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The instructions on a container of antifreeze state, "A 50/50 mixture of antifreeze and water protects against freeze-ups down to $-34^{\circ} \mathrm{F},$ while a $60 / 40 \mathrm{mix}$ protects against freeze-ups down to one and one-half times that temperature." To what temperature does the $60 / 40$ mixture protect?
$$-51^{\circ} \mathrm{F}$$
Algebra
Chapter 1
An Introduction to Algebra
Section 6
Multiplying and Dividing Real Numbers; Multiplication and Division Properties
The Language of Algebra
Equations and Inequalities
Functions
Linear Functions
Quadratic Functions
Polynomials
Campbell University
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Idaho State University
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this problem tells us that a 50 50 mixture of antifreeze I will protect down to 34 degrees. Now. Then it says a 60 40 that will protect down to 1.5 times that amount. So, uh, Thio what? Asked what temperature is done. So we just have to multiply a negative 34 and 1 25 So 34 times 1.5 is gonna be negative because it's a negative times positive. And then half of 34 17. So 1.5 times 34 is negative. 51 degrees. Okay, So the 60 40 ratio of antifreeze I give us protection down to negative 51 F. Thank you very much.
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