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Question Number 144736 Answers: 3 Comments: 0
$${P}\left({z}\right)={az}^{\mathrm{3}} +{z}^{\mathrm{2}} −\left({a}+\mathrm{6}\right){z}+{b}−\mathrm{6} \\ $$$${P}\left({z}\right)=\left({z}^{\mathrm{2}} +\mathrm{4}\right)\centerdot{Q}\left({z}\right) \\ $$$${Find}\:\:{a}\centerdot{b}=? \\ $$
Question Number 144734 Answers: 0 Comments: 0
Question Number 144733 Answers: 0 Comments: 0
Question Number 144727 Answers: 2 Comments: 0
$$\:\:\mathrm{Given}\:\begin{cases}{\mathrm{m}=\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}\\{\mathrm{n}=\mathrm{cos}\:\theta+\mathrm{sin}\:\theta}\end{cases} \\ $$$$\:\:\mathrm{then}\:\sqrt{\frac{\mathrm{m}}{\mathrm{n}}}\:+\sqrt{\frac{\mathrm{n}}{\mathrm{m}}}\:=\:? \\ $$
Question Number 144724 Answers: 0 Comments: 0
Question Number 144721 Answers: 0 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:.........\:\mathrm{Nice}\:......\ast\ast\ast......\mathrm{Calculus}......... \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{f}\:\left(\:\mathrm{x}\:\right)\::\:=\:\left[\:\mathrm{tan}\:\left(\mathrm{x}\right)\:+\:\mathrm{cot}\:\left(\mathrm{x}\right)\:\right] \\ $$$$\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{R}_{\:\mathrm{f}\:\:} \:=\:? \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Hint}::\:\:\:\left[\:\mathrm{x}\:\right]\::=\:\mathrm{Max}\:\left\{\:\mathrm{m}\:\in\mathbb{Z}\:\mid\:\mathrm{m}\:\leqslant\:\mathrm{x}\:\right\}\: \\ $$
Question Number 144720 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.....\mathrm{calculus}..... \\ $$$$\: \\ $$$$\Omega\::=\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sech}\left(\pi{x}\right)}{\mathrm{1}+\mathrm{4}{x}^{\:\mathrm{2}} }\:{dx}\:\overset{?} {=}\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{Ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$ \\ $$
Question Number 144716 Answers: 1 Comments: 0
$$\int\left(\frac{{dx}}{{e}^{\mathrm{2}{x}} +\mathrm{1}}\right)=? \\ $$
Question Number 144708 Answers: 2 Comments: 0
$${Simplify}:\:\:\left(\frac{\mathrm{1}}{\left({x}-\mathrm{2}\right)!}\:-\:\frac{\mathrm{1}}{\left({x}-\mathrm{1}\right)!}\right)\centerdot{x}! \\ $$
Question Number 144705 Answers: 1 Comments: 0
$$\:\:\int\:\frac{\mathrm{x}^{\mathrm{n}−\mathrm{1}} }{\mathrm{x}^{\mathrm{3n}+\mathrm{1}} \:\left(\mathrm{x}^{\mathrm{n}} −\mathrm{a}\right)}\:\mathrm{dx}\:? \\ $$
Question Number 144702 Answers: 1 Comments: 0
$$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)=\mathrm{log}\left(\mathrm{cosx}\:+\mathrm{2sinx}\right) \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$$$ \\ $$
Question Number 144701 Answers: 2 Comments: 0
$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{cos}\left(\mathrm{n}\theta\right)}{\mathrm{n}^{\mathrm{2}} } \\ $$
Question Number 144700 Answers: 1 Comments: 0
$$\mathrm{find}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$
Question Number 144699 Answers: 3 Comments: 0
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{log}\left(\mathrm{cht}\right) \\ $$$$\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$
Question Number 144697 Answers: 2 Comments: 0
$$\mathrm{Evaluate}\:\left(\frac{\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{10}}−{i}\mathrm{sin}\:\frac{\pi}{\mathrm{10}}}{\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{10}}+{i}\mathrm{sin}\:\frac{\pi}{\mathrm{10}}}\right)^{\mathrm{15}} . \\ $$
Question Number 144693 Answers: 1 Comments: 0
Question Number 144691 Answers: 1 Comments: 0
Question Number 144684 Answers: 1 Comments: 0
$$ \\ $$
Question Number 144683 Answers: 1 Comments: 0
$${Determiner}\:{l}'{original}\:{de}\:{laplace} \\ $$$${F}\left({p}\right)=\frac{\mathrm{1}}{\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Question Number 144682 Answers: 1 Comments: 0
$${Compare}:\:\:{x}=\frac{{sin}\left(\mathrm{3}\right)}{{sin}\left(\mathrm{5}\right)}\:\:{and}\:\:{y}=\frac{{cos}\left(\mathrm{3}\right)}{{cos}\left(\mathrm{5}\right)} \\ $$
Question Number 144663 Answers: 1 Comments: 0
$${x}\in\left(\mathrm{0};\pi\right)\:{and}\:\left({a};{b}\right)\:{real}\:{numbers}\:{fixed}. \\ $$$${Find}\:{the}\:{range}\:{of}\:{function}: \\ $$$${g}\left({x}\right)=\:\frac{\left(\mathrm{1}+{a}^{\mathrm{2}} +{cot}^{\mathrm{2}} {x}\right)\centerdot\left(\mathrm{1}+{b}^{\mathrm{2}} +{cot}^{\mathrm{2}} {x}\right)}{\mathrm{1}\:+\:{cot}^{\mathrm{2}} {x}} \\ $$
Question Number 144662 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:...........\:\:\mathrm{Calculus}........... \\ $$$$\:\mathrm{In}\:\:\mathrm{A}\overset{\Delta} {\mathrm{B}C}\:\:: \\ $$$$\hat {\mathrm{B}}\:=\:\mathrm{2}\:\hat {\mathrm{C}}\:\:\:\:,\:\:{a}\:\:=\:\lambda\:{b}\:\:\:{then}\:{specify} \\ $$$$\:{the}\:\:{limits}\:{of}\:{the}\:{changes}\:\:\:'\:\:\lambda\:\:'\:\:: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Question Number 144676 Answers: 1 Comments: 0
$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:\left({a}+{b}\right)\left({b}+{c}\right)\:=\:\mathrm{4}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}+\frac{{b}}{{ca}}\:\geqslant\:\frac{\mathrm{27}}{\mathrm{8}} \\ $$$$\left(\mathrm{Found}\:\mathrm{by}\:\mathrm{WolframAlpha}\right) \\ $$
Question Number 144645 Answers: 2 Comments: 0
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}\:+\:{tanx}}{\mathrm{1}\:+\:{sinx}}\right)^{\frac{\mathrm{1}}{\boldsymbol{{sinx}}}} =\:? \\ $$
Question Number 144639 Answers: 0 Comments: 0
Question Number 144638 Answers: 1 Comments: 0
$$\mathrm{Triangle}\:\mathrm{AOC}\:\mathrm{inscribed} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{region}\:\mathrm{cut}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{parabola}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} \:\mathrm{by}\:\mathrm{the} \\ $$$$\mathrm{line}\:\mathrm{y}=\mathrm{a}^{\mathrm{2}} \:.\mathrm{Find}\:\mathrm{the}\:\mathrm{limit} \\ $$$$\mathrm{of}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{triangle}\:\mathrm{to}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{parabolic}\:\mathrm{region}\:\mathrm{as}\:\mathrm{a}\:\mathrm{approaches} \\ $$$$\mathrm{zero}\: \\ $$
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