Hexagon ijklmn is shown on the coordinate plane below:

hexagon ijklmn on a coordinate plane with ordered pairs at i negative two, six, at j four, four, at k six, negative one at, l n

if hexagon ijklmn is dilated by a scale factor of two fifths from the origin to create hexagon i'j'k'l'm'n', what is the ordered pair of point i'? (4 points)

a) (−0.8, 2.4)

b) (−5, 15)

c) (−2.5, 5)

d) (−0.4, 0.8)

Option: A is the correct answer.

The ordered pair of point I' are:

         A)   (-0.8,2.4)

We know than if any point A with vertices as (a,b) are dilated by a scale factor of k then the vertices of the transformed point is:

       A(a,b) → A'(ka,kb)

We are given vertices of point I as:

                   I(-2,6)

Hence, the vertices of the transformed point I' after dilating by a scale factor of 2/5 is:

         I(-2,6) → I'(-2×(2/5),6×(2/5))

⇒  I(-2,6) → I'(-0.8,2.4)

      Hence, option: A is the answer.

                 A)  (-0.8,2.4)                              

(-0.8,2.4)

Step-by-step explanation:

We know than if any point (a,b) are dilated by a scale factor of k from the origin then the vertices of the image point is :

Given : Hexagon IJKLMN on a coordinate plane with ordered pairs at I (-2,6).

Then, the coordinates of the image point I' after dilating by a scale factor of will be :

Hence, the ordered pair of point I' =(-0.8, 2.4)

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !