Below is a diagram showing a regular hexagon and a regular pentagon. each shape has a lengths of two sides shown in the terms of x or y. find which of the two shapes has the greatest perimeter. you must clearly show how you found your answer.

To solve this, we work out the perimeter of the hexagon, by finding x,  and we work out the perimeter of the pentagon by find y, and the compare them:

Perimeter of the Hexagon:
Since it is a regular hexagon, all the sides are the same: This means we can say that:

4x - 2 = x + 4       (Now lets solve this)
4x = x + 6             ( Add both sides by 2 to get 4x alone)
3x = 6                   (Subtract both sides by x, to collect the x values)
x = 2                     (Divide both sides by 3 to get what just x is)

Now we work out the length of one side by substituting the value for x in
x + 4 = 2 + 4
= 6 

Now that we know the length of one side, we can multiply that by 6 to get the perimeter, because a hexagon has six sides.

6 * 6 = 36

Perimeter of the Pentagon
Since it is a regular Pentagon, all the sides are the same, just like the hexagon. This means we can say that:

5y - 8 = 2y + 1         (Now we solve this like we did before)
5y = 2y + 9               ( Add both sides by 8 to get 5y alone)
3y = 9                       (Subtract both sides by 2y, to collect the y values)
y = 3                         (Divide both sides by 3 to get what just y is)

Now we work out the length of one side by substituting the value for x in
2y + 1 = (2 * 3) + 1
= 7

Now that we know the length of one side, we can multiply that by 5 to get the perimeter, because a pentagon has 5 sides.

5 * 7 = 35

Since the perimeter of the hexagon is 36, and the perimeter of the pentagon is 35, we can see that:

The hexagon has the larger perimeter.

In all honesty, there really is no “work” per say to show. mainly it’s just using the process of elimination. the question is asking which domain(s) would be the most appropriate for the function c(n). n is the number of observed vehicles so right away, 1, 2, and 3 can be eliminated.why? 1, 2 start in the negatives and therefore it is impossible to observe negative cars. so, #4 would be the best answer to the given problem.

Where is the qeustion