Suppose R = {(x, y) | 0 ≤ x ≤ 1,0 ≤ y ≤ 1}. (enter as a reduced fraction). Then 11₂(120² + R (x² + y² + xy) dx dy = =

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answerhappygod
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Suppose R = {(x, y) | 0 ≤ x ≤ 1,0 ≤ y ≤ 1}. (enter as a reduced fraction). Then 11₂(120² + R (x² + y² + xy) dx dy = =

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Suppose R X Y 0 X 1 0 Y 1 Enter As A Reduced Fraction Then 11 120 R X Y Xy Dx Dy 1
Suppose R X Y 0 X 1 0 Y 1 Enter As A Reduced Fraction Then 11 120 R X Y Xy Dx Dy 1 (19.46 KiB) Viewed 16 times
Suppose R = {(x, y) | 0 ≤ x ≤ 1,0 ≤ y ≤ 1}. (enter as a reduced fraction). Then 11₂(120² + R (x² + y² + xy) dx dy = =
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