PLEASE HELPP In Triangle XYZ, A is the midpoint of XY, B is the midpoint of YZ, and C is the midpoint of XZ.
Also, AY = 7. BZ = 8, and XZ = 18. What is the perimeter of Triangle ABC?
Make sure you explain why and show all your work.

The perimeter of triangle ABC is 33.

Explanation:

AY = 7 is half of the line of XY so we would then multiply 7 by 2:

7 x 2 = 14

and we would then do this to all of our numbers:

8 x 2 = 16

18 x 2 = 36

now we make a full equation for the answer of the perimeter of triangle XYZ:

14 + 16 + 36 = 66

66 is the perimeter of triangle XYZ.

We would then divide this answer by 2 to find the perimeter of ABC because it is half of XYZ:

66 ÷ 2 = 33

The perimeter of triangle ABC is 33

Hope this helps! :)

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!

sites: