Question Number 172356 by Mikenice last updated on 25/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 172354 by Mikenice last updated on 25/Jun/22 Answered by mahdipoor last updated on 25/Jun/22 $${log}_{{a}} {b}={m}\:\:\:\:{log}_{{a}} {c}={n} \\ $$$$\Rightarrow\begin{cases}{{log}_{{n}^{\mathrm{2}} } \left(\mathrm{1}/{m}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{log}_{{n}} {m}=\frac{−\mathrm{3}}{\mathrm{2}}}\\{\frac{{log}_{{a}} {b}}{{log}_{{a}}…
Question Number 106815 by pticantor last updated on 07/Aug/20 $$\:\:\:\boldsymbol{{please}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{to}}\:\boldsymbol{{show}} \\ $$$$\boldsymbol{{that}}\:\:\boldsymbol{{the}}\:\boldsymbol{{equation}}\: \\ $$$$\:\boldsymbol{{X}}^{\boldsymbol{{n}}} +\boldsymbol{{aX}}+\boldsymbol{{c}}=\mathrm{0}\:\boldsymbol{{can}}\:\boldsymbol{{not}}\:\boldsymbol{{have}} \\ $$$$\boldsymbol{{more}}\:\boldsymbol{{than}}\:\mathrm{3}\:\boldsymbol{{reals}}\:\boldsymbol{{solutions}} \\ $$$$ \\ $$ Answered by 1549442205PVT last…
Question Number 172347 by Mikenice last updated on 25/Jun/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 106792 by bobhans last updated on 07/Aug/20 $$\:\:\:\:\:\:\succ\mathrm{bobhans}\prec \\ $$$$\mathrm{From}\:\mathrm{a}\:\mathrm{batch}\:\mathrm{containing}\:\mathrm{6}\:\mathrm{boys}\:\mathrm{and}\:\mathrm{4}\:\mathrm{girls} \\ $$$$\mathrm{a}\:\mathrm{group}\:\mathrm{of}\:\mathrm{4}\:\mathrm{students}\:\mathrm{is}\:\mathrm{tobe}\:\mathrm{selected}\:. \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{group}\:\mathrm{formations}\:\mathrm{will}\:\mathrm{have} \\ $$$$\mathrm{exactly}\:\mathrm{2}\:\mathrm{girls}? \\ $$ Answered by bemath last updated…
Question Number 172008 by Mikenice last updated on 23/Jun/22 $${solve}: \\ $$$$\mathrm{15}\left(\mathrm{2}{n}\right)_{{C}_{\left({n}−\mathrm{1}\right)} } =\mathrm{28}\left(\mathrm{2}{n}−\mathrm{1}\right)_{{C}_{{n}} } .\:{find}\:{n} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 171845 by Mikenice last updated on 21/Jun/22 $${n}_{{c}_{{r}+\mathrm{1}} } +\:{n}_{{c}_{{r}} } ={n}+\mathrm{1}_{{c}_{{r}\:\:.{solve}\:{for}\:\:{n}\:{and}\:{r}.} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 106305 by bemath last updated on 04/Aug/20 $$\mathrm{what}\:\mathrm{is}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{5}\:\mathrm{coming}\:\mathrm{up}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{if}\:\mathrm{a}\:\mathrm{die}\: \\ $$$$\mathrm{is}\:\mathrm{rolled}\:\mathrm{3}\:\mathrm{times}\: \\ $$ Answered by Ar Brandon last updated on 04/Aug/20 $$\mathrm{P}\left(\mathrm{Atleast}\:\mathrm{one}\:\mathrm{5}\right)=\mathrm{1}−\mathrm{P}\left(\mathrm{No}\:\mathrm{5}\right) \\ $$$$\mathrm{P}\left(\mathrm{No}\:\mathrm{5}\right)=\left(\frac{\mathrm{5}}{\mathrm{6}}\right)^{\mathrm{3}}…
Question Number 106292 by bemath last updated on 04/Aug/20 $$\mathrm{a}\:\mathrm{box}\:\mathrm{contains}\:\mathrm{4}\:\mathrm{blue},\:\mathrm{3}\:\mathrm{green}\:\mathrm{and}\:\mathrm{2}\:\mathrm{red}\:\mathrm{identicall}\:\mathrm{balls}.\: \\ $$$$\mathrm{if}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{are}\:\mathrm{selected}\:\mathrm{at}\:\mathrm{random}\:\mathrm{without}\: \\ $$$$\mathrm{replacement}\:,\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{two}\:\mathrm{balls}\:\mathrm{be}\:\mathrm{of}\:\mathrm{the}\:\mathrm{same}\:\mathrm{colours}? \\ $$ Answered by bobhans last updated on 04/Aug/20…
Question Number 171761 by Mikenice last updated on 20/Jun/22 $${solve}: \\ $$$$\frac{{sin}\left(\mathrm{10}+{x}\right)}{{sin}\left(\mathrm{13}+{x}\right)}=\frac{{sin}\mathrm{27}{sin}\mathrm{39}}{{sin}\mathrm{24}{sin}\mathrm{57}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com