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Category: Operation Research

Solve-x-2-y-3xy-5y-0-

Question Number 100327 by bobhans last updated on 26/Jun/20 $$\mathcal{S}\mathrm{olve}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}''−\mathrm{3xy}'−\mathrm{5y}=\mathrm{0} \\ $$ Commented by bobhans last updated on 26/Jun/20 $$\mathrm{n}\left(\mathrm{n}−\mathrm{1}\right)−\mathrm{3n}−\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{n}^{\mathrm{2}} −\mathrm{4n}−\mathrm{5}=\mathrm{0},\:\left(\mathrm{n}−\mathrm{5}\right)\left(\mathrm{n}+\mathrm{1}\right)=\mathrm{0} \\…

prove-by-contradiction-that-2-is-irrational-

Question Number 78333 by Lontum Hans last updated on 16/Jan/20 $$\mathrm{prove}\:\mathrm{by}\:\mathrm{contradiction}\:\mathrm{that}\:\sqrt{\mathrm{2}\:}\:\mathrm{is}\:\mathrm{irrational}. \\ $$ Answered by MJS last updated on 16/Jan/20 $$\sqrt{\mathrm{2}}\in\mathbb{Q}\wedge\sqrt{\mathrm{2}}>\mathrm{0}\:\Rightarrow\:\sqrt{\mathrm{2}}=\frac{{p}}{{q}} \\ $$$${p},\:{q}\:\in\mathbb{N}\wedge\mathrm{gcd}\:\left({p},\:{q}\right)\:=\mathrm{1}\:\Leftrightarrow\:{p}\nmid{q}\wedge{q}\nmid{p} \\ $$$$\left(\sqrt{\mathrm{2}}=\frac{{p}}{{q}}\right)^{\mathrm{2}}…

The-n-th-term-of-a-progression-is-np-q-and-the-sum-of-n-terms-is-denoted-by-S-n-Given-that-the-6-th-term-is-4-times-2-nd-term-and-that-S-3-12-find-the-value-of-p-and-q-Express-S-n-in-t

Question Number 11261 by 786786AM last updated on 18/Mar/17 $$\mathrm{The}\:\mathrm{n}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{progression}\:\mathrm{is}\:\mathrm{np}+\mathrm{q}\:\mathrm{and}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{n}\:\mathrm{terms}\:\mathrm{is}\:\mathrm{denoted}\:\mathrm{by}\:\mathrm{S}_{\mathrm{n}} . \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{6}^{\mathrm{th}} \:\mathrm{term}\:\mathrm{is}\:\mathrm{4}\:\mathrm{times}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{term}\:\mathrm{and}\:\mathrm{that}\:\mathrm{S}_{\mathrm{3}} \:=\mathrm{12},\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}. \\ $$$$\mathrm{Express}\:\mathrm{S}_{\mathrm{n}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{n}. \\ $$ Terms of Service…