Question

5) Each table represents a proportional relationship. (From Unit 2 Lesson 2) a) Fill in the missing parts of the table. b) Draw a circle around the constant of proportionality. a х у a b т n 2 10 12 3 15 20 10 3 735 5 10 18 1 1 1

Answer 1

Given:

The table represents a proportional relationship.

a) To find the missing values of table,

For first table,

$\begin{gathered} \frac{10}{2}=\frac{15}{x} \\ 10x=15\times2 \\ 10x=30 \\ x=\frac{30}{10}=3 \\ \frac{15}{3}=\frac{y}{7} \\ 15\times7=3y \\ 3y=105 \\ y=\frac{105}{3}=35 \\ \frac{35}{7}=\frac{y}{1} \\ 35=7y \\ y=\frac{35}{7}=5 \end{gathered}$

For second table,

$\frac{3}{12}=\frac{b}{20}=\frac{10}{a}=\frac{b}{1}$

$\begin{gathered} \frac{3}{12}=\frac{b}{20} \\ 3\times20=12y \\ b=\frac{60}{12}=5 \\ \frac{5}{20}=\frac{10}{a} \\ 5a=10\times20 \\ a=\frac{200}{5}=40 \\ \frac{10}{40}=\frac{b}{1} \\ 40b=10 \\ b=\frac{10}{40}=\frac{1}{4} \end{gathered}$

For third table,

$\begin{gathered} \frac{3}{5}=\frac{n}{10}=\frac{18}{m}=\frac{n}{1} \\ \frac{3}{5}=\frac{n}{10} \\ 30=5n \\ n=\frac{30}{5}=6 \\ \frac{3}{5}=\frac{18}{m} \\ 3m=90 \\ m=30 \\ \frac{3}{5}=\frac{n}{1} \\ 5n=3 \\ n=\frac{3}{5} \end{gathered}$

b) To draw the circle around the constant of proportionality.

For first table the constant of proportionality is 5.

For second table the constant of proportionality is 1/4.

For third table the constant of proportionality is 3/5 .