Question Number 55918 by imamu222 last updated on 06/Mar/19 $${Two}\:{similar}\:{spheres}\:{of}\:{the}\:{same}\:{material}\:{have}\:{masses}\:{of}\:\mathrm{12}{kg}\:{and}\mathrm{250}{kg}\:{respectively}.\:{find}\:{the}\:{radius}\:{of}\:{the}\:{smaller}\:{sphere}\:{if}\:{the}\:{radius}\:{of}\:{the}\:{bigger}\:{shere}\:{is}\:\mathrm{12}.\mathrm{5}{cm} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19 $$\frac{{M}_{{small}} }{{M}_{{large}} }=\frac{\frac{\mathrm{4}}{\mathrm{3}}\pi{r}_{{small}} ^{\mathrm{3}} \rho}{\frac{\mathrm{4}}{\mathrm{3}}\pi{r}_{{large}} ^{\mathrm{3}}…
Question Number 55913 by peter frank last updated on 06/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19 $${O}\overset{\rightarrow} {{A}}=−\mathrm{2}{i}+\mathrm{2}{j}+\mathrm{3}{k} \\ $$$${O}\overset{\rightarrow} {{B}}=\mathrm{2}{i}−\mathrm{3}{j}+{k} \\ $$$${O}\overset{\rightarrow} {{C}}=\mathrm{4}{i}+\mathrm{5}{j}+\mathrm{3}{k}…
Question Number 186924 by Mingma last updated on 12/Feb/23 Answered by mr W last updated on 12/Feb/23 Commented by mr W last updated on 12/Feb/23…
Question Number 186868 by Mingma last updated on 11/Feb/23 Commented by Mingma last updated on 11/Feb/23 Shaded area =? Answered by HeferH last updated on 11/Feb/23 Commented…
Question Number 186784 by Rupesh123 last updated on 10/Feb/23 Answered by a.lgnaoui last updated on 11/Feb/23 $$\mathrm{1}\bullet \\ $$$${soit}\:{ABCD}\:\left({Rectangle}\right) \\ $$$${BC}={BM}+{MC}\:\:\:\:\left({AB}={a}\:\:\:{AC}={b}\right) \\ $$$${AB}^{\mathrm{2}} ={AM}^{\mathrm{2}} +{MC}^{\mathrm{2}}…
Question Number 55473 by pooja24 last updated on 25/Feb/19 Commented by mr W last updated on 25/Feb/19 $${too}\:{less}\:{information}\:{to}\:{define}\:{unique} \\ $$$${shaded}\:{area},\:{please}\:{check}\:{the}\:{question}. \\ $$ Terms of Service…
Question Number 55433 by peter frank last updated on 24/Feb/19 Answered by mr W last updated on 24/Feb/19 Commented by mr W last updated on…
Question Number 186442 by Mingma last updated on 04/Feb/23 Answered by HeferH last updated on 04/Feb/23 Commented by HeferH last updated on 04/Feb/23 $$\left(\frac{\mathrm{11}}{\mathrm{2}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}}…
Question Number 186401 by Rupesh123 last updated on 04/Feb/23 Answered by ajfour last updated on 04/Feb/23 Commented by ajfour last updated on 04/Feb/23 $${Red}_{{area}} =…
Question Number 120790 by peter frank last updated on 02/Nov/20 Commented by TANMAY PANACEA last updated on 02/Nov/20 $${eqn}\:{big}\:{circle}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={R}^{\mathrm{2}} \:\:{and}\:{area}\:\pi{R}^{\mathrm{2}} \\ $$$${small}\:{circle}\:{centre}\left(−{a},\mathrm{0}\right)\:{radius}\:{r} \\…