Question Number 46813 by peter frank last updated on 31/Oct/18 Commented by peter frank last updated on 02/Nov/18 $$\mathrm{help}\:\mathrm{please} \\ $$ Answered by ajfour last…
Question Number 46814 by peter frank last updated on 31/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18 $${cos}\theta=\frac{{OP}^{\mathrm{2}} +{OQ}^{\mathrm{2}} −{PQ}^{\mathrm{2}} }{\mathrm{2}×{OP}×{OQ}} \\ $$$$\:\:\:=\frac{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+\left({c}^{\mathrm{2}}…
Question Number 46809 by MJS last updated on 31/Oct/18 $$\mathrm{an}\:\mathrm{early}\:\mathrm{question}\:\mathrm{for}\:\mathrm{the}\:\mathrm{new}\:\mathrm{year} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{rectangular}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{sides} \\ $$$${a},\:{b},\:{c}\:\in\mathbb{N}^{\bigstar} \:\mathrm{exist}\:\mathrm{with}\:\mathrm{one}\:\mathrm{side}\:=\mathrm{2019} \\ $$ Answered by MrW3 last updated on 02/Nov/18 $${a}^{\mathrm{2}}…
Question Number 46737 by peter frank last updated on 30/Oct/18 Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18 $${Q}\left({asin}\theta_{\mathrm{1}} ,{bcos}\theta_{\mathrm{1}} \right) \\ $$$${R}\left({asin}\theta_{\mathrm{2}} ,{bcos}\theta_{\mathrm{2}} \right) \\…
Question Number 46735 by peter frank last updated on 30/Oct/18 Commented by peter frank last updated on 31/Oct/18 $$\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} \\…