Question Number 48559 by behi83417@gmail.com last updated on 25/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 25/Nov/18 $${z}\left({x}^{\mathrm{2}} −{yz}\right)={a}^{\mathrm{2}} {z} \\ $$$${x}\left({y}^{\mathrm{2}} −{xz}\right)={b}^{\mathrm{2}} {x} \\ $$$${y}\left({z}^{\mathrm{2}}…
Question Number 48553 by Cheyboy last updated on 25/Nov/18 $$\mathrm{In}\:\mathrm{a}\:\mathrm{hospital}\:\mathrm{unit},\mathrm{there}\:\mathrm{are}\:\mathrm{8}\:\mathrm{nurses} \\ $$$$\mathrm{and}\:\mathrm{5}\:\mathrm{physicians}.\:\mathrm{7}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are} \\ $$$$\mathrm{female}\:\mathrm{nurses}\:\mathrm{and}\:\mathrm{3}\:\mathrm{of}\:\mathrm{them}\:\mathrm{are} \\ $$$$\mathrm{male}\:\mathrm{physicians}. \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of} \\ $$$$\mathrm{selecting}\:\mathrm{a}\:\mathrm{staff}\:\mathrm{who}\:\mathrm{is}\:\mathrm{Nurse}\:\mathrm{or} \\ $$$$\mathrm{Male}? \\ $$$$\mathrm{plzz}\:\mathrm{help} \\…
Question Number 114077 by bemath last updated on 17/Sep/20 Answered by john santu last updated on 17/Sep/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 179593 by mathlove last updated on 30/Oct/22 $${pleas}\:{proof}\:{the}\:{elipse}\:{environment} \\ $$$${formullah} \\ $$$${p}=\pi\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} } \\ $$ Commented by mr W last updated on…
Question Number 48509 by naka3546 last updated on 24/Nov/18 $${For}\:\:{every}\:\:{natural}\:\:{numbers}\:\:{n}\:\: \\ $$$${Find}\:\:\:{the}\:\:{value}\:\:{of} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{\mathrm{0}\leqslant{j}\leqslant{i}\leqslant{n}} {\sum}\:\:\frac{\left(−\mathrm{1}\right)^{{j}} }{\left({n}\:−\:{i}\right)!\:{j}!} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 179570 by infinityaction last updated on 30/Oct/22 $$\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{function} \\ $$$$\boldsymbol{\mathrm{f}}\left(\boldsymbol{\mathrm{x}}\right)=\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\frac{\mathrm{4}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }−\mathrm{8}\boldsymbol{\mathrm{x}}−\frac{\mathrm{12}}{\boldsymbol{\mathrm{x}}}+\mathrm{25}}\:+\:\sqrt{\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\frac{\mathrm{4}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }+−\mathrm{16}\boldsymbol{\mathrm{x}}−\frac{\mathrm{16}}{\boldsymbol{\mathrm{x}}}+\mathrm{80}} \\ $$ Answered by a.lgnaoui last updated on 01/Nov/22…
Question Number 114023 by Khalmohmmad last updated on 16/Sep/20 $$\underset{\mathrm{n}=\mathrm{1}} {\overset{\propto} {\sum}}\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} −\mathrm{1}}=? \\ $$ Commented by Dwaipayan Shikari last updated on 16/Sep/20 $${Diverges} \\…
Question Number 48482 by ajfour last updated on 24/Nov/18 $$\left({at}−{h}\right)^{\mathrm{2}} +\left(\frac{{a}}{{t}}−{k}\right)^{\mathrm{2}} ={R}^{\:\mathrm{2}} \\ $$$${where}\:\:\:{a},\:{h},\:{k},\:{R}\:{are}\:{constants}. \\ $$$${Then}\:{find}\: \\ $$$$\:\:\:{s}^{\mathrm{2}} \:=\left({t}_{\mathrm{1}} −{t}_{\mathrm{2}} \right)^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{t}_{\mathrm{1}} ^{\mathrm{2}} {t}_{\mathrm{2}} ^{\mathrm{2}}…
Question Number 113997 by Aina Samuel Temidayo last updated on 16/Sep/20 Answered by bobhans last updated on 16/Sep/20 $${consider}\::\:\sqrt{\mathrm{2}}\:+\sqrt[{\mathrm{4}}]{\mathrm{2}}\:+\mathrm{1}\:=\:\left(\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}.\left(\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)+\mathrm{1}^{\mathrm{2}} \:=\:\frac{\left(\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\:\sqrt[{\mathrm{4}}]{\mathrm{2}}−\mathrm{1}} \\ $$$${then}\:\frac{\mathrm{7}}{\left(\sqrt[{\mathrm{4}}]{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 113996 by Aina Samuel Temidayo last updated on 16/Sep/20 Commented by MJS_new last updated on 16/Sep/20 $$\frac{\mathrm{ln}\:\frac{\mathrm{81}}{\mathrm{8}}}{\mathrm{2ln}\:\frac{\mathrm{9}}{\mathrm{2}}}=\frac{\mathrm{ln}\:\sqrt{\frac{\mathrm{81}}{\mathrm{8}}}}{\mathrm{ln}\:\frac{\mathrm{9}}{\mathrm{2}}}=\frac{\mathrm{ln}\:\frac{\mathrm{9}}{\:\sqrt{\mathrm{8}}}}{\mathrm{ln}\:\frac{\mathrm{9}}{\mathrm{2}}}=\mathrm{log}_{\frac{\mathrm{9}}{\mathrm{2}}} \frac{\mathrm{9}}{\:\sqrt{\mathrm{8}}} \\ $$ Answered by bobhans…