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Question-69644

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-69645

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}…

The-chord-of-contact-of-tangents-fromP-to-a-cicle-pass-through-Q-If-lengths-of-tangents-from-P-Q-are-l-1-l-2-then-PQ-is-l-1-2-l-2-2-how-kindly-tell-

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-69643

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…

How-many-distinct-ways-are-there-for-a-knight-to-reach-from-bottom-left-corner-of-chessboard-to-top-right-corner-knight-going-from-square-a1-to-h8-

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…

Prove-that-m-1-n-1-m-2-m-1-n-C-m-m-m-1-n-1-m-m-

Question Number 4103 by prakash jain last updated on 28/Dec/15 $$\mathrm{Prove}\:\mathrm{that} \\ $$$$\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} \:\centerdot\left(\mathrm{2}^{{m}} −\mathrm{1}\right)\:\centerdot\:^{{n}} {C}_{{m}} }{{m}}\:\:=\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}} \\ $$ Commented…

x-2-9-3x-5-x-3-x-1-x-3-x-1-Find-solution-

Question Number 135172 by bemath last updated on 11/Mar/21 $$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{3x}+\mathrm{5}} \:=\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} .\left(\mathrm{x}+\mathrm{3}\right)^{\mathrm{x}−\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{solution} \\ $$ Answered by john_santu last updated on 11/Mar/21 $$\Rightarrow\left({x}^{\mathrm{2}}…