definition is valid since center is exactly 1radius away from any point on the outersurface.
other figures do not fit this defn
Give a counterexample to disprove the statement all squares are congruent
You take (8x - 6) then add 80 =90 then solve for x
8x-6+80 =90
x= 2
Step-by-step explanation:
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Slope of a line contatining the points (x1,y1) and (x2,y2) is
(y2-y1)/(x2-x1)
so slope between (-6,3) and (2,-5) is
(-5-3)/(2-(-6))=(-8)/(2+6)=-8/8=-1
the slope is -1
