Question Number 32637 by mondodotto@gmail.com last updated on 29/Mar/18 Answered by Joel578 last updated on 30/Mar/18 $${f}\left(−\mathrm{1}\right)\:=\:−\mathrm{2} \\ $$$${f}\left(\mathrm{0}\right)\:=\:\mathrm{2} \\ $$$$\mathrm{Range}:\:\left\{{x}\:\mid\:−\mathrm{2}\:\leqslant\:{x}\:\leqslant\:\mathrm{2}\right\} \\ $$$$ \\ $$$${f}\left(\mathrm{0}\right)\:=\:\mathrm{2}…
Question Number 163707 by SANOGO last updated on 09/Jan/22 $${etudier}\:{suivant}\:{les}\:{reels}\:\:{a}\:{et}\:{b}\:{la}\:{convergence} \\ $$$$\underset{{n}=\mathrm{1}} {\sum}\frac{{a}^{{n}} }{{b}^{{n}} +{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 32610 by naka3546 last updated on 29/Mar/18 $$\mathrm{2}.\:\:{Find}\:\:{the}\:\:{number}\:\:{of}\:\:{ordered}\:\:{triples}\:\:\left({a},\:{b},\:{c}\right)\:\:{of}\:\:{integers}\:\:{satisfying}\:\:\:\:\mathrm{0}\:\leqslant\:\:{a},\:{b},\:{c}\:\:\leqslant\:\:\mathrm{1000}\:\:\:{for}\:\:{which} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} \:\:\equiv\:\:\mathrm{3}{abc}\:+\:\mathrm{1}\:\:\:\left({mod}\:\:\mathrm{1001}\right)\: \\ $$ Commented by Rasheed.Sindhi last updated on 30/Mar/18 $$\left(\mathrm{1},\mathrm{0},\mathrm{0}\right),\left(\mathrm{0},\mathrm{1},\mathrm{0}\right)\:\&\:\left(\mathrm{0},\mathrm{0},\mathrm{1}\right)\:\mathrm{are}\:\mathrm{satisfying}…
Question Number 32609 by naka3546 last updated on 01/Apr/18 $$\mathrm{1}.\:{Suppose}\:\:{that}\:\:{a},\:{b},\:{and}\:\:{c}\:\:{are}\:\:{real}\:\:{numbers}\:\:{such}\:\:{that}\:\:{a}\:<\:{b}\:<\:{c}\:\:{and}\:\:\:{a}^{\mathrm{3}} \:−\:\mathrm{3}{a}\:+\:\mathrm{1}\:\:=\:\:{b}^{\mathrm{3}} \:−\:\mathrm{3}{b}\:+\:\mathrm{1}\:\:=\:\:{c}^{\mathrm{3}} \:−\:\mathrm{3}{c}\:+\:\mathrm{1}\:=\:\:\mathrm{0}\:. \\ $$$$\:\:\:\:{Then}\:\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} \:+\:{b}}\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} \:+\:{c}}\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{2}} \:+\:{a}}\:\:\:\:{can}\:{be}\:\:{written}\:\:{as}\:\:\:\frac{{p}}{{q}}\:\:\:{for}\:\:{relatively}\:\:{prime}\:\:{of}\:\:{positive}\:\:{integers}\:\:\boldsymbol{{p}}\:\:{and}\:\:\:\boldsymbol{{q}}.\:\:\:{Find}\:\:\:\mathrm{100}{p}\:+\:{q} \\ $$ Terms of Service Privacy Policy…
Question Number 98137 by Algoritm last updated on 11/Jun/20 Answered by Farruxjano last updated on 11/Jun/20 $$\left(\mathrm{1}−\boldsymbol{{a}}\right)\left(\mathrm{1}−\boldsymbol{{b}}\right)\left(\mathrm{1}−\boldsymbol{{c}}\right)=\left(\mathrm{1}−\left(\boldsymbol{{a}}+\boldsymbol{{b}}\right)+\boldsymbol{{ab}}\right)\left(\mathrm{1}−\boldsymbol{{c}}\right)= \\ $$$$=\mathrm{1}−\boldsymbol{{c}}−\boldsymbol{{a}}−\boldsymbol{{b}}+\boldsymbol{{ab}}+\boldsymbol{{ac}}+\boldsymbol{{ab}}−\boldsymbol{{abc}}= \\ $$$$=\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}−\boldsymbol{{abc}}=\left(\ast\right) \\ $$$$\mathrm{1}=\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{abc}}}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{\boldsymbol{{abc}}}\leqslant\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ac}}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{\boldsymbol{{a}}^{\mathrm{2}}…
Question Number 32569 by naka3546 last updated on 28/Mar/18 $${Compute}\:\:{the}\:\:{number}\:\:{of}\:\:\:{ordered}\:\:{quadruple}\:\:\left({a},\:{b},\:{c},\:{d}\right)\:\:{of}\:\:{distinct}\:\:{positive}\:\:{integers}\:\:\:{so}\:\:{that}\:\:\begin{pmatrix}{\begin{pmatrix}{{a}}\\{{b}}\end{pmatrix}}\\{\begin{pmatrix}{{c}}\\{{d}}\end{pmatrix}}\end{pmatrix}\:\:\:=\:\:\mathrm{21}\:. \\ $$ Commented by MJS last updated on 28/Mar/18 $$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}=\mathrm{21}\:\Rightarrow\:\left({n};{k}\right)\in\left\{\left(\mathrm{7};\mathrm{2}\right);\left(\mathrm{7};\mathrm{5}\right);\left(\mathrm{21};\mathrm{1}\right);\left(\mathrm{21};\mathrm{20}\right)\right\} \\ $$$$ \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}=\mathrm{1}\:\Rightarrow\:\left({n};{k}\right)=\left({m};{m}\right)\mid{m}\in\mathbb{N} \\…
Question Number 98104 by bemath last updated on 11/Jun/20 $$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}+\mathrm{y}\: \\ $$$$\mathrm{if}\:\begin{cases}{\mathrm{x}^{\mathrm{3}} +\mathrm{y}^{\mathrm{3}} =\mathrm{1}}\\{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{y}+\mathrm{1}\right)=\mathrm{2}}\end{cases} \\ $$$$ \\ $$ Commented by bemath last updated on 12/Jun/20…
Question Number 163638 by mkam last updated on 08/Jan/22 $$\int\:\frac{\boldsymbol{{dx}}}{\:\sqrt{\mathrm{2}\boldsymbol{{c}}_{\mathrm{1}} +\boldsymbol{{e}}^{−\mathrm{2}\boldsymbol{{x}}} }} \\ $$ Commented by mkam last updated on 09/Jan/22 $$????? \\ $$ Commented…
Question Number 98096 by Algoritm last updated on 11/Jun/20 Answered by mathmax by abdo last updated on 11/Jun/20 $$\mathrm{S}\:=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{3n}−\mathrm{1}}{\mathrm{2}^{\mathrm{n}−\mathrm{1}} }\:=_{\mathrm{n}−\mathrm{1}=\mathrm{p}} \:\:\sum_{\mathrm{p}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3p}+\mathrm{2}}{\mathrm{2}^{\mathrm{p}}…
Question Number 163627 by nurtani last updated on 08/Jan/22 Answered by mr W last updated on 08/Jan/22 $${u}=\frac{\mathrm{1}}{{x}+{y}}\:>\mathrm{0} \\ $$$${v}={xy} \\ $$$$\sqrt{\mathrm{21}{v}}\left(\mathrm{1}−{u}^{\mathrm{2}} \right)=\mathrm{8}\sqrt{\mathrm{2}} \\ $$$${v}=\frac{\mathrm{128}}{\mathrm{21}\left(\mathrm{1}−{u}^{\mathrm{2}}…