On a linear X temperature scale water freezes at -119.0°X and boils at 339.0°X. On a linear Y temperature scale, water freezes -84.00°Y and the boils at -39.00°Y. A temperature of 42.000°Y corresponds to what temperature on the X scale?

We are dealing with two linear temperature scales, \( X \) and \( Y \), where the freezing and boiling points of water are given. The problem asks to convert a specific temperature from the \( Y \) scale to the \( X \) scale.

### Step 1: Set up the linear relationship between the two scales
Since the temperature scales are linear, the relationship between \( X \) and \( Y \) can be expressed in the form:
\[
X = mY + b
\]
where \( m \) is the slope and \( b \) is the intercept.

### Step 2: Determine the slope \( m \)
We know the freezing and boiling points on both scales:
- Freezing point: \( -119.0^\circ X \) corresponds to \( -84.00^\circ Y \),
- Boiling point: \( 339.0^\circ X \) corresponds to \( -39.00^\circ Y \).

Using these two points, we can calculate the slope \( m \) as:
\[
m = \frac{\Delta X}{\Delta Y} = \frac{339.0 - (-119.0)}{-39.00 - (-84.00)} = \frac{458.0}{45.00} = 10.18
\]

### Step 3: Determine the intercept \( b \)
We can use one of the known points to solve for \( b \). Using the freezing point \( (-119.0^\circ X, -84.00^\circ Y) \):
\[
-119.0 = 10.18(-84.00) + b
\]
\[
-119.0 = -855.12 + b
\]
\[
b = 736.12
\]

### Step 4: Find the temperature on the \( X \) scale corresponding to \( 42.000^\circ Y \)
Now that we have the equation \( X = 10.18Y + 736.12 \), we can substitute \( Y = 42.000 \) into this equation:
\[
X = 10.18(42.000) + 736.12
\]
\[
X = 427.56 + 736.12 = 1163.68^\circ X
\]

### Conclusion
A temperature of \( 42.000^\circ Y \) corresponds to \( 1163.68^\circ X \).

Would you like more details on any step?

---

Here are 5 related questions:
1. How can you derive the conversion formulas between other temperature scales?
2. What are other temperature scales similar to \( X \) and \( Y \)?
3. How do you find the slope of a line given two points?
4. Why do we use a linear equation for temperature conversions?
5. How do temperature scales like Celsius and Fahrenheit compare to linear scales like \( X \) and \( Y \)?

**Tip:** When dealing with temperature scales, always check if they are linear or logarithmic, as this affects the conversion method.

Learn how to convert a temperature of 42.000°Y to the X scale using a linear relationship between two temperature scales. This problem covers the calculation of slope, intercept, and application of the slope-intercept formula. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation