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 $$\frac{\mathrm{200}}{\mathrm{200}+\mathrm{50}+\mathrm{x}}=\mathrm{0}.\mathrm{05}=\frac{\mathrm{1}}{\mathrm{20}}…
Question Number 63769 by aliesam last updated on 08/Jul/19 Commented by mathmax by abdo last updated on 09/Jul/19 $${let}\:{prove}\:{by}\:{recurence}\:{n}=\mathrm{0}\:\:\:\:{A}_{\mathrm{0}} =\mathrm{0}\:{is}\:{divisible}\:{by}\:\mathrm{6} \\ $$$${let}\:{suppose}\:{A}_{{n}} ={n}^{\mathrm{3}} \:+\mathrm{5}{n}\:{is}\:{divisible}\:{by}\:\mathrm{6}\:\Rightarrow \\…
Question Number 63763 by aliesam last updated on 08/Jul/19 Commented by Prithwish sen last updated on 08/Jul/19 $$\frac{\mathrm{2}}{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:+\:\mathrm{i}\:\mathrm{sin}\left(\mathrm{2x}\right)}\:=\:\frac{\mathrm{2}\left\{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)\right\}}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\:+\:\mathrm{sin}^{\mathrm{2}} \left(\mathrm{2x}\right)} \\ $$$$=\:\frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2x}\right)\:−\:\mathrm{isin}\left(\mathrm{2x}\right)}{\mathrm{1}+\:\mathrm{cos}\left(\mathrm{2x}\right)}\:=\:\mathrm{1}\:−\:\mathrm{i}\frac{\mathrm{2sinx}.\mathrm{cosx}}{\mathrm{2cos}^{\mathrm{2}} \mathrm{x}} \\ $$$$=\:\mathrm{1}\:−\mathrm{itanx}\:\mathrm{Hence}\:\mathrm{proved}.…
Question Number 63751 by Tawa1 last updated on 08/Jul/19 $$\mathrm{Where}\:\mathrm{is}\:\mathrm{sir}\:\mathrm{tanmay} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 129271 by MathSh last updated on 14/Jan/21 $${In}\:{the}\:{I}\:{park}\:\mathrm{50\%}\:{of}\:{the}\:{schools} \\ $$$${students},\:\mathrm{80\%}\:{in}\:{the}\:{II}\:{park}, \\ $$$$\mathrm{90\%}\:{has}\:{partic}\imath{pated}\:{in}\:{the}\:{III}\:{park}. \\ $$$$\mathrm{160}\:{students}\:{have}\:{partic}\imath{pated}\:{in} \\ $$$${all}\:\mathrm{3}\:{parks},\:{the}\:{rest}\:{have} \\ $$$${partic}\imath{pated}\:{in}\:\mathrm{2}\:{park}. \\ $$$${How}\:{many}\:{students}\:{are}\:{there}\:{in} \\ $$$${the}\:{school}\:?\: \\…