Circle 1 is centered at (-4, 5) and has a radius of 2 centimeters. circle 2 is centered at (2, 1) and has a radius of 6 centimeters. what transformations can be applied to circle 1 to prove that the circles are similar? the circles are similar because you can translate circle 1 using the transformation rule ( x, x) and then dilate it using a scale factor of ( ).

we know that
Figures can be proven similar if one, or more, similarity transformations (reflections, translations, rotations, dilations) can be found that map one figure onto another. 
In this problem to prove circle 1 and circle 2 are similar, a translation and a scale factor (from a dilation) will be found to map one circle onto another.

we have that
 

step 1
Move the center of the circle 1 onto the center of the circle 2
the transformation has the following rule
(x,y)> (x+6,y-4)
so
(-4,5)> (-4+6,5-4)> (2,1)
so
center circle 1 is now equal to center circle 2 
The circles are now concentric (they have the same center)

step 2
A dilation is needed to increase the size of circle 1 to coincide with circle 2

scale factor=radius circle 2/radius circle 1> 6/2> 3

radius circle 1 will be=2*scale factor> 2*3> 6 cm
radius circle 1 is now equal to radius circle 2 

A translation, followed by a dilation will map one circle onto the other, thus proving that the circles are similar

the answer is

b: |12 – x| ≤ 0.25

step-by-step explanation:      

let x be the possible lengths in inches, of the ribbons jaime cut.

we have been given that jaime is cutting strips of ribbon for a project. he tries to cut each piece to be 12 inches long, but sometimes he cuts it too long or too short by as much as 0.25 inch.  

we can represent this information in an inequality as:

by the definition of absolute value function we will get,

upon combining our inequalities we will get,

therefore, the inequality   describes all the possible lengths (x), in inches, of the ribbons jaime cut.

step-by-step explanation: