Question Number 126825 by mohammad17 last updated on 24/Dec/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 192349 by 073 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${I}=\int\frac{{cos}^{\mathrm{4}} {x}\sqrt{\mathrm{1}+{sinx}}}{{cosx}}\:{dx}=\int\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)\sqrt{\mathrm{1}+{sinx}}\:{cosxdx} \\ $$$${let}\::\:\mathrm{1}+{sinx}={u}^{\mathrm{2}} \:\Rightarrow{cosxdx}=\mathrm{2}{udu} \\ $$$$\Rightarrow{I}=\mathrm{2}\int\left(−{u}^{\mathrm{2}} +\mathrm{2}{u}\right){u}^{\mathrm{2}}…
Question Number 61283 by naka3546 last updated on 01/Jun/19 $$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\left({x}^{\mathrm{3}} \:+\:{y}^{\mathrm{3}} \right)\:\:=\:\:\mathrm{2} \\ $$$$\left({x}^{\mathrm{4}} +{y}^{\mathrm{4}} \right)\left({x}^{\mathrm{6}} +{y}^{\mathrm{6}} \right)\left({x}^{\mathrm{8}} \:+\:{y}^{\mathrm{8}} \right)\:\:=\:\:\mathrm{4} \\ $$$$\left({x}^{\mathrm{3}} +\:{y}^{\mathrm{3}}…
Question Number 192348 by leandrosriv02 last updated on 15/May/23 Answered by mehdee42 last updated on 15/May/23 $${AB}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${M}\:\:{is}\:{midpoint}\:<{AB}>\Rightarrow{M}\left(\frac{{a}}{\mathrm{2}}\:,\:\frac{{b}}{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{0}{M}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow{OM}={BM}={AM}…
Question Number 61270 by necx1 last updated on 31/May/19 Answered by Askash last updated on 31/May/19 $$\mathrm{24} \\ $$ Answered by Rasheed.Sindhi last updated on…
Question Number 192338 by josemate19 last updated on 15/May/23 $${y}=\:\left(\frac{\left(\underset{{h}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\left({x}+{h}\right)^{\mathrm{3}} −{x}^{\mathrm{3}} }{{h}}\right)\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\right)}{\int_{\mathrm{0}} ^{\:{x}} {lnt}\:{dt}}\right) \\ $$$$ \\ $$$$\frac{{dy}}{{dx}}? \\ $$ Answered…
Question Number 126758 by bounhome last updated on 24/Dec/20 $${solve}: \\ $$$$\mathrm{1}.{y}'^{\mathrm{3}} +{y}^{\mathrm{2}} −{y}'^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}.\mathrm{4}{y}={x}^{\mathrm{2}} +{y}'^{\mathrm{2}} \\ $$ Answered by liberty last updated…
Question Number 61215 by naka3546 last updated on 30/May/19 Answered by ajfour last updated on 30/May/19 $${let}\:{bottom}\:{left}\:{corner}\:{of}\:{square} \\ $$$$\left({i}\:{assume}\:{so},\:{should}\:{be}\:{so}\right)\:{be} \\ $$$${origin}.\:\:{square}\:{side}\:\boldsymbol{{a}}\:,\:{radius}\:\boldsymbol{{r}} \\ $$$${centre}\:{of}\:{circle}\:{C}\left(\boldsymbol{{h}},\boldsymbol{{k}}\right) \\ $$$$\:\:\:{Then}\:{we}\:{have}…
Question Number 192276 by York12 last updated on 13/May/23 $$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left[\frac{{k}\left({n}−{k}\right)!+\left({k}+\mathrm{1}\right)}{\left({k}+\mathrm{1}\right)!\left({n}−{k}\right)!}\right]\right) \\ $$ Answered by witcher3 last updated on 14/May/23 $$\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}} {\sum}}\frac{\mathrm{1}}{\mathrm{k}!\left(\mathrm{n}−\mathrm{k}\right)!}=\frac{\mathrm{1}}{\mathrm{n}!}\underset{\mathrm{k}=\mathrm{0}}…
Question Number 192277 by York12 last updated on 13/May/23 $$ \\ $$$${Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} ,\:{a}_{\mathrm{3}} ,…,\:{a}_{{n}} \:{be}\:{real}\:{numbers}\:{such}\:{that}: \\ $$$$\sqrt{{a}_{\mathrm{1}} }\:+\:\sqrt{{a}_{\mathrm{2}} −\mathrm{1}\:\:}\:+\sqrt{{a}_{\mathrm{3}} −\mathrm{2}\:}\:+…+\sqrt{{a}_{{n}} −\left({n}−\mathrm{1}\right)\:}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}_{\mathrm{1}} +{a}_{\mathrm{2}} +{a}_{\mathrm{3}} +…+{a}_{{n}}…