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Question Number 92474 Answers: 0 Comments: 0
$$\mathrm{please}\:\mathrm{anyone}\:\mathrm{wanna}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{Q91948} \\ $$
Question Number 92468 Answers: 1 Comments: 0
$$\mathrm{y}\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}}−\mathrm{9y}\:=\:\mathrm{3} \\ $$
Question Number 92465 Answers: 1 Comments: 7
$$\mathrm{for}\:\mathrm{a}\:\mathrm{2d}\:\:\mathrm{vectors}\:\mathrm{if}\:\mid{a}\:+\:{b}\mid\:=\:\mid{a}−{b}\mid\:\mathrm{what}\:\mathrm{relationship}\:\mathrm{does}\:{a}\:\mathrm{and}\:{b}\:\mathrm{have}? \\ $$$$ \\ $$
Question Number 92464 Answers: 0 Comments: 4
$$\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{e}^{\mathrm{y}} }{\mathrm{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$
Question Number 92461 Answers: 0 Comments: 1
$$\mathrm{cos}\:\left(\frac{\mathrm{3x}}{\mathrm{4}}−\pi\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}−\mathrm{2x}\right) \\ $$$$\mathrm{x}\:\in\:\left[\mathrm{0},\:\pi\:\right]\: \\ $$
Question Number 92453 Answers: 0 Comments: 2
$$\sqrt[{\mathrm{i}\:\:}]{\:\mathrm{i}}\:? \\ $$
Question Number 92452 Answers: 0 Comments: 0
$$\left(\mathrm{2xy}^{\mathrm{2}} −\mathrm{y}\right)\mathrm{dx}\:=\:\left(\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \mathrm{y}\right)\mathrm{dy}\: \\ $$
Question Number 92448 Answers: 4 Comments: 1
$$\begin{cases}{\mathrm{5}^{\mathrm{x}} .\mathrm{6}^{\mathrm{y}} \:=\:\mathrm{150}}\\{\mathrm{5}^{\mathrm{y}} .\mathrm{6}^{\mathrm{x}} \:=\:\mathrm{180}\:}\end{cases} \\ $$
Question Number 92445 Answers: 1 Comments: 0
Question Number 92438 Answers: 0 Comments: 1
$$\mathrm{find}\:\mathrm{the}\:\mathrm{domaine}\:\mathrm{and}\:\mathrm{simplify} \\ $$$$\mathrm{the}\:\mathrm{function} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{arcos}\left(\frac{\mathrm{1}−\mathrm{x}^{\mathrm{2}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right) \\ $$
Question Number 92426 Answers: 0 Comments: 4
$$\mathrm{If}\:\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}=\mathrm{0}\:\mathrm{find}\:\mathrm{y} \\ $$
Question Number 92425 Answers: 0 Comments: 1
$$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{two} \\ $$$$\mathrm{different}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{A}+\mathrm{B}=\mathrm{C}\:\mathrm{and}\:\mathrm{A}×\mathrm{B}=\mathrm{C} \\ $$$$\mathrm{find}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}. \\ $$
Question Number 92424 Answers: 0 Comments: 3
$$\mathrm{If}\:\mathrm{2x}−\mathrm{0i}=\varrho^{\pi\mathrm{i}} \: \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x} \\ $$
Question Number 92423 Answers: 1 Comments: 0
$${lim}_{{x}\Rightarrow\infty} \frac{\mathrm{4}\left({x}+\mathrm{3}\right)!−{x}!}{{x}\left[\left({x}+\mathrm{2}\right)!−\left({x}−\mathrm{1}\right)!\right]} \\ $$
Question Number 92422 Answers: 0 Comments: 0
$${solve}\:\left({y}^{\mathrm{2}} +{yz}\right){dx}+\left({z}^{\mathrm{2}} +{xy}\right){dy}+\left({y}^{\mathrm{2}} −{xy}\right){dz}=\mathrm{0} \\ $$$$ \\ $$$${help}\:{me}\:{sir} \\ $$
Question Number 92421 Answers: 1 Comments: 0
$$\int\frac{{cscx}}{{cos}\left(\mathrm{2}{x}\right)+\mathrm{2}{cos}^{\mathrm{2}} {x}}{dx} \\ $$$${pleas}\:{sir}\:{help}\:{me} \\ $$
Question Number 92420 Answers: 2 Comments: 2
Question Number 92447 Answers: 0 Comments: 1
$${find}\:\int_{\frac{\mathrm{1}}{\mathrm{6}}} ^{\frac{\mathrm{1}}{\mathrm{5}}} \:\:\frac{{dx}}{\sqrt{\mathrm{1}−\mathrm{3}{x}}+\sqrt{\mathrm{1}+\mathrm{3}{x}}} \\ $$
Question Number 92410 Answers: 0 Comments: 3
$${find}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\:\:\frac{{dx}}{\sqrt{\mathrm{1}+\mathrm{3}{x}}−\sqrt{\mathrm{1}−\mathrm{3}{x}}} \\ $$
Question Number 92407 Answers: 0 Comments: 0
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\mathrm{1}+{x}}+{a}\sqrt{\mathrm{1}−{x}}\right){dx}\:\:\:{with}\:\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){explicite}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{g}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{\mathrm{1}−{x}}}{\sqrt{\mathrm{1}+{x}}+{a}\sqrt{\mathrm{1}−{x}}}\:{dx} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\mathrm{1}+{x}}+\mathrm{2}\sqrt{\mathrm{1}−{x}}\right){dx} \\ $$$${and}\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\mathrm{1}−{x}}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\sqrt{\mathrm{1}+{x}}+{sin}\theta\:\sqrt{\mathrm{1}−{x}}\right){dx}\: \\ $$$$\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$
Question Number 92399 Answers: 0 Comments: 1
$$\mathrm{sin}\:^{\mathrm{3}} \left(\mathrm{x}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\mathrm{x}\right)\:=\:\mathrm{0} \\ $$
Question Number 92398 Answers: 1 Comments: 0
$$\mathrm{x}^{\mathrm{3}} \:\left(\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\right)\:+\mathrm{x}^{\mathrm{2}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{2}} \:=\:\mathrm{ln}\:\left(\mathrm{x}\right)\: \\ $$
Question Number 92397 Answers: 0 Comments: 2
$$\int\:\frac{\mathrm{1}}{{x}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:{dx}\: \\ $$$$\left[\:{x}\:=\:\mathrm{sin}\:{w}\:\right]\: \\ $$$$\int\:\frac{\mathrm{cos}\:\mathrm{w}\:\mathrm{dw}}{\mathrm{sin}\:\mathrm{w}−\mathrm{cos}\:\mathrm{w}}\:=\:\int\:\frac{\mathrm{dw}}{\mathrm{tan}\:\mathrm{w}−\mathrm{1}} \\ $$$$=\:\int\:\frac{\mathrm{sec}^{\mathrm{2}} \:\mathrm{w}\:\mathrm{dw}}{\left(\mathrm{tan}\:\mathrm{w}−\mathrm{1}\right)\mathrm{sec}^{\mathrm{2}} \:\mathrm{w}} \\ $$$$=\:\int\:\frac{\mathrm{du}}{\left(\mathrm{u}−\mathrm{1}\right)\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)}\:;\:\left[\:\mathrm{u}\:=\:\mathrm{tan}\:\mathrm{w}\:\right]\: \\ $$$$=\:\int\:\frac{\mathrm{du}}{\mathrm{2}\left(\mathrm{u}−\mathrm{1}\right)}−\int\:\frac{\mathrm{u}\:\mathrm{du}\:}{\mathrm{2}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{u}−\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{u}^{\mathrm{2}} +\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{u}\right)\:+\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{tan}\:\mathrm{w}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{tan}\:^{\mathrm{2}} \mathrm{w}+\mathrm{1}\mid− \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{tan}\:\mathrm{w}\right)\:+\mathrm{c} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}−\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\mid\mathrm{1}−{x}^{\mathrm{2}} \mid− \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left({x}\right)\:+\:{c} \\ $$
Question Number 92394 Answers: 0 Comments: 2
$$\int\:\mathrm{ln}\:\left(\sqrt{\mathrm{1}−{x}}\:+\:\sqrt{\mathrm{1}+{x}}\:\right)\:{dx}\: \\ $$
Question Number 92390 Answers: 0 Comments: 2
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{meaning}\:\mathrm{of}\:\mathrm{this}\:\mathrm{symbol}\:\:\left(\varepsilon\right)\:\mathrm{in}\:\mathrm{limit}\:\mathrm{please}. \\ $$$$\mathrm{or}\:\mathrm{as}\:\mathrm{used}\:\mathrm{in}\:\mathrm{convergent}/\mathrm{divergent}\:\mathrm{series} \\ $$
Question Number 92366 Answers: 0 Comments: 3
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