Question Number 204511 by Ghisom last updated on 20/Feb/24 $$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}\lfloor{x}\rfloor+\frac{\mathrm{57}}{\mathrm{4}}=\mathrm{0} \\ $$ Answered by mr W last updated on 20/Feb/24 $$\lfloor{x}\rfloor={n}\:\in\:\mathbb{Z}^{+} \\…
Question Number 204521 by explorerapollo last updated on 20/Feb/24 Commented by TonyCWX08 last updated on 20/Feb/24 $${I}\:{don}'{t}\:{understand}\:{your}\:{request}\:{here}. \\ $$ Answered by mr W last updated…
Question Number 204522 by universe last updated on 20/Feb/24 Commented by witcher3 last updated on 20/Feb/24 $$\phi'\left(\mathrm{y}\right)\:\:\mathrm{not}\:\mathrm{mor}\:\mathrm{information}\:\mathrm{about}\:\phi\:\mathrm{finction}? \\ $$$$ \\ $$ Commented by universe last…
Question Number 204533 by bello6646 last updated on 20/Feb/24 Answered by MM42 last updated on 20/Feb/24 $${ln}\frac{{ax}}{{b}}={u}\Rightarrow{I}={ln}\left({ln}\frac{{a}}{{b}}{x}\right)+{c} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204517 by Lindemann last updated on 20/Feb/24 Answered by witcher3 last updated on 20/Feb/24 $$\left.=\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)\mathrm{cos}^{−\mathrm{1}} \left(\mathrm{x}\right)\right]_{−\mathrm{1}} ^{\mathrm{1}} +\int_{−\mathrm{1}} ^{\mathrm{1}} \mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right).\frac{\mathrm{dx}}{\:\sqrt{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }}._{=\mathrm{0}}…
Question Number 204518 by Abdullahrussell last updated on 20/Feb/24 Answered by TonyCWX08 last updated on 20/Feb/24 $$ \\ $$$${let}\:{us}\:{solve}\:{y}^{{y}} =\mathrm{2}\:{first} \\ $$$${y}^{{y}} =\mathrm{2} \\ $$$${y}=\mathrm{2}^{\frac{\mathrm{1}}{{y}}}…
Question Number 204512 by Ghisom last updated on 20/Feb/24 $$\mathrm{solve}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{3}^{\mathrm{2i}{x}} −\mathrm{3}^{\mathrm{i}{x}} \mathrm{2}+\mathrm{5}=\mathrm{0} \\ $$ Answered by Rasheed.Sindhi last updated on 20/Feb/24 $$\left(\mathrm{3}^{{ix}} \right)^{\mathrm{2}}…
Question Number 204509 by JohnSmith last updated on 19/Feb/24 Commented by Rasheed.Sindhi last updated on 20/Feb/24 $$\mathcal{N}{ot}\:{clear}! \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 204505 by cherokeesay last updated on 19/Feb/24 Answered by mr W last updated on 19/Feb/24 $$\left[\sqrt{\left(\mathrm{2}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }−\mathrm{1}\right]^{\mathrm{2}} +\left(\mathrm{1}−{r}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\left[\mathrm{2}\sqrt{\mathrm{1}−{r}}−\mathrm{1}\right]^{\mathrm{2}} =\mathrm{2}{r}−\mathrm{1}…
Question Number 204500 by DeArtist last updated on 19/Feb/24 $$\mathrm{Given}\:\mathrm{that}\:{I}\:=\:\int\int_{{R}} \left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)^{\frac{\mathrm{5}}{\mathrm{2}}} {dxdy}\:\mathrm{where}\:{R} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{region}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:\leqslant\:{a}^{\mathrm{2}} \\ $$$$\mathrm{use}\:\mathrm{a}\:\mathrm{suitable}\:\mathrm{transformation}\:\mathrm{to}\:\mathrm{evaluate}\:{I} \\ $$ Answered by witcher3…