ted-s
ted-s
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you know that the factors can have the form [2x -a][x-b] , since the quadratic starts with 2 x²....now you also need : ab = -3{ thus a & b have opposite signs} and -2b-a = -5......if you are looking for integers then the only pairs are [1,-3] , …
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you know that the factors can have the form [2x -a][x-b] , since the quadratic starts with 2 x²....now you also need : ab = -3{ thus a & b have opposite signs} and -2b-a = -5......if you are looking for integers then the only pairs are [1,-3] , …
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converges to 0....top is bounded by 30 , bottom is not bounded
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solve y = 1 / [ 2x + 6] for x---> 2x + 6 = 1 / y ---> x = - 3 + 1 / [ 2y] if you have to have y as the dependent variable then g inverse (x) = - 3 + 1 / [ 2x ]
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try Emmy Noether, a women who was a mathematician when women were not allow in graduate training Alan Turing , developer of the idea behind the basis of computers
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D = r t , r = sum of the 2 rates , D is given , find t and then 3.5 t
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Come on , you know that the ln function if 1 to 1... thus check if the 1st " bunch of junk " = 2nd " bunch of junk "
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actually none are appropriate PT ╪ TQ so A & B are discarded.. and what does + mean here....if + means Union then D is valid, and C is also valid.
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actually none are appropriate PT ╪ TQ so A & B are discarded.. and what does + mean here....if + means Union then D is valid, and C is also valid.
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your eigenvectors are fine for eigenvalue of 0..the 1st could also be < -3 , 1 , 1 , 1 > ; on my TI 85 one of the vectors is { 4 dec. } < 0.2606 , -0.2919 , 0.0695 , 0.9305 > ; remember that if W is an eigenvector then so is [a W]…
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likely because it is not valid ......but u(x + δx ) ≈ u(x) + δx ( du/dx ) , if du/dx exists
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likely because it is not valid ......but u(x + δx ) ≈ u(x) + δx ( du/dx ) , if du/dx exists
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p = k / r ---> k = 24,000 ---> r = k / p...finish it
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p = k / r ---> k = 24,000 ---> r = k / p...finish it
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x = 3 is the line equation
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x = 3 is the line equation
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your " statement " is impossible to occur...if g(x) = 0 then p(x) = r(x) whose degree > degree of g(x) { which is 0 }
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"T" s comments are true...but not the computations...bottom determinant = 14 , top = 42...thus x = 3
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SORRY...the graph has no oblique asymptote... degree of the top = 1 more than that of the bottom for oblique HA is y = 0 , VAs are x = -2 & 3/2
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SORRY...the graph has no oblique asymptote... degree of the top = 1 more than that of the bottom for oblique HA is y = 0 , VAs are x = -2 & 3/2
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d + q = 24 & 10 d + 25 q = 435...solve...{ 13 ,11 }
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d + q = 24 & 10 d + 25 q = 435...solve...{ 13 ,11 }
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SORRY...the graph has no oblique asymptote... degree of the top = 1 more than that of the bottom for oblique HA is y = 0 , VAs are x = -2 & 3/2
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"T" s comments are true...but not the computations...bottom determinant = 14 , top = 42...thus x = 3
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your " statement " is impossible to occur...if g(x) = 0 then p(x) = r(x) whose degree > degree of g(x) { which is 0 }
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your " statement " is impossible to occur...if g(x) = 0 then p(x) = r(x) whose degree > degree of g(x) { which is 0 }
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x = 3 is the line equation
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p = k / r ---> k = 24,000 ---> r = k / p...finish it
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likely because it is not valid ......but u(x + δx ) ≈ u(x) + δx ( du/dx ) , if du/dx exists
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your eigenvectors are fine for eigenvalue of 0..the 1st could also be < -3 , 1 , 1 , 1 > ; on my TI 85 one of the vectors is { 4 dec. } < 0.2606 , -0.2919 , 0.0695 , 0.9305 > ; remember that if W is an eigenvector then so is [a W]…