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Category: Algebra

if-x-x-26-x-5-find-x-5-x-

Question Number 151307 by mathdanisur last updated on 19/Aug/21 $$\mathrm{if}\:\:\mathrm{x}\sqrt{\mathrm{x}}\:-\:\mathrm{26}\sqrt{\mathrm{x}}\:=\:\mathrm{5} \\ $$$$\mathrm{find}\:\:\mathrm{x}\:-\:\mathrm{5}\sqrt{\mathrm{x}}\:=\:? \\ $$ Answered by MJS_new last updated on 20/Aug/21 $${x}\sqrt{{x}}−\mathrm{26}\sqrt{{x}}−\mathrm{5}=\mathrm{0} \\ $$$$\left(\sqrt{{x}}+\mathrm{5}\right)\left({x}−\mathrm{5}\sqrt{{x}}−\mathrm{1}\right)=\mathrm{0} \\…

5-6-02-10-23-solution-a-3-01-10-24-b-3-01-10-22-

Question Number 151300 by mathdanisur last updated on 19/Aug/21 $$\mathrm{5}\:\centerdot\:\mathrm{6},\mathrm{02}\centerdot\mathrm{10}^{\mathrm{23}} \:=\:?\:\left(\mathrm{solution}\right) \\ $$$$\left.\mathrm{a}\left.\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{24}} \:\:\:\mathrm{b}\right)\mathrm{3},\mathrm{01}\centerdot\mathrm{10}^{\mathrm{22}} \\ $$ Answered by puissant last updated on 19/Aug/21 $$\mathrm{5}×\mathrm{6},\mathrm{02}.\mathrm{10}^{\mathrm{23}} =\mathrm{3},\mathrm{1}.\mathrm{10}^{\mathrm{24}}…

if-a-3-x-1-b-3-y-2-c-3-z-1-1-5-find-abc-

Question Number 151284 by mathdanisur last updated on 19/Aug/21 $$\mathrm{if}\:\:\frac{\mathrm{a}}{\mathrm{3}^{\boldsymbol{\mathrm{x}}-\mathrm{1}} }\:=\:\frac{\mathrm{b}}{\mathrm{3}^{\boldsymbol{\mathrm{y}}+\mathrm{2}} }\:=\:\frac{\mathrm{c}}{\mathrm{3}^{\boldsymbol{\mathrm{z}}-\mathrm{1}} }\:=\:\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\mathrm{find}\:\:\mathrm{abc}\:=\:? \\ $$ Answered by Rasheed.Sindhi last updated on 19/Aug/21 $$\mathrm{a}=\frac{\mathrm{3}^{\mathrm{x}−\mathrm{1}}…

e-x-1-x-1-dx-

Question Number 151278 by mathdanisur last updated on 19/Aug/21 $$\int\:\frac{\boldsymbol{\mathrm{e}}^{\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}} }{\:\sqrt{\boldsymbol{\mathrm{x}}\:-\:\mathrm{1}}}\:\mathrm{dx}\:=\:? \\ $$ Answered by Ar Brandon last updated on 19/Aug/21 $$\mathrm{I}=\int\frac{{e}^{\sqrt{{x}−\mathrm{1}}} }{\:\sqrt{{x}−\mathrm{1}}}{dx},\:{u}=\sqrt{{x}−\mathrm{1}}\Rightarrow{du}=\frac{{dx}}{\mathrm{2}\sqrt{{x}−\mathrm{1}}} \\ $$$$\:\:=\mathrm{2}\int{e}^{{u}}…