Question

1. each table represents a proportional relationship. for each table:

a. fill in the missing parts of the table.
b. draw a circle around the constant of proportionality.
(from unit 2. lesson 2.)

Explanation:

Step1: Find the constant of proportionality for each table

• For the \(x - y\) table:

The constant of proportionality \(k=\frac{y}{x}\). Given \(x = 2,y=10\), then \(k=\frac{10}{2}=5\).

• For the \(a - b\) table:

Given \(a = 12,b = 3\), the constant of proportionality \(k=\frac{b}{a}=\frac{3}{12}=\frac{1}{4}\).

• For the \(m - n\) table:

Given \(m = 5,n = 3\), the constant of proportionality \(k=\frac{n}{m}=\frac{3}{5}\).

Step2: Fill in the missing values

• For the \(x - y\) table:
• If \(y = 15\), then \(x=\frac{y}{k}=\frac{15}{5}=3\).
• If \(x = 7\), then \(y=kx=5\times7 = 35\).
• If \(x = 1\), then \(y=kx=5\times1=5\).
• For the \(a - b\) table:
• If \(a = 20\), then \(b=ka=\frac{1}{4}\times20 = 5\).
• If \(b = 10\), then \(a=\frac{b}{k}=\frac{10}{\frac{1}{4}}=40\).
• If \(a = 1\), then \(b=ka=\frac{1}{4}\times1=\frac{1}{4}\).
• For the \(m - n\) table:
• If \(m = 10\), then \(n=km=\frac{3}{5}\times10 = 6\).
• If \(n = 18\), then \(m=\frac{n}{k}=\frac{18}{\frac{3}{5}}=18\times\frac{5}{3}=30\).
• If \(n = 1\), then \(m=\frac{n}{k}=\frac{1}{\frac{3}{5}}=\frac{5}{3}\).

Answer:

• \(x - y\) table:

\(x\)	\(y\)
\(3\)	\(15\)
\(7\)	\(35\)
\(1\)	\(5\)

• \(a - b\) table:

\(a\)	\(b\)
\(20\)	\(5\)
\(40\)	\(10\)
\(1\)	\(\frac{1}{4}\)

• \(m - n\) table:

\(m\)	\(n\)
\(10\)	\(6\)
\(30\)	\(18\)
\(\frac{5}{3}\)	\(1\)

• The constants of proportionality are \(5\) (for \(x - y\) table), \(\frac{1}{4}\) (for \(a - b\) table), and \(\frac{3}{5}\) (for \(m - n\) table).