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Question Number 217858 Answers: 0 Comments: 0

Question Number 217857 Answers: 2 Comments: 0

Prove that: ((cos20°))^(1/3) + ((cos80°))^(1/3) + ((cos160°))^(1/3) = (((3 ∙ (9)^(1/3) − 6)/2))^(1/3)

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{cos20}°}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{cos80}°}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{cos160}°}\:=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{3}\:\centerdot\:\sqrt[{\mathrm{3}}]{\mathrm{9}}\:−\:\mathrm{6}}{\mathrm{2}}} \\ $$

Question Number 217855 Answers: 2 Comments: 0

i need help ∫_0 ^1 x^n (e^(−(x^2 /2)) )dx ,(n∈N)

$${i}\:{need}\:{help}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}} \left({e}^{−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}} \right){dx}\:\:,\left({n}\in\mathbb{N}\right) \\ $$

Question Number 217849 Answers: 0 Comments: 0

Question Number 217843 Answers: 0 Comments: 5

Question Number 217842 Answers: 2 Comments: 0

f(x)=(√(x^2 −6x+10)) f(3+2(√6))=?

$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{10}} \\ $$$${f}\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{6}}\right)=? \\ $$

Question Number 217837 Answers: 1 Comments: 0

do you guys know about Three Body problem?? when 2−Body problem we can solve equation of motions x_1 ^→ (t) , x_2 ^→ (t) but why we can′t solve 3−body problem? The reason why we can′t solve 3−Body problem is because this Equation isn′t Linear equation???

$$\mathrm{do}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{know}\:\mathrm{about}\:\mathrm{Three}\:\mathrm{Body}\:\mathrm{problem}?? \\ $$$$\mathrm{when}\:\mathrm{2}−\mathrm{Body}\:\mathrm{problem}\: \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{motions}\:\overset{\rightarrow} {\boldsymbol{\mathrm{x}}}_{\mathrm{1}} \left({t}\right)\:,\:\overset{\rightarrow} {\boldsymbol{\mathrm{x}}}_{\mathrm{2}} \left({t}\right) \\ $$$$\mathrm{but}\:\mathrm{why}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{3}−\mathrm{body}\:\mathrm{problem}? \\ $$$$\mathrm{The}\:\mathrm{reason}\:\mathrm{why}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{3}−\mathrm{Body}\:\mathrm{problem} \\ $$$$\mathrm{is}\:\mathrm{because}\:\mathrm{this}\:\mathrm{Equation}\:\mathrm{isn}'\mathrm{t}\:\:\mathrm{Linear}\:\mathrm{equation}??? \\ $$

Question Number 217821 Answers: 0 Comments: 0

Question Number 217822 Answers: 1 Comments: 0

Question Number 217819 Answers: 0 Comments: 0

Question Number 217818 Answers: 1 Comments: 0

Question Number 217813 Answers: 1 Comments: 0

∫_0 ^∞ [(xp(2+x)]^(−1) dx p∈R

$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\left[\left({xp}\left(\mathrm{2}+{x}\right)\right]^{−\mathrm{1}} {dx}\:\:\:\right. \\ $$$${p}\in\mathbb{R} \\ $$

Question Number 217802 Answers: 1 Comments: 2

Question Number 217797 Answers: 3 Comments: 0

Solve: 5x^2 y′′ + x(1 + x) y′ − y = 0

$$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{5x}^{\mathrm{2}} \:\mathrm{y}''\:\:+\:\:\:\mathrm{x}\left(\mathrm{1}\:\:+\:\:\mathrm{x}\right)\:\mathrm{y}'\:\:−\:\:\mathrm{y}\:\:\:=\:\:\:\mathrm{0} \\ $$

Question Number 217795 Answers: 0 Comments: 1

Question Number 217789 Answers: 0 Comments: 6

Question Number 217804 Answers: 0 Comments: 0

Question Number 217783 Answers: 0 Comments: 11

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Question Number 217772 Answers: 1 Comments: 1

Where is the error (−1)=(−1)^1 =(−1)^(2/2) =[(−1)^2 ]^(1/2) =[1]^(1/2) =(√1) =1

$${Where}\:{is}\:{the}\:{error} \\ $$$$\left(−\mathrm{1}\right)=\left(−\mathrm{1}\right)^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\left(−\mathrm{1}\right)^{\frac{\mathrm{2}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\left[\left(−\mathrm{1}\right)^{\mathrm{2}} \right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left[\mathrm{1}\right]^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{1} \\ $$

Question Number 217769 Answers: 1 Comments: 1

Given a consumer with the utility function U = X_1 ^(1/4) + X_2 who faces a budget constraint of B=P_1 X_1 P_2 X_2 Show that the expemditure function facing the consumer is B = 2P_1 ^(1/2) P_2 ^(1/2) U^(1/2)

$${Given}\:{a}\:{consumer}\:{with}\:{the}\:{utility} \\ $$$${function}\:{U}\:=\:{X}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{4}}} +\:{X}_{\mathrm{2}} \:{who}\:{faces} \\ $$$${a}\:{budget}\:{constraint}\:{of}\:{B}={P}_{\mathrm{1}} {X}_{\mathrm{1}} {P}_{\mathrm{2}} {X}_{\mathrm{2}} \\ $$$${Show}\:{that}\:{the}\:{expemditure}\:{function} \\ $$$${facing}\:{the}\:{consumer}\:{is} \\ $$$${B}\:=\:\mathrm{2}{P}_{\mathrm{1}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {P}_{\mathrm{2}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {U}^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$

Question Number 217766 Answers: 4 Comments: 0

Solve for x & y (1/x)+(1/y)=5 (1/x^2 )+(1/y^2 )=13

$${Solve}\:{for}\:{x}\:\&\:{y} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\mathrm{5} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{y}^{\mathrm{2}} }=\mathrm{13} \\ $$

Question Number 217764 Answers: 1 Comments: 0

Let a, b, c be distinct real numbers such that (a/(b−c))+(b/(c−a))+(c/(a−b))=0 then prove that (a/((b−c)^2 ))+(b/((c−a)^2 ))+(c/((a−b)^2 ))=0

$$ \\ $$$$\mathrm{Let}\:\mathrm{a},\:\mathrm{b},\:\mathrm{c}\:\mathrm{be}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{{a}}{{b}−{c}}+\frac{{b}}{{c}−{a}}+\frac{{c}}{{a}−{b}}=\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\frac{{a}}{\left({b}−{c}\right)^{\mathrm{2}} }+\frac{{b}}{\left({c}−{a}\right)^{\mathrm{2}} }+\frac{{c}}{\left({a}−{b}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$

Question Number 217755 Answers: 2 Comments: 2

∫ ((cos(sin^(− 1) x) + cos^(− 1) (sin x))/(ln(ln(ln(1 + (√(x + (√x)))))) dx

$$\int\:\frac{\mathrm{cos}\left(\mathrm{sin}^{−\:\mathrm{1}} \mathrm{x}\right)\:+\:\mathrm{cos}^{−\:\mathrm{1}} \left(\mathrm{sin}\:\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{ln}\left(\mathrm{1}\:+\:\sqrt{\mathrm{x}\:+\:\sqrt{\mathrm{x}}}\right)\right.\right.}\:\mathrm{dx} \\ $$

Question Number 217761 Answers: 1 Comments: 0

prove that : I=∫_0 ^( ∞) ((sin(πx)sin(2πx)sin(3πx))/x^3 ) = π^3