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Author: Tinku Tara

An-aqeous-solution-of-tetraoxosulphate-IV-has-a-density-of-1-80-g-cm-3-and-98-purity-level-what-volume-of-this-solution-must-be-diluted-to-give-250-cm-3-of-0-500-mol-dm-3-H-2-SO-4-solution-

Question Number 6583 by Tawakalitu. last updated on 04/Jul/16 $${An}\:{aqeous}\:{solution}\:{of}\:{tetraoxosulphate}\left({IV}\right)\:{has}\:{a}\:{density}\: \\ $$$${of}\:\mathrm{1}.\mathrm{80}\:{g}/{cm}^{\mathrm{3}} \:{and}\:\mathrm{98\%}\:{purity}\:{level}.\:{what}\:{volume}\:{of}\:{this}\: \\ $$$${solution}\:{must}\:{be}\:{diluted}\:{to}\:{give}\:\mathrm{250}\:{cm}^{\mathrm{3}} \:{of}\:\mathrm{0}.\mathrm{500}\:{mol}/{dm}^{\mathrm{3}} \\ $$$${H}_{\mathrm{2}} {SO}_{\mathrm{4}} \:{solution}\:? \\ $$$$ \\ $$ Answered…

Given-that-are-the-three-roots-of-the-equation-x-59-3-2x-64-3-3x-123-3-find-the-value-of-

Question Number 137652 by ZiYangLee last updated on 05/Apr/21 $$\mathrm{Given}\:\mathrm{that}\:\alpha,\beta,\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{three}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{equation}\:\left({x}−\mathrm{59}\right)^{\mathrm{3}} +\left(\mathrm{2}{x}−\mathrm{64}\right)^{\mathrm{3}} =\left(\mathrm{3}{x}−\mathrm{123}\right)^{\mathrm{3}} , \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\alpha+\beta+\gamma. \\ $$ Answered by mitica last updated on…

Question-72112

Question Number 72112 by A8;15: last updated on 24/Oct/19 Answered by mind is power last updated on 24/Oct/19 $$\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\Rightarrow\mathrm{dx}=\frac{−\mathrm{dt}}{\mathrm{t}^{\mathrm{2}} } \\ $$$$\frac{\frac{\mathrm{1}}{\mathrm{t}^{\mathrm{899999999}} }}{\frac{\left(\mathrm{1}+\mathrm{t}^{\mathrm{9}} \right)^{\mathrm{1000000001}} }{\mathrm{t}^{\mathrm{900}\:\mathrm{000009}}…

Question-137650

Question Number 137650 by bemath last updated on 05/Apr/21 Answered by bemath last updated on 05/Apr/21 $${By}\:{Langrange}\:{multiplier} \\ $$$${f}\left({x},{y},\lambda\right)=\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\lambda\left(\mathrm{6}{x}^{\mathrm{2}} +\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$$\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{2}{x}+\lambda\left(\mathrm{12}{x}+\mathrm{2}{y}\right)=\mathrm{0}…