Question Number 135181 by liberty last updated on 11/Mar/21 $$\mathrm{A}\:\mathrm{bag}\:\mathrm{has}\:\mathrm{4}\:\mathrm{red}\:\mathrm{marbles},\:\mathrm{5}\:\mathrm{white}\: \\ $$$$\mathrm{marbles}\:,\:\mathrm{and}\:\mathrm{6}\:\mathrm{blue}\:\mathrm{marbles}.\:\mathrm{Three} \\ $$$$\mathrm{marbles}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{from}\:\mathrm{the}\:\mathrm{bag},\:\left(\mathrm{without}\right. \\ $$$$\left.\mathrm{replacement}\right)\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability} \\ $$$$\mathrm{that}\:\mathrm{they}\:\mathrm{are}\:\mathrm{all}\:\mathrm{the}\:\mathrm{same}\:\mathrm{color}\:?\: \\ $$ Answered by EDWIN88 last updated…
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Question Number 69644 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by MJS last updated on 26/Sep/19 $$\mathrm{let} \\ $$$${x}=\alpha \\ $$$${y}=\beta−\sqrt{\gamma} \\ $$$${z}=\beta+\sqrt{\gamma} \\ $$$$\begin{cases}{\alpha+\mathrm{2}\beta−\mathrm{12}=\mathrm{0}}\\{\alpha^{\mathrm{2}}…
Question Number 69645 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by Rasheed.Sindhi last updated on 26/Sep/19 $$\sqrt{{x}+\sqrt{{x}+\sqrt{{x}….}}}=\mathrm{2} \\ $$$$\left(\sqrt{{x}+\sqrt{{x}+\sqrt{{x}….}}}\right)=\left(\mathrm{2}\right)^{\mathrm{2}} \\ $$$${x}+\sqrt{{x}+\sqrt{{x}+\sqrt{{x}…}}}=\mathrm{4} \\ $$$${x}+\mathrm{2}=\mathrm{4} \\ $$$${x}=\mathrm{2}…
Question Number 135176 by SLVR last updated on 11/Mar/21 $${The}\:{chord}\:{of}\:{contact}\:{of}\:{tangents}\:{fromP} \\ $$$${to}\:{a}\:{cicle}\:{pass}\:{through}\:{Q}.{If}\:{lengths}\:{of}\:{tangents}\:{from}\:{P},{Q} \\ $$$${are}\:{l}_{\mathrm{1}} ,{l}_{\mathrm{2}} \:{then}\:{PQ}\:{is}\:\sqrt{{l}_{\mathrm{1}} ^{\mathrm{2}} +{l}_{\mathrm{2}} ^{\mathrm{2}} }\:{how}…{kindly}\:{tell} \\ $$ Terms of Service…
Question Number 69643 by ahmadshahhimat775@gmail.com last updated on 26/Sep/19 Answered by MJS last updated on 26/Sep/19 $${l}+{w}=\mathrm{8} \\ $$$${l}^{\mathrm{2}} +{w}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow\:{l}=\mathrm{4}\pm\sqrt{\mathrm{2}}\wedge{w}=\mathrm{4}\mp\sqrt{\mathrm{2}} \\ $$$$\mathrm{perimeter}\:{P}=\mathrm{10}+\mathrm{3}\pi…
Question Number 4105 by prakash jain last updated on 28/Dec/15 $$\mathrm{How}\:\mathrm{many}\:\mathrm{distinct}\:\mathrm{ways}\:\mathrm{are}\:\mathrm{there}\:\mathrm{for} \\ $$$$\mathrm{a}\:\mathrm{knight}\:\mathrm{to}\:\mathrm{reach}\:\mathrm{from}\:\mathrm{bottom}\:\mathrm{left}\:\mathrm{corner}\:\mathrm{of} \\ $$$$\mathrm{chessboard}\:\mathrm{to}\:\mathrm{top}\:\mathrm{right}\:\mathrm{corner}. \\ $$$$\left(\mathrm{knight}\:\mathrm{going}\:\mathrm{from}\:\mathrm{square}\:\mathrm{a1}\:\mathrm{to}\:\mathrm{h8}\right). \\ $$ Commented by Filup last updated on…
Question Number 135178 by bemath last updated on 11/Mar/21 $$ \\ $$The top of a 29 feet ladder, leaning against a vertical wall is slipping…
Question Number 69641 by Henri Boucatchou last updated on 26/Sep/19 $$\boldsymbol{{Solve}}\:\:\boldsymbol{{arctg}}\left(\mathrm{1}−\boldsymbol{{x}}\right)\:+\:\frac{\mathrm{1}}{\boldsymbol{{arcctg}}\left(\mathrm{1}+\boldsymbol{{x}}\right)}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 135173 by bemath last updated on 11/Mar/21 Answered by john_santu last updated on 11/Mar/21 $${Homogenous}\:{problem} \\ $$$${y}_{{h}} '''−{y}_{{h}} ''−\mathrm{2}{y}_{{h}} '\:=\:\mathrm{0} \\ $$$${let}\:{y}_{{h}} \:=\:{e}^{{mx}}…