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Question Number 42191    Answers: 0   Comments: 1

let A_p =∫_0 ^∞ ((sin(px))/(e^x −1)) dx with p>0 1)give A_p at form of serie 2) give A_1 at form of serie .

$${let}\:{A}_{{p}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({px}\right)}{{e}^{{x}} −\mathrm{1}}\:{dx}\:\:{with}\:{p}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right){give}\:{A}_{{p}} \:\:{at}\:{form}\:{of}\:{serie} \\ $$$$\left.\mathrm{2}\right)\:{give}\:{A}_{\mathrm{1}} \:{at}\:{form}\:{of}\:{serie}\:. \\ $$

Question Number 42190    Answers: 0   Comments: 1

let u_n =Σ_(k=1) ^n (((−1)^k )/(√k)) 1) prove that (u_n )is convergente 2) find a equivalent of u_n when n→+∞

$${let}\:\:\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\left(−\mathrm{1}\right)^{{k}} }{\sqrt{{k}}} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:\left({u}_{{n}} \right){is}\:{convergente} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{a}\:{equivalent}\:{of}\:{u}_{{n}} \:{when}\:{n}\rightarrow+\infty \\ $$

Question Number 42189    Answers: 0   Comments: 0

calculate f(x) = ∫_0 ^∞ e^(−t^2 ) arctan(xt^2 )dt

$${calculate}\:\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{t}^{\mathrm{2}} } \:{arctan}\left({xt}^{\mathrm{2}} \right){dt} \\ $$

Question Number 42188    Answers: 0   Comments: 1

let x>0 calculate f(x) =∫_0 ^(+∞) e^(−t) ∣sin(xt)∣ dt

$${let}\:{x}>\mathrm{0}\:\:\:{calculate}\:\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−{t}} \:\mid{sin}\left({xt}\right)\mid\:{dt} \\ $$

Question Number 42187    Answers: 0   Comments: 1

study the convergence of Σ_(k=0) ^∞ e^(−i((kπ)/x)) and find its sum

$${study}\:{the}\:{convergence}\:{of}\:\sum_{{k}=\mathrm{0}} ^{\infty} \:\:{e}^{−{i}\frac{{k}\pi}{{x}}} \:\:\:\:{and}\:{find}\:{its}\:{sum}\: \\ $$

Question Number 42182    Answers: 0   Comments: 3

integrate ∫e^x^2 x^2 dx

$$\boldsymbol{\mathrm{integrate}}\:\:\int\boldsymbol{\mathrm{e}}^{\boldsymbol{{x}}^{\mathrm{2}} } \boldsymbol{{x}}^{\mathrm{2}} \:\boldsymbol{{dx}} \\ $$

Question Number 42232    Answers: 1   Comments: 0

calculate ∫_0 ^(2π) (dθ/((1+cosθ)^3 ))

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{1}+{cos}\theta\right)^{\mathrm{3}} } \\ $$

Question Number 42180    Answers: 1   Comments: 0

The median AD of triangle ABC is bisected at E and BE meets AC at F. Find AF:FC .

$$\mathrm{The}\:\mathrm{median}\:\mathrm{AD}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{ABC}\:\mathrm{is}\: \\ $$$$\mathrm{bisected}\:\mathrm{at}\:\mathrm{E}\:\mathrm{and}\:\mathrm{BE}\:\mathrm{meets}\:\mathrm{AC}\:\mathrm{at}\:\mathrm{F}. \\ $$$$\mathrm{Find}\:\mathrm{AF}:\mathrm{FC}\:. \\ $$

Question Number 42176    Answers: 0   Comments: 1

81(√)

$$\mathrm{81}\sqrt{} \\ $$

Question Number 42175    Answers: 0   Comments: 0

sin3x+sin5x=2(cos^2 2x−sin^2 3x)

$${sin}\mathrm{3}{x}+{sin}\mathrm{5}{x}=\mathrm{2}\left({cos}^{\mathrm{2}} \mathrm{2}{x}−{sin}^{\mathrm{2}} \mathrm{3}{x}\right) \\ $$

Question Number 42174    Answers: 1   Comments: 0

cos^2 3xcos2x−cos^2 x=0

$${cos}^{\mathrm{2}} \mathrm{3}{xcos}\mathrm{2}{x}−{cos}^{\mathrm{2}} {x}=\mathrm{0} \\ $$

Question Number 42170    Answers: 1   Comments: 0

(2tanx−5)tanx+(2cotx−5)cotx−8=0

$$\left(\mathrm{2}{tanx}−\mathrm{5}\right){tanx}+\left(\mathrm{2}{cotx}−\mathrm{5}\right){cotx}−\mathrm{8}=\mathrm{0} \\ $$

Question Number 42169    Answers: 0   Comments: 0

solve 2sin3x(sinx+(√3)cosx)+1+2cos2x=0

$${solve} \\ $$$$\mathrm{2}{sin}\mathrm{3}{x}\left({sinx}+\sqrt{\mathrm{3}}{cosx}\right)+\mathrm{1}+\mathrm{2}{cos}\mathrm{2}{x}=\mathrm{0} \\ $$

Question Number 42157    Answers: 0   Comments: 3

∫_( 0) ^( ∞) (dx/((1 + x^n )))

$$\:\int_{\:\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{n}} \right)} \\ $$

Question Number 42154    Answers: 0   Comments: 5

Question Number 42150    Answers: 0   Comments: 0

Expand (x^x + 5^x )^(3x) up to the term in x^5

$${Expand}\:\:\:\left({x}^{{x}} +\:\mathrm{5}^{{x}} \right)^{\mathrm{3}{x}} \:{up}\:{to}\:{the}\:{term}\:{in}\:{x}^{\mathrm{5}} \\ $$

Question Number 42145    Answers: 1   Comments: 2

Question Number 42133    Answers: 1   Comments: 0

Let a,b,cε R such that a+2b+c=4. Find maximum value of (ab+bc+ca).

$$\mathrm{Let}\:\mathrm{a},\mathrm{b},\mathrm{c}\epsilon\:\mathrm{R}\:\mathrm{such}\:\mathrm{that}\:\mathrm{a}+\mathrm{2b}+\mathrm{c}=\mathrm{4}. \\ $$$$\mathrm{Find}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of}\:\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right). \\ $$

Question Number 42131    Answers: 1   Comments: 1

Determine the total translational energy of two litres of oxygen gas held at a temperature of 0°C and pressure of one atmosphere.

$${Determine}\:{the}\:{total}\:{translational} \\ $$$${energy}\:{of}\:{two}\:{litres}\:{of}\:{oxygen}\:{gas} \\ $$$${held}\:{at}\:{a}\:{temperature}\:{of}\:\mathrm{0}°{C}\:{and}\: \\ $$$${pressure}\:{of}\:{one}\:{atmosphere}. \\ $$$$ \\ $$

Question Number 42128    Answers: 0   Comments: 2

Calculate the density of 1 mole of oxygen at a pressure of 4×10^4 Nm^(−2) and temperature of 273.2K.

$${Calculate}\:{the}\:\:{density}\:{of}\:\mathrm{1}\:{mole}\:{of} \\ $$$${oxygen}\:{at}\:{a}\:{pressure}\:{of}\:\mathrm{4}×\mathrm{10}^{\mathrm{4}} {Nm}^{−\mathrm{2}} \\ $$$${and}\:{temperature}\:{of}\:\mathrm{273}.\mathrm{2}{K}. \\ $$

Question Number 42125    Answers: 0   Comments: 6

If a petrol tank costs σ2000.00 to fill it up at 10°C,how much extra must I pay to fill up the tank at 50°C if the petrol costs σ50.00 per litre and its cubic expansivity is 9.5×10^(−4) K^(−1) ?

$${If}\:{a}\:{petrol}\:{tank}\:{costs}\:\sigma\mathrm{2000}.\mathrm{00}\:{to} \\ $$$${fill}\:{it}\:{up}\:{at}\:\mathrm{10}°{C},{how}\:{much}\:{extra} \\ $$$${must}\:{I}\:{pay}\:{to}\:{fill}\:{up}\:{the}\:{tank}\:{at}\:\mathrm{50}°{C} \\ $$$${if}\:{the}\:{petrol}\:{costs}\:\sigma\mathrm{50}.\mathrm{00}\:{per}\:{litre} \\ $$$${and}\:{its}\:{cubic}\:{expansivity}\:{is} \\ $$$$\mathrm{9}.\mathrm{5}×\mathrm{10}^{−\mathrm{4}} {K}^{−\mathrm{1}} ? \\ $$

Question Number 42118    Answers: 1   Comments: 0

x−4=−(1/x) then . prove x^4 −194=−(1/x^4 )

$$\mathrm{x}−\mathrm{4}=−\frac{\mathrm{1}}{\mathrm{x}}\:\mathrm{then}\:.\:\mathrm{prove}\:\mathrm{x}^{\mathrm{4}} −\mathrm{194}=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} } \\ $$

Question Number 42112    Answers: 1   Comments: 1

Question Number 42110    Answers: 1   Comments: 1

Question Number 42107    Answers: 0   Comments: 0

Question Number 42106    Answers: 1   Comments: 0

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