Mathlamaredutermsaspx college algebra do not get

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Unformatted text preview: x + 10 gallons of the 40% solution once we’re done mixing the two. Here is the basic work equation for this problem. æ Amount of alcohol ö æ Amount of alcohol ö æ Amount of alcohol ö ç ÷+ç ÷=ç ÷ è in 50% Solution ø è in 35% Solution ø è in 40% Solution ø æ Volume of ö æ Volume of ö æ Volume of ö ÷ + ( 0.35 ) ç ÷ = ( 0.4 ) ç ÷ è 50% Solution ø è 35% Solution ø è 40% Solution ø ( 0.5 ) ç Now, plug in the volumes and solve for x. 0.5 x + 0.35 (10 ) = 0.4 ( x + 10 ) 0.5 x + 3.5 = 0.4 x + 4 0.1x = 0.5 0.5 x= = 5gallons 0.1 So, we need 5 gallons of the 50% solution to get a 40% solution. Example 11 We have a 40% acid solution and we want 75 liters of a 15% acid solution. How much water should we put into the 40% solution to do this? Solution Let x be the amount of water we need to add to the 40% solution. Now, we also don’t how much of the 40% solution we’ll need. However, since we know the final volume (75 liters) we will know that we will need 75 - x liters of the 40% solution. Here is the word equation for this problem. æ Amount o...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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