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Question Number 37915    Answers: 1   Comments: 0

If y=4x^2 −1 , then find ((85)/(169))+Σ_(i=1) ^(84) (1/(y(i)))

$$\mathrm{If}\:{y}=\mathrm{4}{x}^{\mathrm{2}} −\mathrm{1}\:,\:\mathrm{then}\:\mathrm{find} \\ $$$$\frac{\mathrm{85}}{\mathrm{169}}+\underset{{i}=\mathrm{1}} {\overset{\mathrm{84}} {\Sigma}}\:\frac{\mathrm{1}}{{y}\left({i}\right)}\: \\ $$

Question Number 37914    Answers: 1   Comments: 0

In △ABC, if sin A=sin^2 B then prove 4 cos 2A−4 cos 2B=1−cos 4B

$$\mathrm{In}\:\bigtriangleup{ABC},\:\mathrm{if}\:\mathrm{sin}\:\mathrm{A}=\mathrm{sin}^{\mathrm{2}} {B}\: \\ $$$$\mathrm{then}\:\mathrm{prove} \\ $$$$\mathrm{4}\:\mathrm{cos}\:\mathrm{2}{A}−\mathrm{4}\:\mathrm{cos}\:\mathrm{2}{B}=\mathrm{1}−\mathrm{cos}\:\mathrm{4}{B} \\ $$

Question Number 37913    Answers: 1   Comments: 0

Solve the diferential equatuion (dy/dx)=((2x+y+1)/(x−2y+3))

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{diferential}\:\mathrm{equatuion} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}{x}+{y}+\mathrm{1}}{{x}−\mathrm{2}{y}+\mathrm{3}}\: \\ $$

Question Number 37912    Answers: 1   Comments: 1

Evaluate : the Integral ∫_(-(π/2)) ^(π/2) ∫_0 ^(3 cos θ) r^2 sin^2 θ. dr dθ

$$\mathrm{Evaluate}\::\:\mathrm{the}\:\mathrm{Integral} \\ $$$$\int_{-\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\mathrm{3}\:\mathrm{cos}\:\theta} {r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta.\:{dr}\:{d}\theta\: \\ $$

Question Number 37911    Answers: 0   Comments: 1

the function f(x) is defined by f(x) = { ((−x + 1 , for x≤3)),((kx − 8 , for x ≥ 3)) :} find the value of k .

$$\mathrm{the}\:\mathrm{function}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by} \\ $$$${f}\left({x}\right)\:=\begin{cases}{−{x}\:+\:\mathrm{1}\:,\:{for}\:{x}\leqslant\mathrm{3}}\\{{kx}\:−\:\mathrm{8}\:,\:{for}\:{x}\:\geqslant\:\mathrm{3}}\end{cases} \\ $$$${find}\:{the}\:{value}\:{of}\:{k}\:. \\ $$

Question Number 37906    Answers: 1   Comments: 0

Question Number 37902    Answers: 2   Comments: 1

ind the value of f(a) =∫_0 ^(+∞) (dx/(x^2 +(√(a^2 +x^2 )))) dx witha>0 2)calculate f^′ (a) .

$${ind}\:{the}\:{value}\:{of}\:{f}\left({a}\right)\:\:=\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:+\sqrt{{a}^{\mathrm{2}} \:+{x}^{\mathrm{2}} }}\:{dx} \\ $$$${witha}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{'} \left({a}\right)\:. \\ $$

Question Number 37901    Answers: 0   Comments: 1

let f(x)= (1+e^(−x) )^n 1) calculate f^((p)) (x) and f^((p)) (o) 2)calculate f^((n)) (0) 3)developp f at integr serie .

$${let}\:{f}\left({x}\right)=\:\left(\mathrm{1}+{e}^{−{x}} \right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({p}\right)} \left({x}\right)\:\:{and}\:{f}^{\left({p}\right)} \left({o}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right){developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$

Question Number 37898    Answers: 1   Comments: 1

calculate f(λ) = ∫_0 ^(+∞) e^(−λx) cos(π[x])dx withλ>0

$${calculate}\:{f}\left(\lambda\right)\:=\:\int_{\mathrm{0}} ^{+\infty} \:\:{e}^{−\lambda{x}} \:{cos}\left(\pi\left[{x}\right]\right){dx} \\ $$$${with}\lambda>\mathrm{0} \\ $$

Question Number 37896    Answers: 2   Comments: 1

let I_n = ∫_0 ^n (((−1)^([x]) )/((2x+1)^2 ))dx 1) calculate I_n interms of n 2) find lim_(n→+∞) I_n

$${let}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{{n}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{\left[{x}\right]} }{\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}_{{n}} \:\:{interms}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+\infty} \:{I}_{{n}} \\ $$

Question Number 37895    Answers: 1   Comments: 0

calculate A_n =∫_0 ^n (x−[(√x)])dx and lim_(n→+∞) A_n

$${calculate}\:\:\:{A}_{{n}} =\int_{\mathrm{0}} ^{{n}} \left({x}−\left[\sqrt{{x}}\right]\right){dx}\:{and} \\ $$$${lim}_{{n}\rightarrow+\infty} \:{A}_{{n}} \\ $$

Question Number 37894    Answers: 0   Comments: 0

find nature of the serie Σ_(n=1) ^∞ (Σ_(k=0) ^n (1/C_n ^k ))x^n

$${find}\:{nature}\:{of}\:{the}\:{serie} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\frac{\mathrm{1}}{{C}_{{n}} ^{{k}} }\right){x}^{{n}} \\ $$

Question Number 37893    Answers: 1   Comments: 1

calculate ∫_0 ^1 (√(x+(√(x+1)))) dx .

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\:{dx}\:. \\ $$

Question Number 37892    Answers: 0   Comments: 3

let f(x)=(√(x+(√(x+1)))) 1) find D_f 2) give the equation of assymtote at point A(0,f(o)) 3) if f(x)∼ a(x−1) +b (x→1) determine a andb 4) calculate f^′ (x) 5) find f^(−1) (x) and (f^(−1) )^′ (x)

$${let}\:\:\:{f}\left({x}\right)=\sqrt{{x}+\sqrt{{x}+\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:\:{give}\:{the}\:{equation}\:{of}\:{assymtote}\:{at}\:{point} \\ $$$${A}\left(\mathrm{0},{f}\left({o}\right)\right) \\ $$$$\left.\mathrm{3}\right)\:{if}\:{f}\left({x}\right)\sim\:{a}\left({x}−\mathrm{1}\right)\:\:+{b}\:\:\left({x}\rightarrow\mathrm{1}\right)\:{determine}\:{a}\:{andb} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{find}\:\:{f}^{−\mathrm{1}} \left({x}\right)\:\:{and}\:\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right) \\ $$

Question Number 37891    Answers: 0   Comments: 1

calculate B_n = Σ_(k=0) ^n (−1)^k (2k^2 +1) interms of n.

$${calculate}\:{B}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \left(\mathrm{2}{k}^{\mathrm{2}} \:+\mathrm{1}\right)\:{interms}\:{of}\:{n}. \\ $$

Question Number 37890    Answers: 0   Comments: 2

calculate A_n = Σ_(k=0) ^n (−1)^k (2k+3) interms of n

$${calculate}\:{A}_{{n}} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \left(\mathrm{2}{k}+\mathrm{3}\right)\:{interms}\:{of}\:{n} \\ $$

Question Number 37889    Answers: 0   Comments: 0

calculate ∫_0 ^∞ ((arctan(2x))/x) e^(−tx) dx with t ≥0

$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left(\mathrm{2}{x}\right)}{{x}}\:{e}^{−{tx}} \:{dx}\:{with}\:{t}\:\geqslant\mathrm{0} \\ $$

Question Number 37888    Answers: 1   Comments: 2

find f(α) = ∫_0 ^1 arctan(e^(−αx) )dx with α≥0

$${find}\:{f}\left(\alpha\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctan}\left({e}^{−\alpha{x}} \right){dx}\:{with}\:\alpha\geqslant\mathrm{0}\: \\ $$

Question Number 37887    Answers: 0   Comments: 0

find f(α) = ∫_0 ^1 arctan(1+e^(−αx) )dx with α≥0

$${find}\:{f}\left(\alpha\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{arctan}\left(\mathrm{1}+{e}^{−\alpha{x}} \right){dx}\:{with}\:\alpha\geqslant\mathrm{0} \\ $$

Question Number 37886    Answers: 0   Comments: 0

finf f(α) = ∫_0 ^1 ln(1+e^(−αx) )dx with α≥0

$${finf}\:\:{f}\left(\alpha\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{e}^{−\alpha{x}} \right){dx}\:\:{with}\:\alpha\geqslant\mathrm{0} \\ $$

Question Number 37885    Answers: 0   Comments: 0

find Π_(k=1) ^n cos(((kπ)/(2n+1))) and Π_(k=1) ^n sin(((kπ)/(2n+1)))

$${find}\:\:\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{cos}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:\:{and}\:\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right) \\ $$

Question Number 37884    Answers: 0   Comments: 1

let I = ∫_0 ^∞ e^(−[x]) cos^2 (2πx)dx and J =∫_0 ^∞ e^(−[x]) sin^2 (2πx) dx 1) calculate I +J and I −J 2) find the value of I and J .

$${let}\:{I}\:\:=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left[{x}\right]} \:{cos}^{\mathrm{2}} \left(\mathrm{2}\pi{x}\right){dx}\:{and}\: \\ $$$${J}\:=\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left[{x}\right]} \:{sin}^{\mathrm{2}} \left(\mathrm{2}\pi{x}\right)\:{dx} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{I}\:+{J}\:{and}\:{I}\:−{J} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\:{I}\:{and}\:{J}\:. \\ $$

Question Number 37871    Answers: 0   Comments: 2

A regular pyramid has for its base polygon of n sides, and each slant face consist of an isosceles triangle of vertical angle 2α. If the slant faces are each inclined at angle β to the base , and at an angle 2γ to one another show that cosβ = tan α cot(π/n) , and sinγ = sec α cos(π/n)

$$\mathrm{A}\:\mathrm{regular}\:\mathrm{pyramid}\:\mathrm{has}\:\mathrm{for}\:\mathrm{its}\:\mathrm{base}\:\mathrm{polygon} \\ $$$$\mathrm{of}\:{n}\:\mathrm{sides},\:\mathrm{and}\:\mathrm{each}\:\mathrm{slant}\:\mathrm{face}\:\mathrm{consist}\:\mathrm{of}\:\mathrm{an}\: \\ $$$$\mathrm{isosceles}\:\mathrm{triangle}\:\mathrm{of}\:\mathrm{vertical}\:\mathrm{angle}\:\mathrm{2}\alpha.\:\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{slant}\:\mathrm{faces}\:\mathrm{are}\:\mathrm{each}\:\mathrm{inclined}\:\mathrm{at}\:\mathrm{angle}\:\beta\:\mathrm{to}\: \\ $$$$\mathrm{the}\:\mathrm{base}\:,\:\mathrm{and}\:\mathrm{at}\:\mathrm{an}\:\mathrm{angle}\:\mathrm{2}\gamma\:\mathrm{to}\:\mathrm{one}\:\mathrm{another} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{cos}\beta\:=\:\mathrm{tan}\:\alpha\:\mathrm{cot}\frac{\pi}{\mathrm{n}}\:,\:\mathrm{and}\:\mathrm{sin}\gamma\:=\:\mathrm{sec}\:\alpha\:\mathrm{cos}\frac{\pi}{\mathrm{n}} \\ $$

Question Number 37864    Answers: 2   Comments: 0

prove that cos (π/(15)) cos ((2π)/(15)) cos ((3π)/(15)) cos ((4π)/(15)) cos ((5π)/(15)) cos ((6π)/(15)) cos ((7π)/(15)) = (1/2^7 )

$$\mathrm{prove}\:\mathrm{that} \\ $$$$\mathrm{cos}\:\frac{\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{2}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{4}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{6}\pi}{\mathrm{15}}\:\mathrm{cos}\:\frac{\mathrm{7}\pi}{\mathrm{15}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{7}} } \\ $$

Question Number 37862    Answers: 1   Comments: 7

Question Number 37861    Answers: 0   Comments: 1

which is the chain rule? A. (dy/dx) = (dy/dx) × 1 B. (dy/dx) = (du/dx) × (dy/dx) C. (dy/dx) = (dy/du) × (du/dx) D. (dy/dx) = (dy/du) × (dy/dx)

$$\:\:\:{which}\:{is}\:{the}\:{chain}\:{rule}? \\ $$$${A}.\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{dx}}\:×\:\mathrm{1} \\ $$$${B}.\:\frac{{dy}}{{dx}}\:=\:\frac{{du}}{{dx}}\:×\:\frac{{dy}}{{dx}} \\ $$$${C}.\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}\:×\:\frac{{du}}{{dx}} \\ $$$${D}.\:\frac{{dy}}{{dx}}\:=\:\frac{{dy}}{{du}}\:×\:\frac{{dy}}{{dx}} \\ $$

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