Question

Matter is in a liquid state when its temperature is between its melting point and its boiling point. Suppose that some substance has a melting point of -48.39°C and a
boiling point of 313.31°C. What is the range of temperatures in degrees Fahrenheit for which this substance is not in a liquid state? (Hint: C=5/9(F-32)) Express the range
as an inequality
Let x represent the temperature in degrees Fahrenheit. What is the range of temperatures for which this substance is not in a liquid state?

(Type an inequality or a compound inequality. Simplify your answer. Use integers or decimals for any numbers in the expression Round to three decimal places as needed)

Answer 1

$Answer:\ -55.102^0F\leq t^0\leq 595.958^0F$

Step-by-step explanation:

$\displaystyle\\C^0=\frac{5}{9} (F^0-32^0)$

Let's multiply both parts of the equation by 9/5:

$\displaystyle\\\frac{9}{5} C^0=F^0-32^0\\\\\frac{9}{5} C^0+32^0=F^0-32^0+32^0\\\\\frac{9}{5} C^0+32^0=F^0\\\\Thus,\\\\F^0=1.8C^0+32^0$

$t=-48.39^0C\\Hence,\\t^0(F)=1.8(-48.39^0)+32^0\\t^0(F)=-87.102^0+32^0\\t^0(F)=-55.102^0$

$t=313,31^0C\\Hence,\\t^0(F)=1.8(313,31^0)+32^0\\t^0(F)=563.958^0+32^0\\t^0(F)=595.958^0$

$Thus,\\-55.102^0F\leq t^0\leq 595.958^0F$