Question Number 166063 by cortano1 last updated on 12/Feb/22 Commented by blackmamba last updated on 12/Feb/22 $$\:\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{3}}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\:\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{2}\left(\mathrm{15}+\mathrm{7}\sqrt{\mathrm{3}}\right)}{\mathrm{13}} \\ $$$$\:\mathrm{tan}\:\left(\mathrm{90}°−\mathrm{2}\alpha\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\alpha}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$$$\:{m}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$…
Question Number 100479 by mathocean1 last updated on 26/Jun/20 $${Please}\:{How}\:{to}\:{calculate}\:{the}\:{focusing} \\ $$$${latitude}\:{of}\:{a}\:{microscope}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 100378 by bemath last updated on 26/Jun/20 Commented by bemath last updated on 26/Jun/20 $$\mathrm{find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{OACB}\:.\:\mathrm{where}\:\mathrm{radius}\:=\:\mathrm{4} \\ $$ Answered by john santu last updated…
Question Number 34803 by Tinkutara last updated on 11/May/18 Commented by ajfour last updated on 11/May/18 Commented by ajfour last updated on 11/May/18 $$\mathrm{2}{x}+{y}=\mathrm{0} \\…
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Question Number 165431 by ajfour last updated on 01/Feb/22 Answered by mr W last updated on 01/Feb/22 Commented by mr W last updated on 01/Feb/22…
Question Number 34326 by 33 last updated on 04/May/18 $${find}\:{the}\:{equation}\:{of}\:{the}\:\mathrm{2}{D} \\ $$$${curve}\:{such}\:{that}\:{the}\:{lines} \\ $$$$\:\frac{{x}}{{t}}\:+\:\frac{{y}}{\left({a}−{t}\right)\:}\:=\:\mathrm{1} \\ $$$$\:{are}\:{always}\:{tangent}\:{to} \\ $$$${the}\:{curve}. \\ $$$${given}\:'{a}'\:\:{is}\:{a}\:{positive}\:{real} \\ $$$${constant}\:{and}\:'{t}'\:{is}\:{a} \\ $$$${parameter}.\:\left(\:\mathrm{0}\:<\:{t}\:<\:{a}\:\right) \\…
Question Number 165272 by cortano1 last updated on 28/Jan/22 Answered by ajfour last updated on 28/Jan/22 $$\mathrm{2}{a}=\mathrm{10}{cm} \\ $$$${a}^{\mathrm{2}} =\left(\mathrm{2}{r}−\mathrm{2}{a}\right)\left(\mathrm{2}{a}\right) \\ $$$${r}=\frac{\mathrm{5}{a}}{\mathrm{4}}=\frac{\mathrm{5}}{\mathrm{8}}\left(\mathrm{10}{cm}\right)=\mathrm{6}.\mathrm{25}{cm} \\ $$ Answered…
Question Number 165024 by cortano1 last updated on 25/Jan/22 Commented by cortano1 last updated on 25/Jan/22 $$\:{find}\:{a}. \\ $$$$\left({A}\right)\mathrm{2}\sqrt{\mathrm{46}}\:\:\:\:\:\:\:\left({C}\right)\mathrm{12} \\ $$$$\left({B}\right)\:\mathrm{2}\sqrt{\mathrm{42}}\:\:\:\:\:\:\:\left({D}\right)\:\mathrm{2}\sqrt{\mathrm{39}} \\ $$ Commented by…
Question Number 33927 by Tinkutara last updated on 27/Apr/18 Commented by Tinkutara last updated on 27/Apr/18 $${Why}\:{here}\:{x}\:{is}\:{substituted}\:{x}=\frac{{x}+{y}}{\:\sqrt{\mathrm{2}}}? \\ $$ Commented by Tinkutara last updated on…