Find the equation of the linear function L⁡(t) that goes through the points on the graph of C⁡(t) corresponding to the baseline CO2 levels in 1985 and 2010, where again t=0 represents 1965. Use the levels 345 ppm for 1985 and 389 ppm for 2010, and round the slope to 2 decimal places. Write your answer in point-slope form L⁡(t)=L0+m⁢(t−t0) with no spaces between characters.

We are asked to find the equation of the linear function \( L(t) \) that passes through two points on the graph of CO2 levels: the year 1985 (345 ppm) and 2010 (389 ppm). 

Given:
- \( t = 0 \) corresponds to 1965.
- In 1985, the CO2 level is 345 ppm, and the time \( t = 1985 - 1965 = 20 \) years.
- In 2010, the CO2 level is 389 ppm, and the time \( t = 2010 - 1965 = 45 \) years.

Now, we use these two points \((20, 345)\) and \((45, 389)\) to find the slope \( m \).

### Step 1: Calculate the slope \( m \)

The slope \( m \) of a line passing through two points \((t_1, C_1)\) and \((t_2, C_2)\) is given by:

\[
m = \frac{C_2 - C_1}{t_2 - t_1}
\]

Substituting the values:

\[
m = \frac{389 - 345}{45 - 20} = \frac{44}{25} = 1.76
\]

### Step 2: Use the point-slope form

The point-slope form of a line is:

\[
L(t) = L_0 + m(t - t_0)
\]

We can use either of the points; let's use \( (t_0, L_0) = (20, 345) \). So, the equation becomes:

\[
L(t) = 345 + 1.76(t - 20)
\]

Thus, the final equation of the linear function is:

\[
L(t) = 345 + 1.76(t-20)
\]

Let me know if you want more details or have any questions!

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Here are 5 related questions:
1. How do you convert from point-slope form to slope-intercept form?
2. What does the slope represent in this context of CO2 levels?
3. How would the equation change if the CO2 level in 2010 were higher than 389 ppm?
4. How do you find the intercept of the linear function \(L(t)\)?
5. What is the CO2 level predicted by this linear function in the year 2025?

**Tip**: When solving linear equations, always make sure to clearly define your reference points (like years and values) to avoid mistakes in calculation!

Find the equation of the linear function L(t) that goes through the points on the graph of C(t) corresponding to the baseline CO2 levels in 1985 and 2010, where again t=0 represents 1965. Use the levels 345 ppm for 1985 and 389 ppm for 2010, and round the slope to 2 decimal places. Write your answer in point-slope form L(t)=L0+m(t−t0) with no spaces between characters.

Learn how to find the equation of a linear function passing through CO2 levels in 1985 (345 ppm) and 2010 (389 ppm), using point-slope form. This problem is suitable for students in Grades 9-11 and covers algebraic concepts and slope calculation.

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