Question Number 46735 by peter frank last updated on 30/Oct/18 Commented by peter frank last updated on 31/Oct/18 $$\mathrm{please}\:\mathrm{sir}\:\mathrm{show}\:\mathrm{it}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 46736 by peter frank last updated on 30/Oct/18 Commented by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18 $${i}\:{have}\:{seen}\:{the}\:{proof}…{hereby}\:{i}\:{am}\:{sharing}\:{it}.. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 46715 by Necxx last updated on 30/Oct/18 $${Given}\:{that}\:{the}\:{first}\:{two}\:{terms}\:{of} \\ $$$${a}\:{G}.{P}\:{is}\:{x}\:{and}\:{the}\:{last}\:{two}\:{terms} \\ $$$${is}\:{y}.\:{Find}\:{the}\:{common}\:{ratio}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 177768 by Linton last updated on 08/Oct/22 $${use}\:{pertfectly}\:{square}\:{numbers}\:{to}\:{find} \\ $$$$\sqrt{\mathrm{11}} \\ $$$${no}\:{calculators}\:{allowed} \\ $$ Commented by Linton last updated on 09/Oct/22 $${a}^{\frac{\mathrm{1}}{\mathrm{2}}} =\:\frac{{a}+{b}}{\mathrm{2}\sqrt{{b}}}\:{where}\:{b}\:{is}\:{a}\:{perfect}\:{square}…
Question Number 46686 by peter frank last updated on 30/Oct/18 Commented by peter frank last updated on 30/Oct/18 $$\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…