Question Number 204505 by cherokeesay last updated on 19/Feb/24 Answered by mr W last updated on 19/Feb/24 $$\left[\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}\right]^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}\sqrt{\mathrm{1}−{r}}−\mathrm{1}\right]^{\mathrm{2}} =\mathrm{2}{r}−\mathrm{1}…
Question Number 204433 by emilagazade last updated on 17/Feb/24 Commented by emilagazade last updated on 17/Feb/24 $${find}\:{max}\:{value}\:{for}\:\mid{DB}\mid+\mid{BC}\mid \\ $$ Answered by deleteduser1 last updated on…
Question Number 204374 by mr W last updated on 14/Feb/24 Commented by mr W last updated on 15/Feb/24 $${The}\:{radius}\:{of}\:{the}\:{big}\:{sphere}\:\left({say}\:\right. \\ $$$$\left.{the}\:{Earth}\right)\:{is}\:{R}\:{and}\:{the}\:{radius}\:{of} \\ $$$${the}\:{small}\:{sphere}\:\left({say}\:{theMoon}\right)\:{is}\:{r}. \\ $$$${find}\:{the}\:{maximum}\:{area}\:{of}\:{the}…
Question Number 204310 by BaliramKumar last updated on 12/Feb/24 Answered by mr W last updated on 12/Feb/24 Commented by mr W last updated on 12/Feb/24…
Question Number 204197 by ajfour last updated on 08/Feb/24 Answered by mr W last updated on 08/Feb/24 $$\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\frac{{a}}{{R}−{a}} \\ $$$${OC}={OB}+{BC} \\ $$$$\frac{{R}}{\mathrm{cos}\:\theta}=\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\frac{\pi}{\mathrm{2}}−\theta}{\mathrm{2}}\right)} \\…
Question Number 204171 by cherokeesay last updated on 07/Feb/24 Commented by AST last updated on 07/Feb/24 $${Are}\:{those}\:{points}\:{on}\:{the}\:{circle}\:{the}\:{vertices}\:{of}\:{a} \\ $$$${regular}\:{hexagon}?\:{Is}\:{H}\:{the}\:{midpoint}\:{of}\:\Gamma\:{and}\:{A}? \\ $$ Answered by AST last…
Question Number 203969 by mr W last updated on 03/Feb/24 Answered by Rasheed.Sindhi last updated on 03/Feb/24 $$?={x} \\ $$$$\bigtriangleup\mathrm{ABC}\sim\bigtriangleup\mathrm{ADE} \\ $$$$\frac{\mathrm{3}}{{r}}=\frac{\mathrm{4}}{\mathrm{4}−{r}}=\frac{\mathrm{5}}{{x}+{r}}\Rightarrow\mathrm{4}{r}=\mathrm{12}−\mathrm{3}{r}\Rightarrow{r}=\frac{\mathrm{12}}{\mathrm{7}} \\ $$$$\frac{\mathrm{3}}{{r}}=\frac{\mathrm{5}}{{x}+{r}}\Rightarrow\frac{\mathrm{3}}{\frac{\mathrm{12}}{\mathrm{7}}}=\frac{\mathrm{5}}{{x}+\frac{\mathrm{12}}{\mathrm{7}}} \\…
Question Number 203803 by ajfour last updated on 28/Jan/24 Answered by ajfour last updated on 28/Jan/24 $$\mathrm{2}{s}^{\mathrm{2}} =\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left(\frac{{h}−{k}}{{h}+{k}}\right)\left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{2}{hk}}{\left({h}+{k}\right)^{\mathrm{2}} }\left({a}^{\mathrm{2}}…
Question Number 203778 by mr W last updated on 27/Jan/24 Answered by mr W last updated on 27/Jan/24 Commented by mr W last updated on…
Question Number 203613 by mr W last updated on 23/Jan/24 Commented by mr W last updated on 23/Jan/24 $${find}\:\frac{{R}}{{r}}=? \\ $$ Commented by esmaeil last…