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Question-197874

Question Number 197874 by yaslm last updated on 01/Oct/23 Answered by aleks041103 last updated on 02/Oct/23 $${v}:\begin{cases}{\mathrm{0}\leqslant{z}\leqslant\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\\{\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} }\end{cases} \\ $$$${in}\:{cylimdrical}\:{coordinates} \\…

Question-197771

Question Number 197771 by sonukgindia last updated on 28/Sep/23 Answered by som(math1967) last updated on 28/Sep/23 $${B}=\frac{\mathrm{1}}{\mathrm{2}}×{r}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{45}}{\mathrm{360}}×\pi{r}^{\mathrm{2}} \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right){squnit} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}}…

a-3b-a-b-1-a-3b-1-a-b-3-4-a-b-

Question Number 197683 by sulaymonnorboyev140 last updated on 26/Sep/23 $$\frac{{a}+\mathrm{3}{b}}{{a}+{b}−\mathrm{1}}+\frac{{a}+\mathrm{3}{b}−\mathrm{1}}{{a}+{b}−\mathrm{3}}=\mathrm{4} \\ $$$${a}+{b}=? \\ $$ Commented by mr W last updated on 26/Sep/23 $${no}\:{unique}\:{solution}\:{for}\:{a}+{b}. \\ $$$${please}\:{check}\:{your}\:{question}!…