Answer:
67,500 m²
Step-by-step explanation:
ASSUMING the fields look like this __________________
| | |
| | | W
|_________|_________|
L
Let L be the length of the combined field and W be the width
2L + 3W = 1800
2L = 1800 - 3W
L = 900 - 1.5W
A = LW
A = (900 - 1.5W)W
A = 900W - 1.5W²
Area will be maximized when the derivative equals zero.
dA/dW = 900 - 3W
0 = 900 - 3W
3W = 900
W = 300 m
L = 900 - 1.5(300)
L = 450 m
A = LW = 450(300) = 135,000 m²
so each sub field is 135000/2 = 67,500 m²
a. The difference between two outputs that are 1 unit apart.
You need to Use y2 - y1 / x2 - x1 to find the difference
I will choose x2 as 1 and x1 as 0
(29 - 21) / (1 - 0)
8/1 so The difference is 8 per 1 unit.
b. Use the same formula
I will choose -3 as x2 and -5 as x1
(5 - (-11)) / (-3 - (-5))
(5 + 11) / (-3 + 5)
16 / 2 so the difference is 16 per 2 units.
c. I will choose 2 as x2 and -1 as x1
(45 - 21) / (2 - (-1))
24/3 so the difference is 24 per 3 units.
d. The ratios of the differences to the input intervals reduced all equal each-other, which is 8 per 1 unit.
Let's say g(x) = x^4 => f(g)(x) = 4*( x^4) +1.