Question Number 166481 by ajfour last updated on 20/Feb/22 Commented by ajfour last updated on 20/Feb/22 $${Find}\:{the}\:{equation}\:{of}\:{the}\:{parabola}. \\ $$ Answered by MJS_new last updated on…
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Question Number 166208 by ajfour last updated on 15/Feb/22 Commented by ajfour last updated on 15/Feb/22 $${If}\:{both}\:{circles}\:{have}\:{equal}\:{radii}, \\ $$$${find}\:{it}. \\ $$ Answered by Eulerian last…
Question Number 166210 by oustmuchiya@gmail.com last updated on 15/Feb/22 Answered by mr W last updated on 16/Feb/22 $$\mathrm{3}{y}=\mathrm{5}×\mathrm{0}−\mathrm{6} \\ $$$$\Rightarrow{y}=−\mathrm{2} \\ $$$${L}\:{cuts}\:{the}\:{y}−{axis}\:{at}\:\left(\mathrm{0},−\mathrm{2}\right). \\ $$ Terms…
Question Number 166143 by Tawa11 last updated on 13/Feb/22 Answered by som(math1967) last updated on 14/Feb/22 $$\:{y}=\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{4}} \\ $$$$\:\frac{{dy}}{{dx}}=−\mathrm{4}\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}=−\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{3}} }{\:\sqrt{{x}}} \\ $$$$\left[\frac{{dy}}{{dx}}\right]_{\left(\mathrm{1},\mathrm{1}\right)} =\frac{\mathrm{2}\left(\mathrm{2}−\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{1}}=−\mathrm{2}…
Question Number 100581 by Dwaipayan Shikari last updated on 27/Jun/20 $${If}\:\alpha=\frac{\mathrm{2}\pi}{\mathrm{7}} \\ $$$${then}\:{prove}\:{that} \\ $$$${tan}\alpha{tan}\mathrm{2}\alpha+{tan}\mathrm{2}\alpha{tan}\mathrm{4}\alpha+{tan}\mathrm{4}\alpha{tan}\alpha=−\mathrm{7} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 166104 by oustmuchiya@gmail.com last updated on 13/Feb/22 Answered by nikif99 last updated on 13/Feb/22 $${A}:\:\left({A}_{{x}} ,{A}_{{y}} \right)=\left(−\mathrm{5},\mathrm{1}\right),\:{B}:\:\left({B}_{{x}} ,{B}_{{y}} \right)=\left(−\mathrm{7},−\mathrm{8}\right),\: \\ $$$${C}:\:\left({C}_{{x}} ,{C}_{{y}} \right),\:{D}:\:\left({D}_{{x}}…
Question Number 166063 by cortano1 last updated on 12/Feb/22 Commented by blackmamba last updated on 12/Feb/22 $$\:\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…