Q: If a boat travels due East for a distance of 15 miles then it travels due North for a distance of 20 miles at which point it drops anchor then how many miles is the boat from its starting point?

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A trick question! The boat is still quite close to the point where it dropped the anchor! ============================ The boat is approximately 25 miles from its starting point, and exactly zero miles from the point where the anchor was dropped.

25 miles

25 miles

This is where you apply the pythagorean theorem: d = √(152 + 202) ∴d = √(225 + 400) ∴d = √(625) ∴d = 25 So the boat is 15 miles from it's starting point

25 miles.

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25

A trick question! The boat is still quite close to the point where it dropped the anchor! ============================ The boat is approximately 25 miles from its starting point, and exactly zero miles from the point where the anchor was dropped.

25 miles

25 miles

5 miles

This is where you apply the pythagorean theorem: d = √(152 + 202) ∴d = √(225 + 400) ∴d = √(625) ∴d = 25 So the boat is 15 miles from it's starting point

That is incorrect. The distance travelled north cancels out the distance travelled south. Therefore - he only travels three blocks east.

25 miles.

you have to use quadratcic equation= A2 + B2 = C2 152 + 202 = C2 625= C2 C= 25 you have to use quadratcic equation= A2 + B2 = C2 152 + 202 = C2 625= C2 C= 25

25 miles

Using the Pythagorean theorem, the distance from the starting point is the square root of (3^2 + 5^2), which is √34 kilometers or approximately 5.83 kilometers.