Question

Linear Change Practice
Complete this assessment to review what you’ve learned. It will not count toward your grade.
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Question
Use the table to answer the question.

Let x
represent the time in minutes and let y
represent the distance in feet.

Time (Minutes), x


1 2 3 4
Distance (Feet), y


75 150 225 300
Use the table to create a formula for the given information.

(1 point)

Answer 1

To create a formula based on the provided table, we first observe the relationship between \( x \) (time in minutes) and \( y \) (distance in feet).

Given the data:

- When \( x = 1 \), \( y = 75 \)
- When \( x = 2 \), \( y = 150 \)
- When \( x = 3 \), \( y = 225 \)
- When \( x = 4 \), \( y = 300 \)

We can see that the distance \( y \) increases by 75 feet for each additional minute. This indicates a linear relationship, where the rate of change (slope) is constant.

To find the slope \( m \):

\[
m = \frac{\Delta y}{\Delta x} = \frac{150 - 75}{2 - 1} = \frac{75}{1} = 75
\]

This shows that for every 1 minute (for each increment of \( x \)), the distance \( y \) increases by 75 feet.

Next, we can use the slope-intercept form of a linear equation, which is given by:

\[
y = mx + b
\]

Where:
- \( m = 75 \) (the slope)
- \( b \) is the y-intercept (the value of \( y \) when \( x = 0 \)).

To find \( b \), we can look at one of the given points. When \( x = 1 \) and \( y = 75 \):

\[
75 = 75(1) + b
\]
\[
75 = 75 + b
\]
\[
b = 75 - 75 = 0
\]

Now we can plug in the values of \( m \) and \( b \) into the equation:

\[
y = 75x + 0
\]

Thus, the formula relating distance to time in this case is:

\[
y = 75x
\]