Question Number 197874 by yaslm last updated on 01/Oct/23 Answered by aleks041103 last updated on 02/Oct/23 $${v}:\begin{cases}{\mathrm{0}\leqslant{z}\leqslant\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }}\\{\mathrm{0}\leqslant{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \leqslant{a}^{\mathrm{2}} }\end{cases} \\ $$$${in}\:{cylimdrical}\:{coordinates} \\…
Question Number 197822 by mokys last updated on 30/Sep/23 $${find}\:{maximum}\:{of}\:\mid{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{3}\mid\:? \\ $$ Commented by AST last updated on 30/Sep/23 $$\infty \\ $$ Commented by…
Question Number 197808 by mokys last updated on 29/Sep/23 $${prove}\:\underset{{n}\rightarrow\infty} {{lim}}\:{x}^{{n}} \:=\:\mathrm{0}\:\:\:\:{when}\:\mid{x}\mid\:<\:\mathrm{1} \\ $$ Commented by Noorzai last updated on 30/Sep/23 $${please}\:{give}\:{more}\:{information}\: \\ $$ Answered…
Question Number 197771 by sonukgindia last updated on 28/Sep/23 Answered by som(math1967) last updated on 28/Sep/23 $${B}=\frac{\mathrm{1}}{\mathrm{2}}×{r}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{45}}{\mathrm{360}}×\pi{r}^{\mathrm{2}} \\ $$$$=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}+\frac{\pi{r}^{\mathrm{2}} }{\mathrm{4}}=\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\mathrm{1}+\frac{\pi}{\mathrm{2}}\right){squnit} \\ $$$${A}=\frac{\mathrm{1}}{\mathrm{2}}\pi{r}^{\mathrm{2}}…
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Question Number 197753 by tri26112004 last updated on 27/Sep/23 $${Solve}\:{the}\:{equation}: \\ $$$$\sqrt{\mathrm{5}{x}^{\mathrm{2}} +\mathrm{14}{x}+\mathrm{9}}−\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{20}}=\mathrm{5}\sqrt{{x}+\mathrm{1}} \\ $$ Answered by Frix last updated on 27/Sep/23 $$\mathrm{Squaring},\:\mathrm{transforming},\:\mathrm{solving},\:\mathrm{testing} \\…
Question Number 197716 by josemate19 last updated on 26/Sep/23 $${sen}\left({x}\right){y}^{''} +{cos}\left({x}\right){y}'+\mathrm{3}{x}^{\mathrm{3}} {y}={tan}\left(\sqrt{{x}}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 197683 by sulaymonnorboyev140 last updated on 26/Sep/23 $$\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 $${no}\:{unique}\:{solution}\:{for}\:{a}+{b}. \\ $$$${please}\:{check}\:{your}\:{question}!…
Question Number 197639 by mokys last updated on 25/Sep/23 $${show}\:{that}\:\frac{\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{{n}} }\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left({n}+{k}−\mathrm{1}\right)!}{{k}!\left({n}−\mathrm{1}\right)!}\:{z}^{{k}} \\ $$ Commented by mr W last updated on 25/Sep/23 $$\frac{\mathrm{1}}{\left(\mathrm{1}−{z}\right)^{{n}} }=\underset{{k}=\mathrm{0}}…
Question Number 197666 by mondli212 last updated on 25/Sep/23 $$ \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{past}\:\mathrm{10}\:\mathrm{years}\:\mathrm{M}\:\mathrm{has}\:\mathrm{deposited} \\ $$$$\$\mathrm{40}\:\mathrm{at}\:\mathrm{the}\:\mathrm{end}\:\mathrm{of}\:\mathrm{each}\:\mathrm{month}\:\mathrm{in}\:\mathrm{as} \\ $$$$\mathrm{saving}\:\mathrm{bank}\:\mathrm{paying}\:\mathrm{3\%}\:\mathrm{p}.\mathrm{a}.\: \\ $$$$\mathrm{compoundedsemi}\:\mathrm{annually}.\:\mathrm{If}\:\mathrm{they} \\ $$$$\mathrm{polic}\:\mathrm{of}\:\mathrm{the}\:\mathrm{bank}\:\mathrm{is}\:\mathrm{to}\:\mathrm{place}\:\mathrm{eacht} \\ $$$$\mathrm{deposi}\:\mathrm{at}\:\mathrm{3}\:\%\mathrm{p}.\mathrm{a}.\:\mathrm{simple}\:\mathrm{interest}\:\mathrm{on}\:\mathrm{ther} \\ $$$$\mathrm{fist}\:\mathrm{of}\:\mathrm{each}\:\mathrm{month}\:\mathrm{and}\:\mathrm{compound}\:\mathrm{semi}\: \\…