 $${OA}=\left(\overset{{x}} {\mathrm{4}}\right)\:{OB}=_{\mathrm{7}} ^{\mathrm{5}} \:{and}\:{AB}=\mathrm{5}\:{units} \\ $$ Answered by A5T last updated on 09/Apr/24 $$\sqrt{\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +\left(\mathrm{7}−\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{5}\Rightarrow\mathrm{25}−\mathrm{16}=\mathrm{3}^{\mathrm{2}}…
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Question Number 206160 by MATHEMATICSAM last updated on 08/Apr/24 $$\mathrm{If}\:{x}\:\mathrm{and}\:{y}\:\mathrm{are}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{then}\:\mathrm{is}\:\mathrm{it} \\ $$$$\mathrm{possible}\:\mathrm{that}\:\mathrm{sec}\theta\:=\:\frac{\mathrm{2}{xy}}{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} }\:? \\ $$ Commented by mr W last updated on 08/Apr/24 $${only}\:{if}\:{x}=\pm{y}.…
Question Number 206156 by cortano21 last updated on 08/Apr/24 Commented by mr W last updated on 08/Apr/24 $${what}\:{i}\:{think}\:{is}\:{clear}: \\ $$$${in}\:{the}\:{package}\:{there}\:{must}\:{be}\:{exactly} \\ $$$${one}\:{egg}.\:{there}\:{must}\:{be}\:{at}\:{least}\:{three} \\ $$$${lettuces}.\:{there}\:{must}\:{be}\:{at}\:{least}\:{one} \\…
Question Number 206157 by cortano21 last updated on 08/Apr/24 Commented by cortano21 last updated on 08/Apr/24 $$\:\:\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$ Commented by A5T last updated…
Question Number 206155 by dimentri last updated on 08/Apr/24 $$\:\:\:\:\mathrm{x}\:=\:\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{2}+\frac{\mathrm{2}}{\mathrm{x}}}}}} \\ $$$$\:\:\:\mathrm{find}\:\mathrm{x}.\: \\ $$ Answered by MM42 last updated on 08/Apr/24 $$\mathrm{2}+\frac{\mathrm{2}}{{x}}=\frac{\mathrm{2}{x}+\mathrm{2}}{{x}}={a} \\ $$$$\mathrm{2}+\frac{\mathrm{2}}{{a}}=\mathrm{2}+\frac{{x}}{{x}+\mathrm{1}}=\frac{\mathrm{3}{x}+\mathrm{2}}{{x}+\mathrm{1}}={b} \\…
Question Number 206150 by Samuel12 last updated on 08/Apr/24 $$\mathrm{Calcul}\:\:\:\mid\mathrm{x}−\alpha\mid\:=\:????? \\ $$$$\:\:\bullet\:\:\:\:\:\:\mathrm{x}=−\mathrm{1}\:\:;\:\:\:\alpha\:\in\:\left[−\mathrm{1};\:−\frac{\mathrm{3}}{\mathrm{4}}\right] \\ $$$$\:\:\bullet\:\:\:\:\mathrm{x}=\mathrm{2}\:\:;\:\:\:\:\alpha\:\in\:\left[\mathrm{0}\:;\:\mathrm{3}\right]\:\:\:\:\:\:\:\mathrm{help}\:\mathrm{please}\:\: \\ $$ Answered by Frix last updated on 08/Apr/24 $$\mid{x}−\alpha\mid=\begin{cases}{{x}−\alpha;\:{x}−\alpha\geqslant\mathrm{0}}\\{−\left({x}−\alpha\right);\:{x}−\alpha<\mathrm{0}}\end{cases} \\…
Question Number 206151 by cortano21 last updated on 08/Apr/24 Answered by dimentri last updated on 08/Apr/24 $$\:\:\mathrm{x}=\mathrm{2}\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\mathrm{4}\: \\ $$$$\:\:\mathrm{x}=\mathrm{1}\Rightarrow\mathrm{f}\left(\mathrm{3}\right)=\:\mathrm{g}^{−\mathrm{1}} \left(\mathrm{1}\right) \\ $$$$\:\:\mathrm{g}\left(\mathrm{4}\right)=\:\mathrm{g}\left(\mathrm{f}\left(\mathrm{3}\right)\right)\:=\:\mathrm{g}\left(\mathrm{g}^{−\mathrm{1}} \left(\mathrm{1}\right)\right)\:=\:\mathrm{1} \\ $$…