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AllQuestion and Answers: Page 62

Question Number 217783 Answers: 0 Comments: 11

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

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$ \\ $$$$\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\pi{x}\right){sin}\left(\mathrm{2}\pi{x}\right){sin}\left(\mathrm{3}\pi{x}\right)}{{x}^{\mathrm{3}} }\:=\:\pi^{\mathrm{3}} \:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 217735 Answers: 1 Comments: 1

a + b + c= abc a^2 + b^2 + c^2 = 49 ab+bc+ca=? . .

$$\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}=\:\mathrm{abc} \\ $$$$\mathrm{a}^{\mathrm{2}} \:+\:\mathrm{b}^{\mathrm{2}} \:+\:\mathrm{c}^{\mathrm{2}} \:=\:\mathrm{49}\: \\ $$$$\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=?\:\:\:.\:\:\:\:\:\:\:\:\:\:\:.\:\:\:\:\:\:\:\: \\ $$

Question Number 217733 Answers: 3 Comments: 0

Question Number 217732 Answers: 2 Comments: 1

{ ((x+y=xy)),((x^2 +y^2 =25)) :}; x^4 +y^4 =?

$$\begin{cases}{{x}+{y}={xy}}\\{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{25}}\end{cases};\:\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} =? \\ $$

Question Number 217730 Answers: 3 Comments: 0

x+(1/x)=3 ; x^5 +(1/x^5 )=?

$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:;\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }=? \\ $$

Question Number 217725 Answers: 0 Comments: 2

Question Number 217723 Answers: 1 Comments: 1

Question Number 217717 Answers: 0 Comments: 0

Question Number 217694 Answers: 0 Comments: 0

Question Number 217691 Answers: 3 Comments: 0

Question Number 217690 Answers: 1 Comments: 3

Question Number 217683 Answers: 1 Comments: 0

Prove:∫_0 ^1 (√((((√(K^2 +36K′^2 ))+6K^′ )/(K^2 +36K^(′2) )) ))(dk/( (√k)(1−k^2 )^(2/3) ))=(√π)((√2)−(√((4−2(√2))/3)))

$$\mathrm{Prove}:\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\sqrt{{K}^{\mathrm{2}} +\mathrm{36}{K}'^{\mathrm{2}} }+\mathrm{6}{K}^{'} }{{K}^{\mathrm{2}} +\mathrm{36}{K}^{'\mathrm{2}} }\:}\frac{{dk}}{\:\sqrt{{k}}\left(\mathrm{1}−{k}^{\mathrm{2}} \right)^{\frac{\mathrm{2}}{\mathrm{3}}} }=\sqrt{\pi}\left(\sqrt{\mathrm{2}}−\sqrt{\frac{\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{3}}}\right) \\ $$

Question Number 217676 Answers: 1 Comments: 0

Find a ral root of the equation x^3 −x−1=0 by fixed point iteration method

$${Find}\:{a}\:{ral}\:{root}\:{of}\:{the}\:{equation}\:{x}^{\mathrm{3}} −{x}−\mathrm{1}=\mathrm{0}\:{by}\:{fixed}\:{point}\:{iteration}\:{method} \\ $$

Question Number 217674 Answers: 0 Comments: 3

Question Number 217669 Answers: 1 Comments: 0

Question Number 217664 Answers: 1 Comments: 0

If f(x) = (√(2x + 3)) prove that: f ′(x) = (1/( (√(2x + 3))))

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}} \\ $$$$\mathrm{prove}\:\mathrm{that}:\:\:\:\mathrm{f}\:'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2x}\:+\:\mathrm{3}}}\: \\ $$

Question Number 217660 Answers: 1 Comments: 0

f(x) + f(y)=f(x+y)+xy f(x)=?

$$\:{f}\left({x}\right)\:+\:{f}\left({y}\right)={f}\left({x}+{y}\right)+{xy}\: \\ $$$${f}\left({x}\right)=? \\ $$

Question Number 217659 Answers: 3 Comments: 0

x+y=7 ∧ x^3 +y^3 =133; x,y=?

$${x}+{y}=\mathrm{7}\:\wedge\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} =\mathrm{133};\:{x},{y}=? \\ $$

Question Number 217686 Answers: 1 Comments: 3

Question Number 217685 Answers: 0 Comments: 0

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