Question Number 178375 by doline last updated on 16/Oct/22 $${calculer}\:{une}\:{primitive}\:{de}\:−\mathrm{3}{x}/\sqrt{{x}^{\mathrm{2}} +\mathrm{3}} \\ $$ Commented by CElcedricjunior last updated on 16/Oct/22 $$\int\frac{−\mathrm{3}\boldsymbol{\mathrm{xdx}}}{\:\sqrt{\mathrm{3}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} }}=−\frac{\mathrm{3}}{\mathrm{2}}\int\frac{\mathrm{2}\boldsymbol{{x}}}{\:\sqrt{\mathrm{3}+\boldsymbol{{x}}^{\mathrm{2}} }}\boldsymbol{{dx}}=−\frac{\mathrm{3}}{\mathrm{2}}\sqrt{\mathrm{3}+\boldsymbol{{x}}^{\mathrm{2}} }\:+\boldsymbol{{K}} \\…
Question Number 178349 by doline last updated on 15/Oct/22 $${etudier}\:{la}\:{fonction}\:{suivante}\:\mathrm{4}{x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{1} \\ $$$${calculer}\:{les}\:{limites}\:{puis}\:{dresser}\:{son}\:{tableau}\:{de}\:{variation} \\ $$ Answered by Acem last updated on 16/Oct/22 $$ \\…
Question Number 112813 by Aina Samuel Temidayo last updated on 09/Sep/20 $$\mathrm{A}\:\mathrm{binary}\:\mathrm{operation}\:\mathrm{has}\:\mathrm{the}\:\mathrm{property} \\ $$$$\mathrm{a}\ast\left(\mathrm{b}\ast\mathrm{c}\right)\:=\:\left(\mathrm{a}\ast\mathrm{b}\right)\bullet\mathrm{c}\:\mathrm{and}\:\mathrm{that}\:\mathrm{a}\ast\mathrm{a}=\mathrm{1}\:\mathrm{for} \\ $$$$\mathrm{all}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\mathrm{b}\:\mathrm{and}\:\mathrm{c}. \\ $$$$\left('\bullet'\:\mathrm{here}\:\mathrm{represent}\:\mathrm{multiplication}\right). \\ $$$$\mathrm{The}\:\mathrm{solution}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{2016}\ast\left(\mathrm{6}\ast\mathrm{x}\right)=\mathrm{100}\:\mathrm{can}\:\mathrm{be}\:\mathrm{written}\:\mathrm{as}\:\frac{\mathrm{p}}{\mathrm{q}} \\ $$$$\mathrm{where}\:\mathrm{p}\:\mathrm{and}\:\mathrm{q}\:\mathrm{are}\:\mathrm{relatively}\:\mathrm{prime} \\…
Question Number 47235 by Rio Michael last updated on 06/Nov/18 $${the}\:{force}\:{acting}\:{on}\:{a}\:{particle}\:{P}\:{of}\:{mass}\:\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}{ti}\:+\mathrm{4}{j}\right){N}. \\ $$$${P}\:{is}\:{initially}\:{at}\:{rest}\:{at}\:{point}\:{with}\:{position}\:{vector}\:\left({i}+\mathrm{2}{j}\right). \\ $$$${Find}\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}=\mathrm{2}\:{and}\:{the}\:{position}\:{vector}\: \\ $$$${when}\:{t}=\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 47234 by Rio Michael last updated on 06/Nov/18 $${the}\:{force}\:{acting}\:{on}\:{a}\:{particle}\:{P}\:{of}\:{mass}\:\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}{ti}\:+\mathrm{4}{j}\right){N}. \\ $$$${P}\:{is}\:{initially}\:{at}\:{rest}\:{at}\:{point}\:{with}\:{position}\:{vector}\:\left({i}+\mathrm{2}{j}\right). \\ $$$${Find}\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}=\mathrm{2}\:{and}\:{the}\:{position}\:{vector}\: \\ $$$${when}\:{t}=\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 47233 by Rio Michael last updated on 06/Nov/18 $${the}\:{force}\:{acting}\:{on}\:{a}\:{particle}\:{P}\:{of}\:{mass}\:\:\mathrm{2}{kg}\:{is}\:\left(\mathrm{2}{ti}\:+\mathrm{4}{j}\right){N}. \\ $$$${P}\:{is}\:{initially}\:{at}\:{rest}\:{at}\:{point}\:{with}\:{position}\:{vector}\:\left({i}+\mathrm{2}{j}\right). \\ $$$${Find}\:{the}\:{velocity}\:{of}\:{P}\:{when}\:{t}=\mathrm{2}\:{and}\:{the}\:{position}\:{vector}\: \\ $$$${when}\:{t}=\mathrm{2}. \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 47138 by Tawa1 last updated on 05/Nov/18 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 176723 by doline last updated on 25/Sep/22 $${donner}\:{la}\:{forme}\:{trigonometrique}\left({i}−\mathrm{1}\right)^{\mathrm{5}} /\left({i}−\mathrm{4}\right)^{\mathrm{4}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 176719 by doline last updated on 25/Sep/22 $$ \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 111105 by bobhans last updated on 02/Sep/20 $$\mathrm{y}''+\mathrm{2y}'+\mathrm{y}=\mathrm{e}^{−\mathrm{2x}} +\mathrm{2x}+\mathrm{3} \\ $$ Commented by mohammad17 last updated on 02/Sep/20 $${r}^{\mathrm{2}} +\mathrm{2}{r}+\mathrm{1}=\mathrm{0}\Rightarrow\left({r}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\Rightarrow{r}_{\mathrm{1}} ={r}_{\mathrm{2}} =−\mathrm{1}…