Question

(01.05 mc)
a truck driver is driving from nome, alaska to death valley, california. because he is traveling between locations with extreme temperatures, he needs to check the weather continuously to make sure the gas in his truck remains in liquid form. the gas he uses freezes at -40°f and evaporates at 140°f.
part a: write an inequality to represent the temperatures at which the gas in the truck will remain in liquid form. (2 points)
part b:describe the graph of the inequality completely from part a. use terms such as open/closed circles and shading directions. explain what the solutions to the inequality represent. (4 points)
part c: in january 1989, the temperature in nome, alaska dropped to -49°f. would the gas in the drivers truck have remained in liquid form so he could have driven on this day? why or why not? (4 points)

Explanation:

Part A:

The gas freezes at - 40°F and evaporates at 140°F. For the gas to remain in liquid - form, the temperature $t$ must be greater than - 40°F and less than 140°F.
The inequality is $-40 < t<140$.

Part B:

On a number - line, for the inequality $-40 < t<140$, we use open circles at - 40 and 140. Open circles are used because the gas does not remain liquid at exactly - 40°F (it freezes) and exactly 140°F (it evaporates). The shading is between the two open circles. The solutions to the inequality represent all the temperatures at which the gas in the truck will remain in liquid form.

Part C:

We check if $t=-49$ satisfies the inequality $-40 < t<140$.
Since $-49<-40$, the value $t = - 49$ does not satisfy the inequality $-40 < t<140$. So the gas in the driver's truck would not have remained in liquid form on this day because the temperature was below the freezing point of the gas.

Answer:

Part A: $-40 < t<140$
Part B: Open circles at - 40 and 140, shading between them. Solutions represent liquid - state temperatures.
Part C: No, because $-49 < - 40$, the temperature was below the freezing point of the gas.