A:
2t
t is the number of toys
b:
(p+12)/16
p is the price
y = x² - 4x + 4
0 = x² - 4x + 4
0 = x² - 2x - 2x + 4
0 = x(x) - x(2) - 2(x) - 2(-2)
0 = x(x - 2) - 2(x - 2)
0 = (x - 2)(x - 2)
0 = (x - 2)²
0 = x - 2
+ 2 + 2
2 = x
(x, y) = (2, 0)
28 divided by 1.0000 is 20
Remember
a^3-b^3=(a-b)(a^2+ab+b^2)
(11x)³-(2y)³=(11x-2y)(121x²+22xy-4y²)
Answer:
Maximum area possible
f(max) = 3906,25 ft²
Dimensions:
a = 62,5 ft
w = 62,5 ft
Step-by-step explanation:
Perimeter of the rectangular fencing P = 250 feet
And sides of the rectangle a and w (width of rectangle)
Then
A = a*w
2a + 2w = 250 ⇒ a = (250 -2w)/ 2 ⇒ a = 125 - w
f(w) = (125 - w ) *w f(w) = 125w - w²
Taking derivatives both sides of the equation
f´(w) = 125 - 2w f´(w) = 0 125 - 2w = 0
w = 125/2
w = 62,5 ft ⇒ a = 125 - 62,5
a = 62,5 ft
f(max) = ( 62,5)²
f(max) = 3906,25 ft²
