Question

lesson 4 assignment

use a separate piece of paper for your journal entry.

journal
define constant of
proportionality in your own
words. provide a specific
example with your definition.
remember
if y is proportional to x, you can represent the
relationship by the equation y = kx, where k is the
constant of proportionality.
practice
1 analyze each table or problem situation to determine whether the relationship is
proportional. state a constant of proportionality if possible. show your work.
2 given a value for the input variable, x, and the output variable, y, calculate the
constant of proportionality.

Explanation:

Step1: For part 1a

Check the ratio of dogs to cats.
For the first row: \(\frac{14}{7} = 2\)
For the second row: \(\frac{21}{9}=\frac{7}{3}\approx2.33\)
Since the ratios are not equal, the relationship is not proportional.

Step2: For part 1b

Let \(x\) be the number of days and \(y\) be the weight.
At \(x = 0\), \(y=5520\). For a proportional relationship \(y = kx\), when \(x = 0\), \(y = 0\). But here \(y
eq0\) when \(x = 0\). So the relationship is not proportional.

Step3: For part 2a

Given \(y=kx\), so \(k=\frac{y}{x}\). Substitute \(x = 21\) and \(y = 6\), \(k=\frac{6}{21}=\frac{2}{7}\)

Step4: For part 2b

Substitute \(x = 60\) and \(y = 18\) into \(k=\frac{y}{x}\), \(k=\frac{18}{60}=\frac{3}{10}\)

Step5: For part 2c

First, convert \(x = 2\frac{2}{5}=\frac{12}{5}\) and \(y = 7\frac{1}{2}=\frac{15}{2}\)
Then \(k=\frac{y}{x}=\frac{\frac{15}{2}}{\frac{12}{5}}=\frac{15}{2}\times\frac{5}{12}=\frac{75}{24}=\frac{25}{8}\)

Step6: For part 2d

Convert \(x = 4\frac{8}{11}=\frac{52}{11}\) and \(y = 3\frac{6}{11}=\frac{39}{11}\)
Then \(k=\frac{y}{x}=\frac{\frac{39}{11}}{\frac{52}{11}}=\frac{39}{52}=\frac{3}{4}\)

Answer:

1a. Not proportional.
1b. Not proportional.
2a. \(\frac{2}{7}\)
2b. \(\frac{3}{10}\)
2c. \(\frac{25}{8}\)
2d. \(\frac{3}{4}\)