A fish is swimming at a constant rate toward the ocean floor. The fish starts 45
 meters below sea level and ascends 7
 meters per second.

Which equation can be used to find y
 the fish’s elevation after x
 seconds?

To model the fish’s elevation after \( x \) seconds, we need to consider both its initial position and its rate of movement.

- The fish starts at \( 45 \) meters below sea level. Below sea level is typically considered negative, so its initial elevation is \( -45 \) meters.
- The fish is ascending (moving upwards) at a rate of \( 7 \) meters per second, meaning that its elevation increases by \( 7 \) meters for every second that passes.

We can now form an equation for \( y \), the fish's elevation after \( x \) seconds:
\[
y = -45 + 7x
\]

Here:
- \( -45 \) is the starting elevation (45 meters below sea level).
- \( 7x \) represents the increase in elevation over time, with \( x \) being the number of seconds.

This equation can be used to find the fish's elevation at any given time.

Would you like more details or have any questions?

Here are 5 related questions:
1. What will the fish’s elevation be after 10 seconds?
2. How long will it take for the fish to reach sea level?
3. What would the equation be if the fish were descending instead of ascending?
4. How does the fish's speed affect the equation?
5. How would the equation change if the fish started at a different depth?

**Tip:** Always be mindful of positive and negative signs when dealing with direction in word problems (e.g., ascending or descending).

A fish is swimming at a constant rate toward the ocean floor. The fish starts 45 meters below sea level and ascends 7 meters per second. Which equation can be used to find y the fish’s elevation after x seconds?

Discover how to model the elevation of a fish ascending from 45 meters below sea level at a rate of 7 meters per second. This problem explores key algebraic concepts, including linear equations and the slope-intercept form, making it ideal for students in Grades 6-8.

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