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Draw each vector on a coordinate plane. Find the direction of the vector to the nearest degree. 24. A boat's velocity is given by the vector \( \langle 4,1.5\rangle \). 25. The path of a submarine is given by the vector \( \langle 3.5,2.5\rangle \). 26. The path of a projectile is given by the vector \( \langle 2,5\rangle \).

Ask by Carroll Knight. in Northern Mariana Islands
Mar 12,2025

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Answer

**Directions of the Vectors:** 1. **Boat's Velocity Vector \( \langle 4,1.5 \rangle \):** - Direction: \( 21^\circ \) 2. **Submarine's Path Vector \( \langle 3.5,2.5 \rangle \):** - Direction: \( 36^\circ \) 3. **Projectile's Path Vector \( \langle 2,5 \rangle \):** - Direction: \( 68^\circ \)

Solution

**24. Boat's velocity vector \( \langle 4,1.5 \rangle \)** 1. Plot the point \( (4,1.5) \) on the coordinate plane with the tail at the origin. 2. The direction angle \( \theta \) is given by \[ \theta = \arctan\left(\frac{1.5}{4}\right) \] 3. Calculate \[ \theta \approx \arctan(0.375) \approx 20.6^\circ. \] 4. Rounding to the nearest degree, the direction is \( 21^\circ \). --- **25. Submarine's path vector \( \langle 3.5,2.5 \rangle \)** 1. Plot the point \( (3.5,2.5) \) on the coordinate plane with the tail at the origin. 2. The direction angle \( \theta \) is given by \[ \theta = \arctan\left(\frac{2.5}{3.5}\right) \] 3. Calculate \[ \theta \approx \arctan(0.7143) \approx 35.5^\circ. \] 4. Rounding to the nearest degree, the direction is \( 36^\circ \). --- **26. Projectile's path vector \( \langle 2,5 \rangle \)** 1. Plot the point \( (2,5) \) on the coordinate plane with the tail at the origin. 2. The direction angle \( \theta \) is given by \[ \theta = \arctan\left(\frac{5}{2}\right) \] 3. Calculate \[ \theta \approx \arctan(2.5) \approx 68.2^\circ. \] 4. Rounding to the nearest degree, the direction is \( 68^\circ \).

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Beyond the Answer

To draw the vectors, start by plotting points on a coordinate plane. Each vector begins at the origin (0,0). For the boat's velocity vector \( \langle 4, 1.5 \rangle \), draw an arrow to point to (4,1.5). For the submarine's path \( \langle 3.5, 2.5 \rangle \), point an arrow to (3.5,2.5). Lastly, for the projectile \( \langle 2, 5 \rangle \), draw an arrow to (2,5). To find the direction, you can use the tangent function: - For the boat, \( \tan(\theta) = \frac{1.5}{4} \), leading to a direction of approximately 21 degrees. - For the submarine, \( \tan(\theta) = \frac{2.5}{3.5} \), yielding around 37 degrees. - For the projectile, \( \tan(\theta) = \frac{5}{2} \), which gives you about 68 degrees.

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