Problem 74 - Introductory Algebra for College Students

Question

The temperature, in degrees Fahrenheit, of a dessert placed in a freezer for \(t\) hours is modeled by $$ \frac{t+30}{t^{2}+4 t+1}-\frac{t-50}{t^{2}+4 t+1} $$ a. Express the temperature as a single rational expression. b. Use your rational expression from part (a) to find the temperature of the dessert, to the nearest hundredth of a degree, after 1 hour and after 2 hours.

Step-by-Step Solution

Answer

The temperature, in degrees Fahrenheit, of a dessert placed in a freezer for \(t\) hours is given by \(\frac{80}{t^{2}+4 t+1}\). After 1 hour, the temperature of the dessert will be 13.33 degrees Fahrenheit, and after 2 hours, it will be 5.71 degrees Fahrenheit, each to the nearest hundredth of a degree.

Step 1: Simplify the Expression

In the expression given, \(\frac{t+30}{t^{2}+4 t+1}-\frac{t-50}{t^{2}+4 t+1}\), two fractions with the same denominators are subtracted. To express that as a single rational expression, these fractions can be combined by subtracting the numerators. You get: \(\frac{(t+30)-(t-50)}{t^{2}+4 t+1}\). This simplifies further to \(\frac{80}{t^{2}+4 t+1}\).

Step 2: Substitute t = 1

With expression simplified as \(\frac{80}{t^{2}+4 t+1}\), calculate the temperature after 1 hour by substituting \(t=1\). The result is \(\frac{80}{1^{2}+4(1)+1} = \frac{80}{6} = 13.33\) degrees Fahrenheit, to the nearest hundredth of a degree.

Step 3: Substitute t = 2

Next, calculate the temperature after 2 hours by substituting \(t=2\) into the simplified equation. The result is \(\frac{80}{2^{2}+4(2)+1} = \frac{80}{14} = 5.71\) degrees Fahrenheit, to the nearest hundredth of a degree.