For the equation F = (9/5)C + 32,F and C are | vedclass

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(D) Given the equation $F = (9/5)C + 32$. Since $F$ and $C$ are numerically equal,let $F = C = x$. Substituting $x$ for $F$ and $C$ in the equation: $

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$-40$

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(D) Given the equation $F = (9/5)C + 32$. Since $F$ and $C$ are numerically equal,let $F = C = x$. Substituting $x$ for $F$ and $C$ in the equation: $x = (9/5)x + 32$ Subtracting $(9/5)x$ from both sides: $x - (9/5)x = 32$ $(5x - 9x) / 5 = 32$ $-4x / 5 = 32$ $-4x = 32 	imes 5$ $-4x = 160$ $x = 160 / -4$ $x = -40$ Therefore,$F$ and $C$ are numerically equal when $F = -40$.

For the equation $F = (9/5)C + 32$,$F$ and $C$ are numerically equal when $F = \ldots$

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