Question Number 154476 by SANOGO last updated on 18/Sep/21 $${soit}:{y}''−\mathrm{3}{y}'−\mathrm{4}{y}=\mathrm{3}{e}^{\mathrm{3}{x}\:} \:{avec}\:, \\ $$$${f}\left({o}\right)=−\frac{\mathrm{1}}{\mathrm{2}}\:{et}\:{f}'\left(\mathrm{0}\right)=\mathrm{4} \\ $$$${alors}\:{f}\left(\mathrm{1}\right)=? \\ $$$$ \\ $$ Answered by ARUNG_Brandon_MBU last updated on…
Question Number 88943 by jagoll last updated on 14/Apr/20 Commented by john santu last updated on 14/Apr/20 $$\left.\mathrm{1}\right)\:\mathrm{20}+\mathrm{22}−\mathrm{39}\:=\:\mathrm{3} \\ $$$$\Rightarrow\mathrm{3}^{\mathrm{4}−\mathrm{1}} \:=\:\mathrm{3}^{\mathrm{3}} \:=\:\mathrm{27} \\ $$$$\left.\mathrm{2}\right)\mathrm{7}+\mathrm{7}−\mathrm{13}\:=\:\mathrm{1} \\…
Question Number 88940 by jagoll last updated on 14/Apr/20 Answered by john santu last updated on 14/Apr/20 $$\mathrm{3}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\mathrm{3}×\mathrm{9}+\mathrm{3}×\mathrm{7}\:\right]\:=\: \\ $$$$\mathrm{27}\:+\:\mathrm{12}\:=\:\mathrm{39}\: \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{4}}\:\left[\:\mathrm{4}×\mathrm{5}+\mathrm{4}×\mathrm{3}\right]\:=\: \\…
Question Number 154478 by mnjuly1970 last updated on 18/Sep/21 $$ \\ $$$$ \\ $$$$\:\:{prove}\:{that}\:# \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{\:{sin}^{\:\mathrm{3}} \left(\:{x}\:\right).{ln}\left(\:{x}\:\right)}{{x}}\:{dx}\:\overset{?} {=}\:\frac{\pi}{\mathrm{8}}\:\left(−\mathrm{2}\gamma\:+{ln}\left(\mathrm{3}\right)\right)\:…..\blacksquare\:{m}.{n}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\: \\ $$$$ \\…
Question Number 23402 by RoachDN last updated on 29/Oct/17 Commented by Tinkutara last updated on 29/Oct/17 $${I}=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{sin}\:{xdx}}{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$ Commented by RoachDN…
Question Number 154475 by mr W last updated on 18/Sep/21 Commented by mr W last updated on 18/Sep/21 $${find}\:{the}\:{area}\:{of}\:{the}\:{big}\:{triangle}\:{whose} \\ $$$${sides}\:{have}\:{the}\:{distances}\:\boldsymbol{{d}}_{\mathrm{1}} ,\boldsymbol{{d}}_{\mathrm{2}} ,\boldsymbol{{d}}_{\mathrm{3}} \:{to}\: \\…
Question Number 88936 by naka3546 last updated on 14/Apr/20 Answered by ajfour last updated on 14/Apr/20 $${let}\:\:\:{a}=\sqrt{\mathrm{196}−{c}^{\mathrm{2}} }\:\:\:\:{from}\:\mathrm{1}^{{st}} \\ $$$${b}^{\mathrm{2}} =\:\mathrm{2}{c}\sqrt{\mathrm{196}−{c}^{\mathrm{2}} }−\mathrm{27}\:\:\:{from}\:\mathrm{2}^{{nd}} \\ $$$${and}\:\:\:\:{b}^{\mathrm{2}} =\left({c}+\sqrt{\mathrm{225}−{c}^{\mathrm{2}}…
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Question Number 154469 by mnjuly1970 last updated on 18/Sep/21 $$ \\ $$$${prove}:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\:\left(\mathrm{4}−\:\mathrm{2}{x}\:+{x}^{\:\mathrm{2}} \right){dx}\:=\mathrm{2}{ln}\left(\frac{\mathrm{2}}{{e}}\right)\:+\frac{\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$ \\ $$ Answered by ARUNG_Brandon_MBU last updated on…
Question Number 23399 by Tinkutara last updated on 29/Oct/17 $$\mathrm{Two}\:\mathrm{blocks}\:\mathrm{of}\:\mathrm{masses}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{and}\:\mathrm{3}\:\mathrm{kg} \\ $$$$\mathrm{are}\:\mathrm{kept}\:\mathrm{on}\:\mathrm{a}\:\mathrm{smooth}\:\mathrm{inclined}\:\mathrm{plane}. \\ $$$$\mathrm{A}\:\mathrm{constant}\:\mathrm{force}\:\mathrm{of}\:\mathrm{magnitude}\:\mathrm{20}\:\mathrm{N}\:\mathrm{is} \\ $$$$\mathrm{applied}\:\mathrm{on}\:\mathrm{2}\:\mathrm{kg}\:\mathrm{block}\:\mathrm{parallel}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{inclined}.\:\mathrm{The}\:\mathrm{contact}\:\mathrm{force}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{two}\:\mathrm{blocks}\:\mathrm{is} \\ $$ Commented by Tinkutara…