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calculus  please help!
Find the slope of the tangent line to the curve (3x+3y−27)^(1/2)+(2xy−39)^(1/2)=8 at the point (8,4). 
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Sorry but I've got a lot of problems that I don't understand. 1) Let f(x)= (3x1)e^x. For which value of x is the slope of the tangent line to f positive? Negative? Zero? 2) Find an equation of the tangent line to the oven curve 
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Find the slope of the tangent line to the curve 0x^2−2xy+2y^3=10 at point (4,1) Help please? Thanks in advance 
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Consider the curve defined by 2y^3+6X^2(y) 12x^2 +6y=1 . a. Show that dy/dx= (4x2xy)/(x^2+y^2+1) b. Write an equation of each horizontal tangent line to the curve. c. The line through the origin with slope 1 is tangent to the 
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original curve: 2y^3+6(x^2)y12x^2+6y=1 dy/dx=(4x2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of 
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original curve: 2y^3+6(x^2)y12x^2+6y=1 dy/dx=(4x2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of 
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The line that is normal to the curve x^2=2xy3y^2=0 at(1,1) intersects the curve at what other point? Please help. Thanks in advance. We have x2=2xy  3y2 = 0 Are there supposed to be 2 equal signs in this expression or is it x2 + 
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original curve: 2y^3+6(x^2)y12x^2+6y=1 dy/dx=(4x2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the curve b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of 
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Suppose that f(x) is an invertible function (that is, has an inverse function), and that the slope of the tangent line to the curve y = f(x) at the point (2, –4) is –0.2. Then: (Points : 1) A) The slope of the tangent line to