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Question Number 154395 by liberty last updated on 18/Sep/21 Commented by Tawa11 last updated on 21/Sep/21 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$ Answered by mr W last updated…
Question Number 88723 by M±th+et£s last updated on 12/Apr/20 $${prove}\:{that} \\ $$$$ \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({k}+\mathrm{2}\right)^{\mathrm{2}} {x}^{{k}} }{\left({k}+\mathrm{3}\right)!}=\frac{{e}^{{x}} }{{x}^{\mathrm{3}} }\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)−\frac{{x}^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}{x}^{\mathrm{3}} } \\ $$…
Question Number 23110 by kasiulka202 last updated on 26/Oct/17 $${In}\:{which}\:{apptent}\:{give}\:{when}\:{to}\:\mathrm{10}\:{grams}\:{stop}\:{in}\:\mathrm{800}−{th}\:{apptent}\:{add}\:\:\mathrm{0}.\mathrm{1}\:{gram}\:{of}\:{copper}? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 154065 by peter frank last updated on 13/Sep/21 Answered by Rasheed.Sindhi last updated on 14/Sep/21 $${Area}\:{of}\:{the}\:{badge}={Two}\:{semi}-{circles} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{One}\:{rectangle} \\ $$$$\mathrm{2}\left(\frac{\pi{x}^{\mathrm{2}} }{\mathrm{2}}\right)+{xy}=\mathrm{20} \\ $$$$\pi{x}^{\mathrm{2}}…
Question Number 22939 by selestian last updated on 24/Oct/17 Answered by ajfour last updated on 24/Oct/17 $$\left({B}\right)\:\mathrm{1}/\mathrm{8} \\ $$$$\frac{{a}}{\mathrm{1}−{r}}=\mathrm{4}\:\:\:\:;\:\:\frac{{a}^{\mathrm{3}} }{\mathrm{1}−{r}^{\mathrm{3}} }=\frac{\mathrm{64}}{\mathrm{7}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{7}{a}^{\mathrm{3}} }{\mathrm{1}−{r}^{\mathrm{3}} }=\frac{{a}^{\mathrm{3}}…
Question Number 88471 by M±th+et£s last updated on 10/Apr/20 Answered by mr W last updated on 10/Apr/20 $${a}_{{n}+\mathrm{1}} =\mathrm{2}+\frac{\mathrm{5}}{{a}_{{n}} } \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}{a}_{{n}+\mathrm{1}} =\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\mathrm{2}+\frac{\mathrm{5}}{{a}_{{n}}…
Question Number 22928 by selestian last updated on 24/Oct/17 Answered by $@ty@m last updated on 24/Oct/17 $${Let}\:{x}^{\mathrm{18}} ={y}^{\mathrm{21}} ={z}^{\mathrm{28}} ={k} \\ $$$$\Rightarrow{x}={k}^{\frac{\mathrm{1}}{\mathrm{18}}} ,\:{y}={k}^{\frac{\mathrm{1}}{\mathrm{21}}} ,{z}={k}^{\frac{\mathrm{1}}{\mathrm{28}}} \\…
Question Number 22929 by selestian last updated on 24/Oct/17 Answered by ajfour last updated on 24/Oct/17 $$\left({A}\right) \\ $$$${r}\left(\mathrm{cot}\:\frac{{A}}{\mathrm{2}}+\mathrm{cot}\:\frac{{B}}{\mathrm{2}}\right)={c} \\ $$$${r}\left(\mathrm{cot}\:\frac{{B}}{\mathrm{2}}+\mathrm{cot}\:\frac{{C}}{\mathrm{2}}\right)={a} \\ $$$${r}\left(\mathrm{cot}\:\frac{{C}}{\mathrm{2}}+\mathrm{cot}\:\frac{{A}}{\mathrm{2}}\right)={b} \\ $$$$\mathrm{2}{b}={a}+{c}\:,\:{so}…
Question Number 22923 by selestian last updated on 24/Oct/17 Answered by $@ty@m last updated on 24/Oct/17 $${I}_{{n}} =\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\mathrm{tan}\:^{{n}} {x}\mathrm{sec}\:^{\mathrm{2}} {xdx} \\ $$$${I}_{{n}} =\left[\frac{\mathrm{tan}\:^{{n}+\mathrm{1}}…