Question

4.6 module 4 post-test

the table represent a direction variation:

\

$$\begin{array}{|c|c|c|c|} \\hline x & 1 & 2 & 3 \\\\ \\hline y & 4 & 5 & 6 \\\\ \\hline \\end{array}$$

true
false

Explanation:

Define direct variation

Using the Direct Variation knowledge point, a relationship between two variables, \(x\) and \(y\), is a direct variation if it can be expressed in the form:

$$y = kx$$

where \(k\) is a non-zero constant called the constant of variation. This means the ratio of \(y\) to \(x\) must be constant for all data points:

$$\frac{y}{x} = k$$

Calculate the ratios

We test the given data points from the table to see if they share a constant ratio \(\frac{y}{x}\).

For the first data point \((1, 4)\):

$$\frac{y_1}{x_1} = \frac{4}{1} = 4$$

For the second data point \((2, 5)\):

$$\frac{y_2}{x_2} = \frac{5}{2} = 2.5$$

For the third data point \((3, 6)\):

$$\frac{y_3}{x_3} = \frac{6}{3} = 2$$

Compare the ratios

We compare the calculated ratios to check for consistency:

$$4 eq 2.5 eq 2$$

Since the ratio \(\frac{y}{x}\) is not constant, the relationship shown in the table does not represent a direct variation. Therefore, the statement is false.

Answer:

• True
• False (Correct answer)