Question

1. the following tables are from a linear relationship. fill in the blanks

a)
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$$\begin{tabular}{|c|c|c|c|c|c|} \\hline x & 0 & 1 & 2 & 3 & 4 \\\\ \\hline y & & 5 & & & 20 \\\\ \\hline \\end{tabular}$$

b)
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$$\begin{tabular}{|c|c|c|c|c|c|} \\hline x & 0 & 5 & 10 & 15 & 20 \\\\ \\hline y & 40 & & & 28 & \\\\ \\hline \\end{tabular}$$

the point (1,6) lies on a line that has a rise of 4 and a run of 3. determine two

Explanation:

Part a)

Step1: Identify the relationship

From the table, when \( x = 1 \), \( y = 5 \); when \( x = 4 \), \( y = 20 \). We can see that \( y = 5x \) (since \( 5\times1 = 5 \) and \( 5\times4 = 20 \)).

Step2: Find \( y \) when \( x = 0 \)

Substitute \( x = 0 \) into \( y = 5x \), we get \( y = 5\times0 = 0 \).

Step3: Find \( y \) when \( x = 2 \)

Substitute \( x = 2 \) into \( y = 5x \), we get \( y = 5\times2 = 10 \).

Step4: Find \( y \) when \( x = 3 \)

Substitute \( x = 3 \) into \( y = 5x \), we get \( y = 5\times3 = 15 \).

Step1: Find the slope

First, find the slope \( m \) using the points. When \( x = 0 \), \( y = 40 \); when \( x = 15 \), \( y = 28 \). The slope \( m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{28 - 40}{15 - 0}=\frac{- 12}{15}=-\frac{4}{5}\). So the equation is \( y - 40=-\frac{4}{5}(x - 0) \), which simplifies to \( y = -\frac{4}{5}x + 40 \).

Step2: Find \( y \) when \( x = 5 \)

Substitute \( x = 5 \) into \( y = -\frac{4}{5}x + 40 \), we get \( y=-\frac{4}{5}\times5 + 40=- 4 + 40 = 36 \).

Step3: Find \( y \) when \( x = 10 \)

Substitute \( x = 10 \) into \( y = -\frac{4}{5}x + 40 \), we get \( y=-\frac{4}{5}\times10 + 40=- 8 + 40 = 32 \).

Step4: Find \( y \) when \( x = 20 \)

Substitute \( x = 20 \) into \( y = -\frac{4}{5}x + 40 \), we get \( y=-\frac{4}{5}\times20 + 40=- 16 + 40 = 24 \).

Answer:

When \( x = 0 \), \( y = 0 \); when \( x = 2 \), \( y = 10 \); when \( x = 3 \), \( y = 15 \). So the filled table is:

\( x \)	0	1	2	3	4

Part b)