PLEASE PLEASE HELP I REALLY NEED IT!! I KEEP GETTING WRONG ANSWERS You can transform ⨀T to ⨀T' by translating it and then performing a dilation centered at T'. Find the translation rule and the scale factor of the dilation.

Simplify the scale factor and write it as a proper fraction, improper fraction, or whole number.

Translation: (x, y)↦ (___ ) , (_)
,
Scale factor:

(x,y)↦ ( x + 1 ) , ( y + 9 )

scale factor 2  

Step-by-step explanation:

Step 1 identify the translation.

To do this we find the center of ⨀T and the center of ⨀T' and find the difference between their x and y values.

The center of ⨀T is at ( -1 , -7 )

The center of ⨀T' is at ( 0 , 2 )

we now find the difference between the x and y values. ( note that you must subtract the x and y values from the preimage from the translated image to get the correct translation. )

For the x values: 0 - ( - 1 ) = + 1

For the y values:  2 - ( -7 ) = + 9

So translation was (x,y)↦ ( x + 1 ) , ( y + 9 )

step 2 Find the scale factor

We can identify the scale factor by dividing the radius of the preimage by the radius of the dilated image.

The radius can be identified by finding the distance between the center and any point on the circle

The radius of the preimage is 3

The radius of the dilated image is 6

So scale factor = 6 / 3 or 2

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

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step-by-step explanation: could you give me 5 stars