Question Number 63981 by ajfour last updated on 11/Jul/19 Answered by mr W last updated on 11/Jul/19 $$\left[\sqrt{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{b}^{\mathrm{2}} }+\sqrt{\left({a}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }\right]^{\mathrm{2}} =\left({b}+{r}\right)^{\mathrm{2}} −\left({b}−{r}\right)^{\mathrm{2}}…
Question Number 129449 by Gulnoza last updated on 15/Jan/21 Answered by MJS_new last updated on 16/Jan/21 $$\mathrm{yes}. \\ $$ Commented by Gulnoza last updated on…
Question Number 63750 by Tawa1 last updated on 08/Jul/19 Commented by Prithwish sen last updated on 08/Jul/19 $$\mathrm{perimeter}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{A}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\pi\mathrm{r}\:\mathrm{unit} \\ $$$$\mathrm{perimeter}\:\mathrm{of}\:\mathrm{circle}\:\mathrm{B}\:=\:\mathrm{2}\pi\mathrm{r}\:\mathrm{unit} \\ $$$$\mathrm{Circle}\:\mathrm{A} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{covers}\:\frac{\mathrm{2}}{\mathrm{3}}\pi\mathrm{r}\:\mathrm{in}\:\mathrm{1}\:\mathrm{time} \\…
Question Number 129244 by ajfour last updated on 14/Jan/21 Commented by ajfour last updated on 14/Jan/21 $${In}\:{terms}\:{of}\:{the}\:{sides}\:{of}\:\bigtriangleup{ABC}, \\ $$$${find}\:{largest}\:{radius}\:{sphere}\:{that} \\ $$$${can}\:{be}\:{placed}\:{against}\:{the}\:{room} \\ $$$${corner}\:{and}\:{triangle}.\:\:\:\:\: \\ $$…
Question Number 63663 by Tawa1 last updated on 07/Jul/19 Answered by mr W last updated on 07/Jul/19 $$\angle{ADB}=\mathrm{360}−\left(\mathrm{180}−\mathrm{6}−\mathrm{30}\right)−\left(\mathrm{180}−\mathrm{6}−\mathrm{24}\right)=\mathrm{66}° \\ $$$$\frac{{AD}}{{DC}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{6}°} \\ $$$$\frac{{BD}}{{DC}}=\frac{\mathrm{sin}\:\mathrm{24}°}{\mathrm{sin}\:\mathrm{6}°} \\ $$$$\Rightarrow\frac{{AD}}{{BD}}=\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{sin}\:\mathrm{24}°} \\…
Question Number 129174 by Study last updated on 13/Jan/21 Commented by Study last updated on 13/Jan/21 $${x}=?\:\:\:\:\:\:\:\left[{y}=???\right. \\ $$ Answered by mr W last updated…
Question Number 129105 by I want to learn more last updated on 12/Jan/21 $$\int_{\:\:\mathrm{0}} ^{\:\frac{\pi}{\mathrm{4}}} \:\sqrt{\mathrm{tan}\:\mathrm{x}\left(\mathrm{1}\:\:−\:\:\mathrm{tan}\:\mathrm{x}\right)}\:\:\mathrm{dx} \\ $$ Answered by Lordose last updated on 13/Jan/21…
Question Number 63564 by Tawa1 last updated on 05/Jul/19 Answered by MJS last updated on 05/Jul/19 $$\mathrm{the}\:\mathrm{flower}\:\mathrm{has}\:\mathrm{got}\:\mathrm{12}\:\mathrm{arcs},\:\mathrm{each}\:\mathrm{one}\:\mathrm{of}\:\mathrm{length} \\ $$$$\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{perimeter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\Rightarrow\: \\ $$$${p}=\mathrm{12}×\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{2}\pi{r}=\mathrm{4}\pi{r} \\ $$$${r}=\mathrm{10}\:\Rightarrow\:{p}=\mathrm{40}\pi \\ $$…
Question Number 129096 by Gulnoza last updated on 12/Jan/21 Answered by talminator2856791 last updated on 15/Jan/21 $$\:\mathrm{2}\left(\frac{\mathrm{300}−\mathrm{75}\pi}{\mathrm{4}}\right) \\ $$ Commented by Gulnoza last updated on…
Question Number 129077 by Study last updated on 12/Jan/21 Commented by benjo_mathlover last updated on 12/Jan/21 $$\frac{\mathrm{FG}}{\mathrm{8}}\:=\:\frac{\mathrm{5}}{\mathrm{6}}\Rightarrow\mathrm{FG}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$\frac{\mathrm{DE}}{\mathrm{8}}=\frac{\mathrm{3}}{\mathrm{6}}\Rightarrow\mathrm{DE}=\mathrm{4} \\ $$$$\mathrm{Area}\:\mathrm{FGED}\:=\:\frac{\mathrm{8}×\mathrm{6}}{\mathrm{2}}−\frac{\frac{\mathrm{20}}{\mathrm{3}}+\mathrm{8}}{\mathrm{2}}×\mathrm{1}−\frac{\mathrm{3}×\mathrm{4}}{\mathrm{2}}\: \\ $$$$\:=\:\mathrm{24}−\left(\frac{\mathrm{10}}{\mathrm{3}}+\mathrm{4}\right)−\mathrm{6} \\ $$$$\:=\:\mathrm{18}−\frac{\mathrm{22}}{\mathrm{3}}\:=\:\frac{\mathrm{32}}{\mathrm{3}}…