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)
Implicit differentiation
chain rule is important here
I'll show the steps partially
now evaluate for (-2,2)
x=-2 and y=2
that's it, simplest form
Answer:
c = -6
d = 2
Step-by-step explanation:
After reflection about the x-axis:
A --> A'
(2,3) --> (2,-3)
(4,3) --> (4,-3)
(2,6) --> (2,-6)
After translation:
(2 + c, -6 + d) --> (-4, -4)
2+c = -4
c = -6
-6+d = -4
d = 2
steps
(x) = 5x^2-3x+7
refine
x=5x^2-3x+7
Switch sides
5x^2-3x+7-x=x-x
Subtract x from both sides
5x^2-4x+7=0
Solve with the quadratic formula
Answer:
C. Quadrant III
Step-by-step explanation:
