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2/3 - 10/9and5/3 and 7/9

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Sagot :

reinhard10158

Step-by-step explanation:

always Pythagoras with the coordinate differences as sides and the distance the Hypotenuse.

c² = (2/3 - 5/3)² + (-10/9 - -7/9)² = (-3/3)² + (-10/9 + 7/9)² =

= (-1)² + (-3/9)² = 1 + (-1/3)² = 1 + 1/9 = 10/9

c = sqrt(10)/3

RexRelative

Answer:

Step-by-step explanation:

Point 1  ([tex]\frac{2}{3}[/tex] , [tex]\frac{-10}{9}[/tex])   in the form (x1,y1)

Point 2 ( [tex]\frac{5}{3}[/tex] , [tex]\frac{-7}{9}[/tex])  in the form (x2,y2)

use the distance formula

dist = sqrt[ (x2-x1)^2 + (y2-y1)^2 ]

dist = sqrt [ [tex]\frac{5}{3}[/tex] -[tex]\frac{2}{3}[/tex])^2 + (  [tex]\frac{-7}{9}[/tex] - ( [tex]\frac{-10}{9}[/tex] ) )^2 ]

dist = sqrt [ ([tex]\frac{3}{3}[/tex])^2 + ([tex]\frac{3}{9}[/tex])^2 ]

dist = sqrt [  1 + ([tex]\frac{1}{3}[/tex])^2 ]

dist = sqrt [  [tex]\frac{9}{9}[/tex] + [tex]\frac{1}{9}[/tex] ]

dist = [tex]\sqrt{\frac{10}{9} }[/tex]

dist = [tex]\sqrt{10}[/tex] *[tex]\sqrt{\frac{1}{9} }[/tex]

dist = [tex]\sqrt{10}[/tex]  * [tex]\frac{1}{3}[/tex]

dist = [tex]\frac{\sqrt{10} }{3}[/tex]

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