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Question-188218

Question Number 188218 by pascal889 last updated on 26/Feb/23 Answered by cortano12 last updated on 27/Feb/23 $$\:\mathrm{let}\:\mathrm{log}\:_{\mathrm{x}} \mathrm{10}=\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{10}} \mathrm{x}}\:=\mathrm{p}\:,\:\mathrm{x}>\mathrm{0}\:,\mathrm{x}\neq\mathrm{1} \\ $$$$\Rightarrow\mathrm{2}.\mathrm{log}\:_{\mathrm{x}} \mathrm{10}−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{log}\:_{\mathrm{x}} \left(\frac{\sqrt{\mathrm{x}}}{\mathrm{100}}\right)\right]−\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{1}−\mathrm{log}\:_{\mathrm{x}} \mathrm{10}}=−\frac{\mathrm{7}}{\mathrm{2}} \\…

Esta-has-a-directory-it-contains-n-n-lt-1000-pages-where-999991-subscribers-are-registered-and-each-page-contains-the-same-numbers-of-subscribers-How-many-pages-does-this-directory-has-

Question Number 122668 by mathocean1 last updated on 18/Nov/20 $${Esta}\:{has}\:{a} \\ $$$$\:{directory}.\:{it}\:{contains}\: \\ $$$${n}\:\left({n}<\mathrm{1000}\right)\:{pages}\:{where}\: \\ $$$$\mathrm{999991}\:{subscribers}\:{are}\:{registered} \\ $$$${and}\:{each}\:{page}\:{contains}\:{the}\: \\ $$$${same}\:{numbers}\:{of}\:{subscribers}. \\ $$$${How}\:{many}\:{pages}\:{does}\:{this} \\ $$$${directory}\:{has}? \\…

Question-188203

Question Number 188203 by moh777 last updated on 26/Feb/23 Answered by mr W last updated on 26/Feb/23 $${g}=\mathrm{32}.\mathrm{2}\:{ft}/{s}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right) \\ $$$${h}_{\mathrm{1}} ={v}_{\mathrm{1}} \left({t}+\mathrm{3}\right)−\frac{{g}\left({t}+\mathrm{3}\right)^{\mathrm{2}} }{\mathrm{2}}…

Question-122638

Question Number 122638 by Backer last updated on 18/Nov/20 Commented by Backer last updated on 18/Nov/20 $$\mathrm{Hi} \\ $$$$\mathrm{Can}\:\mathrm{somebody}\:\mathrm{do}\:\mathrm{me}\:\mathrm{a}\:\mathrm{favor}\:\mathrm{in}\:\mathrm{this} \\ $$$$\mathrm{problem}? \\ $$ Terms of…

Question-57075

Question Number 57075 by Tawa1 last updated on 30/Mar/19 Commented by Tawa1 last updated on 30/Mar/19 $$\mathrm{Please}\:\mathrm{i}\:\mathrm{want}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{how}\:\mathrm{the}\:−\mathrm{1}\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{becomes}\:−\:\mathrm{13} \\ $$$$\mathrm{and}\:\mathrm{the}\:+\:\mathrm{60}\:\mathrm{outside}.\:\mathrm{and}\:\mathrm{how}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{is}\:\:\mathrm{8}.\:\:\mathrm{Using}\:\mathrm{the}\:\mathrm{same}\:\mathrm{method}. \\ $$ Commented by 121194 last…

Question-188107

Question Number 188107 by pascal889 last updated on 25/Feb/23 Answered by CElcedricjunior last updated on 25/Feb/23 $$\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}−\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}+\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}\: \\ $$$$\exists\boldsymbol{{ssi}}\:\boldsymbol{{x}}\neq−\mathrm{2}\:\boldsymbol{{et}}\:\boldsymbol{{x}}\neq\mathrm{2} \\ $$$$=>\boldsymbol{{x}}^{\mathrm{2}} \left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{4}\boldsymbol{{x}}+\mathrm{4}\right)+\boldsymbol{{x}}^{\mathrm{2}}…