Question content area top

Question

Part 1
Find the domain and the range of the following relation. Also determine whether the relation is a function.
203 degrees
95 degrees
248 degrees
212 degrees
95 degrees
35 degrees
100 degrees
120 degrees
Input:
Output:
Degrees
Degrees
Farenheit
Celsius
Question content area bottom
Part 1
Choose the domain below.
A.StartSet 203 degrees comma 95 degrees comma 248 degrees comma 212 degrees EndSet
StartSet 203 degrees comma 95 degrees comma 248 degrees comma 212 degrees EndSet
B.95 degrees to 248 degrees
95 degrees to 248 degrees
C.StartSet 95 degrees comma 35 degrees comma 120 degrees comma 100 degrees EndSet
StartSet 95 degrees comma 35 degrees comma 120 degrees comma 100 degrees EndSet
D.
None of the above

Answer 1

To determine the domain and range of the given relation, we first need to establish the input and output values. The inputs appear to be degrees Fahrenheit and the outputs are degrees Celsius.

The input values given are:

• 203°F
• 95°F
• 248°F
• 212°F
• 95°F (repeated)
• 35°F
• 100°F
• 120°F

Let's summarize the input and output:

Input (Domain)

The unique input values (in degrees Fahrenheit) are:

• 203°F
• 95°F
• 248°F
• 212°F
• 35°F
• 100°F
• 120°F

So the domain is: \[ \text{Domain} = { 203, 95, 248, 212, 35, 100, 120 } \]

Looking at the options:

• A. \( { 203, 95, 248, 212 } \) (missing some)
• B. 95 degrees to 248 degrees (not appropriate as it suggests a continuous range)
• C. \( { 95, 35, 120, 100 } \) (missing some)
• D. None of the above (this appears to be the correct choice)

So, the correct answer for the domain is D. None of the above.

Output (Range)

For the output values (which need temperatures in Celsius), we would need to convert the given Fahrenheit values to Celsius. The conversion formula from Fahrenheit (F) to Celsius (C) is:

\[ C = \frac{5}{9} \times (F - 32) \]

Let's perform the conversions:

1. 203°F:
   \[ C = \frac{5}{9} \times (203 - 32) = \frac{5}{9} \times 171 = 95 \text{°C} \]

2. 95°F:
   \[ C = \frac{5}{9} \times (95 - 32) = \frac{5}{9} \times 63 = 35 \text{°C} \]

3. 248°F:
   \[ C = \frac{5}{9} \times (248 - 32) = \frac{5}{9} \times 216 = 120 \text{°C} \]

4. 212°F:
   \[ C = \frac{5}{9} \times (212 - 32) = \frac{5}{9} \times 180 = 100 \text{°C} \]

5. 35°F:
   \[ C = \frac{5}{9} \times (35 - 32) = \frac{5}{9} \times 3 \approx 1.67 \text{°C} \]

6. 100°F:
   \[ C = \frac{5}{9} \times (100 - 32) = \frac{5}{9} \times 68 \approx 37.78 \text{°C} \]

7. 120°F:
   \[ C = \frac{5}{9} \times (120 - 32) = \frac{5}{9} \times 88 \approx 48.89 \text{°C} \]

From these conversions, the unique Celsius outputs are approximately:

• 95°C from 203°F
• 35°C from 95°F
• 120°C from 248°F
• 100°C from 212°F
• 1.67°C from 35°F
• 37.78°C from 100°F
• 48.89°C from 120°F

Thus, the range is: \[ \text{Range} = { 95, 35, 120, 100, 1.67, 37.78, 48.89 } \]

Function Check

A relation is a function if each input has exactly one output. Here, all input temperatures (even with the repeated 95°F) correspond to unique outputs, so this relation is a function.

Summary

• Domain: \( D. \text{None of the above} \)
• Range: \( { 95, 35, 120, 100, 1.67, 37.78, 48.89 } \)
• Function: Yes, it is a function.