Chapter ISG, Problem 3.1.2.3P

a.

To determine

To write a function for the given situation

Answer to Problem 3.1.2.3P

The function is, $h(x)=80+10x$

Explanation of Solution

Given:

The initial temperature is 80°F

10°F temperature is increased every minute

Concept Used: function formation

Calculation:

Let $h(x)$ be the function

Where “x” is the time, in minutes.

Hence, substituting the values and forming the function.

  $h(x)=80+10x$

Conclusion:The function is, $h(x)=80+10x$

b.

To determine

To calculate the temperature, at 22mins.

Answer to Problem 3.1.2.3P

The temperature at 22mins is, 300°F

Explanation of Solution

Given:

The function is $h(x)=80+10x$

Time (x) is 22mins.

Concept Used:

Solving the function

Calculation:

Substituting the values in the function,

  $h(22)=80+(10\times 22)$

So,

  $h(22)=80+220$

On solving,

  $h(22)=300$

Hence, temperature is 300°F

Conclusion:The temperature at 22mins is, 300°F

c.

To determine

To find a function for cooling of the temperature.

Answer to Problem 3.1.2.3P

The equation is $c(x)=300-5x$

Explanation of Solution

Given:The temperature after which cooling starts is 300°F

5°F is reduced each minute.

Concept Used:Function formation

Calculation:

Let $c(x)$ be the function

Where “x” is the time, in minutes.

Hence, substituting the values and forming the function.

  $c(x)=300-5x$

Conclusion:The equation for cooling is $c(x)=300-5x$

d.

To determine

To calculate the time the chamber takes to reach 80°F

Answer to Problem 3.1.2.3P

It takes 44mins to reach at the 80°F

Explanation of Solution

Given:The function is $c(x)=300-5x$

The final temperature( c(x)) is 80°F

Concept Used:Solving the function

Calculation:

Substituting the values,

  $80=300-5x$

Exchanging the right-hand side and left- hand side values,

  $5x=300-80$

On solving,

  $5x=220$

Cross-multiplying,

  $x=\frac{220}{5}$

On solving,

  $x=44$

Hence, time is 44mins

Conclusion:It takes 44mins to reach at the 80°F