Question Number 200019 by hardmath last updated on 12/Nov/23 Answered by mr W last updated on 12/Nov/23 Commented by hardmath last updated on 12/Nov/23 $$\mathrm{very}\:\mathrm{nice}\:\mathrm{dear}\:\mathrm{pfofessor}……
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Question Number 200013 by hardmath last updated on 12/Nov/23 Commented by mr W last updated on 12/Nov/23 $${geometry}\:{is}\:{not}\:{like}\:{a}\:{novel}.\:{geometry} \\ $$$${is}\:{best}\:{with}\:{diagram},\:{not}\:{with}\:{only} \\ $$$${text}. \\ $$ Commented…
Question Number 199932 by hardmath last updated on 11/Nov/23 $$\mathrm{1}. \\ $$$$\mathrm{If}\:\:\:\mathrm{3}\:\centerdot\:\overline {\mathrm{ab}}\:+\:\overline {\mathrm{bc}}\:=\:\mathrm{115} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{max}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)=? \\ $$$$ \\ $$$$\mathrm{2}. \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c}\in\mathbb{N} \\ $$$$\mathrm{If}\:\:\:\frac{\mathrm{a}}{\mathrm{2}}\:\:+\:\:\frac{\mathrm{b}}{\mathrm{3}}\:\:=\:\:\frac{\mathrm{c}}{\mathrm{4}} \\…
Question Number 199988 by hardmath last updated on 11/Nov/23 $$\mathrm{ABCD}\:-\:\mathrm{trapezium} \\ $$$$\mathrm{EF}\:-\:\mathrm{middle}\:\mathrm{line} \\ $$$$\mathrm{EG}\:=\:\mathrm{2}\:\:,\:\:\mathrm{GF}\:=\:\mathrm{4}\:\:,\:\:\mathrm{DC}\:=\:\mathrm{y}\:\:\mathrm{and}\:\:\mathrm{AB}\:=\:\mathrm{x} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{x}\:-\:\mathrm{y}\:=\:? \\ $$$$\left(\mathrm{Here}\:\mathrm{point}\:\mathrm{G}\:\mathrm{is}\:\mathrm{the}\:\mathrm{point}\:\mathrm{of}\:\mathrm{intersection}\right. \\ $$$$\left.\mathrm{of}\:\mathrm{the}\:\mathrm{extension}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{AB}\:\mathrm{and}\:\mathrm{CD}\right) \\ $$ Terms of Service…
Question Number 199923 by Rupesh123 last updated on 11/Nov/23 Answered by AST last updated on 11/Nov/23 $${a}+\mathrm{1}+\mathrm{1}\geqslant\mathrm{3}\sqrt[{\mathrm{3}}]{{a}}\Rightarrow\Sigma\frac{{a}}{{a}+\mathrm{2}}\leqslant\Sigma\frac{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} }{\mathrm{3}} \\ $$$${Power}\:{mean}\Rightarrow\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{3}}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \:\geqslant\left(\frac{{a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}^{\frac{\mathrm{2}}{\mathrm{3}}}…
Question Number 199900 by Calculusboy last updated on 11/Nov/23 Answered by Rasheed.Sindhi last updated on 11/Nov/23 $$\frac{\sqrt[{{x}}]{\mathrm{3}^{{x}} +\mathrm{7}^{−{x}} }\:+\sqrt[{{x}}]{\mathrm{3}^{−{x}} +\mathrm{7}^{{x}} }\:}{\:\sqrt[{{x}}]{\mathrm{21}^{{x}} +\mathrm{1}}} \\ $$$$=\frac{\sqrt[{{x}}]{\frac{\mathrm{21}^{{x}} +\mathrm{1}}{\mathrm{7}^{{x}}…
Question Number 199898 by Calculusboy last updated on 11/Nov/23 Answered by Frix last updated on 11/Nov/23 $${P}=\mathrm{50}×\mathrm{2}^{\frac{{n}}{\mathrm{12}}} \\ $$ Commented by Calculusboy last updated on…
Question Number 199862 by universe last updated on 10/Nov/23 $$\mathrm{consider}\:\mathrm{the}\:\mathrm{taylor}\:\mathrm{expansion}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}^{\mathrm{3}} } \\ $$$$\mathrm{centered}\:\mathrm{at}\:\mathrm{x}\:=\:\mathrm{1}/\mathrm{2}\:\mathrm{then}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{convergence} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{power}\:\mathrm{series}\:\mathrm{repersentation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\:\mathrm{is} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199840 by Calculusboy last updated on 10/Nov/23 Answered by AST last updated on 10/Nov/23 $${Let}\:\frac{{y}+\sqrt{\mathrm{1}+{y}^{\mathrm{2}} }}{{y}}=\mathrm{1}+\sqrt{\frac{\mathrm{1}}{{y}^{\mathrm{2}} }+\mathrm{1}}={x} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −\mathrm{2}{x}=\frac{\mathrm{1}}{{y}^{\mathrm{2}} }\Rightarrow{y}=\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}}} \\…