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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 
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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= ????
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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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which will become a buffer in 1L of H2O? a) 0.3 mole KOH & 0.3 mole HBr b) 0.5 mole NH3 & 0.5 mole HCl c) 0.2 mole NaCl & 0.3 mole HCl d) 0.2 mole NaOH & 0.2 mole HBr e) 0.4 mole HC2H3O2 & 0.2 mole NaOH I think that the answer is
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Consider the equation: x^2  2xy + 4y^2 = 64 Write an expression of the slope of the curve at any point. (y^p)= y prime My work 2x  2(xy^p + y) + 8yy^p = 0 2x 2xy^p  2y + 8yy^p = 0 2xy^p + 8yy^p = 2y 2x factored out y^p and
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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" =
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