Answer:
The value of time for first row is 3. The value of distance for second row is 6. The value of distance for fourth row is 12.
Step-by-step explanation:
it is given that the table has equivalent ratios of time to distance.
[tex]k=\frac{distance}{time}[/tex]
From the third row we get
[tex]k=\frac{8}{12}=\frac{2}{3}[/tex]
Let the time for first row be t₁.
[tex]\frac{2}{3}=\frac{2}{t_1}[/tex]
[tex]t_1=3[/tex]
The value of time for first row is 3.
Let the distance for second row be d₂.
[tex]\frac{2}{3}=\frac{d_2}{9}[/tex]
[tex]d_2=6[/tex]
The value of distance for second row is 6.
Let the distance for second row be d₄.
[tex]\frac{2}{3}=\frac{d_4}{18}[/tex]
[tex]d_4=12[/tex]
The value of distance for fourth row is 12.
