Yes. If the segment only has 2 points, it can be named both ways.
Side A’B’ and D’C’ are related because they are both parallel which I think means that they would be the same length
F(x)= 9x^3 + 2x^2 - 5x + 4
g(x)= 5x^3 - 7x + 4
f(x) - g(x)
9x³ + 2x² - 5x + 4 - (5x³ - 7x + 4)
9x³ + 2x² - 5x + 4 - 5x³ +7x - 4
9x³ - 5x³ + 2x² -5x + 7x + 4 - 4
4x³ + 2x² + 2x
Answer:
a) see below
b) 40x20 meters
Step-by-step explanation:
Write down what you know:
- The area of the enclosure is length*width, so
![A = x \cdot y](https://tex.z-dn.net/?f=A%20%3D%20x%20%5Ccdot%20y)
- The length of the fencing is 80 meters, so
![2y + x = 80](https://tex.z-dn.net/?f=2y%20%2B%20x%20%3D%2080)
Now we have to combine these two equations above, and get rid of y in the process.
First rewrite the second as:
![y = 40 - \frac12 x](https://tex.z-dn.net/?f=y%20%3D%2040%20-%20%5Cfrac12%20x)
Then substitute for y in the first:
![A = x (40 - \frac12 x) = 40x - \frac12x^2](https://tex.z-dn.net/?f=A%20%3D%20x%20%2840%20-%20%5Cfrac12%20x%29%20%3D%2040x%20-%20%5Cfrac12x%5E2)
b) To maximize A, find the zero of the first derivative:
![A'(x) = 40 - x = 0 \implies x=40](https://tex.z-dn.net/?f=A%27%28x%29%20%3D%2040%20-%20x%20%3D%200%20%5Cimplies%20x%3D40)
So y = (80-40)/2 = 20 meters.