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Question Number 214350 Answers: 2 Comments: 0

Question Number 214340 Answers: 1 Comments: 0

evaluate ((∮_C (z/(2∙sin(z)−2z∙cos(z)−π))dz)/(∮_( C) (2/(2∙sin(z)−2z∙cos(z)−π))dz)) where C is the circle∣z−((3π)/4)∣=(π/4).

$$\mathrm{evaluate} \\ $$$$\frac{\oint_{{C}} \:\frac{{z}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}}{\oint_{\:{C}} \:\frac{\mathrm{2}}{\mathrm{2}\centerdot\mathrm{sin}\left({z}\right)−\mathrm{2}{z}\centerdot\mathrm{cos}\left({z}\right)−\pi}\mathrm{d}{z}} \\ $$$$\mathrm{where}\:{C}\:\mathrm{is}\:\mathrm{the}\:\mathrm{circle}\mid{z}−\frac{\mathrm{3}\pi}{\mathrm{4}}\mid=\frac{\pi}{\mathrm{4}}. \\ $$

Question Number 214322 Answers: 0 Comments: 2

find the errors 7x+4=10x−6 7x+4−10x=10x−6−10x −3x+4=−6 −3x+4+6=−6+6 −3x+10=0 −3x+10−10=0−10 −3x=−10 x=((−3)/(−10)) x=(3/(10)) if this error show the real one

$$\mathrm{find}\:\mathrm{the}\:\mathrm{errors} \\ $$$$\mathrm{7}{x}+\mathrm{4}=\mathrm{10}{x}−\mathrm{6} \\ $$$$\mathrm{7}{x}+\mathrm{4}−\mathrm{10}{x}=\mathrm{10}{x}−\mathrm{6}−\mathrm{10}{x} \\ $$$$−\mathrm{3}{x}+\mathrm{4}=−\mathrm{6} \\ $$$$−\mathrm{3}{x}+\mathrm{4}+\mathrm{6}=−\mathrm{6}+\mathrm{6} \\ $$$$−\mathrm{3}{x}+\mathrm{10}=\mathrm{0} \\ $$$$−\mathrm{3}{x}+\mathrm{10}−\mathrm{10}=\mathrm{0}−\mathrm{10} \\ $$$$−\mathrm{3}{x}=−\mathrm{10} \\ $$$${x}=\frac{−\mathrm{3}}{−\mathrm{10}} \\ $$$${x}=\frac{\mathrm{3}}{\mathrm{10}} \\ $$$$\mathrm{if}\:\mathrm{this}\:\mathrm{error}\:\mathrm{show}\:\mathrm{the}\:\mathrm{real}\:\mathrm{one} \\ $$

Question Number 214321 Answers: 1 Comments: 0

15x(((6x)/(10))+((12x)/(20))+((15x)/(30))+...+((60x)/(100)))=160 solve for x

$$\mathrm{15}{x}\left(\frac{\mathrm{6}{x}}{\mathrm{10}}+\frac{\mathrm{12}{x}}{\mathrm{20}}+\frac{\mathrm{15}{x}}{\mathrm{30}}+...+\frac{\mathrm{60}{x}}{\mathrm{100}}\right)=\mathrm{160} \\ $$$$\mathrm{solve}\:\mathrm{for}\:{x} \\ $$

Question Number 214319 Answers: 1 Comments: 0

f(((10x+3)/(10x−3)) )= ((10)/3) x f(4).f(6).f(8).f(10)...f(2024)=?

$$\:\:\:\:\mathrm{f}\left(\frac{\mathrm{10x}+\mathrm{3}}{\mathrm{10x}−\mathrm{3}}\:\right)=\:\frac{\mathrm{10}}{\mathrm{3}}\:\mathrm{x} \\ $$$$\:\:\:\mathrm{f}\left(\mathrm{4}\right).\mathrm{f}\left(\mathrm{6}\right).\mathrm{f}\left(\mathrm{8}\right).\mathrm{f}\left(\mathrm{10}\right)...\mathrm{f}\left(\mathrm{2024}\right)=? \\ $$

Question Number 214317 Answers: 1 Comments: 0

x^4 +x^3 −11x^2 +x−12=f(x)×g(x) f(x)=? g(x)=?

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{11}{x}^{\mathrm{2}} +{x}−\mathrm{12}={f}\left({x}\right)×{g}\left({x}\right) \\ $$$${f}\left({x}\right)=?\:\:\:\:{g}\left({x}\right)=? \\ $$

Question Number 214329 Answers: 2 Comments: 0

what is the coefficient of x^(50 ) in (1+2x+3x^2 +...+101x^(100) )(1+x+x^2 +...+x^(25) )

$${what}\:{is}\:{the}\:{coefficient}\:{of} \\ $$$${x}^{\mathrm{50}\:\:} \:{in} \\ $$$$\left(\mathrm{1}+\mathrm{2}{x}+\mathrm{3}{x}^{\mathrm{2}} +...+\mathrm{101}{x}^{\mathrm{100}} \right)\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} +...+{x}^{\mathrm{25}} \right) \\ $$

Question Number 214326 Answers: 1 Comments: 0

lim_(x→∞) (((x)^(1/4) −(x)^(1/6) )/( (x)^(1/4) +(x)^(1/6) ))=?

$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{4}}]{{x}}−\sqrt[{\mathrm{6}}]{{x}}}{\:\sqrt[{\mathrm{4}}]{{x}}+\sqrt[{\mathrm{6}}]{{x}}}=? \\ $$

Question Number 214335 Answers: 1 Comments: 0

For what values of k does the equation e^(kx) =3(√x) have only one solution in R?

$$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:{k}\:\mathrm{does}\:\mathrm{the}\:\mathrm{equation} \\ $$$${e}^{{kx}} =\mathrm{3}\sqrt{{x}}\:\mathrm{have}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{in}\:\mathbb{R}? \\ $$

Question Number 214310 Answers: 2 Comments: 0

(1/(2!)) + (2/(3!)) + (3/(4!)) + ... + ((99)/(100!))

$$\frac{\mathrm{1}}{\mathrm{2}!}\:+\:\frac{\mathrm{2}}{\mathrm{3}!}\:+\:\frac{\mathrm{3}}{\mathrm{4}!}\:+\:...\:+\:\frac{\mathrm{99}}{\mathrm{100}!} \\ $$

Question Number 214297 Answers: 1 Comments: 3

Question Number 214293 Answers: 1 Comments: 0

P(x) ⋮(x^2 +3) mod(5x−1) P(x) ⋮(x−2) mod(16) P(x) ⋮(x^2 +3)(x−2) mod(?)

$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}^{\mathrm{2}} +\mathrm{3}\right)\:\:\:\:\:{mod}\left(\mathrm{5}{x}−\mathrm{1}\right) \\ $$$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}−\mathrm{2}\right)\:\:\:\:\:\:\:\:\:{mod}\left(\mathrm{16}\right) \\ $$$${P}\left({x}\right)\:\:\:\:\:\:\vdots\left({x}^{\mathrm{2}} +\mathrm{3}\right)\left({x}−\mathrm{2}\right)\:\:\:{mod}\left(?\right) \\ $$

Question Number 214302 Answers: 1 Comments: 0

Σ_(n=1) ^∞ (1/(n(4n−1)^2 ))= ?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}\left(\mathrm{4}{n}−\mathrm{1}\right)^{\mathrm{2}} }=\:? \\ $$

Question Number 214301 Answers: 1 Comments: 1

∫_0 ^(π/2) sin^2 (sin x)+ cos^2 (cos x) dx =?

$$\:\:\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{2}}} {\int}}\:\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{sin}\:\mathrm{x}\right)+\:\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{cos}\:\mathrm{x}\right)\:\mathrm{dx}\:=? \\ $$

Question Number 214280 Answers: 1 Comments: 1

Question Number 214342 Answers: 1 Comments: 3

why differantiable f → f is continious but f is continous ↛ differantiable ??

$$\mathrm{why} \\ $$$$\mathrm{differantiable}\:{f}\:\rightarrow\:{f}\:\mathrm{is}\:\mathrm{continious}\: \\ $$$$\mathrm{but}\:{f}\:\mathrm{is}\:\mathrm{continous}\:\nrightarrow\:\mathrm{differantiable}\:?? \\ $$

Question Number 214341 Answers: 2 Comments: 0

∫(dx/(3+cosx))=?

$$\int\frac{{dx}}{\mathrm{3}+{cosx}}=? \\ $$

Question Number 214264 Answers: 1 Comments: 1

Question Number 214258 Answers: 3 Comments: 0

lim_(x→∞) ((x^6 −x^5 ))^(1/6) −((x^6 +5x^5 ))^(1/6) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} −\mathrm{x}^{\mathrm{5}} }−\sqrt[{\mathrm{6}}]{\mathrm{x}^{\mathrm{6}} +\mathrm{5x}^{\mathrm{5}} }\:=? \\ $$

Question Number 214256 Answers: 0 Comments: 0

lim_(x → 1) ∝.arctan((2/(1 +x)) − 1) • Calculons la limite a l′intrieur: (2/(1 + x)) − 1 = 0 lim_(x → 1) arctan((2/(1 + x)) − 1)= arctan(0) = 0 lim_(x→1) ∝.arctan((2/(1+ x)) − 1) =∝.0 lim_(x→1) ∝.arctan((2/(1 + x)) −1) = 0

$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\:\rightarrow\:\mathrm{1}} \propto.\boldsymbol{{arctan}}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\boldsymbol{{x}}}\:−\:\mathrm{1}\right) \\ $$$$\bullet\:\boldsymbol{{Calculons}}\:\boldsymbol{{la}}\:\boldsymbol{{limite}}\:\boldsymbol{{a}}\:\boldsymbol{{l}}'\boldsymbol{{intrieur}}: \\ $$$$\frac{\mathrm{2}}{\mathrm{1}\:+\:\boldsymbol{{x}}}\:−\:\mathrm{1}\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\:\rightarrow\:\mathrm{1}} {arctan}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\:{x}}\:−\:\mathrm{1}\right)=\:\boldsymbol{{arctan}}\left(\mathrm{0}\right)\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{1}} \propto.{arctan}\left(\frac{\mathrm{2}}{\mathrm{1}+\:{x}}\:−\:\mathrm{1}\right)\:=\propto.\mathrm{0}\: \\ $$$$\: \\ $$$$\boldsymbol{{lim}}_{\boldsymbol{{x}}\rightarrow\mathrm{1}} \propto.\boldsymbol{{arctan}}\left(\frac{\mathrm{2}}{\mathrm{1}\:+\:\boldsymbol{{x}}}\:−\mathrm{1}\right)\:=\:\mathrm{0} \\ $$

Question Number 214251 Answers: 1 Comments: 1

Question Number 214248 Answers: 3 Comments: 1

Question Number 214247 Answers: 1 Comments: 0

If 2a = 1 − 2(√a) Find ((2a^2 + (√a))/(2a)) = ?

$$\mathrm{If}\:\:\:\mathrm{2a}\:=\:\mathrm{1}\:−\:\mathrm{2}\sqrt{\mathrm{a}} \\ $$$$\mathrm{Find}\:\:\:\frac{\mathrm{2a}^{\mathrm{2}} \:+\:\sqrt{\mathrm{a}}}{\mathrm{2a}}\:=\:? \\ $$

Question Number 214244 Answers: 1 Comments: 0

4=(√(x−7))

$$\mathrm{4}=\sqrt{{x}−\mathrm{7}} \\ $$$$ \\ $$

Question Number 214228 Answers: 5 Comments: 1

Question Number 214220 Answers: 1 Comments: 0

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