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Knowing-that-k-0-n-n-k-2-2n-n-We-have-p-tokens-and-n-boxes-Each-box-is-labeled-with-a-number-from-1-to-n-Each-box-is-enough-big-to-receive-all-p-tokens-In-how-many

Question Number 125413 by Hassen_Timol last updated on 11/Dec/20 $$\mathrm{Knowing}\:\mathrm{that}\:: \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}^{\mathrm{2}} \:=\:\begin{pmatrix}{\mathrm{2}{n}}\\{\:\:{n}}\end{pmatrix} \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{have}\:{p}\:\mathrm{tokens}\:\mathrm{and}\:{n}\:\mathrm{boxes}. \\ $$$$\mathrm{Each}\:\mathrm{box}\:\mathrm{is}\:\mathrm{labeled}\:\mathrm{with}\:\mathrm{a}\:\mathrm{number}\:\mathrm{from}\:\mathrm{1}\:\mathrm{to}\:{n}. \\ $$$$\mathrm{Each}\:\mathrm{box}\:\mathrm{is}\:\mathrm{enough}\:\mathrm{big}\:\mathrm{to}\:\mathrm{receive}\:\mathrm{all}\:{p}\:\mathrm{tokens}. \\ $$$$…

a-b-c-a-a-2-b-2-c-2-a-a-2-b-2-c-2-prove-

Question Number 59861 by ANTARES VY last updated on 15/May/19 $$\sqrt{\boldsymbol{{a}}+\boldsymbol{{b}}\sqrt{\boldsymbol{{c}}}}=\sqrt{\frac{\boldsymbol{{a}}+\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}+\sqrt{\frac{\boldsymbol{{a}}−\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{c}}}}{\mathrm{2}}}. \\ $$$$\boldsymbol{{prove}} \\ $$ Commented by Kunal12588 last updated on…

P-1-1-i-P-n-1-P-n-1-P-n-1-lim-n-Im-P-n-

Question Number 59858 by jimful last updated on 15/May/19 $$\mathrm{P}\left(\mathrm{1}\right)=\frac{\mathrm{1}}{{i}}\: \\ $$$$\mathrm{P}_{\mathrm{n}+\mathrm{1}} \mathrm{P}_{\mathrm{n}} =\mathrm{1}−\mathrm{P}_{\mathrm{n}+\mathrm{1}} \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}Im}\left(\mathrm{P}_{\mathrm{n}} \right)\:=? \\ $$ Terms of Service Privacy Policy…

x-y-5pi-3-sin-x-2sin-y-

Question Number 125359 by weltr last updated on 10/Dec/20 $$\begin{cases}{{x}\:+\:{y}\:=\:\frac{\mathrm{5}\pi}{\mathrm{3}}}\\{{sin}\:{x}\:=\:\mathrm{2}{sin}\:{y}}\end{cases} \\ $$ Answered by Ar Brandon last updated on 10/Dec/20 $$\begin{cases}{\mathrm{x}+\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}}&{…\left(\mathrm{i}\right)}\\{\mathrm{sinx}=\mathrm{2siny}}&{…\left(\mathrm{ii}\right)}\end{cases} \\ $$$$\mathrm{From}\:\left(\mathrm{i}\right),\:\mathrm{x}+\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}\:\Rightarrow\:\mathrm{y}=\frac{\mathrm{5}\pi}{\mathrm{3}}−\mathrm{x} \\ $$$$\mathrm{substituting}\:\mathrm{the}\:\mathrm{above}\:\mathrm{result}\:\mathrm{in}\:\left(\mathrm{ii}\right)\:\mathrm{we}\:\mathrm{have}\:;…

6-3-3-1-3-9-1-3-simplify-

Question Number 59812 by ANTARES VY last updated on 15/May/19 $$\frac{\mathrm{6}}{\mathrm{3}+\sqrt[{\mathrm{3}}]{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{9}}}\:\:\:\boldsymbol{\mathrm{simplify}}. \\ $$ Commented by Kunal12588 last updated on 15/May/19 $${let}\:\sqrt[{\mathrm{3}}]{\mathrm{3}}={a} \\ $$$$\Rightarrow\sqrt[{\mathrm{3}}]{\mathrm{9}}={a}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}={a}^{\mathrm{3}}…