Solve by graphing if possible: 12n² + 11 = 0
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32 gallons of 25 % antifreeze should be combined with 8 gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze
<h3><u>Solution:</u></h3>From given,
Final mixture is 40 gallon
Let "x" be the gallon of 25 % antifreeze
Then, (40 - x) is the gallon of 50 % antifreeze
Therefore, according to question,
"x" gallons of 25 % antifreeze should be combined with (40 - x) gallons of 50 % antifreeze to get 40 gallon of 30% antifreeze
<h3><u>Therefore, we frame a equation as:</u></h3>25 % of x + 50 % of (40 - x) = 30 % of 40
Solve for "x"
Thus, 32 gallons of 25 % antifreeze is used
Then, (40 - x) = (40 - 32) = 8
Thus 8 gallons of 50 % antifreeze is used