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What is the absolute value of -(A-B-C)?
Because abc is the three sides of a triangle
So a+b-c >0 a-b-c
Given that the perimeter of the triangle is 20, if a, b, c represents the length of the three sides of the triangle ABC, find the absolute value of a-b-c + the absolute value of b-c-a + the absolute value of c-a-b
||A-b-c||b-c-a||c-a-b||
Since that sum of the two sides of the triangle is lar than the third side,
||A-b-c||b-c-a||c-a-b||
=|(B+c-a)+(c+a-b)+(a+b-c)|
=|A+b+c|
=A+b+c
=20
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