Question Number 163656 by mnjuly1970 last updated on 09/Jan/22 Answered by mr W last updated on 09/Jan/22 Commented by mr W last updated on 09/Jan/22…
Question Number 163610 by amin96 last updated on 08/Jan/22 Answered by mr W last updated on 08/Jan/22 Commented by mr W last updated on 08/Jan/22…
Question Number 163556 by Ari last updated on 07/Jan/22 Answered by ajfour last updated on 07/Jan/22 $$\frac{{x}}{\mathrm{8}}+\frac{{y}}{\mathrm{15}}=\mathrm{1}\:\:\Rightarrow\:\:\mathrm{15}{x}+\mathrm{8}{y}=\mathrm{120} \\ $$$${r}=\frac{\mathrm{15}{r}+\mathrm{8}{r}−\mathrm{120}}{\:\sqrt{\mathrm{225}+\mathrm{64}}} \\ $$$$\Rightarrow\:\:{r}=\mathrm{20} \\ $$ Answered by…
Question Number 97942 by pranesh last updated on 10/Jun/20 Answered by john santu last updated on 10/Jun/20 Answered by Aniruddha Ghosh last updated on 10/Jun/20…
Question Number 97929 by bobhans last updated on 10/Jun/20 $$\mathrm{Find}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{line}\:\mathrm{through}\:\mathrm{A}\left(\mathrm{1},\mathrm{2},\mathrm{3}\right) \\ $$$$\mathrm{and}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{y}\:\mathrm{axis}\:?\: \\ $$ Commented by mr W last updated on 10/Jun/20 $$\frac{{x}−\mathrm{1}}{\mathrm{0}}=\frac{{y}−\mathrm{2}}{\mathrm{1}}=\frac{{z}−\mathrm{3}}{\mathrm{0}} \\ $$$${or}…
Question Number 163413 by mr W last updated on 06/Jan/22 Commented by mr W last updated on 06/Jan/22 $${AB}={AC}={DC}=\mathrm{1},\:{say} \\ $$$${BC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{40} \\ $$$${BD}=\mathrm{2}\:\mathrm{cos}\:\mathrm{40}−\mathrm{1} \\ $$$${ED}=\frac{{BD}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{40}}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{40}−\mathrm{1}}{\mathrm{2}\:\mathrm{cos}\:\mathrm{40}}…
Question Number 163411 by amin96 last updated on 06/Jan/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 163384 by amin96 last updated on 06/Jan/22 Answered by mr W last updated on 06/Jan/22 Commented by mr W last updated on 06/Jan/22…
Question Number 163292 by amin96 last updated on 05/Jan/22 Answered by mr W last updated on 05/Jan/22 $${say}\:{CD}={AD}=\mathrm{1} \\ $$$${AC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{10}° \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{20}}=\frac{{AC}}{\mathrm{sin}\:\left(\mathrm{30}+\mathrm{20}\right)} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{50}} \\…
Question Number 163280 by mr W last updated on 05/Jan/22 Commented by mr W last updated on 05/Jan/22 $${maybe}\:{impossible}\:{to}\:{solve}: \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{n}^{{th}} \:{circle}. \\ $$ Commented…