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: In particular, let’s focus our attention on the behavior of each graph at and around . 2 and x= -1 for x < 2. There are open circles at both endpoints (2, 1) and (-2, 1). The third is h (x) = 1 / (x-2)^2, in which the function curves asymptotically towards y=0 and x=2 in quadrants one and two."
Step-by-step explanation: I think this is the problem ur on
Answer:
Resulting equation will be 5x²=36.
Step-by-step explanation:
The given equations are 5y = 10x-----(1)
and x²+y²=36------(2)
Now we substitute the value of y from equation (1) into equation (2).
5y = 10x ⇒ y = 2x
Then equation (2) will be x²+(2x)²=36
x²+4x²=36 ⇒ 5x² = 36
So the equation after substitution of the value of y from equation 1 will be 5x² = 36.
Not circle for sure. Line graph is for plotting not really demonstrating a change. Picto graph is just like line plotting. Probably the bar graph works the best to see the smoothness and flow of how it increases and decreases.