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Category: Algebra

Why-x-x-2x-Explain-by-properties-laws-

Question Number 5782 by Rasheed Soomro last updated on 27/May/16 $$\mathrm{Why}\:\mathrm{x}+\mathrm{x}=\mathrm{2x}\:\:? \\ $$$$\mathrm{Explain}\:\mathrm{by}\:\mathrm{properties}/\mathrm{laws}. \\ $$ Answered by FilupSmith last updated on 27/May/16 $$\mathrm{if}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{basket}\:\mathrm{of} \\ $$$${x}\:\mathrm{apples}\:\mathrm{and}\:\mathrm{i}\:\mathrm{have}\:\mathrm{an}\:\mathrm{identical}\:\mathrm{basket}\:…

x-Z-x-p-1-p-2-p-n-p-x-p-Z-x-factors-of-p-1-fact-p-2-fact-p-n-How-could-you-formally-write-the-number-of-ways-you-can-write-x-in-terms-of-the-product-of-n-integers-

Question Number 5777 by FilupSmith last updated on 27/May/16 $$\exists{x}\in\mathbb{Z}^{+} :{x}={p}_{\mathrm{1}} {p}_{\mathrm{2}} …{p}_{{n}} \wedge\forall{p}\neq{x},\:{p}\in\mathbb{Z}^{+} \\ $$$$ \\ $$$${x}=\left({factors}\:{of}\:{p}_{\mathrm{1}} \right)\left({fact}.\:{p}_{\mathrm{2}} \right)…\left({fact}.\:{p}_{{n}} \right) \\ $$$$\mathrm{How}\:\mathrm{could}\:\mathrm{you}\:\mathrm{formally}\:\mathrm{write}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{you}\:\mathrm{can}\:\mathrm{write}\:{x}\:\mathrm{in}\:\mathrm{terms}…

x-1-3x-2-2-3x-2-x-1-3-Find-the-value-of-x-

Question Number 5765 by sanusihammed last updated on 26/May/16 $$\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}}\:\:\:+\:\:\mathrm{2}\:\left(\sqrt{\frac{\mathrm{3}{x}+\mathrm{2}}{{x}−\mathrm{1}}}\:\right)\:\:=\:\:\mathrm{3} \\ $$$$ \\ $$$${Find}\:{the}\:{value}\:{of}\:{x}\: \\ $$ Commented by prakash jain last updated on 27/May/16 $$\sqrt{\frac{{x}−\mathrm{1}}{\mathrm{3}{x}+\mathrm{2}}}={u}…

2-x-4x-Solution-2-x-4x-This-can-be-re-write-as-1-1-x-4x-Using-combination-to-epand-from-the-identity-1-x-n-1-nx-n-n-1-2-x-2-n-n-1-n-2-3-x-3-nCrx-r-Therefore

Question Number 5764 by sanusihammed last updated on 26/May/16 $$\mathrm{2}^{{x}} \:=\:\mathrm{4}{x} \\ $$$${Solution} \\ $$$$\mathrm{2}^{{x}} \:=\:\mathrm{4}{x} \\ $$$${This}\:{can}\:{be}\:{re}\:{write}\:{as} \\ $$$$\left(\mathrm{1}+\mathrm{1}\right)^{{x}} \:=\:\mathrm{4}{x} \\ $$$${Using}\:{combination}\:{to}\:{epand} \\ $$$${from}\:{the}\:{identity}.…