Question Number 70066 by Maclaurin Stickker last updated on 30/Sep/19 $${Solve} \\ $$$$\left.\mathrm{a}\right)\:{e}^{\mathrm{2}{x}} −{e}^{{x}+\mathrm{1}} −{e}^{{x}} +{e}<\mathrm{0} \\ $$$$\left.\mathrm{b}\right)\mathrm{4}.\mathrm{2}^{\mathrm{2}{x}} −\mathrm{9}.\mathrm{2}^{{x}} <−\mathrm{2} \\ $$$$\left.{c}\right)\mathrm{9}^{{x}} −\mathrm{4}.\mathrm{3}^{{x}+\mathrm{1}} +\mathrm{27}>\mathrm{0} \\…
Question Number 135602 by Algoritm last updated on 14/Mar/21 Commented by Dwaipayan Shikari last updated on 14/Mar/21 $${Transendental}\:{Numbers}\:{can}'{t}\:{be}\:{An}\:{Algebraic}\:{solution} \\ $$ Terms of Service Privacy Policy…
Question Number 70044 by naka3546 last updated on 30/Sep/19 $$\sqrt{\mathrm{2016}\:+\:\mathrm{2007}\sqrt{\mathrm{2018}\:+\:\mathrm{2009}\sqrt{\mathrm{2020}\:+\:\mathrm{2011}\sqrt{\mathrm{2022}\:+\:\ldots}}}}\:\:=\:\:… \\ $$ Commented by Prithwish sen last updated on 30/Sep/19 $$\left(\mathrm{a}−\mathrm{6}\right)=\sqrt{\left(\mathrm{a}−\mathrm{6}\right)^{\mathrm{2}} }=\sqrt{\mathrm{a}+\left(\mathrm{a}^{\mathrm{2}} −\mathrm{13a}+\mathrm{36}\right)} \\ $$$$\:\:=\:\sqrt{\mathrm{a}+\left(\mathrm{a}−\mathrm{9}\right)\left(\mathrm{a}−\mathrm{4}\right)}\:=\:\sqrt{\mathrm{a}+\left(\mathrm{a}−\mathrm{9}\right)\sqrt{\left(\mathrm{a}−\mathrm{4}\right)^{\mathrm{2}}…
Question Number 135573 by malwan last updated on 14/Mar/21 $$ \\ $$ Commented by malwan last updated on 14/Mar/21 $${happy}\:{pi}\:{day} \\ $$ Commented by malwan…
Question Number 135572 by learner001 last updated on 14/Mar/21 $${does}\:{the}\:{sequence}\:\left\{{a}_{{n}} \right\}\:{defined}\:{by},\:{a}_{{n}} =\begin{cases}{\mathrm{0},\:{if}\:{n}\:{is}\:{odd}}\\{\frac{\mathrm{1}}{{n}},\:{if}\:{n}\:{is}\:{even}}\end{cases}\:\:\:{converge}. \\ $$ Answered by dhgt last updated on 05/May/21 Terms of Service Privacy…
Question Number 70021 by naka3546 last updated on 30/Sep/19 $$\sqrt[{\mathrm{3}}]{\sqrt{\sqrt{\mathrm{12345689654321233}\:+\:\mathrm{5333334096}\:\sqrt{\mathrm{12345679}}}}\:−\:\sqrt{\sqrt{\mathrm{12345689654321233}\:−\:\mathrm{5333334096}\:\sqrt{\mathrm{12345679}}}}}\:\:=\:\:… \\ $$ Commented by MJS last updated on 30/Sep/19 $$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve} \\ $$$$\sqrt{{a}+{b}\sqrt{{c}}}={p}+{q}\sqrt{{c}}\:\Leftrightarrow\:{a}={p}^{\mathrm{2}} +{cq}^{\mathrm{2}} \wedge{b}=\mathrm{2}{pq} \\…
Question Number 135544 by 0731619177 last updated on 13/Mar/21 Answered by Olaf last updated on 13/Mar/21 $${f}\left({x}\right)\:=\:\mathrm{sin}{x}+\mathrm{cos}{x}\:=\:\sqrt{\mathrm{2}}\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−{x}\right) \\ $$$${x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\:\Rightarrow\:{f}\left({x}\right)\in\left[\mathrm{1},\sqrt{\mathrm{2}}\right] \\ $$$$\Rightarrow\:\forall{x}\in\left[\mathrm{0},\frac{\pi}{\mathrm{4}}\right]\:\lfloor{f}\left({x}\right)\rfloor\:=\:\mathrm{1} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{cos}\left(\mathrm{2}{x}\right)×\mathrm{2}^{\mathrm{1}}…
Question Number 135543 by 0731619177 last updated on 13/Mar/21 Answered by Olaf last updated on 13/Mar/21 $$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}\left(\mathrm{2}{x}\right)\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}}\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right){dx}…
Question Number 135540 by mohammad17 last updated on 13/Mar/21 Answered by Olaf last updated on 13/Mar/21 $$\left.\mathrm{Q1a}\right) \\ $$$$\mathrm{1}/\:\:\:\mathrm{S}_{{n}} \:=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left[\frac{\mathrm{1}}{\mathrm{5}^{{k}} }−\frac{\mathrm{1}}{{k}\left({k}+\mathrm{1}\right)}\right] \\ $$$$\mathrm{S}_{{n}}…
Question Number 135539 by mohammad17 last updated on 13/Mar/21 Terms of Service Privacy Policy Contact: info@tinkutara.com