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The problem asks for the resultant speed of a motorboat traveling east across a river that has a northward current. This is a vector problem where we need to combine the eastward speed of the boat and the northward current of the river.

### Given:
- The boat's speed to the east = 8 meters per second.
- The river's current to the north = 5 meters per second.

These two speeds are perpendicular to each other, so we can find the resultant speed using the Pythagorean theorem:

\[
\text{Resultant speed} = \sqrt{(8^2 + 5^2)} = \sqrt{64 + 25} = \sqrt{89} \approx 9.43 \, \text{m/s}
\]

Rounding to the nearest tenth, the resultant speed is **9.4 m/s**.

### Answer: 
The resultant speed of the motorboat is **9.4 m/s**.

Would you like more details or have any questions? Here are five related questions you could explore:

1. How do we calculate the resultant vector when the boat moves at an angle other than 90° to the current?
2. What is the boat's effective speed in the northward direction due to the current?
3. How far will the boat travel north due to the current after 20 seconds?
4. What if the boat's speed increased to 10 m/s, how would the resultant speed change?
5. How do you calculate the time it takes for the boat to cross the river if the width is 112 meters?

**Tip**: Always break down vector problems into components (x and y axes) to solve them easily using the Pythagorean theorem or trigonometry.

A boat is traveling east across a river that is 112 meters wide at 8 meters per second. If the river has a northward current of 5 meters per second, what is the resultant speed of the motorboat rounded to the nearest tenth?

Learn how to calculate the resultant speed of a motorboat traveling east across a river with a northward current. The boat moves at 8 m/s and the current at 5 m/s. This problem demonstrates vector addition and the use of the Pythagorean theorem. Suitable for students in Grades 8-10.

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