Ignore my response. I completely misunderstood the question.(:
I think I read the question wrong. I thought the question wanted the slope of the tangent of f(x) at (1,-5) and the two equations at this point.
y = x^2 - 2 f' = 2x P(1,-5) f' = slope m = 2(1) = 2 Equation of the line tangent at P(1,-5) m = 2 y = mx + b -5 = 2(1) + b -5 = 2 + b b = -7 Equation of the tangent line is, y = 2x - 7 2x - y = 7 Equation of the normal line at P(1,-5) (The normal line is the line that ...
Post your question, but I've noticed that there are not many experts here for accounting questions (depending on the specific question). Depending, I might be able to help, depending on how involved and the specific area involved. Someone might be able to point you in the ...
You have to do polynomial long division. I can't layout the problem here because of the alignment of columns and such. See this website. It explains in detail how to do this. I'll check your answer if you post it. Good luck :) http://www.purplemath.com/modules/polydiv2...
See this post. Only the side length is different. Same method is used. http://www.jiskha.com/display.cgi?id=1298927532
You need parenthesis. I'll assume your problem is, x/(x+1) + 5/x = 1/(x^2 + x) x/(x+1) + 5/x = 1/(x(x + 1)) Multiply both sides by x(x + 1) x^2 + 5(x + 1) = 1 x^2 + 5x + 5 = 1 x^2 + 5x + 4 = 0 (x + 1)(x + 4) = 0 x = -1 or x = -4 Make sure you check both answers in the orig...
| = integration symbol | u dv = uv - | v du | arctan(4t) dt u = arctan(4t) du = 4/(16t^2 + 1) dt dv = dt v = t It should look like this, = t arctan(4t) - | 4t/(16t^2 + 1) dt After you set that up, substitute w = 16t^2 + 1 dw = 32t dt 1/32 dw = t dt Then, the integration should...
x = Janet's distance to office 2x - 1 = Lynn's distance to office x + 2x - 1 = 8 3x = 9 x = 3 x = 3 mi from Janet's house to office 2x - 1 = 5 miles Lynn's house to office
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