Question Number 46686 by peter frank last updated on 30/Oct/18 Commented by peter frank last updated on 30/Oct/18 $$\mathrm{ABCDEFGH}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}.\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{projections}\:\mathrm{of} \\ $$$$\left.\mathrm{a}\right)\mathrm{AF}\:\mathrm{on}\:\:\mathrm{ABCD} \\ $$$$\left.\mathrm{b}\right)\mathrm{AG}\:\mathrm{on}\:\mathrm{ABCD}…
Question Number 177648 by BaliramKumar last updated on 07/Oct/22 $${If}\:\:\frac{{a}\:+\:{b}}{\:\sqrt{{ab}}}\:=\:\frac{\mathrm{4}}{\mathrm{1}}\:{then}\:\:\:\:{a}\::\:{b}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{a}>{b}\:\right] \\ $$ Answered by Rasheed.Sindhi last updated on 07/Oct/22 $${a}+{b}=\mathrm{4}\sqrt{{ab}}\: \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{16}{ab}−\mathrm{2}{ab}=\mathrm{14}{ab} \\…
Question Number 46567 by Tawa1 last updated on 28/Oct/18 $$\mathrm{Solve}:\:\:\:\:\:\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\boldsymbol{\sum}}}\:\:\left(\frac{\mathrm{log}\:\boldsymbol{\mathrm{n}}}{\boldsymbol{\mathrm{n}}}\right)^{\mathrm{2}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 46552 by peter frank last updated on 28/Oct/18 Commented by MJS last updated on 28/Oct/18 $$\left.{b}\right)\:\mathrm{is}\:\mathrm{not}\:\mathrm{clear}.\:\mathrm{what}\:\mathrm{does}\:\mathrm{this}\:\mathrm{mean}\:“\mathrm{if}\:\left(\mathrm{it}?\right) \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{y}=\mathrm{2}{x}+\mathrm{3}\:\mathrm{at}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}''?\:\mathrm{if}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{curve}\:\begin{pmatrix}{\mathrm{0}}\\{\mathrm{4}}\end{pmatrix}\:\mathrm{cannot}\:\mathrm{be}\:\mathrm{or}\:\mathrm{vice}\:\mathrm{versa}. \\ $$$$\mathrm{please}\:\mathrm{check}\:\mathrm{this} \\…
Question Number 46482 by peter frank last updated on 27/Oct/18 Commented by peter frank last updated on 27/Oct/18 $$\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} \\…