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Category: Mensuration

Question-187176

Question Number 187176 by Rupesh123 last updated on 14/Feb/23 Answered by cortano12 last updated on 14/Feb/23 $$\frac{\mathrm{2}}{\mathrm{5}}\:=\:\frac{{r}+\mathrm{2}}{{r}+\mathrm{9}}\Rightarrow\mathrm{2}{r}+\mathrm{18}=\mathrm{5}{r}+\mathrm{10} \\ $$$$\Rightarrow\mathrm{3}{r}=\mathrm{8}\:;\:{r}=\frac{\mathrm{8}}{\mathrm{3}}\: \\ $$ Terms of Service Privacy…

Question-55914

Question Number 55914 by peter frank last updated on 06/Mar/19 Answered by tanmay.chaudhury50@gmail.com last updated on 06/Mar/19 $$\left({k}−\mathrm{1}\right)\mid\frac{{x}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\mid_{\mathrm{0}} ^{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\left({k}−\mathrm{1}\right)\left(\frac{\mathrm{4}^{{k}+\mathrm{1}} }{{k}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{4}} \\…

Two-similar-spheres-of-the-same-material-have-masses-of-12kg-and250kg-respectively-find-the-radius-of-the-smaller-sphere-if-the-radius-of-the-bigger-shere-is-12-5cm-

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-55913

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-186784

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}}…