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

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!