x^2+6x-5=0
x^2=2x
2x+6x-5=0
2x+6x=8x
8x-5=0
+5 +5
(if you don’t know what I’m doing here is that I’m removing the negative 5 by putting a positive 5 since both of them make 0. Also the “x” always stay alone and the unit stays as the unit)
(I putted the positive under the 0 because what u do to the unit you need to do the same thing with the other.)
8x/8=5/8
(If ur lost here it means dividing.)
The answer
X=0.625
(btw I’m not really sure if this is right so um don’t really trust me in this lol)
The answer would be 50 and 37.
50-37 = 13
50+37 = 87
Proportional because it is constantly adding .99¢
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!
X=8
y=6x-4
y=6(8)-4
y=48-4
y=44
