Question Number 198161 by Blackpanther last updated on 12/Oct/23 Answered by Rasheed.Sindhi last updated on 12/Oct/23 $$\blacktriangle\mathrm{ABC}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{AC}.\mathrm{BC} \\ $$$$\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{12}×{y}=\mathrm{6}{y} \\ $$$$\:\:\:\:\:\:\:\mathrm{48}\leqslant\mathrm{6}{y}\leqslant\mathrm{60} \\ $$$$\:\:\:\:\:\:\:\mathrm{8}\leqslant{y}\leqslant\mathrm{10} \\ $$$$\because\:{y}\in\mathbb{Z}…
Question Number 198136 by sonukgindia last updated on 11/Oct/23 Answered by Frix last updated on 11/Oct/23 $$\frac{\mathrm{3}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{16}}{{x}^{\mathrm{5}} +\mathrm{2}{x}^{\mathrm{4}} +\mathrm{16}{x}+\mathrm{32}}=\frac{\left({x}+\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}{\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{4}} +\mathrm{16}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{2}\right)}−\frac{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right){x}+\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)}{\mathrm{8}\left({x}^{\mathrm{2}}…
Question Number 198151 by sonukgindia last updated on 11/Oct/23 Answered by Mathspace last updated on 12/Oct/23 $${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{zlnz}}{\mathrm{1}+{z}^{\mathrm{3}} }{dz}\:\:\:\:\left({z}={t}^{\frac{\mathrm{1}}{\mathrm{3}}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}}…
Question Number 198077 by SANOGO last updated on 10/Oct/23 Answered by Mathspace last updated on 10/Oct/23 $${par}\:{recurrence}\:{sur}\:{n} \\ $$$${on}\:\Phi\left(\mathrm{0}\right)\geqslant\mathrm{0}\:{vraie}\:{puisque}\:\Phi\left({N}\right)\subset{N} \\ $$$${supposons}\:{que}\:\Phi\left({n}\right)\geqslant{n}\:{et}\:{montrons} \\ $$$${que}\:\Phi\left({n}+\mathrm{1}\right)\geqslant{n}+\mathrm{1} \\ $$$${on}\:{n}+\mathrm{1}>{n}\:{et}\:\Phi\:{strictement}\:…
Question Number 198092 by SteveX last updated on 10/Oct/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 198093 by SteveX last updated on 10/Oct/23 Answered by Safojon last updated on 10/Oct/23 $$\sqrt{{x}}=\frac{−\mathrm{4}+\mathrm{108}−\mathrm{24}}{\mathrm{8}} \\ $$$$\sqrt{{x}}=\frac{\mathrm{80}}{\mathrm{8}}\:\:\left(\sqrt{{x}}\right)^{\mathrm{2}} =\left(\mathrm{10}\right)^{\mathrm{2}} \\ $$$${x}=\mathrm{100}. \\ $$$${answer}\:{x}=\mathrm{100} \\…
Question Number 198084 by SANOGO last updated on 10/Oct/23 Answered by witcher3 last updated on 10/Oct/23 $$\mathrm{Ex}\:\mathrm{1} \\ $$$$\mathrm{soit}\:\mathrm{u}_{\mathrm{n}} \:\mathrm{de}\:\mathrm{cachy}\:\Rightarrow\forall\epsilon>\mathrm{0}\:\exists\mathrm{N}\:\forall\:\mathrm{n}>\mathrm{m}>\mathrm{N}\:\mathrm{d}\left(\mathrm{u}_{\mathrm{n}} ,\mathrm{u}_{\mathrm{m}} \right)<\epsilon \\ $$$$\mathrm{pour}\:\epsilon=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{on}\:\:\mathrm{a}\:\mathrm{d}\left(\mathrm{u}_{\mathrm{n}} ,\mathrm{u}_{\mathrm{m}}…
Question Number 198074 by sonukgindia last updated on 09/Oct/23 Answered by a.lgnaoui last updated on 10/Oct/23 $$\mathrm{x}^{\mathrm{20}} +\mathrm{2x}^{\mathrm{10}} +\mathrm{1}=\left(\mathrm{x}^{\mathrm{10}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{8}} =\frac{\mathrm{x}^{\mathrm{10}} }{\mathrm{x}^{\mathrm{2}} }\Rightarrow…
Question Number 198062 by NANIGOPAL last updated on 09/Oct/23 Commented by mr W last updated on 10/Oct/23 $${equality}\:−\mathrm{2}\:{can}\:{not}\:{hold}! \\ $$$${equality}\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:{holds}\:{at}\:{A}={B}=\mathrm{20}°,\:{C}=\mathrm{140}°. \\ $$$$−\mathrm{2}<\mathrm{sin}\:\mathrm{3}{A}+\mathrm{sin}\:\mathrm{3}{B}+\mathrm{sin}\:\mathrm{3}{C}\leqslant\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$ Terms…
Question Number 198036 by mathlove last updated on 08/Oct/23 $${if}\:{A}\boldsymbol{\div}\mathrm{6}\:{reminder}\:{is}\:\mathrm{4}\:{and}\:{B}\boldsymbol{\div}\mathrm{6} \\ $$$${reminder}\:{is}\:\mathrm{5}\:{then}\:\left({A}+{B}\right)\boldsymbol{\div}\mathrm{6} \\ $$$${reminder}\:{is}\:? \\ $$ Answered by AST last updated on 08/Oct/23 $${A}\overset{\mathrm{6}} {\equiv}\mathrm{4},{B}\overset{\mathrm{6}}…