Answer:
1. 180-(46+14)=120
2.46+14+120=180
Step-by-step explanation:
Basically, what this asks you is to maximize the are A=ab where a and b are the sides of the recatangular area (b is the long side opposite to the river, a is the short side that also is the common fence of both corrals). Your maximization is constrained by the length of the fence, so you have to maximize subject to 3a+b=450 (drawing a sketch helps - again, b is the longer side opposite to the river, a are the three smaller parts restricting the corrals)
3a+b = 450
b = 450 - 3a
so the maximization max(ab) becomes
max(a(450-3a)=max(450a-3a^2)
Since this is in one variable, we can just take the derivative and set it equal to zero:
450-6a=0
6a=450
a=75
Plugging back into b=450-3a yields
b=450-3*75
b=450-225
b=215
Hope that helps!
The answer is a.
x^2 -10x+25 = x*x-2*5*x+5*5 => (x-5)(x-5)
Answer:
C
Step-by-step explanation:
(2*(-5)^-1*3^5)^2/(3*-5^0*3^4)^2 =4/25
