If you have learned how to find the line of best fit manually, then you can do it that way. Perhaps you may want to just find a line that can connect at least two of the points and I believe that that line will be able to represent the other points because, in general, the points are pretty close to one another.
If you don't want to do it manually and have a graphing calculator (which I recommend) then you can use that to find the line of best fit (and if you want then you can see how precise your points are with your r^2 value). Or there is a website (http://illuminations.nctm.org/Activity.aspx?id=4186), which you can use to help you to find the equation of that particular line.
Once you have that done, then you can substitute 2009 for the x value in the equation and then see what y value the equation produces. That will then be your answer :)
Log(7x+1) - log(x-2) =1
log[(7x+1) / (x-2)] =1 or log[(7x+1) / (x-2)] =log10 (log10=1)
it means (7x+1) / (x-2) =10
Solving ===> x=7
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).
