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How much of a 50% alcohol solution should we mix with 10 gallons of a 35% solution to get a 40% solution?
Solution
Let x is the amount of 50% solution which we need. It means that there will be gallons of the 40% solution once we're done mixing the two.
Following is the basic work equation for this problem.
Now, plug in the volumes & solve for x.
0.5x + 0.35 (10) =0.4 ( x + 10)
0.5x + 3.5 + 0.4x + 4
0.1x + 0.5
x =0.5/0.1 = 5 gallons
Thus, we required 5 gallons of the 50% solution to get a 40% solution.
I get how to do it
Actually these problems are variants of the Distance/Rate problems which we just got done working. The standard equation which will be required for these problems is, As y
f(x)=1\2x+3
(xy)-3/(x^-5y)3
To find the calorie density (D) in calories per ounce of food that contains c carories and weight w ounces is given by D=C/w
Example Solve out each of the following equations. 7 x = 9 Solution Okay, although we say above that if we contained a logarithm in fron
graph the following and find the point of intersection 2x+y=-4 y+2x=3
2x-1=10 I got 9/2 but i''m not sure if that is really right
27^(3X)-3 = (1/81)^(10X)-9
are these like terms
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