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Solve each equation for the specified variable.$$F=\frac{9}{5} C+32 \text { for } C$$
$C=\frac{5}{9}(F-32)$
Precalculus
Algebra
Chapter 11
Quadratic Equations, Inequalities, and Functions
Section 4
Equations Quadratic in Form
Introduction to Conic Sections
Equations and Inequalities
Functions
Polynomials
Missouri State University
McMaster University
University of Michigan - Ann Arbor
Idaho State University
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a salvage equation for the given variable. So we're being asked to solve for F. Okay, so in the equation, the first thing we're gonna do is draw a balance line. We're going to box off our variable the f. So here we want to get everything away from that f. And since the F minus 32 is in parentheses, we're going to try to get rid of that F over nine. So what we're actually going to dio is multiply minnery rate this by nine over five on both sides. So if I multiply by nine over five over here and by nine over five over here, the reciprocal this and this are going to cancel. And I am left with nine over five times. See equals F minus 32. And now it's simple. I want to get up by itself. So we're going to add 32 to both sides. Well, we get on the left side is nine over five C plus 32 equals
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