Question Number 166818 by ajfour last updated on 28/Feb/22 Commented by som(math1967) last updated on 28/Feb/22 $${R}=\frac{\mathrm{10}}{\mathrm{3}},\:{r}=\frac{\mathrm{15}}{\mathrm{8}}\:\:{sir}? \\ $$ Commented by ajfour last updated on…
Question Number 166779 by ajfour last updated on 27/Feb/22 Answered by mr W last updated on 27/Feb/22 Commented by mr W last updated on 28/Feb/22…
Question Number 166749 by Tawa11 last updated on 27/Feb/22 Commented by mr W last updated on 27/Feb/22 $${this}\:{question}\:{is}\:{n}\:{times}\:{repeated}. \\ $$ Answered by mr W last…
Question Number 166715 by alephzero last updated on 26/Feb/22 $$\mathrm{The}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{24}.\:\mathrm{Find}\:\mathrm{it}'\mathrm{s}\:\mathrm{area}. \\ $$ Answered by Rasheed.Sindhi last updated on 26/Feb/22 $${Let}\:{each}\:{of}\:{sides}:\: \\ $$$${a}={b}={c}={l} \\…
Question Number 166701 by ajfour last updated on 25/Feb/22 Answered by mr W last updated on 26/Feb/22 $${BC}=\mathrm{1} \\ $$$${CD}={x} \\ $$$${CE}=\frac{{x}}{\mathrm{2}} \\ $$$${DE}=\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{2}} \\…
Question Number 166676 by ajfour last updated on 24/Feb/22 Answered by ajfour last updated on 24/Feb/22 $${R}\mathrm{sin}\:\theta+\mathrm{2}{r}={R} \\ $$$$\mathrm{sin}\:\theta=\frac{{r}}{{R}−{r}} \\ $$$$\Rightarrow\:\:{Rr}=\left({R}−\mathrm{2}{r}\right)\left({R}−{r}\right) \\ $$$$\Rightarrow\:\:{R}^{\mathrm{2}} −\mathrm{4}{rR}+\mathrm{2}{r}^{\mathrm{2}} =\mathrm{0}…
Question Number 35606 by ajfour last updated on 21/May/18 Commented by ajfour last updated on 21/May/18 $${Hence}\:{find}\:{volume}\:{of}\:{the} \\ $$$${triangular}\:{pyramid}. \\ $$ Commented by MrW3 last…
Question Number 166670 by amin96 last updated on 24/Feb/22 Commented by Laters last updated on 24/Feb/22 I think the awnser is 6 Commented by amin96 last updated on 24/Feb/22 $${solution}?…
Question Number 166640 by cherokeesay last updated on 23/Feb/22 Answered by som(math1967) last updated on 24/Feb/22 $${Area}\:{with}\:{circle} \\ $$$$=\mathrm{2}×\left\{\frac{\pi}{\mathrm{6}}×\mathrm{4}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}×\mathrm{4}^{\mathrm{2}} \right\}{cm}^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{16}\pi}{\mathrm{3}}\:−\mathrm{4}\sqrt{\mathrm{3}}\right){cm}^{\mathrm{2}} \: \\…
Question Number 166628 by Nimatullah last updated on 23/Feb/22 Terms of Service Privacy Policy Contact: info@tinkutara.com