Math
posted by Ronnie .
To the nearest tenth, find the perimeter of ABC with vertices A(22), B(0,5) and C(3,1).

AB=sqrt{(x₂x₁)²+(y₂y₁)²} =
=sqrt{(0(2))²+(5(2)) ²} =
=sqrt(4+49)=7.3
BC = sqrt{(x₃x₂ )²+(y₃y₂)²} =
=sqrt{(30)²+(15) ²} =
=sqrt(9+36)=6.7
CA = sqrt{(x₁x₃ )²+(y₁y₃)²} =
=sqrt{(23)²+(2+1) ²} =
=sqrt(25+1)=5.1
P=AB+BC+CA = 7.3+6.7+5.1 =14.1 
19.1 units
AB= 7.3
BC=6.8
CA=5
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