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Spin-only-magnetic-moment-of-25-Mn-x-is-15-B-M-Then-the-value-of-x-is-i-did-following-1s-2-2s-2-2p-6-3s-2-3p-6-4s-2-3d-5-so-to-get-3-unpaired-electron-we-need-to-2-electron-so-x-2-book-

Question Number 17209 by sushmitak last updated on 02/Jul/17 $$\mathrm{Spin}\:\mathrm{only}\:\mathrm{magnetic}\:\mathrm{moment} \\ $$$$\mathrm{of}\:_{\mathrm{25}} \mathrm{Mn}^{\mathrm{x}+} \:\mathrm{is}\:\sqrt{\mathrm{15}}\mathrm{B}.\mathrm{M}.\:\mathrm{Then} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{is}? \\ $$$$\mathrm{i}\:\mathrm{did}\:\mathrm{following} \\ $$$$\mathrm{1s}^{\mathrm{2}} \mathrm{2s}^{\mathrm{2}} \mathrm{2p}^{\mathrm{6}} \mathrm{3s}^{\mathrm{2}} \mathrm{3p}^{\mathrm{6}} \mathrm{4s}^{\mathrm{2}}…

Question-148276

Question Number 148276 by 0731619 last updated on 26/Jul/21 Answered by mathmax by abdo last updated on 26/Jul/21 $$\mathrm{y}\left(\mathrm{n}\right)=\mathrm{e}^{\frac{\mathrm{n}^{\mathrm{2}} +\mathrm{n}}{\mathrm{n}^{\mathrm{3}} −\mathrm{4n}^{\mathrm{2}} }\mathrm{log2}} \:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dn}}=\frac{\mathrm{d}}{\mathrm{dn}}\left(\mathrm{log2}×\frac{\mathrm{n}^{\mathrm{2}} \:+\mathrm{n}}{\mathrm{n}^{\mathrm{3}} −\mathrm{4n}^{\mathrm{2}}…

proof-that-the-equation-of-a-ellipse-at-a-center-0-0-is-x-2-a-2-y-2-b-2-1-

Question Number 148270 by abdurehime last updated on 26/Jul/21 $$\mathrm{proof}\:\mathrm{that}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{a}\:\mathrm{ellipse}\:\mathrm{at} \\ $$$$\mathrm{a}\:\mathrm{center}\left(\mathrm{0}.\mathrm{0}\right)\mathrm{is}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }=\mathrm{1} \\ $$ Commented by abdurehime last updated on 26/Jul/21…

f-x-x-2-x-1-x-1-where-x-1-find-the-range-of-the-function-

Question Number 148241 by 7770 last updated on 26/Jul/21 $$\:{f}:{x}\rightarrow\frac{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}−\mathrm{1}}\:{where}\:{x}\neq\mathrm{1} \\ $$$$\:{find}\:{the}\:{range}\:{of}\:{the}\:{function} \\ $$ Answered by Olaf_Thorendsen last updated on 26/Jul/21 $${f}\left({x}\right)\:=\:\frac{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{{x}−\mathrm{1}}\:=\:{x}+\mathrm{2}+\frac{\mathrm{1}}{{x}−\mathrm{1}} \\…

Question-148221

Question Number 148221 by 0731619 last updated on 26/Jul/21 Answered by Olaf_Thorendsen last updated on 26/Jul/21 $$\mathrm{N}\left({x}\right)\:=\:\mathrm{tan}^{\mathrm{2}} \left(\mathrm{tan}{x}\right)−\mathrm{tan}^{\mathrm{2}} {x} \\ $$$$\mathrm{N}\left({x}\right)\:\underset{\mathrm{0}} {\sim}\:\mathrm{tan}^{\mathrm{2}} \left({x}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+\frac{\mathrm{2}{x}^{\mathrm{5}} }{\mathrm{15}}\right)−\left({x}+\frac{{x}^{\mathrm{3}}…