Question

a mars rover descended from the rim of a crater at a constant rate. after 2 days, the rover’s elevation was 106 meters. after 6 days, its elevation was 48 meters. what elevation did the rover start at? \boxed{} meters complete the equation that describes the relationship between the rovers elevation in meters, e, and the elapsed time in days, t. write your answer using whole numbers or decimals rounded to the nearest tenth. e = \boxed{} t + \boxed{} submit

Explanation:

Step1: Find the rate of change (slope)

The two points are \((t_1, E_1) = (2, 106)\) and \((t_2, E_2) = (6, 48)\). The slope \(m\) is calculated as \(\frac{E_2 - E_1}{t_2 - t_1}=\frac{48 - 106}{6 - 2}=\frac{- 58}{4}=-14.5\).

Step2: Find the initial elevation (y - intercept)

Use the point - slope form \(E - E_1=m(t - t_1)\). Substitute \(m=-14.5\), \(t_1 = 2\) and \(E_1 = 106\) into the formula:
\(E-106=-14.5(t - 2)\)
\(E-106=-14.5t + 29\)
\(E=-14.5t+29 + 106\)
\(E=-14.5t + 135\)
When \(t = 0\) (start time), \(E = 135\) meters.

Answer:

The rover started at 135 meters. The equation is \(E=-14.5t + 135\)