Question Number 206958 by hardmath last updated on 01/May/24 $$\mathrm{If}\:\:\:\frac{\mathrm{1}}{\mathrm{3}^{−\boldsymbol{\mathrm{x}}} }\:=\:\mathrm{5}\:\:\:\mathrm{find}:\:\:\mathrm{9}^{\boldsymbol{\mathrm{x}}+\mathrm{1}} \:=\:? \\ $$ Answered by Frix last updated on 01/May/24 $$\mathrm{3}^{{x}} =\mathrm{5} \\ $$$$\left(\mathrm{3}^{{x}}…
Question Number 206939 by Tawa11 last updated on 01/May/24 In how many ways can the word KINECTIC be arranged so that no vowels can be…
Question Number 206879 by mr W last updated on 29/Apr/24 $${solve}\:{for}\:{positive}\:{integers} \\ $$$$\frac{{x}}{{y}+{z}}+\frac{{y}}{{z}+{x}}+\frac{{z}}{{x}+{y}}=\mathrm{4} \\ $$ Commented by Frix last updated on 30/Apr/24 $$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{impossible}. \\ $$…
Question Number 206888 by universe last updated on 29/Apr/24 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 206881 by efronzo1 last updated on 29/Apr/24 $$\:\:\cancel{\underline{\underbrace{\boldsymbol{{x}}}}} \\ $$ Answered by Skabetix last updated on 29/Apr/24 $$\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{x}^{{n}} }{{n}!}={e}^{{x}} \rightarrow{here}\:{x}\:=\:\mathrm{1}\:{so}\:\mathrm{2}\:+\:\frac{\mathrm{1}}{\mathrm{2}!}+…={e} \\…
Question Number 206862 by hardmath last updated on 28/Apr/24 Commented by SWPlaysMC last updated on 29/Apr/24 $$\mathrm{Easiest}\:\mathrm{way}\:\mathrm{I}\:\mathrm{thought}\:\mathrm{of}\:\mathrm{proving}\:\mathrm{this}\:\left(\mathrm{I}\:\mathrm{partially}\:\mathrm{solved}\:\mathrm{this}\right) \\ $$$$\mathrm{Let}\:{a}=\mathrm{1},\:{b}=\mathrm{1},\:\mathrm{and}\:{c}=\mathrm{1} \\ $$$$\left(\mathrm{Hint}:\:\mathrm{if}\:\mathrm{any}\:\mathrm{one}\:\mathrm{of}\:\mathrm{these}\:\mathrm{is}\:\mathrm{0}\:\mathrm{then}\:\mathrm{expression}\:\mathrm{is}\:\mathrm{theoretically}\:\mathrm{undefined}\:\mathrm{due}\:\mathrm{to}\:\boldsymbol{\div}\:\mathrm{by}\:\mathrm{0}\right) \\ $$$$\frac{\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}}+\frac{\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}}+\frac{\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{1}}}{\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}+\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}}\geqslant\mathrm{3} \\ $$$$\mathrm{so}\:\mathrm{all}\:\sqrt{\frac{\mathrm{1}}{\mathrm{1}}}=\mathrm{1}\:\mathrm{and}\:\frac{\mathrm{1}}{\mathrm{1}}=\mathrm{1},\:\mathrm{thus}\:\mathrm{substitute}\:\mathrm{all}\:\mathrm{nums}\:\mathrm{and}\:\mathrm{denoms}\:\mathrm{with}\:\mathrm{1}+\mathrm{1}\:\mathrm{like}\:\mathrm{so}…
Question Number 206868 by depressiveshrek last updated on 28/Apr/24 $$\mathrm{If}\:{A},\:{B}\:\mathrm{and}\:{A}+{B}\:\mathrm{are}\:\mathrm{non}−\mathrm{singular} \\ $$$$\mathrm{square}\:\mathrm{matrices},\:\mathrm{prove}\:\mathrm{that}\:{A}^{−\mathrm{1}} +{B}^{−\mathrm{1}} \\ $$$$\mathrm{is}\:\mathrm{also}\:\mathrm{non}−\mathrm{singular}. \\ $$ Answered by aleks041103 last updated on 28/Apr/24 $${A}^{−\mathrm{1}}…
Question Number 206869 by universe last updated on 28/Apr/24 Commented by Frix last updated on 28/Apr/24 $$\mathrm{Answer}\:\mathrm{is}\:\mathrm{e}^{{a}_{\mathrm{1}} } −\mathrm{1}\:\mathrm{for}\:{a}_{\mathrm{1}} \in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$ Commented by universe…
Question Number 206845 by hardmath last updated on 27/Apr/24 $$\mathrm{Find}:\:\:\:\frac{\infty!}{\infty^{\infty} }\:=\:? \\ $$ Commented by A5T last updated on 27/Apr/24 $${Do}\:{you}\:{wish}\:{to}\:{find}\:{this}:\:\underset{{n}\rightarrow\infty} {{lim}}\frac{{n}!}{{n}^{{n}} }\:? \\ $$…
Question Number 206839 by efronzo1 last updated on 27/Apr/24 Answered by A5T last updated on 27/Apr/24 $${k}^{\mathrm{3}} +{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{2}\leqslant{k}^{\mathrm{2}} +{k}+\mathrm{300}\Leftrightarrow{k}^{\mathrm{3}} +{k}\leqslant\mathrm{298} \\ $$$$\Rightarrow\mathrm{0}\leqslant{k}\leqslant\mathrm{6};\:{checking}\:{implies}\:{k}=\mathrm{0}\:{or}\:\mathrm{2}. \\ $$$${So},{sum}=\mathrm{2}…