Question Number 185151 by mathlove last updated on 17/Jan/23 Answered by mr W last updated on 20/Jan/23 $${say}\:{the}\:{touching}\:{point}\:{with}\:{the} \\ $$$${parabola}\:{is}\:\left(−{p},\frac{{p}^{\mathrm{2}} }{\mathrm{2}}\right). \\ $$$${the}\:{center}\:{of}\:{circle}\:{is}\:\left(−{R},{h}\right). \\ $$$$\mathrm{tan}\:\theta={p}…
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Question Number 185137 by emmanuelson123 last updated on 17/Jan/23 Answered by CElcedricjunior last updated on 17/Jan/23 $$\boldsymbol{{montrons}}\:\boldsymbol{{que}}\:\forall\boldsymbol{{n}}\in\boldsymbol{{IN}} \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\boldsymbol{{n}}} −\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\boldsymbol{{n}}} \right]\equiv\mathrm{0}\left[\mathrm{6}\right] \\ $$$$\boldsymbol{{posons}}\:\boldsymbol{{p}}\left(\boldsymbol{{n}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left[\left(\mathrm{7}+\mathrm{4}\sqrt{\mathrm{3}}\right)^{\boldsymbol{{n}}} −\left(\mathrm{7}−\mathrm{4}\sqrt{\mathrm{3}}\right)^{\boldsymbol{{n}}} \right]…
Question Number 185128 by Shrinava last updated on 17/Jan/23 Answered by Frix last updated on 17/Jan/23 $$\mathrm{Formulas}\:\mathrm{for}\:\mathrm{incircle}\:\mathrm{and}\:\mathrm{3}\:\mathrm{excircles}: \\ $$$${r}=\frac{\mathrm{2}{A}}{{a}+{b}+{c}} \\ $$$${r}_{{a}} =\frac{\mathrm{2}{A}}{{b}+{c}−{a}} \\ $$$${r}_{{b}} =\frac{\mathrm{2}{A}}{{a}+{c}−{b}}…
Question Number 185127 by Shrinava last updated on 17/Jan/23 $$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{2}} \:−\:\frac{\mathrm{5}}{\mathrm{9}}\:+\:\sqrt{\mathrm{y}\:+\:\mathrm{x}^{\mathrm{2}} \:−\:\frac{\mathrm{5}}{\mathrm{9}}}}\:=\:\mathrm{y}}\\{\sqrt{\mathrm{y}^{\mathrm{2}} \:−\:\frac{\mathrm{4}}{\mathrm{9}}\:+\:\sqrt{\mathrm{x}\:+\:\mathrm{y}^{\mathrm{2}} \:−\:\frac{\mathrm{4}}{\mathrm{9}}}}\:=\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}\:,\:\mathrm{y}\:=\:? \\ $$ Commented by Frix last updated on 17/Jan/23…
Question Number 185126 by Shrinava last updated on 17/Jan/23 $$\left(\mid\mathrm{y}\mid−\mid\mathrm{x}\mid−\mathrm{2},\mathrm{8}\right)\left(\mid\mathrm{x}\mid−\mid\mathrm{y}\mid−\mathrm{1},\mathrm{8}\right)\:=\:\mathrm{0} \\ $$$$\mid\mathrm{y}\mid\:=\:\mathrm{a}\:\mid\mathrm{x}\mid\:+\:\mathrm{a}\:\mid\mid\mathrm{x}\mid\:−\:\mathrm{1},\mathrm{1}\mid\:+\:\mathrm{5} \\ $$$$\mathrm{At}\:\mathrm{what}\:\mathrm{value}\:\mathrm{of}\:\:\boldsymbol{\mathrm{a}}\:\:\mathrm{does}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{have}\:\:\mathrm{4}\:\:\mathrm{roots}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 119580 by mathocean1 last updated on 25/Oct/20 $$\mathrm{Given}\:\mathrm{k}\:\in\:\mathbb{N}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{justify}\:\mathrm{these}\:\mathrm{relations}: \\ $$$$\mathrm{3}^{\mathrm{2k}} +\mathrm{1}\equiv\mathrm{2}\left[\mathrm{8}\right]\:\mathrm{and}\:\mathrm{3}^{\mathrm{2k}+\mathrm{1}} +\mathrm{1}\equiv\mathrm{4}\left[\mathrm{8}\right]. \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Given}\:\left(\mathrm{E}\right):\:\mathrm{2}^{\mathrm{n}} −\mathrm{3}^{\mathrm{m}} =\mathrm{1}.\:\mathrm{n}\:\mathrm{and}\:\mathrm{m}\:\mathrm{are}\:\mathrm{unknowed}. \\ $$$$\bullet\:\mathrm{Show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{m}\:\mathrm{is}\:\mathrm{even}\:,\:\left(\mathrm{E}\right)\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\: \\ $$$$\mathrm{solution}. \\…
Question Number 185122 by Shrinava last updated on 17/Jan/23 $$\begin{cases}{\sqrt{\mathrm{x}^{\mathrm{2}} \:−\:\frac{\mathrm{5}}{\mathrm{9}}\:+\:\sqrt{\mathrm{y}\:+\:\mathrm{x}^{\mathrm{2}} \:−\:\frac{\mathrm{4}}{\mathrm{9}}}}\:=\:\mathrm{y}}\\{\sqrt{\mathrm{y}^{\mathrm{2}} \:−\:\frac{\mathrm{4}}{\mathrm{9}}\:+\:\sqrt{\mathrm{x}\:+\:\mathrm{y}^{\mathrm{2}} \:−\:\frac{\mathrm{4}}{\mathrm{9}}}}\:=\:\mathrm{x}}\end{cases} \\ $$$$\mathrm{find}:\:\:\:\mathrm{x}\:,\:\mathrm{y}\:=\:? \\ $$ Commented by Frix last updated on 17/Jan/23…
Question Number 119576 by bemath last updated on 25/Oct/20 Answered by TANMAY PANACEA last updated on 25/Oct/20 $${x}=\frac{\mathrm{2}{m}}{\mathrm{3}}\:\:{y}=\frac{\mathrm{2}{n}}{\mathrm{3}}\:\:\:{z}=\frac{\mathrm{2}{k}}{\mathrm{3}} \\ $$$$\frac{\mathrm{4}{m}}{\mathrm{3}}+\mathrm{2}{n}+\frac{\mathrm{2}{k}}{\mathrm{3}}=\mathrm{22} \\ $$$$\mathrm{4}{m}+\mathrm{6}{n}+\mathrm{2}{k}=\mathrm{66} \\ $$$$\mathrm{4}\left(\mathrm{13}−{k}\right)+\mathrm{6}{n}+\mathrm{2}{k}=\mathrm{66} \\…