f $0.30 folders and n $1.50 notebooks total $12 Graph the equation and use the graph to determine three different combinations of folders and notebooks that total $12.

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f $0.30 folders and n $1.50 notebooks total $12 Graph the equation and use the graph to determine three different combinations of folders and notebooks that total $12.

Mathematics
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f(.30) + n(1.50) = 12. Looks like we can make a line out of this. make y-axis = n-axis, and x-axis = f-axis. solve for n I guess: n = -(1/5)f + 8 any value of "f" will give us a value of "n" that will satisfy the problem. Perferable we want whole number tho. When f=0, n=8; 8(1.50) = 12.00 is a true statement. Now, for any increment of 5 folders, we will get -1 notebooks. For example: If we add 5 folders, we should have (8-1) notebooks. n =-(1/5)(5) + 8 = -1+8 = 7 5(.30) + 7(1.50) ?=? 12.00 1.50 + 10.50 ?=? 12.00 12.00 = 12.00 true. Now there is a limit, a domain of folders if you will that we cannot stray from. Anything that makes "n" negative has no meaning. It is impossible to sell less than 0 notebooks right? And it makes no sense to sell less than 0 folders. What makes -(1/5)f + 8 < 0? 8 < (1/5)f 8(5) < f 40 < f when f > 40, the numbers no longer make any sense. So lets restrict the domain to [0,40] The combinations that will make 12.00 will be: f n ----- 0 8 5 7 10 6 15 5 20 4 25 3 30 2 35 1 40 0 And this concludes our problem :) i think....what am I forgetting?

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