Answer:
Step-by-step explanation:
(x) = (1100 + x) (100 - .05(x-1100))
This is a quadratic, graphs as a parabola that opens downward. A maximum cam be found.
The zeros of the function are
(1100 + x) = 0 ..... or ..... [100 - .05(x-1100)] = 0
x = -1100 is the left x-intercept.
[100 - .05(x-1100)] = 0
100 = .05(x-1100)
2000 = x - 1100
x = 3100 is the right intercept.
Maximization of profits is at the mid point of the zeros (x-intercepts)
(3100 + -1100)/2 = 1000
1100 + 1000 = 2100 trees should be planted to maximize profits.
f(x) = (1100 + 1000) (100 - .05(1000-1100))
f(x) = (2000) (105) = 220,500 is the maximum profit.
I hope this helps!
First factor the numerator
b^2-6b+8/b-4 * b+8/b-2
u will get
(b-2)(b-4)/(b-4)*(b-2)*b+8
those will cancel out each other will left only
b+8<span />
Answer:
The answer is $40.
Step-by-step explanation:
According to the equation given in the question, we can assume that 550 is constant and was there when Mai started saving into a checking account.
Then as x gets increased by 1 each week, the amount of change in the account per week is $40.
I hope this answer helps.
I believe 10% because 390 and 39 are exactly the same numbers almost
