Given: Segment AC is perpendicular to segment BD Prove: ΔACB ~ ΔECD Reflect ΔECD over segment AC. This establishes that ∠ACB ≅ ∠E'C'D'. Then, . This establishes that . Therefore, ΔACB ~ ΔECD by the AA similarity postulate. translate point E' to point A; ∠E'D'C' ≅ ∠ABC
translate point D' to point A; ∠E'D'C' ≅ ∠BAC
translate point D' to point A; ∠D'E'C' ≅ ∠BAC
translate point E' to point A; ∠D'E'C' ≅ ∠BAC

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step-by-step explanation:

step-by-step explanation:

milk=16 large

soda=280 kids

juice box=82 regular

63 soda +63 regular

124 large total

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1