Question Number 115489 by ruwedkabeh last updated on 26/Sep/20 $${find}\:{range}\:{for} \\ $$$$\mathrm{1}.\:{f}\left({x}\right)=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{9}}}{{x}−\mathrm{1}} \\ $$$$\mathrm{2}.\:{f}\left({x}\right)=\mathrm{ln}\:\left(\sqrt{\mathrm{4}−\mathrm{9}{x}^{\mathrm{2}} }\right) \\ $$ Answered by PRITHWISH SEN 2 last updated…
Question Number 49952 by maxmathsup by imad last updated on 12/Dec/18 $$\left.\mathrm{1}\right)\:{simplify}\:{A}_{{n}} =\:\frac{\mathrm{1}}{\left(\mathrm{2}+{i}\sqrt{\mathrm{3}}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left(\mathrm{2}−{i}\sqrt{\mathrm{3}}\right)^{{n}} } \\ $$$$\left.\mathrm{2}\right)\:{smplify}\:\:{B}_{{n}} =\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} }\:+\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{{n}} } \\ $$$${n}\:{integr}\:{natural}. \\ $$ Commented…
Question Number 49937 by turbo msup by abdo last updated on 12/Dec/18 $${let}\:{u}_{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:+\:\mathrm{3}{n}+\mathrm{1} \\ $$$${find}\:\sum_{{n}=\mathrm{0}} ^{\mathrm{49}} \:{u}_{{n}} \\ $$ Commented by Abdo…
Question Number 180901 by alcohol last updated on 19/Nov/22 $$\int_{\mathrm{1}} ^{\:{n}} \frac{\lfloor{x}\rfloor}{{x}^{\mathrm{2}} }{dx}\:=\: \\ $$ Answered by srikanth2684 last updated on 19/Nov/22 $$\int_{\mathrm{1}} ^{\:\mathrm{2}} \frac{\lfloor{x}\rfloor}{{x}^{\mathrm{2}}…
Question Number 49810 by Abdo msup. last updated on 10/Dec/18 $${calculate}\: \\ $$$${S}_{{n}} \left({x}\right)=\left[\frac{{x}+\mathrm{1}}{\mathrm{2}}\right]\:+\:\left[\frac{{x}+\mathrm{2}}{\mathrm{4}}\right]\:+\left[\frac{{x}+\mathrm{4}}{\mathrm{8}}\right]+…\left[\frac{{x}+\mathrm{2}^{{n}} }{\mathrm{2}^{{n}+\mathrm{1}} }\right] \\ $$ Commented by maxmathsup by imad last updated…
Question Number 49809 by Abdo msup. last updated on 10/Dec/18 $${solve}\:{the}\:{system}\:\:\:\begin{cases}{\frac{\mathrm{4}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}}\:=\frac{\mathrm{5}\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}=\frac{\mathrm{6}\sqrt{\mathrm{1}+{z}^{\mathrm{2}} }}{{z}}}\\{{x}+{y}+{z}={xyz}.}\end{cases} \\ $$$$\left.\begin{matrix}{}\\{}\end{matrix}\right\} \\ $$ Commented by MJS last updated on 12/Dec/18…
Question Number 49804 by maxmathsup by imad last updated on 10/Dec/18 $${let}\:{f}\left({x}\right)\:=\frac{{e}^{−{x}} }{{x}+\mathrm{1}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:\:{f}^{\left({n}\right)} \left({o}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{1}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$ Commented by Abdo msup.…
Question Number 115261 by bobhans last updated on 24/Sep/20 $${Equation}\:{of}\:{circle}\:{touching}\:{the}\:{line}\: \\ $$$$\mid{x}−\mathrm{2}\mid+\mid{y}−\mathrm{3}\mid\:=\:\mathrm{4}\:{will}\:{be}\: \\ $$ Answered by john santu last updated on 24/Sep/20 $${centre}\:{of}\:{circle}\:{is}\:{a}\:{point}\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${with}\:{radius}\:=\:\frac{\mid\mathrm{2}+\mathrm{3}−\mathrm{9}\mid}{\:\sqrt{\mathrm{2}}}\:=\mathrm{2}\sqrt{\mathrm{2}}…
Question Number 49703 by Rio Michael last updated on 09/Dec/18 $${Given}\:{f}\left({x}\right)=\:{sin}\mathrm{2}{x}+\mathrm{2}{cos}\mathrm{2}{x} \\ $$$${find}\:{f}'\left(\frac{\pi}{\mathrm{4}}\right) \\ $$$${hence}\:{given}\:{g}\left({x}\right)\:=\:\begin{cases}{{x},\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}}\\{\mathrm{3}{x}−{x}^{\mathrm{2}} ,\:\:\mathrm{2}\leqslant{x}\leqslant\mathrm{3}}\end{cases} \\ $$$${for}\:{a}\:{total}\:{range}\:{of}\:\mathrm{0}\leqslant{x}\leqslant\mathrm{3}\:\:{sketch}\:{the}\:{graph}\:{for}\:{y}={g}\left({x}\right) \\ $$$${and}\:{find}\:{the}\:{area}\:{with}\:{makes}\:{with}\:{the}\:{x}−{axis} \\ $$$${otherwise},\:{find}\:\:\:{the}\:{composite}\:{function}\:\:{gf}\:\:{in}\:{the}\:{range} \\ $$$$\mathrm{2}\leqslant{x}\leqslant\mathrm{3}. \\…
Question Number 49640 by maxmathsup by imad last updated on 08/Dec/18 $${if}\:{x}\:\in\left[{p},\sqrt{{p}^{\mathrm{2}} \:+\mathrm{2}}\right]\:\:{calculate}\:\left[{x}\right] \\ $$ Commented by maxmathsup by imad last updated on 08/Dec/18 $${p}\:{from}\:{N}…