Chapter 3.4, Problem 58PPS

a.

To determine

To find:Graph represents this drop from 0 to 8 hours.

Explanation of Solution

Given information:

Staring temperature is ${80}^{0}F$

Drop of temperature per hour = $3^0$

From given information

Staring temperature is ${80}^{0}F$

  $1$ hour $\to -3^0$

Multiplying with $t$ on both sides

We get

  $\begin{array}{l}\Rightarrow 1\times t\to -3\times t\\ \Rightarrow t\to -3t\end{array}$

Therefore

For $t$ hours drop of temperature $=-3t$

Therefore

Final temperature = Drop of temperature for $t$ hours +Starting temperature.

Let $F$ be the final temperature.

  $t$ be the time in hours.

  $F=80-3t$

Graph of above equation between time $t=0$ to $t=8$ hours.

b.

To determine

To find:Equation that describes the given situation.

Answer to Problem 58PPS

Equation that represents the given situation is $F=80-3t$

F represents the final temperature after $t$ hours

  $t$ represents time in hours.

Slope = 80

Y-intercept = -3

Explanation of Solution

Given information:

Staring temperature is ${80}^{0}F$

Drop of temperature per hour = $3^0$

From given information

Staring temperature is ${80}^{0}F$

  $1$ hour $\to -3^0$

Multiplying with $t$ on both sides

We get

  $\begin{array}{l}\Rightarrow 1\times t\to -3\times t\\ \Rightarrow t\to -3t\end{array}$

Therefore

For $t$ hours drop of temperature $=-3t$

Therefore

Final temperature = Drop of temperature for $t$ hours +Starting temperature.

Let $F$ be the final temperature.

  $t$ be the time in hours.

  $F=80-3t$

Equation that represents the given situation is $F=80-3t$

Comparing $F=80-3t$ with $y=mx+c$

We get

  $\begin{array}{l}m=80\\ c=-3\end{array}$

Slope = 80

Y-intercept = -3