What is e^{-2}(x^2+2y^2+4xy-2y-4x) ?

The solution to e^{-2}(x^2+2y^2+4xy-2y-4x) is 1/(e^2)(x^2+4xy-4x+2y^2-2y)

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simplify e^{-2}(x^2+2y^2+4xy-2y-4x)

simplify

e^{- 2} (x^{2} + 2 y^{2} + 4 x y - 2 y - 4 x)

\frac{1}{e ^{2}} (x^{2} + 4 x y - 4 x + 2 y^{2} - 2 y)

Solution steps

e^{- 2} (x^{2} + 2 y^{2} + 4 x y - 2 y - 4 x)

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \frac{1}{e ^{2}} (x^{2} + 4 x y - 4 x + 2 y^{2} - 2 y)

8.82 x^{0.4} x sin (45. 5^{\circ})

(a^{2} + b^{2})^{2} \cdot (a^{2} - b^{2})^{2}

e^{x} \cdot e^{x} d x

5 v^{2} w^{2} (8 v^{2} - v w + 3 w^{2})

1.2 \cdot 1 0^{- 12}

What is e^{-2}(x^2+2y^2+4xy-2y-4x) ?
The solution to e^{-2}(x^2+2y^2+4xy-2y-4x) is 1/(e^2)(x^2+4xy-4x+2y^2-2y)