simultaneous equation
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precal
Use Pascal's triangle to expand the expression. (x − y)5 so I tried to work this out: (xy)^5 = [(xy)(xy)][(xy)(xy)](xy) (xy)^2 = (xy)(xy) = x^2  2xy + y^2 x^2  2xy + y^2)(x^2  2xy + y^2) = x^4 4x^3y + 6x^2y^2 
asked by James on August 12, 2015 
Algebra
Simplify (2xy^2+3)((2xy^2)^2(2xy^2)(3)+(3)^2) This reminds me of (U+a)(U^2 aU + a^2). When mulipltiplied out, one gets a cubic equation. Do you remember the formula for factoring the sum of two cubes? a^3+b^3= ????
asked by Juri on August 9, 2007 
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I thought number 4 on this list was incorrect but didn't know how to fully explain it here is the list that I have. Please look at the following simplification of an algebraic expression. Which line contains the mistake and why?
asked by Linda on March 8, 2013 
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Can 2xy  y  1 be factored out? How? about all you can do is take the common y out of the first two terms; that in itself is not much help. dont know if this is rite or not..but just a guess.. assuming 2xy  y  1 = 0 2xy  y = 1
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Please look at the following simplification of an algebraic expression. Which line contains the mistake and why? Line 1: (2x + 6x) + 4y + 3x  2xy Line 2: (8x) + 4y + 3x  2xy Line 3: 8x + 3x + 4y  2xy Line 4: 11x + 4y  2xy Line
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Find y" by implicit differentiation. x^3+y^3 = 1 (x^3)'+(y^3)' = (1)' 3x^2+3y^2(y') = 0 3y^2(y') = 3x^2 y' = 3x^2/3y^2 y' = x^2/y^2 y" = [(y^2)(x^2)'(x^2)(y^2)']/y^4 y" = [(y^2)(2x)(x^2)(2y)]/y^4 y" =
asked by Justin on September 27, 2015 
ap calc
find the equations of ALL horizontal tangents to the curve y^2=(x^2+4)/x, if any exist This is what I have so far: xy^2= x^2+4 2xyDy/Dx =2xy^2 dy/dx = (2x y^2)/2xy (2x y^2)/2xy = 0 2xy^2 I don't know what to do next
asked by Anonymous on October 8, 2014 
ap calc
find the equation of all horizontal tangents to the curve y^2 = (x^2+4)/x, if any exist this is what I have so far: xy^2= x^2+4 2xydy/dx = 2xy^2 dy/dx = (2xy^2)/2xy (2xY^2)/2xy = 0 2x Y^2 = 0 I don't know what to do after this
asked by Anonymous on October 8, 2014 
AP Calc
Find the equation of ALL horizontal tangents to the curve y^2=x^2+4/x, if any exists. My work: Derivative First I multiply across and got x^2+4=xy^2 2x+0=(x)(2y(dy/dx)+(y^2) 2xy^2=2xy(dy/dx) dy/dx=2xy^2/2xy=xy^2/xy I don't
asked by David on October 6, 2014