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Question Number 50175 Answers: 1 Comments: 0
Question Number 50171 Answers: 0 Comments: 3
$$\mathrm{Sir}\:\mathrm{l}'\mathrm{m}\:\mathrm{sorry}\:\mathrm{l}\:\mathrm{dont}\:\mathrm{understand}\: \\ $$$$\mathrm{you} \\ $$
Question Number 50163 Answers: 2 Comments: 0
Question Number 50161 Answers: 1 Comments: 0
$${Find}\:{the}\:{function}\:{whose}\:{first}\: \\ $$$${derivative}\:{is}\:\mathrm{8}−\frac{\mathrm{5}}{\sqrt[{\mathrm{3}}]{{x}^{\mathrm{2}} }}\:{the}\:{initial}\: \\ $$$${conditions}\:{f}\left(\mathrm{8}\right)=−\mathrm{20} \\ $$
Question Number 50158 Answers: 7 Comments: 1
Question Number 50156 Answers: 0 Comments: 1
Question Number 50155 Answers: 0 Comments: 1
$$\mathrm{plz}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir} \\ $$
Question Number 50140 Answers: 5 Comments: 0
$${Solve}\:{the}\:{differential} \\ $$$${equation} \\ $$$$\left.{a}\right){x}\left({x}+{y}\right)\frac{{dy}}{{dx}}={x}^{\mathrm{2}} +{xy}−\mathrm{3}{y}^{\mathrm{2}} \\ $$$$\left.{b}\right){y}+{xy}^{\mathrm{2}} −{x}\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$${c}\:\left[\:\:{x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}\:} }−\mathrm{2}{x}\frac{{dy}}{{dx}}+\mathrm{2}\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right){y}=\mathrm{24}{x}^{\mathrm{3}} \right. \\ $$$${given}\:{that}\:\frac{{dy}}{{dx}}=\mathrm{6}\:\:\:,\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$
Question Number 50138 Answers: 0 Comments: 1
$$\mathrm{D} \\ $$
Question Number 50135 Answers: 2 Comments: 2
Question Number 50132 Answers: 0 Comments: 1
$$\mathrm{Let}\:{f}\:\:\mathrm{be}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{function}.\:\mathrm{Let} \\ $$$${I}_{\mathrm{1}} =\underset{\mathrm{1}−{k}} {\overset{{k}} {\int}}{x}\:{f}\left\{{x}\left(\mathrm{1}−{x}\right\}\:{dx},\:\right. \\ $$$${I}_{\mathrm{2}} =\underset{\mathrm{1}−{k}} {\overset{{k}} {\int}}{x}\:{f}\left\{{x}\left(\mathrm{1}−{x}\right\}\:{dx},\:\right. \\ $$$$\mathrm{where}\:\mathrm{2}{k}−\mathrm{1}>\mathrm{0}.\:\mathrm{Then}\:\frac{{I}_{\mathrm{1}} }{{I}_{\mathrm{2}} }\:\:\mathrm{is} \\ $$
Question Number 50126 Answers: 0 Comments: 0
Question Number 50125 Answers: 0 Comments: 2
$$\mathrm{help}\:\mathrm{me}\:\mathrm{plz}\:\mathrm{sir} \\ $$
Question Number 50101 Answers: 2 Comments: 4
Question Number 50093 Answers: 2 Comments: 1
Question Number 50089 Answers: 1 Comments: 2
$$\underset{{x}\rightarrow\mathrm{8}} {\mathrm{lim}}\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{6x}−\mathrm{16}}{\mid\mathrm{x}−\mathrm{8}\mid} \\ $$
Question Number 50085 Answers: 0 Comments: 0
$$\mathrm{5}^{{x}+\mathrm{2}} −\mathrm{95}^{{x}} =\mathrm{2}^{{x}+\mathrm{9}} +\mathrm{1132}^{{x}} \\ $$
Question Number 50084 Answers: 0 Comments: 0
$$\mathrm{could}\:\mathrm{you}\:\mathrm{help}\:\mathrm{me}\:\mathrm{sir}? \\ $$
Question Number 50083 Answers: 0 Comments: 0
Question Number 50080 Answers: 1 Comments: 0
$$\left.{a}\right)\:{if}\:{f}\left({x}\right)={log}\left({x}+\mathrm{2}\right),\:{solve}\:{the}\:{equation}: \\ $$$$\mathrm{2}^{{f}\left({x}−\mathrm{2}\right)} ×\mathrm{2}^{{f}\left(\mathrm{2}{x}+\mathrm{2}\right)} =\mathrm{4}^{{logf}\left({x}\right)} \\ $$
Question Number 50065 Answers: 3 Comments: 0
Question Number 50064 Answers: 3 Comments: 0
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{y}}^{\boldsymbol{\mathrm{x}}} =\boldsymbol{\mathrm{x}}^{\boldsymbol{\mathrm{y}}} \: \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{are}}\:\boldsymbol{\mathrm{both}}\:\boldsymbol{\mathrm{not}} \\ $$$$\boldsymbol{\mathrm{equal}}\:\boldsymbol{\mathrm{to}}\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{l}} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{value}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}} \\ $$
Question Number 50052 Answers: 5 Comments: 1
Question Number 50047 Answers: 1 Comments: 0
Question Number 50048 Answers: 3 Comments: 4
Question Number 50017 Answers: 0 Comments: 4
$$\mathcal{CONGRATULATIONS}\:! \\ $$$${Tinkutara}-\:{our}\:{forum}\:{on} \\ $$$${having}\:{above}\:\mathrm{50000}\:{question}\:{posts}. \\ $$
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