triangle J'K'L' shown on the grid below a dilation of triangle JKL using the origin as the center of dilation:

The scale factor used to create triangle ΔJ'K'L', is 3

The reason for arriving at the above scale factor is as follows;

The known parameters:

The vertices of the triangle ΔJKL are:

J(2, 4), K(3, 2), and L(4, 3)

The vertices of the triangle ΔJ'K'L' are:

J'(6, 12), K'(9, 6), and L'(12, 9)

The required parameter:

The scale factor used to create ΔJ'K'L'

The strategy:

Find and divide the lengths of the sides of ΔJ'K'L', by the calculated lengths of the sides of the triangle ΔJKL as follows;

The lengths of the segments in ΔJKL are;

JK = √((3 - 2)² + (2 - 4)²) = √5

KL = √((4 - 3)² + (3 - 2)²) = √2

JL = √((4 - 2)² + (3 - 4)²) = √5

The lengths of the segments in ΔJ'K'L' are;

J'K' = √((9 - 6)² + (6 - 12)²) = √(45) = 3·√5

K'L' = √((12 - 9)² + (9 - 6)²) = √(18) = 3·√2

J'L' = √((12 - 6)² + (9 - 12)²) = √(45) = 3·√5

Therefore, we get;

The scale factor used to create triangle triangle ΔJKL from triangle ΔJ'K'L' , S.F. = 3

Learn more about scale factors here:

link

 b) 3

Step-by-step explanation:

From K to L the y-length is 2 and the x-length is 1

From K' to L' the y-length is 6 and the x-length is 3

Therefore, J'K'L' is 3 times as large as JKL so the dilation is 3.

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!