Expanding the limit, we get (x^2+2x∆x+∆x^2-2x-2∆x+1-x^2+2x-1)/<span>∆x
Crossing the 1s , the 2xs, and the x^2s out, we get
(2x</span>∆x+∆x^2-2∆x)/<span>∆x
Dividing the </span><span>∆x, we get
2x+</span><span>∆x-2.
Making the limit of </span><span>∆x=0, we get 2x-2.</span>
Answer:
-40 square root of 10k²
Step-by-step explanation:
Answer:
- 50 ft by 75 ft
- 3750 square feet
Step-by-step explanation:
Let x represent the length of the side not parallel to the partition. Then the length of the side parallel to the partition is ...
y = (300 -2x)/3
And the enclosed area is ...
A = xy = x(300 -2x)/3 = (2/3)(x)(150 -x)
This is the equation of a parabola with x-intercepts at x=0 and x=150. The line of symmetry, hence the vertex, is located halfway between these values, at x=75.
The maximum area is enclosed when the dimensions are ...
50 ft by 75 ft
That maximum area is 3750 square feet.
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<em>Comment on the solution</em>
The generic solution to problems of this sort is that half the fence (cost) is used in each of the orthogonal directions. Here, half the fence is 150 ft, so the long side measures 150'/2 = 75', and the short side measures 150'/3 = 50'. This remains true regardless of the number of partitions, and regardless if part or all of one side is missing (e.g. bounded by a barn or river).
Answer:
fyi b's answer has imaginary numbers in it...
Imaginary: 1 +
Imaginary: 1 -
Step-by-step explanation:
= 
... the negative root will produce imaginary solutions
