Question

What is the rate of change of the table below?

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COURSE OUTLINE

COURSE TOOLS

esc

xy

112

2 19

3 26

4 33

5 40

6 47

(1 point)

3

07

12

O-12
1/7

Answer 1

To determine the rate of change from the table provided, we need to look at how the values change in relation to the input values.

From the data:

- When \( x = 2 \), \( y = 19 \)
- When \( x = 3 \), \( y = 26 \)
- When \( x = 4 \), \( y = 33 \)
- When \( x = 5 \), \( y = 40 \)
- When \( x = 6 \), \( y = 47 \)

To find the rate of change, calculate the change in \( y \) divided by the change in \( x \).

For example:
- From \( x = 2 \) to \( x = 3 \):
\[
\Delta y = 26 - 19 = 7, \quad \Delta x = 3 - 2 = 1 \quad \Rightarrow \quad \text{Rate of change} = \frac{\Delta y}{\Delta x} = \frac{7}{1} = 7
\]

- From \( x = 3 \) to \( x = 4 \):
\[
\Delta y = 33 - 26 = 7, \quad \Delta x = 4 - 3 = 1 \quad \Rightarrow \quad \text{Rate of change} = \frac{7}{1} = 7
\]

- From \( x = 4 \) to \( x = 5 \):
\[
\Delta y = 40 - 33 = 7, \quad \Delta x = 5 - 4 = 1 \quad \Rightarrow \quad \text{Rate of change} = \frac{7}{1} = 7
\]

- From \( x = 5 \) to \( x = 6 \):
\[
\Delta y = 47 - 40 = 7, \quad \Delta x = 6 - 5 = 1 \quad \Rightarrow \quad \text{Rate of change} = \frac{7}{1} = 7
\]

The rate of change is consistent across all intervals and is equal to **7**.