Question Number 168441 by mokys last updated on 10/Apr/22 $${find}\:{the}\:{liength}\:{of}\:{x}\:=\:{y}^{\frac{\mathrm{3}}{\mathrm{2}}} \:{from}\:\left(\mathrm{1},\mathrm{1}\right){to}\:\left(\mathrm{2},\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$ Answered by kowalsky78 last updated on 10/Apr/22 $${There}'{s}\:{something}\:{strange}\:{about}\:{this}\:{point}\:\left(\mathrm{2},\mathrm{2}\sqrt{\mathrm{2}}\right). \\ $$$${Did}\:{you}\:{mean}\:\left(\mathrm{2}\sqrt{\mathrm{2}},\mathrm{2}\right)? \\ $$…
Question Number 168429 by mathocean1 last updated on 10/Apr/22 $${calculate}: \\ $$$${P}=\int_{\mathrm{1}} ^{{e}^{\frac{\pi}{\mathrm{2}}} } \frac{{cos}\left({lnx}\right)}{{x}}{dx} \\ $$ Answered by qaz last updated on 10/Apr/22 $$\int_{\mathrm{1}}…
Question Number 168400 by leicianocosta last updated on 10/Apr/22 Commented by cortano1 last updated on 10/Apr/22 $$\:\:\:\:\begin{cases}{\mathrm{5}\sqrt{\mathrm{7}}\:\pi{x}+\mathrm{2}\pi{y}\:=\:\frac{\mathrm{2}\pi\sqrt{\mathrm{7}}}{\mathrm{7}}}\\{{x}\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\pi{y}\:=\:\mathrm{10}\pi^{\mathrm{2}} }\end{cases} \\ $$$$\:\:\:\:{x}\left(\mathrm{5}\sqrt{\mathrm{7}}\:\pi+\mathrm{1}\right)=\frac{\mathrm{2}\pi\sqrt{\mathrm{7}}}{\mathrm{7}}+\mathrm{10}\pi^{\mathrm{2}} \\ $$$$\:\:\:{x}=\frac{\mathrm{2}\pi\sqrt{\mathrm{7}}+\mathrm{70}\pi^{\mathrm{2}} }{\mathrm{35}\pi\sqrt{\mathrm{7}}+\mathrm{7}}\:;\:\left[\pi=\frac{\mathrm{22}}{\mathrm{7}}\right] \\ $$…
Question Number 102854 by Sontsaronald last updated on 11/Jul/20 Answered by Ar Brandon last updated on 11/Jul/20 $$\mathrm{arctan}\left(\mathrm{ch}\left(\mathrm{x}\right)\right)=\mathrm{arctan}\left(\frac{\mathrm{1}}{\mathrm{sh}\left(\mathrm{x}\right)}\right) \\ $$$$\Rightarrow\:\mathrm{ch}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{sh}\left(\mathrm{x}\right)}\:\Rightarrow\:\mathrm{2sh}\left(\mathrm{x}\right)\mathrm{ch}\left(\mathrm{x}\right)=\mathrm{2}\:\Rightarrow\:\mathrm{sh}\left(\mathrm{2x}\right)=\mathrm{2} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{e}^{\mathrm{2x}} −\mathrm{e}^{−\mathrm{2x}} \right)=\mathrm{2}\:\Rightarrow\:\mathrm{e}^{\mathrm{4x}} −\mathrm{4e}^{\mathrm{2x}}…
Question Number 102857 by mathocean1 last updated on 11/Jul/20 Commented by PRITHWISH SEN 2 last updated on 11/Jul/20 $$\frac{\mathrm{11}}{\mathrm{2}}\left\{\mathrm{2a}+\left(\mathrm{11}−\mathrm{1}\right)\mathrm{4}\right\}=\mathrm{374} \\ $$$$\mathrm{a}=\mathrm{14} \\ $$$$\therefore\:\:\mathrm{It}\:\mathrm{takes}\:\mathrm{14}\:\mathrm{mins}.\:\mathrm{to}\:\mathrm{sweep}\:\mathrm{the}\:\mathrm{first}\:\mathrm{class}\:\: \\ $$…
Question Number 168388 by leicianocosta last updated on 09/Apr/22 Answered by Tibo last updated on 09/Apr/22 $$\left.{a}\right)\:\:\pi{r}^{\mathrm{2}} =\pi×\mathrm{9}\approx\mathrm{28}.\mathrm{26}{cm}^{\mathrm{2}} \\ $$$$\left.{b}\right)\:\:\mathrm{2}\pi{rh}=\mathrm{2}×\pi×\mathrm{3}×\mathrm{8}\approx\mathrm{150}.\mathrm{72}{cm}^{\mathrm{2}} \\ $$$$\left.{c}\right)\:\:\mathrm{2}\left(\mathrm{28}.\mathrm{26}\right)+\mathrm{150}.\mathrm{72}=\mathrm{297}.\mathrm{24}{cm}^{\mathrm{2}} \\ $$$$\left.{d}\right)\:\:\pi{r}^{\mathrm{2}} {h}=\pi×\mathrm{9}×\mathrm{8}\approx\mathrm{226}.\mathrm{08}{cm}^{\mathrm{3}}…
Question Number 102843 by Sontsaronald last updated on 11/Jul/20 Commented by Sontsaronald last updated on 11/Jul/20 $${s}\:{il}\:{vous}\:{plait}\:{ohhhhh} \\ $$ Commented by mr W last updated…
Question Number 168361 by leicianocosta last updated on 08/Apr/22 Answered by som(math1967) last updated on 09/Apr/22 $$\boldsymbol{{x}}=−\boldsymbol{{kz}}−\mathrm{5} \\ $$$$\:\boldsymbol{{x}}+\mathrm{5}=−\boldsymbol{{kz}} \\ $$$$\:\frac{\boldsymbol{{x}}+\mathrm{5}}{−\boldsymbol{{k}}}=\boldsymbol{{z}} \\ $$$$\boldsymbol{{y}}\:=−\mathrm{3}\boldsymbol{{z}}\Rightarrow\frac{{y}}{−\mathrm{3}}=\boldsymbol{{z}} \\ $$$$\boldsymbol{{equn}}.\boldsymbol{{of}}\:\boldsymbol{{r}}\:\:\frac{\boldsymbol{{x}}+\mathrm{5}}{−\boldsymbol{{k}}}=\frac{\boldsymbol{{y}}}{−\mathrm{3}}=\frac{\boldsymbol{{z}}}{\mathrm{1}}…
Question Number 168324 by leicianocosta last updated on 07/Apr/22 Answered by mr W last updated on 08/Apr/22 Commented by mr W last updated on 08/Apr/22…
Question Number 37221 by MrW3 last updated on 11/Jun/18 $${Hi}!\:{MrW}\:{is}\:{back}! \\ $$$${I}'{m}\:{sorry}\:{for}\:{having}\:{been}\:{absent}\:{for} \\ $$$${months}\:{without}\:{giving}\:{you}\:{a}\:{message}. \\ $$$${It}'{s}\:{alright}\:{with}\:{me}.\: \\ $$$${The}\:{last}\:{few}\:{months}\:{I}\:{was}\:{very}\:{very} \\ $$$${busy}\:{with}\:{our}\:{new}\:{house}\:{and}\:{I}\:{had}\:{no} \\ $$$${time}\:{for}\:{other}\:{things}.\:{I}\:{changed}\:{my} \\ $$$${smartphone}\:{and}\:{lost}\:{my}\:{old}\:{ID}. \\…