Question Number 149667 by mnjuly1970 last updated on 06/Aug/21 $$\: \\ $$$${f}\:\left({x}\:\right)=\:\frac{\mathrm{1}}{\:\sqrt{\:\mathrm{1}\:+\:{sin}\:\left({x}\:\right)}\:+\sqrt{\:\mathrm{1}\:+\:{cos}\:\left({x}\right)}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{find}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Min}\left(\:{f}\:\left({x}\right)\right)\:=? \\ $$$$ \\ $$ Answered by iloveisrael last updated…
Question Number 18593 by Joel577 last updated on 25/Jul/17 Answered by 433 last updated on 25/Jul/17 $$ \\ $$$$ \\ $$$$ \\ $$$$\frac{\mathrm{2016}\left(\frac{\mathrm{1}^{\mathrm{2015}} }{{n}^{\mathrm{2014}} }+\frac{\mathrm{2}^{\mathrm{2015}}…
Question Number 84126 by Roland Mbunwe last updated on 09/Mar/20 $$\int\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{\left({x}−\mathrm{4}\right)}}\boldsymbol{{dx}} \\ $$ Answered by MJS last updated on 09/Mar/20 $$−\int\frac{{x}−\mathrm{5}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx}=\int\left(\mathrm{1}−\sqrt{{x}−\mathrm{4}}\right){dx}= \\ $$$$={x}−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{4}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +{C} \\…
Question Number 149660 by fotosy2k last updated on 06/Aug/21 Answered by JDamian last updated on 06/Aug/21 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{a}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }+\frac{{b}^{{n}+\mathrm{1}} }{{a}^{{n}} +{b}^{{n}} }\right)= \\…
Question Number 18590 by Joel577 last updated on 25/Jul/17 $$\int\:\frac{\mathrm{sin}\:{x}}{\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \:{x}}\:{dx} \\ $$ Answered by Arnab Maiti last updated on 25/Jul/17 $$\mathrm{put}\:\mathrm{cos}\:\mathrm{x}=\mathrm{z} \\ $$$$\:\:\:\:\:−\mathrm{sin}\:\mathrm{x}\:\mathrm{dx}=\mathrm{dz} \\…
Question Number 84125 by redmiiuser last updated on 09/Mar/20 $${is}\:{it}\:{possible}\:{to}\:{find}\: \\ $$$${the}\:{no}.\:{of}\:{positive}\: \\ $$$${integral}\:{solutions} \\ $$$${of}\:{x}+{y}+\mathrm{2}{z}+\mathrm{3}{c}=\mathrm{7} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 18588 by Joel577 last updated on 25/Jul/17 $$\mathrm{sin}\:{x}\:=\:\frac{\mathrm{2}{a}\:+\:\mathrm{3}}{{a}\:+\:\mathrm{1}} \\ $$$$\mathrm{How}\:\mathrm{many}\:{a}\:\mathrm{that}\:\mathrm{can}\:\mathrm{satisfy}\:\mathrm{the} \\ $$$$\mathrm{equation}\:\mathrm{above}? \\ $$ Answered by ajfour last updated on 25/Jul/17 $$−\mathrm{1}\leqslant\frac{\mathrm{2a}+\mathrm{3}}{\mathrm{a}+\mathrm{1}}\leqslant\mathrm{1}\:\:\:;\:\:\mathrm{a}\neq−\mathrm{1} \\…
Question Number 18587 by Joel577 last updated on 25/Jul/17 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{{x}\:\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:+\:\mathrm{1}} \\ $$ Commented by Joel577 last updated on 25/Jul/17 $$\Rightarrow\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{{x}\:\mathrm{sin}\:{x}\:+\:\mathrm{2sin}^{\mathrm{2}} \:\frac{\mathrm{1}}{\mathrm{2}}{x}}\:=\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{{x}\:.\:\mathrm{tan}\:{x}}{\mathrm{sin}\:{x}\:\left({x}\:+\:\mathrm{2sin}\:\frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\…
Question Number 84123 by Roland Mbunwe last updated on 09/Mar/20 $$\int\frac{\mathrm{5}−\boldsymbol{{x}}}{\mathrm{1}+\sqrt{\left(\boldsymbol{{x}}−\mathrm{4}\right)}} \\ $$ Commented by mathmax by abdo last updated on 09/Mar/20 $${I}=\int\:\frac{\mathrm{5}−{x}}{\mathrm{1}+\sqrt{{x}−\mathrm{4}}}{dx}\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}−\mathrm{4}}={t}\:\Rightarrow{x}−\mathrm{4}={t}^{\mathrm{2}} \\ $$$${I}\:=\int\:\frac{\mathrm{5}−\left(\mathrm{4}+{t}^{\mathrm{2}}…
Question Number 84121 by Rio Michael last updated on 09/Mar/20 $$\mathrm{given}\:\:\mathrm{that} \\ $$$$\:\mathrm{g}\left({x}\right)\:=\:\begin{cases}{{x}\:+\:\mathrm{2}\:,\:\mathrm{if}\:\:\mathrm{0}\:\leqslant\:{x}\:<\:\mathrm{2}}\\{{x}^{\mathrm{2}} \:,\:\mathrm{if}\:\:\mathrm{2}\:\leqslant\:{x}\:<\:\mathrm{4}}\end{cases} \\ $$$$\mathrm{is}\:\mathrm{periodic}\:\mathrm{of}\:\mathrm{period}\:\mathrm{4}.\: \\ $$$$\mathrm{sketch}\:\mathrm{the}\:\mathrm{curve}\:\mathrm{for}\:\mathrm{g}\left({x}\right)\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\:\:\mathrm{0}\leqslant\:{x}\:<\:\mathrm{8} \\ $$$$\mathrm{evaluate}\:\:\mathrm{g}\left(−\mathrm{6}\right). \\ $$ Commented…