Question

juanita is testing her new model airplane. she starts its descent at an elevation of 2,500 feet. it descends at a constant rate of 500 feet per minute. now, write an equation that represents this line. you can think of the equation as the elevation equals the starting elevation plus the change in elevation as time passes. use e to represent the elevation, and t to represent the time since the plane started to descend. the slope is -500. the vertical intercept is 2,500. \\(\square = \square + \square\\) elevation = starting elevation + change in elevation as time passes

Explanation:

Step1: Identify variables and components

We know \( e \) is elevation, \( t \) is time (minutes). Starting elevation is 2500, rate of descent (change per minute) is -500 (negative because it's descending). The formula structure is \( \text{elevation} = \text{starting elevation} + \text{change in elevation as time passes} \). The change in elevation over time is rate \( \times \) time, so that's \( -500t \).

Step2: Form the equation

Putting it together: \( e = 2500 + (-500t) \), which simplifies to \( e = 2500 - 500t \). Filling in the boxes, the left box is \( e \), middle is \( 2500 \), right is \( -500t \) (or \( (-500)t \)).

Answer:

\( e = 2500 + (-500t) \) (or in the box format: left box \( e \), middle box \( 2500 \), right box \( -500t \))