Question 1029981
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Let *[tex \Large x] represent the amount of 40% solution.  Then the amount of 70% solution must be *[tex \Large 150\ -\ x].  The amount of pure acid in the 40% solution is then *[tex \Large 0.40x] and the amount of pure acid in the 70% solution is *[tex \Large 0.70(150\ -\ x)].  The amount of pure acid in the 46% solution is *[tex \Large 0.46(150)].  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.40x\ +\ 0.70(150\ -\ x)\ =\ 0.46(150)]


Solve for *[tex \Large x] and then calculate *[tex \Large 150\ -\ x]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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