Evaluate the function at the given point
2 answers:
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She has 10lbs of 25% syrup... so, in the 10lbs, 25% of that is syrup, the rest, namely the 75% remaining is water or other substances.
let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.
how much is 25% of 10lbs? well, (25/100) * 10, or 2.5.
the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs? well, (0/100) * x, or 0.00x, which is just 0.
how much is 10% of "y" lbs? well (10/100) * y, or 0.10y.
whatever "x" and "y" are, we know that 10 + x = y.
we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus
let's say she adds "x" lbs of water, to get "y" lbs for the 10% mixture.
how much is 25% of 10lbs? well, (25/100) * 10, or 2.5.
the water has no sugar syrup in it, so is just pure water, so the amoun of syrup in it is 0%, how much is 0% of "x" lbs? well, (0/100) * x, or 0.00x, which is just 0.
how much is 10% of "y" lbs? well (10/100) * y, or 0.10y.
whatever "x" and "y" are, we know that 10 + x = y.
we also know that the syrup amount in that is also 2.5 + 0.00x = 0.10y, thus