Question Number 103478 by mohammad17 last updated on 15/Jul/20 Answered by bemath last updated on 15/Jul/20 $${HE}\:\Rightarrow\lambda^{\mathrm{2}} +\mathrm{3}\lambda−\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{2}=\mathrm{0} \\ $$$$\left(\lambda+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{17}}{\mathrm{4}}\:\Rightarrow\lambda=−\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\…
Question Number 169013 by 0731619 last updated on 23/Apr/22 Commented by cortano1 last updated on 23/Apr/22 $$\:=\:\sqrt{\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{{x}}{{x}+\sqrt{{x}+\sqrt{{x}}}}}\: \\ $$$$=\:\sqrt{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}}{{x}}+\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{3}} }}}}} \\ $$$$=\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}\:=\:\mathrm{1} \\…
Question Number 103477 by smartsmith459@gmail.com last updated on 15/Jul/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 103479 by mohammad17 last updated on 15/Jul/20 $${by}\:{ussing}\:{laplace}\:{find}\:{cosh}\left(\mathrm{2}{t}\right){cos}\left(\mathrm{4}{t}\right)\:? \\ $$ Answered by Worm_Tail last updated on 15/Jul/20 $${L}\left(\:{cosh}\left(\mathrm{2}{t}\right){cos}\left(\mathrm{4}{t}\right)\right)={L}\left(\left(\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right)\left(\frac{{e}^{{i}\mathrm{4}{t}} +{e}^{−{i}\mathrm{4}{t}} }{\mathrm{2}}\right)\right) \\…
Question Number 37930 by zesearho last updated on 19/Jun/18 $${n}\in\mathbb{N} \\ $$$${U}_{{n}+\mathrm{1}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}+\mathrm{1}} +{U}_{{n}} \\ $$$${U}_{{n}} =? \\ $$ Commented by abdo.msup.com last updated on…
Question Number 37871 by kunal1234523 last updated on 18/Jun/18 $$\mathrm{A}\:\mathrm{regular}\:\mathrm{pyramid}\:\mathrm{has}\:\mathrm{for}\:\mathrm{its}\:\mathrm{base}\:\mathrm{polygon} \\ $$$$\mathrm{of}\:{n}\:\mathrm{sides},\:\mathrm{and}\:\mathrm{each}\:\mathrm{slant}\:\mathrm{face}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{an}\: \\ $$$$\mathrm{isosceles}\:\mathrm{triangle}\:\mathrm{of}\:\mathrm{vertical}\:\mathrm{angle}\:\mathrm{2}\alpha.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{slant}\:\mathrm{faces}\:\mathrm{are}\:\mathrm{each}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{angle}\:\beta\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{base}\:,\:\mathrm{and}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{2}\gamma\:\mathrm{to}\:\mathrm{one}\:\mathrm{another} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{cos}\beta\:=\:\mathrm{tan}\:\alpha\:\mathrm{cot}\frac{\pi}{\mathrm{n}}\:,\:\mathrm{and}\:\mathrm{sin}\gamma\:=\:\mathrm{sec}\:\alpha\:\mathrm{cos}\frac{\pi}{\mathrm{n}} \\ $$ Commented…
Question Number 37864 by kunal1234523 last updated on 18/Jun/18 $$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} } \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18 $${cosx}.{cos}\mathrm{2}{x}.{cos}\mathrm{3}{x}.{cos}\mathrm{4}{x}.{cos}\mathrm{5}{x}.{cos}\mathrm{6}{x}.{cos}\mathrm{7}{x}={k} \\ $$$${x}=\frac{\Pi}{\mathrm{15}}…
Question Number 103400 by abony1303 last updated on 14/Jul/20 $$\mathrm{A}\:\mathrm{particle}'\mathrm{s}\:\mathrm{trajectory}\:\mathrm{is}\:\mathrm{described}\:\mathrm{by} \\ $$$$\mathrm{x}=\mathrm{e}^{\mathrm{t}} +\mathrm{e}^{−\mathrm{t}} \:\:\:\:\:\mathrm{y}=\mathrm{2t} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{that}\:\mathrm{the}\:\mathrm{particle} \\ $$$$\mathrm{traveled}\:\mathrm{for}\:\mathrm{0}\leqslant\mathrm{t}\leqslant\mathrm{2} \\ $$ Commented by abony1303 last updated…
Question Number 103393 by otchereabdullai@gmail.com last updated on 14/Jul/20 Answered by som(math1967) last updated on 14/Jul/20 $$\mathrm{x}+\mathrm{y}+\mathrm{r}=\mathrm{110}° \\ $$$$\mathrm{x}+\mathrm{2x}+\mathrm{2x}=\mathrm{110}\:\:\bigstar \\ $$$$\mathrm{5x}=\mathrm{110} \\ $$$$\mathrm{x}=\mathrm{22}°\:\mathrm{ans} \\ $$$$\bigstar\:\because\mathrm{2x}=\mathrm{r}=\mathrm{y}…
Question Number 37859 by kunal1234523 last updated on 18/Jun/18 $$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\underset{{i}=\mathrm{0}} {\overset{{k}−\mathrm{1}} {\prod}}\left({n}−{i}\right)}{{k}!} \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{really}\:\mathrm{mean}\:\mathrm{something} \\ $$ Commented by math khazana by abdo last…