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Question Number 61181 Answers: 0 Comments: 0

sove (1+e^(−x) )y^(′′) +(2+e^x )y^′ =(x+1)e^x

$${sove}\:\left(\mathrm{1}+{e}^{−{x}} \right){y}^{''} \:+\left(\mathrm{2}+{e}^{{x}} \right){y}^{'} \:=\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$

Question Number 61169 Answers: 1 Comments: 1

Question Number 61165 Answers: 1 Comments: 0

Question Number 61162 Answers: 1 Comments: 3

if sin(x) = ((x − 20)/(20)) , find x

$$\mathrm{if}\:\:\:\:\:\mathrm{sin}\left(\mathrm{x}\right)\:\:=\:\:\frac{\mathrm{x}\:−\:\mathrm{20}}{\mathrm{20}}\:\:,\:\:\:\mathrm{find}\:\:\mathrm{x} \\ $$

Question Number 61157 Answers: 1 Comments: 3

if tan 5θ + tan 4θ =1 find 3θ

$${if} \\ $$$${tan}\:\mathrm{5}\theta\:+\:{tan}\:\mathrm{4}\theta\:=\mathrm{1} \\ $$$${find}\:\mathrm{3}\theta \\ $$$$ \\ $$$$ \\ $$

Question Number 61151 Answers: 0 Comments: 0

Question Number 61147 Answers: 1 Comments: 0

prove ∫((1+cos x)/(1−cos x))dx=−2cot (x/2)−x+c

$$\boldsymbol{{prove}} \\ $$$$\int\frac{\mathrm{1}+{cos}\:{x}}{\mathrm{1}−{cos}\:{x}}{dx}=−\mathrm{2}{cot}\:\frac{{x}}{\mathrm{2}}−{x}+{c} \\ $$$$ \\ $$

Question Number 61142 Answers: 0 Comments: 0

Question Number 61140 Answers: 1 Comments: 1

can we find an exact solution? t^6 +4t^4 −12t^3 +24t^2 −24t+8=0

$$\mathrm{can}\:\mathrm{we}\:\mathrm{find}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}? \\ $$$${t}^{\mathrm{6}} +\mathrm{4}{t}^{\mathrm{4}} −\mathrm{12}{t}^{\mathrm{3}} +\mathrm{24}{t}^{\mathrm{2}} −\mathrm{24}{t}+\mathrm{8}=\mathrm{0} \\ $$

Question Number 61137 Answers: 1 Comments: 0

What is the sum of first 3n term of an AP , if the sunm of first n term is 2n and sum of first 2n term is 5n

$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{3n}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:,\:\mathrm{if}\:\mathrm{the}\:\mathrm{sunm}\:\mathrm{of}\:\mathrm{first}\:\mathrm{n}\:\mathrm{term}\:\mathrm{is} \\ $$$$\mathrm{2n}\:\:\mathrm{and}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{2n}\:\mathrm{term}\:\mathrm{is}\:\:\mathrm{5n} \\ $$

Question Number 61117 Answers: 2 Comments: 0

The 2nd, 4th and 8th term of an AP are the consecutive term of a GP. If the sum of the 3rd and 4th term of the AP is 20. Find the sum of the first four terms of the AP.

$$\mathrm{The}\:\mathrm{2nd},\:\mathrm{4th}\:\mathrm{and}\:\mathrm{8th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{an}\:\mathrm{AP}\:\mathrm{are}\:\mathrm{the}\:\mathrm{consecutive}\:\mathrm{term}\:\mathrm{of}\:\mathrm{a}\:\mathrm{GP}. \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{3rd}\:\mathrm{and}\:\mathrm{4th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}\:\mathrm{is}\:\mathrm{20}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{first}\:\mathrm{four}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{AP}. \\ $$

Question Number 61116 Answers: 1 Comments: 0

Question Number 61186 Answers: 0 Comments: 3

Question Number 61112 Answers: 2 Comments: 4

Question Number 61107 Answers: 0 Comments: 0

solve Cauchy′s problem x′= t + (((μx)^2 )/(1+(μx)^2 )), μ∈R x(0)=0

$${solve}\:{Cauchy}'{s}\:{problem} \\ $$$${x}'=\:{t}\:+\:\frac{\left(\mu{x}\right)^{\mathrm{2}} }{\mathrm{1}+\left(\mu{x}\right)^{\mathrm{2}} },\:\mu\in\mathbb{R} \\ $$$${x}\left(\mathrm{0}\right)=\mathrm{0} \\ $$

Question Number 61096 Answers: 0 Comments: 0

∀ a, n ∈ N : ∣a−n∣=1 pour a, n ≥3 a^m ≡1modn (∗) posons : m=n−1 (∗′) subtituons cette valeur dans (∗). on a: a^(n−1) ≡1modn. Mais n n′est pas forcement premier. Test de primalite ∀ n ∈ N, n ≥3. (n−2)^(n−1) ≡1modn ⇒ n est premier.

$$\forall\:{a},\:{n}\:\in\:{N}\::\:\mid{a}−{n}\mid=\mathrm{1}\:{pour}\:{a},\:{n}\:\geqslant\mathrm{3} \\ $$$${a}^{{m}} \equiv\mathrm{1}{modn}\:\left(\ast\right) \\ $$$${posons}\::\:{m}={n}−\mathrm{1}\:\left(\ast'\right) \\ $$$${subtituons}\:{cette}\:{valeur}\:{dans}\:\left(\ast\right). \\ $$$${on}\:{a}:\:{a}^{{n}−\mathrm{1}} \equiv\mathrm{1}{modn}.\:{Mais}\:{n}\:{n}'{est}\:{pas}\:{forcement}\:{premier}. \\ $$$${Test}\:{de}\:{primalite} \\ $$$$\forall\:{n}\:\in\:{N},\:{n}\:\geqslant\mathrm{3}. \\ $$$$\left({n}−\mathrm{2}\right)^{{n}−\mathrm{1}} \equiv\mathrm{1}{modn}\:\Rightarrow\:{n}\:{est}\:{premier}. \\ $$

Question Number 61178 Answers: 0 Comments: 1

solve (1+x^2 )y^′ +(1−x^2 )y =x e^(−3x)

$${solve}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{'} \:+\left(\mathrm{1}−{x}^{\mathrm{2}} \right){y}\:={x}\:{e}^{−\mathrm{3}{x}} \\ $$

Question Number 61177 Answers: 0 Comments: 0