Question Number 208264 by efronzo1 last updated on 09/Jun/24 $$\:\:\:\cancel{\underline{\underbrace{ }}} \\ $$ Answered by mr W last updated on 10/Jun/24 $${let}\:{y}'={p} \\ $$$${p}'−\mathrm{2}{p}=\mathrm{2}\left({e}^{{x}} \mathrm{cos}\:{x}−\mathrm{1}\right)…
Question Number 208251 by yaslm last updated on 09/Jun/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 208263 by efronzo1 last updated on 09/Jun/24 Answered by Ghisom last updated on 09/Jun/24 $${a}=\mathrm{e}^{\alpha} \:\Rightarrow\:{b}=\mathrm{e}^{\lambda} {a}\wedge{c}=\mathrm{e}^{\mathrm{2}\lambda} {a} \\ $$$${A}=\mathrm{log}_{{c}} \:{a}\:=\frac{\alpha}{\alpha+\mathrm{2}\lambda} \\ $$$${B}=\mathrm{log}_{{b}}…
Question Number 208259 by hardmath last updated on 09/Jun/24 Answered by cherokeesay last updated on 09/Jun/24 Answered by mr W last updated on 09/Jun/24 Commented…
Question Number 208238 by alcohol last updated on 08/Jun/24 $$\mathrm{S}{how}\:{that} \\ $$$$\frac{\pi}{\mathrm{4}}\:<\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}\:{using}\:{x}\:=\:{sint} \\ $$$${show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}^{\mathrm{4}} }{dx}<\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}} \\ $$$${using}\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){g}\left({x}\right){dx}\right)^{\mathrm{2}} <\int_{\mathrm{0}}…
Question Number 208235 by efronzo1 last updated on 08/Jun/24 Answered by som(math1967) last updated on 08/Jun/24 $$\:{here}\:{f}\left({x}\right)={f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\:\underset{\mathrm{2}} {\overset{\mathrm{4}} {\int}}\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}{dx} \\ $$$$=\underset{\mathrm{2}} {\overset{\mathrm{4}}…
Question Number 208245 by Shrodinger last updated on 08/Jun/24 $${K}=\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\pi}} {ln}\left({cosx}\right){dx} \\ $$ Answered by mathzup last updated on 09/Jun/24 $${K}=\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\pi}} {ln}\left(\frac{{e}^{{ix}} +{e}^{−{ix}}…
Question Number 208241 by Frix last updated on 08/Jun/24 $$\mathrm{Solve}\:\mathrm{for}\:{p},\:{q},\:{r} \\ $$$${p}+{q}+{r}=\alpha \\ $$$${p}^{\mathrm{2}} +{q}^{\mathrm{2}} +{r}^{\mathrm{2}} =\beta \\ $$$${pq}={r} \\ $$ Answered by mr W…
Question Number 208242 by alcohol last updated on 08/Jun/24 Commented by alcohol last updated on 08/Jun/24 $${please}\:{help}\:{me}\:{translate}\:{and}\:{solve} \\ $$ Answered by mr W last updated…
Question Number 208217 by mr W last updated on 07/Jun/24 $$\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} +\mathrm{13}^{\mathrm{2}} +\mathrm{21}^{\mathrm{2}} =? \\ $$ Answered by A5T last updated…