Question Number 63904 by naka3546 last updated on 11/Jul/19 Commented by Prithwish sen last updated on 13/Jul/19 $$\frac{\mathrm{6}^{\mathrm{k}} }{\left(\mathrm{3}^{\mathrm{k}} −\mathrm{2}^{\mathrm{k}} \right)\left(\mathrm{3}^{\mathrm{k}+\mathrm{1}} −\mathrm{2}^{\mathrm{k}+\mathrm{1}} \right)}\:=\:\frac{\mathrm{3}^{\mathrm{k}} }{\mathrm{3}^{\mathrm{k}} −\mathrm{2}^{\mathrm{k}}…
Question Number 63900 by Rio Michael last updated on 10/Jul/19 Commented by Rio Michael last updated on 10/Jul/19 $${Inthe}\:{Diagram}\:{above}\:{ABCD}\:{and}\:{CDEF}\:{are}\:{parrallelograms} \\ $$$${which}\:{lie}\:{in}\:{thesame}\:{plane}. \\ $$$${A}\overset{\rightarrow} {{B}}=\boldsymbol{{p}}\:,\:{B}\overset{\rightarrow} {{C}}=\boldsymbol{{q}}\:\:{and}\:{C}\overset{\rightarrow}…
Question Number 129430 by MathSh last updated on 15/Jan/21 $$\mathrm{0}<{x}<\mathrm{1}\:\:\:{if}, \\ $$$${Compare}\:{it}: \\ $$$$\frac{\mathrm{1}}{{x}−\mathrm{1}}\:;\:\frac{{x}}{{x}−\mathrm{1}}\:;\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\:;\:\frac{{x}}{\mathrm{1}−{x}}\:;\:\frac{{x}}{\mathrm{2}{x}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129435 by Tinku Tara last updated on 15/Jan/21 $$\mathrm{For}\:\mathrm{any}\:\mathrm{app}\:\mathrm{related}\:\mathrm{question}\:\mathrm{please}\:\mathrm{write}\:\mathrm{in} \\ $$$$\mathrm{English}.\:\mathrm{Thank}\:\:\mathrm{You}. \\ $$$$\mathrm{You}\:\mathrm{can}\:\mathrm{also}\:\mathrm{send}\:\mathrm{us}\:\mathrm{an}\:\mathrm{email} \\ $$ Commented by mr W last updated on 16/Jan/21…
Question Number 129419 by mathocean1 last updated on 15/Jan/21 $${Calculate} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\mathrm{2}{ln}−\mathrm{1}+\frac{\mathrm{1}}{{x}}\:{and}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {{lim}}\frac{{x}}{\mathrm{2}}\:+\:\frac{\mathrm{1}+{lnx}}{{x}} \\ $$$${Detail}\:{if}\:{possible}\:{sirs} \\ $$ Terms of Service Privacy Policy…
Question Number 129409 by Algoritm last updated on 15/Jan/21 Commented by soumyasaha last updated on 15/Jan/21 $$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{9}.\mathrm{11}}+\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{2}} .\mathrm{11}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{9}^{\mathrm{3}} .\mathrm{11}^{\mathrm{3}} }+…\right] \\ $$$$\:\:\:=\:\frac{\mathrm{4}}{\mathrm{9}}\left[\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{99}}+\left(\frac{\mathrm{1}}{\mathrm{99}}\right)^{\mathrm{2}} \:+\left(\frac{\mathrm{1}}{\mathrm{99}}\right)^{\mathrm{3}} +…\right]…
Question Number 129413 by ZiYangLee last updated on 15/Jan/21 $$\mathrm{Let}\:\mathrm{a}\:\mathrm{sequence}\:\left\{{a}_{{n}} \right\}\:\mathrm{satisfies} \\ $$$$\:\:\:\:\begin{cases}{\:\:\:\:\:\:\:\:\:\:\:\:{a}_{\mathrm{1}} =\mathrm{1}}\\{{na}_{{n}} ={n}+\mathrm{2}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{a}_{{k}} ,\:{n}>\mathrm{2}}\end{cases} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{a}_{\mathrm{2021}} . \\ $$ Commented by…
Question Number 129400 by SOMEDAVONG last updated on 15/Jan/21 $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$ Answered by MJS_new last updated on 15/Jan/21 $$\mathrm{in}\:\mathrm{cases}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}\:\mathrm{I}\:\mathrm{use}\:{Ostrogradski}'{s}…
Question Number 129380 by mohammad17 last updated on 15/Jan/21 Answered by physicstutes last updated on 15/Jan/21 $$\left\{{a}_{{n}} \right\}\:=\:{n}^{\mathrm{2}} \\ $$$${a}_{{n}+\mathrm{1}} \:=\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:{a}_{{n}+\mathrm{1}} −{a}_{{n}} \:=\:\left({n}+\mathrm{1}\right)^{\mathrm{2}}…
Question Number 129323 by harckinwunmy last updated on 14/Jan/21 $${In}\:{a}\:{competition}\:{there}\:{are}\:\mathrm{200}\:{children},\:\mathrm{50}\:{are}\:{men} \\ $$$${the}\:{rest}\:{are}\:{women}.\:{If}\:{the}\:{probability}\:{of}\:{selecting}\:{a}\: \\ $$$${child}\:{is}\:\mathrm{0}.\mathrm{05},\:{how}\:{many}\:{are}\:{the}\:{competitor}? \\ $$ Answered by Ar Brandon last updated on 15/Jan/21 $$\frac{\mathrm{200}}{\mathrm{200}+\mathrm{50}+\mathrm{x}}=\mathrm{0}.\mathrm{05}=\frac{\mathrm{1}}{\mathrm{20}}…