Category: Coordinate Geometry

Question Number 208756 by efronzo1 last updated on 22/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com

Question Number 208554 by efronzo1 last updated on 18/Jun/24 Answered by liuxinnan last updated on 18/Jun/24 Commented by efronzo1 last updated on 18/Jun/24 Answered by…

Question Number 208327 by efronzo1 last updated on 12/Jun/24 Answered by A5T last updated on 12/Jun/24 $${DE}=\mathrm{5}{x},{DF}=\mathrm{12}{x} \\ $$$${Let}\:{the}\:{perpendicular}\:{from}\:{A}\:{to}\:{BC}\:{meet}\:{it}\:{at}\:{H}; \\ $$$${AH}×{BC}=\mathrm{3}×\mathrm{4}=\mathrm{12}\Rightarrow{AH}=\frac{\mathrm{12}}{\mathrm{5}} \\ $$$$\Rightarrow{BH}=\sqrt{\mathrm{9}−\frac{\mathrm{144}}{\mathrm{25}}}=\frac{\mathrm{9}}{\mathrm{5}} \\ $$$$\frac{{AD}}{{AB}}=\frac{{DE}}{{BC}}={x}\Rightarrow{AD}=\mathrm{3}{x}\Rightarrow{AE}=\mathrm{4}{x}…

Question Number 207879 by efronzo1 last updated on 29/May/24 Answered by mr W last updated on 29/May/24 $${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{12}=\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{3}\geqslant\mathrm{3} \\ $$$${f}\left({a}\right)=\mathrm{65539}={a}^{\mathrm{2}} −\mathrm{6}{a}+\mathrm{12} \\ $$$$\Rightarrow{a}^{\mathrm{2}}…

Question Number 207231 by efronzo1 last updated on 10/May/24 Answered by Sutrisno last updated on 11/May/24 $$\bullet{f}\left({x}\right)+{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)=\mathrm{1}{x}+\mathrm{1}\:…\left(\mathrm{1}\right) \\ $$$$\bullet{x}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$$$\:\:{f}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)+{f}\left(\frac{−\mathrm{1}}{{x}−\mathrm{1}}\right)=\mathrm{2}−\frac{\mathrm{1}}{{x}}\:…\left(\mathrm{2}\right) \\ $$$$\bullet{x}=\frac{−\mathrm{1}}{{x}−\mathrm{1}} \\ $$$${f}\left(\frac{−\mathrm{1}}{{x}−\mathrm{1}}\right)+{f}\left({x}\right)=\frac{−\mathrm{1}}{{x}−\mathrm{1}}+\mathrm{1}\:…\left(\mathrm{3}\right)…