Question Number 193356 by leandrosriv02 last updated on 11/Jun/23 Answered by AST last updated on 11/Jun/23 $${m}=\mathrm{4} \\ $$$$\frac{{d}}{{dm}}\mathrm{3}^{{m}} =\mathrm{3}^{{m}} {In}\left(\mathrm{3}\right)>\:\frac{{d}}{{dm}}\left(\mathrm{2}^{{m}} +\mathrm{65}\right)=\mathrm{2}^{{m}} {In}\left(\mathrm{2}\right)\:\left({when}\:{m}>\mathrm{0}\right) \\ $$$${So},{there}\:{exists}\:{no}\:{other}\:{solution}\:{after}\:{their}\:…
Question Number 193346 by pascal889 last updated on 11/Jun/23 $$\sqrt{\mathrm{4x}−\mathrm{3}}−\sqrt{\mathrm{2x}−\mathrm{5}}=\mathrm{2} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$ Answered by AST last updated on 11/Jun/23 $${u}−{v}=\mathrm{2};\:{u}^{\mathrm{2}} −\mathrm{2}{v}^{\mathrm{2}} =\mathrm{7} \\…
Question Number 193339 by Rajpurohith last updated on 10/Jun/23 $${Prove}\:{that}\:{a}\:{group}\:{G}\:{of}\:{prime}\:{order}\:{is}\:{cyclic}. \\ $$$$ \\ $$ Answered by witcher3 last updated on 10/Jun/23 $$\mathrm{let}\:\mathrm{g}\in\mathrm{G}−\mathrm{e},\mathrm{existe}\:\mathrm{since}\:\mathrm{ord}\left(\mathrm{G}\right)=\mathrm{p}\geqslant\mathrm{2} \\ $$$$\mathrm{g}^{\mathrm{p}} =\mathrm{e}…
Question Number 193314 by York12 last updated on 10/Jun/23 $$ \\ $$$$\boldsymbol{{a}}\:,\boldsymbol{{b}},\:\boldsymbol{{c}}\:\:>\:\mathrm{0}\:\&\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} =\mathrm{3}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}}\:\right)^{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} } \:}\leqslant\left(\mathrm{1}+\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}\right) \\ $$ Answered by witcher3 last…
Question Number 193296 by MATHEMATICSAM last updated on 09/Jun/23 $$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:=\:\mathrm{16},\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$$\mathrm{and}\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{20}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:? \\ $$ Commented by kapoorshah…
Question Number 193284 by universe last updated on 09/Jun/23 $$\:\:\:\mathrm{6}{y}\:−\mathrm{2}{xy}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\mathrm{8}{z}\:−\:{yz}\:=\:\mathrm{9} \\ $$$$\:\:\:\mathrm{10}{x}\:−\:\mathrm{4}{xz}\:=\:\mathrm{8}\: \\ $$$${find}\:{x}+{y}\:+{z}\:=\:? \\ $$ Answered by AST last updated on 09/Jun/23…
Question Number 193238 by lmcp1203 last updated on 08/Jun/23 $${s}={a}+{b}+{c}+{d}+….. \\ $$$${number}\:{terms}\::{n} \\ $$$$\left\{{a};{b};{c};{d}…..\right\}>\mathrm{0} \\ $$$$\left.{then}\:{E}={s}/\left({s}−{a}\right)+{s}/\left({s}−{b}\right)+{s}/{s}−{c}\right)+…. \\ $$$$\left.{a}\left.\right)\:{E}>={n}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:{b}\right){E}>={n}^{\mathrm{2}} /\left({n}−\mathrm{1}\right) \\ $$$$\left.{c}\left.\right)\:{E}>={n}/\left({n}+\mathrm{1}\right)\:\:\:\:\:\:{d}\right)\:{E}>={n}^{\mathrm{2}} /\left({n}+\mathrm{1}\right) \\ $$$$\left.{e}\right)\:{E}>={n}^{\mathrm{2}}…
Question Number 193213 by simplifiedmaths965 last updated on 07/Jun/23 $$\left(\frac{{a}^{{m}} }{{a}^{−{n}} }\right)^{{m}−{n}} \\ $$ Answered by aba last updated on 07/Jun/23 $$\left(\frac{\mathrm{a}^{\mathrm{m}} }{\mathrm{a}^{−\mathrm{n}} }\right)^{\mathrm{m}−\mathrm{n}} =\left(\mathrm{a}^{\mathrm{m}+\mathrm{n}}…
Question Number 193197 by qaz last updated on 07/Jun/23 $$\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{666}} {mod}\mathrm{1000}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 193195 by qaz last updated on 07/Jun/23 $$\left(−\mathrm{3}\right)^{\mathrm{666}} {mod}\mathrm{1000}=? \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com