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Question Number 182632 Answers: 1 Comments: 0
$$\:\:\mathrm{If}\:\mathrm{log}\:_{\mathrm{11}} \left(\mathrm{3p}\right)=\mathrm{log}\:_{\mathrm{13}} \left(\mathrm{q}+\mathrm{6p}\right)\:=\:\mathrm{log}\:_{\mathrm{143}} \left(\mathrm{q}^{\mathrm{2}} \right) \\ $$$$\:\mathrm{find}\:\frac{\mathrm{p}}{\mathrm{q}}. \\ $$
Question Number 182629 Answers: 0 Comments: 0
$$\:\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{2}} {\overset{\infty} {\prod}}\:\boldsymbol{\mathrm{e}}\left(\mathrm{1}−\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\right)^{\boldsymbol{\mathrm{n}}^{\mathrm{2}} } \:=\:\:? \\ $$
Question Number 182627 Answers: 1 Comments: 0
$$\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{15}} =\frac{{a}+{b}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${find}\:{a}+{b} \\ $$
Question Number 182623 Answers: 2 Comments: 1
Question Number 182620 Answers: 1 Comments: 0
Question Number 182613 Answers: 2 Comments: 0
Question Number 182608 Answers: 2 Comments: 1
Question Number 182600 Answers: 3 Comments: 0
Question Number 182595 Answers: 0 Comments: 1
Question Number 182593 Answers: 1 Comments: 1
Question Number 182586 Answers: 0 Comments: 0
Question Number 182584 Answers: 1 Comments: 0
$$\:\:{Transform}\:{into}\:{factorial}\:{form}\:{the}\:{following} \\ $$$$\:\:{A}\:=\:\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}×\mathrm{9}×...×\left(\mathrm{2}{n}−\mathrm{1}\right)\:.\: \\ $$$${Example}:\:\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{4}! \\ $$
Question Number 182582 Answers: 2 Comments: 0
Question Number 182581 Answers: 1 Comments: 0
$${f}\left({x}\right)=\frac{{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{3}}\:\:\: \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=? \\ $$
Question Number 182576 Answers: 0 Comments: 0
Question Number 182575 Answers: 1 Comments: 0
$$\sqrt[{{x}}]{{x}+\mathrm{1}}>\sqrt[{{e}}]{{e}} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Solve}\:\mathrm{for}\:{x}\left({x}\in\mathbb{N}^{+} \right). \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Solve}\:\mathrm{for}\:{x}\left({x}>\mathrm{0}\right). \\ $$
Question Number 182571 Answers: 1 Comments: 0
Question Number 182566 Answers: 0 Comments: 0
$$\mathrm{Solve} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} \:+\left(\:\mathrm{xy}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \right)\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}^{\mathrm{2}} \\ $$
Question Number 182560 Answers: 1 Comments: 0
$$\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:−\:\mathrm{cot}^{\mathrm{2}} {x}\:=\:\:? \\ $$
Question Number 182552 Answers: 0 Comments: 0
$${Find}\:{the}\:{period}\:{of}\:{the}\:{following}: \\ $$$$\:{a}\bullet\:\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\:{b}\bullet\:\mathrm{sin}\:\pi{x}+\:\mathrm{cos}\:{x} \\ $$$$\:{c}\bullet\:\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{3}{x}−\:\mathrm{3}\:\mathrm{tan}\:\mathrm{4}{x}+\:\mathrm{4}\:\mathrm{cot}\:\mathrm{6}{x}}{\mid\mathrm{cosec}\:\mathrm{8}{x}\mid−\:\mathrm{sec}^{\mathrm{3}} \:\mathrm{10}{x}+\:\sqrt{\mathrm{cot}\:\mathrm{12}{x}}} \\ $$
Question Number 182546 Answers: 2 Comments: 1
Question Number 182542 Answers: 1 Comments: 1
Question Number 182534 Answers: 1 Comments: 3
$$\mathrm{xy}\:+\:\left(\mathrm{xy}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\:\mathrm{y}^{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Solve} \\ $$
Question Number 182533 Answers: 1 Comments: 0
$${If}\:{you}\:{walk}\:{around}\:{a}\:{triangle}\:{with}\:{sides}\:\mathrm{3},\:\mathrm{6},\:\mathrm{8} \\ $$$${respectively}\:{such}\:{a}\:{way}\:{that}\:{you}\:{keep}\:{a}\:{distance} \\ $$$$\:{of}\:\mathrm{3}{m}\:{from}\:{it},\:{then}\: \\ $$$$\:{how}\:{much}\:{distance}\:{will}\:{you}\:{travel}? \\ $$
Question Number 182532 Answers: 0 Comments: 0
$${Q}.\mathrm{181494} \\ $$$$\:{Let}'{s}\:{try}\:{it} \\ $$
Question Number 182530 Answers: 3 Comments: 0
$${If}:\:\:\:\frac{{a}}{{x}^{\mathrm{9}} }\:+\:\frac{{x}^{\mathrm{9}} }{{a}}\:=\:\mathrm{7} \\ $$$$\:{find}:\:\sqrt[{\mathrm{4}}]{\frac{{a}}{{x}^{\mathrm{9}} }}\:+\:\sqrt[{\mathrm{4}}]{\frac{{x}^{\mathrm{9}} }{{a}}} \\ $$
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