Math

QuestionHow far did a submarine travel from 106.3 feet to 481.2 feet below sea level in 5 hours? (1 point) The submarine traveled 481.2106.3481.2 - 106.3 feet.

Studdy Solution

STEP 1

Assumptions1. The initial depth of the submarine below sea level is106.3 feet. . The final depth of the submarine below sea level is481. feet.
3. The submarine only moved vertically, either up or down.
4. The distance traveled by the submarine is the absolute difference between the final and initial depths.

STEP 2

To find the distance traveled by the submarine, we need to subtract the initial depth from the final depth.
Distance=FinaldepthInitialdepthDistance = Final\, depth - Initial\, depth

STEP 3

Now, plug in the given values for the initial and final depths to calculate the distance.
Distance=481.2feet106.3feetDistance =481.2\, feet -106.3\, feet

STEP 4

Calculate the distance traveled by the submarine.
Distance=481.2feet106.3feet=374.9feetDistance =481.2\, feet -106.3\, feet =374.9\, feetThe submarine traveled374.9 feet.

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