Since, the polygon is a trapezoid made up of a rectangle and a right triangle. Therefore, according to the question, the figure of the polygon is attached.
Since, perimeter is the total length of the outer boundary of the figure. Therefore,
Perimeter of the polygon is


Area of the polygon = Area of Rectangle + Area of Triangle
![=[(18) \times (15)] + [(\frac{1}{2}) \times (8) \times (15)]](https://tex.z-dn.net/?f=%3D%5B%2818%29%20%5Ctimes%20%2815%29%5D%20%2B%20%5B%28%5Cfrac%7B1%7D%7B2%7D%29%20%5Ctimes%20%288%29%20%5Ctimes%20%2815%29%5D)
![=270 + [(\frac{8}{2}) \times (15)]](https://tex.z-dn.net/?f=%3D270%20%2B%20%5B%28%5Cfrac%7B8%7D%7B2%7D%29%20%5Ctimes%20%2815%29%5D)
![=270 + [4 \times (15)]](https://tex.z-dn.net/?f=%3D270%20%2B%20%5B4%20%5Ctimes%20%2815%29%5D)


Answer:
No, I don't think so.
Step-by-step explanation:
I don't think this is a function because the triangle is made of multiple points. Some of these points share the same <em>x</em> point, so this concludes that a triangle on a graph is not a function.
Hope it helps!
![x= \sqrt[8]{y} \\ y = z^{16} ](https://tex.z-dn.net/?f=x%3D%20%5Csqrt%5B8%5D%7By%7D%0A%5C%5C%20y%20%3D%20z%5E%7B16%7D%0A%0A%20)
Plug "y" into the first equation.
Answer:
B is the Y-Intercept of the graph/equation
Step-by-step explanation: