Question Number 191961 by Shlock last updated on 04/May/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191950 by Shrinava last updated on 04/May/23 $$\sqrt{\mathrm{a}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\:\:\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\sqrt{\mathrm{a}}\:=\:? \\ $$ Answered by AST last updated on 04/May/23 $$\frac{\mathrm{1}}{\:\sqrt{{a}}−\mathrm{1}}=\frac{\sqrt{{a}}+\mathrm{1}}{{a}−\mathrm{1}}={a}\Rightarrow{a}^{\mathrm{2}} −{a}−\sqrt{{a}}=\mathrm{1} \\ $$ Terms…
Question Number 191947 by universe last updated on 04/May/23 $${prove}\:{that} \\ $$$$\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{…}}}}}\:\:=\:\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }+{b}}{\mathrm{2}} \\ $$ Answered by ajfour last updated on 05/May/23 $${Let}\:{s}=\sqrt{{a}+{b}\sqrt{{a}−{bs}}}=\frac{{b}+\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }}{\mathrm{2}} \\…
Question Number 191946 by mustafazaheen last updated on 04/May/23 $$\frac{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)+\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{15}\right)}{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)}=? \\ $$ Answered by horsebrand11 last updated on 04/May/23 $$=\:\mathrm{4} \\ $$…
Question Number 126391 by mathocean1 last updated on 20/Dec/20 $$ \\ $$$${show}\:{that}\: \\ $$$${m}^{\mathrm{2}} +{m}\:{and}\:\:\mathrm{2}{m}+\mathrm{1}\:{are}\:{prime}\:{betwen} \\ $$$${them}. \\ $$ Answered by akornes last updated on…
Question Number 60854 by Askash last updated on 26/May/19 $$\left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)/\left({x}−\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$ Commented by Askash last updated on 26/May/19…
Question Number 126363 by ajfour last updated on 19/Dec/20 $${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0}\:\:\:\:\:\:\:;\:\:\:\:\left[{c}<\mathrm{2}/\left(\mathrm{3}\sqrt{\mathrm{3}}\right)\right] \\ $$$$\left({Solve}\:{by}\:{a}\:\:{method}\:{other}\:{than}\right. \\ $$$$\left.{trigonometric}\:{solution}\right). \\ $$ Answered by mathmax by abdo last updated on…
Question Number 60816 by ANTARES VY last updated on 26/May/19 $$\boldsymbol{\mathrm{S}}=\mathrm{4}\boldsymbol{\pi\mathrm{R}}^{\mathrm{2}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$ Answered by tanmay last updated on 26/May/19 $${ds}={Rd}\theta×{Rsin}\theta{d}\phi \\ $$$${S}={R}^{\mathrm{2}} \int_{\mathrm{0}}…
Question Number 60817 by ANTARES VY last updated on 26/May/19 $$\boldsymbol{\mathrm{V}}=\frac{\mathrm{4}}{\mathrm{3}}\boldsymbol{\pi\mathrm{R}}^{\mathrm{3}} \:\:\:\boldsymbol{\mathrm{prove}} \\ $$ Commented by Prithwish sen last updated on 26/May/19 http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html Commented by…
Question Number 191867 by Spillover last updated on 02/May/23 $${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$ Answered by qaz last updated on 02/May/23…