Question

a snail is moving away from a rock at a constant rate. this table shows the distance the snail is from the rock at certain times.

distance (in.)	time (min)
1	$1\frac{1}{3}$

select all the equations that represent the relationship between the distance in inches, $d$, and time in minutes, $t$.
$d=\frac{4}{3}t$ $d=\frac{3}{4}t$ $t=\frac{4}{3}d$ $t=\frac{3}{4}d$

Explanation:

Step1: Find the rate of the snail's movement

The snail moves from distance \(d = 0\) at \(t = 0\) to \(d = 1\) at \(t = 1\frac{1}{3}=\frac{4}{3}\) minutes. The rate (slope) of the linear relationship between \(d\) and \(t\) is \(\frac{d}{t}\) (since it's a proportional relationship, \(d = kt\) where \(k\) is the rate). So \(k=\frac{d}{t}=\frac{1}{\frac{4}{3}}=\frac{3}{4}\). So the equation \(d=\frac{3}{4}t\) is valid.

Step2: Derive the inverse relationship

From \(d=\frac{3}{4}t\), we can solve for \(t\) by multiplying both sides by \(\frac{4}{3}\). So \(t=\frac{4}{3}d\) is also valid. Let's check the other equations: For \(d = \frac{4}{3}t\), if \(t=\frac{4}{3}\), then \(d=\frac{4}{3}\times\frac{4}{3}=\frac{16}{9}
eq1\), so it's invalid. For \(t=\frac{3}{4}d\), if \(d = 1\), then \(t=\frac{3}{4}
eq\frac{4}{3}\), so it's invalid.

Answer:

\(d=\frac{3}{4}t\), \(t=\frac{4}{3}d\)