Question Number 208111 by mr W last updated on 05/Jun/24 Commented by mr W last updated on 05/Jun/24 $${yellow}\:{area}\:=? \\ $$ Answered by A5T last…
Question Number 208100 by efronzo1 last updated on 05/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208049 by mr W last updated on 03/Jun/24 Commented by Frix last updated on 03/Jun/24 $${a}=\mathrm{1}\wedge{b}=\mathrm{3} \\ $$$${r}={a}+{b}−\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{4}−\sqrt{\mathrm{10}} \\ $$ Answered…
Question Number 208023 by efronzo1 last updated on 02/Jun/24 Answered by mr W last updated on 02/Jun/24 $${x}={a}+\mid{r}\mid\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{y}={b}+\mid{r}\mid\:\mathrm{sin}\:\theta \\ $$$$\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\…
Question Number 208021 by efronzo1 last updated on 02/Jun/24 Answered by A5T last updated on 02/Jun/24 $$\frac{{CD}}{{CD}+{r}}=\frac{{r}}{{r}+{AB}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{{r}}{{CD}} \\ $$$$\frac{{r}+{AB}}{\mathrm{8}+{r}}=\frac{{r}+{CD}}{\mathrm{3}+{r}}\Rightarrow\frac{{r}+{AB}}{{r}+{CD}}=\frac{\mathrm{8}+{r}}{\mathrm{3}+{r}}=\frac{{r}}{{CD}} \\ $$$$\Rightarrow{CD}=\frac{{r}\left(\mathrm{3}+{r}\right)}{\mathrm{8}+{r}} \\ $$$$\frac{{AB}}{{AB}+{r}}=\frac{{r}}{{CD}+{r}}={AB}\left({CD}+{r}\right)={r}\left({AB}\right)+{r}^{\mathrm{2}} \\ $$$${AB}=\frac{{r}^{\mathrm{2}}…
Question Number 207985 by efronzo1 last updated on 02/Jun/24 Answered by mr W last updated on 02/Jun/24 $${say}\:{AC}={s}={AB}={CB} \\ $$$${AP}^{\mathrm{2}} ={s}×{AM}\:\Rightarrow{AP}=\sqrt{{s}×{AM}} \\ $$$${CP}^{\mathrm{2}} ={s}×{CN}\:\Rightarrow{CP}=\sqrt{{s}×{CN}} \\…
Question Number 208011 by necx122 last updated on 02/Jun/24 Commented by necx122 last updated on 02/Jun/24 $${If}\:{the}\:{ratio}\:{of}\:<{ARS}\::\:<{BRT}\:=\:\mathrm{9}:\mathrm{4}, \\ $$$${find}\:{the}\:{ratio}\:{of}\:{area}\:{ABC}\::\:{area}\:{ARS} \\ $$ Commented by A5T last…
Question Number 208003 by efronzo1 last updated on 02/Jun/24 Answered by A5T last updated on 02/Jun/24 Commented by A5T last updated on 02/Jun/24 $${PM}=\frac{\mathrm{18}}{{s}};{Let}\:{line}\:{through}\:{Q}\:{parallel}\:{to}\:{KN}\:{meet} \\…
Question Number 207980 by mnjuly1970 last updated on 01/Jun/24 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{In}\:\:\:{A}\overset{\Delta} {{B}C}\::\:\:{B}=\:\mathrm{90}^{\:\mathrm{o}} \: \\ $$$$\:\mathrm{BB}'\:\:\bot\:\mathrm{CC}'\:\left(\:\mathrm{BB}'\:{and}\:\mathrm{CC}'\:{are}\:{medians}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\frac{\boldsymbol{{m}}_{\boldsymbol{{c}}} }{\boldsymbol{{m}}_{\boldsymbol{{a}}} \:}\:=\:? \\ $$$$\boldsymbol{{note}}:\:\:\mid\:\mathrm{CC}'\:\mid\:=\:\boldsymbol{{m}}_{\boldsymbol{{c}}} \:\:,\:\mid\mathrm{AA}'\mid=\:\boldsymbol{{m}}_{\boldsymbol{{a}}}…
Question Number 207950 by mnjuly1970 last updated on 31/May/24 Terms of Service Privacy Policy Contact: info@tinkutara.com