Find the perimeter of the polygon with the given vertices. round your answer to the nearest hundredth.
the perimeter is .

The perimeter is about 16.94 units.

Step-by-step explanation:

we know that

The perimeter of the polygon is the sum of its length sides

so

the formula to calculate the distance between two points is equal to

step 1

Find the distance AB

A(0,4),B(2,0)

substitute in the formula

step 2

Find the distance BC

B(2,0),C(2,-2)

substitute in the formula

step 3

Find the distance CD

C(2,-2),D(0,-2)

substitute in the formula

step 4

Find the distance DE

D(0,-2),E(-2,2)

substitute in the formula

step 5

Find the distance EF

E(-2,2),F(-2,4)

substitute in the formula

step 6

Find the  distance FA

F(-2,4),A(0,4)

substitute in the formula

step 7

Find the perimeter

answer

see attachment

explanation

the given function is

this function has y-intercept at

-4 and a slope of 2.

note that, the graph has a positive slope.

the graph that represents this function is the straight line that moves upward from left to right.

the graph touches the y-axis at -4.

the correct choice is the graph shown in the attachment.

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.