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Question Number 161505 Answers: 1 Comments: 3
$$\sqrt[{\mathrm{3}}]{{a}\:+\:\frac{{a}+\mathrm{8}}{\mathrm{3}}\:\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:+\:\sqrt[{\mathrm{3}}]{{a}\:−\:\frac{{a}+\mathrm{8}}{\mathrm{3}}\:\sqrt{\frac{{a}−\mathrm{1}}{\mathrm{3}}}}\:\:=\:\:? \\ $$
Question Number 161504 Answers: 1 Comments: 0
$${Montrer}\:\grave {{a}}\:{partir}\:{du}\:{crit}\grave {{e}re}\:{de}\: \\ $$$${Cauchy}\:{que}\:{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:{est}\:{une} \\ $$$${de}\:{Cauchy}. \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${Show}\:{by}\:{using}\:{Cauchy}'{s}\:{sequence} \\ $$$${definition}\:{that}\:{U}_{{n}} =\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\:{is}\:{a}\: \\ $$$${sequence}\:{of}\:{Cauchy}. \\ $$
Question Number 161500 Answers: 1 Comments: 0
$$\mathrm{x}^{\mathrm{6}} \:-\:\mathrm{6x}^{\mathrm{5}} \:+\:\mathrm{ax}^{\mathrm{4}} \:+\:\mathrm{bx}^{\mathrm{3}} \:+\:\mathrm{cx}^{\mathrm{2}} \:+\:\mathrm{dx}\:+\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\mathrm{all}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{are}\:\mathrm{positive} \\ $$$$\mathrm{find}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=? \\ $$
Question Number 161485 Answers: 0 Comments: 0
$$\:\mathrm{Find}\:\mathrm{range}\:\mathrm{of}\:\mathrm{function}\:\mathrm{y}=\frac{\mathrm{cos}\:\mathrm{4x}+\mathrm{4sin}\:\mathrm{4x}+\mathrm{1}}{\mathrm{cos}\:\mathrm{4x}+\mathrm{2}} \\ $$
Question Number 161484 Answers: 2 Comments: 0
$$\:\:\begin{cases}{\frac{\mathrm{1}}{\mathrm{a}}+\frac{\mathrm{1}}{\mathrm{b}}=\mathrm{9}}\\{\left(\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{b}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{a}}}\right)\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{b}}}\right)=\mathrm{18}}\end{cases} \\ $$$$\:\:\:\mathrm{8a}+\mathrm{4b}=? \\ $$
Question Number 161482 Answers: 3 Comments: 1
Question Number 161476 Answers: 0 Comments: 2
$$\:{Between}\:\frac{\mathrm{3}}{\mathrm{6}}\:\:{and}\:−\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\:{How}\:{do}\:{i}\:{list}\:{two}\:{rational}\: \\ $$$$\:{numbers}\:{please}? \\ $$
Question Number 161538 Answers: 2 Comments: 0
$$ \\ $$$${lim}_{\:{x}\:\rightarrow\:−\mathrm{2}\:\:} \left(\frac{\mathrm{2}+\:\mathrm{3}{x}\:+\:\mathrm{3}{x}^{\:\mathrm{2}} \:+\:{x}^{\:\mathrm{3}} }{\:{sin}\:\left(\:\frac{\pi{x}}{\mathrm{2}}\:\right)}\:\right)=? \\ $$$$\:\:\:\:−−−− \\ $$
Question Number 161491 Answers: 0 Comments: 0
Question Number 161464 Answers: 0 Comments: 1
$$\mathrm{Given}\:\mathrm{that}\:\mathrm{in}\:\Delta\mathrm{ABC}, \\ $$$$\left(\mathrm{sin}\:\mathrm{A}+\mathrm{sin}\:\mathrm{B}\right):\left(\mathrm{sin}\:\mathrm{B}+\mathrm{sin}\:\mathrm{C}\right):\left(\mathrm{sin}\:\mathrm{C}+\mathrm{sin}\:\mathrm{A}\right)=\:\mathrm{6}:\:\mathrm{4}:\:\mathrm{5} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{A}. \\ $$
Question Number 161463 Answers: 2 Comments: 0
$$\mathrm{Let}\:{f}\left({x}\right)=\:{x}+\mid{x}\mid−\mathrm{1}.\: \\ $$$$\mathrm{Find}\:\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}\:\mathrm{and}\:\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\frac{{f}\left({x}\right)−{f}\left(\mathrm{0}\right)}{{x}−\mathrm{0}}. \\ $$$$\mathrm{Hence},\:\mathrm{determine}\:\mathrm{whether}\:{f}\left({x}\right)\:\mathrm{is}\: \\ $$$$\mathrm{differentiable}\:\mathrm{at}\:{x}=\mathrm{0}. \\ $$
Question Number 161462 Answers: 1 Comments: 0
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{differential} \\ $$$$\mathrm{equation}\:\mathrm{2}{y}''+\mathrm{5}{y}'+\mathrm{2}{y}=\mathrm{0}\:\mathrm{subject}\:\mathrm{to}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{conditions}\:\:{y}\left(\mathrm{0}\right)=\mathrm{2}{y}\:,\:\:{y}'\left(\mathrm{0}\right)=\mathrm{1}\:. \\ $$
Question Number 161461 Answers: 0 Comments: 3
$$\mathrm{Let}\:\mathrm{the}\:\mathrm{region}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$${y}={x}\left(\mathrm{1}−{x}\right)\:\mathrm{and}\:\mathrm{the}\:{x}-\mathrm{axis}\:\mathrm{be}\:\mathrm{R}. \\ $$$$\mathrm{The}\:\mathrm{line}\:{y}={mx}\:\mathrm{divides}\:\mathrm{R}\:\mathrm{into}\:\mathrm{two}\:\mathrm{parts}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{1}−{m}\right)^{\mathrm{3}} . \\ $$
Question Number 161456 Answers: 0 Comments: 0
Question Number 161450 Answers: 0 Comments: 0
$$\mathrm{A}\: \mathrm{ector}\:\mathrm{field}\:\mathrm{is}\:\mathrm{given}\:\mathrm{by}\:\mathrm{v}= \\ $$$$\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} +\mathrm{x}\right)\mathrm{i}−\left(\mathrm{2xy}+\mathrm{y}\right)\mathrm{j}.\:\mathrm{Show}\:\mathrm{that} \\ $$$$\mathrm{vector}\:\mathrm{v}\:\mathrm{is}\:\mathrm{irrotational}\:\mathrm{hence}\:\mathrm{find} \\ $$$$\:\mathrm{the}\:\mathrm{scalar}\:\mathrm{potential} \\ $$
Question Number 161444 Answers: 0 Comments: 2
Question Number 161443 Answers: 1 Comments: 0
Question Number 161442 Answers: 1 Comments: 0
$$\mathrm{Use}\:\mathrm{the}\:\mathrm{binomial}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{write} \\ $$$$\mathrm{the}\:\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{expansion} \\ $$$$\mathrm{of}\:\sqrt{\mathrm{2}+\mathrm{3}{x}−{x}^{\mathrm{2}} } \\ $$
Question Number 161439 Answers: 0 Comments: 1
Question Number 161437 Answers: 2 Comments: 0
$$\:\:\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{10}}]{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{1}}\:+\sqrt[{\mathrm{5}}]{\mathrm{7}{x}^{\mathrm{2}} +\mathrm{3}{x}+\mathrm{1}}−\mathrm{2}}{{x}}\:=? \\ $$
Question Number 161433 Answers: 0 Comments: 0
$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}>\mathrm{0}\:\:\mathrm{such}\:\mathrm{that}\:\:\mathrm{x}+\mathrm{y}+\mathrm{z}=\mathrm{3} \\ $$$$\mathrm{and}\:\:\boldsymbol{\lambda}\geqslant\mathrm{0}\:\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}}{\mathrm{y}^{\mathrm{3}} +\lambda\mathrm{y}^{\mathrm{2}} }\:+\:\frac{\mathrm{y}}{\mathrm{z}^{\mathrm{3}} +\lambda\mathrm{z}^{\mathrm{2}} }\:+\:\frac{\mathrm{z}}{\mathrm{x}^{\mathrm{3}} +\lambda\mathrm{x}^{\mathrm{2}} }\:\geqslant\:\frac{\mathrm{3}}{\lambda+\mathrm{1}} \\ $$
Question Number 161429 Answers: 0 Comments: 1
$$\mathrm{help}\:\mathrm{me}\:! \\ $$$$\begin{cases}{{x}+\mathrm{3}{y}+{z}=\mathrm{2}}\\{−\mathrm{3}{x}+\mathrm{4}{y}+\mathrm{2}{z}=\mathrm{3}}\\{−\mathrm{2}{x}+\mathrm{7}{y}+\mathrm{3}{z}=\mathrm{5}}\end{cases} \\ $$$$\boldsymbol{\mathrm{G}}\mathrm{auss}\:\mathrm{Method}... \\ $$
Question Number 161424 Answers: 1 Comments: 0
Question Number 161418 Answers: 0 Comments: 0
$$\mathrm{4}^{\mathrm{2021}\:} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} ,\:\left({a}:{b}:{c}\right)\Rightarrow\:{natural}\:{numbers} \\ $$
Question Number 161416 Answers: 0 Comments: 0
Question Number 161412 Answers: 0 Comments: 0
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