Question Number 46482 by peter frank last updated on 27/Oct/18 Commented by peter frank last updated on 27/Oct/18 $$\mathrm{help} \\ $$$$ \\ $$ Answered by…
Question Number 46475 by peter frank last updated on 27/Oct/18 Commented by Joel578 last updated on 27/Oct/18 $${pls}\:{recheck}\:{question}\:\left({b}\right) \\ $$ Commented by peter frank last…
Question Number 46467 by Tawa1 last updated on 26/Oct/18 Commented by MrW3 last updated on 26/Oct/18 $${if}\:{you}\:{write}\:{the}\:{formula}\:{in}\:{this}\:{way}: \\ $$$$\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{ln}\:\left({x}+\mathrm{1}\right)}+\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{ln}\:\left({x}+\mathrm{2}\right)}+\frac{\mathrm{ln}\:\mathrm{4}}{\mathrm{ln}\:\left({x}+\mathrm{3}\right)}=\mathrm{3} \\ $$$${you}\:{would}\:{see}\:{that}\:{the}\:{solution} \\ $$$${is}\:{x}=\mathrm{1}. \\ $$…
Question Number 46460 by Tawa1 last updated on 26/Oct/18 $$\mathrm{If}\:\:\mathrm{k}\:\mathrm{is}\:\mathrm{odd},\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:\:\:\:\mathrm{1}^{\mathrm{k}} \:+\:\mathrm{2}^{\mathrm{k}} \:+\:\mathrm{3}^{\mathrm{k}} \:+\:…\:+\:\mathrm{n}^{\mathrm{k}} \:\:\mathrm{is}\:\mathrm{divisible}\:\mathrm{by}\:\:\: \\ $$$$\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:…\:+\:\mathrm{n},\:\:\:\:\:\mathrm{for}\:\mathrm{every}\:\:\:\mathrm{n}\:\in\:\mathrm{N} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 46461 by Tawa1 last updated on 26/Oct/18 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nth}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{4}}\:+\:\frac{\mathrm{7}}{\mathrm{8}}\:+\:\frac{\mathrm{15}}{\mathrm{16}}\:+\:… \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 26/Oct/18 $${S}_{{n}} =\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }+… \\…
Question Number 46382 by peter frank last updated on 24/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 25/Oct/18 $$\frac{{dy}}{{dx}}=−\frac{{y}^{\mathrm{2}} +{y}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$\frac{{dy}}{{y}^{\mathrm{2}} +{y}+\mathrm{1}}+\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}=\mathrm{0} \\…
Question Number 46383 by peter frank last updated on 24/Oct/18 Commented by maxmathsup by imad last updated on 24/Oct/18 $${z}=\mathrm{0}\:{is}\:{not}\:{solution}\:{for}\:{z}\neq\mathrm{0}\:\:\left({e}\right)\:\Leftrightarrow\:\left(\frac{{z}+\mathrm{1}}{{z}}\right)^{{n}} =\mathrm{1}\:\Leftrightarrow\left(\mathrm{1}+{z}^{−\mathrm{1}} \right)^{{n}} =\mathrm{1}\:\:{the}\:{roots}\:{of} \\ $$$${Z}^{{n}}…
Question Number 46378 by rahul 19 last updated on 24/Oct/18 $${Factorise}\:: \\ $$$$\mathrm{3}{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{3}} +\mathrm{8}{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}=\mathrm{0}. \\ $$ Answered by MJS last updated on 24/Oct/18…
Question Number 46358 by peter frank last updated on 24/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18 $$\left.{b}\right)\left({x}+{y}\right)^{\mathrm{2}} \left({xdy}+{ydx}\right)={xy}\left({dx}+{dy}\right) \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} {d}\left({xy}\right)={xy}\left({dx}+{dy}\right) \\ $$$$\frac{{d}\left({xy}\right)}{{xy}}=\frac{{d}\left({x}+{y}\right)}{\left({x}+{y}\right)^{\mathrm{2}} }…
Question Number 46168 by peter frank last updated on 21/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 22/Oct/18 $$\mathrm{1}+{w}+{w}^{\mathrm{2}} =\mathrm{0}\:\:\:{w}^{\mathrm{3}} =\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\left(−{w}^{\mathrm{2}} −{w}^{\mathrm{2}} \right)^{\mathrm{3}} −\left(−{w}−{w}\right)^{\mathrm{3}}…