The coordinates of the vertices of a triangle are a (-1, 2), b (3, 1), and c (-3, -2). the coordinates of the vertices of its image are a '(3, -2), b (7, -3), and c '(1, -6). what transformations would result in this image? (a) δ abc was translated up 4 units and right 4 units. (b) δ abc was translated up 4 units and left 4 units. (c) δ abc was translated right 4 units and down 4 units. (d) δ abc was translated left 4 units and down 4 units.

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

-12 > a

step-by-step explanation:

-11 > 1 + a

subtract 1 from each side

-11 -1> 1 -1+ a

-12 > a

explanation:

previous steps have made pa≅ pb ≅ st ≅ su. setting the compass width to ab allows you to make tu the same length. then δstu ≅ δpab by the sss congruence postulate, and ∠p ≅ ∠s by the reason cpctc.