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(.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:
And this concludes our problem :) i think....what am I forgetting?