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.
\(k=\frac{distance}{time}\)
From the third row we get
\(k=\frac{8}{12}=\frac{2}{3}\)
Let the time for first row be t₁.
\(\frac{2}{3}=\frac{2}{t_1}\)
\(t_1=3\)
The value of time for first row is 3.
Let the distance for second row be d₂.
\(\frac{2}{3}=\frac{d_2}{9}\)
\(d_2=6\)
The value of distance for second row is 6.
Let the distance for second row be d₄.
\(\frac{2}{3}=\frac{d_4}{18}\)
\(d_4=12\)
The value of distance for fourth row is 12.
